UC -NHLF ti!ti ll4iti!t ll ti I i B 4 2?48l 673 I I I 1 T RE C OM MEND ATIONS. Whilst this work was passing through the press, the sheets were submitted for inspection to several teachers and gentlemen capable of judging of the merits of the work, and the following testimonials have been received. The reader will bear in mind that these notices are all from practical Instructors, who are now, or have been, until recently, engaged ia teaching; and also, that the majority of them have been furnished without solicitation, and may therefore be regarded as in every sense of the word impartial. i From the Rev. J. H. GOOD, A. M., of Columbus, formerly Rector j of the Preparatory Department of Marshall College, Pa., andfor many years a practical teacher. Mr. C. H. MATTOON —Dear Sir: I have examined your "Common Arithmetic" with considerable care, having perused nearly the whole of it as it passed through the press. It is an excellent practical work. From its arrangement, its fullness of Rules and Examples, its clearness of explanation-above all, its eminently practical character, I regard it as peculiarly well adapted to the wants of that large class of persons, who usually pass through but one work on Arithmetic preparatory to their entrance upon public life. Every teacher can at once perceive, by examination, how well it is suited to the wants of ourCommon Schools. I can cordially recommend it to the careful examination of teachers. Yours, Respectfully, J. H. GOOD. From the Rev. E, W. CLARK, formerly an Agent for Granville College, Ohio. GENOA, Delaware Co., 0., December 24th, 1849. Mr. C. H. MATTOON-Sir: I have examined those sheets of your Arithmetic which you sent me, and would say that I am well pleased with the same. I consider the definitions accurate-the arrangement natural and scientific-the introduction and application of the later improvements in this science judicious -the rejection of much old and useless matter, and the introduction in its stead of a large number of practical Rules, an improvement-and, in short, considering its clearness and simplicity of style, together with its thoroughneess and comprehensiveness of subjects, and its full and extensive application to practical questions, I would say that I think it better adapted to meet the wants both of our Common and High Schools, than any other work of the kind with which I am acquainted. "I am respectfully yours, "E. W. CLARK. RECOMMENDATIONS. Extract from a letter from Mr. BARTON MOORE, a thorough practical instructor, near Sunbery, O. " I received the copies of your Arithmetic which you sent me, and in those I have examined, I believe the principles of numbers can be more easily understood by the youth, and the method of explanation is far superior to any with which I am acquainted. "Yours, &c., "B. MOORE." From MR. S. L. WALLACE, a practical teacher near Kingston, Ross county, Ohio. KINGSTON, Ross Co., O., Dec. 28, 1849. MR. C. H. MATTOON:-I received those sheets of your book, which you were kind enough to send me, and after a careful perusal of them, I am satisfied that the work is better adapted to the use of common schools, and academies, than any other work of the kind with which I am acquainted. 1st. Because it comprises more useful and practical matter than any other work on Arithmetic extant. 2nd. On account of its perfect adaptation to the purposes of instruction, together with its thoroughly progressive character; one subject, and one difficulty only, at a time, being presented, and that made perfectly familiar before proceeding to the next step. 3d. The style is clear and easy, thus fitting it for the lower as well as for the higher classes of learners. 4th. Although it is plain and simple, it is also full and compreh ensive, and the continual reference to preceding principles, &c., together with many original and important remarks interspersed throughout the work, form a material help to the learner. 5th. In the precision and accuracy of the definitions; in the natural and scientific arrangement of subjects; in the careful and judicious introduction and application of the later improvements; in the clear and lucid explanation of hard and difficult subjects; and in the tersenesss and comprehensiveness of the rules, it far excels any other work upon this subject that has yet fallen under my observation, and supplies an important desideratum long wanted in our Arithmetics. 6th. The rejection of much ol;dand useless matter, and the introduction of many rules applicable to practical' business, which heretofore have receivd but little or no attention, speak well in its favor. In short, I consider it a work of sterling merit, and superior excellence, and decidedly preferable to a majority of the trash that at this day is being crowded upon our schools. I remain, yours respectfully, S. L. WALLACE. From MR. WM. C. GILDERSLEEVE, an old and experienced teacher near Chillicothe, Ohio. CHILLICOTHE, 0., Dec. 29th,.1849, MR. CHARLES H. MATTOON-Sir:-I have carefully examined your work entitled "COMMON ARITHMETIC" and unhesitatingly say. that in my opinion, it i is far superior to any other work of the kind, I have yet perused. The rules, (some of which are new to me) are better calculated, and more nearly related to the nature and uses of mathematical computations, in all ordinary business, than those usually given. Therefore, on account of its simplicity, comprehensiveness and intrinsic merit, I would reccommend it to the Schools and Academies of our Western country; Yours respectfully, WM C. G1LDERSLEEVE. ANALYTIC SERIES, NO. 2, COMMON ARITHMETIC, UPON THE ANALYTIC METHOD OF INSTRUCTION. ALSO: THE PRINCIPLES OF CANCELATION, AND OTHER MODERN IMPROVEMENTS. ILLUSrRATED AND APP LIEI), The wllole made simple and easy by numerous practical examples. DESIGNED FOR THE IUSE OF SCHOOLS. BY CHARLES H. MATTOON. "There's nothing so hard, but search will find it out." AEAM PRESS OF S. MEDARY. 1850.' Jv W538 i 9 Entered, according to Act of Congress, in the year 1849, BY CHARLES H. MATTOON, In the Clerk's Office of the District Court of Ohio. PREFACE. IN presenting this treatise to the public, the Author does not think it necessary either to give his reasons for so doing, or to apologise for its appearance. For the former, the student cares but little, and for the latter, still less. Neither will he enumerate what he considers to be its exc llencies being convinced that if it really possesses merit, its worth will be discovered by others, without his blazing it abroad. He will, however, present a few of its most prominent characteristics for the consideration of the public, and leave it for them to decide whether they add to, or detract from the merits of the work. 1, Each section commences with a number of Mental Exercises, in order to give the pupil an idea of the subject before he is burdened with learning abstract rules and principles. 2. Dzfinitions are then given, and the principles demonstrated and explained separately; the first being proved from self-etuent propositions, and those following, by those already explained, In no case is one principle used to explain another, until it has itself been explained. 3. The principles are then summed up into a general rule, for ref-* erence and review. After the rule follow a. number of Exercises for the Slate, and these, as well as the Mental Exercises are chiefly of a practical nature, and strictly confined to the principles already explained. 4. Those Rules most used in practical business have received more particular attention, than those of less importance. 5. Pounds, shillings and pence are rejected entirely, and their places supplied with Federal Money. If we have a national currency, why not adopt it exclusive of all others? iv PREFACE. 6. The tables of Weights and Measures have been prepared with direct reference to existing laws and present uses. 7. The subject of Cancelation is explained, and the method of applying its principles to business distinctly shown. Unlike some modern writers, however, the Author does not make it the chief of all rules. Other modern improvements are also illustrated and applied. 8. The Rules of the Cu be Root, Progression, Alligation, &c., are omitted for three reasons: 1st. They seldom if ever occur in common business, and those who have occasion to use them in their particular pursuits, are generally men of science, who are acquainted with the higher departments of mathematics, in which they are explained more clearly and satisfactorily than they can be in an elementary work on Arithmetic. 2nd. Their place can be supplied with rules of more importance; that is, rules that frequently occur in ordinary transactions. 3d. They increase the size of a book, and the labors of the learner unnecessarily. The Author has yet to be convinced of the necessity of burdening the mind with abstruse and metaphysical questions, or tedious and complicated demonstrations, which in nine cases out of ten can never be of any practical utility to the pupil. 9. The rules in Mensuration are demonstrated, with but few exceptions. The language, however, is chosen more for perspicuity than for mathematical precision. 10. The language adopted throughout the work, is simple, terse and definite. Childishness and puerility are avoided on the one hand, as much as complication and mystery on the other, The aim has been to unite brevity with perspicuity. 11. The arrangement of subjects is such as seems the most consistent with the natural order of the sciences, Therefore Fractions follow immediately after the Fundamnetal Rules; because, 1st, they partake of the nature of Division; and 'nd. They are frequently used in Compound.Numbers; hence, it is necessary to understand Fractions before the operations in Ccmpound iNumbers can be thoroughly understood. Also, Federal Money is placed with Decimarl.ractions, because the same rules are applicable to both, since PREFACE. V they are based on the same scale of notation. Likewise, Commission, Insurance, Interest, &c. are placed under Percentage, upon the principles of which they are based. 12. The analogies and relations of numbers are clearly pointed out and illustrated; and from these, the scholar is taught to reason -to think for himself. He is required to believe no theory without proof and to take no principle without demonstration. He is every where taught that there is no guess work about results-that all questions in Arithmetic can be solved upon true principles-that there is no uncertainty about mathematical conclusions-in short that every theory or principle connected either directly or indirectly with mathematics, that is not capable of demonstration, must be erroneous. Thus he is led to investigate, and examine-to exercise his reasoning faculties-to work intellectually, instead of mechanically-and to believe a theory or principle because it is clearly and conclusively proven, and not merely because "the book says so." Such is a brief outline of the present work. We have endeavored to present every principle that a business man may have occasion to use, in a clear and systematic manner, and also, to prepare it in such a way as to be adapted to precede the study of Algebra, and the higher branches of mathematics. Our object has not been to produce an abstruse and metaphysical treatise, but simply to present a plain, practical, and comprehensive work, which should meet the wants of our business men, and arithmetical students. We have rejected many old, useless, and obsolete rules, and supplied their place with such modern improvements as were deemed of any importance. In short, we have endeavored to say everything that was necessary, and no more. The work is designed for use, and not merely for show. It was prepared by the Author whilst engaged as a teacher, and is the result of arduous labor, and hard study, aided by practical observation and experience in the school room. It may be proper to remark here, that this is designed as the second book of a serEts, now in course of preparation by the Author, The first work, entitled "Arithmetic; Mental and Practical" for beIA vi PREFACE. ginners, is in progress and will be published probably some time the ensuing season. How successfully our work has been executed, time alone will decide. For the present we submit it to teachers and scholars for examination, convinced that it must stand or fall, upon its own merits alone. And should it be found to lessen the labors of teachers, or to promote the intellectual attainments of scholars, its highest aims will be accomplished. THE AUTHOR. 4 CONTENTS Page Suggestions for Teachers- xvi SECTION 1. Arithmetic defined ------------------------- NOTATION- --------------------------------------- 1 " the Roman method ----- - -------- 2 " the Arabic method ------------------ 3 NUMERATION- 4 Numeration Table --- —------------------------- 5 Exercises in Numeration-. 7 Exercises in Notation.- ------------ 8 SECTION II. ADDITION-Mental Exercises --- —--------- 9 Definitions-Addition Table - -----------— 10 Proof of Addition ---- ---- 11 Carrying in Addition _ 12 General Rule-Exercises for the slate ---- ----- - 13 SECTION III. SUBTRACTION-Mental Exercises-Definitions - ---- - 16 Subtraction Table- -17 Proof of Subtraction ------- --------- 18 Borrowing and carrying in Subtraction illustrated 19 General Rule-Exercises for the slate --- ----- - 20 SECTION IV. MULTIPLICATION-Mental Exercises —Definitions -23 Multiplication Table — 24 Proof of Alultiplic:ition -------------— _ 26 Methol of carrying illastrated-Rule when the multiplier is less than 12 --- —-27 General Rte-Exeroises for the slate --- 30 viii CONTENTS. Page. Contractions in Multiplication-When the multiplier is 10, 100, 1000, &c.-To multiply by a Composite number --- —--- 32 When there are ciphers at the right of either orboth factors_ 34 SECTION V. DIVISIoN-Mental Exercises - - ------- 36 Definitions-Division Table ------------------- 37 Proof of Division-Proof of Multiplication by Division 39 Remarks respecting the Remainder - __ --- —-- 40 Method of carrying illlustrated -41 Rule for Short Division --------------—... ---- - 42 Exercises for the Slate ------ --- ------- - 43 1o.g Division -- — __ -_ ------- - 44 Rule for Long Division-Exercise for the Slate ------ 46 Contractions in Division-T'o divide by 10, 100, 1000, &c- 48 To divide by a composite number ---------- - 49 When there are ciphers at the right of the divisor -. 50 SECTION VI.: lEcAPITuLATIoN-Remarks and inferences-Notation ------- 51 Numeration-Addition --------------—. --- —------- 52 Subtraction - ---------------- - - 53 Multiplication -- ---- 5 -Proof by casting out the 9's_ 55 Divisioan. --— 57 Proof by casting out the 9's-___ -------- - 59 Principles deduced from Division ---- ------ - 60-62 Contractions and Abbreviations-To multiplyby 25. By 50. By 125 --- —---------------- 62 -When part of the multiplier is a factor of another part - - 62 -By aid of exponents —. 63 To divide by 25 ---------------- -- 64 — By 125. When there is a remainder after dividing by several divisors, to find the true remainder_ -- _ _ — 65 To contract operations in Long Division.- 66 SECTION VII. CANCELATION-Definitions and Illustrations ------- 67 General Rule -..._....-.. 70 1 Exercises for the slate ----------------— 71 il reatest Common Divisor — __-___- ------—,- 72 Rqle-for two numbers --------------- 73,I -For more than two numbers-two rules ------ 74 geast Common fMultiple --------- --- -76 Aule — Exercises for the slate-,... 80 CONTENTS. ix Page. SECTION VIII. FRACTIONs-Mental Exercises ------- — _. - - 81 COMMON FRACTIONS-Definitions --- 85 Fundamental Propositions_ --- — - 87 Reduction of Fractions-To reduce fractions to its lowest terms -------—.. --- ------------ --- g90 -An improper fraction to a whole or mixed number -------- 91 -A whole or mixed number to an improper fraction -------- 92 -A compound fraction to a simple one -93 -Ditto, by cancelation _ 94 To reduce fractions to their least common denominator --- 94 General Rules for the Reduction of Fractions ------------ 96 Exercises for the slate.-. 97 Addition and Subtraction of Friactions-Mental Exercises _ 98 General Rule-Exercises for the slate --- — ----- 100 Multiplication of Fractions —A ental Exercises ------ 102 Exercises for the slate-To multiply a fraction and a whole number together --- ___.._- ------— 103 To multiply a whole number and a mixed number together_ 105 — Fractions together --- ___- — __-,_ —106 — Mixed numbers together. _-..107 General Rules for the Multiplication of Fractions. ------ 108 Exercises for the slate- 108 Division of Fractions -- ____ — _____ — 109 Mental Exercises-Exercises for the slate-To divide a fraction by a whole number - 110 To divide a whole number bv a fraction -------— 111 -One fraction by another -1 --- —-_. 113 Method of proceeding when mixed numbers occur --- 114 Complex fractions- 116 T( change a fraction to any required denominator ------- 118 General Rules for the Division of Fractions- 120 Exercises for the slate.-___ ---_____ -— 121 Miscellaneous exercises for the ste ----------— 123 DECIMAL FRACTIONs-Definition, &c. -—.-...-. —. 124 Decimal Numeration Table.. — 125 Numeration, and Notation of Decimals- 127 Federal Money-Table --------- -----— 128 To read sums in Federal Money ----- --- ---- 130 To write sums in Federal Money --- -------- 131 Reduction of Federal Money, &c.- _-.._ --- —------- -- 131 Exercises for the slate-Reduction of Decimal Fractions —. 132 To change a Decimal to a Common Fraction.. 132 To change a Common fraction to a Decimal —. ---- ---- - 133 ; X CONTENTS. Addition and Subtraction of Decimals and Federal Money- - 135 Rule_ 137 Rule --- —----------------------------------------- 137 Exercises for the slate -------- ----------------- 138 Multiplication of Decimals and Federal Money -----— 139 Rule-Exercises for the slate 140 Method of buying and selling articles by the 100 or 1000-. 141 Division of Decimals and Federal Money ------— 142 Rule --- —------------- ------------— 143 Exercises for the slate- 144 Miscellaneous exercises for the slate --------— 145 Bills, Accounts, &c. 146 SECTION IX. COMPOUND NTuMBERS-Definitions — 148 Reduction — ---------- --------------------— 149 WEIGHTs- Troy Weight -- 150 Exercises in Troy weight 151 Rules for Reduction, Ascending and Descending ------ 152 Avoirdupois Weight - -- - -- -- 152 Exercises in Avoirdupois Weight -153 Apothecaries Weight. --- —-.. --- —-—. ---. -- - 153 Exercises in Apothecaries Weight -154 MEASURES OF CAPACIIY-DryfleaSsure- -------- 154 Exercises in Dry Measure — 155 Wine Measure - ---- ------- ----- 156 Exercises in Wine Measure ------------ --—. 156 Ale or Beer Measure --------------------- 157 Exercises in Ale or Beer Measure ----------— 157 MEASURES -OF EXTENSION-Long ieasure - - 158 Exercises in long Measure.- -159 Cloth Measure ----------------------------- 59 Exercises in Cloth Measure ---.. --... ------- 160 Land or Square Measure.._..-.._ 161 Exercises in,and or Square Measure --- —----- 162 Solid or Cubic Measure... ---—.164 Exercises in Solid or Cubic Measuie.-.... - 165 Time,. —166 Exercises in Time- 167 Circular Measure or Motion _ ----- -- -168 Exercises in Circular Measure or Miotion - -------- 169 l Miscellaneous Exercises in Reduction -169 Factions of Compound umers ----------------------- 174 l To reduce higher to lower fractions-Rule_ - -- - 174 To reduce fractions to integers, and integers to fractionsRule - -- 175 1. i. CONTENTS. xi Page. To reduce decimals to fractions and fractions to decimals_- 177 Addition of Conpound Numbers ----------— 179 Rule.180 Exercises in Addition of Compound Numbers- 181 Subtraction of Compound Numbers __- -------- 183 Rule- 184 Exercises in Subtraction of Compound Numbers ----- 185 Multiplication and Division of Compound Naumbers ---- 187 Rules- 187 Exercises in Multiplication and Division of Compound Numbers._ —..____-. __......_. 188 Difference in time of two places-Rule-..-..._ — 193 SECTION X. EXCHANGE —lDefinitions, &-c. _ - ----- - -195 Table of Foreign coins_ 196 Rule-To change foreign coins to Federal Money ----- 196 Rule-To change Federal Money to Foreign Coins.. 197 SECTION XI. ANALYSIS.-Definitions, &c. ----- -------— 198 Mental Exercises in Analysis ---- ---- ---- 199 Exercises for the Slate _ 201 Practice-Definitions, &c. -------------— 207 Exercises in do. - ----- ---— 208 Barter, Definition and Exercises. 212 Alligation-Medial.-. 213 Porportional Division 214 Exercises in Analysis. -------- -------— 215 SECTION XII. RATIO AND PROPORTION-Definition of Ratio - ---- 219 Mental Exercises in do ---- — _- 220 Rules and Exercises in Ratio -----—. --- —. --- —---—.. 220 Simple and Compound Ratio ------ -----— 223 Proportion-Definition, &c._.224 Difference of Ratio and Proportion ---__ - ---- 225 Extremes and Afeans in Proportion ---- -----— 225 Rules in Proportion ----— _ __ --- ------------- 226 Mental Exercises 227 SIMPLE PROPORT:ON, Definitions, &c._ 227 Rules for working questions in Simple Proportion.229 General Rule for Simple Proportion -- --- ----- 232 Exercises for the slate ------—, --- 233 Xli CONTI'NTs. Page. COMPOUND PRlOPORTION, Definitions, c. --- —--------— 235 Solution of questions in do -__ — --------- 236 How to prove Compound Proportion_ -. - _ —_ --- —--- 237 Genera.l Rh'le' /or Comnpouncd Proportion -- 238 Exercises for tlhe slate n do -. --- --- ------..- 239 PARrTNERSIPI, Definition, Solutions, &c. ------------ 240 Rule Jor IJarttnership, Exercises --— 241 Bankruptcy, Exercises ---------------------- ----—. 242 General Average, Exercise --- —-------------— 42 SECTION X1II. PERCEN'TAGE-Definiltions- - -_.. - -- -------------- 244 Mental Exercises ---------- --------- 245 Rate for -'ercentage, Exercises - -247 COMIMISSION AND INSIURALNCE _-_ --- —--- ---- 249 Exercises in Commission ---------------------- 250 Exercises in Insurance — 251 PROFIT AND Loss, Definitions --- —---- ---- 254 Exerci-es in do. --- —--- ---------------- — 255 STOCKS, Definitioins_ ---------------- -260 Exercises in do. -----------------— 261 BROKAGE, Definitions, Exercises. --- —------------ - 262 DUTIES, Definitions. --- ——. --- —-- - 262 Specific Dzuties, Definitions -.. —__ --- —---------- --— 263 Exercises in do. -- - --- ------------- 264 AD VALORAM DUTIES, Definitions ---------------- 264 TAXES, Definitions ------------------ 265 General Rulefor Taxes --- —266 Table for Taxes --- —------------------------- ---- 267 INTEREST. Definitions and Illustrations —28 Simple Interest, Illustration ---- ------- 269 Rule-Exercises ----- ------- -270 Rule for years and months. _ _ — ----- - -— 271 Rule for days. --- —----------- -. - -. --- —---— 272 General Rule for Interest -.- - - - - -- -- - - - 272 Table for Interest ---— _ ------------- 273 Exercises in Interest.. ---- 274 Interest at ANY rate-Rule -------- - ----—.. 276 &Second methodfor " -277 Rule for Interest on NVotes and Bonds ---- ---- ---- 279 Vermont Rule ( ( -— 281 Connecticut Rule " " ------— 282 Exercises in the above -----------------— 283 Problems in Interest - - ^ COINTL-N;T'r. Xiii To find the Interest, To find the rate --------- 285 To find the time. 286 To find the Principal -237 --- —-------- 27 To find the Principal and Interest 288 DISCOUNT, Definitions ---------------— 89 Exercises in._ —. _.. ---_ 290 BANKING, Definitions -—. --- —--------- 291 Exercises in -- --------- ------ -- 2c 22 Equation of Payments, illustrations ------ ____ 293 Rule for finding equated Time.294 COMPOUND INTEREST, Illustrations - -- 296 Table for Compound Interest ------------— 97 Exercises in ".298 SECTION XIV. DUODECIMALS, Table, Multiplication of --- —, ___ 299 Rulefor Multiplication of Decimals --.-. ---..301 Painters' Powers and Plasterers' Work. --- —-.-. 301 Masons' Work -, _.._____303 SECTION XV. POWERS AND ROOTS, Definitions 303 INVOLUTION, Definitions, Exercises -- - -__ __304 Rule for Involution, Exercises --------- -___305 Involution of Fractions ----—., 306 EVOLUTION, Definitions. --- —-—. --- _________308 Mental Exercises, Illustrations. ----.- --.. - -. 309 Extract;on of the Square Root, Illustrations ---— __ 311 Illustrations, Diagrams, — ____ —_ _ ___ _____ 313 Rule for Extracting the Square Root -- — __ __31L5 Exercises in " " -. 317 Rule to Extract the Square Root of Fractions ----—... 318 TRIANGLES, Illustrations --- - - -_____319 Mean Proportionals ---- - — _______321 CIRCLES, Definitions, Illustrations ---- ________ 323 SECTION XVI. MENSURATION, Mensuration of Surfaces, 324 Area of Triangles, Rule 325 Area of Trapezoids, Rule ---------------.326 Area of Circles and Polygons 327 Circumscribed and Inscribed Polygons —.._- _ _ 328 Quadrature of the Circle 329 Ratio of the Diameter to Circumference 330 _ ________, X1V VON TENTS. To find the Circumference, Diameter -330 Rule for finding the Area of the Circle — 331 Area of a Square inscribed in a Circle._ 332 Dimensions of the Circumscribed Circle 336 MENSURATION OF SOLIDS, Prisms, Area of a Prism ---- 338 To find the solidity of a Prism -----— 339 Cylinders, Definitions -339 To find the solidity of a Cylinder --------------------— 340 Boilers, To find the contents of ---------— 340 To find the contents of a cistern ---------— 341 Parallelopiped, to find the surface and solidity ------- 342 Pyramids, Definitions ----------—. --- —------ 342 To find the surface of a Pyramid or Cone ----— _343 To find the surface of the Frustrum of a Pyramid orCone - 344 To find the solidity of a Pyramid or Cone ----- - 344 To find the solidity of the Frustrum of a Pyramid or Cone - 345 SPHERES OR GLOBES. To find the surface of a Sphere or Globe -..346 To find the solidity of a Globe or Sphere -------- 346 TONNAGE OF VESSELS, Definitions ---------— 346 Rule for calculating tonnage.- 347 To find the solidity of Irregular Bodies... -- 348 Guaging. To find the number of bushels a vessel will contain --------— 348 To find the number of wine and beer gallons in a cask -- 349 SECTION XVII. PHILOSOPHICAL AND MISCELLANEOUS QUESTIONS........ 350 THE MECHANICAL POWERS, Definitions -------—. —.. 350 The Lever, Definitions and Rules ----------------—.. — 351 The Wheel and Axle, Definitions and Rules ----- - 352 The Pully, Definitions and Rules --------------------..352 The Inclined Plane, Definitions and Rules ------ 353 The Wedge, Definitions and Rules --- -------- 354 The Screw, Definitions aud Rules.354 Machinery, Defininitions and Rules ---- ----- - 355 Book Keeping, Definitions and Forms - 356 Book Keeping for drovers and farmers --- — _ --- 357 FORMS. Form of Orders-Of Receipts -- ------- 358 Form of Due-bills-Of Votes —. 359 Remarks on Notes --------------—. 356 Miscellaneous Problems and Rules --- ------— 360 To find the heighth of an object-its distance -261 To find the time a time a body has been falling-Its velocity 361 To find the space fallen through...................... 362 PREFACE. Problem relating to the right-angled triangle ------- 362 To measure standing timber-... --- —------- 362 To find the temperature of any day-De Witt's Rule 363 Various Problems — 363 Analytic Questions ------------------- 365 SECTION XVIII. MISCELLANEOUS EXAMPLES FOR THE SLATE — --------- 368 ERRATA ------------------------ 387 I i, INTRODUCTORY REMARKS: FOR TEACHERS. There are at least three different departments of science with which every person should be familiar, and they are, leading, Writing, and Arithmetic. These, in consequence of their universal application to all kinds of business, might with propriety be termed the Golden Branches of Science. Other departments may be desirable for ornament, or necessary in some particular profession or| occupatio:n, or useful in storing the mind with general information ' but there are none that will apply so universally to all classes of individuals, and to every department of business, as the three branches above named. Such being the case, how necessary is it that they should be taught correctly, thoroughly, and completely: that the learner should be taught correct theories, correct principles, and the correct method of applying these theories and principles to practical usein short, that he should be taught to work intellectually, instead of 'I echanically. in Arithmetic, especially, is it necessary to be particular, as much is often at stake by a single operation; and the greater the risk, the more the need of care. In teaching Arithmetic, many things are to be considered, the principal of which are1st. The age of the pupil; and 2nd. His natural abilities for receiving instructions. With young pupils, the subject under consideration should be presented in clear, simple, and definite language-free, on the one hand, from childishness and puerility, and on the other hand from complication and mystery. If possible, technical terms should be Ii rNTRODUCTORY REMA RK!. XVII avoided; but if it is absolutely necessary to make use of any, they should be fully and thoroughly explained. As he advances, additional principles may be presented to his view, and new theories explained for his consideration; and thus he may be led on, step by step, gradually and thoroughly, until he has mastered the first elementary rudiments of mathematical science, and is able to grap - ple successfully with the more abstruse and difficult propositions. As his mind enlarges, and his intellect expands, and his ability for receiving instruction increases, the explanatory language may be proportionably altered, and some of the technicalities of the science may be introduced; the pupil has now learned the first rudiments and is able to think and reason to some extent for himself, and needs not the particular demonstration of every simple problem, or illustration of every common term. As he advances still farther, the illustrations may be in strict sci - entfic language, and he may. be required to investigate theories, examine principles, demonstrate abstruse propositions, unravel difficult questions, or solve knotty problems-to prove why a theory or statement is, or is nrot correct, and in short, to reason, think, examine, and demonstrate a truth by his own investigating powers, independent of the " say so " of the book. Practical Applications of principles to business should be introduced throughout the course, thus uniting practice with theory, and science with art, and making the pupil both a scientific scholar, and a judicious man of business. To accomplish this would of course require years of labor, care, and study; but the pupil thus educated would finally become an honor to himself, a credit to his teacher, and an ornament to society. Throughout this series we have endeavored to follow the plan here laid down, with what success, remains yet to be decided. But whilst a work may be prepared with care, and the principles presented clearly and lucidly, yet the instructor must second the aim of the author, and enforce theories and the method of application upon the attention of the scholar, and be sure to xviii INTrRODUCTORY REMARKS. have them correctly understood, To teach with success, however, the teacher must possess1st. A tnhrouh, kn.-oedrle of h.s sucject; and 2nd. T,.e,ctcire co:f'idenc, and go;d will of his pupils. If in addition to this, he has a good black-board and numerical frame, he need not fear, but push on, certain of success. The first book of these series will propably be published the ensuing season. It is recommended that this or some other mental work should be studied before studying written Arithmetic. A few errors were discovered in this work after the forms were printed, and are mentioned in an Errata at the end of the volume. It may be possible some errors have escaped detection. Those discovered will all be corrected in future editions. Should some portions of this work be thought too difficult at first, they may be omitted, until a review, It is left to the judgment of the teacher, however, what part to omit, and what to study. In teaching, the scholar should be tanght to depend upon his own resources. Examples should be explained indirectly, and the method adopted by some, of working sums for scholars on the slate, without illustration, should be discarded at once. The teacher, however, should not permit the learner to pass by a question or principle until it is perfectly familiar to him. In conclusion, the writer would say to Teachers-Your success must depend principally upon your own exertions. You have an important charge committed to your keeping, and therefore yours is a heavy responsibility. Let your motto then be " ONWARD AND UPWARD." And whilst you instil the principles of science into the youthful mind, endeavor to train it in such a manner that it shall reflect credit upon its teacher, and honor upon itself. Excite in the minds of your pupils a noble spirit of generous rivalry and emulation; and not only excite this spirit, but keep up the excitement, and your success will astonish even yourself. To Students he would say-Persevere. Never give up a question because it is hard, or pass by a demonstration because you cannot I I i i Ir XiX INTRODUCTORY REMARKS. comprehend it at once, Remember, " Where there is a will there is a way," and " Nothing so hard but search will find it out." If you ever expect to gain a knowledge of Arithmetic without close Vr e vI., [r,.. I.;., l vot, ee "l.b(iCi c,e dlutely, and success will evenuaally crown your efforts. I COMMON ARITHMETIC. SECTION I. NOTATION AND NUMERATION. A single thing is called a unit, or one; one and one more are called two; two and one more are called three; three and one more are called four; four and one more are called five; five and one more are called six, &c. The expressions one, two, three, four, &c., are called NUMBERS.Hence: Observation 1. Number is an expression signifying a unit, or a collection of units. It always answers to the question, "How many?" Numbers are subject to certain laws and regulations, and are applied to practical business. These laws and relations, when arranged in systematic order, form what is called an ARITHMETIC.Hence: Obs. 2. Arithmetic is the science of numbers. It is both theoretical and practical. Obs. 3. The theory of Arithmetic consists in the analization of the laws and principles of numbers. Obs. 4. The practice of Arithmetic consists in the application of its principles to common business transactions. REMARK.-Science is knowledge reduced to order. ARTICLE I. yOTATION. Obs. 1. Notation is the method of expressing numbers by characters or signs. Obs. 2. There are two different methods of notation in use, called the ROMAN and ARABIc methods, because it is supposed that these nations first invented them. What is a single thing called? What does unit mean? What are ope and more called? Two and one more? Three and one more?. Four and oje more? Five and one more? Six and one more? Seven and one more? Eight and one more? Nine and one more? What are the expressions one, two, three, &c., called? What then is number? To what question dbea it answer? To what are numbers subject? To what are they applied?,What do these laws and regulations form when arranged in order? What th n is Arithmetic? How is Arithmetic divided? What is the theory of Arlttntttc? The practice of Arithmetic? What is Science? What is NotationE -How many methods of Notation are there in un? What ar they? Why at they o called? 2 COMMON ARITHMETIC. Sect. I CASE I. The Roman Method.-By the Roman method we express numbers by the use of the seven following letters, viz: I. V. X. L. C. D. and M. When standing alone, their values are as follows: I., one; V., five; X., ten; L., fifty; C., one hundted, and M., one thousand; and by various repetitions and combinations, may express any other number. Their manner of use is clearly shown in the following TABLE. I denotes —. ------------- ---— one II -.two III. — -- ----— three IV. ------- four V -------------------------- five VI " ----------------- six VII —.seven VII ----------------------- eight IX- - --- ------— nine X -— ten XI " -- ---------- -eleven XII " twelve XII " ----------- -— thirteen XIV " fourteen XV -------------------- fifteen XVI " ------------- sixteen XVII ----------— " seventeen XVI1I ------------- -------- eighteen XIX.. nineteen XX --------------— " -— twenty XX ------------------------- twenty-one XXII ---- twenty-two &c. XXX " thirty XL -— ". --- — forty L "... fifty LX " -... —... - sixty LXX ".. --------.,-. seventy LXXX " -eighty XC ".. --- —. __ ninety C --— one hundred C -- — two hundred CCC.. three hundred CCCC " -—, —,., --- — four hundred How do we express numbers by the Roman method? What value have these letters when standing alone? Can any number be expressed by the lettors? How? Art. 1. NOT ATION. D denotes..... five hundred DC " -- ------------. six hundred DCC --------------------—.. ---- seven hundred D C " ------ -----. eight hundred DCCCC " - nine hundred M --------------— one thousand MDCCCXLVIII denotes —..one thousand, eight hundred and forty-eight. By examining this table, it will be perceived that when a letter is repeated, its value is repeated; as II two, XX twenty &c; also, when two letters of different values are joined together, if the less be placed before the greater, the greater is diminished by a sum equal to the value of the less; but it the less be placed after the greater, the value of the greater is increased by a sum equal to the value of the less. Thus: X denotes ten; but IX denotes nine, and XI denotes eleven. Obs. 3. A line (-) placed over a letter increases its value a thousand times. Thus: V denotes five; but V denotes five thousand. NOTE.-This method of expressing numbers is but little used, except in denoting Chapters, Sections, &c. CASE 2. The Arabic iMethod.-Numbers are expressed by the common, or Arabic method, by the following characters or figures, viz 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. one, two, three, four, five, six, seven, eight, nine, cipher. Obs. 4. 'The first nine characters are called significant figures, because they always express some number. '' hey are also termed digits, from the Latin word digitus, signifyingfinger. Obs. 5. The last character (0), when standing alone, has no value; therefore it is called a cipher, which means nothing; but when placed at the right of a significant figure, it increases the value of this figure ten times. Numbers larger than nine are expressed by different combitlations of the foregoing characters. Thus: ten is expressed by 1 and 0, thus, 10; twenty is expressed by 2 and 0, thus, 20, &c. What effect does it have upon its value to repeat a letter? What effect does it have to place a letter expressing a smaller number before a letter expressing a larger number? To place the smaller after the larger? How many times doea a line (-) placed over a letter itcrease its value? For what is this method of expressing numbers used? How are numbers expressed by the Common, or Arabic method? What are the names of these characters? What are the first nine characters called? Why? What other name are they termed? From what? What value has the character (0) when standing alone? What is it called? What does cipher mean? What effect does it have to place a cipher at the right of a significan figure? How are numberB larger than nine expressed? How is ten expremed 4 COMMON ARITHMETIC. Sect. 1. TABLE OF NUMBERS FROM TEN TO ONE THOUSAND. ten --- — -- - is expressed 10 eleven -- " ---- ---- -.- 11 twelve —. --- — -. " -12 thirteen ---- -- " --------- ---— 13 fourteen --- —--------- -------- 14 fifteen --- —------ --- -— 15 sixteen --- —--------- --------------- 16 seventeen.. --- —-- " --- —----------- --- 17 eighteen. -- 18 nineteen - - --------- 19 twenty - - - - 20 twenty-one --- — --- -------------- 21 twenty-two - ------ " 22 twenty-three — "- --- --- -------- 23&c thirty - ------------------------- 30 thirty-one-. " -.31&c forty --- -------- --------------------- 40 fifty. --—... ------- --- -- ------— 50 sixty ---- - ------------------- 60 seventy ---------- ------------------------- 70 eighty --------- ---------------------- 80 ninety --------- ---------- --- 90 one hundred --- — " --- —----- 100 one hundred and one- - 101 one hundred and ten -.-... " ---,... 110 one hundred and fifty - -, 150 two hundred " ------------------- 200 three hundred. ------------------— 300 four hundred --------------------- 400 five hundred --- 500 six hundred ------ " --- — ------- 600 seven hundred. -------------------------- 700 eight hundred — ---- -------- " 800 nine hundred - " 900 nine hundred and ninety-nine is expressed. --- —. ------- 999 one thousand is expressed --- —----------- - 000 ARTICLE 2. NUMERATION. Obs. 1. Numeration is the art of expressing numbers by words, or of reading numbers expressed by figures. In Numeration, different places are assigned for figures, and a name given to these places. Hence Twenty? Thirty? Fifty? One hundred? Five hundred? What is Numetrtion? Art. 2. NUMERAT ION. 5 Obs. 2. The value of afigure depends upon the place it occupies. Thus: 4, standing alone, expresses simply 4 things, or 4 units, and is written thus- ----------------- 4 But if we annex a cipher to 4, we increase its value ten times, and make it 4 tens, or forty, written thus...... ------- 40 If we annex two ciphers to 4, we increase its value one hundred times, and make it 4 Jundred, written thus --— 400 If we annex three ciphers to 4, we increase its value one thousand times, and make it 4 thousand, written thus ---— 4000 Obs. 3. Hence-Every removal of a figure towards the left increases its value ten times; or, it always takes 10 units of any one order to make 1 of the next higher order. It matters not whether the figures at the left are significant figures or ciphers, the increase is the same. Thus: the above expressions may be written 4444; the right hand figure being units, the second tens, the third hundreds, and the fourth thousands; and as the figure in each order is 4, the whole expression is read four thousand, four hundred,four tens (forty) and Jour. The different places, or orders for figures, and their names, may be exhibited in tile following TABLE. a a '- a 4a a ~ o a a Ce a m a a a 1 4 5 7 9 6 2 0 5 7 9 3 5 4 2 8 a s ~ a ^ ~ NOTE.-This is according to the FRENCH method. The ENGLIS, after hundreds of millions, hundreds of billions, hundreds of trillions, &c., reckon thousands of millions, tens of thousands of millions, hundreds of thousands of millions, and the same of billions, trillions, &c., assigning sixplaces to millions, billions, &c., instead of three. Upon what does the value of a figure depend? What does the number 4, standing alone, express? If we annex a cipher to 4, what does it express? If we annex two ciphers, what does it express? Three ciphers? What effect does every removal of a figure towards the left have upon its value? How many units does it take of one order to make a unit of the next higher order? Does it make any difference whether the figures at the jight are significant figures or ciphers? How s tihe expis.silo 4444 read? Repeat the Table. COMMON ARITHMETIC. Sect. I. Obs. 4. Every number must occupy one or more places. Thus: o 5... The number six occupies the units place only: thus_ - - 6 The number twenty-three occupies two places, viz: units and tens; thus. - - 2 3 The number two hundred and sixty-seven occupies three places, viz: units, tens and hundreds; thus - 2 6 7 The nnuber two thousand, two hundred and twenty-two occupies four places, viz: units, tens, hundreds and thousands; thus --- —------------ 2 2 2 2 From the example we perceive Obs. 5. That the same figure has different values, according to the place it occupies. Hence, we infer that figures have two values, viz: simple and local. Obs. 6. The simple value is the value expressed by thefigure when standing alone: as 5 expresses simply five, and 6 expresses simply six. The local value is the value expressed by the figure when combined with other figures, andz depends upon the place it occupies. Thus: In the number 55, the 5 in the unit's place is simply 5; but the 5 in the ten's place, if considered with reference to the other 5, becomes 5 tens, orfifty. The term local is derived from the Latin word locus, signifying place. To facilitate the reading of numbers, the orders are pointed of into periods of three figures each. 'I le first right hand period is called the unit's period; the second, the thousand's period; the third, the million's period, &c. '1 his may be illustrated as follows: O: X3 o t 0o o 3 Trillion's Billion's Million's Thousand's Unit's perio period. period. period period. r", 02 ^, "._^ - ~4- ci C *' tO_ -^ > How many places will the number six occupy? Twenty-three? Two hundred and sixty-seven? Two thousand, two hundred and twenty-two? Name Art. 2,. N UMERATION* As we have said before, every number must occupy one or more places. Thus: The number 23 consists of 2 tens and 3 units, and is called twenty-three. 365 consists of 3 hundreds, 6 tens and 5 units, and is called three hundred and sixty-five. 1848 consists of 1 thousand, 8 hundreds, 4 tens and 8 nits, and reads one thousand, eight hundred andforty-eight.. How many units and tens are there in 63? Ans. 3 units and 6 tent. How many units and tens are there in 47? 54? 27? 80? 36? r How many units, tens, and hundreds are there in 126? Ans. 6 units, 2 tens, and 1 hundred. How many units, tens, and hundreds are there in 216? 423? 108? 240? 300? 789? 972? How many units, tens, hundreds, and thousands are there in 1827? 2362? 4786? 6721? How many units, tens, &c., in the following numbers: 5? 18? 20? 17? 60? 58? 40? 120? 111? 107? 209? 780? 2820? 2006? 7460? 7000? 12067? Obs. 7. To numerate figures-Commence at the right hand, and call the orders as they stand in the Table. To read figures-Commence at the right hand, and point them of into periods of three figures each; then commence at the left hand and read each period as if it stood alone, add.ng the name of the period. Numerate and read the number 623. Ans. 3 unts, 2 tens, 6 hundreds, whiclh reads six hundred and twenty-three. In like manner let the pupil numerate and read the following numbers: 46 126 1026 16932 61728421 78 150 1428 214975 74500112 96 197 2764 379681 345678907 47 265 4829 127962 763945316 54 390 7008 987943 5438659123 67 411 9060 7237684 29685734120 75 578 10857 4623794 149769004721 87 976 12764 2776547 691239468721623 the places or orders occupied by each of these numbers. Can the same figure have different values? How? How many values have figures? What e they? What is the simple value? The local value? Upon what does the local value depend? In the number 55, what is the value of the 5 in the unit's place? Of the 5 in the ten's place, when considered with reference to the other figure? How do we facilitate the reading of numbers? How many figures does each period embrace? What is the first period called? The second? The third? The fourth? The fifth? How many places does te number 23 occupy? 365? 1848? What are the names of the places e.ch of these figures occupy? How do we numerate figures? Where do we commence to point off figures into periods? Where do we commence to readt How dX we read each peri.d? a Sect. 1. COMMON A iITPTIMBTIq. The number forty-six consists of 6 units and 4 tens; therefore to write forty-six, we place a 6 in the unit's place, and a 4 in the ten's place, thus. --- ------ 46 The number two hundred and six consists of 2 hundreds, 0 tens, and 6 units, and is written thus — 206 The number one thousand and sixty consists of 1 thousand, 0 hundred, 6 tens, and 0 units, and is written thus —.-.. 1060 NOTZ.-When units only are mentioned, the right hand place is always understood. Hence, to write numbers, we have this RuLE.- Write each significant figure in the order which it belongs, and place a cipher in all vacant orders. EXERCISES FOR THE SLATE. Should the learner fiud any difficulty in reading the following numbers, it might assist him to proceed thus: Draw a number of lines, and write the names A of the orders in;4 their respective spaces; then by -e -* disti n g u ishing the periods by g e double lines, the *' 2 e numbers can be 3 4 6 1 0 4 0 2 7 easily written and read. Thus: to write three hundred and forty six millions, one hundred and four thousand, and twenty-seven, we write each figure in its order, and place a cipher in the orders left vacant, and then it is easily read, as above. 1 Write twenty-five. 12 Five hundred and ninety-nine. 2 Forty-eight, 13 Seven hundred. 3 Fifty-four. 14 Eight hundred and three. 4 Seventy-six. 15 Nine hundred and fifty one. 5 Eighty-nine. 16 One thousand and four. 6 Ninety-seven. 17 Three thousand, four hun7 One hundred and seven. dred. 8 One hundred and thirty-two. 18 Six thousand, four hundred 9 Two hundred and forty. and thirty. 10 Three hundred and sixty- 19 Nine thousand, three hunfive. dred and twenty-six. 11 Four hundred and eighty-six. 20 Ten thousand, one hundred. How do we write the numberforty-six? Two hundred and six? One thousand and sixty? When units only are mentioned, what place is understood? What is the rule for writing numbers? SIMPLE ADDITION. 9 21 Twenty-two thousand andl 28 Three millions, two hundred seventy-three. I and seventy-six thousand. 22 Sixty-four thousand, nine 29 Eight hundred and forty milhundred and seventy-seven. lions. 23 Eighty thousand, three hun- 30 Eight billions, ninety-seven dred and one. millions, six thousand, seven 24 One hundred and four thou- hundred and nine. sand and eight. 31 Six hundred and ninety-seven 25. Seven hundred thousand, trillions, two hundred and four nine hundred. billions, seventy millions, four 26. Nine hundred thousand, two hundred and six thousand and hundred and forty-seven. one 27. One million, sixty thousand and nine. NoTE.-The method of reading and writing numbers should be clearly explained on the black-board by the teacher, until the scholar thorougkly understands it, and can read or write any number readily. A little labor and exertion on the part of the teacher will be found to be mere beaeficial and satisfactory than an elaborate treatise on the subject. SECTION I[. SIMPLE ADDITION. ARTICLE 1. MENTAL EXERCISES. 1. Charles bought a book for 8 cents, a slate for 6 cents, tan a pencil for I cent. How many cents did he pay for the whole Ans. 15 cents. 8 cents, and 6 cents, and I cent are how many cents? 2. John gave 6 cents for a writing-book, 6 cents for an ink-stand, and 4 cents for some quills. How many cents did he give for all? 6 and 6 and 4 are how many? 3. A poor boy met some ladies, one of whmn gave him 3 cents, another 5 cents, amother 7 cents, and another 9 cents. How many cents did he get? 4. A man paid 8 dollars for some sheep, 7 dollars for some h.lgs, and 2 dollars for some fowls. How many dollars did he spend? 5. A lady bought 4 yards of cloth mi one piece,;6 in another, 5 in another, 7 in another, and 9 in another. How many yards did she buy? 6. John has 8 marbles, William 6, Thomas 9, Henry 5, Charles 7, James 3, and Robert 4. How many have they all? 7. One boy has 6 apples, another 7 apples, another 4.apples, 'another 8 apples, and anotherS4 apples. How many have they all? 8. One man has 6 horses, another 4, another 9, another 3,:amother 8, and another 7. How many have they all? A3 10 COMMON ARITHMETIC. Sect. II. 9. On c.e she11t are.F9 b; 3, Lunod;er 7, o, ano ther 8, 1o anoLher 6, ndl on ai_.oaer 5. o'w V many are tflle in all? 10. I bought a book for 4 shillings, a slate for 2 shillings, some paper for 3 shillings, and a map for 9 shillings. How many shillings did I spend? 11. A man sold some wheat for 8 dollars, some corn for 5 dollars, and some oats for 7 dollars. How much money did he get? 12. 4, and 7, and 5, and 3, and 9, and 2, and 1, and 6, and 8, are how many? ARTICLE 2. DEFINITIONS. Obs. 1. The putting together two or more numbers to find one number, as in the preceding examples, is called ADDITION; and the number thus obtained is called the SUM.* Addition may be of Simple or of Compound numbers. Obs. 2. Simple Addition is when the numbers all express things of the same name or kind; as all dollars, all yards, &c. Obs. 3. SINs..-A cross (-), one line horizontal and the other perpendicular, is the sign of addition. It is called plus, which is a Latin word, meaning more. It signifies, that the numbers between which it stands are to be added together. Two parallel horizontal lines (=) are the sign of equality. It shows that the number before it is equal to the number after it. Thus: 6+4=10. To facilitate the addition of numbers, we subjoin the following ADDITION TABLE. 2 and I and 4 and 5 and 6 and 7 a.l 8 al 9.:(i | 1 ae 3 1 are 4 1 are 5 1 a;e G 1 are 7 1 a'e 8 1 a-e 9 1 arrt 10 4 2. 4 " 5 2 ";) 2 '* 8 9 10 2 11 3: 5 6 3 7 3 " 8:3 " 1 3 't 1 ' 'i 1 ' 4 6 4 4 8 4 9 4 " 1 4 " 11 4 ' 12 4 " 5 ', 5 8 5 9 5 ' 10 5 11 5 " 12 5:13. 6 " 8 6 9 i "1 10 6 11 6i 12 6; 13 6 1- 6 15 7 9 7 " 1 7 " 1 1 7 ' 12 7: '3 7 " 14 7 " 14 7 2 7 16 8 10 8 11 8 12 8 1 8"143 8 15 8 '; 17 9 " 11 9 12 9 " 13 " 14 9 " 15 9 " 1; 9 17 9 18 10) 12 10 " 13 10' 14 10 " 15 1) ' 16 13"1 7 " 1 18 10 1' The pupil should be frequently exercised in mental addition, un il he is peTfectlyfamlldiar with the operation. Care should be taken, however, that the questions are not answered mechanically, from a knowledge of the regular increase of numbers. This may be prevented by asking promiscuous questions. 7 + 6 = how many? 9 + 7 =- how many? 9 + 4 = how many? 4 + 3 + 5 = how many? What is Addition? What is the sum? How may Addition be divided? What is Simple Addition? What is the sign of Addition? What does plus mean? *By many Authors, the result obtained by adding two or more numbers together, is ca!led the amount t hut the term Amount, we think properly belongs to Interest, and ahou:d not be applied to Addit'ion. Art. 2. SIMPLE ATDITION. 11 --- 4 -- 7 bhow many? '; - 3 i -- I tow imany? 6 4- 2 — }- 7 -- 8 = how many? 7 8+ 5 + 2 how many? 6 + 8 + 2 4 -+ = how many? 7 + 2 + 1 -1- 4 -]- 3 = how many? 3 - 2 — 1 -- 9 - 7 = how many? 9 -_- 7 -1- 5 -+- 6 -"- 2 -1- 1 -- 3 = how many? Example 1. A man has 2 orchards; one has 24 trees, and the other has 32 trees. How many trees in both orchards? Solution.-In 24 are 2 tens and 4 units.; in 32 are 3 tens and 2 units. Now units and tens cannot be added together, because it takes ten units to make one ten; but we can add units to units, and tens to tens; and 4 units and 2 units are 6 units, and 2 tens and 3 tens are 5 tens, and 5 tens and 6 units arc 56. Ans. 66. Or we may set the numbers down on the slate, and 32 add them, recollecting to place units under units, and 24 tens under tens. Thus: 56 Obs. 4. We place units under units, tens under tens, &c., because it is more convenient to add them when placed in this manner, as we can only add together numbers of the same name. HenceTo add numbers: Obs. 5. Write units under units, tens under tens, hundreds under hundreds, d'c. Conmmence at the right hand. Add each column separately, and place the result beneath it. Ob.. 6. PRooF.-Begin at the top and add downwards; if the two results are alike, the work is correct. 1. We add the numbers from the top downwards, in order to take them in a different manner, it being hardly probable that the same mistake would occur by both processes, unless it was intentional. 2. The learner must not think when he takes up the slate, that he can dispense with thinking and reasoning. This is not the case. He should exercise his mental faculties as much when he uses the slate as when he does not. His slate is merely used for convenience in setting down the operation when it is too long, or the numbers too large to be easily retained in the mind. That is, it is used to assist, and not to take the place of the mental faculties. What does it signify? What is the sign of equality? What does it show? How do we set the numbers down to add? Why do we place units under units, tels under tens, &c.? Where do we commence to add? Where do we place the sum of each column? How do we prove the operation? Why do we add the numbers from the top downwards? Should the scholar dispense with thinking and reasoning when he takes up the slate? What is the use of the slate? 12 COMMON ARITHMETIC. Sect. II. 2. In one field are 127 sheep, in another 231, and in another 341. How many are there in all the fields? Ans. 699. 3. In one school are 62 scholars, in another 123, and in another 114. How many are there in all? Ans. 299. 4. One man has 324 dollars, another 232 dollars, another 2123 dollars, and another 6210 dollars. How many dollars have they all? Ans. 8889. 5. In one book are 240 pages, in another 320 pages.;n another 1112 'ages, in another 1023 pages, and in another 1304 pages. How many pages in all? Ans. 3999. 6. 241 -1- 122 -j- 1234 -- 21201 --- 45101 = how many? Ans. 67899. 7. 1234 -1- 20412 -I- 30211 -+- 12132 -I- 24000 = how many? Ans. 87989. ARTICLE 3. CARRYING IN ADDITION. Ex 1. A man has 76 bushels of grain in one box, and 58 in another box. How many bushels has he in both? Solutionm.-Here a difficulty presents itself, as 8 units -- 6 units - 14 units, which cannot be expressed by one figure; but the learner must recollect that 14 units = 1 ten and 4 units; we will therefore write the 4 units under the units, and reserve the 1 ten for the next column. Thus: 8 units-1-6 units = 14 units, which is 1 ten and 4 units; 5 tens -1- 7 tens = 12 tens, -I- 1 ten, which was reserved = 13 tens, which is I hundred, and 3 tens; and the whole result is, 1 hundred, 3 tens, and 4 units, or 134. Ans. 134. Hence-When the sum of any column exceeds 9 -Obs. 1. Set down the unit figure only, and carry the tens to the next column; or in other words, carry 1 for every 10. This principle may be illustrated as follows: 2. Let it be required to add together 446 and 678. The sum of 6 units -- 8 Operation. units = 14 units, or 1 ten 678 and 4 units, which we set 446 down accordingly: 4 tens -1- 7 tens = 11 tens, or 1 14 = the sum of the units. hundred and 1 ten, which 11- d" " " tens. we write in their respective 10 -= ' " " hundreds. orders: 4 hundreds and 6 hundreds = 10 hundreds, 1124 = the total sum. or 1 thousand, 0 hundreds, which we also place in their proper places; then adding the several sums together, as they stand, we obtain 1124 as the total sum or answer. Ans. 1124. How do we proceed when the sum of any order exceeds 97 Explain the prinoiple of carrying. Art. 3. SIMPLE ADDITION. 3. Add together 2467, 1924, 2761, and 6279, and prove the operation. 2467 1924 2761 6279 Sum 13431 As the two results are alike, the work is supposed to be correct. Proof. 13431 From the preceding illustration, we derive the following GENERAL RULE FOR ADDITION. I. lVrite the numbers to be added so that the figures of the same order may stand directly under each other, and draw a line beneath. (Art. 2. Obs. 5.) 1I. Begin at the right hand and add each column separately; and if the sum be 9, or less, write it under the column added. (Art. 2. Obs. 5.) III. But if the result should exceed 9, set down the unitfigure of the result, and carry the tens to the next column. (Art. 3. Obs. 1.) IV. Set down the whole sum of the left hand column. PROOF.-Begin at the top and add downwards; if the two results are alike, the work is correct. (Art. 2. Obs. 6.) EXERCISES FOR THE SLATE. 1. A man bought a horse for 85 dollars, and a wagon for 67 dollars. What did they both cost him? Ans. 152 dollars. 2. A man has 64 sheep in one field, and 148 in another field. How many sheep has he? Ans. 212. 3. A man being asked his age, answered: " I was 27 years of age when I married, and have been married 49 years." How old was he? Ans. 75 years. 4. A man has owing to him-from A. 463 dollars, from B. 798 dollars, from C. 840 dollars, from D. 56 dollars, and from E. 15 dollars. How much has hie owing him? Ans. 2172 dollars. 5. A man has 423 bushels of wheat, 291 bushels of oats, 1479 bushels of corn, and 276 bushels of barley. How many bushels of grain has he in all? Ans. 2469. 6. A man received 1243 dollars for horses, 972 dollars for cattle, 672 dollars for grain, 427 dollars for sheep, 197 dollars for hogs, and has 327 dollars at home. How many dollars has he in all? Ans. 3838. 7. A man has 4 children: one is 18 years old, another is 22 years old, another is 16 years old, and the other is 12 years old; What is the goneral rule for Addition? The proof? 14 COMMON ARITHIMETIC. Sect. II. thne,fthier's a.- i, 21 v<.rs more than t11.e cl of ci hid e c:l'n.rc's, lqt'ir't'-th' ^ of ti).'t te o thf t er..A; 9 yea,., 8. A. ow:es one man 642 dollars, anotier 927 dollars, another 745 dollars, and another 2457 dollars. What is the sum of his debts? Ans. 4771 dollars. 9. A merchant commenced trading with 14768 dollars; the first year he gained 1794 dollars, the second year he gained 2986 dollars, the third year he gained 3789 dollars, and the fourth year he gained 4697 dollars. How much had he then? Ans. 28034 dollars. 10. A land-holder has in one farm 1217 acres, in another 976 acres, in another 840 acres, and in another 679 acres. How many acres has he? Ans. 3712. 11. A boy walked-one day 8 miles, another day 12 miles, another day 17 miles, and another day 23 miles. How many miles had he walked in all? Ans. 60. 12. A drover has in one field 329 sheep, and 142 lambs; in another field he has 476 sheep, and 216 lambs; and in a third field he has 512 sheep, and 319 lambs. How many sheep has he?How many lambs? How many sheep and lambs together? Ans. to the last. 1994. 13. Four men traded in partnership. A. put in 1269 dollars, B. put in 7642 dollars, C. put in 3768 dollars, and D. put in 5429 dollars. How much did they all put in? Ans. 18108 dollars. 14. A gentleman owns a farm worth 8478 dollars, a store worth 12694 dollars, a house andl lot worth 8621 dollars, a vessel worth 10216 dollars, and has 14376 dollars worth of other property.How much is lie worth in ali? Ans. 54385 dollars. 15. A man left his property to his wife, two sons, and three daughters: to his wife he gave 14612 dollars, to each of his sons he ave 8279 dollars, and to each of his daughters he gave 6297 dollars. What was the value of his property? Ans. 50061 dollars. 16. A man has paid on a, note-at one time 417 dollars, at another time 312 dollars, at another time 512 dollars, and at another time 794 dollars; he now has 238 dollars to pay. How much was the note? Ans. 2273 dollars. 17. The distances on the Ohio canal are as follows: from Portsmouth to Chillicothe 51 miles; from Chillicothe to Circleville 22 miles; from Circleville to Newark 60 miles; from Newark to Roscoe 41 miles; from Roscoe to Dover 42 miles; from Dover to Massilon 28 miles; from Massilon to Akron 27 miles; from Akron to Cleveland 38 miles. Reqiired-the length of the Ohio canal. Ans. 309 miles. 18. A man bought a span of horses for 312 dollars, a carriage for 497 dollars, and a harness for 75 dollars. What did they all cost? Ans. 884 dollars. Art. 3. SIMPLE ADDITION, 15 19. A merchant bourrht cali,(eo the amf i m:nt of 64 - do!' —l bro.i clolh t, t he amount of 837 ldo:'lir, and other gooi< t-> t, amount of 497 dollars. What did they all cost? Ans. 1983 dollars. 20 The above merchant sold his goods so as to gain 178 dollars on the calico, 279 dollars on the broadcloth, and 125 dollars on the other goods. How much did he gain, and for what sum did lie sell the lot? Ans He gained 582 dollars. He sold the lot for 2565 dollars. 21. There are four numbers: the first is 1248, the second is 2397, the third is as much as the first and second, and the fourth is as much as the second and third. Required-the third and the fourth numbers, and the sum of the four. Ans. to the last. 13332. 22. After a battle it was found that 7620 men were killed, 9276 were wounded, 792 had deserted, 5874 had been taken prisoners, 1892 were missing, and 32716 were left fit for action. Of how many men did the army consist at first? Ans. 58170. 23. The second Punic war commenced 529 years after the founding of Rome; Carthage was destroyed 78 years later; the Christian Era commenced 146 after this; the death of Nero was 69 years later; 242 years afterwards Constantine ascended the Roman throne, and the dissolution of the Roman Empire occurred about 165 years after. Required-the age of the Roman Empire. Ans. 1229 years. 24. Find the sum of the following' numb)ers: two hundred and four; seven hundred an(l thirty-nine; one thousand and seventeen; three thousand, seven hundred and sixty; fifteen thousand, nine hundred and nine; three hundred and six thousand, one hundred and eiglht. Ans. 337737. 25. Aill the following numbers: two thousand and one; seven thousanl, four hunlllred and seventy-nine; eleven thousand, one hundred and eleven; two hundred and twelve thousand and ninetyfive; eight hundred thousand and one; nine hundred and ninetynine thousand, nine hundred and ninety-nine. Ans. 2032686. 26. In one book are two hundred and thirty-two pages, in another are two hundred and sixty-four pages, in another are three hundred and forty-six pages, and in another are two hundred and ninety five pages. How many pages in all? Ans. 1137. 27. A man's farm cost him eighteen thousand, five hundred dollars; a store cost him twenty thousand, two hundred and fifty dollars; a house and lot ten thousand and fifty dollars, and a boat twelve thousand, five hundred dollars. What was the cost of the whole? Ans. 61300. 28. Required-the sum of 12345 -H- 67890 -- 96432 -1- 7456 -j- 6217 -1- 149321 -I- 819360. Ans. 1159021. 16 COMMON ARITHMETIC. Seat. III. 29. Required-the sum of 97452 -- 68714 -- 127983 -1- 15791 -- 6829467 -[- 8932164-I- 187437621 -I- 19734653 -I- 890072 -- 6214573 -- 9876543210. Ans. 10106891700. 30. In 1840 the New England States contained 2234822 inhabitants, the Middle States contained 4604345, the Southern States contained 5067843, the Western States contained 4984097, the Territories contained 128534, the District of Columbia contained 43712, and on board vessels of war were 6100. Required-the population and Naval service of the United States in 1840. Ans. 17069453. SECTION III. SIMPLE SUBTRACTION. ARTICLE 1. MENTAL EXERCISES. 1. John had 8 apples, and gave 3 of them to his sister. How many had he left? Take 3 from 8, and how many remains? Ans. 5. 2. Mary had 12 pins, and lost 5 of them. How many had she left? 3. A boy having 18 cents, bought a slate for 10 cents. How many cents had he remaining? 4. A boy had 9 marbles, and gave his brother 5. How many had he left? 5. A lady having 16 yards of cloth, cut off 9 yards. How many yards were left? 6. A man having 14 dollars, paid away 7 dollars. How many dollars had he then? 7. James went 15 miles, and John 5. How much farther did James go than John? 8. Henry had 12 cents, and spent 6 of them. How many had he left? 9. William made 17 marks on his slate, and rubbed out 8 of them. How many marks were there left? 10. John's book had 20 pages in it, and he tore out 10 of them. How many pages were left? 11. Fourteen boys were standing together, and 5 of them went away. How many remained? 12. A man having 22 sheep, sold 11 of them. How many had he left? ARTICLE 2. DEFINITIONS, &C. Obs. 1 The finding the di@ference between two numbers, as in the preceding examples, is called SUBTRACTION. It may be defined -the takin qf a le wrmbOrfrom a grecatr. Art. 2. SIMPLE SUBTRACTION. 17 Subtraction may be either of Simple or of Compound numbers. Obs. 2. Simple Subtraction is when the numbers all express things of the same name, or kind-as all dollars, all cents, all yards, &c. Obs. 3. The greater number is called the MINUEND, (which means to be diminished.) The lesser number is called the SUBTRAHIEND, (which means to be subtracted.) The result, or answer, is called the DIFFERENCE, or REMAINDER. Obs. 4. SIGN.-The sign of Subtraction is a short horizontal line (-), called MINUS, which means less, and signifies that the number after it is to be taken from the number before it. Thus: 8-5=3 shows that 5 taken from 8 leaves 3, and reads, 8 minus 5 is equal to 3. To facilitate the progress of the learner, we subjoin the following SUBTRACTION TABLE. 1 from 2 from 3 from 4 from 5 from 1 leaves 0 2 leaves 0 3 leaves 0 4 leaves 0 5 leaves 0 2 " 1 3 " 1 4 " 1 5 1 6 " 1 3 " 2 4 " 2 5 " 2 6 ' 2 7 " 2 4 " 3 5 " 3 6 " 3 7 " 3 8 ' 3 5 " 4 6 " 4 7 " 4 8 < 4 9 " 4 6 " 5 7 " 5 8 5 9 " 5 10 " 5 7 " 6 8 " 6 9 " 6 10 " 6 11 6 8 " 7 9 " 7 10 " 7 11 " 7 12 7 9 " 8 10 " 8 11 " 8 12 " 8 13 " 8 10 " 9 11 9 12 " 9 13 " 9 14 "- 9 6 from 7 from 8 from 9 from 10 from 6 leaves 0 7 leaves 0 8 leaves 0 9 leaves 0 10 le'ves0 7 " 1 8 " 1 9 " 1 10 " 1 11 " 1 8 " 2 9 " 2 10 " 2 11 " 2 12 " 2 9 3 10" 3 11" 3 12 " 3 13 3 10 " 4 11 4 12 " 4 13 " 4 14 4 11 " 5 12 " 5 13 5 14 5 15 " 5 12 6 13 " 6 14 15 6 16 " 6 13 " 7 14 7 15 " 7 16 " 7 17 " 7 14" 8 15" 8 16 " 8 17 "8 18 " 8 15 " 9 16 " 9 17 9 18 " 9 19 " 9 NOTE.-This Table should be thoroughly committed to memory before proceeding farther. What is Subtraction? How may it be defined? How divided? What is Simple Subtraction? What is the greater number called? What does minuend mean? What is the lesser number called? What does subtrahend mean? What is the result, or answer, called? What is the sign of Suhtraction? What does minus mean? What does it signify? What does 8-5=;3 show? How is it read? 18 COMMON ARITHMETIC, Sect. II. 6 - 4 = how many? 18 - 7 -- 3 = how many? 14 - 3 = how many? 17 - 8- 9 = how many? 17 - 7 = how many? 14 - 5-1- 6. how many? 12 - 5 = how many? 12 - 3 - 9 = how many? 18 - 10 = how many? 16 - 4 -- 8 how many? 17 - 6 = how many? 1.9 - 10 -i- 7 = how many? Obs. 5. A line, or vinculum (-), drawn over two or more numbers, signifies that they are to be taken together as one number. A parenthesis ( )is also sometimes used for the same purpose. Thus:18 - 4 -i- 2, or 18- (4 -1- 2) signifies that 4 -- 2, or 6, is to be taken from 18, which leaves 12; but 18-4 -- 2 signifies that 4 is to be taken from 1 8, and 2 added to the remainder, which makes 16. 18- 4 — 7 how many? 7-1 6 3-j- 8 =how many? 7 -- - [- 3 —6 - (9 -- 4-I- 7) = how many? 12 -3 - 2 -4 -- 7 -j- 6 = how many? 7-i- 8 -- 4- (10 -1-7-1- 6 - 4) == how many? 6 - 8 - -I- 2-1-7-l-6 - (9 — 8)= how many? Obs. 6. Subtraction is the reverse of Addition. Addition is finding the sum, and Subtraction is finding the difference of numbers. Now if 3 taken from 9 leaves 6, it is evident that 3 added to 6 will equal 9. Hence: To prove SubtractionObs. 7. Add the remainder and subtrahend together, and if their sum is equal to the minuend the work is correct. Ex. 1. A man having 98 dollars, paid 52 dollars for a horse.How much had he left? Ans. 46 dollars. Solution.-98 is composed of 9 tens and 8 units; 52 is composed of 5 tens and and 2 units: now we cannot take units from tens, nor tens from units, because it takes 10 of the one (units), to make 1 of the other (tens); but we can take units from units, and tens from tens. Thus: 2 units from 8 units leave 6 units, and 5 tens from 9 tens leave 4 tens, and 4 tens and 6 units -= 46. Or, we may set the numbers down, the less under the greater, What does a line drawn over two or more numbers signify? What other mark is used forthe same purpose? Of what is Subtraction the reverse?Show the comparison. How do wo prove Subtraction? Why cannot we take units from tens, or tens from units? How do we set the numbers down to subtract? 0' hy do we place units under units, &c.? (Sect. 11, Art. 2, Obs. 4.) Where do we commence to subtract? Where do we set the remainder? Art. 3. SIMPLE SU BT.RACTION. 19 and subtract one from the other, commencing at the right hand. Thus: Proof. 98 52 In subtracting, we say. 2 from 8 leaves 6, and 52 46 place the 6 under the 2; then 5 from 9 leaves 4, and - - place the 4 under the 5. 46 98 Hence-To subtract numbers: Obs. 8. Place t/he less number under the greater, setting units under units, tens under tens, &c*., and draw a line beneath. Commence at the right hand. Take successively, each figure in the lower line from the figure above it, and write the difference below. 2. A man had 288 sheep, and sold 184 of them. How many had he left? Ans. 104. 3. A man has lived 97 years, and was married when he was 25 years of age. How many years was he married? Ans. 72. 4. A gentleman having 948 dollars, paid away 736 dollars. How many dollars had he remaining? Ans. 212. 5. One book contains 489 pages, and another contains 372 pages. How ihany more pages in the one than in the other? Ans. 117. 6. A man's property is worth 6894 dollars, and his debts amount to 4080 dollars.. HIow much will remain after paying his debts? Ans. 2814 dollars. ARTICLE 3. BORROWING AND CARRYING IN SUBTRACTION. Ex. 1. A man gave 95 dollars for a horse, and 68 dollars for a wagon. How much more did he give for the horse than for the wagon? Ans. 27 dollars. Soluti6to.-Here a difficulty occurs; for we cannot take 8 units from 5 units. But 95 is composed of 9 tens and 5 units; now if we take 1 ten from the 9 tens, and add it to the 5 units, we shall have 8 tens and 15 units; then 8 units from 15 units leaves 7 units, and 6 tens from 8 tens leaves 2 tens; and 2 tens and 7 units -- 27. 2. Required-the difference between 124 and 86. In this example we cannot take 6 Operation. units from 4 units, therefore we bor- tens. units. row 1 ten from the 2 tens, and add 124 — 11 -- 14 the 4 units, making 14 units, and 14 86 8-1- 6 — 6- 8 units. Again, we cannot take 8 tens from 1 ten, so we bor- Ans. 3 -]- 8 38 rem. row the I hundred and add it to the 1 ten, which makes 11 tens; then 11 - 8 -- 3 tens, and 3 tens and 8 units -38. Explain the method of separating a number into its numerical parts, as in Examples 1st and 2nd. Explain the principle of borrowing ten. Wlien the lower figure is the largest, can we subtract without resolving the numbers into their numerical parts? 20 COMMON ARITHMETIC, Sect. III. Obs. 1. The learner will perceive in both these examples, that we add ten to the figure of the minuend when it is less than the corresponding figure of the subtrahend. Indeed, it cannot be otherwise, when we borrow 1 of the next higher order, as 1 unit of any order is equal to 10 units of the next lower order. (Sect. 1. Art. 2, Obs. 3.) This is called borrowing ten. 3. Required-the difference between 526 and 328. Operation. In the solution of this example, instead of resolving 526 the numbers into their numerical parts, as above, we 328 will set them down according to Obs. 8., Art. 2. To - take 8 from 6 is impossible; therefore we will borrow 198 1 (ten) from the 2 (tens), and add it to the 6, making 16; then 16- 8 =8. Now to compensate for the one we borrowed, we will add 1 to the next figure of the subtrahend, (which is the same as taking one from the next figure of the minuend,) and 2 -1- 1 - 3; then 2 - 3 is impossible; hence we will borrow 1 (hundred), and add it to the 2 (tens), making 12 (tens), and 12- 3 =9. Then 3 1 —1 = 4, and 5 -4 1. Ans. 198. REMARK 1.-We add 1 to the next figure of the subtrahend, because it is more convenient than to take 1 from the minuend. 2.-The reason of carrying 1 is evident from this fact: As we borrow 1 from the minuend, we must either make the figure of the minuend, (which we borrowed from) 1 less, or the figure of the subtrahend immediately beneath it 1 greater, to pay for that we borrowed. From the foregoing remarks and illustrations, we derive the following GENERAL RULE FOR SUBTRACTION. I. Write the less number under the greater, that the figures of the same order may stand under each other. (Art. 2, Obs. 8.) II. Commence at the right hand. Take successively each figure of the lower number from the one above it, and write the remainder below. (Art. 2, Obs. 8.) III. When the figure in the lower number is the largest, add ten to the figure above it, after which subtract as usual, remembering to add, or carry 1 to the next figure in the lower number before the next subtraction. (Art. 3, Obs. 1. Rem.) PROOF.-Add the remainder and subtrahend together; if their sum is equal to the minuend, the work is correct. (Art. 2, Obs. 7.) EXERCISES 'OR THE SLATE. 1. A man had 1000 dollars, and paid away 478 dollars. How many dollars had he left? Ans. 522. 2. A man bought a farm for 7864 dollars, and afterwards sold it Explain the process. Why do we add 1 to the next figure of the subtrahend? Explain the reason of carrying. What is the general rule for Sub. traction? The proof? Art. 8. SIMPLE SUBTRAC('ON. for 975 dollars less than he gave for it. For how much did he sell it? Ans. 6889 dollars. 3. A merchant bought 8473 dollars worth of goods, and sold them for 10312 dollars. How many dollars did he gain? Ans. 1839. 4. A man paid 6767 dollars for land, and 12843 dollars for merchandise. [low much more did he pay for the merchandise than for land? Ans. 6076 dollars. 5. A merchant sold goods to the amount of 15784 dollars; he paid 2897 dollars less for them. How much did they cost him? Ans. 12887 dollars. 6. America was discovered in 1492. How many years since, it now being 1849? Ans. 357. 7. The United States declared their Independence in the year 1776. How many years since, the present year being 1849? Ans. 73. 8. A gentleman's income is 4742 dollars a year, and his expenses are 3953 dollars a year. How much does he save per year? Ans. 789 dollars. 9. A farmer raised 1747 bushels of grain one year, and 1699 bushels the next year. How many bushels did he raise the first year more than the second? Ans. 48. 10. At a certain school are 423 students, of which 137 are young ladies. How many gentlemen are there? Ans. 286. 11. From Columbus, (Ohio,) to Cincinnati, by way of Springfield. it is 127 miles; and from Springfield to Cincinnati it is 85 miles.- - How far is it from Columbus to Springfield? Ans. 42 miles. 12. In the year 1800, the population of the United States was 5305925, and in 1840 it was 17069453. Required-the increase of population in 40 years. Ans. 11763528. 13. In 1830 the population was 12861192. Required-the increase in ten years. Ans. 4208261. 14. A drover having 1468 sheep, sold 948; he then bought 467, and sold 987. How many had he left? Ans. None. 15. A vessel, the cargo of which was valued at 84000 dollars, in a storm lost part of her cargo valued at 27212 dollars. What was the value of the remaining part? Ans. 56788 dollars. 16. A man bought 58 dollars worth of wheat, 97 dollars worth of pork, 73 dollars worth of cheese, and a fine horse and buggy for 347 dollars. He gave his note for 497 dollars, and paid the rest in money. How much money did he pay? Ans. 78 dollars. 17. A man borrowed at one time 217 dollars, at another time 313 dollars, and at another time 428 dollars; he afterwards paid 869 dollars. How much did he then owe? Ans. 89 dollars. 18. A man at his death left each of his two sons 5732 dollars, and each of his three daughters 4784 dollars, and his widow the COMMON ARITHMETIC. Sect. III. balance of his property. How much did the widow receive, the estate being worth 40000 dollars? Ans. 14184 dollars. 19. A gentleman's property was worth 34768 dollars; but a store worth 4762 dollars, and 14796 dollars worth of goods were destroyed by fire. How much had he left? Ans. 15210 dollars. 20. A man owning 4821 acres of land, gave one son 623 acres to another son 427 acres, and to another son 873 acres. How many acres had he left? Ans. 2898. 21. A man's income is 6000 dollars a year. He spends 372 dollars for clothing, 724 dollars, for house rent, 892 dollars for provisions, 429 dollars for servants, and 527 dollars for traveling.How much has he left at the end of the year? Ans. 3056 dollars. 22' A man owing 1894 dollars, paid at one time 723 dollars, at another time 674 dollars, and at another time 500 dollars. How did the account then stand? Ans. He overpaid 3 dollars. 23. In 1840 the population of the New England States was 2234822; of the Middle States 4604345; of the Southern States, including Florida, 5122320; and of the Western States, including Wisconsin and Iowa, 5058154. Required-the excess of the population of the Southern and Western States over that of the New England and Middle States? Ans. 3341307 inhabitants. 24. In the last question, required-the excess of the population of the Middle States over that of the New England States. Ans. 2369523. 25. Also, required-the excess of the population of the Southern States over that of the Western' States. Ans. 64166. 26. By the last census, the population of the Northern States* was 9807007, and the population of the Southern States* was 7256346. Required-the excess of the population of the Northern over that of the Southern States. Ans. 2550661. 27. A man bought 27 dollars worth of sugar, 19 dollars worth of spice, 43 dollars worth of coffee, and 10 dollars worth of tea. He gave in payment a one hundred dollar bill. Had he ought to receive any thing back? If so, how much? Ans. He should receive 1 dollar. 28. A certain store-house contained 5970 bushels of wheat, 3752 bushels of corn, 5978 bushels of rye, and 9847 bushels of oats. It caught fire, and 19768 bushels of grain were saved. How many bushels were destroyed? Ans. 5779. 29. A ship of war sailing with 650 men, lost in one battle 29 men, in another 37, and by sickness 19 more. How many were still living? Ans. 565. 30. Four men bought a lot of land for 978 dollars. The first man paid 386 dollars; the second paid 97 dollars less than the *In this example, by the Northern are jmeant the Free States, and by the Southern are meant the Slave States. 81MTLE MULTIPLICATION. 23 first; the third paid 73 dollars less than the second; and the fourth paid the balance. How much did the second, third and fourth men pay? i The second man paid 289 dollars. Ans. < " third " 216 " fourth " 87 31. From six trillions, eilghty-seven millions, twelve thousand and three, take twenty billions, one hundred millions, two hundred and sixteen thousand and nine. Ans. 5979986795994. SECTION IV. SIMPLE MULTIPLICATION. ARTICLE 1. MENTAL EXERCISES. 1. If one lemon cost 4 cents, how much will 3 lemons cost? Solution.-If one lemon costs 4 cents, 2 lemons will cost 4-1- 4 8 cents, and three lemons will cost 8 — 4, or 4-[-4 — 4 = 12 cents. Ans. 12 cents. 2. What costs 4 oranges at 6 cents apiece? 6 — 6- -6-1-6 -- how many? 3. At 2 dollars a pair, what will 3 pair of boots cost? 4. In one yard are 3 feet; how many feet in 4 yards? 5. How much will 5 yards of ribon cost at 8 cents a yard? 6. How much will 7 lead pencils cost at 5 cents apiece? 7. What cost 9 pounds of saleratus at 6 cents a pound? 8. At 10 cents a pound, how much will 10 pounds of sugar cost? 9. What cost 8 slates at 6 cents apiece? 10. At 4 dollars a barrel, how much will 7 barrels of flour cost? 11. If one sheep costs 2 dollars, how much will 11 sheep cost? 12. At 12 cents a pound, how much will 6 pounds of butter cost? ARTICLE 2. DEFINITIONS, &C. There is an orchard containing 4 rows of trees, and 6 trees in a row. How many trees are there in the orchard? Instead of adding the 4 six times, or the 6 four times, as 6 in the preceding examples, we will shorten the operation by 4 multiplying one by the other. Thus: 24 To illustrate this, we will suppose these * * * * ** stars to be the orchard. We perceive * * * * ** that there are 4 rows of stars, and 6 * * * * * stars in a row, and 24 stars in all. There- * * * * * fore 4 times 6 are 24. Now count the stars the other way. We perceive that there are 6 rows, and 4 stars in a row, and 24 stars 24 COMMON ARITHMETIC, Sect. IV. in all, as before. Hence, 6 times 4 are 24. Thus, we perceive it makes no difference whether we multiply 6 by 4, or 4 by 6, the result will still be the same-that is, 24. Ans. 24. At 5 dollars a yard, what will 5 yards of cloth cost? 5 times 5 are how many? Obs. 1. This method of repeating a number is called MULTIPLICATION. Multiplication may therefore be definedA method of repeating any number a given number of times. Multiplication may be either of Simple or Compound numbers. Obs. 2. Simple Multiplication is when the number to be multiplied expresses things of but one name or kind-as dollars, yards, &c. Obs. 3. The number which we multiply, or which is to be REPEATED, is called the MULTIPLICAND. The number which we multiply by, or which shows how many times the other is to be repeated, is called the MULTIPLIER. The result, or answer, is called the PRODUCT. The multiplicand and multiplier taken together are called the FACTORS, or producers of the product. Obs. 4. SIGN.-The sign of multiplication is a cross (X) like the letter X, and signifies that the numbers between which it stands are to be multiplied together. Thus: 4 X 6 signifies that 4 and 6 are to be multiplied together, the product of which is 24. Before proceeding farther, the learner must commit accurately to memory the following MULTIPLICATION TABLE. 2 times 3 times 4 times 5 times 6 times 7 times 1 are 2 1 are 3 1 are 4 1 are 5 lare 6 1 are 7 2 "4 2 " 6 2 " 8 2 10 2"12 2 "14 3 " 6 3 " 9 3 "12 3 "15 3 18 3 21 4 " 8 4 "12 4 "16 4 "20 4 "24 4 "28 5 "10 5 "15 6 " 20 5 25 5 "30 5 "35 6 "12 6 "18 6 "24 6 "30 6 "36 6 "42 7 "14 7 "21 7 "28 7 35 7 "42 7 49 8 "16 8 "24 8 "32 8 " 40 8 "48 8 "56 9 "18 9 "27 9 " 36 9 "45 9 "54 9 "63 10 "20 10 "30 10 " 40 10 "50 10 "60 10 "70 11 "22 11 "53 11 "44 55 1 1 "5 6 11 " 77 12 ' 24 12 "36 12 "48 12 "60 12 "72 12 84 What is Multiplication? How is it divided? What is Simple Multiplication? What is the number we multiply called? What is the number we multiply by called? What is the result, or answer, called? What are the multiplicand and multiplier, taken together, called? What is the aiga of multiplicat94n? What does it *ignify7 Art. 3. SIMJPLE MULTIPICATION. 25 MULTIPLICATION TABLE-CONTINUED. 8 times 9 times 10 times 11 times 12 times 1 are 8 1 are 9 1 are 10 1 are 11 1 are 12 2 " 16 2 18 2 " 20 2 24 3 " 24 3 " 27 3 " 30 3 " 36 4 " 32 4 " 36 4 " 40 4 " 44 4 " 48 5 " 40 5 " 45 5 " r5 5 " 55 5 60 6 " 48 6 " 54 6 " 60 6 " 66 6 " 72 7 " 56 7 " 63 7 " 70 7 '" 77 7 " 84 8 " 64 3 " 7l 8 " 80 8 " 81 8 " 96 9 " 72 ) " 0~, 90 9 9 9 " 108 10 " 80 10 90 100 " 01 10 10 120 11 " 88 11 " 99 1 11 " 121 11 " 132 12 " 96 12 " 108 12 " 120 12 " 132 12 A 144 NOTE.-The pupil shold tnt leave this Table until he has thoroughly committed it to memory. Without knowing it perfectly, no one can becone a good arithmetician. 7 X 6 = how many? 6 X 9 how many? 9 X 5 = how many? 7 X 8= how many? 7 X 4 =.~w n.Wnv? 9 X P = vv L;!n,? 8 X 7 = how many? 9 X 7 how many? 11 X 6 how many? 11 X 12= how many? 12 X 8 how many? 11 X 7 _ how many? In each of these questions we took one number as many times as there were units in another. HenceObs. 5. To multiply one number by another is to take the mudtiplicand as many times as there are units in the multiplier. The product of any two numbers is the same, whichever factor is taken as the multiplier. Obs. 6. As it does not alter the name of a number to repeat it, and the multiplicand is always the number repeated, it is evi lent that the product will always be of the same name as the multiplicand.REMARK 1.-The multiplier must always be an abstract number, as it ex presses the number of times the multiplicand is taken. 2.-An abstract number is one that has no relation to any particular object whatever, but merely expresses the number: as 14, 29, &c. 3-When a number has some relation to a particular object, it is called a concrete number: as 5 dollars, 8 miles, &c. To multiply one number by another, is to take the multiplicand how many times? Does it make any difference in the result which factor isused as the multiplier? Does it alter tthe name of a number to repeat it? Which factor is repeated? Of what name then is the product? What must the multiplier always be? Why? Whatis an abstract number? A concrete number? What numbers can be multiplied together? If the multiplicand is an abstract number, what is the product? If the multiplicand is a concrete number, what is the product? 3 26 COMMON ARITHMETIC. Sect. IV. 4.-As we cannot multiply by a concrete number, it follows that we can only multiply two abstract numbers together, in. which case the product is abstract, Or an abstract and concrete number together, in which case the product is concrete. Ex. 1. If 1 yard of cloth cost 5 dollars, what will 6 yards cost? Here we have two concrete numbers, viz: 5 dollars and 6 yards. Now the multiplier should be an abstract number, for it would be equally as absurd to multiply 5 dollars by 6 yards, as it would be to multiply 5 houses by 6 gardens. But we can resolve the question in this manner: if 1 yard costs 5 dollars, 6 yards will cost 6 times as much as 1 yard, or 6 tims 5 dollars, which is 30 dollars. In this way we make the 6 an abstract number. Let the learner resolve the following questions in the same manner: 2. A man bought 12 cows, paying 11 dollars apiece for them. How much did they cost him? 3 How much would 9 calves come to, at 4 dollars apiece? 4. How much would 8 pair of boots cost, at 5 dollars a pair? 5. How much would 32 yards of cloth cost, at 3 dollars a yard? Solution.-To obtain the result in this question, we set one number under the other, and multiply each figure 32 of the multiplicand separately. Thus- 3 3 times 2 units are 0 units, and 3 times 3 tens are 9 tens, and 9 tens and 6 units are 96. Ans. 96 dollars. Obs. 7. PRooF.-Take 1 from the multiplier; multiply the multiplicand by the remainder, and to the product add the multiplicand; if this result is the same as the first, the work is correct. Proof of the last Example. 32 3-1=, 2 As this result is the same as the first, the work is supposed 64 to be correct. 32 96 The reason of this rule is too evident to need demonstration.Let the learner see if he cannot demonstrate it himself. 6. Multiply 321 by 4. Ans. 1284. 7. If 1 pound of butter cost 8 cents, what will 16 pounds cost? 6 X 8 = 48 = 4 tens and 8 units; we set down Operation. the 8 units; then 8 times 1 ten = 8 tens, and 4 tens 16 make 12 teas. 8 Ans. 128 cents. Art. 2. SIMPLE MULTIPLICATION. 27 This principle may be illustrated as follows: 8. Required-the product of 247 by 9. 1st Operation. 2d Operation. 247 247 =200 + 40 + 7. 9 200 X 9=1800. ~ --- -~- ~40 X 9 360. 63 = product of 9 into 7 units. 7 X 9 = 63. 36 = " " " 4 tens. 18 = " " " 2 hundreds. Ans. 2223 Ans. 2223_ " " " 247. By each of these methods it will be perceived that the ten's figure of each partial product is added to the next partial product.Thus, in 63, the 6 is added to the next partial product, (36); and in 36, the 3 is added to the next partial product, (18.) Now as this is always the case, we may shorten the operation by not setting down all the work, but merely carrying the tens, as follows: 247 Here we say 9 times 7 are 63, and set down the 3, 9 and carry the 6; then, 9 times 4 are 36, and 6 to carry = 42; we set down 2 and carry 4; finally, 9 times Ans. 2223 2 are 18, and 4 to carry = 22. The learner will perceive that each figure of the product occupies the same place, (numerically,) as the figure multiplied; that is, if the figure multiplied is units, the product is units; if tens, the product is tens, &c. HenceObs. 8. When the multiplier occupies the unit's place, each fgure of the product occupies the same place, (numerically,) as the figure multiplied. From the preceding illustrations, we deduce the following RULE —WHEN THE MULTIPLIER IS LESS THAN 12. I. Write the multiplier under the unit gure of the multiplicand. II. Commence at the right hand. Miultiply successively each figure of the multiplicand by the multiplier, setting down the unit figure of each product, and carrying the tens to the nextproduct, as in Addition. PROOF.-Take 1 from. the multiplier; multiply the nmltiplicand by the remainder, and to- the product add the multiplicand; if this result is the same as the first, the teork is correct. (Obs. 7.) Give the solution of Example 1, and show how the multiplier is made au abstract number. How do we prove mattiplication? Explain the prihciple of carrying, and show why it is correct. Is there any shorter wy -than to set down the whole product of each figure multiplied, or of separating the multiplicand into its nnierie;tl parts? 'Etplai-. it.: Whtt place does each figure of the product occupy, numerically?, What inference is deduced from this? What is the Rule When the multiplier is less tiau 12? The proof? 28 COMMON ARITHMETIC. Sect. IV. EXERCISES FOR THE SLATE. 1. How much would 356 pair of shoes cost, at 2 dollars a pair? Ans. 712 dollars. 2. How much would 431 yards of cloth cost at 5 dollars a yard? Ans. 2155 dollars. 3. If a man rides 120 miles in one day, how far can he ride in 6 days? Ans. 720 miles. 4. In one mile are 320 rods. How many rods in 8 miles? Ans. 2560. 5. How much would 9 horses cost at 140 dollars apiece? Ans. 1260 dollars. 6. How much would 398 pounds of salaratus cost, at 6 cents a pound? Ans. 2388 cents. 7. How much would 554 hats cost, at 5 dollars apiece? Ans. 2770 dollars. 8. There are 8 quarts in a peck. How many quarts in 389 pecks? Ans. 3112. 9. How much would 674 yards of cloth cost, at 4 dollars a yard? Ans. 2696 dollars. 10. How much would 6 carriages cost, at 494 dollars apiece? Ans. 2964 dollars. 11. How much would 738 barrels of flour cost, at 6 dollars a barrel? Ans. 4428 dollars. 12. At 8 shillings a day, how many shillings can a man earn in 325 days? Ans. 2600. 13. How many yards will 12 pieces of cloth contain, each piece containing 42 yards? Ans. 504. 14. How much would the above cloth cost, at 9 cents a yard? Ans. 4536 cents. 15. How much would 386 hats cost, at 8 dollars apiece? Ans. 3088 dollars. 16. How much would 273 pair of boots cost, at 5 dollars a pair? Ans. 1365 dollars. 17. How much would 12 stoves cost, at 37 dollars apiece? Ans. 444 dollars. 18. How much would 198 pounds of coffee cost, at 11 cents a pound? * Ans. 2178 cents. 19 How much would 177 pounds of sugar cost, at 7 cents a pound? Ans. 1239 cents. 20. How much would 789 acres of land cost, at 12 dollars an acre?.; Ans. 9468 dollars. ARTICLE 3. WHEN THE MULTIPLIER EXCEEDS 12. Ex. 1. What will 243 acres of land cost, at 24 dollars an acre? Solution.-We meet with a difficulty in this example, as our mul Art. 3. SIMPLE MULTIPLICATION. tiplier is greater than 12. But 24 is composed of 2 tens and 4 units. We will therefore work this question as follows: In the first place we multiply by the Operation. 4 units, as usual. We next multi- 243 ply by the 2 tens. Now in reality, 24 we multiply by 20, because 2 tens =20. Then 20 times 3 are 60; 972 = productby 4 units. or, if we reject the cipher, 2 times 486 = " 2 tens. 3 are 6; but as the 2 is 2 tens, the 6 is also 6 tens, and should there- 5832 = 24 fore occupy the ten's place, the same as if the cipher were added. We multiply the remaining figures as usual. Finally, we add the two products together, (because 20 + 4 = 24,) as they stand, and their sum is the product required. REMARK.-In writing our factors, we place figures of the same order under each other. 2. Multiply 5297 by 642. The 6 in the multiplier, in this example, occu- Operation. pies the hundred's place, and to multiply by it is 5297 the same as to multiply by 600. Therefore, we 642 place the first figure of the product in the hun- - dred's place. Otherwise, we proceed as in the 10594 last example. 21188 31782 Ans. 3400674 Obs. 1. From the preceding illustrations we perceive that the first figure of each partial product occupies the same place, numerically, as the figure by which we multiply. Hence-When the multiplier exceeds 12: Obs. 2. We first multiply by each figure of the multiplier separately, remembering to place the first figure of the product directly under the figure by which we multiply; and then add together the several partial products, as they stand, for the required product. 3. Multiply 49832 by 148. Ans. 7375136. 4. How much would 1927 tons of iron cost, at 108 dollars a ton? Explain the solution of Example 1st, and show the several steps of the operation. What place, numerically, does the first figure of each partial product occupy? Give the reason of this. How do we proceed when the multiplier exceeds 12? When there are ciphers between the significant figures of the multiplier how do we proceed? What is the general rule for Multiplication? The proof? COMMON ARITHMETIC. Sect. IV. Operation. In this example there are no tens, but as 1927 0 multiplied into any number produces only 108 0, we may omit the 0, and multiply by the significant figures, only, observing to place 15416 the first figure of each partial product accor- 1927 ding to Obs. 2. Hence- Ans. 208116 dollars. When there are ciphers between the significant figures of the multiplier: Obs. 3. We multiply by the significant figures only, and in all other respects proceed according to Obs. 2. 5. Multiply 1776 by 305. Ans. 541680. From the preceding remarks and illustrations, we derive the following GENERAL RULE FOR MULTIPLICATION. I. Write the multiplier under the multiplicand, placing figures of the same order under each other. II. Commence with the units, and multiply each figure of the multiplicand by each sign.ficant figure of the multiplier, placing the first figure of each partial product directly under the figure by which we multiply. (Obs. 2.) III. Add together the several partial products as they stand; their sum will be the product required. (Obs. 2.) PRooF.-Take 1 from the multiplier; multiply the multiplicand by the remainder, and to the product add the multiplicand; if this result is the same as the first, the work is correct. (Art. 2, Obs. 7.) EXERCISES FOR THE SLATE. 1. How much would 596 sheep cost, at 3 dollars apiece? Ans. 1788 dollars. 2. How much would 1487 yards of cloth cost, at 5 dollars a yard? Ans. 7435 dollars. 3. How much would 276 horses cost, 98 dollars apiece? Ans. 27048 dollars. 4. How much would 138 acres of land cost, as 47 dollars an acre? Ans. 6486 dollars. 5. How much would 976 yoke of oxen cost, at 67 dollars a yoke? Ans. 65392 dollars. 6. How much would 235 tons of hay cost, at 16 dollars a ton? Ans. 3760 dollars. 7. How much would 2798 acres of land cost, at 23 dollars an acre?:. Ans. 64354 dollars. 8. If a man can raise 47 bushels of corn on one acre of land, how many bushels can he raise on 179 acres? Ans. 8413. Art. 3. SIMPLE MULTIPLICATION. 31 9. How much would 329 carriages come to, at 574 dollars apiece? Ans. 188846. 10. How many days has a man lived who is 67 years old, allowing 365 days to the year? Ans. 24455. 11. Allowing he has lived but 69 years, how many days would it be? Ans. 21535. 12. If he has lived 84 years, how many days is it? Ans. 30660. 13. If it takes 872 men 204 days to dig a canal, how long will it take 1 man to dig it? Ans. 177888 days. 14. How much would 279 horses cost, at 207 dollars apiece? Ans. 57753 dollars. 15. How much would tih above horses cost, at 196 dollars apiece? Ans. 54684 dollars. 16. How much would they cost, at 175 dollars apiece? Ans. 48825 dollars. 17. How many dollars would 4072 men receive, if each received 408 dollars? Ans. 1661376. 18. If a vessel should sail 129 miles a day, how many miles would she sail in 437 days? Ans. 56373 miles. 19. How much would 19407 pounds of opium cost, at 52 shillings a pound? Ans. 1009164 shillings. 20 How much would 96 thousand feet of boards cost, at 13 dollars a thousand? Ans. 1248 dollars. 21. How much would 284 hogsheads of molasses cost, at 29 dollars a hogshead? Ans. 8236 dollars. 22. If 47 men should form a partnership, each man paying 7684 dollars, what amount of capital would be invested? Ans. 361148 dollars. 23. How much would 923 pieces of broadcloth cost, at 312 dollars a piece? Ans. 287976 dollars. 24. How much would 867 town lots cost, at 489 dollars apiece? Ans. 423963 dollars. 25. Suppose 476 men were to receive 237 dollars apiece for doing a pit ce of work, how much would they all receive? Ans. 112812 dollars. 26. If a carriage wheel turn over 348 times in going a mile, how many times will it turn over in going 957 miles? Ans. 333036 times. 27. Multiplyone hundred and two thousand, four hundred and seven, by threethousand and seventeen. Ans. 308961919. 28. Multiply four hundred and ninety-six thousand, five hundred and fourteen, by thirty-three thousand, nine hundred and ninetynine. Ans. 16880979486. 29. Multiply two million, seven hundred and five, by six hundred and seventy-two thousand, two hundred and fifteen. Ans. 1344903911575. 32 COMMON ARITHMETIC. Sect. IV. 30. Multiply sixty-six trillion, six hundred and sixty-six billion, six hundred and sixty-six, by one hundred and forty-four million, one hundred and forty-four. Ans. 9599913599999904095904. ARTICLE 4. CONTRACTIONS IN MULTIPLICATION. CAsE 1.-When the Mdutiplier is 10, 100, 1000, &c.Ex. 1. How much must I pay for 47 cows, at 10 dollars apiece? Ans. 470 dollars. As ieery remnoval of a figure tm^,trd.s th p 7lei ii-....,se, its value ten times, (Sect. 1. Art. 2. Obs. 3.) it follows, that to multiply any number by 10, we need only remove it one place to the left, or simply annex a cipher; to multiply by 100, remove it two places, or annex two ciphers; to multiply by 1000, remove it three places, or annex three ciphers, &c. Hence-To multiply by 1 with any number of ciphers annexed: Obs. 1. Annex to the multiplicand as many ciphers as there are at the right of the multiplier. NoTE.-Annex means to place after, or at the right hand. 2. There are 10 cents in one dime; how many cents are there in 327 dimes? 3. There are 100 cents in one dollar; how many cents in 385 dollars? 4 There are 1000 mills in ore dollar; how many mills in 476 dollars? 5. There are 10000 mills in an eagle; how many mills in 835 eagles? 6. How many pages are there in 575 books, each book having 1000 pages? CASE 2. —Multiplication by Composite lNumbers. Obs. 2. A Composite number is a number which can be obtained by multiplying two or more factors together, each of which is greater than unity. Thus, 12 can be obtained by multiplying 3 and 4, or 2 and 6 together; therefore, 12 is a composite number. Also, 24 ismcomposed of 8 X 3, or 4 X 6, or 12 X 2, or 4 X 2 X 3, or 2 X 2 X 2 X 3, either of which expressions multiplied together will produce 24. Hence, 24 is a Composite number. Is 48 a composite number? Is 49? 54? 81? 120? 144? 132? Name the factors, or parts, of each of the above numbers. The learner will find all these, as well as all the composite numWhat is it necessary to do to multipiy by 10? Why? By 100? Why? By 1000? Why? How then can we multiply by 1 with any number of ciphers annexed? What does annex mean? What is a Composite number? Give examples. Will any number admit of more than one set of factors? Name some such numbers. Art. 4, CONTRACTIONS IN MULTIPLICA-IIONS bers following, in the Mlitiplication Table, if he examines it attenive, l:'.,yv>.'hi~i*'vhb cn<-^ of I'^'tor:; ^" ^ f:' hey "ire (o JJosed. *MX.,, *- +i?, v'o + 81>*. Kril 4xs -t fac1tors tithA those l',tImId ami! tile Tiabl. Let the h:ar:l.l' see if he caenot ascertain what they a.e. Obs. 3. The numbers, or factors, which are multiplied togeter to produce any number, are called the component parts of that number. Thus: 3 and 4 are the component p-rts of 12, because 3 X 4 - 12; and 3, 2 and 5 are the component parts of thirty, because 3 2 X 5 = 30. What are the component parts, or factors, of 36? 42? 56? 72? 84? 90? 96? 100? 121? 132? 144? REIARK 1.-A number which cannot be produced by the multiplication of two or more factors, is called a PRIME number. Thus: 2, 3, 5, 7, 11, &c., are prime numbers. 2.-Every number must be either a prime or a composite number. Every even number is a composite number. An even number always ends with 0, 2, 4, 6, or 8, and can always be divided by 2-that is, 2 is one factor of all scli numbers All numbers ending with 5, or 0, are composite also, because 5 is one factor of all such numbers. 3.-A prime number is always odd. An odd number always ends with 1, 3, 5, 7, or 9, and cannot be divided by 2-that is, 2 is not a factor of such numbers. A composite number may be either odd or even. 4.-2 is a primt ntunber. altiough it is even. This is the only exception. The learner will unders and the nature of odd and even numbers better when he has become acquainted with l)ivision. Tell which are prime and which are composite numbers in the following: 4, 7, 12, 15, 17, 19, 20, 21, 23, 25, 27, 29, 51, 56, 59, 63, 79, 84, 121, 143, 156, 108, 120. 'he following is a list of all the prime numbers, from 1 to 100: 2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. 1. How much would 18 cows cost, at 23 dollars apiece? Ist Operation. 2nd Operation. 23 23 18 =6 X3. 18 6 one factorof 18. 184 138 = cost of 6 cows. 23 3 the other factor of 18, Ans. 414 dollars. Ans. 414 = cost of 3 times 6, or 18 cows. In this case 18 is composed of two factors, 6 and 3. Then if we Wrhat are the component parts of a number? Give examples. What isa prime number? Give examples. What must every nmrrber be? What are ull even numbers? With what dees an even number always end? What are all numbers ending with 5? What is a prime number always? With what does au odd uumber always end? What may a composite number be? What even number ib a prillo? Are ere ay other exceptions? AS 34 COMMON ARITHMEIT1C. Sect. IV. mlnltipl by 6, we shlall fin how mucll 6 cows would cost. Now 18 cows will evidently cost 3 times as much 6 cows, because 6 X 3 18. Therefore, we multiply the price of 6 cows by 3, which gih es the price of 18 cows. Hence —To multiply by a composite number: Obs. 4. Resolve the multiplier into its component parts, or factors, then multiply the multiplicand first by one of these parts, and the product thence arising by the other, and so on. The last product will be the answer required. REMARK 1.-It matters not by which factor we multiply first; if no mistake is made, the final result will be the same. The reason of this is, because the product of all thefactors of any number, in whatever order taken, is equal to that number. 2. —The learner will observe that there is a difference between resolving a number intofactors, and separating it intoparts. The former has reference ta Multiplication, and the latter to Addition. The product of the factors of any number is the s.ame as the sum of its parts; and it is an established axiom in mathematics, that the whole is equal to the sum of all its parts; or, the product of all itsfactors.* Thus: 2 and 8, or 4 and 4, or 2 and 2 and 4, or 2 and 2 and 2 and 2, are the factors of 16; but 14 and 2, 12 and 4, 10 and 6, 3 and 13, &c., are its parts. 2. How much would 26 tons of hay cost, at 14 dollars a ton? [14 = 7 X 2.1 Ans. 364 dollars. 3. At 75 dollars apiece, how much would 27 horses cost? [27 = 3 X 3 X 3.] Ans. 2025 dollars. 4. How much would 147 acres of land cost, at 36 dollars an acre? 36 =4 X 3 X 3.] Ans. 5292 dollars. 5. How much would 41 books cost, at 25 shillings apiece? [25 = 5 X 5.] Ans. 1025 shillings. 6. How much would 54 town lots cost, at 378 dollars apiece? [54 = 9 X 6.] Ans. 20412 dollars. 7. How much would 478 sheep cost, at 24 shillings apiece? [24 = 6 X 4.] Ans. 11472 shillings. 8. How much would 123 wagons cost, at 72 dollars apiece? [72 = 9 X 8.] Ans. 8856 dollars. CASE 3.-When there are ciphers at the right of either, or loth factors. Ex. 1. How much would 64 acres of land come to, at 20 dollars per acre? Operation. 20 is a composite number, the factors of which 64 are 2 and 10. Then 64 X 2 = 128; and annex20 ino a cipher-which is the same as multiplving by Ans. 1280 ten, (Obs. 1.)-we have 1280 dollars as our answer. How do we multiply by a composite number? Does it make any difference with the final result which factor we multiply by first? Why not? Is there any difference between resolving a number into factors, and separating it into parts? Explain the difference, and' give exampled - *T"'hat is, all of one set of parts, or factors. Art 4. CONTRACTIONS IN MULTIPLICATION. 2. How much would 40 horses cost, at 60 dollars apiece? Operation. 60 and 40 are both composite numbers. The 40 component parts of the one are 6 and 10, and of 60 the other 4 and 10. Now 6 X 4 X 10 X 10 is the - same as 6 X 10 X (4 X 10.) (Obs. 4, Rem. 1.) Ans. 2400 doll. Then 6 X 4 = 24, and 10 X 10 = 100, and 24 X 100 = 2400. (Obs. 1.) The learner will perceive that the significant figures only are multiplied, and the ciphers at the right of both factors are merely annexed to the product. He will also perceive that 10 is one factor of all numbers ending with a cipher, and the significant figures another factor. HenceWhen there are ciphers at the right of either or both factors: Obs. 5. Multiply the significant figures together, only, and to the proguct annex as many ciphers as there are at the right of both factors. 3. There are 50 cents in a half dollar. How many cents in 320 half dollars? Ans. 16000. 4. There are 320 rods in a mile. How many rods are there in 1820 miles? Ans. 582400. 5. How much would 3900 pounds of tea come to, at 80 cents per pound? Ans. 312000 cents. 6. An army of 13000 men having plundered a city, each man's share of the spoil was 480 dollars. What was the amount of plunder taken? Ans. 6240000 dollars. 7. There are 160 square rods in an acre; how many square rods in 3900 acres? Ans. 624000. 8. How many pages would there be in 270 books, if there were 360) pages in.<ach? Ans. 97200. 9. Sound moves at the rate of 1130 feet per second; how far will it move in 4716 seconds? Ans. 5329080 feet. 10. In a certain building there are 24 rooms; 12 of these rooms have one window each, and 12 have 2 windows each; each window has 24 lights. How many lights in the building? Ans. 864. 11. It has been computed that an elm produces annually, at an average; three hundred and twenty-nine thousand grains, or seeds, each of which would produce a tree. How many trees might be produced from 178 elms, in 97 years? Aus. 5680514000.. 12. Multiply 956802700 by 324007020. Ans. 31001379 1549540)0. 13, Multiply 1234567890 by 9876543210. Ans. 12193263111263526900. Of what numbers is 10 always a factor? What is another factor? Wheri there are ciphers at the.right of either, or both factors, how do *e proceed? Explain this.... 86 COMMON ARtTHLMEIC. Sect. V. 14. Multiply two million, three hundred and fifty-seven thousand, by iV., hundret i.i ii.; teen thousand, three hundred,.Ans. t505 i':10510000. 15. Multiply seven hundred and seventy-seven thousand, sevhundred and seventy, by fifty-five thousand, five hundred and fifty. Ans. 43205123500. 16. Multiply eight hundred million, nine hundred and forty thousand, three hundred, by twenty million, one hundred and forty thousand, six hundred. Ans. 16131418206180000. SECTION V. SIMPLE DIVISION. ARTICLE 1. MENTAL EXERCISES. 1. At 4 cents apiece, how many lemons can you buy for 12 cents? Solution.-The first lemon costs 4 cents, and you have 2 — 4 = 8 cents left. The second lemon costs 4 cents, and you have 8-4 - 4 cents left, with which you can buy but I lemon. Therefore, you can buy 3 lemons for 12 cents. 2. How many oranges, at 5 cents apiece, can you buy for 15 cents? 5 is one factor of 15; what is the other factor? (See Multiplication Table.) Ans. 3. 3. How many hats, at 5 dollars apiece, can be bought for 20 dollars? 4. How many yards of c'oth can be bought for 24 dollars, at 4 dollars a yard? 6. At 3 shillings a bushel, how many bushels of potatoes may be bought for 18 shillings? 6. At 6 shillings a pair, how many pair of gloves may be bought for 36 shillings? 7. At 3 dollars apiece, how many sheep can be bought for 33 dollars? 8. At 7 shillings apiece, how many books may be bought for 49 shillings? 9. If a pound of tea costs 7 shillings, how many pounds may be bought for 63 shillings? 10. If I pay 6 cents a mile for riding on the stage, how many miles can I ride for 54 cents? 11. In one peck are 8 quarts. How many pecks are there in 72 quarts? I. Tbehe are 7 days in a wek; how many weeks are there in aOdqW Art. 2. SIMPLE DIVISION. ARTICLE 2. DEFINITIONS, &eC. The separ. tin r a number iro 1qM] pal)r.,ts, s in the prccedii, exarnples, is cailed DIVIIrro. iec i. --- Obs. 1. Division may be defined-the separacting a m'vmber into any given number of equal parts; or, finding how often one number is contained in another. Division may be either of Simple or Compound numbers. Obs. 2. Simple Division is when the number to be divided expresses things of but one name, or kind —as dollars, bushels &c. Obs. 3. The number which is to be divided is called the DIVIDEND. The number we divide by is called the DIVISOR. The result, or answer is called the QUOTIENT, which is derived from the Latin word quoties, signifying how many. Obs. 4. SIGN.-The sign of Division is a short horizontal line between two dots. (-4-). It shows that the number before it, is to be divided by the number after it. Thus: 24-+-3 shows that 24 is to be divided by 3. a. Division is also sometimes expressed by writing the divisor under the dividend, with a line between them; thus: 2-4 shows that 24 is to be divided by 3, us before. Before proceeding farther, the learner must commit accurately to memory the following DIVISION TABLE. 2 in 3 in 4 in 5 in 6 in 7 in 2 once 3 once 4 once 5 once 6 once 7 once 4 2 6 2 8 2 10 2 12 2 14 2 6 3 9 3 12 3 15 3 18 3 21 3 8 4 12 4 16 4 4 20 4 24 4 28 4 10 5 15 5 20 5 25 5 30 5 35 5 12 6 18 6 24 6 30 6 36 6 42 6 14 7 21 7 28 7 35 7 42 7 49 7 16 8 24 8 32 8 40 8 48 8 56 8 18 9 27 9 36 9 45 9 54 9 63 9 8 in 9 in 10 in 11 in 12 in 8 once 9 once 10 once 11 once 12 once 16 2 18 2 20 2 22 2 24 2 24 3 27 3 30 3 33 3 36 3 32 4 38 4 40 4 44 4 48 4 40 5 45 5 50 5 55 5 60 5 48 6 54 6 60 6 66 6 72 6 56 7 63 7 70 7 77 7 84 7 64 8 72 8 80 8 88 8 96 8 72 9 81 9 90 9 99 9 108 9 How I DIvheiok dfinub? How divided t What is Simpi% Divin 7 COMNMON ARITHMETIC. Sect. V. 21 -- 7 -- how many? 63 - 7 = how many? 48 - G6 = how many? 64- 8 = how many? 54 - 9 = how mLany? 32 - 4 =how many? 27 - 9 = how many? 49 - 7 = how many? 35 - 5 = how many? 24 -3 = how many? 54 6 = how many? 72. 9 = how many? Ex. 1. A person paid 369 dollars for 3 horses; how much was that apiece? Solution.-369 = 300 + 60 - 9. Then 3 in 300 goes 100 times; 3 in 60 goes 20 times; and 3 in 9 goes 3 times; and 100 + 20 + 3 =123. Ans. 123 dollars. We divide thus by 3, because it is evident that if 3 horses cost 369 dolldrs, one horse will cost one-third as much as 3 horses, or one-third of 369 dollars. But there is a method of dividing numbers without separating them into their numerical parts, which we will illustrate as follows: Operation. By this method we place the divisor at Divisor Dividend the left of the dividend, with a line between 3 ) 369 them. We next draw a line below the div-- dend to separate it from the quotient, which 123 Quotient. is to be placed below. Then we say 3 in 3 (hundreds) goes 1 (hundred) times, and write the 1 below the 3; then 3 in 6 (tens) goes 2 (tens) times, and we write the 2 under the 6; then 3 in 9 (units) goes 3 (units) times, and we write the 3 under the 9, and the work is done. Obs. 5. The pupil will observe that theform used is of comparatively little importance, it being used merely for the sake of convenience. In short, there is no form by which any operation in Arithmetic is performed, that is absolutely necessary, if the pupil 'till only keep the principles of numbers firmly fixed in his mind, and without this he cannot succeed if he has a book full of forms. They are used only for looks and convenience in illustrating our operations. It will be noticed that each figure of the quotient occupies the same place, numerically, as the figure of the dividend that was divided to produce it. This is always the case. What is the number to be divided called? The number we divide by? The result or answer? From what? What does Quoties signify? What is the sign of Division? What does it show? How is division otherwise expressed? Give the solution of Ex. 1st. Is there any other method of working such questions? How do we write the numbers by this method? How proceed next? Is this form of any particular importance? Why then is it used? Is any form absolutely necessary in arithmetic? What must a scholar do if he dispenses with forms? Can he succeed well without this? Why not? What then is the use of forms? What place does each figure occupy numerically? I. thii alwayj the ease?'; - Art. 2. SIMPLE DIVISION. 2. A man spent 488 cents for ribon, at 4 cents a yard; how many yards did he get? Ans. 122. 3. Four men received 845 dollars; how much was that apiece? Ans. 211 dollars. In this example, after dividing by 4, there was 1 dollar left.Now this 1 dollar is to be divided between 4 men; but as I will not contain 4, we will merely express the division; thus:!. This is called one-fourth, and annexing it to the right of our quotient, we find that each man received 211 - dollars. The 1 which was left after the division was performed, is called a REMAINDER. HenceObs. 6. When there is a remainder, it is written over the divisor, and annexed to the quotient. Obs. 7. It will be perceived that Ditision is exactly the reverse of Miultiplication. In the latter we have two factors given to find their produtct: in the former we have the product and one factor given to find the other factor, the quotient and divisor answering to to the two factors, and the dividend answering to to the product, in Multiplication. HenceObs. 8. If we multiply the divisor and quotiezt together, we shall obtain the dividend. But if, after a division has been performed, there is a remainder, it is evident that the product of the quotient and divisor will want just this remainder to make the dividend. HenceTo prove Division: Obs. 9. Bfdtiply the divisor and quotient together, and to the product add the remainder, (if any); if the result is equal to the dividend, the work is correct. From the preceding remarks, we derive the following method of proving Multiplication by Division: Obs. 10. Divide the product by one factor, and if the result s the other factor, the work is correct. NoTE.-In this case there is never any remainder.* 4. At 3 dollars a yard, how many yards of cloth can be bought for 697 dollars? Ans. 232-. Let the learner prove this and the following examples: 5. At 5 dollars an acre<, how many acres of land can be bought for 558 dollars? Ans. 111. If there is a number left after dividing, what is it called? What is done with it? Of what is Division the reverse? Show the comparison. To whatdo the different terms of Division answer in Multiplication? What inference is dez duced from this? But suppose after the division has been performed, there ii a remainder; what then is the conclusion? How then do we prove Division? How can we prove Multiplication by - Division? Is there any rejnainder in this case? Why not? '-iThi-l, In Inttral numberS. 40 COUNIONS ARtTILMUM.lC' Sect. VO. 6. I wish to set out 126 trees in 6 rows; how many must I set in a row? Ans. 21. i: tiis Cx;m.iple the ^ ii: o t. e di ' idnd di lut oi."i t'h divisor, therefore;rc Look ito i uL,'c;; if tio ligUiCes ilad nut contained it, we should have taken three figures, &c. HenceObs. 11. If the left hand figure of the dividend will not contain the divisor, take two figures; if two fiures will not contain it, take three figures; and universally take as many figures at the left of the dividend, as will contain the divisor once or more, to obtain the first quotient figure, which we write under the right hand figure of the part taken. 7. Six men had 366 dollars to divide between them. How much was that apiece? Ans. 61 dollars. 8. A man had 976 acres of land, which he wished to divide between his 4 suns. How many acres would each receive? Operation. 976 = 9 (hundreds) + 7 (tens) + 6. 9 4)976 (hundreds) = 8 (hundreds) + 1 (hundred). -- Now 4 will go in 8(hundreds) just 2(hundreds) Ans. 244 acres. times; consequently it will go in 9 (hundreds) 2 (hundreds) times, and have 1 (hundreds) remainder. Adding the 1 (hundreds) remainder, (which is 10 tens,) to the 7 (tens), we have 17 (tens,) which divided by 4 gives 4 (tens) for a quotient, and 1 (ten) remainder. This I (ten) is equal to 10 units, which added to 6 (units) gives 16 (units) This divided by 4 gives 4 (units), and our work is done. Hence, onr quotient is 2 (hundreds), 4 (tens), and 4 (units), or 244. Obs. 12. By attentively examining the above solution, the learner will perce:ive the following considerations, viz: a. 1st. The remainder each time is of the same name as the dividend. This is always the case, whether the remainder occurs after each quotient figure is obtained, or only at the end of the operation; because the remainder is always a part of the dividend. b. 2nd. The remainder each time should be less than the divisor; because, if it viere equal to, or greater than the divisor, the divisor could be contained another time in the dividend. c. 3d. The remainder each time has ten times the value of the next lower order. This is evident, according to Sect. 1, Art. 2, Obs. 3. Hence, we must multiply it by 10 before adding it to the If the first figure of the dividend will not contain the divisor, how do we proceed? How many figures should we take, in all cases from the dividend, to obtain our first quotient figure? Explain the solution of Ex. 8. What is the first consideration deduced from this? Is this always the case? Why? What is the second consideration? Whiy? What is the third consideration? Why? What then inual be done before it can be added to the next lower figure? Art. 2. SIMPLE DIVISIOoN, 41 next figure, (Sect. II., Art, 2, Obs. 4.) It has the same effect, however, to prefix it to the next figure, as the cipher always occupies the place filled by this figure. NOTE.-Prefix means to place before. or at the left hand. This may be illustrated as follows: 0. Divide 1074 by 3. Operation. In the first case we have 1 remainder, 3)10 7 4 which, multiplied by 10, equals 10, and 7 10 20 equals 17. We next have 2 remainder, which, - - multiplied by 10, equals 20, and 4 makes 24, 17 24 which contains 3 without a remainder.: — Hence3 5 8 358 Obs. 13. When there is a remainder after dividing any figure of the dividend, prefix it (mentally) to the next figure of the dividend, and proceed as before. 10. At 8 dollars an acre, how many acres of land can be bought for 8984 dollars? Ans. 1123. 11. At 5 dollars a pair, how many pair of boots may be bought for 685 dollars? Ans. 137. 12. At 4 shillings a bushel, how many bushels of potatoes can be obtained for 1628 shillings. Operation. In this case the divisor is not contained in the 4)1628 ten's figure, therefore we write a cipher in the quotient; and prefix-the 2 to the next figure, as 407 if it were a remainder, which it really is. Then 4 in 28 goes 7 times. HenceObs. 14. When any partial dividend will not contain the divisor, write a cipher in the quotient, call the partial dividend a remainder, and proceed as before. RzMARK.-rhe partial dividend is that part taken to find each quotient figure. Obs. 15. Also-After the first quotient figure is obtained, either a cipher or a significant figure must be put in the quotie7l for each figure or cipher in the dividend. 13. A man traveled 1624 miles in 8 weeks; how many miles was that a week? Ans. 203. Why? By what other method can we effect the same result? Why is this correct? What does prefix mean? Give the solution of Ex. 9, and show why it has the same effect to prefix the remainder to the next lower figure as to multiply it by 10, and then add it. What inference is deduced from this? — When the divisor cannot be contained in any partial dividend, how do we proceed? What is the partial dividend? How many figures, or ciphers, must be put id the quotient after the first quotient figure is obtained? When the divisor is small,/and the opbration is carried on chiefly in the mind, what is it called? 42 COMMON ARITHMETIC. Sect. V. When the divisor is small, as in the preceding examples, the operation is performed chiefly in the mind; it is then called SIORT DIVISION. From the remarko and illuetrations already given, we derive the following RULE FOR SHORT DIVISION. 1. Write the divisor at the left of the dividend, with a line between them; also, draw a line below the dividend. 1. Take from the left of the dividend as many figures as will contain the divisor once or more; divide them, write the quotient under the right hand figure of the part taken, (Obs. 11); prefix the remainder (if any,) to the next figure of the dividend, and divide as before, until all the figures of the dividend have been divided. (Obs. 13.) IlI. If the divisor is not contained in any partial dividend, write a cipher in the quotient, prefix this partial dividend to the next figure, as if it were a remainder, and proceed as usual. (Obs. 14.) IV. If a remainder occurs after dividing the last figure of the dividend, write it over the divisor, and annex it to the quotient.(Obs. 6.) PRooF. — fultiply the divisor and quotient together, and to the product add the remainder (if any); if the sum is equal to the dividend, the work is correct. (Obs. 9.) 1. We commence at the left hand to perform division, because the remainders which occur must be reduced to lower orders before the division can be continued. (Obs. 12, 3d.) Besides, it is evident that if we divide any order ot figures by any number, and there is a remainder, that this remainder cannot be reduced to a higher order, and still contain the same divisor. Hence, it must be reduced lower; and as numbers decrease from the left hand towards the right, we commence at the left to divide. The learner will better understand the reasons for carrying in the Fundamental Rules, when he has become acquainted with Compound numbers. 2. The divisor, dividend and quotient may any of them be either abstract or concrete. The divisor and quotient, however, cannot both be concrete in the same question, and when either of them is concrete, the dividend is also concrete, and of the same name. NeiWhat is the rule for Short Division? Why do we commence at the left hand to divide? Why must the remainders be reduced to lower orders before we Can continue the operation? What is said respecting the divisor, dividend, and quotient being abstract or concrete? Canuthe divisorand quotient both be concrete in the same question? When one of them is concrete, what must the dividend be? Can the diridnd be abstract, without the divisor and quotient both being abstract? What nate, then) must the dividend always be? Give examples. Art. 2. SIMPLE DIVISION. 43 ther can the dividend be abstract, without the divisor and quotient both being abstract also. In short, the dividend must always be of the same name as the divisor, or quotient, or both. Thus: in the 6th question, we reason-If 126 trees are put in 6 rows, each row will contain one-sixth of 126 trees, which is 21 trees, and not 21 times, or merely 21, as is often asserted. Here, the divisor (6) is abstract, whilst the dividend and quotient are both concrete, and of the same kind. In the 1 th example we divide 685 dollars by 5 dollars; but ihe learner must not infer that 685 dollars divided by 5 dollars gives 137 boots for an answer. In this case, as the boots cost 5 dollars a pair, we can buy as many pair as 5 dollars s contained in 685 dollars, which is 137 times. Here our divisor and dividend are concrete, and of the same name, and the quotient is abstract. It would be as absurd to make the quotient concrete in this case, as it would be to say, (in the 8th Ex.) that 976 acres contain 4 sons 244 times, or that one fourth of 976 acres is any ihing else than 244 acres. EXERCISES FOR TIHE SLATE. 1. At 6 cents a ball, how many balls of tape may be bou. ht for 186 cents? Ans. 31. 2. How many sheep can I buy for 786 dollars, at 3dollars ap.iece? Ans. 262. 3. How many pair of gloves can I buy for 1728 shillings, at 9 shillings a pair? Ans. 192. 4. There are 3 feet in a yard; how many yards in 479 feet? Ans. 1592. 5. A mali bought 9 horses for 1382 dollars. How much was that apiece? Ans. 153- dollars. 6. A man sold 6 yoke of oxen for 492 dollars. How much was that a yoke? Ans. 82 dollars. 7. If 7 pieces of cloth cost 438 dollars, how much is that apiece? Ans. 624 dollars. 8. If 8 wagons cost 548 dollars what is that apiece? Ans. 68' dollars. 9. At 4 shillings apiece, how many books can be bought for 6488 shillings? Ans. 1622. 10. At 11 miles a day, how many days would it take to travel 483 miles? Ans. 43~~. 11. There are 8 quarts in a peck; how many pecks in 9952 quarts? Ans. 1244. 12. At 5 cents a mile, how far can I ride for 8730 cents? Ans. 1746 miles. 13. At 12 dollars a month, how many months would it take to earn 4623 dollars? Ans. 385 3. 14. How many calves, at 7 dollars apiece, can I buy (or 368 dollar~?. Ans. 52, aud have. 4 dollars left. 44 COMMON ARITHMETIC. Sect. V. 15. A man has 2832 pair of shoes in 8 boxes. How many are there in each box? Ans. 354 pair. 16. At 5 dollars a barrel, how many barrels of flour can I buy for 2540 dollars? Ans. 508. 17. If 8 laborers should perform a piece of work for 1628 dollars, how much would that be apiece? Ans. 203- dollars. 18. 16 times 489271 is how many times 2? Ans. 3914168. 19. 15 iimes 14962 is how many times 5? Ans. 44886. 20. 17 times 6217 is how many times 11? Ans. 9608-!. ARTICLE 3. LONG DIVISION. Obs. 1. In Short Division the operation is performed mentally to a considerable extent; but in Long Division the operation is all set down upon the slate. Ihe principles of both, however, are the same. Tlheformer is gene ally used when the divisor is less than 12, and the latter when the divisor exceeds 12. 1. A man paid 4096 dollars for some land, at 16 dollars an acre. How many acres did lie buy? Operation.. We write our numbers as before, except 16)4096(256 Ans. the quotient, which we place at the right 32 * hand. We do this tecau.se it is more convenient than to place it below. We next say, 89 16 in 40 (hundreds) goes 2 (hundreds) 80 times, and write our 2 in the quotient. We -- then multiply our divisor (16) by our quo- 96 tient figure (2), and place the product (32) 96 under the part taken of the dividend.- This is done to ascertain whether there is 00 any remainder after dividing. Next we subtract this product (32) from the dividend, (40,) which leaves 8(hundreds) as our remainder. We then bring down the 9 (tens) at the right of the 8 (hundreds,) (which is the same as prefixing the remainder to the 9 (tens), which makes 89 (tens). Then 16 in 89 (tens) goes 5 (tens) times, which we write in the quotient. We place the 5 at the right of the 2 because the 2 is 4undreds, and the 5 is tens, and tens occupy the place at the right of hundreds. Then 5 times 16 is 80, which we place under the 89, and subtracting, we have 9 (tens) as our second remainder. To this we bring down our next figure in the dividend (6), making 96 (units), which divided by 16, gives 6 (units) What is the difference between the operation of Long and Short Division? What is the difference between the principles of the two? When is Short Division generally used? Long Division? How do we write the numbers in Long Division? Why do we place the quotient on the right? What is the first step of the operation? The second? Third? Fourth? Why do we multiply the divisor by the quotient figure? What is meant by bringing down? Why do we place our second quotient figure at the right of the first? Art. 3. LOXG DIV.S10ON 45 for our next quotient figure. Finally, multiplying and subtracting as before, we find there is no remainder; and as all the figures of the dividend have been brought down and divided, our work is done. NoTE.-The dots are placed under figures of the dividend when they are brought down, to prevent their being brought down again. They are not necessary if the pupil is observing. Obs. 2. By examining this solution attentively, the learner will observe the two following considerations, viz: 1st. Each figure in the quotient occupies the same place, numerically, as the right hand figure of that part of the dividend taken to produce it. (Art. 2, Obs. 5.) 2nd. The operations of Long Division consist of four steps, viz: 1st, to divide; 2nd, to multiply; 3rd, to subtract; and 4 h, to bring down. This is the whole process. 2. A man bought 15 acres of land for 375 dollars. How much was that per acre? Ans. 25 dollars. 3. If a man travel 625 miles in 25 days, how many miles is that per day? Ans. 25. 4. A man paid 32 dollars an acre for some land, and bought enough to come to 34576 dollars. How many acres did he buy? Operation. 32)34576(1080^. 32 in 34 goes 1 time, and 2 remainder. 32 Bring down the 5, we fi;d that 25 will not conta n 32; therefore we write a ci2o7 pher in the quotient, (Art. 2, Obs. 14), 2;Q and bring down the next figure, 7. 32 - in 257 goes 7 times, and 1 remainder. 1~ Bringing down the 6, we fin l that i6 will not contain the 32, and therefore write another cipher in the quo ient; then as there are no more figures in the dividend to bring down, 16 is our remainder, which we write over our divisor, and annex to the quotient, (Art. 2, Obs. 6), and thus obtain for our Ans. 1080. 5. How many times will 6144 contain 24? 1st Operation. 24)6144(257 48 -~- ~ Here, we supposed our last quotient figure 134 to be 7; but by multiplying, we find that our 120 product is greater than our dividend; therefore --- the quotient figure is too large. 144 168 ~:~ 46 COMM 3N ARITHMETIC. Sect. V. 2d Operation. 24)6144(256 48 134 Here, we supposed it to Le 6, which we 120 find to be correct. --- ~ Hence144 144 000 Obs. 3. TVhen the product of the divisor by any quot'ient figure is greater than the dividend, the quotient figure must be made less. The rem rks respecting the remainder, (Art. 2, Obs. 12,) aie equ.lly as applicable to Longr as to Short Division. From the preceding remarks and illustrations, we derive the following RULE FOR LONG DIVISION. I. Write the divisor and dividend as before directed, and draw a line at the right of the dividend to separate it from the quotient. II. Take as many figures at the left of the dividend as wil contain the divisor once or more, and place the number of times they contain it at the right of the first quotient figure. III. Multiply the divisor by this quotient igure, and subtract the product from that part of the dividend taken. IV. To the righAt of the remainder bring down the next figure of the dividend, and proceed as before. So continue to do till all the figures of the dividend have been brought down and divided. V. If any partial dividend will not contain the divisor, write a cipher in the quotient, bring down the next figure of the dividend, and proceed as before. (Art. 2, Obs. 14.) PROOF. —Multiply together the.divisor and quotient, and to the product add the remainder (if any); if the result is equal to the dividend, the work is correct. (Art. 2, Obs. 9.) EXERCISES FOR THE SLATE. 1. If a man walk 4 miles an hour, how many hours will it take him to walk 12S miles? Ans. 32. For what are the dots used? Are they really necessary? What place does each quotient figure occupy numerically? Of how many steps does the operation in Long Division consist? Name them. When the dividend will not contain the divisor, how do we proceed? What is done with the final remainder? When the product of the divisor by any quotient figRre is greater than the dividend, what is to be done? Are the remarks respecting the remainder, (Art. 2, Obs. 8.) applicable to Long as well as Short Division? What were these remarks? What is the rule for Long Divi.ion? Art. 3-, LONG DIVISION. 47 2. There are 12 months in a year; how many years in 20176 months? Ans. 1681 -. 3. There are 8 quarts in a peck. How many pecks in 32024 quarts? Ans. 4003..4 There are 6 working days in a week. How many weeks in 1256 working days? Ans. 209o. 5. A halt:eagle is worth 5 dollars. How many half eagles would it take to be worth 394820 dollars? Ans. 78964. 6. In one square yard are 9 square feet. How many square yards in 42786 square feet? Ans. 4754. 7. How many calves, at 14 dollars apiece, can be bought for 476 dollars? Anls. 34. 8. At 17 dollars a month, how many months would it take to earn 969 dollars? Ans. 57. 9. The steamer Great Western was 15 days in crossing the Atlantic Ocean, a d stance of 3000 miles. How far did she sail per day? Ans. 200 miles. 10. Suppose 21528 dollars were equally divided between 104 men; how much would each receive? Ans. 207 dollars. 11. Suppose a vessel has sailed 15340 miles, at the rate of 126 miles per day; how many days has she sailed? Ans. 121 -. 12. There are 63 gallons in a hogshead. How many gallons in 7560 hogsheads? Ans. 120. 13. At 16 dollars a month, how long would it take to earn 624 dollars? Ans. 89 months. 14. How m ny days would it take to walk 576 miles, walking 24 miles per day? Ans. 24 days. 15. In every day are 24 hours. How many days in 4096 hours? Ans. 170w. 16. At 18 dollars a ton, how many tons of hay could be bought for 450 dollars? Ans. 25. 17. How many casks of wine could be bought for 4950 dollars, at 25 dollars a cask? Ans. 198.. 18. If a man has traveled 6255 miles in 15 weeks, how far is that per week? Ans. 417 miles. 19. If 75 wagons cost 5625 dollars, how much is that apiece? Ans. 75 dolla s. 20. If 53 horses cost 3286 dollars, how much is that for each?. Ans. 62 dollars. 21. If 32 pounds of raisins cost 800 cents, how nuch is that per pound?.. Ans. 25 cen s. 22. A man has a note to pay, amounting to 1204 dollars. If he pays 86 dollars at a time, how many, payments will he have to make to pay up his note? Ans. 14. 23. There are 52 weeks in a year. How many years in 1248 weeks? Ans, 4. 48 COMMON ARITHMETIC. Sect. V. 24. Suppose a firm to consist of 37 partners, and their capital to amount to 1178450 dollars. How much is that for each? Ans. 31850 dollars. 25. At 144 shillings a month, how long would it take a man to earn 7488 shillings? Ans. 52 months. 26. 25 times 8469 is how many times 9? Ans. 23525. 27. 144 times 7324 is how many times 48? Ans. 21972. 28. 1728 times 20736 is how many times 288? Ans. 124416. 2.. Two numbers multiplied together produce one billion, thirty million, six hundred and three thousand, six hundred and fifteen, end one of the numbers is three thousand, two hundred and fifteen. What is the other number? Ans. 320561. 30. Divide th rty-seven billion, four hundred and twenty-three million, eilht hundred;ind ihirty-four thousand, five hundred and sixty, by one hundred and twenty-three thousand, four hundred and fifty-six. Ans. 303135. ARTICLE 4. CONTRACTIONS IN DIVISION. CASE 1.-To divide by 10, 100, 1000, &c. 1. At 10 dollars a month, how many months would it take to earn 534 dollars? Operation. As a cipher annexed to any figure increases 110 I 5314 its value 10 times, (Sect. 1, Art. 1, Obs. 5,) -- and two ciphers annexed increases its value 100 Ans. 53-4-. times, &c.. so to cut off one figure from the right of any number, diminishes the value of that number 10 times, to cut off-two figures diminishes its value 100 times, to cut off three figures diminishes its value 1000 times, &c. Hence-To divide by I with any number of ciphers annexed: Obs. 1. Cut off from the right of the dividend as many figures as there are ciphers at the right of the divisor; the figures at the left will be the quotient, and those at the right the remainder. 2. Suppose you wished to divide 4267 dollars equally between 100 men, how much would you give to each man? Ans. 42-8o dollars. 3. There are 10 mills in a cent. How many cents in 150 mills? 4. There are 100 cents in a dollar. How many dollars in 4875 cents? 5. There are 10 dollars in an eagle. How many eagles in 3854 dollars? 6. There are 1000 mills in a dollar. How many dollars in 65832 mills? 7. There are 10000 mills in an eagle. How many eagles in 489621 mills? How do we divide by 10,100,1000, &c.? Explain why this rule is correct. CONTRACTIONS IN DIVISION, 49 CASE 21- To divide by a Composite Number. If we divide 12 by 6, we obtain 2. Again: if we divide 12 by 3, we obtain 4; and if we divide 4 by 2, we also obtain 2. Now 3 and 2, (our latter divisors,) are the factors of 6, and to divide 12 by them produces the same result as to divide 12 by 6. HenceObs. 2. To divide the dividend by the factors of the divisor, will produce the same result as to divide it by the divisor itself. a. Therefore, conversely, Any dividend that will contain a divisor will also contain the factors of that divisor. 1. If I pay 240 dollars for 48 yards of cloth, what is that per yard? Ans. 5 dollars. 1st Operation. 2d Operation. 48)240(5 dollars. Ans. 8)240 240 -- ~- -~- ~ 6)30 000 Ans. 5 dollars. 48 is composed of two factors, 6 and 8, (8 X 6 = 48,) and according to the above remark, we may divide 240 first by one factor, and tie quotient thence arising by the other, and it will produce the same result as to divide it by 48, that is, 5 dollars a yard. Hence-When the divisor is a composite number: Obs. 3. Divide the dividend first by one factor, and the quotient thence arising by another factor, and so on; the last quotient will be the answer. NOTE.-It makes no difference in any case which factor we divide by first; if no mistake is made, the final result will be the same. 2. If a man travel 32 miles per day, how many days will it take him to travel 1472 miles? (32 = 4 X 8.) Ans. 46. 3. If a man pay 2835 dollars for horses, at the rate of 63 dollars apiece, how many can he buy? (63 = 9 X 7.) Ans. 45. 4. How many acres of land, at 25 dollars an acre, can be bought for 875 dollars? (25 = 5 X 5.) Ans. 35. 5. How many yoke of oxen can I buy for 1904 dollars, at 56 dollarsayoke? (56 =-8 X 7.) Ans. 34. 6. How many watches can be bought for 1875 dollars, at 75 dollars apiece? (75 = 5 X 5 X 3.) Ans. 25. What is the difference in the result of dividing 12 by 6, or by 3 and 2, the factors of 6? What is the first inference deduced from this? The second? (ive the solution of example 1st, and show why it is correct. How do we prow feed when the divisor is a composite number? Does it make any difference in the inal result which factor we divide by first? Give the solution of Ex. 1I Case 3, End show why it is correct. What is done with the remainder, 3, in Ex. T Hew do we proceed when there are ciphers at the right of te diisr? -... 4.. COMMON ARITHMETIC. Sect. V. 7. A school teacher wishes to divide 256 apples between 32 pupils. How many must he give to each't (32 = 4 X 8.) Ans. 8. 8. If 175 watches cost 4375 dollars, how much is that apiece? (175 = 5 X 5 X 7.) Ans. 25 dollars. CASE 3. When there are ciphers at the right of the divisor. 1. There are 40 rods in a furlong. How many furlongs in 689 rods? 1st Operation. - 2d Operation. 40)689(17-. Ans. 410)6819 40 -- Ans. 17f. 289 280 40 is composed of two factors, 4 and 10, - (10 X 4 = 40); therefore, we may first di9 vide by ten, (which is only to cut off one figure at the right-Obs. 1,) and then divide the quotient thence arising by 4, (Obs. 3;) this is done by cutting off the cipher at the right o' the divisor, and a figure at the right of the dividend, and then dividing the remaining figures of the dividend (68,) by the remaining figure of the divisor (4); the figure that was cut off from the dividend is the remainder. 2. There are 60 minutes in an hour. How many hours in 754 minutes? Operation. 6 0)75|4 In this example, after dividing by 6, there was - 3 remainder, to which we annexed the 4 that Ans. 12|4 was cut off, making 34 as our true remainder. Hence-When there are ciphers at the right of the divisor: Obs. 4. Cut off these ciphers; also, cut of as many places at the right of the dividend; then divide the remaningfigures of the dividend by the remaining figures of the divisor, and to the last remainder annex the figures, or ciphers, that were cut of from the dividend,for the true remainder. 3. How many horses can be bought for 6400 dollars, at 80 dollars apiece? Ans. 80. 4. At 30 dollars an acre, how much land can be bought for 1568 dollars? Ans. 5287 acres. 5. How many barrels will 9700 pounds of pork pack, allowing 200 pounds to the barrel? Ans. 48T. 6. How. many bins would it take to contain 43200 bushels of Wheat, allowing each bin to contain 320 bushels? Ans. 135. 7. Allowing each bin to contain 360 bushels, how many Woiwtd it take? An. 120. RECAPITULATION, 51 8. At 60 dollars a yoke, how many yoke of oxen could be bought for 480 dollars? Ans. 8. 9. At 80 dollars a yoke, how many yoke could be bought? Ans. 6. 10. An army of 13500 men having plundered a city, took 5130000 dollars. What was each man's share? Ans. 380 dollars. SECTION VIL RECAPITULATION. ARTICLE 1. REMARKS AND INFERENCES DEDUCED FROM THE PRECEDING RULES. Obs. 1. The preceding Rules, viz.-Notation and Numertion Addition, Subtraction, Multiplication and Division-are called the FUNDAMENTAL RULES OF ARITHMETIC, because they are the basis, or foundation of a l the other Rules, and without the aid of one or more of these, no operation in numbers can be performed. In this Section it is intended to review these Rules, and make a few remarks, and present a few ideas and suggestions in addition to what has already been said. The rules, as a general thing are so evident, that they need no demonstration. If the pupil cannot demonstrate them himself, he should turn back and study over the preceding Sections again. It is absolutely necessary that he should understand the preceding Rules thoroughly before he can ever become an arithmetician Obs. 2. NOTATION.-All numbers are expressed (by the common, or Arabic method,) by nine digits and a cipher. As the cipher, however, is only used to give value to significant figures, (it having no value itself,) it might with propriety be called an auxiliary digit. Obs. 3. Our system of Notation is also called the decimal system, because the numbers increase in a ten-fold ratio. The term decimal is derived from the Latin word decem, which signifies ten. Obs. 4. The names, as well as the characters which represent What are the preceding rules called? Why? Is it necessary for the learner to understand these rules, before he can become an arithmetician? Why so? How are numbers expressed by the common method of notationi? For what is the cipher used? What, then, might it properly be called? By what other name, is the common system of notation known? Why is it thus called? From what is the term decimal derived? What is said respecting the names and characters that we use to represent numbers? Would any othernamesorcharacters have answered as well bad they been adopted 52 COMMlON ARITHMEETIC. Sect. VI. numbers, are entirely arbitrary. Any other names or characters would have served the same purpose had they been adopted. It is use only that renders them familiar. Obs. 5. The names of numbers below ten, are either primitive words, or derived from the Latin. Eleven and twelve are regarded by some as derivative words, formed from ten and one, ten and two; thus: eleven means ten leave one-that is, if you lake ten from it you leave one; twelve means, ten leave two-that is, if you take ten from it you leave two. This is, however, rather a forced signification, and they are generally regarded as primitive words. Names of numbers higher than twelve (re but repetitions of the preceding names. Thus: tlirtecn, fourteen, &c., signify three and ten, four and ten, &c., and twenty, thirty, &c., signify two tens, three tens, &c., and so on, The different powers of ten, as one hundred, one thousand, &c., areprimitive words. Obs. 6. NUMERATION.-It may be asked by some, why do we commence at the left hand to read numbers, and at the right hand to numerate them? The reason is obvious. In speaking of numbers we always name the highest orders first, as it is more convenient; and therefore, as the higher orders are at the left hand, by our system of notation, we commence at the left hand to read. Again-In numerating, we call the lower orders first; therefore, as numbers increase from the right hand towards the left, the lower orders must be at the right; consequently, we commence there to numerate. Had our system of notation been such, that numbers increased from the left hand towards the right, we should have commenced at the rigd1 hand to read, and at the left hand to numerate. Obs. 7. ADDITION.-Wre cannot add numbers of different names together; because if I have 5 pear trees and 8 cherry trees in an orchard, it cannot be said that their sum is 13 pear trees, or 13 cherry trees. We can, however, say that their sum is 13 trees. Again: if I saw 5 hay stacks, 7 wheat stacks, 12 trees, 2 barns, 12 horses, and 15 cattle in a lot, their sum would not be all stacks, trees, barns, horses, or cattle; but if asked how many objects I s-iw in the Are the names of numbers below ten primitive or derivative words? What are eleven and twelve regarded by some? Formed from what? How? Is this a natural signification? How are they generally regarded? How are numbers higher than twelve formed? Give examples. What is said respecting the different powers of ten? Why do we commence at the left hand to read numbers? Why do we commence at the right to numerate? Had our system of notation been such that numbers increased from the left hand towards the right, where would we have commenced to read, and where to numerate? Explain why we cannot add numbers of different names together. What is necessary in adding se'el;~i a*nllcrs together? Wlheu wo iave givesk several numbers, how do we find their sumi? Art. 1. REC PrT'CrLATION,. 53 field, their sum would express the number. Also, the sum of 2 units and 6 tens, is neither 8 units nor 8 tens; but if we reduce the 6 tens to units, their sum would be 68 units. Thus, we perceive that we cannot add several numbers together, unless they can be classifiee under one order in which one name is common to them all. Obs. 8. When we have given several numbers, to find their sum-Aldd them together. Ex. 1. There arc two numbers; one is 144, and the other is 62. What is their sum? Ans. 206. 2. A. has 1200 dollars, and B. has 433 dollars. How many dollars have both? Ans. 1680 dollars. 3. The distance A. has traveled is 180 miles; B. has traveled 114 miles. How many miles have both traveled? Ans. 294. Obs. 9. It may be asked by some-Why, in Addition, we carry 1 for every ten, instead of one for every eight, nine, twelve, or some other number? The answer is-because our system of notation increases in a ten-fold ratio; consequently, as often as we have ten of one order, we must have 1 of the next higher order. Had the base of our system of notation been eight, nine, twelve, or any other number, instead of ten, we should have carried 1 for every eigld, nine, twelve, or whatever other number was the base of our scale of notation. The same remark applies to Subtraction, Multiplication, and Division. Obs. 10. SUBTRACTION.-The numbers must all be of the same name, or kind in Subtraction as well as in Addition. Thus: if I ha've in a field 13 cherry trees, and 5 pear trees, my taking the pear trees away does not diminish the number of cherry trees, although it lessens the number of trees in the lot. Again: we cannot take 2 units from 8 tens, unless we first reduce the tens to units. Thus, we perceive that we cannot subtract two numbers, unless they have a name common to both; It is not absolutely necessary, in Subtraction, that we place the greater number above the other. This is done only for convenience. Obs. 11. When we have given the sum of two numbers, and cither number, to find the other-F-om the sum subtiact the given number. Demonstrate this rule. Why, in Addition, do we carry 1 for every ten, instead of 1 for every eight, twelve, or some other number? Had the Lase of our system of notation been eight, twelve, or some other number ilstead of ten, how often would 'we have carried 1? Explain why, in Slblraction, the numbers nmst all be of the same namre, or kind. What is necessary in order to subtract two numbers? Is it necessary in Subtraciion that we place the greater number above the other? Why then is it done? ' hen wem have given the sum of two numbers, and either number, how do we find tlie other number? Demnonstrate this rule. COMMON ARITHMETIC. Sect. VI. 4. The sum of two numbers is 40, and one of them is 18. What is the other number? Ans. 22. 5. Two men have 1000 dollars; A. has 572 dollars. How many dollars has B.? Ans. 428. Obs. 12. When the diference and greater of two numbers are given to find the lesser number-From thegreater number subtract the difference. 6. The greater of two numbers is 53, and their difference is 14. Required-the lesser number. Ans. 39. 7. A. has 1000 dollars, which is 346 dollars more than B. has. How many dollars has B.? Ans. 654. 8. A. has traveled 300 miles, which is 163 miles farther than B. has traveled. How far has B. traveled? Ans. 137 miles. Obs. 13. Given, the difference and lesser of two numbers, to find the greater-Add the d~iference and lesser number together. NQF,.-This is the same, in reality, as the proof of Subtraction. 9. The difference of two numbers is 7, and the lesser number is 13. What is the greater number? Ans. 20. 1 0. A. has 560 dollars, and B. has 40 dollars more. How much has B.? Ans. 600 dollars. 11. A. went 140 miles, and B. went 60 miles farther. How far did B. go? Ans. 200 miles. Obs. 14. Given the sum and difference of two numbers, to find those numbers-From their sum subtract their difference, and half the remainder will be the lesser number. 12. The sum of two numbers is 16, and their difference is 4. Required-the numbers. Ans. 6 and 10. 13. The sum of the ages of two men is 66 years, and one of them is 20 years older than the othe. Required-their ages. Ans. 43 and 23 years. 14. Two men together have 500 dollars, and one of them has 240 dollars more than the other. How many dollars have both? Ans. One has 370 dollars, and the other 130 dollars. Obs. 15. MULTIPLCATION.-AS Multiplication is the repeated addition of the same number to itself, it may be thought by some that the multiplier can sometimes be a concrete number, (as we cannot add numbers of different names together). But the learner must When the difference and greater of two numbersare given, how do we find the lesser number? Demonstrate this rule. When we have given the difference and lesser of two numbers, how do we find the greater? Demonstrate this rule When we have given the sum and difference of two numbers, how do we find the numbers? Demonstrate this rule. Explain why the multiplier is always anl aBtclaqt fUtber. When we have given the product of two factors, and Art. 1. RECAPITULATION. 55 recollect that the multiplicand is the number to be added, and that the multiplier only expresses how many times it is to be added, or repeated. Hence, the multiplier is always an abstract number. TO multiply acres by dollars, or cents by yards, &c., is sheer nonsense. For th method of analyzing concrete questions, see Sect. IV, Art. 2, Ex. 1. Obs. 16. Given the p'oduct of two factors, and either factor, to find the other factor-Divide the product by the given factor. (Sect. V., Art. 2, Obs. 10.) 15. The product of two factors is 48, and one of them is 6. What is the other factor? Ans. 8. 16. The product of two factors is 576, and one of them is 48. What is the other factor? Ans. 12. To prove Multiplication by casting out the 9'sObs. 17. Cast the 9's out the multiplier and multiplicand; multiply their excesses together, and cast the 9's out of their product; then, if the excess of 9's in this, is the same as the excess of 9's in the total product, or answer to the question, the work is correct. This method of proof depends upon A property of the number 9, viz: Any number divided by 9 will leave the same remainder as the sum of its digits, or figures, divided by 9. Demonstration.-Take any number, as 876. This separated into its nume ical parts, equals 800 - 70 + 6. But 800 = 8 X 100 = 8 X (99 + 1) = 8 X 99 + 8. Also, 70 = 7 X 9- 7. Hence, 876 = 8 X 99 + 8+(7 X 9 + 7) + 6 = 8 X 99 +(7 X 9) + 8 + 7 +- 6, and 876. 9 = (8 X 99 + 7 X 9 + 8 -7 + 6) - 9. But 8 X 99 + (7 X 9) is evidently divided by 9 without a remainder, (because 9 is one factor of hiis expression); therefore, 876. 9 will leave the same remainder as (8+ 7 + 6) - 9. The same method of reasoning will apply to any other number. Now, from this demonstration, the reason of the rule is evident. Because, if we reject the 9's from any number, and also reject the 9's from the several parts of the same number, add the latter excesses together, and reject the 9's from their sum, this latter excess ought to be equal to the excess of 9's in the number itself; the whole being equal to the sum of all its parts. Now, in Multiplication, the product is a number, of which the multiplicand is a part, taken as many times as there are units in the either factor, how do we find the other factor? Demonstrate this rule. How do we prove multiplication by casting outthe 9's? Upon wiat does this method of proof depend? What is this property? Demonstrate it. Show from the demonstration why the rule is evident, How is this applied to Multiplication 86 COMMON ARITHMETIC. Sect. VI. multiplier. Hence, if we multiply the multiplier by the excess of 9's in the multiplicand, the excess of 9's in the product ought to be equal to the excess of 9's in tife total product, or answer to the question. But it will produce the same result to multiply the excess of 9's in the multiplicand by the excess of 9's in the multiplier, and reject the 9's from this product. Hence, the rule is correct. To cast out the 9's in any number, we add together the digits of that number, and as often as we obtain 9, reject it, take the remainder and proceed as before. Thus, to cast the 9's out of 1657324, we say 4 and 2 are 6, and 3 are 9; (rejecting the 9,) 7 and 5 are 12; (rejecting the 9,) 3 and 6 are 9; (rejecting the 9,) we find the excess in the whole number to be 1. NOTE.-This property of the number 9, belongs to no other digit except 3. 17. Multiply 278 by 745, and prove the operation. OpeMion. 278 Excess of 9's in the multiplicand is 8. 745 " " " multiplier is 7. — " * a" < productof 8X7is2. 1390.( ( " whole product is 2. 1112 Hcn,'(, the operation is correct. 1946 207110 It is customary to write the excesses in the four spaces of a cross; the excesses of the two factors being placed above and below, and the other excesses at the right and left, thus: 2 Then if the excesses at the right and left are alike, the / work is correct. 18. What is the product of 4682 by 378? Ans. 1769796. 19. What is the product of 4781 by 6213? Ans. 29704353. 20. What is the product of 37682 by 26571? Ans. 1001248422. Obs. 18. When we have given the cost of 1, to find the cost of a given quantity, either more or less-Multiply the cost and quantity together. REMARK.-The term quantity applies to any thing capable of increase or diminution-as numbers, lines, cloth, &c. NoTE.-The learner must not infer from the above that we multiply two concrete numbers together, because we say, multiply together the cost and 1Iow do we cast the 9's out of a number? Does this property of the number U belong to any other digit? When we have given the cost of 1, how do we find the cost of a given quantity, either more or less? To what does the trnm iantity apply? Art. 1, RECAPITULATION. 57 quantity. We express the rule in this way for the sake of brevity, but the multiplier should always be abstract. (Obs. 15, and Sect. IV, Art. 2, Obs. 6,Rem. 1.) 21. If 1 bushel of apples cost 42 cents, how much will 25 bushels cost? Ans. 1050 cents. 22. If 1 stove costs 25 dollars, how much will 19 stoves cost? Ans. 475 dollars. 23. If 1 book costs 84 cents, how much will 216 books cost? Ans. 18144 cents. Obs. 19. If we add a unit to the multiplier, it will produce the same result as to add the multiplicand to the product; if we add two units, the result will be the same as to add the multiplicand twice to the product; and universally, to add any number to the multiplier, gnreases the product as many times the multiplicand as there are units in the number added. Again, if we subtract a unit from the multiplier, it will produce the same result as to subtract the multiplicand from the product; to subtract two units, the result will be the same as to subtract twice the multiplicand from the product; and universally, to subtract any number from the multiplier, diminishes the product as many times the multiplicand, as there are units in the number' subtracted. It is for these reasons that the proof in Multiplication is correct. (Sect. IV., Art. 2, Obs. 7.) REMARK.-The learner will bear in mind that in all cases where the multiplier is unity, or 1, the product is equal to the multiplicand; when the multiplier is greater than 1, the product is greater than the multiplicand; and when the multiplier is less than 1, the product is less than the multlplicand. Obs. 20. DIVISION.-Division has been defined by many authors of Arithmetic as a short method of performing many subtractions, when the numbers to be subtracted are all equal; and as a consequence it is urged that the divisor is always of the same name as the dividend, (because we cannot subtract numbers of different names from each other,) and that the quotient is always an abstract number, (because it expresses the number of times the subtractions have been performed.) This, however, is not always the case, because we very often have two concrete numbers given-one as a dividend, and the other If we add a unit to the multiplier, what effect does it have on the product? If we add two units what effect does it have? What inference is deduced from this? What effect does it have on the product to subtract a unit from the multiplier? Two units? What inference is deduced from this? If the multiplier is 1, to what is the product equal? If-the multiplier is greater than 1, to what is the product equal? If the multiplier' is less than 1, to what is the product equal? How is division defined bymany aithots? What inference 's deducel from this? Is this always: tietse?' A4 58 COMMON ARITHMETIC. Sect. VI. as a divisor, which are of entirely different names. For instance, suppose it were required to divide 100 dollars equally between 4 men, and tell how many dollars each would receive. Now, how are we to take 4 men from 100 dollars, and how many dollars will remain after the operation has been performed? Why, the very idea of such a thing is absurd and preposterous in the extreme. The learner will bear in mind that the dividend is equal to the product of the divisor and quotient, plus the remainder, if any.(Sect. V., Art. 2, Obs. 9.) Hence, the divisor or quotient must one of them always be of the same name as the dividend, since the product must always be of the same name as the multiplicand. (Sect IV., Art 2, Obs. 6.) Now in the above example the dividend is dollars, and the answer required is dollars; and we reason-if 4 men receive 100 dollars, one man will receive one-fourth of 100 dollars. But one-fourth of one hundred dollars is 25 dollars, and it cannot be any thing else. Here, our divisor is abstract, whilst our dividend and quotient are both concrete. This example could not be performed by Subtraction; hence, the above definition of Division cannot be true in all cases. The divisor is abstract, and the dividend and quotient are concrete; hence, the inferences deduced from this definition are incorrect. Obs. 21. Given the divisor and quotient to find the dividendMultiply the divisor and quotient together, and to the product add the remainder, if any. 24. The divisor of a certain number is 15, and the quotient is 6. Required-the number. Ans. 90. 25. If the divisor is 24, and the quotient 36, what is the dividend? Ans. 864. 26. If the divisor is 48, the quotient 15, and the remainder 37, what is the dividend? Ans. 757. Obs. 22. Given the dividend and quotient to find the divisorSubtract the remainder (if any,) from the dividend, and divide the result by the quotient. 27. If the dividend is 456, and the quotient is 8, what is the divisor? Ans. 57. 28. If the dividend is 108, and the quotient 9, what is the divisor? Ans. 12. 29. If the dividend is 1736, the quotient 144, and the remainder 8, what is the divisor? Ans. 12. Why not? Give an example. To what is the dividend equal? What inference is deduced from this? Why? In the example given what is the dividend? What answer is required? How do we reason? Of what name is our divisor in this case? Our dividend and quotient? Could this exampie have been performed by Subtraction? What do we conclude from this? When we have given the divisor and quotient, how do we find the dividenld? Art. 1. RECAPITULATION. 30. If the dividend is 4829, the quotient 37, and the remainder 19, what is the divisor? Ans. 130. Obs. 23. To prove division by casting out the 9's —Cast the 9's out of the divisor an] quotient; multiply their excesses together, and cast the 9's out of their product; cast the 9's out of the remainder, add this excess to the last, and cast the 9's out of their sum; and if this latter excess is equal to the excess of 9's in the dividend, the work is correct. This Rule is demonstrated in the same manner as the rule for proving Multiplication by casting out the 9's, the dividend being the number separated, and the parts being the divisor3 (or quotient,) taken as many times as there a'e units in the quotient, (or divisor,) and remainder. It may be proper to remark that Addition and Subtraction can also be proved by casting out the 9's, but as the method of proof given in Sects. II. and II1. is full as short, and easier to be understood, we shall let them suffice at present. An ingenious scholar, however, can study out the proof from the demonstration and remarks under Obs. 17, if he thorougly understands the subject thus far. 31. Divide 207748 by 745, and prove the operation. Operation. 745)207748(278 The excess of 9's in the divisor is 7 1490 " " " " quotient is 8 5874 " " " " product of 8X7 is2 5215 " " " " remainder is 8 6598 " " " ' sumof 8+2 is 1 5960 ' " " " dividend is 1 638 Hence, the work is correct. 32. Divide 456789 by 365. Ans. 1251, and 174 rem. 33. Divide 764218 by 213. Ans. 3587, andl87 rem. 34. Divide 932684 by 416. Ans.2242,andl2 rem. 35. l)ivide 897653by 321. Ans. 2796, and 137 rem. Obs. 24. When we have given a quantity either more or less, to find the cost of unity, or 1-Divide the cost by the quantity. 36. If 8 pounds of coffee cost 100 cents, how much is that per pound? Ans. 124 cents. Demonstrate this rule. When we have given the dividend and qtotient, how do we find the divisor? Demonstrate this rule. Hew do we prove division by casting out the 9's? How is this rule demonstratedi Which is the number separated? Which are the parts? Can Addition and Subtraction be proved by casting out the 9's? Can you study out the rule? When we hate given a quantity either more or less, and its cost, how do we find the eost of 1$ COMMON ARITHMETIC. Sect. VI. 37. If 24 stoves cost 576 dollars, how much is that apiece? Ans. 24 dollars. 38. If 124 horses cost 8308 dollars, how much is that apiece? Ans. 67 dollars. a. When we have given the cost of a quantity, and the cost of unity, or 1, to find the quantity-Divide the cost of the quantity by the cost of unity, or 1. 39. How many acres of land can I buy for 3834 dollars, at 27 dollars per acre? Ans. 142. 40. How many stoves can I buy for 336 dollars, at 28 dollars apiece? Ans. 12. Obs. 25. Although Subtraction and Division both separate numbers into parts, the learner will observe that there is a great difference between them. In Division the parts are always equal, (being factors of the dividend;) but in Subtraction, the parts may be either equal or unequal. Thus 24 may be separated into 4 parts by Division, each of which is 6; and likewise it may be separated into 4 parts by Subtraction, the parts being 7, 3, 9, 5, or 8, 6, 4, 6, or 9, 8, 6, 3, &c. But the former parts (factors,) we multiply together to produce 24, (6 X 4 = 24,) whilst we add the latter to obtain the same result. (7 + 3 + 9 + 5 = 24, &c.) (Sect. IV., Art. 4, Obs. 4, Rem. 2.) Obs. 26. It is plain from the nature of Division, that the value of the quotient depends both on the divisor and dividend. Because, it is self-evident, that with the same dividend, the greater the divisor, the less will be the quotient, and the smallcr the divisor, the greater wll be the quotient. Hencea. If we increase our divisor we diminish our quotient; and conversely, if we diminish our divisor, we increase our quotient, if the dividend remains unaltered. Becauseb. If our divisor is unity, or 1, the quotient is equal to the dividend; if our divisor is greater than 1, the quotient is less than the dividend; and if our divisor is less than 1, the quotient is greater thna the dividend. c. It is also self-evident, that if we increase the dividend, we increase the quotient; and if we diminish the dividend, ee diminish the quotient, if the divisor remains unaltered. BecauseWhen we have given the cost of a quantity, and the cost of unity, how do we find the quantity? Explain the difference between Subtraction and Division. Give an example. Upon what does the value of Ihe quotient depend? Why? Whatinference is deduced from this fact? If our divisor is unity, or 1, to what is the quotient equal? If our divisor is greater than unity, what is the value of the quotient? If our divisor is less than unity, what is the value of the quotient? What effect does it have upon the quotient to increase or decrease our dividend, without altering the divisor? Art. 1. RECAPITULATION. 61 d. When the dividend and divisor are equal, the quotient is unity, or 1; when the dividend is 9reater than the divisor, the quotient is greater than; and when the dividend is less than the divisor, the quotient is less than 1. Obs. 27. From these remarks, it is evident that it produces the same effect on the quotient, to multiply the divisor by any number, as to divide the dividend by the same number; and also, it produces the same effect on the quotient to divide the divisor by any number, as to multiply the dividend by the same number. From these facts, we deduce the following considerations a. 1st. To divide the divisor by any number, or to multiply the dividend by the same number, is in efect multiplying the quotient by this number. Thus: 4 is contained in 12 3 times. 4 -2 " " 12 6 times, or 3 X 2 times. 4 " " 12 X 2 6 times, or3 X 2 times. b. 2d. To multiply the divisor by any number, or to divide the dividend by the same number, is in efect dividing the quotient by this number. Thus: 3 is contained in 12 4 times. 3 X 2 " " 12 2 times, or 4 - 2 times. 3 " 12 - 2 2 times, or 4 - 2 times. Obs. 28. To multiply or divide both the divisor and dividend by the same number, does not alter the quotient. Thus: 8 is contained in 16 2 times. 8X2 " " 16 X 2 2times. 8 - 2 " " 16 2 2 times. Obs. 29. If we add the same number to both the divisor and dividend, we diminish the quotient; and if we subtract the same number from both the divisor and dividend, we increase the quotient. Thus: 6 -2= 3. And 12- 4= 3. 6+2- 2+2=2. 12- 2 (4 -2)==5. REMARK.-In each of these cases the divisor is supposed to be less than the dividend. When the divisor and dividend are equal it does not alter the quoWhy is this correct? What fact is evident from these remarks? What is the first consideration we notice from this fact? Give an example. What is the second consideration? Give an example. What effect does it have upon the quotient to multiply or divide both the divisor and dividend by the same number? Give an example. If we add the same number to both the divisor and dividend, what effect does it have upon the quotient? If we subtract the same number from both the divisor and dividend, what effect does it have on the quotient? Give examples. In the last two cases, what is the value of the divisor, compared with that of the dividend? If thi divisor and dividend are equal, what effect does it have upon the quotient to add the same number to, or subtract the sams number from both? 62 COMMON ARITHMETIC. Sect. VI. tient to add the same number to, or subtract it from both. Also-When the divisor is greater than the dividend, the converse of the above propositions is true. Obs. 30. If a given number be both multiplied and divided by the same number, the finalresult will be the original number. Thus --- 8 X 7 =56. 56 - 7 8. ARTICLE 2. CONTRACTIONS AND ABBREVIATIONS. NOTE.-In addition to the rules for contracting the operations in Multiplication and Division, given in Sects. IV. and V. the following may also be of advantage in some particular cases. Obs. 1. As to multiply by 10, 100, 1000, &c., we have merely to annex ciphers, to the multiplicand, [Sect. IV., Art. 4, Obs. 1.] it follows, that if a multiplier is an exact part of 100, 1000, &c., we can annex two, three, or more ciphers, if necessary, to the multiplicand, and then divide it by such a number;.s the multiplier is a part of, 100, 1000, &c. Thus: To multiply by 25-Annex two ciphers, and divide by 4: because 25 is one-fourth of 100. To multiply by 50-Annex two ciphers, and divide by 2: because 50 is one-half of 100. To multiply by 125-Annex three ciphers, and divide by 8: because 125 is one-eilghth of 1000. Ex. 1. Multiply 48 by 25. Ans. 1200. 2. Multiply 72 by 25. Ans. 1800. 3. Multiply 649 by 25. Ans. 16225. 4. Multiply 288 by 50. Ans. 14400. 5. Multiply 897 by 50. Ans. 44850. 6. Multiply 462 by 125. Ans. 57750. 7. Multiply 3426 by 125. Ans. 428250. 8. Multiply 7894 by 125. Ans. 986750. 9. Multiply 478264 by 12348. Operation. 478264 12348 Proof. 1434792 \4 5739168 0o0 22956672 /0\ Ans. 5905603872. If the divisor is greater than the dividend, what is the effect? If a number be multiplied, and the product divided by the same number, what effect does it have upon the final result? Give an example. How do we multiply by 10, 100, 1000, &c.? What inference is deduced from this? How can we multiply by 25 Why is this correct? By 50? Why is this correctT' By 125? Art, 2. CONTRACrIONS AND ABBREVI.TIO-5TlO 63 We first multiply by 3, and then multiply this product by 4, because 3 X 4 = 12, and consequently 4 times 3 times any number is evidently twelve times that number. Next, we multiply this latter product by 4, because 12 X 4 = 48, and place the first figure of the product under the 8, because, in reality, we multiply the multiplicand by 48. The first figure of our second partial product we place under the 2, because, in reality, we multiply the multiplicand by 12. The several results are added together, as usual. The operation is proved by casting out the 9's. 10. Multiply 14246 by 819. (81 = 9 X 9.) Ans. 11667474. 11. Multiply 327436 by 126721. (21 =7 X 3; 126 = 21 X 6.) Ans. 41493017356. 12. Multiply 9476245 by 648963. (63 = 9 X 7; 64 = 8 X 8.) Ans. 6149732383935. Obs. 2. The learner will perceive from these examples, that it makes no difference which figure of the multiplier we multiply by first, provided we place the first figure of the product directly under the.figure by which we multiply. 13. Multiply 326 by 241. Operation. By this method we number the figures of 33226l each factor, commencing (at the right hand,) 22 4 1 with 1 in the multiplicand, and 0 in the multiplier. This is done to assist the memory. Ans. 7584536261. These small figures we call EXPONENTS. We number each figure of the product, (our first exponent being 1,) to show which figures we multiply together. Thus, we wish in the first place to find a figure in the product, the exponent of which is 1. Now we multiply those figures together, the sum of whose exponents is equal to the exponent required in the product: the exponent of 1 is 0, and the exponent of 6 is 1; 0 +- 1 = 1; then we multiply 6 by 1, and 6 X I = 6, the first figure of the product. Next we wish to find a figure of the product whose exponent is 2. rihe exponent of 1 is 0, and that of 2 is 2; 0 + 2 = 2: also the exponent of 4 is 1, and that of 6 is 1; 1 + 1 = 2; then 2 X 1 = 2, and 6 X 4 = 24; 24+ 2 26. We set down the 6 and carry the 2 as usual. Now we wish a figure in the product whose exponent is 3: 0 +1= 3; 1 +2 = 3; and 2 + 1 == 3; then 3 X 1 +(2 X 4)+ (6 X 2) = 23, and 2 to carry makes 25. Then 5 is our third figure Why is this correct? Explain the operation of Ex. 9. What conclusion is drawn from this and the three following examples? What is our first step in the operation of Ex. 13? Why is this done? What are these small figures called? What is our first exponent in the pro.luct? Why do wae innrmbr the product? What do we wish to obtain first? Which figuresdo woenmJtil!y together? COMMON ARITHMETIC. Sect. VI. in the product, and 2 to carry. The next figure in the product must have 4 for its exponent: 1 + 3 = 4, and 2 - 2 = 4; then 3 X 4 + (2X2) = 16, and 2 to carry makes 18; we set down 8 and carry 1. Our next exponent in the product is 5; 2 + 3 = 5; then 3 X 2 = 6, and 1 to carry makes 7 as our next figure in the product, and as the sum of no two exponents exceeds 5, we conclude that our work is done. Obs. 3. By examining this operation attentively we notice the following considerations: 1st. We add but two exponents together at a time, one of which belongs to eachfactor. 2d. We multiply those figures together, the sum of whose exponents is equal to the exponent sought, add their several products together mentally, and set down and carry as usual. Obs. 4. The only difference between this and the usual method of multiplying is, that by this method the operation is performed mentally, whilst by the other it is all written upon the slate; the mental exercise and discipline of mind, (no contemptible considerations,) being the chief advantages gained by this manner of operating. 14. Multiply 379 by 647. Ans. 245213. 15. Multiply 1846 by 1234. Ans. 2277964. 16. Multiply 27345 by 97216. Ans. 2658371520. 17. Multiply 894832 by 687219. Ans. 614945552208. REMARK.-We have already learned hoav to multiply by 25, 50 and 125, and it is evident that we can divide by these numbers, by merely reversing the operation, as division is the opposite of multiplication. Obs. 5. Hence, to divide by 25 —Multiply the dividend by 4; cut of two figures from the right, and take one-jburth of these for the true remainder. 18. Divide 176 by 25. Ans. 7, and 1 rem. Operation.-176 X 4 = 7104; 4 4 4 1. 19. Divide 275 by 25. Ans. 11. 20. Divide 1284 by 25. Ans. 51, and 9 rem. 21. Divide 4294 by 25. Ans. 171, and 19 rem 22. Divide 36812 by 25. Ans. 1472, and 12 rem. REMARtr. —The reason why we divide the remainder by 4, is because the remainder is one-hundredths, when it should be twenty-fifths. We reduce it to twenty-fifths by dividing it by 4. The learner will understand this better when he has studied Fractions. _S Give the remaining solution of this question. How do we know when oar operation is completed? In examining this operation what is the first consideration we notice? The second? What is the difference between this and the comrmon method of operating? Whvnt are,he chief advantrge!; derived from tiis method? How may we divide by 25? Why do we take one-fourth of the figures cut off for the true remainder? Art. 2. CONTRACTIONS AND ABBREVIATIONS. 05 Obs. 6. To divide by 125: Multiply the dividend by 8; cut off three figures from the right, and take one-eighth of these for the true remainder. REMARK.-The figures cut off are thousandths, whereas they should be 125ths. We reduce them to 125ths by dividing by 8. (See remark above.) 23. Divide 1462 by 125. Ans. 11, and 87 rem. Operation.-1462 X 8 = 11 696; 696 8= 87. 24. Divide 3216 by 125. Ans. 25, and 91 rem. 25. Divide 4781 by 125. Ans. 38, and 31 rem. 26. Divide 4821 by 125. Ans. 38, and 71 rem. 27. Divide 3976 by 125. Ans. 31, and 101 rem. Obs. 7. When there is a remainder after dividing by several numbers, to find the true remainder-JMultiply each remainder by all the preceding divisors, and to the sum of their product add the first remainder. 28. Suppose a teacher had 158 apples to divide between 6 classes, and each class contained 4 scholars. How many should he give to eaeh scholar? Ans. 6, and he would have 14 apples left, or 6 l4 apples each. Operation. 6 158 4 26- 2rem. 2 X 6 =12; 12+ 2 = 14 true rem. 6 - 2 rem. He first divides the apples into 6 piles, one for each class, and finds that each pile will contain 26 apples, and there will be 2 apples left. He then divides each pile into 4 parts, one for each scholar, and finds that there are 2 apples left in each pile. Now as there are 6 piles, there mu t be 2 X 6 = 12 apples left in all the piles, and the two that were left in the first place make 14, The same method of reasoning will apply to any number of divisors; hence, the above rule is correet. How do we divide by 125? Why do we take one-eighth of the figures cut off for the true remainder? When there is a remainder after dividing by several divisors, how do we find the true remainder? Explain the operation of Ex. 28, and show why it is correct. From what divisor is each remainder exempt from being multiplied? How may we often contract operations in Long Division? Are the general rules, Sects. IV. and V., sufficient for all calculations in Multiplication and Division? Why then do we use contractions and abbreviations? 66 COMMON ARITHMETIC. Sect. VI. 29. Divide 2373 by 2, 3, 4, 2 and 3. Ans. 16, and 69 rem. Operation. 2j2373 Last rem. 1 X 2 X 4X 3 X 2 48 311186- 1 rem. Third rem, 3 X3X2= 18 Second rem, 1 X 2 2 41395- 1 rem. First rem,, 1 2j98 - 3 rem. Sum 69 true rem. 3149 16 1 rnm. REMAnK.-The learner will perceive that no remainder is multiplied by th divisor from which this remainder accrued. 30. Divide 1706 by 2, 3,7, 3, and 4. Ans. 3, and 194 rem. 31. Divide 2903 by 3, 4, 2, and 3, Ans. 40, and 23 rem. 32. Divide 3721 by 2, 4, 2, 3, and 2. Ans. 38, and 73 rem. 33. Divide 4973 by 4, 2, 4, 2, and 5. Ans. 15, and 173 rem. 34. Divide 79641 by 4, 7, 6, 8, 9. and 5. Ans. 1, and 19161 rem. Obs. 8. We may often contract operations in Long Division, by rejecting factors; that is, by dividing both the dividend and divisor by any number that will divide the divisor without a remainder. 35. Divide 625 by 25. Ans. 25. Operation. 5)'.5,625 - -| — We first divide both the divisor and dividend by 5 125 5, and then proceed as in Simple Division. Ans. 25 36. Divide 34921 by 1728. Operation. 8)1728i34921 Last rem. I - Second I 8)216, 4365-1 rem. First " 9)271 545-5 3j 60-5 " 20 Ans. 20, and 361 rem. 5 X 8 X 8 320 5X8= 40 361true rem Sum 361 true rem. CANCELATION. 67. 37. Divide 4276 by 288. (288 = 9 X 8 X 4.) Ans. 14, and 244 rem. 38. Divide 6953 by 256. (256 = 8 X 8 X 4.) Ans. 27, and 41 rem. 39, Divide 7491 by 625. (625 X 5 X 5 X 5.) Ans. 11, and 616 rem. 40. Divide 17426 by 1296. (1296 =9 X 6 X 8 X 3.) Ans. 13, and 578 rem. REMARK.-The ingenious pupil, when he becomes acquainted with Fractions, can easily study out more contractions, both in Multiplication and Division, (and can derive both pleasure and profit from his labor); but he must recollect that the General Rules, in Sects. 1V. and V., will apply to any case of either that may occur. The chief advantage derived from abbreviations is, that by the aid of these we can often solve questions mentally, which would otherwise require the aid of the slate. More methods might be given, but is thought unnecessary, as these are sufficient for any calculations in common business. SECTION VII. C ANCELATION. ARTICLE 1. DEFINITIONS AND ILLUSTRATIONS. Obs. 1. CANCELATION is a short method of performing many operations of numbers, when Multiplication and Division are both concerned. To Cancel means to erase, or reject. Obs. 2. It very often happens in arithmetical calculations that we have, in the same question, several numbers that are multipliers, and several that are divisors. Now, if we should so arrange these numbers that all the divisors would be by themselves, and all the other numbers by themselves, it is highly probable that the same number might occur both as a multiplier and divisor; or, if not, there might be factors common to both multipliers and divisors, which might be rejected, and thus very much shorten the operation.This is the object of Cancelation. Obs. 3. The first thing we do towards arranging these numbers, so as to cancel them, is to draw a perpendicular line, and place all our divisors at the left, and all our other numbers at the right hand side of this line. This is done because it is most convenient. Obs. 4. Now from the very manner in which our numbers are placed, it follows, that the numbers at the right are all factors of some number which is a dividend, and the numbers on the left are allfactors What is Cancelation? What does cancel mean? What often happens in arithmetical calculations? How may we often shorten the operation in such cases? What is the first thing we do in order to cancel lnubers? COMMON ARITHMETIC. Sect. VILIl of some number which is a divisor, and the answer to the question is the quotient of the one divided by the other. Therefore, in the following examples, we have used the term Dividend to represent the numbers at the right, when taken collectively, and the term Divisor to represent the numbers at the left, when taken in the same manner, and the term Factor to represent the numbers on either side of thejline when taken separately. Ex. 1. Multiply 24 by 6 and 8, and divide the result by 16 and 6. Operation. 1 4 —12 We first find that the factor 6 is common ~ —X 6 to both our divisor and dividend, and there- fore cancel it. (Sect. VI., Art. 1, Obs. 30.) We next divide both our divisor and divi-Ans. 12 dend by 8; this is done by canceling the 8 in our dividend, into the 16 in our divisor, and setting the other factor of 16 (2) on the side of our divisor (Sect. VI., Art. 1, Obs. 28.) Finally, we cancel the 2 into 24, which gives 12 as our answer. 2. Multiply 4667 by 24, 50 and 63, 28, 26 and 40. and divide the result by 36, Operation. We first divide both the divisor -- t i4 - 359 and dividend by 10; this is done - ^ A 4 — by canceling the cipher in 40 and 4- 5p 50. (Sect. V., Art. 4, Obs. 1.) 4p 1 -- - 3 Then we divide both divisor and divi- - ------ dend by 12; this is done by rejecting 161 538.5(33679a Ans 12 in the factors 24 and 36, canceling 48 these factors, and placing each quotient (from dividing them by 12, )on the 58 side of the number divided to produce 48 it. In the same manner we cancel 28 and 63, dividing both by 7. We also 105 cancel 26 and 4667, by dividing both 96 by 13. The value of the result as yet remains unaltered. (Sect..VI., Art. 1, 9 Obs. 28.) We now have a 2 on each side of the line, which we cancel, and also cancel the 3 into the 9. Now we have 359, 5, and 3 left on the right, and two 4's on the left, and as these will not cancel, we multiply those on the left together for a divisor, and those on khe right for a dividend, and proceed as in Simple Division. Why are they placed in this manner? What conclusion is deduced from the numbers being placed in this way? In the examples given, what do the terms divisor, dividend and factor represent? Explain the operation of Ex. 1. Explain the operation of Ex. 2. Art. 1. CANCELATION 69 3. Multiply 132 by.9, 7, 16 and 12, and divide the result by 11, 144, 63, 8 and 2. Operation. XiX Xt 9 times 7 X 63; therefore, we cancel these XI -- X14 two into 63. 8 times 2 - 16; we cancel these I' into 16. We cancel 12 in 144, which leaves 12 * Xo as the other factor (of 14I). 12 times 11 Xl 132; therefore we cancel these into 132. Thus -- every number cencels, and consequently the diAlis. 1. visor is equal to the dividend, and the answer is 1. (Sect. VI., Art. 1, Obs. 26, d.) 4. Multiply 72 by 32, and 16, and divide the product by 144, 48, and 24. Operation. -X4a 1'/? 72 cancels into 144; 16 into 48; 8 3 -4A - - 4- 2 divides 24 and 32, and 2 cancels into 4. 3 - 14 X Then we multiply the numbers on the... -. — left together, which gives 9 as our di9 2 = Ans. 2. visor, whilst our dividend is 2. In this case our result is fraction, or less than unity, (Sect. VI., Art. 1, Obs. 26, d.,) and we can only express' the divisions by writing the dividend over the divisor. 5. Multiply 48 by 24, and divide the product by 96 and 72. Operation.. 2 -( $ 4 Ih this case all the numbers cancel on the 3 -,7 ~4 right of the line; but when all the numbers cancel on either side of the line, the remain6 a. Ans. ing factor on that side is unity, or 1. Thus, 48 in 48 goes 1 time; 48 in 96 goes 2 times; 24 in 24 goes 1 time; 24 in 72 goes 3 times; 1 X 1 = 1; 2 X 3 = 6; hence, we must divide 1 by 6, and 1 - 6 = (. (Sect. 5. Art. 2. Obs. 4. a.) )bs. 5. From attentively examining these operations, we notice the following considerations a. 1st. In all questions in Cancelation we have an operating number or leading term, which is always placed at the right. b. 2d. It does not alter the result to divide numbers on both sides of the line by the same number, and canceling one number into another is in reality dividing both sides by that number. c. 3d. When we divide two factors, (one of the divisor and the Explain the operation of Ex. 3. Explain the operation of Ex. 4. When all the numbers cancel on both sides, what is the result? Wlea all the numbers cancel on either side of the line what is the remaining factor on that side? Illustrate this. In examitiag thos operations what is the first consideration we notice? The second? 70 COMMON ARITHMETIC. Sect. VII1 other of the dividend,) by any number, we place each result on the side of th!e number divided to produce it. d. 4th. When we have canceled all we can, we multiply together the remaining numbers at the left for a divisor, and those at the right for a dividend, and proceed as in Division of Simple numbers. e. 5th. If all the numbers on both sides of the line cancel, the answer is 1. f. 6th. If, after canceling, the divisor is greater than the dividend, the answer is less than unity, and the division can only be expressed. g. 7th. Unity is a factor on either side of a line, when all the numbers cancel on that side. Obs. 6. The learner must bear in mind, in all his operations, that the principle of cancelation consists, simply, in rejecting thcfactors common to both the divisor and dividend. This being the case, all our canceling is performed by Division. Some, however, attempt to explain it on the principles of Subtraction, saying, that if equals are taken from equals, the remainder will be equal. The principle is correct; but as our divisor and dividend are not always equal, unfortunately, this principle cannot apply to Cancelation; because the subtraction of the same number from unequal quantities, where one is a divisor and the other a dividend, will materially alter the value of the quotient. (Sect. VI., Art. 1, Obs. 29.) From the preceding remarks and illustrations, we derive the following GENERAL RULE FOR CANCELATION. I. Draw a perpendicular line, and place the operating number, or leading term, at the right, together with all the multipliers, and allithe other numbers at the left. (Obs. 3 and 5, a.) II. When the same factor occurs on both sides of the line, cancel it in both places. III. After canceling all common factors, multiply together all the remaining numbers at the rightfor a dividend, and those at the let for a divisor, and thenproceed as in Division of Simple numbers. (Obs. 5, d.) NoTE.-Read carefully the considerations under Obs. 5. Third? Fourth? Fifth? Sixth? Seventh? In what does the principle of Cancelation consist? How is it performed? How do some explain it? Why do they explain it thus? Is this principle correct? Why, then, will it not apply to Cancelation? Will it produce the same final result to subtract the sani, number from unequal quantities, when one is a divisor and the other a dividend? Why not? In the General Rule, what is the first thing we do? What i. the second? The third Art. 1. CANCELATION. 71 EXERCISES FOR THE SLATE. 1. If 6 bushels of potatoes cost 18 shillings, how much would 9 bushels cost? Operation. If 6 bushels cost 18 shillings, 1 bushel ( I X-3 will cost 18 - 6=3 shilllings, and 9 bushels 9 would cost 9 times as much as I bushel.- Hence, in this example we divide by 6 and Ans. 27 shillings. multiply by 9. 2. If 16 yards of cloth cost 64 dollars, how much would 19 yards cost? Ans. 76 dollars. 3. Paid 50 dollars for 16 sheep. how much should I pay for 40 sheep at the same rate? Ans. 125 dollars. 4. How much must I pay for 36 horses, if I pay 450 dollars for 11 horses? Ans. 1472TL dollars. 5. How much must I pay for 32 yards of broadcloth, if 8 yards cost 48 dollars? Ans. 192 dollars. 6. How much must I pay for 18 cows, if 6 cows cost 90 dollars? Ans. 270 dollars. 7. There are 365 days in a year, and 24 hours in a day, and the earth moves around the sun at the rate of 68000 miles an hour. Now how many years would it take a man to travel the distance the earth moves in a year, allowing a man to travel 40 miles a day? Ans. 40800. 8. How long would it take him, if he traveled 60 miles a day? Ans. 272000 years. 9. How long, if he traveled 80 miles a day? Ans. 20400 years. 10. If a man walk 200 miles in 6 days, how many miles can he walk in 28 days? Ans. 933-. 11. If I pay 24 dollars for 12 books, how many dollars must I pay for 19 books? Ans. 38. 12. How much would 40 acres of land cost, if 30 acres cost 150 dollars? Ans. 200 dollars. 13. How much would 60 bushels of wheat cost, if 25 bushels cost 30 dollars? Ans. 72 dollars. 14. How much would 76 bushels of oats cost, if 15 bushels cost 4 dollars? Ans. 20-4 dollar. 15. If I pay 150 dollars for 6 stoves, how much must I pay for 14 stoves? Ans. 350 dollars. 16. 18 times 432 are how many times 12. Ans. 648 times. 17. Multiply 847 by 19, 28, and 54, and divide the result by 57, 36, and 98.. Ans. 121. 18. Multiply 8700 by 91, 46, and 144, and divide the result by 390, 132, and 1740. Ans. 58TT. 72 COMMO0N ARITHIMETIC. Sect. VII. 19. Multiply 50 by 128, 75, 35, and 85, and divide the result by 1200, 1360, and 875. Ans. 1. 20. Multiply 380 by 57, 115, 186, and 323, and divide the result by 760, 171, 92, 589, and 765. Ans. 3a. ARTICLE 2. GREATEST COMMON DIVISOR. Suppose it were required to find some number that would divide 8 and 12 without a remainder. We find that 2 will divide both, because 8 - 2 - 4, and 12 2 = 6. Then 2 is a common divisor of 8 and 12. HenceObs. 1. A common divisor of two or more numbers, is a number that will divide them without a remainder. But suppose we wished to find the greatest number that would divide 8 and 12. This we find to be 4, as 8 -.-4 - 2, and 12 -4 = 3, and 2 and 3 are both prime numbers, and cannot be divided again. Then 4 is the greatest common divisor of 8 and 12. — HenceObs. 3. THE GREATEST COMMON DIVISOR of two or more numbers is the greatest number that will divide them without a remainder. REMARK 1.-One number is said to be a measure of another number, when theformer is contained in the latter without a remainder. Therefor-, a common divisor is often called a COMMON MEASURE; and the greatest common divisor of two or. more numbers, is often called the GREATEST COMMON MEASURE of those numbers. 2. It must be observed, however, that whilst a measure can exist with reference to two numbers, the common measure, and the greatest common measure must both always have reference to at least three, one of which must divide the other two. Thus, 4 is a measure of 16,ja common measure of 20 and 24, and the greatest common measure of 8 and 12. Ex. 1. What is the greatest common divisor of 9 and 126? Solution.-The greatest common divisor cannot be greater than 9, because it must divide 9. 9 will divide itself; let us see if it will divide 126. We find by trial, that 126 contains 9 exactly 14 times; then 9 must be the greatest common divisor of 9 and 12 6. 2. What is the greatest common divisor of 15 and 80? Operation. 15)80(8 We first try if 15 is the greatest com75 mon divisor. After dividing 80 by 15, - we have a remainder of 5. Now if 5 5)15(3 will divide, 15, it will also divide 80; be15 cause if 15 contains 5, 75 will also con-- tain 5, as 75 is divisible by 15; then if 00 75 contains 5, 80 will also contain 5; beWhat is a common divis6r of 2 or more numbers? The greatest common divisor? When is one number said to be a measure of another? What is a common divisor often called? The greatest common divisor? What is the difference between a measure, a common"measure, and the greatest common measure of numbers? Give an example. Explain the solutin of Ex. 1. Art. 2. GREATEST COMMON DIVISOR. T3 cause 80 = 75 - 5; that is, 80 will contain 5, one more time than 75. We find that 15 will contain 5, therefore 5 is the greatest common divisor of 15 and 80. From this example, we perceive that the greatest common divisor of two numbers, must be a common divisor of the least number and their remainder after division; therefore, it cannot be larger than this remainder, because the remainder must be divided by it.HenceTo find the greatest common divisor of two numbers, we have the following RULE.-Divide the greater number by the less, and that divisor by the remainder, and so on, continuingto divide the last divisor by tie last remainder, till nothing remains. The last divisor will be the greatest common divisor. NOTE.-1. The learner will perceive that a common divisor of two or more numbers, is simply a common factor of those numbers; and the greatest common divisor is their greatest common factor. L 2. We learn, (Sect. IV. Art. 4, Obs. 3, Rem. 1.) that a prime number cannot consist of two factors, both greater than unity, or 1. Therefore, a prime number is divisible only by itself and unity. 3. One number is said to be prime to another, when only a unit will divide both of them. HenceObs. 3. Two or more prime numbers cannot have a common divisor, because they have no commonfactors. EXERCISES FOR THE SLATE. 1. What is the greatest common divisor of 18 and 78? Ans. 6; 2. What is the greatest common divisor of 23 and 120? Ans. 1. 3. What is the greatest common divisor of 32 and 122? Ans. 2. 4, What is the greatest common divisor of 69 and 291?, Ans, 3. 5. What is the greatest common divisor of 231 and 517? Ans. 11. 6. What is the greatest common divisor of 5191 and 8497? Ans. 29. To find the greatest common divisor of more than two numbers: Obs. 4. Find the greatest common divisor of two of them; then Explain the operation of Ex. 2. What consideration do we notice in the solution of this example? How do we find the greatest common divisor of two numbers? What is a common divisor? The greatest common divisor? Of what cannot a prime number consist? What inference do we deduce from this? When is a number said to be prime to another? Can two or more prime numbers have a common divisor? Why not? What is the first rule for finding the greatest common divisor of more than two numbers? 5 COMMON ARITHMMET1O. Sect. 11. of this greatest common divisor, and another given number, and so on, through all the given numbers. The last common divisorfound will be the one required. Or, Obs. 5. Write the numbers in a line. Divide them by any number that will divide them all without a remainder, and write the quotients below, as itn Lion. Divide these quotients in the same manner, and so continue to do, till no number greater than I will divide them all without a remainder. Finally, multiply together all the divisors, and their product will be the greatest common divisor. 'Deirnstration.-We learn, Sect. IV, Art. 4, Obs. 2, that every covnposft n'umber is composed of factors; and if we resolve two or more numbers into their several factors, and the same factor occurs inall the numbers, this factor is evidently a common divisor of these numbers, and the product of allthese common factors is their greatest common divisor. Now these common factors are the numbers by which we divide; hence, the above rule is correct. 7. What is the greatest common divisor of 64 and 96? Operation. 8)64 —96 We first divide by 8 and then by 4. 2 and 3 are not divisible by any number greater than 1, hence, 4)8__12 8 X 4 = 32 the greatest common divisor required. 2-_ 3 8X4=32 Ans. 8. What is the greatest common divisor 1197? By thefirst Rule. 2457)3213(1 189)1197(6 2457 1134 756)2457(3 Ans. 63)11 2268 1 189)756(4 0( 756 of 2457, 3213, and By the second Rule. 9 2457,3213 1197 7 273.-357- 133 39 _ 51..19 B9(3 89 30 - -~- 9 X 7 = 63. Ans. 000 Ohs. 6. The following rules may assist the learner in finding common divisors: i. Ary eien *umber may be divided by 2, because as the remainThe second rule? Demonstrate this rule. What numbers can be divided by2? By3? By 9? Art. 2. GRZATEST COMMON P IVISOR. 75 der must always be less than the divisor, (Sect. V, Art. 2, Obs. 12, b.) the last partial dividend must be 0, 2, 4, 6, a, 10, 12, 16, or 18, citherof which will contain 2. b. Any number can be divided by 3 when the sum of its digits can be divided by 3. c. Also, Any number can be divided by 9 when the sum of its dig. its can be divided by 9. - The last two rules are both demonstrated in the same manner, (Sect VI, Art. 1, Obs. 17, demonstration.) d. Any number can be divided by 4, when its two right hand digits can be divided by 4. Because, if the number can be divided by 4, it can be divided by 2 X 2; (Sect. V., Art. 4, Obs. 2, a.) Therefore, the half of such a number must be an even number} and if half of any number will contain 2, the number itself will contain 4. e. Any number can be divided by 5, when its right hand figure is 5 or 0. Because, as the remainder is always less than the divisor, (Sect. V., Art. 2, Obs 12, b.,) the last partial dividend must be 5, 10, 15, 20, 25, 30, 35, 40, or 45, all of which will contain 5. f. Any number ending with 0, 00, 000, &c., can be divided by 10, 100, 1000, &c. (Sect V., Art. 4, Case 1.).q. No even number will divide an odd number, and no number will contain a greater number than its half. Because1st. If an odd number will contain an even number, it will also contain 2, which is a factor of all even numbers; (Sect. V., Art. 4, Obs. 2, a.) But this cannot be according to Sect. IV., Art. 4, Obs. 3, Rem. 3. The reason why an odd number will not contain 2, is because the last partial dividend is always 1, 3, 5, 7, 9, 11, 13, 15, 17, or 19, none of which will contain 2 without a remainder. 2nd. Any number will contain its half just twice. Hence, if the divisor is greater than the half of the number, it must be equal to the number itself, or there will be a remainder after the division is performed; else it must be greater than the number, when the result is a fraction. (Sect. VI., Art. 1, Obs. 26, d.) Find the greatest common divisor of the following numbers: 9. 48 and 60. Ans. 12. 10. 24, 36, and 84. Ans. 12. 11. 36, 96, and 144. Ans. 12. 12. 25, 45, 85, and 115. Ans. 5. 13. 22, 44, 143, 209, and 297. Ans. 1. 14. 18, 63, 99, 117, and 171. Ans. 9. By 4? By 5? By 10, 100, 1000, &c.? Will an even number divide an odd number? Why not? Wha't is the greatest diyisor any.riia hbeir Wi1 contain? Demonstrate these rules. ' - COMMON ARITHMETICc Sect. VII. 15. 48, 54, 75, and 111. Ans. 3. 16. 249, 332, and 415. Ans. 83. 17. 94, 188, 282, and 423. Ans. 47. 18. 78, 117, 143, and 169. Ans. 13. 19. 12, 20, 36, 44, and 48. Ans. 4. 20. 21, 49, 63, 91, and 133. Ans. 7. ARTICLE 3. LEAST COMMON MULTIPLEt Obs. 1. One number is said to be a MULTIPLE of another number, when theformer contains the latter without a remainder. Thus, 4 or 6 is a multiple of 2, because either of them will contain 2. Obs. 2. A COMMON MULTIPLE is any number that will contain two or more numbers without a remainder. Thus, 12 is a common multiple of 3 and 4, because it will contain both 3 and 4 without a remainder. REMARK.-A multiple, whether common of on,, two, or more numbers, is always a composite number, and the numbers contained in it are its factors. Thus, 2 is a factor of 4, and 3 and 4 are factors of 12. Hence-To find a common multiple of two or more numbers: Obs. 3. Multiply them together. Ex. 1. What is a common multiple of 3, 4, and 6. Ans. 72. 2. Of 4, 8, and 10. Ans. 320. 3. Of 2, 4, 6, 8, 10, and 12. Ans. 46080. REMARK 1.-It is sometimes desirable to find the least number that will contain two or more numbers without a remainder. This is called their LEAST COMMON MULTIPLE. Thus, 72 is a common multiple of 3, 4, and 6; but 12 is their least common multiple. 2, A common multiple of two or more numbers must contain all the differentfactors of these numbers. Thus, 48,;a common multiple of 6 and 8, contains]3 and 2, the factors of 6, and also 4 and 2, or 2, 2, and 2, the factors of 8; but 32 will not contain the factors of 6 and 8, therefore it is not a common multiple of these numbers, 4. What is the least common multiple of 3, 5, and 7? These numbers are all prime, and have nofactors; therefore, no number less than their continued product will contain them; that is 3X5X7=105. Ans. HenceTo find the least common multiple of prime numbersObs. 4. Multiply them together. When is one number said to be a multiple of another? Give an example. What is a common multiple? Give an example. What kind of a rumber is a multiple? What are its factors? Give examples. How do we find a common multiple, or two or more numbers? What is the least common multiple of two or more numbers? Give an example. What mustall common multiples of two or more numbers contain? Give an example. What factor of 6 or 8 will not 32 contain? Art. 3. LEAST COMMON MULTIPLE. 77 5. What is the least common multiple of 2, 3, and 5? Ans. 30. 6. What is the least common multiple of 7, 11, and 13? Ans. 1001. 7. What is the least common multiple of 17, 19, 23, and 29? Ans. 215441. To find the least common multiple of composite numbers: Obs. 5. As we have said before, (Sect. V., Art. 4, Obs. 2, a,) if any number will contain a composite number, it will also contain its factors. Thus, 24 will contain 12, and it will also conta'n 2 and 6, or 3 and 4, the factors of 12. 8. What is the least common multiple of 6 and 8? Solution.-The factors of 6 are 2 X 3; of 8, 2 X 2 X 2. Now we wish our multiple to contain all these factors, but it is evident that if the same factor occurs in both numbers it will occur twice in their product, or common multiple, where it need only occur once. Then as the factor 2 occurs in both 6 and 8, we will reject it in one of them, and the product of the other factors 3 X 2 X 2 X 2 = 24, the least common multiple required. 9. What is the least common multiple of 2, 4, 6, 8, 12, and 16? Operation. -_ 4 —0_- _XE16 Any number that we can divide by 3 16, we can divide by 2, 4, and 8, I e16 X 3-= 48 Ans. cause these numbers are factors of 16; also, any number that we can divide by 12, we can divide by 6, as 6 is a factor of 12. (Obs. 5.) Now we have remaining 12 and 16, but these numbers have a common factor, (4.) This we reject in the 12, and multiplying 16 by the other factor of 12, (3,) we obtain 48 as our answer. REMARK.-We write our numbers, placing the largest at the right, because it is more convenient to place them in this manner. 10. What is the least common multiple of 3, 9, 8, 15, 6, 5, 14, 10 and 12? Operation. 1st line, _._ _ __.___ p__ _ 14_ 15 2nd 3"- __ 4 3rd" 4 _ 2 4th 2 2X2X3X14X15 — 2520 Ans. When any number will contain a composite number, what must it also contain? Give an example. Explain the solution of Ex.8. Why do we reject the factor 2 from one of the numbers in this example. Explain the operation of Ex. 9. How do we write our numbers? Why? Explain the operation of Ex..9 78 COMMON AlITtHM.TI Beet. VII. We first cancel 3, 5, and 6, because they are factors of 9, 10, and 12. We next find that the factor 3 is common to 9 and 15; we cancel it in the 9; 5 is a common factor of 10 and 15; we cancel it in the 10; 3 is a common factor of 12 and 15; we cancel it in the 12; the remaining factors of these numbers (9, 10, and 12,) we write below, and these (the factors 3, 2, and 4), form the second line. Next, we find that the factor 2 is common to 8 and 14; we cancel it in the 8; 2 (in the 2nd line,) will divide 14; we cancel it, (2;) 2 is a factor common to 4, (in the 2nd line,) and 14; we cancel it in the 4; we write the remaining factors (of 8 and 4) below, as our third line. Next, we find the factor 2 common to 4 and 2, (in the third line;) rejecting it in the 4, we write the remaining factor (of 4) below as our fourth line. Finally, we multiply together the numbers remaining (in all the lines,) and their product is the least common multiple required. Obs. 6. From examining this operation, we notice the following considerations: 1st. W1e work, in all cases, from the left hand towards the rtght, because it is more convenient, from the manner in which the numbers are written. Thus: a. We cancel a number at the left, when it will divide one at the right, because the number' at the left is afactor of the number at the right. b. When we cancel a commonfactor, we cancel it in the nvmber at the left. This is done, because the number at the left being smaller than the number at the right, the factor that is not canceled (at the left) is smaller than the corresponding factor of the number at the right, and by this means we obtain the smallest numbers as our final multipliers. 2d. When we reject commonfactorsfrom two numbers, we reject the GREATEST FACTOR common to them. This is done to prevent the same factor from occurring more times than is necessary in the multiple. This would frequently be the case if we simply canceled the smaller factors. Besides, by canceling the greater factors, our final multipliers are smaller. 3d. We cannot reject factors twice from the same numbers. Because the first time, we cancel the greatest factor common, and if we should again cancel factors from the same numbers, our multiple might not contain both these numbers. Thus, (in the last example) after canceling the factor 3 from 9, (because 3 is a common fac-. - o~ Whatis the first consideration we notice from examining this operation Why? Show how we do this in the first place. Why? How in the second place? Why do we cancel the factor in the number at the left? What is the second consideration we notice? Why do we reject the greatest factor? What is the third consideration? Why not? Art. 3. LEAST COMMON MULTIPLI. tor of 9 and 15,) some might say that 3 (the other faator of 9) was a factor of 15, and cancel it accordingly. This would, in reality, be canceling 9 into 15, or making 9 a factor of 15, whichis not the case, and the consequence would be that the final result would not contain 9. 4th. We first cancel factors with reference to the rigit hand number; next, with reference to the second number from the right, and then with reference to the third number from the right, and so on. We do this, because the numbers at the right being the largest, it is highly probable that many of the smaller numbers will cancel out entirly in these, and thus we would have fewer numbers to multiply together to obtain our final result. 5th. After canceling all we can, we multiply the remaining numbers together. We do this, because, after all the common factors are canceled, the least common multiple must be the product of the remaining numbers. Some may think that we can reject factors indiscriminately and obtain the same result. The following example will show their error: 11. What is the least common multiple of 3, 9, 12, 16, -18, and 24? First Operation. Second Operation. Third Operation. ~-9 _ _, '24 6- -^ _ -24J^ -j'f _. r,.24 2 3 0 3 4.6 2 24X2X3=144 24X3=72 24X6X2=288. an incorrect result. an incorrect result In the second operation we canceled 2 into the 16, (it being common to 16 and 18;) and 6 into the 18, (it being common to 18 and 24;) and 8 into 24. Then 24 X 3 = 72, a result too small. In the third operation we did not cancel the greatest conmaan factor each time, and the result was too large. To show that the first result is correct, we will resolve 144 into its several factors, and see if all the different factors of the given numbers occur in it. The factors of 144 are 3 X 3 X 2 X 2 X 2 X 2. The factors of the given numbers are 3X 3X3 X 2X2X3 X 2X2X2X2 X 2X3X3 X 2X2X2.X3. Give an example illustrating this, and show the consequence upon the reult by departing from this rule? What is the fourth considerationt Why do we do this? Why will it notdo te reject factors indiscririiately? (,xplain a few examples on the black-board illustrating this point.) tew o i,.t show that the result ef the first operation, ]i. 11, was correct? 80 COMMON ARITHMETIC. Sect. VII. I Canceling superfluous factors --- o X X X o X ~ X X 2X2X2X2X, X 3 X3XXX~X We have 2 X 2 X2 X 2 X 3 X 3= 144. Therofore, it is correct. The braces include the factors of each number. Hence-To find the least common multiple of two or more numbers, we have the following RTULE.-I. Set the numbers down in a line, placing the greatest at the right. (Rem. 3.) II. When the numbers at the left have factors common to the right hand number, cancel the factors at the left only; so proceed till all the common factors at the left are canceled. III. Multiply together the remaining factors, andtheir product will be the least common multiple. (Obs. 6, 5th.) NoTE. —The learner will perceive that when the second and third numbers from the right cancel in the first part of the operation, we take their factors below, and proceed as with the numbers themselves. No numbers are used after being canceled, as all their factors are found in the other numbers. EXERCISES FOR THE SLATE. What is the least common multiple of the following numbers: 1. 6, 8, Ans. 24. 2. 7, 12, Ans. 84.. 3. 9, 12, 18, Ans. 36. 4. 8, 15, 24,.. Ans. 120. 5. 6, 9, 15, 20, Ans. 180. 6. 5, 7, 10, 25, Ans. 350. 7. 10, 12, 16, 18, 24, Ans. 720. 8. 10, 16, 20, 24, 32, Ans. 480. 9. 6, 8,10, 16, 18, 20, Ans. 720. 10. 3, 5, 2, 9, 12, 15, 18, Ans. 180. 11. 3, 4, 8, 12, 16, 20, 24, 28, Ans. 1680. 12. 4, 2, 6, 8, 9, 12, 14, 15, 16, Ans. 5040. 13. 5, 2, 36, 10, 12, 18, 20, 24, Ans. 360. 14. 6, 4, 8, 12, 16, 24, 32, Ans. 96. 15. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Ans. 2520. 16. 8, 12, 14, 18, 24, 32, 36, 40, Ans. 10080. 17. 9, 12, 16, 32, 36, 48, 96, Ans. 288. 18. 10, 3, 8, 12, 32, 16, 27, 18, Ans. 4320. 19. 7, 11, 22,49, 55, 70, 84, Ans. 32340. 20. 12, 8, 14, 18, 24, 15, 36, 28, Ans. 2520. What is the first step in finding the least common multiple of numbers?The second? Third? When the second and third numbers from the right cancel how do we proceed? Do we use a number after it is canceled? Why not? FRACTIONS. 81 SECTION VIII FRACTIONS. ARTICLE 1. MENTAL EXERCISES, DEFINITIONS, &C. Obs. 1. A FRACTION is a part of any thing. Thus if I speak of an apple, I evidently mean a whole apple; but if I cut the apple into two equal parts, each part is called onehalf; if I cut it into three equal parts, each part is called one-third; if I cut it into four equal parts, each part is called one-fourth, and two of these last named parts are called two-fourths; three parts are: called three-fourths, &c., and the expressions, one-half, onethird, one-fourth, two-fourths, three-fourths, &c., are called FRACTIONS. We can divide numbers into parts as well as apples; thus we can take one-half of 2, one-third of 3, one-fourth of 8, or any other part of any number. To take one-half of any number, we divide it by 2; to take onethird of any number, we divide it by 3; to take one-fourth of any number, we divide it by 4, &c. Ex. 1. What is one-half of the following'numbers? 4, 6, 8, 10, 12, 14, 18, 16, 24, 22, 20, 36. 2.. |,What is one-third of the following numbers? 6, 9, 12, 15, 18, 24, 21, 36, 33, 30, 27. 3. What is one-fourth of the following numbers? 4, 8, 12, 16, 24, 20, 36, 32, 28, 40, 44. 4. What is one-fifth of the following numbers? 5, 10, 15, 25, 20, 45, 30, 40, 35, 60, 50, 45, 55. 5. What is one-sixth of the following numbers? 12, 18, 48, 24, 36, 30, 42, 60, 66. 6. What is one-seventh of the following numbers? 14, 35, 42, 21, 28, 49, 56, 63. REMARK.-In each of these examples, the learner will perceive that he divides a number into equal parts, the same as in division. (Sect. V, Art 2, Obs. 1.) Hence-Fractions partake of the nature of Division. If a unit is divided into two parts, each part is called one-half; if into three parts, each part is called one-third, &c. HenceWhat is a Fraction? If I speak of an apple, what do I evidently mean? If I cut the apple into two equal parts, what is one part called? If I cut it into three equal parts, what is one part called? If I cut it into four equal parts, what is one part called? What are two parts of the latter division called? Three parts? What are the expressions one-half, one-third, one-fourth, &c., called? How do we take one-half of anumber? One-third of a number? One-fourth of anumber? In each of the given examples what does the learner do? To which of the Fundamental Rules are fractions similar? If a unit is divided into two parts, what is each part called? If it is divided into three parts, what is each part called? From what then does the fraction take its name? 5A 82 COMMON ARI'HMETIC. Sect. VIIt. Obs. 2. The ftradion takes its name from the number of parts into which the unit is divided. Thus, if a unit is divided into two parts, each part is called one-half; if into three parts, each part is called one-third, &c. Therefore, in a unit, or whole number, there are two halves, three thirds, four fourths, nine ninths, twenty twentieths, &c. 7. How many fifths in a unit or whole number? How many sixths? Tenths? Fifteenths? Twenty-ninths? Fifty-fifths. REMARK.-Every fraction must have some value, and this value also depends upon the number of parts into which the unit is divided. Thus if a unit is divided into 3 parts, or thirds, the parts will evidently be less than if divided into 2 parts, or halves; and if divided into 4 parts, will be less than if divided into only 3. Hence: Obs. 3. The greater the number of parts into which a unit ts divided the less will be the value of each part. 8. Which is of the greatest value, one-half or one-third? onefourth or one-fifth? 2 fourths or 2 fifths? 3 fifths or 3 sixths? 9. Which is of the greatest value, one-sixth or one-seventh? onetenth orone-twelfth? 3 eighths or3 ninths? 2 sevenths or 2 eighths? MENTAL EXERCISES. 1. What is one-half of the following numbers? 10, 18, 26, 40, 84, 100, 150. 2. What part of 2 is 1? Ans. one-half. 3. What part of 3 is I? of 4? 8? 5? 7? 9? 6? 10? 12? 20? 25? 4. What part of 3 is 2? Ans. two-thirds. Solution.-One is one-third of 3, and 2 is twice 1; therefore 2 is twice one-third of 3, and twice one-third of 3 is two-thirds of 3. 5. What part of 5 is 2? of 6? 7? 8? 12? 9? 10? 14? 18? 25? 6. What part ofl2 is 3?is 4? 7? 8? 6? 5? 9? 11? 10? 15? 12? 7. If half a bushel of corn cost 2 shillings what would 1 bushel cost? Ans. 4 shillings. Solution.-There are two halves in a whole number; therefore, a bushel would cost twice as much as half a bushel, and twice 2 are 4. 8. If one-third of a pound of raisins costs 8 cents, what cost twothirds of a pound? What cost a whole pound? 9. If one-fourth of a pound of spice cost 4 cents; what cost twofourths of a pound? 3 fourths? A whole pound? 10. If 1 fifth of a yard of cloth cost 5 shillings what cost 2 fifths of a yard? 3 fifths? 4 fifths? a whole yard? What must every fraction have? Upon what does this value depend? If one apple were divided into two parts, or halves, and another'apple were divided into three parts or thirds, in which would the parts be the least? If one were divided into three parts, and the other into four parts, in which would the vplue of the parts be the least? What do we conclude from this? f Art. 1. FRACTIONS. 83 11. If I sixth of a barrel of aider cost 3 shillings, what cost 2 sixths? 3 sixths? 4 sixths? 5 sixths? a whole barrel. 12. If 1 eighth of a bolt of cloth cost one dollar, what cost 2 eighths? 4 eighths? 6 eighths? 3 eighths? 7 eighths? 5 eighths? a whole bolt? 2 bolts? 3 bolts? 4 bolts? 6 bolts? 9 bolts? 13. If a pound of coffee cost 12 cents, what would a half a pound cost? Solution.-A half a pound would evidently cost half as much as a pound, and one-half of 12 cents is 6 cents. 14. If 1 yard of cloth cost 24 cents, what will a third of a yard cost? 2 thirds? 3 yards? 15. If 1 acre of land cost 20 dollars, what cost 1 fourth of an acre? 3 fourths? 2 fourths? 5 acres? 16. What cost 1 sixth of a ton of hay at 12 dollars a. ton? 3 sixths? 6 sixths? 2 sixths? 4sixths? 17. What costs 3 sevenths of an acre of land at 14 dollars an acre. Direction.-First find what 1 seventh of an acre would cost and then multiply this by 3. 18. What cost 5 ninths of a pound of tea at 72 cents a pound? 3 ninths? 7ninths? 1 ninth? 6 ninths? 2 ninths? 5 ninths? 8 ninths? 19. If 1 book cost half a dollar, what will 2 books cost? Solution.-2 books would cost twice as much as one book; twice 1 half are 2 halves, and two halves are a unit or whole number. Ans. I dollar. 20. If 1 man eat half a pie, how many pies will 2 men eat? 3 men? 4 men? 7 men? 12 men? 20 men? 21. If a boy plants 1 third of an acre of corn in a day, how much will he plant in 2 days? 3 days? 4 days? 5 days? 6 days? 7 days? 22. If a horse eats 1 fourth of a bushel of oats in a day how much will he eat in 3 days? 2 days? 4 days? 7 days? 5 days? 8 days? 6 days. 23. What cost 8 marbles at 1 eighth of a cent a piece? What cost 10? 12? 16? 20? 24? 32? 40? 56? 72? 88? 96? 24. If 6 apples were divided equally between 3 boys, what part of them would each boy receive? How many apples would each receive? Solution.-1 is one-third of 3; therefore, each boy receives onethird of the apples. One-third of 6 is 2, hence each boy receives 2 apples. 25. Four men are to receive 40 dollars, each receiving an equal share, what part of the money must 1 man receive? 2 men? 3 men? How many dollars must 1 man retive? 2 men? 3 mm? COMMON ARITHMETIC. Sect. VIII. 26. If 6 bushels of wheat cost 12 dollars, what part of that does 1 bushel cost? 5 bushels? 3 bushels? 2 bushels? 4 bushels? -How many dollars does 1 bushel cost? 2 bushels,? 6 bushels 3 bushels? 4 bushels? 27. 2 is one-third of what number? Solution.-There are 3 thirds in a whole number, hence, if 2 is one-third, the number must be 3 times 2, or 6. Ans. 6. 28. 3 is 1 third of what number? 1 fourth? 1 fifth? 1 sixth? 1 seventh? 29. 4 is 1 third of what number? 1 sixth? 1 tenth? 1 eighth? 1 twelfth? 30. 5 is 1 fifth of what number? 1 seventh? 1 tenth? 1 eleventh? 31. 4 is 2 thirds of what number? Solution.-If 4 is 2 thirds, one-half of 4, or 2, is 1 third of the number. Then 2 X 3 - 6. Ans. 6. 32. Paid 3 dollars for 3 fourths of a yard of cloth; what was that a yard? 3 is 3 fourths of what number? Ans. 4. 33. If 7 eighths of a bushel of wheat cost 84 cents, what is that a bushel? 84 is 7 eighths of what number? 34. If 5 sixths of an acre of land costs 30 dollars, how much is that per acre? 30 is 5 sixths of what number? 35. 9 tenths of a dollar is 90 cents; how many cents are there in a dollar? 90 is 9 tenths of what number? 36. 7 ninths of a hogshead (wine measure) is 49 gallons; how many gallons in a hogshead? 49 is 7 ninths of what number? 37, 6 is 2 thirds of what number? 38. 9 is 3 fourths of what number? 39. 10 is 2 fifths of what number? 40. 28 is 7 elevenths of what number? 41. What is 3 fourths of 12? Solution.-I fourth of 12 is 3; then 3 fourths is 3 X 3 = 9. Ans. 9. 42. What is 7 eighths of 24? 43. What is 4 sixths of 18? 44. What is 4 tenths of 40? 45. What is 9 twelfths of 24? 46. What is 8 ninths of 27? 47. What is 8 sixths of 18? Ans. 24. 48. What is 12 sevenths of 35? 49. What is 15 twelfths of 24? 50. What is 16 elevenths of 33? Obs. 4. When a number is divided into equal parts, as in the preceding examples, these parts are called FRACTIONS. When a number is divided into equal parts what are.these parts called? I Art. 1. COMMON FRACTIONS. 85 Fractions are of three kinds; COMMON, DECIMAL, and DUODECIMAL. 1. COMMON FRACTIONS. Obs. 6. COMMON FRACTIONS are expressed by two numbers, one written above the other, with a line between them. Thus, |, -, 4, 19-6, &c. Obs. 6. he number below the line is called the DENOMINATOR, because it shows into how many parts a unit or thing is divided. Obs. 7. Thie number above the line is called the NUMERATOR, because it shows the number of parts taken, or expressed by the fraction. Thus, in the fraction 3, the denominator 4 shows that the unit is divided into 4 parts, and the numerator shows that 3 of these parts are expressed by the fraction. Obs. 8. The numerator and denominator taken together are called the TEuMS OF THE FRACTION. REMARK 1. The number below the line is called the denominator, because it gives a name to the parts. 2. The number above the line is called the numerator, because it numbers the parts used. 3. The word Fraction is derived from the Latin, and signifiesbroken; hence, fractions are often called BROKEN NUMBERS. Also, a whole number is often called an INTEGER. NOTE.-The learner will perceive that the fraction is the part, and not the number itself when broken. Thus if I break an apple, it is evidently a broken apple, but only the piece broken off is the fraction of the apple. Obs. 9. It will be perceived that Fractions arise from Division, and they may be regarded as expressions of unexecuted division, the numerator answering to the dividend, and the denominator to the divisor; the value of the fraction being the quotient of the numerator divided by denominator. Thus, in the fraction 4-, 4 is the dividend, 2 the divisor, and 4 - 2 = 2, the quotient; and in the fraction, 1 is the dividend, 3 the 3, divisor, and 1 — 3, or 3 of 1 the quotient. (Sect. V. Art. 2, Obs. 4, a.) Of how many kinds are fractions? What are they? How are common fractions expressed? What is the number below the line called? Why? What is the number above the line called? Why? In the fraction i which is the denominator? Which is the numerator? What are the numerator and denominator taken togetherecalled? Give another reason why the number below the line is called the denominator? Give another reason why the number above the line is called the numerator? From what is the term fRactton derived? What-does it signify? What then are fractions often called? What is a whole numnbea, likewise often called? From what do fractions arise? What may they be egarded? To what do the numerator and denominator answer in division? What is the value of the fraction? In the fraction 4 which is the divisor, which thedividend, and what the quotient? In the fraction i which is the divisorzwhich the dividend and what the quotient? 8G COMMON ARITHMTrI~. Sect. VIII. Obs. 10. COMMON FRACTIONS are either Proper, Improper, Aimple, Comnlound, or Complex. Obs. 11. A PROPER FRACTION is one in which the numerator is less than the denominator; as, -,, C,. Obs. 12. An IMPROPER FRACTION is one in which the numerator is equal to, or greater than the denominator; as -, -, 15, &c. REIMARK.-When a whole number and a fraction are expressed together, it ji called a MIXED NUMBER; as 2, 31-, 54, 8, &C. Obs. 13. A SIMPLE FRACTION is a single expression; as -, X, &c. It may be eitherproper or improper. Obs. 14. A COMPOUND FRACTION is a fraction of a fraction; it consists of severalfractions connected together by the word or; as of, 7 of of 5 &c. Obs. 15. A COMPLEX FRACTION is one which has a fraction in 4 2 32 4. its numerator, or denominator, or in both; as -, -, -, &c. 51 3,1s 0 4 NoTE.-Besides these we have CONTINUED FRACTIONS, but as they are not used in common business,fhey are not treated of in this work. From the preceding remarks and definitions, the demonstration ot the following rules is evident: Obs. 16. To read fractions-First read the number of parts used as shown by the numerator, and then the size of the parts as shown by the denominator. 1. Read the following expressions:- -;; 3.; -, (4 i..; 1 4 a. A7.. 5. 3. 5 3 7. 7 17T. 11. 25. 122. 8 2. 25 4. 5 8-' T' 71' ', T' -' 7 1, T'12 T', T', 20-, TT' TT' 37 T, 4- TT of 87of 146; 164892 A8413. 40551 0& ew01 T7 4; 1o5643; T J To-a2g 6OO0To- 2 1; 32 41 78 2. Read the following cxpresions: - -;-;; --. 14 3 4^ 5' 10T3. A teacher wishing to divide some fruit among his scholars,separated it into a certain number of equal parts, and gave some a greater, and some a less number of these parts, according to their deportment in school. What John received could be expressed by the fraction,. Now can you tell me into how many parts the teacher divided the fruit, and how many of these parts John obtained? 4. What Charles received could be expressed by the fraction s; how many parts did he get? iL!.v air cnmmon fractions divided? What is a proper fraction? Give i.~X i -. W'!i t is an improperfraction? Give examples. What is a nixed nliUb3r? (j;: t.x't1,n l-'. v 'lhat is aSimple fraction? Give examples. May it be proper or ilpyropr? Wh:at it a compound fraction Givye examples. What is a complex fraction? U;ve ex.imtlI. How do w$ t*d fractions? Art. 2. FUNDAMENTAL PROPOSITIONS. 87 5. What Henry received could be expressed by the fraction W-; how many parts did he get? Obs. 17. To write fractions-First write the number of parts used, as the numerator, and then the size of the parts as the denominator. 6. Write two-thirds. One-fifth. Two-fifths. Three-fifths.Four-fifths. Two-sixths. Four-sixths. One-sixth. Three-sixths. Five-sixths. Three-sevenths. Five-eighths. Six-ninths. Threetenths. Eight-tenths. Four-elevenths. Seven-elevenths. Ninetwelfths. 7. Write eleven-twelfths. Seven-twelfths. Eleven-sixteenths. Eight-twentieths. Sixteen-twentieths. Nineteen-twenty-fourths.Forty-eight-seventieths. Fifty units, and thirty-six-eighty-ninths. One hundred units, and four hundred-one-thousandths. Eighty units, and three-ten-thousandths. Art. 2. FUNDAMENTAL PROPOSITIONS. In the following remarks and observations, the learner will remember that the numerator answers to the dividend, the denominator to the divisor, and the value of the fraction to the quotient of the numerator divided by the denominator. (Art. 1, Obs. 9.) Thus the value of - is 1; of 3 is 2; of - is one-fourth of 1, &c. Therefore, Obs. 1. If the denominator remains the same, to multiply the numerator by any number multiplies the value of the fraction by that number. Take the fraction = 2; multiplying the numerator by 2, w obtain - 4, which is the same as 2 X 2. REMARK.-By multiplying the dividend, we multiply the quotient. (Sect. VI, Art. 1, Obs. 27. a.) Obs. 2. If the denominator reimtins the same, to divide the numerator by any number divides the value of the fraction by that number. Take = 2; dividing the numerator by 2, we obtain - = 1, which is the same as 2 2 = 1. How do we write fractions? If the denominator remains the snme,whbt effect does it have upon the value of thlF frnrtin', to m:n'<i,!vi th.;, ii'. ritol byv I)im number? Give an ex::pile. [IIfl do w.\ kuiov thi.i proi)ositioiI to be tria,? ] f t",, 1, 'c.:;,in itor re;i iins tlie aiine what effect does it have on the value of frT;.c:itj[r to diviie the numerator by any number? Givo examples. Why is this proio.'itiooa correct? What inference is deduced from this? Give examples. 88 COMMON ARITHMETIC. Sect. VIII. REMARIC.-By dividitg the dividend, we divide the quotient. (Sect. VI, Art. 1, Obs. 27, b.) Therefore Obs. 3. With a given denominator, the greater the numerator, the greater the value of the fraction. For - is greater than 7, 3 than -, &c. REMARK.-Because, the greater the dividend, the greater the quotient.(Sect. VI, Art. 1, Obs. 26, c.) Obs. 4. If the numerator remains the same, to multiply the denominator by any number divides the value of the fraction by that number. Take = 2; multiplying the denominator by 2, makes it 4 1, which is the same as 2. 2. REMARK.-By multiplying the divisor we divide the quotient. (Sect. VI, Art. 1, Obs. 27.) Obs. 5. If the numerator remains the same, to divide the denominator by any number multiplies the value of the fraction by that number. Take 4 = 1; dividing the denominator by by 2, we obtain - =2, which is the same as 1 X 2. REMARK. By dividing the divisor we multiply the quotient. (Sect. VI, Aft. 1, Obs. 27, a.) Therefore Obs. 6. With a given numerator the greater the denominator, the less will be the value of the fraction. For is l ess than is less than., &c. REMARK.-Because, the greater the divisor, the less the quotient. (Sect. VI, Art. 1, Obs. 26.) Obs. 7. From these observations we notice, a. 1st. It has the same qffect upon the value of the fraction, to multiply the numerator by any number, or to divide the denominator by the same number. Take l; multiplying the numerator by 2, we obtain = 1; dividing the denominator by 2, we obtain = 1, as before. REMARK.-Because, to multiply the dividend, or to divide the divisor, has the same effect upon the quotient. (Sect. VI, Art. 1, Obs. 27, a.) b. 2d. It has the same effect upon the value of the fraction, to divide the numerator by any number, as to multiply the denominator by the same number. Why is this correct? If the numerator remains the same, what effect dods it haVe upoet the value of the fraction, to multiply the denominator by any'number? Give an example. lIow do we know this to be correct? If the numerator remains the same, what effect does it have upon the value of the fraction, to divide the denominator by any number? Give an example. How do we know this to be correct? What inference is deduced from this? Give examples. Why is this correct? What is the first consideration we notice from these prepositions? Givean example-. Why is this correct? What is the second consideration we notice? Give an example? Art. 2. FUNDAMENTAL PROPOSITIONS. 89 Take; dividing the numerator by 2, we obtain 1: multiplying the denominator by 2, we obtain o _ 1; as before. REMARK.-Because, to divide the dividend, or to multiply the divisor, has the same effect upon the quotient. (Sect. VI, Art. 1, Obs. 27,.) Obs. 8. Multiplying or dividing both the numerator and denominator by any number, changes the form of the fraction without altering its value. Take = 1; multiplying both terms of the fraction by 2, we obtain - = 1; dividing both terms of the fraction by 2, we obtain RREMARK.-Beeause to multiply both the divisor and dividend by the same number does not alter the quotient. (Sect. VI, Art. 1, Obs. 28.) Obs. 9. ff the same number be added to both terms of a proper fraction, the resulting fraction will be greater than the former fraction. Take 23; adding 3 to both terms of the fraction, we obtain 5 which is greater than -. a. If the same number be subtracted from both terms of a proper fraction, the resulting fraction will be less than the former fraction. Take 4, subtracting 3 from both terms of the fraction, we obtain, which is less than 4. REMARK.-These two propositions are the converse of the propositions given in Sect. VI, Art. 1, Obs. 29. When the terms of a fraction are equal, however, it does not affect the value of the fraction to add or subtract the same number from both terms of it, because if equals are added to, or subtracted from, equals, their sums, or differences, will be equal. (Sect. VI, Art. 1, Obs. 29, Rem.) Obs. 10. If the numerator is less than the denominator the value of the fraction is less than 1. Thus = one-half of 1, 4 = threefourths of 1, &c. REMARK.-Because when the numerator is less than the denominator, the dividend is less than the divisor. a. If the numerator and denominator are equal, the value of the fraction is 1. Thus, - = 1; 1; = 1, &c. REMARK.-Because when both terms of the fraction are equal, the dividend and divisor are equal. Why is this correct? What effect does it have upon the value of the fraction to multiply or divide both the numerator and denomiator by the same number? Give an example. Why is this correct? If the same number is added to both terms of a proper fraction, what is the value of the resulting fraction compared with that of the former fraction.? Give an example? If the same number be subtracted from both terms of a proper fraction, what is the value of the resulting fraction, compared with that of the former fraction? Give an example. Of what are these two propositions the converse? When the terms of the fraction are equal, what effect does it have upon its value, to add or subtract the same number from both of them? Why? If the numerator is less than the denominator, what is the value of the fraction? Why? If the numerator is greater than the denominator, what is the value of the fraction? Why? If the numerator and denominator are equal, what is the value of the fracti'n? 90 COMMONIO ARITHMEITIC. Sect. VIII. b. If the numerator is greater than the denominator, the value of the fraction is greater than 1. Thus 3 1 = = 3; - ec. RrMARJC. —Because, when the numerator is greater than the denominator the dividend is greater than the divisor. Obs. 11. Fractions may be added, subtracted, multiplied, or divided, as well as whole numbers, but to perform these operations it is necessary to make changes in the form of the fraction. Art. 3. REDUCTION OF FRACTIONS. Obs. 1. The process of changing the terms of a fraction without altering its value, is called REDUCTION OF FRACTIONS. CASE 1. To reduce a fraction to its lowest terms. Obs. 2. A fraction is reduced to its lowest terms, when the numerator and denominator are prime to each other. (Sect. VII, Art. 2, Note 3, under the rule.) Ex. 1. Reduce to its lowest terms. Operation. To divide both terms of a fraction by the same 4) 6 " Ans. number, does not alter its value. (Art. 2, Obs. 8.) Therefore, dividing both 12 and 16 by 4, we obtain 3 as our answer. 2, Reduce 2 to its lowest terms, First Method, I, Second Method,.1 3 8 4 2' 9 4 An2. 6) - 4) 3a 2) Ans. Or, 48) An. By the first method we civide both 192 and 240 by 6, which gives -w; next we divide both 32 and 40 by 4, which gives 8; and dividing both these by 2, we obtain 5 as our answer. By the second method we divide both 192 and 240 by their greatest common divisor, (48,) found according to the rule in Sect, VII, Art, 2; this gives 4- for our answer as before. Hence-To reduce a fraction to its lowest terms: Obs. 3. Divide both terms of the fraction by any number that will divide both of them without a remainder; so continue to do, till no number greater than 1 will divide bqkh numerator and denominator vdithout a remainder. Or, a. Divide both terms of thefractionby their greatest common divisor. NOTE 1, When the terms of the fraction are small, it is generally best to work aceerding to the first method; but when the terms are large, the latter method is the most convenient. Why? Can fractions be aided, subtracted, multiplied and divided? What is itnecessary to do in order to perform these operatiouns What is Reduction of Fractions? When is a fraction reduced to its lowest termnw? When are numbers prite to each other? What is the irst rule for reducjng.frAcons to their iawrt terms? The second rule? When Is it bot ta Ui thi frt sirl? REDUCTION OF F; ACTIONS. 91 2. Tih learner will bear in mind that the value expressed by the fraction does not depend upon the magnitude of its ternas, but upou the re!btion they bear to each other. Thus, J~~~-0~, is much less than i, although its terms are larger. 3. Tie value of a fraction is not altered by reducing it te its lowest terms, because the relation of the terms to each other remains the same. (Art. 2, Obs. 8.) Reduce the following expressions to their lowest terms: 3..1 Ans. 2. 11 28. Ans. v. 4. An50. Ans... Ans. 5. -153 Ans. ns. 6 5. Ans. ~. 1 5 As. 8. As 1 Ans. 136. ~Ans. 3. 1544 7 144 9. 7. Ans.:. 17. i 2 ' 3. Ans. 9. 1.0225 Ans 1043). 4)3 An. 10. '' S. Ans. 4-79. -4 609 Ans. 8.27 1 Ans. '1'ns. CASE 2. To reduce an ienproperfi faion to a whole or mixed num'er Obs. 4. A MIXED NUMBER is the value of an improperfraction, expressed by integers. 9. As, 17. ~: Ans. ~. Exa 1. Reduce 3 to a whole number. Ans. 2. The value of a fraction is the quotient of the numerator, divided by the denominator. (Art. i. Obs. 9.) Therefore, 6. 3 2. Ans. Operation. In this case there was a remainder after divi- 6)21 ding by 6; this we write over our divisor, and - reduce the fraction to its lowest terms, as direct- 33| = 3f. cd in Case 1. HenceC o reduce an improper fraction to a whole or mixed number: Obs. 5. Divide the numerator by the deno inator, and if there is a remainder, write it over the divisor, and anne. it to the quotient. NOTE.-It is generally best to reduce the fraction, (in the result or answer,) to its lowest terms. to its lowest terms. Upon what does the value expressed by the fraction depend? Give examples. Is the value of the fraction altered by reducing it to the lowest terms? Why not? What is a mixed number? To what is the value of a fraction equal? What is the rule for reducing an improper fraction to a whole or mixed number? 92 COMMZION ARITHMETIC.C Sect. VIII. Reduce the following expressions to whole or mixed numbers: 3. 4. Ans. 4. 11. 1728 Ans. 30-L. 4. 3 Ans. 6'. 12. 2,36 Ans. 224. 5. 4. Ans. 42-. 13. l'4. Ans. 8 —V. B ' 1 7 6 17 ' 6. 7 -. Ans. 23. 14. 4567 Ans. 5_1. 7. '149 Ans. 13,. 15. 5102 36 Ans. 8, 8. 36 9 Ans. 52 5 16. 4632 Ans. 10 265 0.7 *~,~.Oj b ^6Y*4567 4567o 9. 4!92 Ans. 611.. 17. 81723 Ans. 51'-4. 10. 576 Ans. 3825. 18. 624156. Ans. 1809. CASE 3. To change a whole or mixed number to an improper fraction. Obs. 6. In changing a whole or mixed number to an improper fraction, its value must not be altered. a. An integer is reduced to an improperfraction by writing a unit under it. Thus, 6 —; 10 = -; 19_ ', &c. Ex. 1. Chance 6 to a fraction, the denominator of which is 3. Ans. '8 Solution.-6 - ~I multiplying both numerator and denominator by 3, we obtain 1-. Ans. Or, there are 3 thirds in a whole number, or unit, and therefore in 6 whole numbers are 6 X 3 = 18 thirds, or, 1-. Ans. HenceTo reduce a whole number to a fraction with any given denominator: Obs. 7. Multiply the whole number by the given denominator: the product will be the numerator of the required fraction, under which write the denominator. 2. Change 12 to a fraction, whose denominator shall be 9. Ans. 18. 3. Change 15 to a fraction whose denominator shall be 12. Ans. 180 4. Change 24 to a fraction whose denominator shall be 36. Ans. b 14 5. Change 48 to a fraction whose denominator shall be 114. Ans. W.74 6. Change 3 to an improper fraction. Solution.-3 X 4 = 12 fourths, and 3 fourths make 15 fourths, or '. Ans. Hence What must be observed in changing a whole or mixed number to an improper fraction? How can we change a whole number to an improper fraction? Give examples. How do we reduce a whole number to a fraction with any given denominator? I! Art. 3. REDUCTION OF FRACTIONS. 93 To change a mixed number to an improper fraction: Obs. 8. Multiply the whole number by the denominator of the fraction, to the product add the numerator, and write the result over the denominator. REMARK.-This rule is directly the reverse of Obs. 5, and each proves the other. Reduce the following expressions to improper fractions: 7. 6k-. Ans. 25 15. 45. Ans. 593. 8. 12. Ans. 8. 16. 85'74 Ans. 1457 9. 18-. Ans. 169 17. 1162. Ans. 54 8 10. 235. Ans. 1'9 18. 240o. Ans. 120z1 11. 31. Ans. 1.1. 19. 242o32 Ans. 7 741 12. 457. Ans. 3 7 20. 356 - o Ans. 15 986 13. 78 4. Ans. s,2. 21. 48236. Ans. 286673 14. 9911 Ans. "99 22. 1848 -2 —760.Ans. 2282343 * ' 'T 1'23 4 234 CASE 4. To change a compound fraction to a simple one: Obs. 9. This process consists in finding a single expression equal in value to several expressions. Ex. 1. Cilange 2 of -7 to a simple fraction, Ans. T Solution.- of - is twice as much as 3 of. - of - is -; (Art. 2. Obs. 4.) and 2 being twice as much, is 1; (Art. 2, Obs. 1.) T= - Ans. It will be perceived that the result is obtained by multiplying tie numerators together, and the denominators. Thus, - X 4 The same process is observed when the compound fraction consists of more than two terms. HenceTo change a compound fraction to a simple one: Obs. 10. Multiply together all the numerators for a new numerator, an] all the denominators for a new denominator. Compound fractions may be reduced to simple ones by CANCELATION, which is often much shorter, as it at once reduces the fraction to its lowest terms. 2. Change 4 of of '2 of 7 of 7 to a single fraction. 80 I eAns.. Ans. T How do we change a mixed number to an improper fraction? Of what is this rule the reverse? What relation do they bear to each other? In what does the process of changing a compound fraction to a simple one consist? How is the result obtained in the solution of the first exanple? 9t COMMON ARmIT.IMTIC. Sect. VIII. Operation. I 4 Since the numerator of a fraction answers to the 9 5 dividend, and the denominator to the divisor, (Art. 1. X, X?- 4 Obs. 9.,) we place the numerator at the right, and 8 f the denominator at the left. in all other respects, we /7 i proceed according to the General Rule, Sect. VIII., - Art. 1. HenceAns. 8. To work fractions by cancelation: Obs. 11. Place the numerators at the right, and the denominators at the left of the line; in all other respects proceed as usual in canceling. In the answer, the numerator is always at the right, aad the denominator at the left. NOTE.-When whole or mixed numbers occur, they must be reduced to improper fractions. Reduce the following expressions to simple fractions: 3. f of o. Ans. 2. 4. 4 of of Ans. 2,. 5. of 6 times - of. Ans. - 1 l4. 6. 9 times 15 times 1 4 of A. Ans. 987. 7. ofof of. Ans. T. 8. 4 of 175 of 15 times 423. Ans. 1316. 9. 4 of 1f of 63 of 24'. Ans, 1-5. 10. Of 5 o f 2 - o f f f of Ans. 2f 4. 1 0 T o T 1 7 6 4-. Ans. es 5 11. of 7 of 0 of _ of I of 4 of of 11. ns. 3o 12. o- of 5- of 21' of — 4 of 7s of 9- of 13'. Ans. 10765. CASE 5.-To change a complex fraction to a simple one. This process cannot be thoroughly understood before the learner has studied Division of Fractions. It is explained in Art. 6, Case 5. CASE 6.-To reduce fractions to their least ccmmon denominatcr, Obs. 12. Fractions have a common denominator, when their denominators are alike, and in reducing them their values must not be altered. Is the same process observed when the compound fraction consists of moro than two terms? What then ithehe rule for reducing a compound fraction to a simpl one? By what other method can the operation be performed? Why is this method preferable? How do we write the terms of the fraction by this method? Why are they thus placed? Hew do we otherwise proceed? What then is the rule? When whole or mixed numbers occur, what must be done with them? When have fractions a common dieroniuator? What must be observed in finding their common denominator? Art. 3. REDUCTION OF FRACTIONS. 95 Ex 1. Change I and - to fiactions having a common denominator. Solution.-If we multiply both terms of a fraction by the same number, its value will not be altered, (Art. 2. Obs. 8.) Therefore if we multiply both terms of the first fraction (2-,) by the denominator of the second fraction, (a,) and both terms of the second fraction, (3,) by the denominator of the first fraction, (1,) we shall obtain _ - =, and -—. Ans. 2. Change -, j, and -- to fractions having their least common denominator. It is evident that each denominator must divide the common denominator; therefore, the least common denominator must be the least common multiple of the denominators of the given fractions. Operation. We find the least common multiple of -. _ 4._6 the denominators of the given fractions 2 to be 12. Then 2X6 =12. -of 12 = 6, X a - ~of 12 = 4, and the fractions are X ' I of 12 0 3 1 2 O of 12 = 2, j 2 =.of ~12=2,J ~5 1. Here we perceive that each result is found by multiplying both terms of each fraction by the quotient arising from dividing the least common denominator, by the denominator of this fraction. The reason why we do this, is, because we wish to multiply both terms of each fraction by such a number as will produce an equivalent fraction, having the least common denominator. Now the denominator of each fraction must be one factor of this denominator, therefore the other factor must be the number by which both terms of the fraction are to be multiplied. HenceTo reduce fractions to their least common denominator: Obs. 13. Find the least common multiple of the denominators of the given fractions, and multiply both terms of each fraction by the quotient arising from dividing the least common multiple by the denominator of this fraction. NoT'.-Compound fractions must be reduced to simele ones before finding their least common denominator. The object of reducing fractions to a common denominator, is to facilitate te addition and subtraction of fractions, (Art. 4.); and as Explain the solution of Ex. 1, and show why it is correct. What must the least common denominator of several fractions always be? Why so? How do we obtain each result in Ex. 2? Why do we proeed&in this manner? How then de we reduce fractions to their leat common daeominater? What turt be done with compound fractions when they occur? 96 COMMON ARITHMETIC. Sect. VIII, it is generally considered most practical to find the least common denominator, we have given but one rule.* 3. Reduce 4, -, and 8 to fractions having their least common denominator. NoTE.-When the fractions are small, the pupil should be taught to reduce them mentally, and not be allowed the use of the slate or black-board at all. This will accustom him to thinking for himself. In the last example we at once discover that 10 will contain 5, but will not contain 3, or any factor of 3; then 10 X 3, or 30, is the least common denominator of the fractions, Now let the learner be questioned thus: How much is 1 of 30? If 6 is x' of 30, how much is? Then if 4 of 30 is 24, - 2 4 Proceed in the same way with the other fractions. The answers are =; and 8 34 Reduce the following expressions to their least common denominators: 4. -,, and 52. Ans. 28 and 3 5. 2I and. Ans. 3, 26 and. 5. 4 2 ~1 2 35 6. and Ans.,. and. 7 1 2 9 1 3 23 and 7 ' 25, 40, O6~ slVo and 1OO' A7to 864 405 390 4360 66 - ns. 0 T 8 o 1T-600 180 T70 -8.5 l0 1 75 1 0 85 - 54 8. 5 10 17 andof. Aus. 75l TiF and -— of. 9. 3of, of 4 of f a, o f f It E, and -, - of Ans. 6048 57 60 1120 22 0 'Ans 71 o 0 8 0T T T' 08, TT'WW ' T 'i-0oWK' For the convenience of the learner, we will now present at one view the following GENERAL RULES FOR THE REDUCTION OF FRACTIONS. To reduce a fraction to its lowest terms: a. Divide both terms of the fraction by any number that will divide them without a remainder; so continue to do till no number greater than 1 will divide them. (Obs. 3.) Or, b. Divide both terms of the fraction by their greatest common divisor. (Obs. 3, a.) 2d, To reduce an improper fraction to a whole or mixed number: What isthe object of reducing fractions to a common denominator? What is the advantage derived from solving questions mentally? How do we reduce a fraction to its lowest terms? *Fractions may be reduced to a common denominator, by multiplying all their tlenotnina tore together for a new denominator, and each numerator into all the denominators except its-own for a new numqrator; but this rule will not always give the least common denominator. Art. 3. REDUCTION OF FRACTIONS. 97 Divide the numerator by the denominator; write the remaina^r, (if any,) over the divisor, and annex it to the quotient. (Obs. E, 3d. To change a whole or mixed number to an improper fraction: Multiply the whole number by the denominator of the fraction, to the product add the numerator, (if it is a mixed number,) and under the result write the denominator. (Obs. 7 and 8.) 4th. To change a compound fraction to a simple one: a. MItltiply together all the'numerators for a new numererator, and all the denominators for a new denominator; then reduce the fraction to its lowest terms. (Obs. 10.) Or, by cancelation: b. Place the numerators at the right, and the denominators at the lett of the line; then proceed as usual. (Obs. 11.) 5th. To reduce fractions to their least common denominator: First.,find the least common multiple of the denominators; then multiply both terms of each fraction by the quotient, arising from dividing this multiple by the denominator of the fraction. (Obs. 13.) EXERCISES FOR THE SLATE. 1. Reduce ^512 to its lowest terms. Ans.. 2. Reduce 4 8 to its lowest terms. Ans. 7T. 3. Reduce. 5 5 to its lowest terms. Ans. 5 4. Change 12 3 to a whole or mixed number. Ans. 253: 5. Change 9 876 to a whole or mixed number. Ans. 182g. 6. Change 144 7 to an improper fraction. Ans. 3 3s29 7. Change 246 — to an improper fraction. Ans. 81879 8. Reduce 4 of of L of 2 T of oof i to a simple fraction. Ans. 25 9. Reduce of ofof 2 8 of 1 of 7 of of9 to a simple 44 2048 10 fraction. Ans. 2 0 - 10. Reduce 5,, 1 and 115, to their least common denominator. Ans. 20 2 1 and 3. 11. Reduce 3 7 and 1 to their least common denomi 180 210 100 135 an 144 nator. Ans. -s2- 2,T 240'~ 20, and t44 12. Reduce 7 of ^1 of w4 and y of 3634 of:1 to their least common denominator s.RAns. c and 3. men denominator. Ans. 03 and ' How do we change an improper fraction to a whole or mixed number? How do we change a whole or mixed number to an improper fraction? How do we change a compound fraction to a simple one?-'How by eatnclatloa? How do we reduce fractions to their least common denominator? ' 6 98 COMMON ARITHMETIC. Sect. VIll. ARTczLE 4. ADDITION AND SUBTRACTION OF FRACTIONS. Obs.. F.ractions are added or subtracted, by adding or subtracting eir nuoer<aors. MENTAL EXERCISES. 1. James had ~ of an apple, John -, and William. How much had they all? a, and 2, and how many sixths? Ans. 1. 2. Susan had 3-of an orange, Mary -, and Jane '. How much had they all? 3.3A man gave one boy a of a pie, another,4, and another (. How much did he give them all? 4. One man planted 17 of an acre of land, another -, and another 14. How much did they all plant? 6. Whatis the sum of 4, 4,, ad,,and -. 2 6 19 40 75 12 5 6. What is the sum of, TA-, TU 1, 40, -1, and 99 9 7. Matthew has 3 of a pie, and John a. How much more has Matthew than John? from 3 leaves how many fourths? Ans. 1 =. 8. One boy has - of a dollar, and another a. How much has the one more than the other? 9. Henry has 7 of a pound of raisins, and Thomas T. How much has Henry more than Thomas? 10. A. traveled g of a mile, and B. traveled -. How much farther did A. travel than B.? 11. From - take 2 I2. From 7 5 take A. Method of adding and sutracting fractions haing diferent denominators, mixed numbers, cc. Ex. 1. John had a of a dollar, and Frank I of a dollar. How much had both? 3 Solution.-If we add a to, the result will neither be nor But _,and =; and + =-5 Ans. 2. How much more had John than Frank in the last example? Solution.-; =; -=; -3 =. Ans. REMAmx.-We perceive in these examples, that the fractions have to be reduced to a common denominator before they can be either added or subtracted. This principle is the same as that laid down in Sect. VI. Art. 1. Obs. 7 and 10. HenceObs. 2. No numbers, whether integral or fractional, can be added r subtracted, unless they are all of the srame rame or kind.. tw dow. w. b~ $ fisio4s? plain tho outiow of. -1. d....e ^ Art. 4. ADDITION AND SUBTRACTION OF FRACTIONS. 99 3. Add together and. Ans. - - 1. 4 and472 21 4. Add together x and 7. Ans. 1. 5. Add together ~,,, and. Ans. 2T2. l I. From 7 take Ans. A 7. From 2 take. Ans. 6. 8. From l take '3. Ans. A-. 9. Charles has 127 dollars, and Henry has 103 dollars. How many dollars have both? 1st Operation. 128 = 1 83 By the first operation 103 3 8 -.. we reduce the ex1 0 + 8 8 23 dollars. Ans. pressions to improper fractions, and then proceed as in Ex. 1. This method is preferable when the numbers ars small. 2d Operation. By the second operation, we add the fractions and integers separately. This method is to be preferred when the numbers are large. 127 A t< 3 1n6 7 + 13 1 5 0 - - 0: 8 it 1r Ans. 23- dollars. 10. In the last example, how much more had Charles than Henry? 1st Operation. 127 = 03. 8 8 103 - 4 3 86 4 4 - 8 - 103 86 17 O) -lH..-. A... 2nd Operation. 1278 10 = 10i. -V -- = -g-, =_ z UulllA. l113 -- Ans. 21- dollars. Explanation-The same as above, except to read Ex. 2, instead of Ex. 1, and subtract instead of add. 11. Theren has 56- cents, ancd Clarence has 371 cents. much has Theren more than Clarence? Operation. How We cannot take ' froml-; therefore we 566 4 borrow 1 (-) from the 6, and add it to 37- = 372 it the -, making 5; then 5 a- 3 - But as we borrowed 1 we must return it; there- Ans. 183- cents. fore I to carry to 7 makes 8, and 8 from 16 leaves 8, &c. 12. From 78~ take 39-. Ans. 38-9. What principle do we notice in these solutions? With what does this principle oincide? What inference is deduced from this? Explain the operations of Ex. 9 and 10? When is the first method to be preerredt The-coad- nat!id? Explain the operation of Ex. 11. 10o COMMON ARITHMETIC. Sect. VIII. From the preceding remarks and illustrations, we derive the following GENERAL RULE FOR THE ADDITION AND SUBTRACTION OF FRACTIONS. Reduce the fractions to their least common denominator, (Rem.,) and then add or subtract their numerators. (Obs. 1.) NoTE.-Compound fractions must be reduced to simple ones before they are added or subtracted. To add or subtract mixed numbers: Obs. 3. Either reduce them to improper fractions, and proceed as above directed. Or, add or subtract the integral and fractional parts separately. When there are but two fractions to be added, and the numerator of each is 1, we ramy add their denominators for a numerator, and multiply them for a denominator, and then reduce the fraction to its lowest terms. Thus. -- 510 1 5- 3 IhU0. T X -f =50 T -1 If we wish to subtract fractions, when the numerator of each is 1, we mny subtract their denominators for a numerator, and multiply them for a denominator, and then reduce the fraction to its lowest terms. Thus: 1 1 10-5 5 1. s l = - =lo X - -5 0 T.. This is in effect multiplying both terms of each fraction by the denominator of the other fraction, and then subtracting the numerators. Thus: ) X ) =(10)+ -) = 1 r = -i-or. EXERCISES FOR THE SLATE. 1. A man owning 38 of a vessel, afterwards bought 4 of the vessel more. How much did he then own? Ans. 5s of the vessel. 2. William had 5 of a dollar, and his father gave him - of a dollar more. What part of a dollar had he then? Ans. 5 117 3. He afterwards spent ' of a dollar. How much had he left? Ans. 5 of a dollar. What is the general rule for the addition and subtraction of fractions? What must be done with compound fractions when they occur? How do we add or subtract mixed numbers? When there are but two fractions to add, and the numerator of each is 1, how do we proceed? When we wish to subtract fractions, and the numerator of each is 1, how may we proceed? How, in reality, are these two last cases performed? Art. 4. ADDITION AND SUBTRACTION OF FRACTIONS. 101 4. A man planted 4' acres to corn, 1-7 acres to potatoes, and 2-4 acres to beans. How many acres did he plant in all? Ans. 72. 5. How much more did he plant to corn than to beans? Ans. 1 2. 6. How much more to corn than to beans and potatoes both? Ans. -T of an acre. 7. John has 2 of 4 of 3 dollars, Charles has 4 of 5 of 5 dollars, 3 5 and Luke has 1 of of 6 dollars. How much have they all? Ans. 7-47- dollars. 8 How much more has Charles than John? Ans. 1 4' dollars. 9 How much more has Luke than John? Ans. 1 -5 dollars. 10. A merchant has tlree pieces of cloth, one containing 24 -yards, another 163 yards, and the other 14' yards. How many yards in all? Ans. 55-, 1. How many yards in the first piece more than in the second? Ans. 75. 12. How many yards in the first p'ece more than in the third? Ans. 97. 13. How many yards in the second and third pieces together more than in the first? Ans. 67 14. A. has traveled 282 miles, and B. 214 miles. How many miles have they both traveled? Ans. 50 4. 15. How much farther has A. traveled than B.? Ans. 7{ miles. 16. Thomas has 62' cents, and Henry has 43- cents. How many cents have both? Ans. 106t 17. How much more has Thomas than Henry? Ans. 184 cents. 18. A. has 1244- dollars, B. has 1194 dollars, and C. has 214 dollars. How many dollars have they all? Ans. 459 ' 19. How much has A. more than B.? Ans. 44 dollars. 20. How much has C. more than A.? Ans. 90k. 21. How much have A. and B. together, more than C.? Ans. 29. 24. A little girl had 1 of an orange, and her mother gave her - of an orange more. Iowv much had she then? Ans. 3 of an orange. 23. Add together ' and. Ans. 4. Tv 4. A 4n5' 24. Add together J- and 4. Ans. 224 25. Add together _- and 25. Ans. r27 26. Add together ' and 3' Ans..V 27. From 'T take 1. Ans, T'28. From ' take 3. Ans. Ta. 102 COMMON ARITHMETIC. Sect. VIII. 29. From UJ take ~v' Ans. 1T.8. 30. From T-f take - Ans. -. ARTICLE 5. MULTIPLICATION OF FRACTIONS. Obs. 1. In multiplying by a proper fraction, the learner will always find that the proauct is less than the multiplicand; but this need not cause him any surprise, if he recollects that the multiplier is less than unity, or 1; and therefore, he only repeats the multiplicand a part of 1 time. (Sect. VI. Art. 1. Obs. 19. Rem.) MENTAL EXERCISES* 1. If an apple costs f of a cent, how much will 3 apples cost? Solution.-3 apples will evidently cost 3 times as much as 1 apple; and 3 times ' equals 3-, or 1 cent. (Art. 2. Obs. 1.) Ans. 1 cent. 2. If 1 marble costs ' of a cent, how much will 5 marbles cost? Ans. v of a cent. 3. If 1 pear costs i of a cent, how much will 8 ears cost? Ans. -= 6 cents. 3. At 1 of a dollar per bushel, how much will 6 bushels of corn cost? 5. At 3 of a dollar per bushel, how much would 9 bushels of apples cost? 6. At I of a dollar apiece, how much would 9 books cost? 7. At 25 cents a pound, how much would 3 of a pound of raisins cost? Solution.-The learner will perceive that this question is just like the preceding, except that the multiplier and multiplicand have changed places. Now as it makes no difference which factor we use as the multiplier, (Sect. IV. Art. 2. Obs. 5. Rem.,) we may proceed as in Ex. 1, and say, 25 times - = 15 cents. Ans. Or, we may prodeed thus: If 1 pound cost 25 cents, I of a pound will cost x of 25 cents, or 5 cents, and x will cost 3 times as much as -, and 5 X 3 15 cents. Ans. It would be a good plan to solve such questions both ways. 8. A tree 60 feet high, had J- of its length broken off by the wind; what was the length of the broken piece? Ans. 25 feet. 9. The remaining part was -7 of the length of the tree. Required -the length of this part. What will always be noticed in multiplying by a proper fraction? Need this cause any surprise? Why not? Art. 5. MULTIPLICATION OF FRACTIONS. 10. At 50 cents a yard, how much would 17 of a yard of cloth cost? 11. At 96 cents a pound, how much will 7 of a pound of tea cost? 12. There are 320 rods in a mile. How many rods in:-Y of a mile? 13. One-eighth of a dollar is 121- cents. How many cents are there in 7 of a dollar? Solution.-There are evidently 7 times as many cents as there axe in I of a dollar. Then 7 times 12 = 84 cents, and 7 times a a cent 2- 2 - 32 cents, and 84 cents and 31 cents make 874 cents. Or, 12- = 2e; 7 times 25 - 175 = 87 cents, as before. 14. At 16;3 cents a yard, how mueh will 6 yards of celico cost? 15. At 183 cents apiece, how much will 8 slates cost? 16. At 6- cents apiece, how much will 12 lead pencils cost? Solution.-Here the multiplier and multiplicand have changed places; but according to Sect. IV. Art. 2. Obs. 5. Rem., we can proceed as in the last three examples. Thus, 6 X 12 = 72; - X 12 = i 2 3; 72 + 3 -75cents. Ans. Or, we can say12 pencils, at 6 cents apiece, will cost 72 cents, and 12 pencils at 4 of a cent apiece, will cost 3 cents; and 72 cents and 3 cents are 75 cents, as before. It would be a good plan to solve such questions both ways. 17. How much would 10 pounds of coffee cost, at 12- cents per pound? 18. How much would 12 pounds of sugar cost, at 8- cents per pound? EXERCISES FOR THE SLATS CASE 1.-To multiply afraction an2J v whole number together. 1. If a man earn s of a dollar a day, how much can he earn in 6 days? Solution.-6 times _ = 3 5 dollars. (Art. 2. Obs. 1.) Or 6 times = 5 dollars. (Art. 2. Obs. 5.) Ans. 5 dollars. 2. How much will of a pound cf raisins cost, at 30 cents per pound? Solution.-They will cost 2 times 30 cents; but this is the same as 30 times 2; (Sect. IV. Art. 2. Obs 5. Rem.) 30 times a = = 20 cents. Ans. Or, 3 of 30 is 10, and 2 is twice 10, or 20 cents, as before. It will be perceived that both these examples can be worked by one rde. Hence 104 ~, COMMON ARITHMETIC. Sect. VIII. To multiply a fraction and whole number together: Obs. 2. MAultiply the numerator of the fraction and the whole number together, and divide the product by the denominator. Or, a. Divide the denominator of the fraction by the whole number, when it can be done without a remainder, and divide the numerator by the quotient. (Art. 2. Obs. 7. a.) REMARK.-Compound fractions must of course be reduced to simple ones before the operation can be performed. 'If we choose, we can work these sums by cancelation. Thus: Ex. 1. Operation Ex. 2. Operation. (p 5 loP_ _ o 10 Ans. 5 dollars. Ans. 20 cents. In both questions, we write the fractions according to Art. 3. Obs. 11., and the whole numbers according to the General Rule, Sect. VII. Art. 1., and cancel as usual. Hence — To multiply fractions and whole numbers together by cancelation: Obs. 3. Write the fraction as directed by Art. 3. Obs. 11., and the whole number as directed by the General Rule, Sect. VII., Art. 1., and cancel as usual. By this method compound fractions may be reduced to simple ones by expression merely. 3. If 7 of a cord of wood last a month, how much will it take to last 18 months? Ans. 14 cords. 4. If a bushel of wheat weigh - of a hundred weight, how much will 24 bushels weigh? Ans. 14' hundred weight. 5. If 1 yard of cloth cost 1 of a dollar, how much will 60 yards cost? Ans. 48- dollars.. 6. 8 men spent each of them 3 of a dollar. How many dollars did they all spend? Ans. 6. 7. If I pay 'l- of a dollar for a bushel of rye, how much must I pay for 28 bushels? Ans. 22 dollars. 8. If a family eat y of a barrel of flour in a month, how many barrels will it take to last them a year, or 12 months? Ans. 104. 9. If a merchant sells tea at 7 of a dollar per pound, how much will 32 pounds cost? Ans. 28 dollars. Hoe,! do we multiply a fraction and whole number together? Explain why ibis r ethod is correct. By what other method can such questions be worked? How do we proceed according to this method? I Art. 5. MULTIPLICATION OF FRACTIONS. 105 10. If a nan spends,-s of a dollar a day, how much will he spend in 48 days? Ans. 45 dollars. 11. How nmuch would G of a pound of tea cost, at 96 cents per pound? Ans. 64 cents. 12. How much would 2- of y of - of a bushel of wheat cost, at 90 cents per bushel? Ans. 72 cents. 13. How much would 3 of 2 of a yard of cloth cost, at 60 cents per yard? Ans. 135 cents. 14. How mucl would 2- of an acre of land cost, at 30 dollars per acre? Ans. 24J- dollars. CASE 2. —To multiply a whole number and a mixed nuinber together. 1. How much would 132' acres of land cost, at 14 dollars an acre? Operation. 132~ We first multipl y by 14, 14 which gives 12' Next we multiply -t:f- 132 by 14, and adding the several 12 = 14 times - products together, we obtain 1860 528 dolla s as our answer. The reason 132 of our working this question thus, is the same as in Ex. 13, Mental Ans. 1860 dollars. Exercises. 2. How much would 48 yards of calico cost, at 18- cents per yard? Operation. ' 48 This question is just i 183 ilike the preceding one, - 48 X =36 except that the multi36 = cost at I cents per yard. plier and multiplicand 384 have chnnged places. 48 In the operation, we first i -- nmultiply 48 by 3-, acAns. 900 cents. cording to Obs. 2 or 3, and then multiply 48 by 18, and adding the several products together, we obtain 900 cents as our answer. HenceTo multiply a whole number and mixed number together: Obs. 4. MIultiply the integral and fractional parts separtely, and add their products together. How do we multiply a whole number and a mixed number together? How may we proceed when the mixed number is small? 6A 106 COMMON ARITIIMET[IC. Sect. VIII ';:z.vAsa..-If the mixed number is small, we may,;f we choosey reduce it to an improper fraction, and proceed according to Obs. 2 or 3. 3 How much would 26- yards of cloth cost, at 25 cents per yard? Ans. 659:- cents. 4. How much will 473 acres of land cost, at 16 dollars an acre? Ans. 764 dollars. 5. If a man travel 144-7 miles in a week, how far can he travel in 48 weeks? Ans. 6934k miles. 6. How much will 37-J yards of cloth cost, at 144 cents per yard? Ans. 5388 cents. 7. If a man read 288k- 5 pages in a month, how many pages can he read in 24 months? Ans. 6934t. 8. At 73| dollars apiece, how much would 124 horses cost? Ans. 9134- dollars. 9. How much would 432 acres of land cost, at 36- dollars per acre? Ans. 15930 dollars. 10. How much would 210 pieces of broadcloth cost, at 97'T dollars apiece?.Ans. 21108 dollars. 11. How much would 86 pounds of tea cost, at 933 cents per pound? Ans, 80062 cents. 12. How much would 74 yoke of cattle cost, at 527- dollars a yoke? Ans. 3912 dollars CASz 3. To multiply fractions together. Ex. 1. A man having - a dollar, spent 2 of it for his dinner. How much did his dinner cost him? Ans. - of a dollar. Solution If we divide '- an apple into two equal parts, each part will evidently be ' of the apple. Also, if we divide ' a dollar into two equal parts, each part will be - of the dollar. Or, If he gives of all lie has for his dinner, and has but I a dollar, his dinner must cost him I of 2 a dollar; but I of.i is a compound fraction and equals '. (Art. 3, Obs. 10.) 2. At o of a dollar a yard, how much would of a yard of lace cost? Solution.-If a yard cost, of a dollar, of a yard will cost I of, which is x of a dollar; (Art. 2. Obs. 2.); and d of a yard will cost 3 times 2, or 4 =2 of a dollar. (Art. 2. Obs. 1.) Or, If 1 yard costs v of a dollar, 4 of a yard will cost of, or 4 c 4 of a dollar, (Art. 2. Obs. 4. and Obs. 7, b.;) and, will cost 24, or - of a dollar, as before. Ans. of a dollar. 315 As How do we multiply fractions together? What is the shortest method? Why so? Art. 5. MULTIPLICATION OF FRACTIONS. 107 It will be perceived that the result in both these questions is obtained by multiplying the numerators together, and the denominators. HenceTo multiply one fraction by another: Obs. 5. Multiply koyether the nmtnerators for a mew nvnerator, and the denominators for a new denominator. NOTr.-The learner will perceive that this is precisely like reducing compound fractions to simple ones. REMARK.-The shortest method is by caneelation, (Art. 3. Obe. 11.), as it at once reduces it to the lowest terms. 3. At s of a dollar a yard, how much would 7- of a yard of linen cost? Ans. 4- of a dollar. 4. At 5 of a dollar a bushel, how much would 1 of a bushel of wheat cost? Ans. 3j of a dollar. 5. At 14 of a dollar a pound, how much would '4 of a pound of tea cost? Ans.. of a dollar. 6. How much would 3 of a gallon of oil cost, at -4 of a dollar per gallon? Ans. 1 dollar. 7. How much would - of a yard of satinet cost, at 7 of a dollar a yard? Ans. ~ of a dollar, 8. Multiply 3 by 7 of U. Ans. T. 9. Multiply -1 of, 2 by T of 3. Ans. -40. 10. Multiply 7 o f by o f, by f f ' of. Ans. 4. CASE 4. To multiply mixed numbers together. 1. Multiply 14| by 124. Solulion.-14 - = 4; 12i =; X t 1 Q Ans HenceTo mu'tiply mixed numbers together: Obs. 6. Reduce the mixed numbers to improper fractions, and then proceed according to Obs. 5. 2. How much would 121 yards of cloth cost, at 5M dollars a yard? Ans. 67- dollars. 3. How much must be paid for 18- acres of land, at 311 dollars per acre? Ans. 5905- dollars. 4. How m much must be paid for 643 yards of cloth, at 6' dollars a yard? Ans. 412 dollars. 5. Multiply 8' by 16&. Ans. 1371. 6. Multiply 183 by 24-. Ans. 465. 7. Multiply 34,7 by 48. Ans. 1692. Mew s 4 n multiply mixed numbers together? 108 COMMON ARITHMETIC. Sect. VIII, 8. Multiply 371- by 62i. Ans. 2343k. 9. Multiply 45X- by 46-,' Ans. 2117. 10. Multiply 504- by 753. Ans. 380714 9 -. AnsT I iT. For the convenience of the learner we now present at one view the following GENERAL RULES FOR THE MULTIPLICATION OF FRACTIONS 1st. To multiply fractions and whole numbers together: a. Either multiply the numerator of the fraction and the whole number together, and divide the product by the denominator, (Obs. 2.) Or, b. Divide the denominator by.the whole number, when it can be done without a remainder, and divide the numerator by the quotient. (Obs. 2. a.) Or, by cancelation: c. Write the fraction as directed by the 4th.Rule, Art. 3, and the whole number as directed by the General Rule, Sect. VII., Art. 1, and cancel as usual. (Obs. 3.) 2nd. To multiply mixed numbers and whole numbers together: a..Multiply the integral and fractional parts separately, and add their lproducts tcyether. (Obs. 4.) 3d. To m;ltiply fractions together: Multiply the numeralors together for a new numerator, and the denominay'or, for a new deomiinator. (Obs. 5.) 4th. To multiply mixed numbers together: First reduce them to imprope ) c fractcions, and then proceed as in fuidtiplication of Fractions. (Obs. 6.) EXERCISES FOR THE SLATE. 1. How much would 24 bushels of wheat cJst, at '- of a dollar per bushel? Ans. 30 dollars. 2. How much would 36 bushels of corn cost, at.- of a dollar per bushel? Ans. 24 dollars. 3. If a store is worth 25000 dollars, how much is ' ' of it worth? Ans. 135713 dollars. 4. How much would 60 bushels of wheat cost, at i of a dollar per bushel? Ans. 822 dollars. What is the first rule for multiplying fractions and whole numbers together? The second? What is the rule by cancelation? How do we multiply whole numbers and mixed numbers together? How do we multiply fractions together? How do we multiply mixed numbers together? Art. 6. DIVISION OF FRACTIONS. 5. How much would 36 yards of cloth cost, at 7 of a dollar per yard? Ans. 251 dollars. 6. How much would 66 books cost, at 7 of a dollar apiece? Ans. 573 dollars. 7. How much would 96 bushels of corn cost, at 5 of a dollar per bushel? Ans. 60 dollars. 8. When land is worth 12 dollars an acre, how much is '- of an acre worth? Ans. 11} dollars. 9. If powder is 75 cents a pound, how much is "5 of a pound worth? Ans. 55 cents. 10. How much would 18- yards of cambric cost, at 16 cents per yard? Ans. 300 cents. 11. How much would 127 cords of wood cost, at 2 dollars per cord? Ans. 25- dollars. 12. At 87' cents a yard, how much must I pay fcr 12 yards of satinet? Ans. 1050 cents. 13. How much would 40 acres of land cost, at 15- dollars an acre? Ans. 632 dollars. 14. How much would 72 yards of cloth cost, at 4' dollars a yard? Ans. 328 dollars. 15. How much would A a pound of coffee cost, at - of a dollar per pound? Ans..-9 of a dollar? 16. How much would -2- of a yard of cloth cost, at 3 of a dollar a yard? Ans. 3 of a dollar. 17. How much would 4127 bushels of corn cost, at -3 of a dollar a bushel? Ans. 123- dollars. 18. Iow much would 121 bushels of apples cost, at 31 cents a bushel? Ans. 390- cents. 19. How much would 252 O acres of land cost, at 1625 dallars an acre? Ans. 414L dollars. 20. If a man travel 283 miles in one day, how far can he travel in 218 days? ADS. 614-7 miles. ARTICLE 6. DIVISION OF FRACTIONS. Obs. 1. In dividing by aproper fraction, the learner will always find that the quotient is larger than the dividend; but this need not cause any surprise, if he recollects that the divisor is less than unity, and consequently the number of parts into which the number is dividel, must be greater than the dividend itself. (Sect. VI. Art. 1. Obs. 26. 6.) What is always noticed iu dividing by a proper fraction? Need this cause any surprise? Why not? How do we divide a fraction by a whole number? 110 COMMON ARITHMETIC. Sect. VIII MENTAL EXERCISES. If 3 pounds of coffee cost 4 of a dollar, how much is that per pound? Solution.-1 pound will evidently cost 3 as much as 3 pounds, and - of 4 is 4. (Art. 2. Obs. 2.) 2. If 5 yards of calico cost 1o of a dollar, how much is that per yard? 3 If 8 pounds of lead cost 24 of a dollar, how much is that per pound? 4. If 6 men own y2 of a vessel, what part of it is owned by each? 5. If 12 lead pencils-cost -j of a dollar, how much is that apiece? 6. At 4 of a dollar a bushel, how many bushels of potatoes can you buy for 2 dollars? Solution.-You can evidently buy as many bushels as - is contained in 2. Now there 4 fourths in a unit, or 1; then in 2 there are twice 4 fourths, or 8 fourths; that is, 2 contains 8 3 times. 7. At I a dollar a yard, how many yards of cloth can I buy for 6 dollars? 8. At 4 of a cent apiece, how many marbles can be bought for 10 cents? 9. At 4 of a dollar a pound, how many pounds of coffee can be bought for 5 dollars? 10. At - of a cent apiece, how many marbles can be bought for 4 of a cent? Ans. 3. Solution.-There can evidently be as many marbles bought as - is contained in, and 3 contains -, 3 times. 11. At 2 of a dollar apiece, how many books can be bought for T of a dollar? 12. At 3 of a dollar a bushel, how many bushels of potatoes can be bought for '5 of a dollar? EXERCISES FOR IHE SLATE. CASE. 1. To divide a fraction by a whole uumber. 1. If 4 yards of cambric cost - of a dollar, how much is that a yard? Solution.-To find the cost of 1, we must divide the cost of the quantity by the quantity. (Sect. VI., Art. 1. Obs. 24.) Then - 4 = 9; (Art. 2, Obs. 2,) or,.4 8; (Art. 2. Obs. 4.) A - a, as before. Art. 6. DIVISION OF FRACTIONS. 11I Or, by cancelation: 9 8 2 We place our numbers according to Art. 5, 4 Obs. 3, and cancel as usual. - - 9 2 2 -Ans.. of a dollar. The several results it will be perceived are alike. HenceTo divide a fraction by a whole number: Obs. 2. Divide the numerator of the fraction by the whole number, when it can be done without a remainder; and under the quotient write the denominator. Or, a. Multiply the denominator of the fraction by the whole number, and over the product write the numerator. b. Or, by Cancelation: Proceed acording to Art. 5, Obs. 3. NOTE.-Compound fractions, both in this, and the following cases, must of course, be reduced to simple ones before the operation can be performed.When the operation is performed by cancelation, however, we may reduce them by expressien, merely. 2. A man divided - of a dollar between his two children; how much did each receive? Ans. 2 of a dollar. 3. If 5 boys can earn '-of a dollar, how much can 1 boy earn? Ans. A of a dollar. 4. If 6 pounds of coffee cost 3 of a dollar, how much is that a poundt Ans. y of a dollar. 5. If 4 bushels of corn cost of a dollar, how much is that per bushel? Ans. 5- of a dollar. 6. If 7 yards of cloth cost 4 of a dollar how much is that per yard'? Ans. 8 of a dollar. 7. If 12 yards of ribbon cost 4 of,o of "- of 10 dollars, how much is that a yard? Ans. ' of a dollar. 8. If 15 bushels of wheat cost f1f of 100 dollars, how much is that per bushel? Ans. 7 of 'a dollar. 9. If 16 pounds of nails cost 4 of a dollar, how much is that per pound? Ans. ~- of a dollar. 10. If 18 lemons cost T of a dollar, how much is that apiece. Ans. -I- of a dollar. CASE 2. To divide a whole number by a fraction. 1. A teacher divided 8 apples among a class at school, giving Why do we divide the fraction by the whole number in Ex. 1, Case 1?How do we divide a fraction by a whole number? What must be done with compound fractions when they occur? If the operation is performed by can, celation how may we proceed 112 COMMON ArltHME TIC. Sect. VIII. each scholar 3 of an apple. How many scholars were there in the class? Ans. 12. Solution.-It is evident that there were as many scholars as 2 is contained in 8, because each scholar receives 4 of an apple, and there are 8 apples to be divided. We will first divide 8 by. In a unit are 3 thirds, therefore in 8 units are 3 X 8 - 21 thirds; that is, 8 contains - 24 times. Now 2 is twice as much as, consequent3i, 8contain ~ ~ ly 8 will not contain but half as many times as it will 4; that is, 12 times. (Sect. VI, Art. 1, Obs. 26, a.) It will be perceived in this example, that we multiply the whole number by the denominator of the fraction, and divide the product by the numerator. It makes no difference, however, whether we perform the multiplication or divission first; because 8 X 3 = 24; 24 2- = 12, and 8 2-= 4; 4 X 3 = 12 as before. HenceTo divide a whole number by a fraction: Obs. 3. Mlultiply the whole numberby the denominator of the fraction, and divide the product by the numerator. Or, when it can be done without a remainder, a. Divide the uhole number by the numerator, and multiply the quotient by the demominator. Obs. 4. The reci)rocal of a number in the quotient arising from dividing a unit by that number. Thus the reciprocal of 2 is 4; of 3, -;of 5,, &c. If we divide a unit by -, the quotient is or 2; if we divide a unit by 4, the quotient is - 1-=. HenceObs. 5. i'he reciprocal of a. f)actio, is the same fraction inverted. Invert, means to turn upside down. Thus, 4 inverted becomes 3, &c. It will appear from these definitions, that the solution of the above example consists merely in multiplying the whole number and the reciprocal of the fraction together, according to Art. 5. General Rules, Ist."-''. Examples of this kind can also be performed by Cancelation.Thus: $-4 We proceed as directed, Art. 5, Obs. 3, except to 3 use the reciprocal of the fraction, instead of the frac[- I tion itself. HenceExplain the solution of Ex. 1, Case 2. Does it make any difference whether we perform the multiplication or the division first? Why not? How do we divide a whole number by a fraction? What is the reciprocal of a number? Give examples. What is the reciprocal of a fraction? Show why this is correct. What does invert mean? In what does the solution of Ex. 1, Case 2 consist? Art. 6. DIVISION OF FRACTIONS. 113 To divide a whole number by a fraction, by cancelation: Obs. 6. Place the numerator of the divisor at the left, and the denominator at the right; in other respects proceed as usual. 2. At - of a dollar abushel, how much corn can be bought for 3 dollars? Ans. 12 bushels. 3. At 2 a dollar a bushel, how much rye can be bought for 7 dollars? Ans. 14 bushels. 4. At 4 of a dollar a yard, how many yards of muslin can be bought for 6 dollars? Ans. 48 yards. 5. At I of a dollar apiece, how many books can be bought for 8 dollars? Ans. 12. 6. At 5 of a dollar a bushel, how much barley can be bought for 12 dollars? Ans. 164 bushels. 7. A man gave 12 dollars to some destitute people, giving each 7 of a dollar. How many were relieved? Ans. 14. 8. At l of 34 of 10 dollars a bushel, how much wheat can be bought for 18 dollars? Ans. 145 bushels. In this example, the expression - of 3 of 10 dollars is our divisor. 9. At 2 of a dollar apiece, how many hats may be bought for 6 dollars? Ans. 21. 10. At 12 f of of of 12 dollars apiece, how many slates can be bought for 9 dollars? Ans. 36. CASE 3. To divide one fraction by another. 1. At 2 of a dollar a yard, how many yards of cloth can be bought for 4 of adollar? Ans. 13. The result is evidently found by dividing 4 by 4. Operation. 2- _. 6 We first reduce the fractions to a common denomina9. - 13. tor, and then divide the numerator of the dividend, by the numerator of the divisor. We can however, divide one fraction by another without reducing them to a common denominator. We learn from the remark under Obs. 5, that to divide a whole number by a fraction, consists merely in multiplying the whole number by the reciprocal of this fraction. Hence, it is evident that to divide one fraction by another consists merely in multiplying thefracHow do we divide a whole number by a fraction, by cancelation? Explain the solution of Ex. 1, Case 3. Is there any other method of performing such operations? In what do such operations consist? 114 ~jOMMON 4ARrrHMETIC. Sect. VIII. tion whichiis the dividend,by the reciprocal of the fraction which is the divisor. That ie, we invertoar divisor and proceed according to Art. 3, Obs. 5, thus: X 3 = = 1. Ans. Or, we may proceed by oancelation, according to Art. 3, Obs. 11, thus: 3__0 _- 4 The learner will perceive that we write our l vdivisor as directed by Obs. 6. In all these op- -- erations the final result is the same. Hence --- 3 4=13.Ans. To'divide one fraction by another: Obs. 7. Invert the divisor and then proceed as in r ltipaltcatton of fractions. Or, by cancelation: a. Write the divisor as directed by Obs. 6, and the other nt7wlrs as directed by Art. 3, Obs. 11, and cancel as usuall 2. If a man mow 7 of an acre of grass per day, how many days will it take him to mow - of an acre? Ans. 3a. 3. If a man travel 3 of a league in an hour, how many hours will it take him to travel 2- of a league? Ans. 7 of an hour. 4. At 7 of a dollar a yard, how many yards of cloth can I buy forl' of a dollar? Ans. 17. 5. If I pay 4 of a dollar a pound, how many pounds of tea can I buy for 4 of a dollar? Ans. X of a pound. 6. How many yards of satinet can I buy for l of a dollar, at T8 of a dollar a yard? Ans. 3 of a yard. 7. Divide of of 7 4b of 4,by of of of 7. Ans. -4. The learner will observe that he must invert all thie terms of the divisor, when it is a compound fraction. 8. Divide of of of 14 of 52 by of f of of of 28. Ans. 1. 9. Divide x of 7 of, of 4 of 23 byf of f of 46. Ans. 10. Divide! of 4 of -7~ of 20, by 3 of Y of 5 of 32. Ans. 1. CAsE 4. Method of proceeding when mixed numbers occur. 1. How many bushels of wheat can I buy for 5, dollars, at 1' dollars per bushel? Ans. 4o. Solution. 51- — V; 1 = = '; V.-9 = 44 bushels, Ans. How do we divide one fraction by another? How by cancelation? When our divisor is a compound fraction, what must be observed in the operation] Art. 6. DIVISION OF FRACTIONS. 115 Hence-When mixed numbers occur in both the dividend end divisor: Obs. 8. Reduce them to improperfractions, and then proaed according to Obs. 7. 2. How much rye can I buy for 4 —27 dollars, at 17 of a doliarper bushel? Ans. 8 bushels 3. How many books can I buy for 1' dollarsj at - ' of a dollar apiece? Ans. 2. 4. At 51 dollars an acre, how many acres of land can I buy for 52- dollars? Ans. 10. 5. At 2' dollars apiece, how many sheep can I buy for 32- dollars? Ans. 15. 6. At 24- dollars per acre, how many acres of land can I buy for 620 dollars? Ans. 25. Solution.-24- 1 4; 620- - 14 = 25 acres, Ans. Hence-When only the divisor is a mixed number: Obs. 9. Reduce it to an inmroper fraction, and then proceed according to Obs. 3, or 6. 7. At 52- dollars apiece, how many horses can I buy for 2220 dollars? Ans. 42. 8. At 174 dollars apiece, how many cows can I buy for 314 dollars? Ans. 18. 9. At 933 cents per bushel, how many bushels of wheat can I Duy for 1875 cents? Ans. 20. 10. At 62' cents per bushel, how many bushels of apples can I buy for 3000 cents? Ans. 48. 11. At 3 dollars a yard, how many yards of cloth can I buy for 187' dollars? Ans. 62.2 Solution.-187' - 375 375 -3 = 25 62'. Or, 187' - 3 - 62, and 1l remainder; 11 — =;. 3 = 1, which annexed to 62 = 62'. HenceWhen only the dividend is a mixed number: Obs 10. Reduce it to an improperfraction, and then proceed according to Obs. 2. Or, a. Divide the integral part as in simple d(ivision; to the remainder (if any) annex the fractionalpart and proceed as above directed. NOTE.-The first rule is the best when the dividend is small, but when the dividendis large, the latter method is preferable. When both dividend and divisor are mixed numbers, how do we proceed?When only the divisor is a mixed number, how do we proceed? When only the dividend is a mixed number, how do we proceed? When is the first rule preferable in this case? The second? What must be observed in all cases of reduction of fractions? 116 COMMON ARITIMETIC. Sect. VIII. 12. How many acres of land can I buy for 1463- dollars, at 8 dollars an acre? Ans. 18-. 13. How many lead pencils can I buy for 37' cents, at 3 cents apiece? Ans. 12,. 14. At 15 dollars an acre, how many acres of land can I buy for 12344 dollars? Ans. 82-. 15. At 45 cents a yard, how many yards of cloth can be bought for 843- cents? Ans. 183, CASE 5. Complex Fractions. NOTE.-This and the following case properly belong to Reduction of Fractions, but it was thought best to defer them until the learner had studied Division of Fractions. REMARK.-The learner will recollect, that in all cases of reduction of fracions, the value of the fraction must not be altered. (Art. 3, Obs. 1.).21 1. Reduce - to a simple fraction. Ans. 1. 41T' 4 -Solution.-This expression is the same as 2 -' 4-. (Art. 1. Obs. 9.) By Obs. 8, we find the quotient of 24 divided by 4' to be 4. HenceTo reduce a complex fraction to a simple expression: Obs. 11. Reduce both numerator and denominator to improper fractions, and then divide the former expression by the latter, as directed by Obs. 7. Change the following expressions to their simplest form: 42 15 82 5 8' 3. Ans. 22 B 3.- s. 7, - Ans, 25. 4,- Ans. 1 24' 8. - Ans. 2. 6 31 9- 5. - Ans.1' 24 ci -56 9, --- Ans, 6. 51 4 41 How do we reduce a complex fraction to a simple expression? Art. 6. DIVISION OF FRACTIONS. 117 4 6 10. - Ans. 24. 12. 5 Ans. 62 71 3 -57 15 4 1 11. -- Ans, 2. 13. Ans. - 127 | 1 REMAKK 1.-Complex fractions may be either added, subtracted, multiplied or divided, by first reducing them to simple expressions. 42 61' 14. Add together -, and. Ans. 4'.s 1! 33 6 4 4 6' 22 15. Add together -, -, and -. Ans. 43^. 9 33 14 9 3 -1- 1 5 6 4 16. Add together -,, and. Ans. l467 2; 7 32 7 1 17, From —take -. Ans. -42387. 9~; 8' 16' 4 18. From take. Ans. 1 -. 9 102 -1 3 3 7 19. From - take. Ans' 'o4. 4 29 -T REMARK 2.-When complex fractions are multiplied or divided by cancelution, we may merely reduce them by expression, and proceed according to Obs. 7, or Art. 5, Obs. 5. 72 62 20. Multiply -- by -- Ans. 43. 91M 7ti 8y 94 56 4 21. Multiply by-. Ans. 245. 51 31 4 2 Can complex fractions be added, subtracted, multiplied, or divided? How? When complex fractions are multiplied or divided by cancelation, how may we proceed? 118 COMMON ARITHMETIC. Sect. VIIi. 22. Multiply by -. Ans, 62. 41 64 21 7 23, Multiply - by -. Ans. 2, 5 42 l1 21 1 1 24, Multiply - by -. Ans. IV. 34 41 243 41 25, Divide - b - Ans, 6, 81 88 12 56 26. Divide - by -. Ans, 3, 41- 6 42 83 27. Divide - by -, Ans. -, 4 3 2 1 28. Divide - by -. Ans, 1I 2' 10 12 21 29. Divide -bv-. Ans. 5~4 31 4 - 2Q3 63 8 30. Multiply - by - and divide product by —. Ans. 1. 54 143, 40 -CASE 6.-To change afraction to a required denominator. This process consists merely in multiplying both the terms of the given fraction, by such a number as will give a resulting fraction having the required denominator. HenceObs. 12. The denominator of the gyvenfraction, is afactor of the denominator of the required fraction. In this case, our multiplier (which must be the other factor,) is found by dividing the denominator of the required fraction, by the In what does the process of changing of a fraction to any required denominator consist? What inference is deduced from this? Art. 6. DIVISION OF FRACTIONS. 119 denominator of the given fradtion. (Sect. VI. Art. 1. Obs. 16.) 1. Change 4 to a'fraction, the denominator of which is 8. Ans. 2. Solution.-Dividing 8 by 4, we obtain 2 as our multiplier, and multiplying both terms of the fraction (-) by 2, we obtain. as our answer. 71. 2. Change | to twelfths. Ans. - 12. As the denominator of the required fraction is always given, our only trouble is in finding the numerator. In the last example, our numerator is found by dividing 12 by 5, and multiplying the quotient by:. It is evident that by first performing the multiplication, and then the division, we shall obtain the same result, and thus avoid a fractional multiplier. HenceTo change a fraction to any required denominator: Obs. 13. Multiply the numerator of the givenfraction by the denominator of the required fraction, and divide the product by the denominator of the given fraction; the result will be the numerator of the required fraction under which write the required denominator. 3. Change 4 to thirty-fifths. Ans. j. 4 4. Change 3- to fifteenths. Ans. 15 5. Change 4 to thirds. Ans.-. 3 6. Change 2 to fifths. 5 91 7. Change W to elevenths. Ans. - 11 2| 8. Change - to fourths. Ans. -. 4 In this case what must our multiplier be? How is it found? In what does the chief difficulty lie, in solving questions of this nature? How do we find the numerator in Ex. 2? 120. COMMON ARITHMETIC. Sect. VIII. 9. Change iy toninths. 9 31 -10. Change.8 to sevenths. 7 For the convenience of the learner, we will now present at one view the following GENERAL RULES FOR THE DIVISION OF FRACTIONS. 1st. To divide a fraction by a whole number: a. Divide the numerator of the fraction by the whole number, when it can be done without a remainder, and under the quotient write the denominator. (Obs. 2.) Or, b. Multiply the denominator of the fraction by the whole number, and over the product write the numerator. (Obs. 2, a.) or bycancelation: c. Proceed in every respect as in multiplication of fractions.(Obs. 2,b.) 2d. To divide a whole number by a fraction: a. Multiply the whole number by the denominator of the fraction and divide the product by the numerator. (Obs. 3.) Or, when it can be done without a remainder. b. Divide the whole number by the numerator, and multiply the quotient by the denominator. (Obs. 3. a.) Or, by cancelation. c. Proceed in every respect as in mudtplication of fractions, except to write the numerator of the divisor at the left, and the denominator at the right. (Obs. 6.) 3d. To divide one fraction by another: Invert the divisor and thenproceed as in multiplication of fractions. (Obs. 7.) The rule by cancelation is the same. (Obs. 7.) 4th. When mixed numbers occur either in the dividend, or divisor, or both. Reduce them to improper fractions, and proceed according to preceding rules. (Obs. 8, 9 and 10.) 5th. To reduce a a complex fraction to a simple one: Consider the numerator a dividend, and the denominator a divisor, andthen proceed according tothe last rule. (Obs. 11.) What is the rule for dividing a whole number by a fraction? The rule by cancelation? What is the rule for dividing one fraction by another? What is the rule when mixed numbers occur? What is the rule for reducing a complex fraction to a simple one? What is the rule for changing a fraction to any required denominator? Art. 6. nIvr[ION OP FRACTIONS. 121 6th. To change a fraction to any required denominator. lMultiply the numerator qf the given fraction, by the denominator of the required fraction, and divide the product by the denominator of the givenfraction; the result will be the numerator of the required fraction, under which write the denominator. (Obs. 13.) EXERCISES FOR THE SLATE. 1. If 6 slates cost 3 of a dollar, how much is that apiece? Ans. - of a dollar. 2. If 14 yards of ribbon cost 2- of a dollar, how much is that per yard? Ans. 5 — of a dollar. 3. At - of a dollar per bushel, how many bushel of corn can I buy for 10 dollars? Ans. 40. 4. At y3 of a dollar per bushel, how many bushel of oats can be bought for 75 dollars? Ans. 400. 5. How many yards of cloth at ~ of a dollar a yard, can be bought for 6 dollars? Ans. 16. 6. How many books at ' of a dollar apiece, can be bought for 10 dollars? Ans. 12, 7. How many tumblers at — L of a dollar apiece, can be bought for 15 dollars? Ans. 48. 8. How much wheat can be bought for 15 dollars, at 1- dollars per bushel? Ans. 12 bushels. 9. At 3 of a dollar apiece, how many slates can I buy for ' of a dollar? Ans. 11. 10. At -Q of a dollar a roll, how many rolls of tape can be bought for 7 of a dollar? Ans. 14. 1 1. At 8- cents a pound, how many pounds of coffee can be bought for 87 - cents? Ans. 101. 12. At 16. cents a pound, how many pounds of spice can be bought for 93- cents? Ans. 5-: 13. How many boxes will it take to contain 1600 pounds of tea, each box containing 44- pounds? Ans. 36. 14. How many barrels of pork can be bought for 4784 dollars, at 132-e dollars per barrel'? Ans. 36. 15. How many bales of velvet in 11663 yards, each bale containing 12 9 17 yards? Ans. 9 1 31 41 16. Add together, — and-. Ans. 1'j a4. 16. Add together,-and. Ans. -1-'' 21 51 63 24 44 15 17. Addtogether -, - and -. Ans. 43y. 3 6 7 122 CO2MMON ARITIMETIC. Sect. VIII. 41 91 45 8' 15 41. 19. Fr om take - Ans. 5. 7 — 3 — 20. MIultiply- by-. Ans. 2a. 4- t3 8 -5 1 21. Multiply- by- Ans. - 4 1) 2. 2 j`3 22. Divide - bv Ans. 21-. 4 4:, 5 u cr! 23. Divide - by. Ans. 4- 1-' 2i I 0: 24. Clhage 4 to tlhirteclitlls. Ans. -- 13 21l 25. Clirng~e to twenty-fifths. Ans. - 25 26. It 4- bushels of wheat cost 3 - dollars, what will 6- bushels cost? Ans. 5j6 dollars. First find the cost of I bushel, and then of 63 27. If 4 men spend 17 dollars in -5 days, how many dollars will 7 men spend in 163 days? Ans. 905. Solution.-If 4 men spend 17 dollars in 5L- days, I man will spend 17 4= 4- dollars in 5'^ days; and hle will spend 4'- 5 --.* of a dollar per day. Then 7 men will spend 7 times 1, or '- dollars per day, and X 16-= 90o dollars in 163 days. 28& A man after spendinq,. and x of his money, found he had 26 dollars left; how much had he at first? Ans. 120 dollars. Solution.-Aldding ',. and ' together, we find, their sum to be Wt. -li i wh ^e spent The'n 1 - — = is what he had I eft. Theua26 -is o wha.. nAumir Art. G. DIViS.ON OF FRACTIONS. 123 '9. If a cierlt.all llumliber is incIrased bv -, '1, ' 3and 1 of itself, an(l 33 nmore, the result will be 4 times the ori'inal number; what is this numlnbe?. Ans. 180. Sohtu./iV,.-Ad dinj, together.l 4,, -,. and we find their sum to be' This added to 1, or,' - 4- 3!`; then 33 is ' of xwhat numnber? MISCELLAN-EOU)r lI,:lCISES OCISES R T'1 S~LATE: Intvolviwq t1 te pri'in,ilcss of, Cmmonu IrctioLs. 1. At 8- of a dollar a yard, Low Tmuch would -- of a yard Iof r:d coest? lAn,. '-A of a dollar. 2. If ' of a yard of cloth c~;t -|- of a dollai, Iow mucl is tha-. per yar.? Ans. a of a dollar. 3. If 6 potunds of coifec cost, of a dollar, I'!o much is that per pound? Ans. - of a dollar. 4. If.2- yards of cloth cost 5 of a dollar 1how much would 6' yards cot? Ans. 144 dollars. 5. How much would 12-, acres of land cost, at 84- dollars an acre? Ains. 105 dollars. 6. If -p of - a ship cost 4728 dollars, lhow much is j of her worul? A s. 59 10 dollars. 7. If -- of -0 of a store is worth 2448 dollars, whlat is the worth of thle store? Ails. 3400 dollars. 8. If 7- busliels of whoeat cost 8 dollars, how miuch will 16 - bushels cost? Ans. 194' dollars. 0. If a pole 91 feet lhigih cast a shadow 12' feet, Low hIigh must that pole be thalt casts a shadow of 147 feet? Ans. 114 feet. 10. If 9 men can do a piece of work iln lt- days, in wlhat time can 14 mten perform it?..Ans. 107 days. 11. If it requires 3'- yards of cloth to make a coat, when it is but 4 of a yard wide, how much will it require whlen the cloth is 1 yard wide? How much whlen the cloth is 1' yards wide? Ans. to tlie last. 2dl- yards. 12. If 9 hnorses consume 53 tons of hay in 7 weeks, how many tons will 16 horses consume in 12 weeks. Ans. 174. 13. If 6 students spend 5o- dollars in 121 days, how many dollars will 15 students spend in 231 days? Ans. 243. 14. If a family of 7 persons drink 18- gallons ot beer in 24 weeks, how many gallons will they drink in 124 weeks, if 7 persons more are added to the family? Ans. 187. 15. A man spent 5, and pi of his money, and had 69 dollars left, how much had he at first? Ans. 180 dollars. 16. A man has an o rd in wohich of the trees bear apples, w Ans. t o ~ ~tlcas.2ryd. m24 COOMMON ARITHMETIC. Sect.. V1II. bear peactSres, ariid 2 trees bear plums. Hew many trees are there in, the orchard? Ans. TOO. 17. In a certain school - of the pupils study arithmetic, J! study grammar, 8 geograplhy, l earl- to write, ald 15 learn to read.How many pupils in the scllool? Ans. 120. 18. If a certain lnmbelr be increased by,:^, b, and o ofitself, and 6 more, the sum will be double the number. Required-the numbe7? Ans. 1 0. i9. A man being asked the time, answered, "If you increase itby, of itself, it will be 1 2 o'clock. What time was it? Ans. 8 o'clock. 2;). I desire toknow the time by knowing that if it is increased by -, -, and -- of itself, it will be half past 12 o'clock. Ans. G o'clock. 21. "If to my age there added be, One-half, ore-third, and three times three, The whole will make six score and ten,.-* Pray t1ll my age now if you can." Ans. 66 ycars, 2. DIECIMIAI FRACTIONS. ARTICLE 7. DIEFINITIONS, &C. RE.ARK.-AS we have saiJ before, if a unit is divided into equal parts these parts are called FRACTIOcs. (Art. 1, Obs. 4.) Obs. 1. A DECIMI.L FIbACTON' is one in wvJd/iC tie delZomainato is 1 with any number of ciphers annexed; as '-, -O, 2-70-), &c. NOTE.- The word decitlwl is derived from tho Latin word decan, which signifies ten. REMARK.-Decimal Fractions are geerrally written without the denominator being expressed, in which case a point (. ) called a SEPARATRIX, or separuting point is placed before it to distinguish it fromn whole numbers. 'Thus-.5,.17,. 479, &c. are read 5 te:itis, 17 haaldredths, 473 thoustutlJths, &e. HenceObs. 2. The denominator of a decimal fraction is 1, with as many ciphers annexed as there are figures in the nmnerator, or decimal. Thus *- 1 1^'- I 132- 19-1.12347 12347.22=.4376= -- 378 |.476892T3 476892 1 0.4 376 -- 4T,': v 476892 — To o —(-oWhat is thedenominator of.3?.7?.19?.54?. 1008?.156?.98?.2007?.9? Suppose it were required tb write 7 without the dcominator. *5W jev'. Art. 7. DEC M \. FR ACTON.S, 12s In this case the inameratur does not contain as many figures as there are ciphers in the deneeminator, but this is remedied by writing a cipher before the 4, thl.,.04. 1n the same manner — 6.-006;.1 0oo o=.0003, &c. HenceTo write decimals, when the numerator does ctot contain as mnany significant figures as thlerv are ciphers in the denominator: Ohs. 3. P1reffix. ciphers tote stin ficant fiqrtres!of the n22mzrato', unil the number, f decirnma place-s is eqawl to the n (Iwer of ciplers in.-ei denomindaor. rThIs-0 is written..<....! 1 — i 4 — is twrittea n,. - 0(G, o <........ 0. 000,6.7r f3 0 (."' -.. 0 * *06 t - < 000006 Obs. 4. From this it appears that the iirst placke at the right of the separatrix is called TsNTI.S, btceatse a.UVit is Cvvi'ded into ten'equal p2rls; the second place is calk d utTNLamiEDTElnr /'iore id4-i(// ent'hs isnt iMn equal parts, or a v4r&it into <a /huzldr/d ecl palIts.; -i&c. This can be easily shown from tIhe, followsing DECIMAL A'TMIERA'TIO(3 TA1'J,,4T.C........ I ~-* C, S: _ 4 ~f*. - - $ s1. ~... 2 I I 7.. 7: 12.137::::.: 3 7 0 3 4 2:.2::: " 1 2 4 5 6.! O ' 19;:: 6 0 00 0 0 004. 4: '3 0 0 0 0 C, 2 7 8 ~ 12..' 4 7 3 2 1 3 '.~ ).0 0,) 0 0.0 0.0.0 'is read 6 tentfls. kS read 9 u nits. 4.3 llun'i ~ read 1 2-tits, 1S7 thonsaLdths. 5 is read 437 uits, -342 tea 8 thtlusaidtll) s. is 'nva I 12-45 units, 6004t llind(liwd tlh.)ue?;ndths. is.ad,1 (GCOCC0 -units, 4 milis -read 300 units, '674841 ' te ll ltli nlit s. is iea(l 1i f units, P9i 73a243 i) undred n2illionths. iis read 200,miV, 9 ilhu 11iF'.ls 126 COlMMON ARITHM3ETIC. Sect. Vii. Obs. 5. By examining this t!able attcntively we notice tlec following' Considera'tions 1st. Decimn ls derelase from the lft hand touwards the rilht in a ten fold ratio. Thus,. 4 is only 4 tenths;. 04 is 4 hundredths;.004 is 4 thousandths, &c. Thelre.)re converseley, D]ecimcls increase from the right hand towrd,, t/ e lef, inr a ten/flld ratio. Thus,. 4 is ten times larger than.04.04 is ten times largerl than.004; &c. IHenceOb3. G. ] 7'.'/ >'mova of a 0 te?'imli /7,re to t,'e rif/ht decreases, and ever?/ renra, / to the lejt increa-ses its lcvlue ten times. REMARK.-Thec p'rpil will perceive that decimafls ipcrease and decrease in the same mnanner as w\.i[tl num11'm rs. 21. The 'vale (f ever/y.fiqre, whether a decimal, or oan integer is de.ermined by its picefror unit. ThereforeOhs. 7. a. 'Prfixinq one c]ipher to cz deeimadl decreases its value ten times; tu'o ciher/s, a hundred tine.s, d'(' Thus.. 3; 03 -,;.003 - &e. B3ut b. Alnne.in, / a c7iphr t)o a deci'f!, howevere, does not alter ils value. Thiu,,.7 =;.70- -; -, ~,,, as betfirc.HIenceOhs. 8. J])e.mnals (f di qretmf eomina ors waf Ie reduced to a commo0 n d;-)om''nator, l'/ anne.rify ciplhers u.tl the n,;mber (of decimal places iZ e(fAh,.' eo/. Tr, s. 4,.G,.0 37, are. eqv. to. 400,.06,. 037, <.\ AI.'Obs. 9. I iole ~::i';. b,.',.;:'s.'W/ be -re4t '.'7 to de'im<.ls 1iy (V2?:c.;''./ i phe == P-,- T ',:-;.-, or 10 ) tentlhs, 160 1 hundredths. (Art. 3, Ob. 7.) RitMAIK.-WXXhe1,n the n-.', norni>er hi}::s reduced is writtei wAlhout the deTnotiinitor, it is best to pl,i:i theo stel.r:i.ltrix\ fore the cip}ers. Tius, 16 = 16.0= — 16. C 0 &c*. WTh:It are Fra-tion:'? Wh'iat is a lt.'eimnil Frtu'o? T ro-n wb.\tt is tile terin decimail derived? How atre1 dclnitii fracliis gen erullv vl ritten? W hat do wet use in ttlls case? W lI-r,- i Iiis pl:int p,;ced? yVli ' W hat 1-i tihe denolninator of a decimal frin'i,,n" H1ow do) we wri ite (' ieci,lal, wlihen the umltinerator does not conti;n:;,s n;i y iJgniflica.t figurets a tit i ree are ( ihers ini ile (ienoirinator? Whlat is t.re iirs plae.;I. lthee riglt ('f i}e spi tlorix callec? Wihv? The second l,lace? h\' X? H1(o'A 4! docim;ils d(ecrease? hlow illcrease? What inilerencl: is ded'1used from tlis.? 1, hiut is (tie d1tt irence betwleen tihe increase and decrease of (ecinals;aI. 'vi.o e nm)lers.? flow is the value of every fitaire determi:l I. Wi, tfiet f does it liave to prefix a cipher toil dcinTa I? Two ckithers'? What ofEfr- t ldoes it. havre to;tiinex a (cillr to dt-eimall? Ilow do we reduce d oimnals of dire e:it deno'nlinatllors to a cotninoa denoaninator? How may -,,hole niiniI crs je re:duced to drciniials? WVheie the whole 11nu1)er is written wi:houtl to denotnlillitor, how do we proceea? How ihy v whole uni}miers and dtecirr;lls bk writtenu 1retler? Whlat are suchl ex)Fressiolu called? Art. 7. IECIMAI, FRACTIONS. 127 3,1. W,'oie na.m' er.. andc.,l de r:;malas wa'y e w r;'.tn. ft,-:/(bc~er iy i iac;nq the.we;nertin point bet'eeen them. 'J Ius, 9 units, 9 miliontlts is written 9.000009. a. A whole number and decimal written toqether is called a MlIXED NUMBER. Thus, 4 7, 63.08, and 4.027 are mixed numbers. 4tA, Th'e rnits place is at the ri/ht of whole nmmbers, an d at the left of decim~als. tHencceTAe ee/fb:t of cInnexeni.q orprefixin.q ciphers to decimals, is the rever.se of anne.xin orpr fixi, g them to whole 2num'lers. 5th. The nam e of tie order of the riqght hadm! fiure of the decimal is given to the whole. HenceTo read decimals: Obs. 10. Read as il, 7J7wole tmers, anid to I/h riq/ht haond fi.tre ad( the name of its ord(er. NOTE.-W-heln mixeld numhers are read, it is prefe.,rahle to plute the word units after the whole nummbrr, to prevent ambiguityl. 'lmT'l, 4()9. 0i16 would hb read by many pipv ils 416 tenl Itlotianulthls. But by reading it 1O00 units, and 16 ten thousandths, ull anlbiguity is removed. RE.ARK 1. —Expressi ng decimals by words is c.;led NUMERATIO. OF D)crIMAL.t S. Read the following decimals: 1..5.!5 12.00021. I 9. 1200.00( 6.'13. 4.07. 2..27. G. 21 7. 12345. 10. 710 8.004. 1. 1. 167. 3. 1 I.1. 1;7. 1 00.00 ()00 1. 1 O.. 1 5. 11.472. 4. 9.3'.. 1.30''0 'I 0)0. 1000.' c 16. 7(,.:)75. REt:IMARK. 2.-ExprePs.sin(r eu'cimai. l, figures is icnSal,' 'N'I'.ATioN Oi Pr.ciMALS. To writ (: l-cirmals Obs. 1. I}'i/c alrch tiqtre in the order inz, "ii';/',t (be;l,!in, (ond place a ciphc~r in tll var.(,ant orders. Write the fr^ettl io,,l prt.s of the fioT!l in lr lie srs n dec imals: W\here is the umnits plac. in whole numlhers7 In deci!mr'nls What is the difference in tle eflect of annIxing and prefixing cipl(irs to whnole numrbers, and -to decirnals.? \V ht n iie is given to the (dcirnul? How th. 11 (lo we readti ecimals? Howv is it best to read mnixed numbers? Why so'? Wlat is numnratiou of dccina's? XVWat is notation of decimals? How do ( e wriltc decimals? 128 COMMON ARITHMETIC. Sect. VIII 2. 27 1 2. 1 69 -8 3. 147-.o Ans. 14. 09. 13. 463 13-3 4. 130 a. 14. 8. 5. 79610.oo 15. l 00 6. 141, 9 16. 400 l. 7. 16,027o0oo0 17. 78 9. 1008 oooooou 19. 60lo 10. 49 lo-6- 20. 1476T!0_. Write the following expr ssions in decimals: 21. 7 teathsl 31. 7 units, 12billionths. 22. 4 hundreths. 32. 14 units, 2 thousandths. 23. 14 thousandths. 33. 612 units, 25 hundredths. 24. 6 millionths. 34. 406 units, 406 thousanths. 25. 18 hundredths. 35. 400 units, 6 thousandths. 26. 36 thousandths. 36. 1000 units, 1 millionth. 27. 9 tenths. 37. 79 units,2001 ten thousandths 28. 112 ten thousandths. 38. 976 ten billionths. 29. 4002 ten millionths. 39. 1001 hundred trillionths. 30. 3 units, 4 thousandths. 40. 100 units, 111 millionths. NOTE.-The reading and writing of decimals is of great importance, and the pupil should be exercised at it until it is perfectly familiar to him. Art. 8. FEDERAL MONEV. Obs. 1. Federal 3foney is the national currency of t/Ie UNITED STATES, as established BY CONGRESS, 1Atust 8tl1, 1776. The denominations are MILL, CENT, DIME, DOLLAR and EAGI.E. TAB1LE. 10 mills (m.).-. _ make 1 cent --- —-- marked ct, 10 cents ---, __ " 1 dime. --- — " d. 10 dimes ---- 1 dollar -. "doll. or $ 10 dollars -— "- 1 eagle " E. NOTE.-This character ($) may be regarded as a contraction of U. S., and signifies United States money. 1st. The Eagle is a gold coin, and contains 10 pennyweights, 18 What is Federal Money? What are the denominations? Repeat the table? What are the national coin of the United States! Alt.. -FEDERAL MONEY-. 1,29 grains 258 grains of standard gold.* Besides 'the eagle, we have the half-eagle and quarter-eagke, which are gold coins, and their value and weight accordingly. 2d. The Dollar is a silver coin, and contains:17 pennyweights, 4: grains = 4112y grains of standard silver.t Besides the dollar there are half-dollars, quarter-dollars, dimes, and half-dimes, which are silver coins, and their value and weight accordingly. 3(1. The Cent is a copper coin, and contains 7 pennyweights = 1-68 grains of p'ire copper. Thle lhalf-ceni accordingly. lfiils are only imaginary, and are not coined; 'Obs. 3. Pure gold is supposed to be divided into-24 -equal'parts called carats, and its fineness depends on the number of parts of some baser metal called alloy that it contains. Thus, if it contains 20 parts pure gold, and 4 parts alloy, itis said to be 20 carats fine:; if 'it contains 6 parts of all6y, it is said to be;1-8 carats fine. Obs. 4. Previous to 1837, the standard for gbld-was 22 parts of pure metal, to 2 parts of alloy, and it was said to be '22 carats fine. Now it is 21.,3 carats fine. Obs. 5. By Act of Oongress, 1837, the legal standard for gold and silver coins in the'UNITED STATES, is 900 parts of pure metal, by weight, to 100 parts alloy. The alloy of gold coin is composed of is ilv r and copper, the silver not to exceed copper in weight. 'rhe alloy of silver coins is pure copper. Obs. 6. Acoounts in the:UNITED STATES ar-, usually 'kept 'in dollars, cents, and mills; eagles being expresssed as dollars, and dimes as cents. Thus, instead of saying 2 eagles, 5 dollars, we say 25 dollars; and insteaidof-saying 7 dimes, 5 cents, 'we say 75~ nts, &c. Five mills are often called -z.a cent; thus 12 cents, 5;anills, are generally'called 1 2} eents. REMARK.-It will be perceived from the table that the denominationsef Fed'eral Money increase and decrease in a ten fold ratio,: in the same manner as What is the eagle? Its woight? What other gold coins have we? Wht is thedollar? Its weight? What other silver coins have we? What isthe-cent? Its weight? What are mills? How is pure geld divided? Upoa whatdoes its fineness depend? If it contains 4 parts alloy, how fine is it.? What.Wts the standard for gold previous to 1837? What is it now? What is the legal stan"dard for gold and silver coins in the United States, by Act of Congressobf '1837? Of what is the alloy of gold coins composd? Of silver coins?. How are.eccounts in the United States genri.lily kept? How are eagles exptressed — Dimes? What do we say instead of 2 eagles, 5 dollars? 7 dimes, aeJd 5 cents? What are 5 mills often called? Ifow is-12 cents, 5 mills,generally read? -How do the denominations of Federal Money increase and decrease? ' Ea'!e~ coine:l before July 31st, ]834, o-ltain' 11 pennyweigihts, t -grains = 70 grains of standard gold. These are worth $10... Hafand quarter ea.ksa':coldintrl'. t The dollar originally contained 17 pennyweights, 8 grains =416 grains of sindllard thior. t Tle cent originally contained 264 grains of pure coppei-. 7A COMMON ARITHliETIC'. Sect. VIII. whole atrmbers. Hence-the dollar being regarded as the wnit, and the cents and mrileas tlh fractional parts of the u-nt or dolar, it follows that Obs. 7. All operations in Federal Joney can be performed precisely as in decimal fractions, the dollar being regarded as the unit, cets. as tenths and hundredths1 (because 100 cents make a dollar) and the mills as thousandths; the separating point being placed between the dollars and cents. Obs. 8. As cents are so many hfundredths of a dollar, (it taking 100 to- make a dollar,) it follows that they must occupy the first two places at'the right of dollars. and if ihe cents are less than 10, a cipher must be placed at the left, or in the tenth place. Thus, 18 dollars, 5 enits is written $18.05. likewise, as mills are so many thousandths of a dollar, (it taking 1000 to nmake a dollar, ) they must occupy the third place at the right of dollars; and if.o cents are given, theplace of cents must be supplied with ciphers. Thus, 9 cts, is written. 09; 18 dolls. 6 ets. 7 m., is writen $18, 067; 15dolls. 37 c[s, 5 m., is written 8$5.375, or $1537j f 6 dolls.. 9 m., is written $6.009 and 7 m., is written $0.007; &cc;. R.N.ARK.-Busiress rremn often write cents-as the fractionat parto of aEdo4lars Tlus, they write $10. 46, $10()6. &c. | To read any sum in Federal Money:. Obs. 9. Call the figures at the left of the separating pont dolars the first two places at the right cents, and the thirdplace' at the righ m7lls; the other places at the' right' are decimals of a mill. ihus, $56. 275 6 is read 56 dolls, 27 ets, 5 mi.,-ad g hundredtlis of amill. loTE.. —Tre decimals at the right of rmi.1Ts-rc seldom eouaned;, the nilla being sufficiently exact for all business calculations. Some however, reckon il his way-if the mills.exceed 5, they count another cevt; if they are less than. 5 they rejeet them entirely; S m. they count &, a Cent.Eow can operations in Federal Money be performed? What is the dollar regarded?' The cents? Why? The mills? Where is the separating point paced?' By what are the first two placesat the right of the separating point ocerpiedT Ifthe cents are less than 10 how do we proceed? What place do mills occupy? If no. cents are given what must e done? How do business men often; write cents? How do we read any sum in Federal Money? Why are not thledecimal* at the right of mills generatly reckoned? How do sonic reckon 'mitlr Art. O. FEDERAL MONEI. 131 Read the following sums in Federal Money: 1. $0.06. 11. $70.005. 2. $1.703. 12. $32671.09999. 3. $17 6954. 13. $14.007. 4. $243.009. 14. $0.001. 5. 78. 805. 15. $0.158. 6. $204. 73021. 16. $12.70. 7. 14. 12-. 17. $9.371. 8. $100.50. 18. $14.181. 9. $1086.375. 19. $0.56$. 10. $12.1875. 20. $100.001. Obs. 10. We write sums in Federal Money accordinglo (Art. 7, Obs. 11.) dollars occupying the place of whole numbers, cents the place of tenths and hundredths, and mills the place of thousandths. Thus, $18.37 cts., 5 m., is written $18.375, &c. Write the following sums in 1. 27 dolls. 49 cts. 7 m. 2. 18 dolls. 18 cts. 8 m. 3. 20 dolls. 27 cts. 6 m. 4. 97cts. 4 m.; 68 cts. 5. 78 cts. 9 m.; 47 cts. I m, 6. 1007 (lolls. 87 cts. 6 m. 7. 947 dolls. 69 cts. 3 m. 8. 7 dolls. 5 cts. 9 m. 9. 18 dolls. 6 cts. 3 m. 10. 19 dolls. 9 cts. 1 m. 11. 34 dolls. 7 cts. 8 in. 12. 19 dolls. 2 cts. 3m. 13. 10 dolls. 9 m.; 4 cts. 14. 74 dolls. 3 cts, 7 m. 15. 143 dolls. 9 m. 16. 9 m. and 12 hundredths a mill. Federal 17. 18. 19. 20. 21.:2. 23. 24. 25. 26. 27. 28. 29. 30. of Money: 1 doll. 1 m.; 3 m. 3 cts. 2 m.; 7 m. 129 dolls. 8 cts. 1 m. 6 m.; 7 m.; 4 m. 8 m.; 1 ct. 1 m. 200 dolls. 14 cts. 2 m. 93- cts.; 4 m. 75 cts. 1 m. 1 doll. I ct. 1 m. 20 dolls, 2 m, 79 dolls. 4 m. 1 m.; 9 m. 5 tenths of a mill, 25 hundredths of a mill. 100 dolls. 1 ct. 75 hundredths of a mill. ARTICLE. 9. REnlDUTION OF FEDERAL MONEY AND DICIMAL FRACTIONS. Obs. 1. As 10 mills make a cent, and 100 cents make a dollar, it follows that a. Dollars are reduced to cents by annexing two ciphers, and to mills by annexing three ciphers. In either case we remove the sign of dollars ($). How do we write sums in Federal Money? How are dollars reduce to ceats? 132 COMMON AWRITHMETIC. Sect. VTlll Dollars and cents are reduced to cents, and dollars, cents, and mills are reduced to mills by erasing the separatinyg aoi'nt, and the sign of dollars. b. Cents are reduced to dollars by pointing off two figzres at the rigkt and prefixing the sign of dollars. Cents arc reduced to mills by annexing a cipher. c. i. lls are reduced to cents by.pointing off three figures ad the right and prefixing the sign of dollars. E-XERCISES FOR TIlE SLATE. 1. Reduce $17 to cents. 2. Reduce $34 to mills. 3. Reduce $18.73 to cents. 4. Reduce $8t.645 to mills. 5. Reduce $1.01 to mills. 6. Reduce $4.20 to cents. 7. Reduce $27.06 to mills. 8. Reduce $0.479 to mills. 9. Reduce $12.06 to cents. 10. Reduce 478 cents to dollars. 11, Reduce 164 cents to mills. 12. Reduce 2080 cents to dollars. 13. Reduce 14000 cents to mills. 14. Reduce 120 mills to c:nts. 15. Reduce 14000 mills to dollars. 16. Reduce 1785 mills to cents. 17. Reduce 800 cents to dollars. 18. Reduce 1768 mills to dollars, 19. Reduce 12435 mills to cents and dollars, Ans. 1700 cts, REDUCTION OF DECIMAL FRACTION.S Obs. 2. The learner will remember that inull cases of Reduction, whether of Common or Decimal Fractions, the value of the given number must not be altered. (Art. 3, Obs. 1. and Art. 6, Case 5, Rem.) CASE 1. To change a Decimal to a Common Fraction. REMARK.-As we have said before, the denominator of a decimal fraction is 1 with as many ciphers annexed as there are places of decima!l in the numerator. (Art. 7, Obs. 2.) Hence-To change a decimal to a common fiaction: To mills? What is necessary in these cases? How are dollars and cents reduced to cents? How aie dollars, cents, and mills reduced to mills? How are cents reduced to dollars? To mills? How are mills reduced to cents? To dollars? What must be observed in all cases of reduction? What is the denominator of a decimal fraction? How then do we change a decimal to a common fraetion? Art. 9. REDUCTION OF DECIMAL FRACTIONS. Obs. 3. Erase the decidal point, and supply the denominator of the decimal: NOTE. —It is generally-cestomary, after the decimal is reduced to a 'bmrnon fraction, to reduceit to its lowest terms. (Art. 3) Obs. 3.) This is the case with the following examples. Ex, 1, Change.25 to a common traction, Ans. -oo =. 2.. Change. 125 to a common fraction. Ans.. 3. Change.0625 to a common fraction, Ans. -1. 4. Change.7425 to a common fraction. Ans. a o 7 5. Change. 75 to a common friction, Ans. 3. 6. Change. 875 to a common fraction, Ans. a. '. Change ~5625 to a common fraction. Ans. 2-, 8, Change.9375 to a common firactioii Ans. f s 9. Change.333 to a common fraction. Ans. 3 -3 10. Change.99 to a common fraction. Ans..11. Change. 1246 to a common fraction. Ans.,52. 12. Change.03125 to a common fraction. Ans. -1-. CASE 2. To change a Common Fraction to a Decimal, NOTE. —This case is exactly the reverse of the.preceding one, and eachl roves the other. Ex. 1. Change 2 to a decimal, 'OperatioL. Annexing a cipher to our numerator, we nake it ~0 6)2. 0 tenths: 20 tenths divided by 5 (the denominator) -- equals 4 tenths. Ans.. 4 The correctness of this operation may be shown as follows: Annexing a cipher to the numerator, is, in reality multiplying it by 10; (Sect. IV, Art. 4, Case 1, Rem.) bit multiplying the numerator is multiplying the value of the fraction; (Art. 2, Obs. 1.) hence, the quotient (after dividing the numerator by the denominaw tor,) is 10 times too large, and we *nust therefore divide it by 10 to obtain a correct result. This is done by simply pointing off a figure at the right, or placing the decimal point before it. (Sect. V, Art. 4, Obs. 1.) By the same course of reasoning, it follows that it we annex two WhatrelationdoesCase 1, andCase 2 bear toeach oter? Explain why the operation of Ex. 1 is correct. If we annex two ciphers to the numerator, What effect does it have on the quotient? If we antnex three ciphers, what is the effect? Why 'o? 184 COMMON~ ARITHhEVCC Sect. VIII. ciphers to the numerator we multiply it by 100, and must point off two decimals in the result; if we annex three ciphers to the numerator, we multiply it by 1000, and must point off three decimals in the result, &c. Hence — To chancge a common fraction to a decimal: Obs. 4. Annex ciphers to the numerator, and divide by the denominator, pointing of' a decimal in the quotient for every cipher annexed; and if there is not a szcficient number of figures in the quotient, supply the deficiency byprefixing ciphers. a. From this and the last case, the learner can often form rules for contracting the operation- in multiplication and division. Thus, to multiply by 25, lie can divide by 4, because.25 =; to multiply by. 125, divide by 8, because. 125 =-; &c.; and conversely to divide by. 25, multiply by 4, and to divide by. 125, multiply by 8, &c., for the same reason, Again, to multiply by. 875, we multiply by 7, and divide by 8,be cause.875 =-, &c.; and conversely, to divide by.875 we multiply by 8, and divide by 7 for the same reason. But when our multiplier is a whole number (and we contract the operation,) we must annex to the product 1, 2, 3, &c, ciphers, because our multiplier is a part of 10, 100, 1000, &c,; and conversely when our divisor is integral, we must point off from the right 1, 2, 3, &c., figures for the same reason. The figures cut off however, must be divided by our multiplier, because these are the remainder, and are as many times too large as there are units in this multiplier. (Sect. VI, Art 2, Rem, under Ex, 22 and 23,) We can also abridge the operation when the multiplier or divisor is partly integral, and partly decimal; thus to multiply or divide by 1. 25, we proceed exactly as if it were 125. In such cases we only annex as many ciphers, or point off as many decimals as there are integral numbers in the multiplier or divisor. When there are decimals in the dividend or multiplicand, we may proceed with them according to the rules in multiplication and division of decimals. It may be proper to remark that the subject of abbreviations can only be useful to a person well acquainted with the properties and How then do we change a common fraction to decimal? To what pdrticular object can these two cases be applied? give an example or two illustrating this point? When our multiplier is a whole number, how do we proceed? Why? When the divisor is integral, how do we proceed? Why? u hat must be done with the figures cut off? Why? When the multiplier or divisor is a mixed number, how do we proceed? What must be observed in such cases? When there are decimals in the divltdeii or multiplicand, how do we proceed? What is aecessary in order to make tho subjectof abberviatioQu useful to W pesou? Art. 10. AD- IToN ANb SULitRAtt;CN OF bECIMALS, &C. 36 relations of numbers,* and with such a few hints are generally sufficient. We will therefore -only present the two following propositions, which are so evident that we omit the demonstration: 1. If we wish to divide one numbel by another, it will produce the same effect to multiply the dividend by the reciprocal of the divisor.A-id 2. If we wish to multiply one nuzcer by another, we can obtain the same result by dividing the multiplicand by the recijprocal of the multiplier. 2. Change 4 to decimal, Ans..75 3. Change - to a decimal, Ans..6. 4. Change - to a decimal, Ans. 05b. t. Change T-jY to a decimal. Ans,.048, 6. Change ~ to a decimal, Ans..375. 7, Change - to a decimal. Ans. 1. 25. 8. Change - to a decimal. Ans,. 36. 9. Change 6-5 to a decimal, Ans.. 008. 10. Change -r7-o toa decimal. Ans,. 007. 11. Change ~ to a decimal, Ans.,333333 +. 12. Change -, to a decimal. Ans. 468468 —. In the last two examples, it will be perceived that the decimal will be continually repeated, the same remainder occuring after a certain number of divisions, Obs, 5., Decimals which consist of the same figure or set of figures continually repeated, are called RtEPEATING DECIMALS, CIRCULATING DECIMALS, or REPETENDS, ARTICLE 10. ADDITION AND SUBTRACTION OF DECIMALS AND FEDERAL MONEY. Obs. 1. Decimal Fractions can be added, subtracted, multiplied, or divided, the same as whole numbers. The only difficulty is in knowing where to place the separating point.. I Obs. 2. As Federal Money is based upon the decimal notation, t is evident Ihat it is suyiect to the same laws as decimal fractions, and therefore the same rules are applicable to both. Ex. 1. A man has several accounts to collect; the first is $12; the If we wish to divide one number by another, by what other way can we obtain the same result? If we wish to multiply one number by another, by What other way cal we obtain the same result? Demonstrate these propositions. What are circulating decimals or repetends What is the only difficulty in the audition, &c., of decimals? Upon what is Federal Money based? What is the inference deduced from this? * The preceding ruls and prin.iples cxp'in all that is necesry to undertand this poiant 'COMifN AITfHRIETIU. Sect. VtI4, second, $9,375; the third, $6,08; and the fourth, $5,403. Required-the amount of the whole, In this example we place dollars tnder dollars, cents u;.der cents, and mills under mills, and add in every respect as il Addition of Simple Numbers. (Sect. II. General Rule.) Operation. $12.000 9.375 6-.080 ~. 403 -Xns. $32.858 2. A man has in one bag 2.3 bushels of oats, in another 3.161 'bushels, in another 1.34 bushels, and in a box 14.132 bushels. }low much has lhe n all? in this example, we place wholo mnmb)ers under whole numbers, tenths under tenths, hundredths under hundreddhs, &c., and add as us';tal. Operatioh. 2.300 3.161 1.340 14.132 Ans. 20. 933 bushels, NOTE-In both these examples, it will be perceived that we reduced all the decimals to the same defibminations by annexing ciphers. This however, is n6t,bsolutely necessary if thet pupil places a.l the figures of the same order directly under each other. That these operations are correct, may be shown as follows: 3. Add together 5. 846 and 7.9, Operation. 5.846 Or, 5.846 = 5To 7 9 79 7 9 9 00O.900 |+ 84 6 /kJ ^/.y ===,/ yy. TO-== T*7 - WV 6 o ooo T o-77 ___74 = 1=tW Tos 10,' - = Z ==0 0 0 1o 1. 7 + 5 12, an(n Ans. 13. 76 l4 340. But 74.746. 1000 -- 1QlO O74 Hence ---13-40 =0 13.746, as before. The same ireasoning will apply to Federal Money, or to any nAmiber of decimals. 4. A man having $18.25i, spent $9.625. How much had he left? fOperation. In this exampte, We write dollars under $18.250 dollars, cents under cents, and mills under 9.625 mills, and subtract precisely as in Subtraction of Simple Numbers. (Sect. III. General Rule.) Ans. 8.625 Explain why the operations given are correct, Art. 10. REDUCTION AND SUBTRACTION OF I ECIMALS, &C. 137 NOTE.-AS there are no mills in the minuend, we write a cipher inthe mi ls' place; but this is not necsssary, as we always proceed as if the cipher was there, whether we express it or not. 5. A man having 18.4 yards of cloth, sold 12.694 yards. How much had he left? Operation. In this example, we write whole numbers 18.4000 under whole numbers, tenths under tenths, 12.694 &c., and subtract as usual. Ans. 5.706 y'ds. It will be perceived that we annex ciphers to the minuend until the number of decimal places are equal to those in the subtrahend. The same example may also be solved thus: 18.4 18,. 18,- =- 18 400 171400 171-0 12o 694 1~.~~~ -— ~~~~1 I - or- l y -i ) o 7 ' T -- ) o) o I 12.694 = 12 ~-n. 5 -6 = 5. 706. This shows the operation to be correct. The same method of reasoning will apply to either Decimals or Federal Money. Obs. 3. From the preceding operations, we notice two considerations:; 1st. The separating 2points, both in the gitren nuzaners and in the result, all fall directly under each other. 2nd. In Subtraction, that number is made the minuend, in which the wole number is the largest, without reference to the Decimals, because the Decimal is only a part of unity, und therefore merely a fraction. HenceTo add or subtract Decimals or Federal Money, we have this RULE.-I. Write the numbers so that the same orders, and also the separating points, shall all fall directly under each other. (Sect. II. Art. 2. Obs. 4.) II. Then add or subtract as in Simple Vumbers. (Sects. II. and III., General Rules ) III, Place the separating point in the result directly under the other points. NOTE.-In subtraction, of course we place the greater number at the top. PROOF.-The same as in Simple Numbers. (Sect. II. Art. 2. Obs. 6. and Sect. III. Art. 2. Obs. 7.) Where do the separating points fall in these examples? Which number do we make the minuend in subtracting? Why? What is the rule for adding or subtracting Decimals or Federal Money? What is the proof? 138 COMMON ARIThMETIC. Sect. VIII.;:'E EXERCISES FOR THE SLATE. 1. A. has $18, B. has$14.06,C has $210.3275, D has $6.005, and E. has $12.3125. How much have they all? Ans. 870.705. 2. How much has A. more than B. in the last example? Ans. 63.94. 3. In the same example, how much has A. B. and C. more.han D and E.. Ans. 34.07. 4. A farmer lihs grain as follows: of wheat 123.0993 bushels: of oats, 109.6 bushel.; of corn. 83.97 bushels; of rye, 75.004.bushels; and of -ibar.ley, 58.3267 bushels-. How much. grain lhas le Ans. 450 bushels. 5. How much more wlheat has he than corn? Ans. 39.1293 bushels. 6 IHowm much morle wheat and oats h hs he than all otler kinds,rain? Ans. 15.398 bushels. 7. A gentleman bought a coat for 1 5.75; a halt for 85.375; a pair of boots for $5.06; and an overcoat for 23.875. M4ow much did he pay for the wholle? Ans. "50.06. 8. Hlow nmch did the coat t ostoe than the hat? Ans. 810. 75. 9. THow much- dii the eaty hlat, and boots cost, more than theovercoat.? Ans. $2. 31. 10. A merchalnt sor s;v ral pieces of cloth, as follows: the tirst piece contained l1t;. yards; tlhe second, 19.006 yards tl' e tirdI. 22.14 yardls; and tl e fol.th, 22.05( yatrds. How U much F s t:here in all? XAns. 80 y, irdls. 11. HXow much didt the thiyd piece cntrai more thli thl fu lr.h?.ns. v. ar.l<. 12. IHowv much did thw second and third pieces tcnt:*;i, I.o\, than the first and fourth? Ans. 2.2'.)2. i3. A rmerc;iant ha; u, tu o hi t he, folli>win, su i;: 1 d7 (ol]ars, 7 cents; 9( dolWlars, 6f cints; 47 doll irs, 5 mill: 57 l3 l:rs, 8 cents; t1) dollars, Is,- cL nts; ant 25 do(l. How m.tch }as he owing him in all? Ans. 71 41. 14. From $10 tak.e I a c-, Ans. 9.995. 15., From 81 take 2 tlhousa dths. An-. 80. 099. 16. Add to,'e her 10 units, 3 huinrllihs'; 5 ten tlocusliandtils 1 2341r56 mirlliontihs; 7862 units,.97 844 m1i illolhs, 1and 12 nils, 8972 ten thousandths. Ans. 785. 419. 17. From 18 units take 4 billionths. Ans. 1799999999. 18. From 18 dollars, 75 cents, take SC} cents. Ans. t$7.86-. 19. From 5 dollars take 3 mills. Ans. $4. 997. 29. Adi. tog;ethler 5 eag.les, 5 dimes; G eagles, 3 dollars; 12 Art. 11. MULTIPTICATION OF DECINIALS, OR FEDERAL MONEY. 139 tents; 2 (dollars, 4 mills; 213 mills 57;689 mills, and 15 eagles. Ans. $ 32 3.531. 21. From 9 eales take 9 dimes. Ans. $89. 1C. ARTICLE 11. MULTIPLICATION OF I)ECIMALS, OR FEDERAL MONEY. Ex. 1. Multiply 9.25 by 6. Operf )'on. 9.2)5 In this example, we multiply as usual, and (f poin of' two decimals in the product, because - there are two decimals in the multipli(cand. The Ans. 55. 53 cotrectness of this ex-mple may be shlown by the followinr llust' t 'on-. 95 - 9-U — -2 - p To5 10U 100s s s n 1 = 55 To _ -- 52.50. 2. A (; enltleman Iought 12 yards of clolh, at $2. 375 a yard. How lmuch was the cost of the w\hol(e?' O('eraf/eio. 2.375 In this example, we multiply and point off cx1 2 actly as in the first example. The correctness of ---- this operationl may be shown in the same( manner Ans.;2,. 5200 Ias in the l;st. 3. Multiply 7.42 by 8.35. O_:eration. 7.42 In tlis Ce raple, we point off four d ci8.35 rnals in the prod- ct, beccause theli are two -- decimals in tle multiplicand, and two in the 37 10 multiplier, n:ilinil 4 in both. 222 2 We pove,; thlis to be correlt, thus: 7.42 = 'st)'36 7-,- *' = 1,, -,3. <> = — 02 3 '5 S 3.; '' 7 ^ X 5936 7: '. o 5 -_ ": 3'0 ' -';- 7 f': _ '*= * Ft 7 - 'r ~,Jo3- i8; ) ' - i i - I o,, - - - ~i-"-0 U, 1. 0 1 Ans. l. 9570 before. 4. Multiply.04 by.07. 2Opera/iou..O04 In tllis exa, le, tlhete are but two fi'ures in the.)7 lprol.ct, thercller wet pretix ciphers to obtain tile - requisite number of (lecimals. The correctne s of Ans..0028 this operation may be shown tlhus:.04 __ Y-;.07 =,,,; (Art. 1. Ohs. 2.),- X o = o-.0028 as before. (Arft. 1. Obs. 3.) This illustration shows why the ( iphers should be prefixed, rather than annexed, to the product. 140 COIMMON ARITIMETIC. Sect. VIII. From these examples and illustrations, we derive the following RULE FOR THE MULTIPLICATION OF DECIMALS, OR FEDERAL MONEY I. WFrie the numbers, and multiply as in Simple Nenembers. (Seet. IV., General Rule.) II. Point (ff as many decimals in the product, as there are decimals in both the multiplicand and multiplier. III. If there are znot as many decimals in the product as in both factors, prefix ciphers to it until the requisite nzmber of decimals are obtained. PROOF.-The same as in Siqple Numbers. (Sect. IV. Art. 2. Obs. 7.) 1. When the multiplier is a composite number, we may often contract the operation, the same as in Simple Numbers. 2. To multiply by 10, 100, 1000, &c., we need only remove the separatrix 1, 2, 3, cfc., places to the right, annexing ciphers if necessary. Thus: 1.6 X 1000 = 1600.0, or simply 1600, EXERCISES FOR THE SLATE. 1. How much would 14.5 yards of cloth cost, at $1.375 per yard? Ans. $19.9375. 2. A farmer has 12 bags, each containing 2.75 bushels of wheat. How much has he in all? Ans. 33 bushels 3. If he sells it at $1. 173 per bushel, how much will it come to? Ans. $38.709. 4. How much w'ould 112 pounds of coffee cost, at 0. 117 per pound? Ans. $13.104. 5. How much would 216 pounds of sugar cost, at $0.071 per pound? Ans. $15. 336. 6. How much grain in 16 boxes, each containing 77.93 bushels? Ans. 1246. 88 bushels. 7. How many yards of cloth in 27 pieces, each containing 32.78 yards? Ans. 885.06. 8. How much would 1000 pounds of sagar cost, at $0.07 per pound? Ans. $70. 9. Multiply.0763 by 2.16. Ans.. 164808. 10. Multiply 2.97 by.(042. Ans..012474. 11. Multiply 12.62 by 81-. Ans. 1031.68'. 12. Multiply.276 by.C03437. Ans..000120612. What is the rule for Multiplication of Decimals or Federal Money? Explain, by Ex. 1, 2, 3, and 4, wliy this rule is correct. 1Iow do we prove the operation? When the multiplier is a composite number, how may we proceed? How may we multiply by 10, 100, 1000, &c.? Art. 1!. MULTIPLICATION OF DECIMALS OR FEDERAL MONEY. 141 13. Iow much would 97 pounds of pork cost, at $3.50 per hundred? Ans. $3.395. Operation. 3.50 The result obtained is evidently the cost of 9700 97 pounds, (because the price is so much per hundred,) - whereas we only wish the cost of 97 pounds. Now 2450 97 is - ' of 9700, therefore the cost ot 97 pounds is 3159 but 7' the cost of 9700 pounds. Then $339.50 - 100 = $3.3950. It will be perceived that the ]at$339.50 ter result is obtained by removing the separating point two additional places to the left. The correctness of this last point may be shown thus: 339 50 =339 5 = 3 -0 339 0 * 100. 3 (Art. 6. Obs. 2. a.) 3 3 = 5 = _ 3 l3 9oo = 3.3950, as before. 14. How much would 794 feet of boards cost, at $5.875 per thousand? Ans. 4. 664-. Operation. $5.875 The result obtained is evidently the cost of 794 794000 feet, (as the price is so much per thou---- sand,) whereas we only wish the cost of 794 23 500 feet. Now 794 feet is 'Tv of 794000, and 528 75 therefore the cost of 794 feet is T of the cost 4112 5 of 794000. Then $4664.750 - 1000.4. 4.664750. It will be perceived that the latter $4664.750 result is obtained by removing the separating point three additional places to the left. fllustration.-4664.750 - 4664 7,5o- 464o75 0; 4 00470 1 n'it'l.'~ 4664'7.50 /. 4,,. 6 fL 4 7 0 4 6 6 6 750 1000= 1 O O o O o O~ ( Art. 6. Obs. 2. a.) 1O 0 '" = 4T6o 0-o 75 4.664750. as before. REMARK -The sign of Addition placed at the right of an answer, signifies that the answer is not complete, or, that there is a remnainder' Itence-When buying or selling articles by the 100: Obs. 1. Point off tw) additional places at the right of the product. When buying or selling articles by the 1000: Point of three additional places at the riglt of the product. REMARK 1.-C. stands for 100, and M. stands for 1000; from two Latin words, Centum and Mille, which signify hundred and thousand. 2.-The learner will remember that in all cases of multiplication, whether of Simple, Compound, or Decimal numbers, the.multiplier is to be considered an abstract number. (Sect. IV. Art. 2. Obs. 6. Rem. 1.) How is the final result obtained in Exs. 13 and 14? Show why this is correct? What does the sign of addition, placed after an answer, signify? How do we proceed in buying and selling articles by the 100? By the 1000? For what does C. stand? M.? From what? What must be observed in all cass of MuP'iplicatidn, whether of Simple, Compounld, or Decimal Numbiers? 142 COMMIGN AR [TIMETIC'. Sect. VII1. 15. HIow much would 18275 brick'cost, at $4.7 - per M. Ans. 79.953- -. 16. HIow mulr]l woullld 276 bun'chels to' sliln,'l's cost, each bunch containirn 1242, at 43. 12', p(r 1 M.? 'Ans. 61071.2?5. 17. fl;Yow muich would 417233 ipounds f1 bftt co-:;t, at $3. 7 per C.? 8 Ans. 177. 1125. 18. 1ow much would 1623 feet of lumber ( ost, at $4.621 per C.? Ans. $7<50.683+. 19. How muclh would 5273 pounds of pork cost, at 8$4.25j per C.? Ans. $224.1025. 20. Ilow much would 2314 feet of mahogany cost, at $;64.25 per M.? Ans. $148.6745. ARTICLE 12. DivISION OL DECI:MALS, OR FEI1IT1AL MOEi'. Ex. 1. Divide.84 by.7. Opieration. Obs. 1. As we have said before, the dividend.7). 84 in Division answers to the product in NMutiplication; (Sect. V. Art. 2. Obs. 7.) and as we point Ans. 1.2 oiff as man decimals in the prod:uct as there aure decinmals in both lactors, it follows that the qc/oficnt in )ivit!iion of IDe'iniu(/s 1must t(dl-?wC/ys contain (ts 'ma0i/(ir decillUl:l(/ces, as t/l decii l.cs in tite dividend e.rceed those in the divisor, the divisor and (quotlient beino the factors which, multiplied together, produce the dividend. 'Ihus, in the last example, the dividend has two decimal places, and the divisor but one; therefore we point off one decimal place in the quotient. Thlis point may also be illustrated as follows: O.8 84. 7 7., Xo 1= 1. v4 _ - =, O =O; -- 1 = -= I 0 1.2. as before. (Art. 6. Obs. 7.) quotient. 1.2 The example may be proved thus: divisor..7 divid'nd.84 2. Divide.1476 by 1.8. Ans..082. Operation. We perform this operation by Long Division. 1.8).1476).082 As the quotient does not contain as many signifi- 144 cant figures, as the decimals in the dividend ex- -- ceed those in the divisor, we supply the deficiency 36 by prefixing ciphers. 36 Analytic llustration.-. 1476 =- ~-4-0o; 1.8 - 1 18 147. 18 1 7 R 10 82 2f TTIT = r>, TTg - IT I== ] U,, X Ts8 -= T0W - U.082. (Art. 7. Obs. 3.) This illustration shows why we prefix, rather than annex ciphers to the quotient. How mnanyv decimal plar.;s should the quotient, ji l[yision of Decimarl contain? Why dd? Art. 12. DIVISION OF DECLMALS OR FEDERAL MOENY. 143 3. Divide 20970 by.12. Operation. In tiis example, we alnnex two cipliers to the.12)2970.00 d vi.le (l to niAke the number of dlecimals equilto tl( Ise in tle divisorn; l and l I accou:nt of this An\. 24750 e quality, o ir quotient is a whole nunber. lllrsfation. —. 12 = '2 2970. 2 237, x -1 X = 207,0 24750, as before. 4. A man bought,0 pair of boots for $42.75. How much was that apiece? Operation. We proceed in Federal Money, in all cases, 9)$42.75 exactly as in Decimal Fractions. (Art. 10. Obs. 2.) Hence- $4.75 To perform operations in Division of Decimals, cr Federal Money, we lave this RI:LT.-I. I Write the ltmibers, uand di ide as in Sieiple unmbhers. ( kS(,1t. V. General Rule.) II. 'Point,fi rs imty (dccit(i1 p/'ces;i; tfe qototient, as the!mtnber f' de(Ic'iws il the divideid ieceed those ti the divtisor. (O1)s. 1.) HlT. I t/here is not (.' setif.cieit itc r *s (i.s'if/,Oiiitifigttrcs in the iqatietd io;o, iit j fj;e/' decidlcs, suppliiy the d(iciency by lprjixing ciphers. (Ex. 2.) Proof.-The same as in SiSple 'Vmr71bers. REMARK K 1.-When the number of the decimals in the dividend is equal to those. in the divisor, the result i. a whole naumber. (Ex. 3.) 2.- hen th)l number of decimal places in the dividend is not equal to thnose in the divisor, annex cipher., to the right of the dividend until they are equal. (Ex. 3 ) The result in this case is also a whole number. NoTE.-Tn both these cases, the divisor is supposed to be less than the dividend, and also that there is no remainder. 3.-If there is a remainder after the division has been performed, ciphers may be annexed to the remainder, and the division continued; but after the number of decimal places in the dividend are equal to those in the divisor, each cipher annexed produces an additional decimal in the quotient. This is evident from Art. 1, Obs. 7 and 9. For 1.4 = 1.40 =1.400= 1. 4000, &c., dnd thus we mav increase the decimals in the dividend without allering the decimals in the divisor, and in this way increase the decimals in the quotient. What is the rule for Division of Decimals, or Federal Money?What is the moehofl (f Proof? When the number of decimals in the dividend are equal to ti as?:n the diviior, what is the result? When the number of decimalk in the -,ividend is not equal to those in the divisor, how do we proceed? What is the result in this case? If there is a remainder after f';, dlivi,'on has been per 'err mted, how imy we proceed? What must be observed il this cage? Explain why this ia corre,.t 144 COMMON ARITHMETIC. Sect. VIIl, NOTE.-In common business calculations, it will be sufficiently exact to extend the quotient to three or four places of decimals; but when great accuracy is required, it should be extended farther. 4.-When the divisor is a composite number, we may often shorten the operation, as in whole numbers. (Sect. V. Art. 4. Obs. 3.) 5.-To divide by 10, 100, 1000, &e., we need only remove the separatingpoint as nmany places to the left, as there are ciphers in the divisor; and if there is not a sufficiency of figures, supply the deficiency by prefixing ciphers. Illestlration.-Divide I by 1000.. 1 = -; v-.1000 = Tra (Art. 6. Obs. 2. a.) -,,0- =.0001. (Art. 7.,0bs. 3.) EXERCISES FOR THE SLATE. 1. If 6 pounds of tea cost $4.50, how much is that per pound? Ans. $0.75. 2. If 12 pounds of coffee cost $1.50, how much is that per pound? Ans. $0.12-. 3. If 18 yards of cloth cost $27, how much is that per yard? Ans. $1.60. 4. If 26 acres of land cost $232.44, how much is that per acre? Ans..8.94. 5. If 150 pounds of but er cost $24, how much is that per pound? Ans. $0.16. 6. If 124 bushels of wheat cost $139.50, how much is that per bushel? Ans. $1.125. 7. If 464 pounds of feathers cost 8116, how much is that per pound? Ans. $0.25. 8. If 24 chairs cost $60, how much is that apiece? Ans. $2.50. 9. If it cost a man $2.375 a week for board, how long will $228 last him? Ans. 96 weeks. 10. A mechanic received $144 for doing a piece of work, which took him 64 days. How much did he receive per day? Ans. $2.25. 11. Divide 45 by. 15. Ans. 300. 12. Divide 2.88 by 1.2. Ans. 2.4. 13. Divide.20736 by 288. Ans..00072. 14. Divide 13.64589 by 2.19. Ans. 6.231. 15. Divide 1329.6 by.24. Ans. 5540. 16. Divide 1.4112 by 21. Ans..0672. 17. Divide 1 tenth by 10. Ans..01. 18. Divide 10 by 1 tenth. Ans. 100. To how many places of decimalsshould the quotient be extended in common business calculations? When should it be extended farther? How may we proceed when the divisor is a composite number? How may we divide by 10, 100, 1i00, &c.? Explain why this is correct? Art. 12. DIVISION OF DECIMALS OR FEDERAL MONEY. 145 19. Divide 1728 billionths by 288 thousandths. Ans..000006. 20. Divide 221 thousandths by 17 billionths. Ans. 13000000. MISCELLANEOUS EXERCISES FOR THE SLATE: Involving the principles of Decimal Fractions and Federal Money 1. A man has owing to him as follows: from A., $17; from B., $6.12-; from C., $27.064; and from D,, $0.81j. How mich has he owing to him in all? Ans. $51. 2. How much does C. owe him more than A. and B.? Ans. $3.93j. 3. How much does C. owe him more than all the rest Ans, $3.12., 4. A gentleman having $100, spent $25 for clothin-, $15 for books, $6.62- for riding in the stage, and $40.87- for jewelry; how much had he left? Ans. $12.50. 5. How much more did he give for his jewelry than for his clothing? Ans. $15.87j. 6. How much more did he spend for clothing than for riding in the stage? Ans. $18.37~. 7. A farmer received for his marketing $27.125; of this, he spent $2.375 for sugar, $1.1875 for coffee, $2.25 for tea, $0.75 for spice, $14.0625 for cloth, and took the rest home. How much did he take home? Ans. $6.50. 8. How much more did he spend for cloth than for all the other things? Ans. $7.50. 9. Add together fourteen units, six-tenths; two-hundredths; nine units, forty-seven thousandths; seventy-six ten-thousandths; and one unit, two ten-thousandths. Ans. 24.6748. 10. Add together 7 units, 602 thousandths; 18 units; 9 hundredths; 43 units, 26 hundredths; 7 units, 8071 ten-thousandths; and 286 units, 4 tenths. Ans. 363.1591. 11. From 18 units, take 63 thousandths. Ans. 17.937. 12. From 1 unit take 3 millionths. Ans..999997. 13. How much would 16 sacks of coffee, each containing 173 pounds, cost, at 7- cents per pound? Ans. $207.60. 14. How many bushels of wheat, at $0 875 per bushel, would it take to pay for the coffee, in the last example? Ans. 2371x". 15. A merchant bought 46 bags of oats, each containing 2.5 bushels, at $0.1875 per bushel. How much did they cost him? Ans. $21.5625. 16. IHi paid for them with coffee, at 10-cents per pound. How many pounds did it take? Ans. 215.626. 8 146 COMMON ARITHMETIC. Sect. V111. 17. How much would 47398 feet of lumber cost, at $3.50 per C.? Ans. $1658.93. 18. How much would 8372 pounds of pork cost, at $3.875 per C.? Ans. $324.415. 19. How much would 2146 feet of boards cost, at $10 per M.? Ans. $21.46. 20. How much would 1623 shingles cost, at $6. 125 per M.? Ans. $9.94+-. 21. If a man earn $0.625 per day, how much can he earn in a year, (365 days.) Ans. $228.125. 22. If he spends $0.375 per day, how much will he have left at the end of the year? Ans. $91.25. 23. How many yards of cloth can he buy with this money, at $3.625 per yard? Ans. $25~-. 24. Multiply 62.7 by 100. Ans. 6270, 25. Divide 8.726 by 100. Ans..08726. 2B, Just fifteen pair of ladies' gloves, For forty dimes had I: How many pair of that same kind Will forty eagles buy? Ans. 1090. ARTICLE 13. BILLS, ACCOUNTS, &C. Obs. 1. A BILL, in dealings with merchants, and others, is a woritten paper, containing a statement of the particulars, and total cost of the goods sold. To find the total cost of a Bill: Obs. 2. Find the cost of each particular at the price mentioned, and the 3sm of these Wil be the cost required. Required-the cost of the several articles, and the total sum in followin bills: (1.) COLUMBUS, Jan. 6th, 1847. Petr PFaywell, Bought of James Freeman, dc Co. 16 pounds Coffee,.......... at $0.121 per pound, 18 ( Sugar,............ at.09 12 Tea,............ at.87i so 26 " Saleratus,....... at.04 25 " Rsis,,.?.....:s. at.16i Total cost,.................. $19.32f: WAllAb dt o we find the total cost of %t ll? t Art. 13. BILLS, ACOUNTS, &C. 147 (2.) CLEVILAND, Feb. 2d., 1848. S. J. Aren, Bouglht of Charles Mantin. 6 yards Broadloth,.......... at $4.62- per yard 4 " Cambri c,............. at 12- " 21 dozen Buttons,............... at.25 per dozen 9 skeins Silk,............ at.06j per skein 4~ yards Wadding,........... at.09 per yard Total cost,...............4 $29. 84. (3.) CINCiNNAI, June 7th, 1848, Henry Plyiard, To Lewis Anderson, & Co. Dr. For 4 copies Davies' Bourdon,........ at $1 37- each ( 3 "( Anthon's Caesar,..........at 1.00 " "4 4" Leverett's Dictionary,...... at 4.87i " g '4 Greek Testament,.......... at I 12, 2 Total cost,............. $38- 212i. (4.) DELAWARE, March 1st, 1847. Awr16oee Pkiner, TJ Augustus Reichawts, d Co. Dr. For 897 feet Boards,...... at $0. 871 per 'C. " 1247 " Plank,........... at 1.12k 4' " 479 " Scantling,......... at.75 " 4' 2479 " Flooring.......... at 1.25 " 8762 " Shingles,........... at 4.37l per MTotal cost,................ $94. 788-i-. Received Payment.....~~~~~~Att S 148 CXMMON ARITHMETIC. Sect IX (5.) PITTSBURG, May 1st, 1848. Thzomas Thrifty, Sold TWi. Trader, & Co. 786 bushels Wheat,........... at $1.12~ per bushel 1423 " Barley,......... at.56j " 4679 " Corn,............at.22 " 3716 " Oats.............. at.25 " 2742 pounds Cheese,........ at.08 per pound In return, he receivedIn Cash,................................. $3000.00 144 yards Satinet,............ at $1.25 per yard 68 "i Silk,................ at 1.06J " 176 " Muslin............. at.11 " 168 " Calico,.............. at.16|- " 1236 pounds Coffee,.............. at. 1 per pound 1374 " Sugar,............. at.08 " Required-the difference between the accounts, and in whose favor. Ans. $308.38- in favor of Thomas Thrifty. SECTION IX. COMPOUND NUMBERS. ARTICLE I. DEFINITIONS, &C. Obs. I. When the ratio of increase is the same, numbers are called SIMPLE. Thus: 156, 28, 5 dollars; 13 yards; 147 miles, &c., are called simple numbers, because each order has ten times the value of the next lower order. Obs. 2. When the ratio of increase varies in the different orders or numbers, they are called C4OMPOUNDa. T'hu, 12 bushels, 3 pecks, 6 quarts; 3 miles, 30 rods, 14 feet, &c., are called compound numbers, because in some of these it takes more, and in some less, than ten units of one order to make one unit of the next higher order. REMARK.- Compound numbers, by somi authors, are called Denominate Numbers. Obs. 3. The only difference between Simple and Compound Whea are numbevr called Simple numbers? Give examples. When are they called Compound numbeasl Give examples. Art. 2. REDUCTION. 149 numbers, is this: In Simple numbers, figures increase uniformly by 10; that is, it takes 10 units of each order to make 1 unit of the next higher order; but in Compound numbers figures increase differently, sometimes taking more, and sometimes less, tJuw 10 units of one order, to make 1 unit of the next.higher order. In the former one table alone is necessary; in the latter many tables are required. Obs. 4. The Tables in Compound numbers teach Cow many units it takes of one order to make a unit of the next higher order. The usumeration Table beaches the same in Simple numbers. REMARK.-The different orders in Compound numbers are called denominations. Obs. 5. In Simple numbers, the units of any order can always be expressed by one figure; in Compound numbers, it sometimes takes two, three, or even four figures to express the units of a single order. Likewise, the sum, or difference of any order, in Simple numbers, can always be expressed by one figure; but in Compound numbers, the sum, or difference, can sometimes be expressed by one figure, and sometimes it requires more than one figure to express it. REMAnK.-The learner will bear in mind, that we carry 1 to the next higher order, as often as we obtain a sufficient number of units of the lower order, to snake I unit of this higher order, whether in Simple or Compound numbers HenceObs. 6. The principles of Simple and Compound numbers are tle saCte. ARTICLE 2, REDUCTION. Obs. 1. Changing numbers from one denomination or kind to another, without altering their value, is called REDUCTION. Thus, there are 4 peeks in a bushel; then in 2 bushels there are 8 pecks; therefore, 8 pecks, or 2 bushels, express the same quantity. REMARK-Reduction is of two kinds-Descending and Ascending. What is the difference between Simple and Compound numbers? How many tables are necessary in the former? Are more tables than this required in the latter? What do the tables in Compound numbers teach? What does it teach in Simple tnumbers? What are the 4differ.en orders in Compoand numbers called? How many figures does it take to express a unit of a single order in Sinmple numbers? How mauy in Compouud numbers'? How many figures does it take to express the sum, or difference, of any order in Simple numbers? How many in Compound numbers? How often do we carry 1 lo the next higher order? Is this the same, both in Simple and Compound numbers? What then is the difference between the principles of the two? What is Reduction? How is it divided? What is Reduction descending? What is tRe duetisn abreading? 4iveaa example illustrating each case. COMrMON ARITHMETICct Sect. IX. 0'bs. 2. When a number is chumged from a denomination of greater value to a denomination of a less value, the proces, is called REDUCTION DDESCENDJNG. When a number is changed from a denomintion of less value to c denomination of greater value, the process is called REDUCTION AsCENDING. Thus, to change busbeTs to pecks, we change a greater denomination to a less, and the process is called Reduction JDescending. But to change pecks to bushels, we change a less denomination to a greater, and the process is called Reductiao Ascending. There are 4 pecks in a bushel; how many pecks in 2 bushels? In 3; 4; 6, 10; 12 bushels? How many bushels in 4 pecks? In 8, 16; 20; 28, 3.9; 32; 44 pecks? Obs. 3. From these examples, the pupil will perive that Reduction Descending is performed by Multiplication, and Beduction Ascending isperformed by dwision. ThereforeReduction Descending and Reduction Ascending mutually prove each other. REMARK.-Compound numbers are chiefly confined to weights and measures. WEIGHTS. 1. TROY WEIGHT. Ohs. 4. This is used in weighing gold, silver, jewels, liquors, &c. The denominations are Grain, Pemnyweiyht, Ounce and Pound. TABLE. 24 grains (grs.) --- make 1 pennyweight, marked pwt. 20 pennyweights ------ ounce, " o. 12 ounces -— " 1 -pound.-.. ' lb. REWrARK 1.-The standard of Weights varies in different States in the Union In 1834 the Government adopted a uniform standard for the use of the sev-' eral Custom-houses, and etherpurposes. It is very desrable that this standard should be adopted throughout the Union. 2.-The standard unit of weight adopted by the United States is the Troy pound of the U. S. Mint, which is the same w the Imperial Troy pound of Englatd, established by Act of Parliament, A. ID. 1826 How is Reduction Descending performed? Reduction Ascending? What relation do they bear to each other? To what are compound numbers chiefly appled7 Forwhat is Troy weight used? What are the denominations? Repeat the Table. What is the standard. unit of Weight in the United States? Art. 2. XLCUCTION. 151 MENTAL EXERCISES. 1. How many ounces in 2 pounds? 4; 9; 7; 12; 13; 6; 5; 11; 8? 2. How many pennyweights in 2 ounces? 4; 9; 6; 3; 5; 8; 10? 3. How many pounds in 24 ounces? 36; 72, 96; 120; 132? EXERCISES FOR THE SLATE. 5. Reduce 3 lbs. 7 oz, 15 pwts. 18 grains to grains. Operation. lbs. oz. pwts. grs. 3_ 7 -— 15-.-18 12 36 7 oz. added. 43 20 860 15 875 24 6. Reduce 21018 grs. to pounds Operation 24)21018 210)8715 + 18 grs. rem. 12)43 + 15 pwts. rem. 3 + 7 oz. rem. Ans. 3 1bs. 7 oz. 15 pwts. 18 grs. We first reduce the 21018 grs. to pwts. by dividing by 24; the result is 875 pwts., and 18 grs. remaining. We next reduce the 875 pwts. to oz., by dividing by 20; the result is 43 oz., 15 pwts. We reduce the 43 oz. to lbs. by dividing by 12; the result is 3 lbs. 7 oz., and the other remainders make 3 lbs., 7 oz., 15 pwts., 18 grs. as our answer, pwts. added. 3500 1750 21000 18 grs. added. 21018 grs. Ans. We first multiply by 12, to reduce the lbs. to oz., and add in the 7 oz., making 43 oz. We multiply this by 20, to reduce it to pwts., and add in the 15 pwts., making 875 pwts. This we multiply by 24 to reduce it to grs., and add in the 18 grs., making 2018 grs., as our answer. If we choose, we can add in the numbers given of the different denominations mentally, and thus shorten the operation. Thus: 3 X 12 = 36; 36+ 7 43; set down 43, &c. 152 COMMON ARITHMETIC. Sec t.IX N TE.-It would be a good plan to write the name of the denominator to whi< h each number belongs, over the number, (as in the 5th example,) to prevent mistakes. From these examples, we derive the following GENERAL RULES FOR REDUCTION. For Reduction Ascending: For Reduction Ascending: I. Multiply the highest denomi-\ I. Divide the given quantity by nation by that number which it takes that number which it takes of this of the next less to make 1 of this denomination to make one of the higher, and add to the product the next higher. rumber given, (if any,) of the low-i II. Proceed in this manner until er denomination. the whole is reduced to the lenomiII. Proceed in the same mannernation required. The last quotient, with the remaining denominators, toether with the several remainders, until the whole is reduced to the de- (f cny,) annexed, wil be the annomination required. swer sought. NOTE.-If, after any division, there NOTE.-If any denomination is want- is no remainder, a cipher should be ing, supply its place with a cipher. iwritten in the place of that denomina~'.~~~~~~~~ Ition. 7. Reduce 3lbs. 8 oz. 10pwts. 8. Reduce 21360 grs to pounds to grains. 9. Reduce 6 lbs. 16 pwts. to 10. Reduce 34968 grains to grains. pounds. 11. Reduce 7 lbs, 19 grs. to 12. Reduce 40339 grains to grains. pounds. 13. Reduce 2 lbs. to penny- 14. Reduce 480 pennyweights weights. to pounds. 2. AVOIRDUPOIS WEIGHT. Obs. 5. This is used in weighing the articles of a coarse, heavy nature, such as tea. cofee, sugar,flour, hogs, grain, &c., and all metals except gold and silver. It is also used in buying and selling medicines. The denominations are DRAM, OUNCE, PO7ND, QUARTER, HUNDRED-WEIGHT and TON. Why do we write the name of the denominations over the numbers? What is the rule for Reduction Descending? For Reduction Ascending? If any denomination is wanting, how do we proceed? If after any division there is no remainder, how do we proceed? For what is Avoirdupois weight used? What are the denominations? Repeat the Table. Art. 2. REDUCTION. 153 16 drams (drs.) -— _16 ounces --- -- 25 pounds -- _ 4 quarters ------- 20 hundred weight -— _ TABLE. make 1 ounce --- — " 1 pound --- —-—.. " 1 quarter —. --- 1 hundred w-eiht - " I ton marked oz. " lb. <" qr. C T. '~ PI', REMARK 1.-The Avoirdupois Pound of the United States is determined from the standard Pound Troy,* and is in the ratio of 5760 to 7000. ['lat is, 1 pound Troy contains 5760 grains; 1 pound Avoirdupois contains 7000 grains Troy. 2. —ls this weight the words gross and net occur. Gross weight is the weight of goods, together with that of the boxes, casks, or b:gs that contain them. Net weight is the weight of the goods alone. 3.-Formerly it was customary to allow 112 pounds for the hundred weight, and 28 pounds for the quarter; but this practice has become nearly, or quite, obsolete. In buying and selling all articles of commerce estimated by weight, the laws of most of the States, as well as general usage, call 100 pounds a hundred weight, and 25 pounds a quarter. In the U. S. Custom!-house, and also in invoices of English goods, and of coal from the Pennsylvania mines, 28 Ibs. are called a quarter, and 2240 lbs. a ton. 1. How many drams in 2 ounces? 3; 5; 4; 6? 2. How many ounces in 2 pounds? 6; 4; 3; 5? 3. How many quarters in 2 cwt.? 4; 9; 3; 6; 5; 7; 12? 4. How many pounds in 32 ounces? 48; 80, 64; 76? 5. How many cwt. in 8 quarters? 20; 27; 36; 24; 48? 6. Reduce 7 lbs. 9 oz. todrams. 7. Reduce 1936 drs. to ounces. 8. Reduce 7 cwt. 3 qrs. 22 lbs. 9. Reduce 12752 oz. to hunto ounces. dred weight. 10. Reduce 3 tons to drans. 11. Reduce 1536000 drs. to tons. 12. Reduce 7 T. 14 cwt. 23l 13. Reduce 246768 oz. to!bs. to ounces. tons. 14. Reduce 5 T. 18 lbs. 9 oz. 15. Reduce 2564752 drs. to to dramns, tons. 3. APOTHECARIES WEIGHT. Obs. 6. This is used by apot/hecaries and physicians in mixing medicines. How is the Avoirdupoispound determined? Whatis the ratiobetween the two? How many grains in a pound Troy? In a pound Avoirdupois? Whatisgross weight? Net weight? For what is Apothecaries weight used? * A pound Avoirdupois is heavier than a pound Troy, liut an ounce Troy is heavier than an unce Avoirdupois. 8A 154 00MtMON, ARITIUMbETIV. 'eect. iA. The denominations are Grain, Scruple, Dram, Ounce, and Pound. TABLE, 20 grains (grs.)- - make 1 scruple, --- marked sc. or D. 3 scruples -"... -- 1 dram,. —. " dr. or 3. 8 drams ------ 1 ounce,, " oz. or 3.; 12 ounces_ ---- " I pound, " lb. REMARK.-The pound and ounce are the same in this weight as they are in Troy weight; but the other denominations are different. 1. How many scruples in 2 drams? 4; 6; 9; 8; 5; 7; 12? 2. How many drams in 2 ounces? 6; 5; 7; 4; 8; 10; 12? 3. How many ounces in 2 pounds? 4; 6; 8; 12; 5; 7; 9? 4. How many drams in 6 scruples? 24; 18; 36; 15; 27? 5. How many ounces in 16 drams? 32; 56; 40; 72; 96? 6. How manypounds in 24 ounces? 72; 60; 96; 144; 108? 7. Reduce I lb. 7 3. 2 3. 1 3.1 8. Reduce 9276 grains to 16 grs. to grains. pounds. 9. Reduce 7 lb. 6 3. to scru- 10. Reduce 2034 scruples to ples. pounds. 11. Reduce 6 lb. 1 D. to 12. Reduce 34580 grains to grains. pounds. MEASURES OF CAPACITY. 1. DRY MEASURE. Obs. 7. This is used to measure grain, fruit, salt, d'c. The denominations are Pints, Quarts, Pecks and Bushels. TABLE. 2 pints (pts.) ------ make 1 quart, marked qt. 8 quarts --- - " 1 peck,. pk. 4 pecks. --- —------- 1 bushel, " bu REMARK 1.-A quart Dry Measure contains 67t cubic, or solid inches. 2 —A Winchester bushel is 184 inches in diameter, (that is, across the top,) and 8 inches deep, and contains 21502 cubic inches. By this is meant a compact bushel, as wheat, oats, salt, shelled corn, &c. A What are the denominations? Repeat the Table. To what other weights are the pound and ounce, Apothecaries weight, equal? For what is Dry measure used? What are the denominations? Repeat the the Table. -low many cubic inches in a quart, dry measure? What is the size of a Winchester bushel? How many cubic inches does it contain? What kind of a bushel is meant by this? Art. 2. REDUCTION. dry bushel, as apples, peaches, coal, potatoes, contaitn 2688 cubic, or sol;d inches. The former is an even bushel, and the latter is a heaped bushel. 3.-The measure used varies in different States. In Connecticut, 2193 cubio inches make a bushel. In New York, 2218.192 cubic inches make a bushel. This is the imperial bushel of Great Britain, and weighs 8 lb3. Avoirdupois, of distilled water, at 62~ Fahrenheit, and 30 inches of the barometer. The Winchester bushel is the United States standard unit of Dry Measure, and contains 77.627413 Ibs. Avoirdupois distilled water, maximum density, (which is about 41" Fahrenheit,) and weighed in air at 30 inches of the barometer. Its true contents are 2150.42 cubic inches, nearly, although 2150.4 is usually given. It is desirable that the United States standard should be adopted by the different States, the same as our currency. This would create uniformity, and thus destroy much trouble and confusion usually attendant upon a varia.ion of standards. 4. Many persons purchase grain by weight, instead of by measure. In Ohio, and also in a I ajority of the States, the weight of grain is established by law, as follows: 1 bushel of Wheat............. weighs......... 60 lbs. Avoirdupois. 1 " Rye or Indian Corn, "......... 56 ( " 1 " Barley............. ".........4 " 1 " Oats,....................... 33 " " Hence-To find the number of bushels of grain in a given quantity: a. Divide the weight of the grain by 60, 56, 43, or 33, according as it is Wheat, Rye, Corn, Barley, or Oats. 1. How many pecks in 2 bushels? 5; 8; 10; 15; 18; 23; 30? 2. How many quarts in 2 pecks? 3; 4; 7; 5; 6; 9; 8' 11? 3. How many pints in 2quarts? 5; 4; 6; 3; 9; 7; 8? 4. How many bushels in 8 pecks? 16; 32; 24; 48; 36? 5. How many pecks in 16 quarts? 32; 64; 48; 72; 96? 6. How many quarts in 4 pints? 12; 24; 16; 36; 18; 7. How many quartsin 1 bushel? 2; 4; 6; 3; 5; 8; 10? 8. H6w many pints in 1 peck? 2; 6; 4; 10; 5; 8; 3? 9. How many pints in 1 bushel? 2; 4; 6; 3; 5; 8; 10? 10. How many bushels in 32 quarts? 64; 96; 128; 160; 192? 11. How many pecks in 16 pints? 32; 48; 80; 61; 96; 112? 12. How many bushels in 64 pints? 128; 256; 192: 320? Each of the following examples proves the one opposite: 13. Reduce 3 bu. 2 pks. to 14. Reduce 14 pks. to bushpecks. els. 15. Reduce 6 bu. 1 pk. 7 qts. 16. Reduee 414 pts. to bushto pints. els. 17. Reduce 10 bu. 3 pks. 6 18. Reduce 701 pts. to bushqts. 1 pt. to pints. els. How many cubic inches does the dry, or heaped bushel, contain? What ir the United States staudard unit of Dry Measure? 156 COMMI',N ARITHMETIC. Sect. IX 19. Reduce 37 bu. 1 pk. 5 qts. 20. Reduce 2395 pts. to bush1 pt. to pints. els. 21. Reduce 68 bu. 1 pt. to 22. Reduce 4353 pts. to bushpints. els. 28. Reduce 121 bu. 2 qts. to 24. Reduce 7748 pts. to bushpints. els. 2. WINE MEASURE. Obs. 8. This is used for measuring all liquids, ale, beer, and milk excepted. The denominations are Gill, Pint, Quart, Gallon, Barrel and Hogshead. TABLE. 4 gills (gi.) --— make 1 pint, ----- " pt. 2 pints ------ 1 quart, ---------- " qt. 4 (uarts.-_ --- —-- 1 gallon, --- —---- " gal. 31 gallons --- - 1 barrel, — bbl. 63 gallons, or 2 bbls. " 1 hogshead, - hhd. ALSO: 42 gallons. ------— make 1 tierce, --- —- marked tier. 84 gallons, or 2 tier. I' 1 puncheon,. --- — - " pun. 126 gallons, or 2 hhd. " 1 pipe, or butt, - - -. " P. 2 pipes " 1 tun, " T. REMARK. —The standard unit of Liquor Measure adopted by tile United States is the wine gallon, containing 231 cubic inches, equal to 8.339 Ibs. Avoir. of distilled water at the maximum density; i. e., 41 P Fahrenheit. 1. How many gills in 2 pints? 3; 5; 9; 7: 12; 8; 18; 4; 11? 2. How many gills in 2 quarts? 3; 7; 9; 5; 12; 10. 6; 8; 11? 3. How many pints in 1 gallon? 2; 5; 3; 9; 6; 5; 7; 12; 10? 4. How many gills in 1 gallon? 2; 4; 3; 5? 5. How many pints in 8 gills? 16; 48; 40; 12; 32; 20; 36? 6. How many quarts in 8 gills? 16, 48; 96; 40; 20; 36; 28? 7. How many gallons in 8 pints? 16; 48; 60; 96; 56; 88; 40? 8. How many gallons in 32 gills? 96; 160; 64; 138? 9. Reduce 10 gal. 2 qts. to 10, Reduce 336 gills to galgills. Ions. 11. Reduce 1 bl. to pints. 12. Reduce 252 p'nts to barrels, 13. Reduce 2 hhds. 27 galls. 14. Reduce 4927 gills to hogs3 qts. 1 pt. 3 gi. to gills, heads. For what is Wine measure used? What are the denominations? Repeat the Table. What is the standard unit of Liquid Measure adopted by the United 6e ates? What does it contain? Xit: 2. REDUCTION. 157 15, Reduce 1 T. 1 P. 1 hhd. 16. Reduce 3785 pts. to tuns. 32 galls. 1 pt. to pints, 17. Reduce 7 hhds, 3 gi, to 18. Reduce 14115 gills to hogsgills, heads, 3. ALE, OR BEER MEASURE. Obs. 9. This is usedfor measuriusg ale, beer, and milk. The denominations are Pint, Quart, Gallon, Barrel, and Ilogshead. TABLE. 2 pints (pts,) ----— make 1 quart,_ - marked qt. 4 quarts -__. " 1 gallon, " gal. 36 gallons -— " 1 barrel, " bar. 54 gallons, or 1- barrels,__ " 1 hogshead,.__,. " hhd. REMARK.-A gallon, beer measure, contains 282 cubic inches. 1. How many pints in 2 quarts? 4; 9; 5; 7; 6; 12; 8; 10; 9? 2. How many quarts in 2 gallons? 3; 5; 4; 9; 12; 6; 8; 11? 3. How many pints in 1 gallon? 2; 4; 5; 3; 7; 12; 9; 10? 4. How many quarts in 8 pints? 4; 16; 12; 24; 14; 18; 22? 5. How many gallons in 8 quarts? 16; 20; 12; 48; 40; 36? 6. How many gallons in 8 pints? 24; 16; 48; 32; 56; 40; 96? 7. Reduce 18 galls. 2 qts. 1 8. Reduce 149 pts. to gallons. pt. to pints. 9. Reduce 3 hhds. 3 qts, to 10. Reduce 1302 pts. to hogspints. heads. 11. Reduce 6 bar. 1 pt. to 12. Reduce 1729 pts. to barpints. rels. 13. Reduce 2 hhds. 3 qts. to 15. Reduce 435 qts. to hogsquarts. heads. MEASURES OF EXTENSION. REMARK 1.-The three kinds of measures we have just given are used for measuring liquids, krain,fruit, &c., for which purpose we have different kinds of vessels, as the half bushel, peck measure, pint cup, quart cup, gallon measure, &c. The kinds of measure we are now about to mention are used to measure extension; that is, distances, surfaces, 4-c. For this purpose we have the Sur For whatis Beer measure used? What are the denominations? Repeat the Table? How many cubic inches does a gallon, Beer measure, contain? For what are the preceding measures used? What do we have for this purpose? For what are the other measures used! What do we have for this purpose? 158 COMMON ARITHMETIC. Sect. IX veyor's chain, the square, or rule, yard measure, tapes, and lines of differen lengths, &c. 2.-Extension has three dimensions: length, breadth, and thickness. 1. LONG MEASURE. Obs. 10. This measure is used when length or distance is considered, without regard to breadth or thickness. It is frequently called Linear, or Lineal measure. The denominations are Inch, Foot, Yard, Rod, Furlong, and Mile. TABLE. 12 inches (in.)........ make 1 foot, ---- marked ft. 3 feet.- I".. 1 yard, " yd. 65 yards, or 161 feet, I r 1 rod. perch, or I ~ ipole,. " rd. 40 rods, or 220 yards, - 1 furlong, " fur. 8 furlongs, or 320 rods, " 1 mile, -... " M. ALSO: 3 barley corns, (bar. c.) 4 inches ----- 6 points. --- —12 lines 6 feet. — ----- 3 miles. --- —-- 60 geographical, or 69L* statute miles,-.. 360. " 1 inch, (but little used.) b 1 hand, used in measuring the height of horses. 1 line; and, 1 inch, used in measuring the length of clock pendulums., 1 fathom, used in measuring the depth of water., 1 league, used in measuring distances at sea. " 1 degree, - deg.or0. the circumference, or distance round the earth. REMARK.-The standard unit of length adopted by the United States is the Yard of 3 feet, or 36 inches, and is the same as the Imperial Yard of Great Britain. It is made of brass, and is determined at the temperature of 60~ Fahrenheit, from the scale of Troughton, a celebrated English artist. For what is Long measure used? What is it frequently called? What are the denominations? Repeat the Table. What is the standard unit of length adopted by the United States? How determined? * This is not exactly correct, although it is what is usually given. On the equator. 69 and one-sixth statute miles make 1, or 60 geographical miles, nearly; and on the meridian, at a mean, 694 statute miles make 1i Art. 2. REDUCTION. 159 1. How rany inohes in 2 feet? 3; 5; 7; 12; 8; 6; 9; 11; 15; 2. How many feet in 3 yards? 4; 9; 12; 6; 5; 8; 22? 3. How many furlongs in 2 miles? 4; 6; 9; 11; 8; 12? 4. How many inches in 1 yard? 2; 5; 3; 4? 5. How many feet in 24 inches? 48; 108; 36; 84; 60? 6. How many yards in 6 feet? 12; 36; 24; 15; 33; 18? 7. How many miles in 16 furlongs? 32f 72; 96; 48; 64? 8. How many yards in 36 iuches? 108; 180; 72; 144? 9. How many barley-corns in a foot? 2; 4; 5; 3? 10. How many feet in3 hands? 4; 7; 8; 5; 9; 6; 12? 11. How many points in 1 inch? 2; 3; 4? 12. How many feet in 2 fathoms? 3; 5; 9; 12; 8; 6; 7? 13. How many feet in 36 barley-corns? 72; 144; 80; 108? 14, How many miles in 2 leagues? 4; 12; 8; 5; 7; 8? 15. How many hands in 12 inches? 1 ft. 4 in.; 3 ft; 2 ft. 8 in. 16. How many inches in 72 points? 144; 216; 288? 17. How many fathoms in 12 feet? 24; 72; 28; 60, 24? 18. How many leagues in 6 miles? 12; 36; 24. 32; 48? 19. How many indies in 2 ft. 3 in.? 3 ft. 6 in.; 3 ft. 9 in.; 5 ft.? 20. How many feet in 25 inches? 33; 40; 44; 57; 70? 21. Reduce 85 miles to inches. 23. Reduce 7 M. 6 fur. 27 rds. 3 vds. 2 ft. to inches. 25. Reduce 7~. 49 M. 7 fur. 16 rds. 12 ft. 9 in. 2 bar. c. to barley corns. NOTE.-Statute miles are understood in this example. 27. How many barley corns would it take to reach around the globe, it being 360 degrees. NoTE.-Multiply by 69i to reduce it to miles. 29. How many times will a carriage wheel turn over in going 360 miles, it being 16 ft. 6 in. in circumference, that is, around the outside? 22. Reduce 5385600 inches to miles. 24. Reduce 496518 inches to miles. 26. Reduce 101964125 barley corns to degrees. 28. How many degrees is it around the globe, it being 4755 -801600 barley corns? 30. How many miles does a carriage wheel, 16 ft. 6 in in. circumference, proceed, in turning over 115200 times. I 8. CLOTH MEASURE. Obs. 11. This is used in measuring cloth, tape, lace, and all kinds b6 goods bought and sold by the yard. 160 COMMON ARITHMETIC. Sect. IX The denominations are 3a/il, Quarter, Yard, and Ell. TABLE. 4 nails (na.) -___-.. _ - make 4 quarters " 3 quarters, or 3 of a yd. " 5 quarters, or 1' yards " 6 quarters, or 1 yards " 1 quarter, -- - — marked qr. 1 yard,. --- —--- " yd. 1 Flemish ell;, - " Fl. e. 1 English ell, " E. e. 1 French ell, - - - - F. e. REMARK.-Cloth measure is a species of Long measure. The yard is tie same in both; thatis, it contains3 feet, or36 inches. Therefore, a quarter contains 9 inches, and a nail 21 incles. Cloths, &c., are bought and sold according to their length, without regard to their width. The nail is but little used, eighths and sixteenths being used in its place. 1. How many nails in 2 quarters? 4; 8; 12; 16; 11; 9? 2. How many quarters in 2 yards? 9; 9; 7; 5; 12; 11? 3. How many nails in 2 yards? 3; 5; 4; 6? 4. How many quarters in 2 Flemish ells? 4; 8; 9; 6; 12? 5. How many quarters in 2 English ells? A, 4; 7; 5; 9; 12? 6. How many quarters in 2 French ells? 12; 9; 8; 7; 5; 6? 7. How many quarters in 8 nails? 16; 48; 32; 12; 44; 36? 8. How many yards in 8 quarters? 48; 12; 16; 32; 36; 44? 9. How many yards in 6 nails? 64; 32; 80; 48; 96? 10. How many Flemish ells in 6 quarters? 36; 18; 30; 27; 52? 11. How many English ells in 10 quarters? 44; 16; 50; 20? - 12. How many French ells in 12 quarters? 24; 42; 54; 72; 48? 13. Reduce 6 yds. 3 qrs. to nails. 15. Reduce 15 yds. 1 qr. 3 na. to inches. 17. Reduce 8 F. e. 3 qrs. to nails. 19. Reduce 10 E. e. 1 qr. 2 na. to nails. 21. Reduce 12 F. e, to Fl. e. First reduce the French ells to quarters, and then to Flemish ells. 23. Reduce '25 F. e. to English ells. 25. Reduce 18 Fl.e. to French ells. 14. Reduce 108 nails to yards. 16. Reduce 5553 inches to yards. 18. Reduce 204 nails to French ells. 20. Reduce 206 nails to English ells. 22. Reduce 24 Fl. e. to F. e. NOTE.-First reduce the Flemish ells to quarters, and then to French ells. 24. Reduce 30 E. e, to French ells. 26. Reduce 9 F. e. to Flemish ells. For what is Cloth measure used? What are the denominations? Repeat he Table. What kind of measure is Cloth measure? What measuie is the ame in both? How many feet and inches does the yard contain? How are loths, &c., bought and sold? Art. 2. REDUCTION. 161 27. Reduce 15 Fl. e. to 28, Reduce 9 E. e. to French French ells. ells. 29. Reduce 36 E. e. to Flem- 30. Reduce 60 Fl. e. to English ish ells. ells. 3. LAND, OR SQUARE MEASURE. Obs. 12. This is used for measuring land, flooring, or anything in which length and breadth are considered, without regard to depth or thickness. The denominations are Square Inch, Square Foot, Square Y'ard, Square Rod, Rood, Acre and Square Mile, A. Obs. 13. When any two lines meet together, the opening at their place of meeting is called an ANGLE. When they meet so as to form a square corner, like the corner A., the angle is called a RIGHT ANGLE. 1 square yard. ~ Obs. 14. A figure having four equal 2 -- --- - sides, and its angles right angles, is called a SQUARE. 9....- ~.. A Square inch is a square, each side of. which measures an inch in length. 3 ft. long. A Square yard is a square, each side of which is a yard, or 3 feet in length, and contains 9 square feet. A square foot contains 12X 12 = 144 square inches. TABLE. 144 square inches(sq. in.) make 1 square foot,.- marked sq. ft. 9 square feet -- 1 square yard,-. " sq.yd 30, square yards,or 272- T I 1 square rod, perch, sq. rd square feet, 5 or pole, --- —-- 40 square rods -" 1 rood, — - " R. 4 roods, or 160 sq. yds. " 1 acre, - " A. 640 acres1 square mile, or sq.. ~640 acres -. —, --- - j section, --- M For what is Land, or Square measure used? What are the denominations? What is an angle? A right angle? A square? A square inch? A square yard? How many square feet does a square yard contain? How many square inches does a square foot contain? Repeat the Table. 162 COMMON ARITHMETIC. Sect. IX. In measuring land, surveyors use a chain 4 rods long, and containining 100 links. Hence79- inches --- —----- make 1 link, --- —------ marked 1. 25 links- 1 rod, ----------- rd. 4 rods or 66 feet, ---- " I chain,- ch. 80 chains --- —- " 1 mile,- M. These, the learner will observe, are all linear measure. But 1 square chain ------— makes 16 perches, marked P. 10 square chains -----— make 1 acre,- A. This chain is generally called GUNTER'S CHAIN, from the name of its inventor. 1. How many square feet in 2 square yards? 3; 6; 12; 9; 2. How many roods in 2 acres? 4; 9; 5; 7; 9; 12; 10? 3. How many square yards in 18 square feet? 27; 54; 90? 3. How many acres in 8 roods? 44; 32; 24: 48; 36; 49? 5- Reduce 1 sq. M. 326 A. 3 6. Reduce 6065153964 sq. R. 27 sq. rd. 16 sq. yds. to square to square miles. inches. 7. Reduce 16 A. 2 R. 17 sq. 8. Reduce 104183011 sq. rd. 124 sq. ft. 127 sq. in. to to acres. _ _.__1__ _! 7 __ 7? in. in. square inches. 9. Reduce 7 A. 120 sq. rd. 212 sq. ft. 114 sq. in. to square inches. 11. If a man's farm measures 120 chains in length, how many rods long is it? How many miles? 10. Reduce 48643602 sq. in. to acres. 12. If a man's farm measures 1 miles in length, how many rods long is it? How many chains? The student will observe that the mile is Long measure. Obs. 15. Although a square foot and afoot square are the same, there is a difference between squarefeet and feet square. 3 ft. sq.=-9 sq, ft. It will be perceived from the figure, that 3 feet square measures 3 feet on -each side, and contains 3 X 3 = 9 square feet. What is used in measuring land? How long is this chain? How many links does it contain? Repeat the Table. What name is usually given to this chain? Why? Art. 2. REDUCTION. 163 3 Fi. ft. On the other hand, 3 square feet measures 3 feet in length, and only 1 foot in width, and contains 3 X 1 = 3 square feet. Therefore, there is a difference of 6 sq. ft. between 3 ft. sq. and 3 sq. ft. Between 2 sq. ft. and 2 ft. sq. there is a difference of 2 sq. ft. Between 4 sq. ft. and 4 ft. sq. there is a difference of 12 sq, ft. 13. What is the difference between 8 sq. ft. and 8 ft. sq.? 14. What ts the difference between 9 sq. ft. and 9 ft. sq,? 15. What is the difference between 12 sq. ft. and 12 ft. sq.? 16. What is the difference between 25 sq. ft. and 25 ft. sq. Ans. 600 sq. ft. Besides the square, we have other four-sided figures, which have different names, according to their forms. The most common are the RECTANGLE, PARALLELOGRAM, and RHOMBUS. Rectangle. Parallelogram. Rhombus. 4 ft. long. 4 ft. long. ' Obs. 16. The space enclosed by tile lines which bound a figure, is called its AREA, or superficial contents. It will be perceived that the area of the above figures, as well as as that of the square, is found by multiplying together their length and breadth. HenceTo find the area of a square, rectangle, parallelogram, or rhombus: Obs. 17. Multiply together their length and breadth: REMARK.-Recollect that feet multiplied by feet produce square feet, and so of all the other denominations of linear measure. 17. What is the difference in the size of two rooms, one being 15 feet square, and the other containing 196 square feet? Ans. The one 15 feet square contains 29 sq. ft. the most. What is the difference between a square foot and a foot square? Between 3 sq. ft. and 3 ft sq.? Show why this is the case. What four other figures have we besides the square? What is the area of a figure? How do we find the area of a square, rectangle, parallelogram, or rhombus? What do the denominations of linear measure, multiplied by the same, produce? 161 COMMON ARITtLMTlIC. Sect. IX. 18. If a floor is 18 feet long, and 12 feet wide, how many square feet does it contain? Ans. 216. 19. If a piece of land is 45 chains long, and 80 rods wide, how many acres does it contain? Ans. 90. Reduce both to the same denomination. 20. How many square feet in a board 18 inches wide, and 24 feet long? Ans. 36. Reduce both to the same denomination. 21. How many square feet in a board 32 feet long, and 16 inches long? Ans. 42 sq. ft. 96 sq. in. 22. How many acres in a square field, each side of which measures 120 rods? Ans. 90. 23. How many acres in a field 46 rods long, and 20 rods wide? Ans, 5 A. 120 sq. rds. 24. The largest of the Egyptian pyramids is a square at the base, each side being about 693 feet in length. How many acres does it cover? Ans. 11 A. 4 sq. rds. 25. What is the difference between a floor 25 feet square, and two others, each 16 feet square? Ans. 113 sq. ft. 26. If a room is 30 feet long, how wide must it be to contain 240 sq. ft.? Ans. 8 feet. NoTE. —The length and breadth are the factors, which, multiplied together, produce the area. Therefore, the area divided by either factor, will give the other.. (Sect. VI. Art. 1. Obs. 16.) 27. If a board is 18 inches wide, how long must it be to contain 18 sq.ft.? - Ans. 12 ft. Reduce both to the same denomination. 28. If a board is 15 feet long, how wide must it be to contain 20 sq. ft.? Ans. 16 in. 29. If a room is 12 ft. long, how wide must it be to contain 216 sq. ft'? Ans, 18 ft. 30. How many yards of muslin, 3- of a yard wi4e, will it take to line 3 yds. satinet, 1 yds. wide? Ans. 5 yds. 4. SOLID, OR CUBIC MEASURE. This is used to measure wood, stone, or anr thing in which length, breadth, and thickness, or depth, are considered. The denominations are Solid Inch, Solid Foot, Solid Yard, Cord, and Ton. Obs. 19. A solid body having six equal square faces, is called a CUBE, or HEX^EDRON. # b *... I i When we have the area, and either the length or breadth given, how do we find the other? Why is this correct? For what is Solid or Cubic measure used? What are the denominations? Art. 2. REDUCTION. 3 ft. If each side of a cube measures....! an inch in length, it is called a cubic, or solid incA. If each side measures!i a yard in length, it is called a cubic, feet; that is, 3 feet in length 3 feet in width, and 3 feet in thickness. A solid foot contains 12 X 12 X 12 = 1728 solid inches. 3 ft. NOTE.-This can best be explained by a number of small cubical blocks, with which any teacher oan supply himself. TABLE. 1728 solid inches (s. in.)...make 1 solid foot,..-marked s. ft. 27 solid feet --- —- - 1 solid yard, c. " s. yd. 50 feet of round, or I ton, " T. 40 feet of hewn timber. 1 ton, 128 solid ft. 8 X 4 Xof wood or 4, that is 8 ft. long, 4. 1 cord of wood or ft. wide, aid 4 ft.high, REMARK 1.-A pile of wood I foot long, 4 feet wide, and 4 feet high, is called a cord foot, and contains 16 solid feet. 8 cord feet make 1 cord. 2.-In estimating the tonnage of ships, 42 solid feet are allowed for a ton. 3. —By a ton of round timber, is meant such a quantity of timber in its rough, or natural state, as when hewn will make 40 cubic feet. 4. —A cubic foot contains 7.48 wine gallons, and 6.127 beer gallons. 5.-A cubic foot of distilled water weighs about 1000 oz. Avoirdupois, or very nearly 62i lbs., at 40~ temperature. At 60~, which is generally used, it weighs only 62.353 Ibs., which is less than 1000 oz. The foot weighs 911.458 oz. Troy, or.5274 oz. per cubic inch. 1. Reduce 6 solid yards to 2. Reduce 279936solidinches solid inches. to solid yards. 3. Reduce 12 tons of round 4. In 1036800 solid inches, timber to solid inches. how many tons of round timber? What is a cube? A cubic inch? A cubic yard? How many solid feet does a solid yard contain? How many solid inches in a solid foot? Repeat the Table. What is a cord foot? How many solid feet does it contain? How many cord feet make a cord? How many solid feet are allowed for a tol in estimating the tonnage of vessels? What is meant by a ton of round timber? How many wine gallons does a cubic foot contain? fleer gallons? What is the weight of a cubic foot distilled water? How do:we find the contents of a solid, the sides of which are squares, rectangles, &c.? 166 COMMON ARITHMETIC. Sect. IX. 5. In 5 cords of wood, how 6. In 540 solid feet of wood, many cord feet? How many how many cord feet? How many solid feet? cords? 7. A ship contains 12600 solid 8. Required-the number of feet. Required-the tonnage. solid feet in a ship of 300 tons. From the remarks under Obs. 17, we conclude that To find the contents of a solid, the sides of which are squares, rectangles, &c.: * Obs. 20. Multiply together its length, breadth, and thickness, or depth. 9. How many solid feet in a room 18 feet long, 16 feet wide, and 12 feet high? Ans. 3456. 10. How many solid feet in a box 3 feet long, 2 feet wide, and 60 inches deep? Ans. 8. Reduce all to the same denomination. 11. How many cubic feet in a pile of wood 14 feet long, 4 feet wide, and 3' feet high? Ans. 196. 12. How many cords of wood in a pile 186 feet long, 6 feet wide, and 4' feet high? Ans. 39 cords, 17 cord feet. 5. TIME. Obs. 21. Time is the measure of duration. It is naturally divided into days and years, the former being caused by the revolution of the earth around its axis, and the latter by revolution around the sun. Other divisions have been made by man; the day being divided into hours, minutes, and seconds; and the year into months, weeks, and days. TABLE. 60 seconds (sec.) make 1 minute..... marked min. 60 minutes --- — - 1 hour. ---- - h. 24 hours-. 1 day- " d. 3651 days -" t year " yr. 100 years - " 1 century --- —---." cen. ALSO: 7 days- ----. makes 1 week marked wk. 4 weeks — -- 1 month " mo. 13 months, 1 day, and 6 hours, (nearly,) or 3654 days, makes 1 common or Julian year..How is Time naturally divided? What causes the day? The year? How b the day divided? The year? Repeat the Table. Art. 2. REDUCTION. 167 REMARK 1.-A Solar year is the exact time in which the earth revolves around the sun, and contains 365 d. 5 h. 48 min. 48 sec. 2. The 6 hours, or: of a day is not added to each year, but reserved until every fourth year, when a whole day is gained. Therefore, every fourth year must contain 366 days. This year is called Bissextile, or Leap Year. 3. Leap Years are those which can be divided by 4 without a remainder; as 1840, 1844, 1848, &c. 4. Exceptions.-It will be perceived that in counting 365I days to the year, we reckon 11 min. and 12 sec. too much. This in 100 years amounts to 18 hrs. 40 min.; dropping the 40 min., it being too small to be noticed, we fiud that each 100 years we count 18 hrs. or i of a day too much, and in 400 years we increase 3 days in time. Therefore, to produce the exact time, we count every fourth centurial year as leap year, and the intermediate centurial years as common years, (365days.) Hence-If the centurial years can be divided by 400 they are leap years, otherwise they are not. Examples.-1200, 1600,2000, &c., are leap years, but 1500,1700, 1800, &c. are not. 5. The year is also divided into 12 calendar months, the order of which, and the number of days in each, are as follows: January, (Jan.) 1st month has 31 days. February, (Feb.) 2d " " 28 " March, (Mar.) 3d " " 31 " April, (Apr.) 4th " 30 May, 5th " " 31 June, 6th " " 30 July, 7th " " 31 August, (Aug.) 8th " " 31 " September, (Sept.) 9th " " 30 October, (Oct.) 10th " " 31 " November, (Nov.) 11th " " 30 December, (Dec.) 12th " " 31 6. The odd day which is added to every fourth year, is added to the month of February, Therefore, every fourth year February has 29 days. 7. The number of days in each month may be easily remembered by committing to memory the following lines: Thirty days hath September, April, June, and November; All the rest have thirty-one, Save February alone; Which hath twenty-eight in store, Till Leap year gives it one day more. 1. How many days in 2 weeks? 5? 6? 4? 7? 3? 9? 8? 12? 2. How many weeks in 2 months? 4? 6? 3? 7? 5? 8? 9? 12? 3. How many minutes in 2 hours? 3? 6? 4? 7? 5? 8? 12? 9? What is a solar year? What is its length? What is done with the 6 hours or of a day? How many days has every fourth year? What is this year called? Which are leap years? Give examples. How is the year otherwise divided? Repeat the names of the months together, with the number of days in each. What is done retpectin the centurial years? To which month is the od day added? yow iiv.j a iysthen ha February il leap yearT Reo peat the lines for remembering the number of 4ua i qe1h minth? 168 COMMON ARITHMETIC. Sect. IX 4. How many weeks in 14 days? 29? 42? 35? 56? 84? 5. How many months in 8 weeks? 24? 48? 16? 40? 32? 44? 6. How many hours in 120 minutes? 300? 720? 360? 480? 7. Suppose a person's age to be 8. Reduce 586845817 seconds 18 yrs., 217 d., 16 h., 43 min.. to years. 37 sec, how many seconds old is he? 9. How many seconds in 3 10. Reduce 10382754780 seecenturies, 57 yrs., 7 mo., 3 wks. onds to centuries. 1 d., 18 h., 33 min.? 11. How many seconds be- 12. Reduce 2272147200 sectween July 4th., 1776, and July cnds to years. 4th., 1848, it being 72 years? 6. CIRCULAR MEASURE, OR MOTION. Obs. 22. This is used in estimating latitude and longitude, and also in measuring the motions of the heavenly bodies. All the calculations by this measure are made in circles, every circle, whether large or small, being supposed to be divided into 360 equal parts called degrees. Each degree is divided into 60 equal parts called minutes, and each minute into 60 equal parts called seconds. TABLE. 60seconds (") --- —-— make 1 minute ----- marked 60 minutes --- — " 1 degree 30 degrees- " 1 sign " s. 12 Signs, or 360~ --. " 1 circle c. REMARK 1.-The Sign is but little used, calculations by this measure being chiefly made in degrees, minutes, and seconds 2. The minute is the same as the geographical mile, 60 of which make a degree. 3. It must not be inferred, however, that 60 geographical, or. 691 satute miles (table long measure,) make a degree in every circle. This is only the case when the circumference of the circle is the same as the circumference of the earth. If the circle is larger or smaller, the distance is greater or less in proportion. For what is circular measure used? How are the calculations made? How is the circle divided? The degree? The minute? Repeat the Table. What denominations are used principally in calculating by this measure? To what does the minute correspond? How large must a circle be to have 6) geographical, or 69i statute miles make a degree? Art. 2. REDUCTION. 169 1. How many signs in 2 circles? 4? 6? 10? 8? 5? 7? 12? 2. How many degrees in 2 signs? 4? 5? 7? 9? 8? 12? 3. How many minutes in 2 degrees? 4? 8? 5? 7? 6? 4. How many circles in 24 signs? 48? 60? 36? 72? 144? 5. How many signs in 60 degrees? 120? 210? 180? 240? 6. How many degrees in 120 minutes? 180? 300? 540? 720? 7. Reduce 17~ 43' 57" to sec- 8. Reduce 63837" to degrees. onds. 9. Reduce 8s. 29~ 39" to sec- 10. Reduce 968439" to signs. onds. 11. Reduce lc. 355~ 59' 49" 12. Reduce 2577589" to cirto seconds. cles. MISCELLANEOUS TABLE. 12 things-.. make 1 dozen --- —.12 dozen -- 1 gross 12 gross — — "-_ 1 great gross-_. 20 things --—. —_. " 1 score. 56 pounds_ - — I 1 firkin of buttei 112pounds --- 1 quintalof fish 196 pounds " 1 barrel of flour 200 pounds - -. " 1 barrel ol pork, 14 pounds of iron or lead " 1 stone. 21~ stone --- — " 1 pig. 8 pigs ----------- 1 fother. 24 sheets of paper _- " 1 quire. 20quires -.. --- 1 ream. BOOKS. -- mark( '~, or beef. ed doz. gr. g. gr. fir. quin. A sheet folded in two leaves is A sheet " four " A sheet c" eight " A sheet " twelve " A sheet "' eighteen " called a folio. " a quarto, or 4 to. " an octavo, or 8 vo, " a duodecimo, or 12 mo. " an 18 mo. MISCELLANEOUS EXERCISES FOR THE SLATE, Involving the principles of Reduction. 1. In 15 bars of silver, each weighing 6 lbs. 8 oz. 15pwtt, 11 grs., how many grains? Ains. 58156. 2. How many thimbles, each weighing 8 pwts. 16 grs. can made from 1 lb. 8 oz. 16 pwts. of silver? Ans. 48. Repeat the Miscellaneous Table. What is a shedt folded;ta two leaves called? In four leave? In eight leaved? In 12 leavy7 In 18 bavt? 9 ]70 COMMON ARITHMETIC Sect. IX 3. How many rings, each weighing 5 pwts, 7 grs., can be made from 2 lbs. 8 oz. 16 pwts. 4 grs. of gold? Ans. 124. 4. How many pounds of hay in 6 loads, each weighing 14 cwt. 2 grs. 18 lbs.? Ans. 8808. 5. How many nails, each weighing 6 drams, can be made from 29 lbs. 3 oz. 4 drs. o? iron? Ans. 1246. 6. How many casks, each holding 68 lbs., can be filled from 16 cwt. 1 gr. 7 lbs of sugar? Ans. 24. 7. If a family consume 4 lbs. 12 oz of coffee in a month, how long will 1 cwt. 3 qrs. 9 lbs. 8 oz. last them? Ans. 38w. months. 8. In 45 pounds avoirdupois, how many pounds troy? Ans. 541. 9. In 36 pounds troy, how many pounds avoirdupois? Ans. 29. 9 10. A physician made 136pills, each containing 4 grains, how much did they all contain? Ans, 13, 13, 4 grs. 11. How many doses of calomel each containing 8 grs,, can be made from lib, 23. 33. 29, 12 grs. Ans, 869. 12. What quantity of rhubarb will it take to put up 324 doses, each containg 24 grs. Ans, lib. 43. 13, 1, 16 grs, 13. A merchant wishes to put 675 bushels of clover seed into casks, each holding 6 bushels, 3 pecks. How many casks are required? Ans, 100, 14. At I of a dollar a peck, what would 18 bu. 1 pk. of wheat cost? Ans, 18- dollars. 15. How many bottles will it take, each holding I a pint, (2 gills,) to put up a hogshead of wine? Ans, 1008, 16, A certain cistern holds 66 barrels of water, how many times will it fill a pail holding 3 galls. 1 qt. 1 pt. Ans. 616 times, 17. How many casks, each holding 7 galls. 3 qts. 1 pt., can be filled from a tun of wine? Ans. 32, 18. How many dozen bottles, each dozen containing 3 gal. 3 qts, Ipt. 2 gi., can be filled from a hogshead of cider? Ans, 16. 19. How many gallon, quart, pint, and gill bottles, of each an equal number, can be filled from 210 galls, 3 qts. 1 pt. 2 gi. of wine? Ans. 150. 20. How many pints in a hogshead of ale? Ans. 432. 21. How many gallon, quart, and pint bottles will it take to put up a barrel of beer, having the same quantity in each kind of bottles? Ains. 12 gallon bottles, 48 quart bottles, and 96 pint bottles. 22. How many casks, each holding 6 glas, 3 qts. 1 pt., will it take tah0old 19 hhds,, 19 galls of ale?.. Ana. 1f2. Art. 2. REDUCTION. 171 23. If a man drink 1 pint of beer each day for 24 years, how much would it amount to? Ans. 20 hhds., 15 galls., 3 qts. 24. How many quarters in 12 pieces of cloth, each containing 15 yds., 2 qrs. Ans. 744 = 186 yds. 25. How many pieces of cloth each containing 12 yds. 3 qrs. 2 na., are there in 579 yds., 1 qr., 2 na? Ans. 45. 26. How many suits of clothes, each containing 4 yds. 1 qr., 3 na., can be made; from 53 yds., 1 qr. of cloth? Ans. 12. 27. How many pair of pantaloons, each conta'ning 1 yd. 2 qrs. 3 na., can be made from 18 Flemish ells of broadcloath? Ans. 8. 28. How many times will a carriage wheel 16 ft. 6 in. in circumference, turn over in going 18 miles? Ans. 57b0. 29 How many steps of 2 ft. 5 in. each, will it a man take in traveling 1 m. 7 fur. 33 rods. 3 yds. 1 ft.? Ans. 4326. 30. If a man travel 20 m. 6 fur. 26 rds. 11 ft. in a day, how long will it take him to travel around the earth, it being about 25000 miles? Ar.s. 1200 days = 3yrs., 104d., 6 hrs. 31. How many times will a ship. 133 ft. 4 in. long, sail her length in crossing the Atlantic ocean, it being 3000 miles. Ans. 118800. 32. How many yards of carpeting I yard wide, will it take to cover the floor of a room 16 feet long, and 12 feet wide? Ans. 21-. 33. A land owner has in one place 329 acres, in another place 870 acres, in another place 1236 acres, and in another place 1405 acres; if it was all together, how many square miles would it make? Ans. 6. 34. A farmer wishing to know the length of his farm, found it to be 146 chains, 662 links. Required, its length in rods? In miles? -Ans. 586- rods = 1- miles. 35. The surface of the planet Venus contains about 9327776422 -03545600 square inches; how many square miles? Ans. 232352736. 36. How many blocks 3 inches long, and 2 inches wide, can be cut from a board 3 feet In length, and 2 feet in width? Ans. 144. 37. A piece of land is 144 rods long, and 95 rods wide; how many acres does it contain? Ans. 85-. 38. A.'s farm is 66 chains, 50 links in length, and 45 chains, 75 links in width; required the number of acres owned by A.? Ans. 304'. 3U. If a room is. 1i fuet log, andu 12 fctwide, how many square oet doesit contain? -... -ans 19a. 172 COlMMON ARITHMETIC. Sect. IX 40. A board is 14 feet long, and 16 inches wide at one end, and 8 inches wide at the other end; required the number of square feet in the board? Ans. 14. The width at one end is 16 inches, and at the other end 8 inches; therefore the widll at both ends is 16 - 8 = 24, and the half of this or 12 inches, must be the mean or average widtth. 12- 12 1 foo,. 14 X 1 = 14 Ans. HenceTo find the mean width of a board, or any other surface whose edg;es are straight, but the width at the ends are different: Obs. 23. Addl togeher the width of the twuo ends, and take half their sum. 41. If a board is 21 feet lonrg, and 32 inches wide at one end, and 18 inches wide at the other end, how many square feet in the board? Ans. 5C-. 42. How many square feet in the surface of a rafter 19i feet long, 3 inches thick, and the width at the ends 4 and 3 inches? Ans. 21 sq. ft., 13 sq. in. 43. There is a room 18 feet in length, 16 feet in width, and 8 feet in heighth; how many rolls of paper, 2 feet wide, and containing 11 yards in each roll will it take to cover the walls. Ans. 833. 44. How many blocks, 2 inches long, 2 inches wide, and 2 inches thick, can be cut from a cubical blo:k measuring 3 feet on each side, allowing no waste for cutting? Ans. 5832. 45. How many cords of wood in a pile 56 feet long, 12 feet wide, and 8 feet high? Ans. 42. 46. A cubic post of oak weighs 950 ounces avoirdupois; how many pounds will 6 cords of oak weigh? Ans. 45600. 47. If a clock tick 60 times a minute, how many times will it tick in a year? Ans. 31557600. 48. How many seconds longer was March than February in the year 1847. Ans. 259200. 49. What are the leap years between 1846 and 1862? Ans. 1848, 1852, 1856, 1860, 50. If a person wastes 3 of an hour each day, how much time will he lose in 28 years. Ans. 319 d. 14 h. 15 min. 51. How much time will a person gain in 50 years by rising 2 an hour (30 min.) earlier than usual, each day? Ans. 1 yr. 15 d. 5 h. 15 min. 52. Sound moves at the rate of 1142 feet in a second of time; How do we find the number of square feet in a board, or any other surface' whosm edges are strait, but the width of the ends different? Art. 2. REDUCTION. 173 required the time it would take sound to pas.; from the sun to the planet Uranus, it being distant 183u00O00U miles? Ans. 26 3 yrs. 261 d. i l hr. 48 min.+ NoTE.-The learner will observe that the remainder after dividing by 3654 is fourths of days, not whvle days. 53. If a person spends 8 hours of the day in sleeping and eating, how many minutes per day does he devote to other business. Ans. 960. 54. If a vessel sails 10 miles an hour, how far will she sail in 4 wks. 5 d. 16 hrs. Ans. 833J miles, 55. The precise length of the tropical year is 365 d. 5 h. 48 min, 48 sec.; how many such years in 1104492480 seconds? Ans. 35. 56. How many seconds from the commencement of the Christian Era, until Oct 7th, 1849, 4 o'clock, P. M., allowance being mode for leap year? Ans. 58341398400. 57. If a planet move through an area of 3~ 36' in one day, how long will it take it to move through an area of 9s. 1600? Ans. 79~- days. 58. How long, would it take a planet to move through a quadrant (900) at the rate of 59' 8" per minute? Ans. 1 hr., 31l- min. 59. How long would it take a comet to move through a semicircle, (180~) at the rate of 8~ 16' daily? Ans. 21- days. 60. What cost 3 dozen dozen yards of cloth, at 2 dollars a yard? Ans. 864 dollars. 61. What would 6 gross of buttons cost, at 6 cents per dozen? Ans. 432 cents. 62. What would 3 great gross of combs cost, at 3 cents apiece? Ans. 15552 cents. 63. What would 4 score of hogs cost, at 2 dollars apiece? Ans. 160 dollars. 64. A merchant purchased 8 fothers of iron; how many pounds was this? Ans. 19264. 6 5. What cost 2 reams of paper, at 1l cents per sheet? Ans. 1440 cents. 66. A farmer started to market with 12 dozen dozen eggs; he broke a half a dozen dozen by the way, and then returned and got 6 dozen dozen more; how many did he take to market at last? Ans. 2520. 174 COMMON ARITHMETIC. Sect. IX ARTICLE 3. FRACTIONS OF COMPOU'D NUMBERS. CASE 1. To reduce a fraction of a higher, to a fraction of a lower denomination, and fractions of lower to fractions of higher denominations. Ex. 1. Reduce 4-J of a bushel to the fraction of a pint. Solution.-We learn from Art. 2, Obs. 3, that we reduce integers from larger to small denominations by multiplication. Fractions are reduced in the same manner as whole numbers.Thus: — X 4 =1; 11 X 8; 3 2 4 Alls. Or, by cancelation: Sect. VILI, Art. 5, Obs. 2 and 3. 3 4=- Ans. 3 4 —1 We multiply by 4, 8 and 2, because 4 pecks make a bushel, &c. 3. Reduce - of a bushel to the fraction of a pint. 5. Reduce 3 012 of a hogshead, wine measure, to the fraction of a gill. 7. Reduce -To of a barrel, wine measure, to the fraction of a pint. 9. Reduce -3 of a barrel, beer measure, to the fraction of a pint. 11. Reduce 2, of a French ell to the fraction of a nail. 2. Reduce 4 of a pint to the fraction of a bushel. Solution.-Integers are reduced from a lower to a higher denomination by division. (Art. 2, Obs. 3). Fractions are reduced in the same manner as whole numbers. Thus- -~ 2; 3 — 8= I ~- 1' 4=J Ans. Or, by cancclation: Sect. VIII, Art. 6, Obs. 2. 3 4 2 8 4 48 l=8Ans. We divide by 2, 8, and 4, because 2 pints make a quart, 8 quarts make a peck, &c. 4. Reduce 8 of a pint to the fraction of a bushel. 6. Reduce - of a gill to the fraction of a hogshead. 8. Reduce I a pint to the fraction of a barrel, wine measure. 10. Reduce of a pint, beer measure, to the fraction of a barrel. 12. Reduce ' of a nail to the fraction of a French ell. / How do we reduce fractions of higher, to fractions of lower denominations? How do we reduce fractions of lower, to fractions of higher denominations Art. 3. REDUCTION. 175 13. A cucumber grew 427 of a mile in len:rth what fraction was that of a foot? 15. Reduce ' 7 of an acre to the fraction of a foot. 17. Reduce — 6' of a solid yard to the fraction of a solid inch. 19. Reduce '^ of a p und Troy, to the fraction of a p:.mnywei:-ht. 21. Reduce 3'3- of a ton to the fraction of an ounce. 23. Ritdd.ilc- 3-of a pound to the frtction of a dram. 25. Reduce y3, of a pound to the fr;ction of a scruple. 27. Reduce ' of a day to the fraction of a minute. 29. Reduce.I of a circle to the fraction of a minute. 14. A cucumber gr"w 7 of a foot in ienl:; what fraction is that o - i m? 16. Reduce 112 of a foot to the fraction of an acr. 18. Reduce 46 of a solid inch to tle fractioa of a solid yari. 20. Reduce s of a pennyweight to the fraction of a pound. 22. Reduce "o of an ounce to the fraction of a ton. 24. Reduce -2 of a dram to the fracti:);i ot a, ound. 26. Rcdu: y27 of a scruple to the fraft.ion of a pound. 28. Reduce I of a minute to the fraction of a day. 30. Reduce - of a minute to the fiaction of a circle. CASE 2. To reduce fractions of compound numbers to integers of the same, and also to reduce the integers back to fractions. Ex. 1. Reduce ~ a bushel to pecks. Solution.-As 4 pecks make a bushel, it is evident that ~ a bushel contains 2 X 4 = 2 pecks, Ans. 3. Reduce 3 of a pint to gills. 5. Reduce t of a peck to quarts. 7. Reduce ' of a bushel to pecks and quarts. Solution.-Multiply 5 by 4, we obtain 2 -= 2 pecks; multiplying 4 by 8, we obtain 3 4 qts. Ans. 2 pks. 4 qts. 2. What part of a bushel is 2 pecks? Solution.-As 4 pecks make a bushel, it is evident that 2 pecks are 2 or a a bushel. 4. What part of a pint is 3 gills? 6. What part (f a peck is 2 quarts? 8. Reduce 2 pks. 4 qts. to the fraction of a bushel. Solution.-We first reduce the 4 qts. to the fraction of a peck by dividing by 8. (4- 8= -.) We then have 2. = '- pecks, which How do we reduce fractions of compound numbers to integers of the same? Explain the operation of Ex. 7 and 8, Case 2. 176 COMMON ARITHMETIC. Sect. IX This operation may be shown thus5 numerator. 4 pecks in a bushel. denom. 8)20 pks. 2+4 8 qts. in a peck. denom. 8)32 4 qts. Or, the following form may be adopted if preferred by the student. denom. 5X4-=2 pks. & rem. I 1X8= 4 qts. We cancel 4 into 8, leaving 2; dividing 5 by 2 we obtain 2 pks. and 1 remainder, which we place below the 5. We multiply 1 by 8, which cancels against the 2 leaving 4, and 1 X4=4. Ans. 2 pks. 4 qts. It will be perceived that the number at the left is not canceled unless it divides the numbers at the right without a remainder, although we divide the numbers at the right by the number at the left. This process is the same as in Art. 2, General Rule for Reduction Descending, except that we divide by the denominator after each multiplication. 9. Reduce x of a hhd., wine measure, to its proper value. we reduce to the fraction of a bushel by dividing by 4. (2 4 = -.) This gives - of a bushel as our answer. The opelat-on may be shown, thus014 4-I1 -- 8 — 2 4 21 - 5 of a bushel Ans. The reasoning is the same as above, HenceTo reduce integers of compound numbers to fractions of the same, we have this RULE.-I. Divide the lowest denomination ly that number which it takes of this to make a un.t of the next higher denomination, and prefix to the result the number given of this higher denomination. II.-Divide this in the same manner as the first; so continue to do until it is reduced to the fraction re ui d REMARK.-If any denomination is wanting, place a cipher in its stead. 10. Reduce 18 galls. 3 qts. 1 p:s. to the fraction of a hogs head. 11. Reduce 2 of a bl., beer 12. Reduce 14 galls. 1 qt. 1V measure, to its proper value. pts. to the fraction of a barrel. 13. What is the value of 3- of 14. Reduce 2 qrs. 2|- na. to a yard, cloth measure? the fraction of a yard. How do we proceed in canceling? To what does this process correspond? How do we reduce integers of compound numbers to fractions of the same? Art. 3. FRACTIOWS OF COMPOONb NYUMnRS, 177 15. What is the value of 4 of a mile? 17. What is the value of of tn acre? 19. What is the value of - of a cord? 21. What is the va'ue of " of. pound Troy? 23. What is the value of - of a ton? 25. What is the value of,5,ofa pound, Apothecaries weight'? 27. I of a month is how many days, hours, &c. 29. What is the value of ~ of a circle? CASE 3. To reduce the decimal of a compound number to its proper value. Ex. 1. Reduce:625 of a bu.shel to its proper value; that is, to pecks and quarts. Operation..625 4 pis. 2.500 8 16. Reduce 4 fur. 22 rds. 4 yds, 2 ft. 14 in. to the fraction of a mihe. 18. Reduce 3 R. 20 sq. rds. to the fraction of an acre. 20. Reduce 85 sq. ft. 576 sq. in. to the fraction of a cord. 22. Reduce 6 oz. 13 pwt. 8 grs. to the fraction of a pound. 24. Reduce 9 cwt. 23 lbs. 1 oz. 3, - drs. to the fraction of a ton. 26. Reduce 55. 33. 1D. 188TR grs to the fraction of a pound. 28. 3 wks. 1 d. 9 hrs. 36 min. is what fraction of a month? 30. Reduce 102~ 51' 254" to the fraction of a circle, CAsa 4.-To chanre numbers of different denominations o a decimal of one denomination. 2. Reduce 2 pks. 4 qts. to the decimal of a bushel. Operation. 8t4 0 4 2.500 Ans..3.25 qts. 4.000 Ans. 2pks. 4 qts.. 625= —6o. (Sect. VIII, Art. As we have before said, (Sect. 7, Obs. 2.) Reducing this as VIII, Art. 7, Obs. 9) a whole in Case 2, Ex. 7, we obtain 2 pks. number can be reduced to a deci4 qts. as our answer. But since mal by annexing ciphers. Hence, t denominator of a decimal 4 qts.=-4.0 qts., and this divided tion is 1 with ciphers annexed by 8, (because 8 qts., make 1 (Sect. VIII., Art. 7, Obs. 2,) peck,)gives. 5 of peck. We we may dispense with the divis- then have 2.5 pks. which divi If any denomination is wanting, how.do we proceed? How doweTeduce the decimal of 4 compound utumbler tt integers? How.do we rcduce the it'ger a^ack tb debimale? gs 178 COMMON ARITHMETIC. Sect. IX ion, and point off as in multipli- ded by 4, (because 4 pks. make cation of decimals. (Sect. VIII, 1 bushel,) gives. 625 of a bushArt. 11. Rule.) el as our answer. HenceREMARK.-The learner will perceige To change numbers of differthat the numbers at tte left of the deci- ent denominations to a decimal of rnal point are not multiplied, they not one denonmination, we have this being decimals, but whole numbers.- RULE I.Begin with the leas liH -i i enee denomination: annex ciphers to the To changre a decimal of a com- T c right and divide it by that number pound number to its proper value. h take of this to make a which it takes of thi s to make a we have this we have this. unit of the nert higher, and point RULE I.- Multiply the given RUJLE I. - Muti.ply the g, a decimal for every cipher andecimal by that number which it neoed. takes of the next less denomination. o al ths 11. To the decimal thus obtained tomake 1 ofthis hiqher, and point, to ie 1 oftiS lcher, and point prefix the number given (if' any,) offas in Multiplication, of Deciof this higher denomination, and mals. Sect. VIII, Art. 11, divide as before, pointing (fas in tule.) zDivision of Decimals. ( Sect. I. Proceed tnthe same manner, l r - t r Il.Proceed in the same manner, VIII, Art. 12, Rule,and Reinm. 3.) nmultdplying Ike decimals only, of So continue to do until it is reduced each succeeding product, until no the den tion requied. to the denomination required. decimals remain, or it is reduced to thelowest denomination of thekind STOE 1.-If any denominatio is to whbch it belongs. The numbers wanting, a cipher may be put in its place at the left of the decimals. at the left of the dferent separating 2. Ciphers mlust be prefixed to the points, will form the required an- several quotients, (not to the integral swer. parts,) whei it is neceasary, to conmNOTE.-If after any multiplication plte te decimal plces. there is no whole nainber at the left of the separating point, a cipher may be put in its stead. 3. Reduce. 75 of a bushel to 4. Redttce 3 pks. to the deciits proper value. mal of a bushel. 5. Reduce. 25 of a hhd., wina 6. Reduce 15 g^sa. 3 qts. to measure to its proper value. the decimal of a hhd., wine measuI'e. 7. Reduce. 8 of a gallon to 8. Recdude 3 qts.. 6 gi. to its proper value. the decimal of a gallon. 9, Reduce. 5625 of ayard to 10. Reduce 2 qrs. 1 na. to he quarters and nails. decimal of a yard. 11 Reduce.8125 of a miTe 12. Re&uce 6 fur. 20 rds. to to furlongs nd rods. the decimal of a mile. 132. Reduce.5 of a yard to 14. Reduce 1 ft. 6 in. to the feet and inches. decimal of a yard. 15- Rprduio. 9375 of an acre p 16;. educe 3 R. 30 so.rds. to. propai~dFer quanfi~tty. the dermal of an ac-e. Art. 4. 179 ADDITION OF COMPOUND NUMBERS. 17. Reduce. 73125 of a pound 18. Reduce 8oz., 15pwts., 1 Troy, to its proper value. grs. to the decimal of a pound. 19. Reduce. 7896875 of a 20. Reduce 15 cwt. 3 grs. 4 ton t.) its proper value. Ibs. 6 oz., to the decimal ot a ton. 21. Reduce.3984375 of a 22. Reduce 43. 63. 15 grs. to pound Apothecaries' weight, to the decimalof a pound. its proper value. 23. Change. 5625 of a year 24. Reduce 6 mos. 3 wks.! to to months and weeks. the decimal of a year. REMARatR.-There are two othercases of Reduction of Decimals of Compound numbers but they are not deemed of sufficient importance to present them in this work. We give below, an example of each, with their solution, by the aid of which the observing pupil can perceive their application and use. 1. Reduce. 5 of a pint to the 2. Reduce.0625 of a gallon decimal of a gallon. to the decimal of a pint. Operation. Operation. 2 50..0625 4 4.2500 I - --. 2500 Ans..0625. 2 We annex ciphers lo our deci- - mal, which does not alter its val- Ans.. 500 te, (Sect. VIII, Art. 1,Obs7. b.) We multiply by 4 and 2, beand divide by,2 and 4 because 2 cause 4 qts. make a gallon, &c., pts. make a quart, &c. We point and point off as in Multiplication off as in Division of Decimals. — of Decimals. (Sect. VIII, Art, (Sect, VIII, Art. 12. Rule.) 1l, Rule.) ARTICLE 4. ADDITION OF COMPOUIND NUMBERS, Ex, 1. There are two bushels of wheat in one bag, one bushel in another, and 2 bu., 3 pks. in another. How much in all? Ans. 5 bu. 3 pks. 2. Thomas has I lb. 4 oz. of raisins, James has 2 lbs. 6 oi., and lliam has 2 lbs. 5 oz. How many pounds have they all? Ans. 5 lbs. 15 oz. 3. A man sold at one time 18 bu. 3 pks. 6 qts.of oats, at another time 23 bu. 2 pks. 4 qts., and at another time 12 bu. I pk. 7 qts. Hlow much did befell ina Jl? COMMON ARITHMlt'rle. Sect. IX Operation, We first write down the numbers, placing bu. pks. qts. those of the same kind under each other.18, 3 - 6 We commence at the right hand, or lowest 23 __ 2 - 4 denomination, and add it as in simple num12 _ 1 7 bers; 7 4 + 6 = 17qts. = 2pks. lqt.;. —... - setting down the 1 qt., we carry the 2 pks. Ans.55 - 0 1 to the next column; 1+2+-3 -6 pks. and 2 pks to carry make 8 pks.; 8 pks = 2 bu. and no remainder; setting down a cipher in the place of pecks, (because we had no pecks remaining,) we carry the 2 bu. to the column of bushels, and add as in simple numbers, setting down the entire sum. From this illustration we derive the following RULE FOR THE ADDITION OF COMPOUND NUMBERS. I. Set down the numbers, placing those of the same denomination unier each other. (Sect. It, Art. 2, Obs. 4.) II. Begin with the lowest denomination; add together the numbers given of this denomination, as in Addition of Simple numbers, and divide their sum by that number which it takes of this denomination to make a unit of the next higher; place the remainder under the column added, and carry the quotient to the next denomination. III. Add the remaining denominations in the same mnanner as the first. PROOF.-2he same as in Addition of Simple numbers, (Sect. It, Art. 2 Obs. 6.) REMARK 1.-If any denomination is wanting, supply its place with a cipher. 2. The principles of Addition of compound and simple numbers are the same. (Art. 1, Obs. 6.) This can be shown by the following example: Add together 2436, 1589, 1612, and 2849. Operation.. 9 -+ 2 + 9 - 6 = 26; 26 - 10 = 2 thou. hund. tens. units. and 6 remainder. 4-+ 1 + 8 + 3 = 2 _4 _3 6 16 and 2 to carry = 18; 18-10 = 1 1 __ 5 __ 8 -- 9 and 8rem. 8+6+5 +4=23and 1 6 1 2 1 to carry = 24, 24- 10 = 2 and 4 2 8 _ 4 9'. rem. 2+ 1+ t 1+-I- 6 and o -------- carry =8. Ans. 8. 4 _ 8 6 What is the rule for addition of compound numbets? What is the proof? If any denomination is wanting, how do we precid?-' What is the diffoieuoe between the principles of addition of simple and compound numbers? Art. 4. ADDIT[ON OF COMP6UNi) NUMBElS. 11 It will be perceived that we divide, &c., the same as in the above example, but as the remainder is always the right hand figure, this formal process of dividing may be dispensed with. 4. A merchant bought at one time 36 bu. 3 pks. 6 qts of wheat; at another time 25 bu. 2 pks. 3 qts.; at another tine 30 blu. 1 pk. 7 qts.; at another time 28 bu. 4 qts.; and at another time 17bu. 2 pks.; how many did he buy in all? Ans. 118 bu. 2 pks. 4 qts. |. Bought 5 loads of oats; the first contained 27 bu. 8 qts.; the second, 37 bu.; the third, 263 bu. 6' qts.; the fourth, 36'- bit. 1 pk. 6 qts.; and the fifth, 35 bu. 37 pks.; how many bushels in all? Ans. 164 btl. 32 qts. 6. A wine merchant has 1 hhd. 27 galls. 3 qts. of Madeira; 2 hhd. 30 galls. 31 qts of Port; 61,- galls. Champaign, and3 hhds. 12- bls. of Claret wine; besides these he has 217 hhds. of Brandy; and 1 hhd. 24 galls. 23 qts. of other liquors. How much has he in all? Ans. 12 hhds. 1 bl. 31 galls. 2 qts. 1 pt. 33 gi. NOTE.-The learner will bear in mind that the remainder, after dividing by 31i is half gallots, not whole gallons. 7. A wine merchant sold at one time 45 galls. 3 qts. 1 pt. 3 gi. of wine; at another time, 36 galls. 2 qts. 1 gi.; at another time 49 galls 1 pt.; at another time 57 galls. 3 gi.; and at another time 38 galls.; how much did he sell in all? Ans. 3 hhds. 37 galls. 2 qts. 1 pt. 3 gi., 8. A brewer sold S casks of beer: the first contained 33 galls. 2 qts. 1 pt.; the second 28 galls. 1 pt.; the thir,, 34- galls.; the fourth, 26 galls. 31 qts.; the fifth, 18 galls. 2 qts.; how much was there in all? Ans. 2 hhds. 33 galls. 3 qts. I pt. 9. Bought 4 pieces of cloth; the first contained 22 yds. 3 qrs. 2 na.; the second, 214 yds.; thi, yds. 2- qrs.; the fourth, 33 yds. 2 qrs. 21 na.; how many yards in all? Ans. 97. 10. If it takes 3 yd. qrs. 3 q of cloth to make a coat, 2 yds. 2 qrs. 3 na. to make a pair of pants, and 24 qrs. to make a vest how many yards will it take for the whole? Ans. 7 yds. 1 qr. 1 na. 11. Thomas walked due north 43 miles, 7 fur. 29 rds. 12 ft., and James walked due south 56 miles, 5 fur. 34 rds. 14 ft.;how far apatt arrley?' Ans. 100 M. 5 fur. 24rds. 91 ft. A gentleman traveling, rode the first week 167 M. 7 fur. 38 rds.; the second week, 180 M. 6 fur. 27 rds.; the third week, 173 M. 5 fur. 36 rds.; and the fourth week 77 M. 3 fur. 19 rds.; how many miles did.e travel? Ans. 600. tiw by the exampte give t why this Is the ose? Why do we not divide' i this manner in additi6h dt' simple numb'ers? 182 COMMON ARMiHMREVC. Sect. IX 13. A man has 4 farms; the first contains 120 A. 3 R. 27 sq, rds.; the second, 112 A, 2- R.; the third, 944 A.; and the fourth, 87A' A.; how many acres does he own in all? Ans. 410 A. 1 R. 9 sq. rds. or perches. 14. There are 6 piles of wood; the first contains 3 C. 64 sq. ft. 1126 s. in.; lhe second, 4 C. 112 s. ft. 1692 s. in.; the third, 5 C. 127 s. ft.. 1724 s. in.; the fourth, 52- C.; the fifth, 6 C. 109k sq. ft.; and the sixth contains 3 C. 123 s. ft. 498 s. in,; how many cords in the whole? ' Ans. 30 C. 7 c. ft. 15. A silversmith purchased 4 bars of silver: the first weighed 2 lbs., 9 ot. 15 pwts. 22 grs; the second, 1a lbs.; the third, 3 Ibs. 1. oz.; and the fourth, 2 lbs. 11 oz. 9 grs.; how much did they all weigh? Ans. 10 lbs. 8 oz. 4 pwts. 23 grs. 16. Sold atone time 1 cwt. 2 qrs. 23 lbs. 14- oz. of sugar; at another time 2 ~ cwt.; at another time 2 cwt. 1 qr. 16- lbs.; and at another time 2 cwt. 3 qrs. 19 lbs. 13 oz. 8 drs.; how much was sold in all? Ans. 9 cwt. 3 qrs. 23 lbs. 17. Bought 4stacks of hay; the first contained 3,T. 19 cwt. 2 qrs. 15 lbs.; the second, 2 T. 14 cwt. 34 qrs.; the third, 3- T.; and the fourth, 4 T. 191 cwt., how many tons in all? Ans. 15. 18. An Apothecary mixed one compound with 6 simples: the first weighed 1 lb. 2 3. 43. 1 3.; the second weighed 9 3. 7 3. 2 B. 16 grs.; the third, 6 5. 53. 19 grs.; the fourth, 3 3. 15 grs., and the fifth, 7 3. 1 B.; what was the weight of the whole? Ans. 2I, 1I 3. 1 3, 1 0 grs. 19. A gentleman attended school until he was 21 yrs. 212 da. of age; he then traveled 15 yrs. 84 da.; he then married and lived with his wife 14 yrs. 176 da., when his wife died; after this he lived 23 yrs. 219 da.; what was his age when he died? Ans. 75 yrs. 240 da. 12 hrs. 20. Upon what day of the year does Christmas come in leap year? It,~~ ~Ansi the 360th. 21. On what day of a common year does the fourth oi July happen? Ans. the 185th. 22. Boston is 71~ 3' west longitude from Greenwich, Wasington 60 40' farther west, Cincinnati 6~ 44' still farther, and th tern line of Vancouvers' Island 380 33' still farther west. 'equired the longitude of Vancouvers' Island from Greenwich? Ans. 123~. 23. Bought 4 casks of butter: the first contained 1 fir. 29 lbs.; 'the second, 2 fir. 43 lbs.; t de third, 1ik*'; nd the fourth, 27 r.; how muth was the in allf' ' ". 8 #r. lt A rt. 5. UlV'TRACTION OF COMPOUND NUMBEfts. 24. Bought 6 boxes of eggs: the first contained 2 dozen dozen; the second. 7 dozen dozen; the third, 12 dozen dozen; the fourth, half a dozen dozen; the fifth, 1 great gross; and the sixth, 12 score. How many eggs in all, and how many dozen? Ans. 5064 egs = 4= 2 dozen. 25. Bought 1 ream, 18 quire, 9 sheets of paper at one time; 2 reams, 10 sheets at anotier time; 1l r ams at another time; 25 reams at another time; and 1 ream, 8 quire, 13 sheets at another time; how much did I buy in all? Ans. 10 reams. 26. To 3 of a bushel, add - a peck. Ans. 3 pks. 4 qts. 27. To 7 of a hhd., wine measure, add - of a barrel, Ans. 1 hhd. 15 galls. 3 qts. NoTE.-Reduce both to thoir value in integers, and then add them. 28. To 5- of a mile, add 2 of a rod. Ans. 6 fur. 27 rds. 5 ft. 0 in. 29. To 17 of a cwt., add 3 of a pound. Ans. 3 qr. 13 lbs. 4 oz. 30. To, a year, add | of a day. Ans. 183 da. 7 hr. ARTICLE 5. SUBTRACTION OF COMPOUND NUMBERS, Ex. 1. There vrere 2 bu, 3 pks of oats in a box, and 1 bu, 2 pks. were taken out; how much was left? Ans. 1 bu. I pk. 2. From a box containing 8 bu. 2 pks. 6 qts., 5 bu. 1 pk, 2 qts. were taken, how muc-h remained? Ans. 3 bu. 1 pk. 4 qts. 3. A man having 64 bu. I pk. 3 qts, of wheat, sold 45 bu. 3 pks. 6 qts.; how much had he lefti Ans. 18 bu. 1 pk. 5 qts. Operation. We cannot take 6 qts. from 3 qts., but we bu. pks. qts. can borrow 1 pk. (8 qts.) from the next or64 -- 1 -- 3 der; then 8 qts. -- 3 qts. - 11 qts., 11 qts. 45 - 3 - 6 -6 qts.- 5 qts.; we set down the 5 qts,, and as we borrowed 1 pk. we must carry Ans. 18 I-1. 5 1 to the column of pecks, to make up for that we borrowed I +3 == 4 pks.; we cannot take 4 pks. from 1 pk., but we can borrow 1 bu. (4 pk,.) from the column of bushels: 4 pks. + 1 pk. 5 pks.-4 pks.- 1 pk. 'We set down the I pk,, and carry 1 (to pay for that we borrowed) to 5 making 6; 6 from 14 =,, &c. This process may also be shown, thus- bu. pks. qts. bu. pks. qts. This operation is so sitnple 64- 1 3 =3 63- 4 _11 that it requires no explana45 _ 3.- 6 tion. AnM. S. 1.. I 184 COMMON ARIYII MEt1C. S'ect. IX From these illustrations we derive the following RULE FOR THE SUBTRACTION OF COMPOUND NUMBERS. I. Write the numbers, the tess under the greater, placing those of the same denomination under each other. II. Commence with the lowest denomination: take successively each denomination from the one above it, placing the result below as in simple numbers. (Sect. III, General Rule.) III. If the lower number should be the laest, add to the upper number a unit of the next higher denomination, after which, subtract as usual, and carry 1 to the next tenomination for that which was borrowed. PRooF.-The same as in Subtraction of Simple,Yumbers. (Sect, III, Art. 2, Obs. 7.) REMARK 1. If aly denomination is wanting, supply its place with a cipher. q. The principles of subtraction of simple and compound numbers are the same. (Art, 1, Obs. 6.) This may be shown by the following example: From 356 take 289. 1st, Operation. 2d Operation, hund. tens. units. hunds. tens, units. 3 _6 5 _ _ 6 356 =2 -- 14. 16 2 -_ 8 -- 9 a_2 8. 9 6 7 6 7 = 67. It will be perceived that the principle is the same in this example as in the one above. 4. A farmer raised 1000 bushels of wheat, and sold 679 bu. 3 pks. 6 qts.; how much had he left? Ans. 320 bu. 2 qts. 5. A wine merchant sold from a hogshead of Madeira, 35 galls. 3 qts. I pt. 2 gi.; he then put in 28 galls. 1 qt. 3 gi., and afterwards sold 43 galls. 3 gi.; how much had he left? Ans. 12 galls. 1 qt. 2 gi. 6. A wine merchant drew 4 bls. I qt. 1 gi. of wine from a cask containing 3 bls. 31 galls. 3 qts. 1 pt. 3 gi. Required-the quantity left in the cask. Ans. 1 pt. 2 gi. i: 7. From 1. hhds. of Port wine was taken 543 galls. How much was left? Ans. 31 galls, 3 qts. 1 pt. What is the rule for the subtraction of compound numbers? What is the proof? If any denomination is wauting, how do we proceed? What is the difference between the plriciples of subtraction of simple and compound num-:betrs? Explain why this is the case. Art. 5. SUBTRACTION OF COMPOUND NUMBERS. 185 8. From 2' bbls. of beer, was taken 34- galls. How much remained? Ans, 1 bl. 19 gals. 1 pt. 9. From apiece of cloth containing 18- yds., was made a coat containing 2 yds. 1 qr. 3 na., a vest containing 2' qrs., and a pair of pants containing 17 yds.; how much was left? Ans. 13 yds. 3 qrs. 1 na. 10, A traveled 140 miles: B followed after him 76 M. 7 fur. 35 rds.; how far apart were they? Ans. 63 M. 5 rds, 11. It is 24 miles from Columbus, 0. to Delaware; if a man travel 15- miles, from Columbus, how much farther must he go to get to Delaware? Ans. 8 M, 2 fur, 26 rds. 11 ft. 12. A man owned a farm containing 134 A. 33 R.; he sold at one time 43 A. 2 R. 27 sq. rds., and at another time 32!- A.; how much had he left? Ans. 58 A. 2 R. 23 sq. rds. 13. A certain pile of wood contains 17 cord, 112 s. feet. 1438 s. in.; another pile contains 14 C. 123 s. ft. 1649 s. in.; how much more in one pile than in the other? Ans. 2 C. 116 s, tt. 1517s. in. 14, A silversmith had a piece of silver weighing 4 lbs. 8 oz. 165 pwts. from this he made a set of spoons weighing 2 lbs, 6- oz., and thimbles enough to weigh 1- lbs.; how much was left? Ans. 9 oz. 10 pwts. 20 grs. 15. A box of goods weighed 62 lbs. 12 oz.; the box alone weighed 18 lbs. 143 oz.; how much did the goods weigh? Ans. 43 lbs. 13 oz. 4 drs. 16. If a waggon loaded with hay weighs 28 cwt. 3- qrs., and the waggon weighs 7 cwt. 3 qrs. 20 lbs., what is the weight of the hay? Ans. 21 cwt. = 1 T. 1 cwt. 17. A merchant having I22 cwt. of sugar, sold at one time 3 cwt. 3 qrs. 21 lbs., at'another time 2 cwt. 1 qr. 15 lbs., and at another time 6-~1 cwt.; how much had he left? Ans. None. 18. From a package of medicine containing 4 lf. 103. 6 3. 19. 15 grs., was taken 2 lb. 11 3. 7 3. 2. 18 grs.; how much was left? Ans. 1 lb. 10 3. 6 3. 1 l. 17 grs. 19. A. worked at a certain place 2 yrs. 186 da. 12 hrs.; B. worked at the same place 1 yr. 310 da. 14 hrs.; how much longer did A. work than B.? Ans. 241{ da. 20. Henry's age is 17 yrs. 6 mo. 23 d.; George's age is 10 yrs., 7 mo. 21 d.; how much older is Heiry than George? Ans. 6 yrs. 11 mo. 2 d. 27. Washington was born Feb. 2d, 1732; he died Dec. 14th. 1799; what was his age at the time of his death? Ans. 67 yrs. 9 mo. 22 d. 186 COM31ON AR[TTHMET10t Sect. IX Operation. We set down the year, the number of the yrs. mo. d. month, (calling Jan. 1, Feb. 2, &c.) and the 1799_ 12 _14 day of the month, and subtract as usual, 1732__ 2 _22 reckoning 30 days to the month, and 12 months to the year. In this example, Dec. Ans. 67 _ 9 -22 is the 12th month, Feb. the 2nd., &c.HenceObs. 1. The period of time between any two dales may be found by subtracting the earlier date from the latter, the montl s being reckoned 30 days. This rule is much used in calculating interest. (Sect. XIV, Art. 5.) 21. A note was givenJan. 24th, 1841, and paid Aug. 12th 1843; what time did it run? Ans. 2 yrs. 6 mo. 18 d. Obs. 2. The time which elapsesfrom the giving of a note until the pcyment of it is called the time it has to run. 22. A man left home Sept. 28th, 1837, arid returned June 18th, 1840; how long was he absent? Ans. 2 yrs. 8 mo. 20 d. 23. How long has a note to run, which is giv. n Nov. 16th, 1828, and paid Jan. 3d. 1847? Ans. 18 yrs. I mo. 17 d. 24. London is 51~ 32', and Gibralter 36~ 6' 30" north latitude, what is the difference of latitude between the two places? Ans. 15~ 25' 30". 25. From an area of 180~, take 57~ 48' 39". Ans. 122~ 11' 21". 26. The longitude of Boston is 71~ 3', and that of the Sandwich Islands 155~ west from Greenwich; what is their difference in longitude? Ans. 83~ 57'. 27. A man having 6 dozen dozen eggs, broke a half a dozen dozen; how many had he left? Ans. 792. 28. Take 149 lbs. from a barrel of flour, and how much remains? Ans. 47 lbs. 29. From a barrel of pork, take 97 lbs. Ans. 103 lbs. 30. From 5 of a bushel, take - of a peck. Ans. 1 pk. 6 qts. 31. From 5- of a hogshead, wine measure, take 3 of a gallon. Ans. 52 galls. 1 pt. 32. From s a mile take -1- of a furlong. Ans. 3 fur. 12 rds. 33. From 4 of a cwt,, take -7 of a quarter. Ans. 2 qrs. 21 lbs. 4 oz. 34. From 2- of a year, take of a day. Ans, 243 days 8 hrs. 35. From -7I of a circle, take: of a degree. Ans. 156~ 37' 30". How is the period of time between two dates found? What is meant by the time a note has to run? Art. 6. MULTIPLICATION AND DIVISION OF COMPOUND NO'S. 187 ARTICLE 6. MULTIPLICATION AND DIVISION OF COMPOUND INUMBERS, Ex. 1. A farmer has 6 bags of wheat eact containing 2 bu. 3 i pks. 6 (qts. how much wheat in all the btags? Operation. bu. pks. qts. 2 __ 3 _ 6 6 Ans. 17 __ 2 __ 4 We writ' our numbers as usual, placing our multiplier u:)der the lowest denomination of the multiplicand, (qts.) Then, 6 X 6 - 36 qts.; 36 qte. = 4 plis. 4 qts.; we set down the 4 qts. and carry the 4 pks., 3 X 6 = 18 pks. and 4 pks. to carry make 22 pks.; 22 pks. = 5 bu. 2 pks.; we set down the 2 pks. and carry the 5 bu.; 2 X 6 = 12 and 5 to carry 17 bu. From this illustration we derive the following RULE FOR THE MULTICATION OF COMPOUND NUMBERS. I. Write the multiplier under the lowest denomination of the multiplicand. II. Commence at the right hand multiply as in simple numbers and carry as in addition of compound numbers. PROOF.-The same as in simple numbers. (Sect. IV, Art 2, Obs. 7.) REMARK 1. If any denomination is wanting in the multiplicand, supply its place with a cipher. 2. A farmer has 17 bu. 2 pks. 4 qts. of wheat put up in 6 bags, each containing an equal quantity; how much is there in each? Op1eration. bu. pks. qts. 6)17 _- 2 _, 4 Ans. 2.. 3 _ 6 17-6 = 2, and 5 remainder. This 5 is bushels, because the dividend, 17, is bushels. (Sect. V, Art. 2, Obs. 12, a.) 5 bu. 20 pks.; 20 pks. + 2 pks.= 22 pks.; 22 pks. - 6 = 3 pks. and 4 pks. remainder. 4 pks. = 32 qts. 32 qts. + 4 qts. = 36 qts.; 36 qts. -6 = 6 qts. From this illustration we derive the following RULE FOR THE DIVISION OF COMPOUND NUMBERS. I. Write the dividend and divisor as usual. (Sect. V, Rule.) II. Commence with the highest denomination; divide, and place the quotient as in division of simple numbers. (Sect. V, Rule.) III. If a remainder occurs reduce it to the next lower denomination and add to it the number given, (if any) of this denomination; after which divivde as usual. PROOF.-The same as in Simple 1Vumbners. (Sect. V, Art. 2 Obs. 9. REMARK 3. If any denomination What is the rule for the multiplication of compound numbers? What is the proof? If any denomination is wanting in the multiplicand, how do we proceed? What is the difference between the principles of the multiplication of simple and compound numbers? Explain by the example given, why this is the case? Why do we not reduce in this way In simple numbers? \ hat is the rule for the division of compound numbers? What is the proof? 188 COMMON ARITIIMETIC. Sect. IX 2. The principles of multiplication of simple and compound numbers are the same as can be shown by the following example. Multiply 256 by 8. Operation. thou. hunds. tens. units. 2 __ 5 _ 6 8 Ans. 2,_ 0 4_ 4 8 6 X 8 - 48 -- 10-4 ens, 8 units. 5 X8= 40, and 4 to carry-= 44; 44 10 = 4 hund. 4 tens. 2 X 8 16 and 4 to carry = 20; 20 — 10 - 2 thou. 0 hun. It will be perceived that there is no difference between this operation, and that of Ex. 1.But as the number that we set down is always the right hand figure of the partial product, it is evident that the above formal process is unnecessary. 3. A farmer took 7 loads of wheat to market, each load containing 22 bu. 2 pks. 5 qts.; how much was there in all? 5. A wine merchant has 9 casks of wine, each containing 20 galls. 1 pt.; how much had he in all? 7. A gentleman has 8 casks of brandy, each containing 45 galls. 1 pt.; how many gallons has he in all? 9. How many gallons in 5 in the divdend is wanting supply its place with a cipher. 4. The principles of the division of simple and compound numbers are the same, as can be shown by the following example: Divide 2048 by 8. Operation. thou. hund. tens. units. 8)2 __ 0 - 4 -- 8 Ans. 2 5 -- 6 2 thou. X 10 = 20 hund.; 20 - 8 = 2 hund. and 4 hunds. rem.: 4 X 10 = 40 tens, and 4 tens = 44 tens. 44- 8 = 5 tens and 4 tens rena.; 4 X 10 = 40 units, and 8 units = 48 units; 48. 8 6 units. It will be perceived that the operation is similar to that of Ex. 2. But the denomination given only occupies the pace of the cipher, when added to the one that is reduced, this process may be dispense:l with. (Sect. VY Art. 2, Ex. 9, and Obs. 13.) 4. If 7 equal loads of wheat contain 158 bu. 2 pks. 3 qts., how much does each contain' 6. If 9 casks of wine contain 181 galls. 1 pt., how many gallons does each contain, each containing an equal quantity? 8. If 8 casks of brandy, containing each an equal quantity, hold 361 galls., how many gallons are there in each? 10. If 5 equal sized flasks con If any denomination is wanting in the dividend, how do wo proceed?What is the difference between the principles of the division of simple and compound numbers? Explain why this is the case. Why is not this process adopted in simple numbers? Art. '6. MULTIPLICATION AND DIVISION OF COMPOUND NO.'S 189 flasks of beer, each containing 6 galls. 2 qts. 1 pt? 11. How many yards of cloth will it take to make 7 coats, each containing 3 yds. 2 qrs. 1 na.? 13. If a man walk 22 M. 6 fur. 29 rds in one day, how far can he walk in 6 days? 15. A man has 4 farms, each containing 54 A. 3R. 23 sq. rds.; how much in all? 17. There are 6 piles of wood each containing 12 C. 110 s. ft. 1129 s. in.; how many cords in all? 19. A silversmith made 12 spoons, each weighing 2 oz. 3 pwts. 8 grs. Required, the weioght of the whole? 21. A farmer sold 11 loads of hay, each weighing- 18 cwt. 3 qrs. 21 lbs. Required, the weight of the whole? 23. A merchant bought 10 sacks of cofee, each weighing 156 lbs. 12oz. 14drs. Required, the weight of the whole' 25. What is the weight of 9 packages of medicine,each weighing 3 fb. 9 3. 5 3. 1 D. 16 grs.? 27. If it takes a man 1 hr. 10 min. 20 sec. to walk a league, how long will it take him to walk 7 leagucs? 2~. The earth moves daily in her orbit, about 59' 8"; how much does it move in 9 days? 31. There are 18 sacks of flaxseed, each containing 9 bu. 3 pks. 6 qts.; how much in all? tain 33 galls. 1 pt., how much does each contain? 12. If it takes 24 yds., 3 qrs. 3 na, to make 7 coats, how much does it take to make one? 14. If a man walks 137 M. 14 rds. in 6 days, how far does he walk in a day? 16. If 4 equal farms contain 219 A. 2 R. 12 sq. rds., how many acres does each contain? 18, If 6 equal piles of wood contain 77 C. 23 s. ft., 1590 s. in,, how much is there in each pile? 20, If 12 spoons weigh 2 lbs. 2 oz,, what is the weight of each, each weighing the same? 22. If 11 equal loids of hay weigh 10 T, 8 cwt. 2 qrs. 6 lbs., what is the weight of each load? 24. If 10 sacks of coffee weigh 15 cwt. 2 qrs., 18 lbs. 12 drs., what is the weight of each sack, all being equal? 26. If 9 equal packages of medicine weighs 34 lb. 3 3. 23. 1 9. 4 grs., what is th- weight of each'! 28. If it takes a man 8 hrs. 12 min. 20 sec. to walk 7 leagues, how long will it take him to walk one league? 30, If the earth moves in her orbit 8~ 52' 12" in 9 days, how far does she move per day? 32. If 18 sacks contain 178 bu. 3 pks. 4 qts. of flaxseed, how much does each contain, all being of the same size? ___ I REMARK 5. When the multiplier is REMARK 6. When the divisor is a a composite number we may first mul- composite number, we may first divide When the multiplier or divisor is a composite number, how mayi we proeed? 190 COMMON ARITHMETIC. Sect. IX tiply by one factor and then by the first by one factor, and then by the other. (Sect. 1V, Art. 4, Obs. 4.) otner. (Sect. V, Art. 4, Ob(. 3.) Operation of the last example. bu. pks. qts. 9 _ 3 _ 6 18=6X3 6 59 2 - 4 = quantity in 3 [6 sacks. Ans. 178 - 3__ 4= quantity in [6 X 3 = 18 sacks. 33. In 36 casks of wine each containing 28 galls. 3 qts. 1 pt. 3 gi.; how many gallons? 35. If it takes 2 yds. 1 qr. 3 na. of cloth to make a pair of pantaloons, how much will it take to make 24 pair? 37. A man took 49 loads of wood to the city, each containing I C. 15s. ft., 124 s. in.; how many cords in all? 39. A merchant bought 63 kegs of butter, each weighing 32 lbs. 14 oz. 12 drs. Required, the weight of the whole? 41. If a man walk 25 M. 6 fur, 37 rds, in a day, how far can he walk in 127 days? REMARK 7. When the multiplier exceeds 12, and is not a composite number, we multiply each denomination, and reduce it oa a part of the slate separate from the multiplier and multiplicand, and merely set down the result as when the multiplier is less than 12. Operation of the last example. bu. pks. qts. 18=3 X6 3)178_, 3.. 4 ----- [of 18, or 6 sks. 6) 59.- 2 4 = quantity in - Ans. 9.. 3. 6 = quantity in l [of 6,or 1 sk. 34. If 36 equal casks of wine contain 16 hhds. 34 galls. 3 qts. 1 pt.; how much does each contain? 36. If it takes 581 yds. of cloth to make 24 pair of pantaloons, how much does it take to make one pair? 38. If 49 equal loads of wood contain 54 C. 98 s. feet. 892 s. in., how much is that per load? 40 If 63 kegs of butter weigh 1 T. 2 qrs. 24 lbs. 1 oz. 4 drs., what is the weight of each, all weighing al ke? 42. If a man travel 3284 M. 7 fur. 19 rds. in 127 days. traveling the same distance each day how far does he travel per day? REMARK 8. When the divisor is greater than 12, and is not a comnposie number, perform the operation by long division. -When the nulitiplier exceeds 12, and is not a composite number. hlow do we proceed? By what other method? What is the third metilofi? When the divisor excads 12, aud is not a oomposite number, how do we proceed? Art. 6. MULTIPLICATION AND DIVISION OF COMPOUND NO'S. 191 Operation of the last example. M. fur. rds. 25.. 6.. 37 127 Ans. 3284.. 7.. 19 Operation of the last example. M. fur. 127)3284.. 7 254 rds. [Ans... 19(25.. 637 rds. 37X 127 = 4699 410),t! i9 fur. 117-19 rds.rem. 6X127 = 762 fur, add. 8)879 fur. M. 109-7 fur. rem. 25X 127 = 3175 M. add. 3284 M. Or, if the student prefers, he may first multiply by 10, and the product thence arising by 10, which will give the product by 100: then multiply the product by 100, by the number of hundreds, the product by 10, by the number of tens; and the first multiplicand by the number of units; and add these last three results together. Thus — 744 635 109 M. remainder. 8 872 furlongs. 7 fur. add. 127)879(6 fur. 762 117 fur. remainder. 40 4680 rods. 19 rds. add. 127)4699(37 rds. 381 889 889 000 It will be perceived that the only difference betwe' n this operation, and that of Ex. 2, is that in this operation the entire work is set down, whilst in Ex. 2 it is not. 192 COMMON ARITHMETIC. Sect. IX Operation 2. 127 = 100 - 20 + 7, or 1 hund., 2 tens, 7 units. M. fur. rds. M. fur. rds. 25. 6.. 37 X 7 181.. 0.. 19 10 258.. 5.. 10 X 2 - 517.. 2.. 20 10 2586.. 4.. 20 = product by 100. Add 517.. 2 __ 20 =proluct by2 tens, or 20. 181.. 0.. 19 -product by 7 units, or 7. Ans. 3284.. 7. 19 product by 100 + 20 + 7 127. Operation 3. 25 M. 6 fur. 37 rds. = 8277 rds. 127 57939 16554 8277 1051179 rds. 1051179 rds. = 3284 M. 7 fur. 19 rds. Ans. Obs. 1. It may be asked by some pupils, why we cannot multiply compound numbers, when the multiplier exceeds 12, in the same manner as simple numbers; that is, by multiplying by each figure of the multiplier separately, and writing the first figure of each product under the figure by which we multiply? We answer, that we can, if the multiplicand is first reduced to but one denomination, as in Operation 3. But when the multiplicand is not thus reduced, each denomination must be multiplied separately by the entire multiplier, on account of the inequality of their increase. Did all the denominations increase by the same ratio, we could then multiply in compound as in simple numbers. What is the difference between the operations of Ex. 41 and 42? When can we multiply compound numbers by multipliers exceeding 12, in the same manner as in simple numbers? Why cannot we proceed in this way when the multiplicand is not thus reduced? What would be necessary, in order to have the same rule apply to compound and simple numbers? Art. 6. MULTIPLICATION AND DIVISION OF COMPOUND NO'B. 193 43. How much wine in 29 casks, each containing 47 galls. 3 qts. 2 pts. 3 gi.? 45. How much wheat can be put in 23 casks, each containing 4 bu. 3 pks. 6 qts.? 47. If it takes 4 yds. 3 qrs. 3 na. of cloth to make a suit of clothes, how much will it take to make 349 suits? 49. How much sugar in 53 hhds.. each containing 12 cwt. 2 qrs. 23 lbs.? 44. If 29 casks of wine, each containing the same -quantity, hold 22 hhds. 8 galls. 2 qts, 1 Pt. 1 3 gi., how much is that for each? 46. If 23 casks hold 113bu. 2 pks. 2 qts. of wheat, how much is that for each, they all being of the same size? 48. If it takes 1723 yds. 3 na: of cloth to make 349 suits of clothes, how much does it take to make one suit? 50. If 53 hhds. of sugar contain 33 T. 14 cwt. 2 qrs. 19 Ibs,, how much is that for each? To find the difference in the time of two places: Obs. 2. The earth in revolving on its axis, performs one entire revolution in 24 hours, and as the circumference of the earth is 360~, she makes 360~ - 24 = 15~ of motion in 1 hour of time, and 60 -15 = 4, i.e. 1~ of motion in 4 minutes of time. Therefore, there is a d fference of 1 hour in the time between two places for every 15" of longitude between them, and of 4 minutes for every degree. HenceTo find the difference in the time of two places: Obs. 3. Multiply the difference in the longitude of the two places, (in degrees and minutes,) by 4: the result will be the difference in time, in minutes and seconds. REMARK. — he student will bear in mind, that as the earth revolves from west to east, places easterly hare the time earlier than places westerly.* HenceObc. 4. To find the time of places easterly ADD the dfference in the time of the two places to the given time, and to find the time of places westerly SUBTRACT the diference in the time of the two places, from the given time. 51. The difference in the longitude between Boston and WasHow often does the earth perform one revolution on its axis? How many degrees of motion does it make in one hour of time? Why is this the case? How long does it take it to make la of motion? What is the inference deduced from these considerations? How then do we find the differences in -ti time of two places? Which has the time the earliest, places easterly, -r pl" westerly? Why so? + Th~ teacher can best explat this by means of an appl0 or otr rod endml 10 194 COMMON ARITHMETIC. Sect. IX ington City is 6~ 40'; now when it is 4 o'clock, P. M. at Washington, what is the time at Boston? Operation. We find the difference in the time of the two 6~ 40' places to be 26 min. 40 sec.; and Boston being 4 east of Washington, the time is earlier, and - consequently it must be 26 min. 40 sec. past 4 2 min. 40 sec. o'clock at Boston, when it is 4 o'clock at Washington. 52. Washington, D. C., being 770 1' 30", and Columbus. 0., 83~ 3' W. longitude from Greenwich, I wish to know what time it is at Columbus, when it is 9, A. M., at Washington? Ans. 35 min. 54 sec. past 8 o'clock, A. M. 53. Reading, Eng., is about 1~, and Buffalo, N. Y., is about 78~ 55' W. longitude from Greenwich. Required-the difference in the time of the two places. Ans. 5 hrs. 11 min. 40 sec. 54. If it is 9 o'clock, 30 min. P. M. at Buffalo, what is the time at Reading? Ans. 41 min. 40 sec. past 2 o'clock, A. M. (on the next morning.) 55. If Buffalo is 78~ 55' W. longitude, and Tunis, in Africa about 10~ E. longitude from Greenwich, when it is 5 min. 20 sec. past 6 o'clock, P. M. at Buffalo; what is the time at Tunis? Ans. 1 min. past 12 o'clock, midnight. RE:MAR 1. When one place is East, and the other place West longitude, we must add their longitude together to find the distance between them. 56. In the last example, when it is 26 min. 10 sec. past 1 o'clock A. M. at Tunis, what is the time at Buffalo? Ans. 30 min. 30 sec. past 7 o'clock P. M. (i.e. the day before.) REMARK 2. The learner will observe in these examples, that we count our time by half days or 12 hours; and therefore, when we obtain more than 12 hours, it is some hour less than 12, of the preceding or succeeding A. M. or P. M. part of the day. Consequently, if our given time isless thau the difference of time between the two places, and we wish to subtract the difference of time, *e must add 12 hours to our given time, and then subtract as usual. 67. If a certain place is 160~ E. longitude, and another place is 120~ W. longitude, and it is 5 o'clock,P. M.,at the latter place,what time is it at the former? Ans. 40 min. past 11 o'clock, A. M. (i. e. before.) How do we find the time of places easterly? Of places westerly? When one place is east, and another place is west longitude, how do we find the distance between them'? In these examples, how do we count our time? What then is the result when we have more than 12 hours? If our given time is less than the difference between the time of the two places, and we wish to subtret, bow do we proceed? EXCHANGE. 195 65. In the last example, when it is 10 o'clock, A. M. at the former place, what time is it at the latter? A us. 20 min. past 3 o'clock, P. M. (afterwards.) 59. What is thle difference of time between the places last mentioned? Ans. 5 h;s. 20 min. 60. Suppose a meteor to appear so high in the heavens as to be visible at the same moment, at Boston 71~ 4' 9", at the city of Washington 770 1' 30," and at the Sandwich Islands 155O' W. longitude, and that its appearance at Washington, be at 45 min. 45 sec. past 11 o'clock in tie evening, what would be the time of its appearance at Boston, and at the Sandwich Islands. At Boston, ) min. 34 sec. past 12 o'clock, midnight (after.) Ans. At the Sandwich Islands, 33 min. 51 sec. past 6 o'clock, P.M. (be'ore.) SECTION X. EXCHAI N GE. Obs. 1. The method of finding the value of the currency of one country in that of another, is called EXCuHANGE. Obs. 2. CURRENCY signilies the circulating medium of trade, and is called MONEY. Obs. 8.. Money is said to have two values: an intrinsic, and a commercial value. Obs. 4, The intrinsic value, is the value of the money of one country, as compared with that of anollier, with respect both to weight, and the purity of the metal of which it is made. Obs. 5. The commercial value is the compr tive value of the money of different countries, with respect to their weight, fineness, and market price. Obs. 6. The relatiVe value of foreign coins, (that is; the value they bear to each other,) is detelrmined by the laws of the country. The value of a few of the most common foreign coins in the What is Exchange? What dJes currency signify? How many values has money? What is the intrinsic value? The commercial value? ilow is the relative value of foreign coins determined? 196 COMMON ARITHMETIC. Sect. X UNITED STATES as established by Act of Congress, 1842, is shown in the following TABLE: The value of a Pound Sterling, or Sovereign, of England, is $4.846. The " a Guinea, " is 5.075. The " a Franc. of France is 0.185. The " a Five Franc piece, " is 0.93. The " a Dollar, of.itxico, Peru, and Chili, is 1.00. The " a Doubloon, of Spain, Mexico, &c., is 15.535. The " a Ducat of Russia, is 2.297. CASE 1. To change Foreign Coins to Federal Pioney. Ex. 1. What is the value of 20 Pounds Sterling, in Federal Money? Operation. One Pound Sterling contains $4.846, therefore 20 $4. 846 Pounds Sterling contains 20 times $4.846. 20 Ans. ~96.920 REMARK.-This character (~.) usually stands for the Pound or Sovereign. The other coins are reduced to Federal Money in the same manner. HenceTo change foreign coins to Federal Money: Obs. 7. Multiply the givenm number of coins, by the number of dollars in one coin as taken fom the table. 2. Reduce 50 Sovereigns to Federal Money. Ans. $24,. 30. 3. Change 75 Guineas to Federal Money. Ans. $380.62/-. 4. Change 90 Guineas to Federal Money. Ans. $456.75. 5. Change 55 Francs to Federal Money. Ans. $10.175. 6. Change 180 Francs to Federal Money. Ans. *33.30. 7. In 84 Five Franc pieces, how many dollars? Ans. $78.12. 8. What is the value of 125 Five Franc pieces in Federal Money? Ans. $116.25. 9. Change 35 Doubloons to Federal Money. Ans. $543.725. 10. Change 18 Doubloons to Federal Money. Ans. $279.63. 11. Change 38 Ducats to Federal Money. Ans. 887.286. 12. Change 22 Ducats to Federal Money. Ans. $50.534. Give the value of some of the foreign coins as shown in the table? How do we change foreign coins to Federal MoUey 7 EXCHANGE. 197 CASE 2. To change Federal Money to other currencies. REMARK-This case is the opposite of tlhe preceding one, aud each proves the other. Ex. 1. Change $72.69 to Pounds Sterling. Ans. 15. Operation. As 1 Pound contains 84.846. there will 4.846)72.690 15 Ans. evidently be as many Pounds in '72.69, as 4846 $4,846 is contained in $72.69. 24233 24230 00000 Federal Money is changed to other currencies in the same manner. HIenceTo change Federal Money to other currencies; Obs. 8. Divide the given sum by the number of dollars it takes to make one of the coin to which it is to be'. reduced, and point off as in Division of Decimals. (Sect. VIii, Art. 12. Rule.) REMARK.-4 farthings make 1 penny; 12 pence make 1 shilling; and29 shillings make 1 Pound or Sovereign, Hence-if lhere isa remainder i) reducing Federal Money to'Pouads St-rling, we can reduce it lower as in'Division of Compound Numbers. (Sect. IX, Art. 6. Rule for the division of Compound Numbers). Farthings are marked, fLr.; pence, d. and shillings, s.; 1, 2, and 3 farthings are generally written as i, i, and - of a penny. 2. Change $121.15 to Sovereigns. Ans. 25. 3. Change $253.75 to Guineas., Ans. 50. 4. In 669.75 how many Five Franc pieces? Ans. 75. 5. In $217.49 how many Doubloons? Ans. 14. 6. In $68.91 how many Ducats? Ans. 30. 7. Change $32.83165 to pounds,'shillings, and pence. Ans. ~6, 15s; 6d. 8. Change $21.807 to Sterling Money. Ans. ~4, 10s. 9. Change $52.3368 to Sterling Money. Ans. ~10, 16s. 10. Cha1ne $27.591 to Sterling Money. Ans. ~5, 13s. Sd. 3 far.+ How do we change Federal Money to other currencies? low is the Pouad divided? If there is a remainderin reducing Federal Money to Pounds Sterling, how may we proceed? 198 COMMON ARITHMETIC. Sect. XI SECTION XI. ANALYSIS. ARTICLE 1. DEFINITIONS AND MENTAL EXERCISES. Obs. 1. Analysis signifies the separation of a body into parts, and is used in mathematics for the dcvelopement and illustration of principles, and may also be applied to the solution of questions with great practical advantage. In the preceding sections we have analyzed Principles; in this section we design to analyze practical questions, or Problems. KToTE.-We often hear people te}l of "doing sums in their head." They in in reality, solve their questions by Analysis, or the commrnon sense rule. Obs. 2. This method is of extensive application, and great uLtility. It strengthens the mental.faculties, (ind accustoms tle pupTil to habits of close thinking, and with a little practice he can become very expert at it. No particular direction can be given for solving every question in analysis. The folowing mnay perhaps answer for general directions, but the learner must try and think for himself and not depend too much on others. Obs. 3. Examine the question attentively; reason according to the nature of the sum, and work as common sense directs. REMARK.-The operation of solving questions by analysi is called an Analytic Solution. Ohs. 4. In solving questions analytically, we generally reasonfrcm the given number to 1, and then from 1 to the required number. MENTAL EXERCISES. 1. If 5 tons of hay cost $30, what will 8 tons cost? Ans. 48. Solution.-If 5 tons cost $30, 1 ton will cost ~ of $30, or $,and 8 tons will cost 8 times e6 or $48. 2. If 3 firkins of butter cost $15, what cost 4 firkins? 6; 8; 10; 12? What does analysis signify? To Y lhat is it used in mathematics? To what may it be applied? How do people who "do sums in their head" in reality solve their questions? What advantages are derived from this method? What directions are given for analyzing? What is the operation of solving a question by analysis called? How do we generally reason in solving quetionu analytically? Art. 1. ANALYSES. 199 3. If a man can earn $14 in 7 days, how much can he earn in 3 days? 4; 5; 6; 8; 10; 12; 15? 4. If a man spend $18 in 6 days, how much does he spend in 4 days? 3; 5; 9; 10; 12? 5. If a b:)y walk 72 miles in 6 days, how far does he walk in 5 days? 8; 10; 12? 6. If 4 men can do a piece of work in 3 days, how long will it take 6 men to do the same work? Ans. 2 days. Solution.-If it takes 4 men 3 days to do the work, it will take 1 man 3X 4==12 days to do it. Again if it takes 1 man 12 days to do it, 6 men can do it in - of the time, or 2 days. 7. If it takes 3 men 8 days to do a piece of work, how long will it take 2 men to do it? 4; 5; 6; 8; 12; 24? 8. If 4 horses consume a stack of hay in 12 weeks, how long would it keep 2 horses? 6; 8; 12; 24? 9. If a certain pasture keeps 6 cows 6 weeks, how long will it keep 4 cows? 8; 9; 12; 18; 36; 1). Two boys were counting their money: one said, "I have 50 cents;" the other said, "I have 4 as much as you." How much had the latter boy? Solution. — of 50 cents is 10 cents, and ^ is 4 times as much as., and 4 times 10 are 40. Ans. 40 cents 11. A. and B. were talking of their ages: A.'s age was 63 years, and B,'s age was - of A.'s age. What was the age of B.? 12. John and Thomas were playing marbles: John had 24, and Thomas had 3 as many as John. How many had Thomas? 13. A. has 96sheep; B. has -7- as many as A. How many has B. 14. One man has 132 hogs, another man has iT as many; how many has the latter man? 15. Tho nas has 120 bushel of oats; of Thomas' is equal to 4 of James'; how many bushels has James? Ans, 126. Solution. —i of 120 is 12, and 9 of 120 is 9 times 12 or 108; then 108 is - of James'; if 108 is a, I is f of 108 or 18, and, or the whole quantity 7 times 18 or 126. 16. Two boys were talking of their money; one says, "I have 12 cents;" the other says,,3j of your money is equal to. of mine;" how much had he? 17. Two boys talking of their ages, one said "I am 15 years old;" the other said, 4 of your age is equal to 7 of mine". How old was he? 18 A. has 14 sheep; 7 of his number is equal to 2 of B.'s; how many has B.? 200 COMMON ARITHMETIC. Sect. XI 19. Charles has $36; 4 of his is equal to 2-L of Francis'; how much has Francis?!0. A man bought a horse and paid $30 down, which was - of the price of him. What was the price of the horse? Ans. $54. Solution.-If 30 is, of a number, l- will be 5 of 30 or 6, and 9i, or the number itself will be 9 times 6, or 54. 21. A note has run 45 days, which is | of the time it has to run; how long has it to run? 22. A man bought a waggon, paying $36 down, which was ' of the price of it. Required, its price? 23. James bought some iron, paying $48 "trade; this was 5 of the cost of it; what did it cost? 24. A lady purchasing some goods, paid $72 down; this was iF of her purchases; how much did she trade? 25. A man in trading. paid down $84; this was 1-7 of his bill; required the amount of his bill, and how much he had yet to pay? 26. Frank and Henry were speaking of their money. Henry says "I have 84 cents;" says Frank "I have v as much as y u, and am going to buy lemons with it, at 4 cents apiece;" how many can he buy? Ans. 18. Solution. — of 84 is 72, the number of cents Frank had. Then as the lemons are 4 cents apiece, he can buy 72 - 4 — 18 lemons for 72 cents. 27. A man has - of $144; how many yards of cloth can he buy at $6 per yard? 28. Henry has ll of $132; how many steers can he buy with it at $9 apiece? 29. A m-,an having 7~- of $120, bought 9 coats with it; how much were they apiece? 30. Charles h:d a note for $96; he obtained 8 of it, and spent it for sheep at $3 apiece; how many did he buy? 31. William spent $14 for books, which Oas T oL all he had; he spent the remainder for 7 yards of cloth; what did the cloth cost him per vard? Ans, $5. Solution.-If $14 was 7 of all he had, 4 would be ' of $14, or $7, aud would be 7 times $7, or $49. Then as he spent $14, he would have $49-$14=8$3 remaining; with this he bought 7 yards of cloth, which must have cost him $35 7-9$5 a yard. 32. A farmer paid $30 for goods, which was 3 of his bill; he paid the balance in wheat at $1- per bushel; how many bushels did it take? 33. A man spent $150 for sheep, which was 3I of all he had: he spent the remainder for land at $10. n acre; how many acres did he buy? Art. 2 ANALYSIS 201 34. A man bought a horse paying $36 down, which was, -of what he was to pay; he paid the remain'der in flour at $5 a barrel; how many barrels didit take? 35. Lewis spent $72 for cloth, which was - of all he had; he spent the remainder for books at $4 apiece; how many books did he buy? 36. Silas spent 24 cents for a slate. which was ' of alUl e had; he bought 10 lead pencils with what was left; how much were the pencils apiece? 37. - of 24 is how many times 10? Ans. 2. Solution. — of 24 is 20; 20- 10=2; hence, 20 is twice 10. 38. of 48 is how many timcs 6? 39. 5- of 54 is how many times 3? 40. - of 56 is how many times 4? '41. 48 is -of how many times 6? Ans. 9. Solution.-We first inquire, 48 is of what number? IT 48 is, one ninth is 8 of 48, or 6; and v or the number itself is 9 times 6, or 54. Then 54 is how many times 6? -1-6 R 9, that is 54 is 9 times 6?9 42. 63 is' -.-of hcdwmany times 15? 43. 108 is ~ of how many times 30? 44. 6 times 6 is of what number? Ans. 81. 45. 8 times 9 is - of how many times 3 times 4? Ans. 7. 46. 3 times 30 is F- of how many times 4 times 5? Ans. 5. 47. - of 36 is L of what number? Ans. 100.,Solution. — of 36 is 30; 30 is -L of 100, 48. ~ of 42 is of what number? '49. 3 of 48 is of how many times 3? 'Ans. i4..6. of 72 is A-: of how many times:5 times 51 Ans. 2. ARTICLE -. EXERCISES FOR '*HE 'SLAY'. '1. If 25 yds. of cloth cost $100, what will 40 ds cpst? Ans. $160. Solution. —If 25 yds. cost $100, 1 yd. will cost $.100- 25 =$4. 'Then 40 yds. will cost 40 times as much as 1 yd., afd $4 X 40 = $160, Arfs. 'REMARK.-II solving such questions, the learhfr can oftth s'heteli lte speration, by merely expressing the work and cauteliig. ThusHow can the operation in analysis oftn 'be rhorted i*: 10A fcOM(OBi ATIUTHMETIC. Sect. XI Whien tle learner performs the operatior by can- kj0-4 celation; (or any other way) he shokuld bIe required 44) to give his reason for every step he takes. It is a matteTrof unimportance, how or where he sets his $160 Ans: figfrResvif he thoroughly understands his subject NoTE.-Questions of this kind are usually worked by 5.r)ple Proportidfi or the' Rul of Three, which r expltahied in, the next section. Business men,hovwevepr gesraly olire them by analysis. 2. What cost 48 cows, if 20 cows cost $240? Ans. $576. 3.. What cost 48 lbs. of coffee, if 60 lbs. cost $6.60? Ans. $5.28. 4. It I pay $3f.20 for the use of $520 a certain time, how mlaach must I pay for the use of $720 the same time? P.ns. $43.20. 5. If a stage travel 90 miles in 15 hours, hew maay mSes will it travel in- 36 hours? Ans. 216. 6 What cost 60 bushels of barley if 600 bushels cost $304! Ans. $425. 7. If 25 horses eat 112' bushels of oats in a week, how many blshels will 14 horses eat? Ans. 63. 8. I;f 416 lbs. of wool cost $104, what wilI 599 as. cost? Ans. $149.75. 9. If 48 tons of hay cost $720, what will 36 tons cost? Ans. $540. 10. If a railroad car travels t54 miles in 1 1 hours, how far wilf it travel in 18 hours? Ans. 252 miles. I;. A man can do a piece of work in 8 days, laboring 12 hours per day: how long wil4 it take him when he labors but 9 hours per day? Ans. 10- days. 12. If 4 lbs. of coffee cost $0.50, what will 40 yds. of calicu cost, if 7' lbs. of coffee are worth 5 yds. of c-clico? An., $7.50. 13:. If 12 men can do a piece of work in 24 days, how long: would it take 30 men to do the same *or.? Suggestion. —If 12 men do the work in 24 days, it will take t man 12 times as long to do it, and 30 men will do it in -L of the time in which one man could perform it.. Ans. 93 days. 14. If 18 men eat a barrel of flour in 8 days how long will it last 45 men? Ans 3- days. r5. If a certain quantity of oats last 16 horses 18 days, how long will it last 28 horses? Ans. 1027 days. rs it a mtatter of importv nce 'wlhre r L fow the figures are set if the learner wnderstands his subject? Art. 2. ANALYStS. 203 16. If a barrel of flour lasts 8 men 30 days, how long will it last 18 men? Ans. 134 days. 17. If 16 men can do a piece of work in 12 days, how long will it take 24 men to do it? Ans. 8 days. 18. If 6 stacks of hay keep 40 cattle 90 days, how long will they keep 60 cattle? Ans. 72 days. The learner will perceive that the number of stacks is not to be regarded in solving this question. 19. If - a bushel of wheat cost $0.50, what will - of a bushel cost? Ans. $0.75. Suggestion.-If a bushel cost $0.50, a bushel will cost twice as much, and 3 of a bushel will cost ass much as a bushel. 20. If; of a yard of cloth cost $3.20, what costs 7 of a yard? Ans, $3.50. 21. If 4 of an acre of land costs $18, what costs 4 of an acre? Ans. 14. 22. If i of a ton of hay costs $10.80, what will i of a ton cost? Ans. 10. 23. If - of a ton of coal costs $4.90 what will 9- of a ton cost? Ans. $5.67. 24. A merchant bought a load of wheat, T of which cost $24, and afterwards sold 1- of his load at cost; how much did he get for what he sold? Ans. $9. 25. A has 420 sheep; - of A's is equal to 4 of B's; how many has B? Ans, 294. 26. A man bought 16 yds. of cloth for $60; if he should sell 4 of it at cost, how much would he get for it? Ans. $37.50. 27. If a man pay $33.50 for 264 days work, how much must he pay for 67k. days work? Ans. $84. 28, If a man ride 645 miles in 18' days, how far can he ride in 29' days? Ans. 1036 miles. 29. If 7 lbs. of tea cost $5f,how much will 12 lbs. cost? Ans. $9. Suggestion-$5 =- $2. If 7 lbs. cost $t, 1 lb will cost I of $^, and 12 lbs will cost 12 times as much as 1 lb. 30. If 15 acres of land cost $131X, what will 43 acres costt Ans. $376.25. 31. If it cost $27 to build 112 rds, of fence, how much will it cost to build 216 rds.? Ans. $5.54.3 32. If 15 pair of side combs cost $0.93X, what cost 27 pair? Ans. 61.684. 33 If 4 of a bushel of wheat costs "' of a dollar, what costs 3 of a bushel? Sotutin.-If 5 of a bushel costs, will cost as much or ~ = of a dollar, and a bushel will cost 6 times as amuh, or I, o04 COMMON ARITHMETIC.. Sect. XI - Again, if a bushel costs$1, of a buushel will cost 3 of a dollar, or $0.75. 34. If - of a yard of cloth costs - of a dollar, what costs 3 of a yard? Ans. $0.62. 35. If A7 of a farm costs $2100, what costs - of the same farm? Ans.,2000. 36. If,8 of a ship costs $48000, what is 2 of her worth? Ans. $56000, 37. Suppose an army of 1500 men have 120000 lbs. of bread; how long will it last them, allowing each soldier 11 lbs per day? Ans, 64 days. 38. Suppose the above named army should lose 120 barrels of bread, each containing 250 lbs.; how much must be taken from each soldier's allowance per day, in order that the remainder may last the same time, and how much will be the allowance of each soldier per day after the reduction is made? Ans. Each soldier loses 6 os. per day, and has 1:5 oz. retaining. 39. Suppose an army of 1500 men, having lost 4 of their bread, were obliged to subsist upon 15 oz. per day f r 64 days; had none of their bread been lost, they would have had 1- lbs. per day for the same time; how much bread had they at first, and how muth was lost? Ai. 5 They had at first 120000 lbs. ns They lost 30000 lbs. 40. Suppose the allowance of an army of 1500 men was shortened 5 oz. per day for 64 days in consequence of a part of their bread being spoiled, and the amount spoiled to be 4 of the whole; what was the whole amount both good and bad, the amount spoiled and the allowance of each soldier per day, both before and after their bread was spoiled? r Total amount 120000 Ibs. j:: Amount spoiled 30000 lbs. 4 Daily allowance of each soldier at first 1- lbs. Daily " " " after a patt wasspoiled 15 oz, NoTE.-Questions similar to the four preceding, and the 21 following ones, are generally solved by reasoning from one statement to another, Without reference to any particular rules. 41. -A: man owning 7 of a farm, sold ' of his share for $2800; required the worth of the farm? Ans. $4800. Solution.-As he sold 3 of his share, he sold j of 7 of the entire farm. of =i S (Sect. VIII, Art. 3, Obs. 10 or t1.) Then $2800 is. of the worth of the farm, and y must be 1 of $2800, or $400, and the entire farm must be worth 12 tiles as.uach as t w $4x00X2=48040. -.. Art. 2. ANALYSIS. 205 4. A merchant owning store, sold of his store, sol for $15 000; what was the worth of the store? Ans. $25000 43. A farmer owning -5 of a field of wheat, sold 4 of his share for $100; what was the field worth at this rate? Ans. Wt70. 44. A cistern holding 400 gallons is to be filled with water. By one pipe 30 gallons run in in an hour, whilst by another pipe 10 gallons run o it in an hour; if both are left open, how long will it take the cistern to fill? Solution.-As 30 galls. run in in an hour, and 10 galls. run out in the same time, it fills 30-10 = 20 galls. per hour. Then as it holds 400 galls. it will take it 400 - 20 = 20 hours to fill, Ans, 20 hours. 45. A boat traveling up stream is driven by steam at the rate of 14 miles per hour, whilst she is retarded by the current 4- miles per hour; how long will it take her to travel 418 miles? Ans. 44 hours. 46. A vessel sprung a leak at sea, and it was not discovered until she had 215 gallons of water in the hold; the pumps empty 5 gallons per minute, whilst the leak lets in 2- gallons per minute; how long at this rate would it take to empty the ship? Ans. 86 min.= 1 hr. 26 min. 47. A man earns $9 per week, and spends $3.50 per week; how much can he sanve in 66 weeks? Ans. $363. 48. A cistern can be filled by one pipe in 10 hours, and by another pipe in 12 hours. If both are left open how long will it take the cistern to fill? Ans. V5 hours. Solutlon.-If a pipe fills it in 10 hours, it will fill L of it in 1 hour. Again, if a pipe fills it in 12 hours it will fill - of it in 1 hour; therefore both will fill to ~ 1 =lof it in an hour. Then to fill the whole cistern, or i, it will'take 60 - 11 l 5,S hours.: 49. A cistern has a pipe which-eill fill it in 8 hours, and another that will empty it in 10- hours. If both are'left open, how long will it take the cistern to fill? Ans. 40 hours. Suggestion,-'It fills - full in 1 hour, and emties.a in the same time; hence, it gains I --- '-T of its contents in an hour. 50. A. can do a piece of work in 6 days,'wiilst it willtake B. 8 days to do the same work. If both work together,i in 'hat time will they do it?: Ans. 3- days. 51. A. can do a piece of work in 9 days; B. in 12 days, and C. in 15 days; how long would it take them all working together, to do it? Ans, 33. 69.i A. can do a certain piece of work in 12 days; B. in 16 days;. in,20 days; andrh in 24 days; in what time will they ll,.woking together, do it?. -.. - -d.. y ax i 4 7 206 COMMON ARtHMETIC. Sect. XI 53. A. anl B. to grther can plow a certain piece of ground in 10 days. A. can do it alone in 16 days; in what time can B do it? Ans. 26- days. 54. A man and his wife eat the flour of a bushel of wheat in 12 days; the woman would eat it alone in 27 days; in what time would the man cat it? Ans. 21 days. 55. A cistern holding 600 gallons, has two pipes which together will let in 140 gallons of water in 3 hours, it has also one pipe which will 'et out 158 gallons in 5 hours; If all are left open, how long will it take the cistern to fill? Ans 39 ti- hours. 56. A. B. and C. can do a piece of work in 15 days; A. and B. can do it in 24 days; in what time can C. do it alone? Ans. 40 days. 57. James and Henry wish to divide 25 cents between them so that Ienry may have 5 cents more than James; how much must each have? Solulon.-If Henry takes the 5 cents, he is to have more than James, it is evident that the remainider should be divided equally between them. Then 25 —5 = 20; 20 — 2 = 10 cts., James' share,:nd, 10-+-5= 15 cts.; Henry's share. (Sect. VI, Art 1, Obs. 14.) Proof, 15 + 10 = 25. 58. Divide 36 marbles between two boys, giving one 10 more than the other? Ans. 13 and 23. 59. A man left $4500 to be divided between two children, giving one $800 more than the other; what was the share of each? Ans. One's share $1850: the other's $2650. 60. A man left his property amounting to $7500 to be divided between his wife, son, and daughter, giving the wife $1200 more thlan to the son, and to the son $800 more than to the daughter.What was the share of each? A j Wife's share, $3566.661; Son's $2366.66-; A" Daughter's $1566661. 61. Divide $1200 between four person, giving A. $200 more than B; B $160 more than C; and C $100 more than D. Ans. A's share $550; B's $350: C's $200; D's $100, 62. What cost 40 yds. of cloth, at $0.25 per yard? Ans. $10. Solution. —Had the cloth been $1 per yard, it would evidently have cost $40. Now $0.26 is \ of $1, therefore it will cost as many dollars a there are yards; and X of 40 is 1O. Art. 24 A^NALYSI-; 267 Obs, 1. The solution of questions by takiny parts, as in this example, is called PRACTICE. Obs. 2. The chief advantage derived from this method of operating, is, that it generally contracts the operation, thus often ennabling the learner to solve questione mentally which would otherwise reedire the use of the slate. Obs. 3. Any number that forms an exact part of another number, is called analiquot part of that number. Thu., 4 is an aliquot part of 12; 60 cts. of $1.00; &c. REMARK.-It will be perceived that the terms aliquot partS, component parts, and factors, are synonymous; that is, they express the same meaning. To obtain the cost of any number of articles, when the price of 1 is an aliquot part of a dollar: Obs. 4. Divide thegiven number of articles by that part of $1 at which a single article is priced: the result will be the answer, in dollars. REMARK,-If there is a remainder after dividingi the quotsiitt may be extended to cents, and mills, by annexing ciphers. Obs. 5. Practice is chiefly used in computing the price of varicu; articles of trade; as groceries, cloth, land, grain, &c. The aliquot parts of $1 are shown in the followi;g TABLE. 12: cents, -, is, of -1. 16 s.-. is of ---- $1. 20 " - is! of.._... --- -- $1. 25 - is O f -,- _$_- -- 1. 33 "-, --- —is of-... $1. 50 -. --- —- is ot --- — $ 1. ALSO 1'2 cents _ is of $0.25. 12 -.. is of -— 0.. 25 -is of $.6 50. 25 -- i of -0.75. 37 2 --- -- - is 1 of --- $0.75. 371 " ----is8= 4+ —. $1. 62 -------- - is 1 + 8 of --- — $1. 75 --------- --- is -2 + of ---- --- $1. 75 S-is + $. -— b6| --- —. 3 is — + of. $1.. 87' is -- - +, or 1-4 of $ What is praclice? What is the advantage derived ftom this method of eperating? What is an aliquot part of a number? Give examples. What isthe difference in the meaning of the terms aliquot parts, component parts, and factors? How do we obtain the cost of any number of articles when the priceof 1 is analiquot part of a dollar? f here is a remainder how 'do we proceed? For what is practice chiefly usedl Give *etapipesi.. G.ivT omn of the aliquot parts of $1t 208 COMMON ARITItME'tI. Sect. XI 63. What cost 72 lbs, of coffee, at 12 cents a pound? Ans. 89. 64. What cost 68 yds, of muslin, at 124 cents a yard? Ans $8. 50. 65. What cost 75 slates, at 162 cents apiece? Ans. $12.50. 66. What cost 110 books) at 20 cents apiece? Ans. $22. 67. What cost 50 palm le:f hats at 25 cents apiece? Ans. $12.50. 68. What cost 45 bushels of potatoes, at 33}- cents per bushel? Ans. $15. 69. What cost 1200 bushels of barley, at 371 cents per bushel? Ans. $450. Operation. The cost at $1 per bushel, would be 4 $1200 cost at $1. $1200, the cost at 25 cts. per bushel - would be - of $1200 s $300; the cost 2 $300 costat 25 cts. at 124 cts. per bushel would be ~ as $150 cost at 12 cts. much as at 25 ct-., or ' of $300= $150; then the cost at 37- ets. per Atns. $450 = cost at 37- cts bushel would be equal to the cost at 25 cts. + the cost at 121 cts,, or $300 - $150 = $450, because 37- cts. = 25 cts. + 121 cts. 701 What cost 400 pair of socks, at 37' cts. a pair? Ans. $150. 71. What cost 247 yds. of cloth, at 50 cts. a yard? Ans. $123. 50 72. WhlA cost t 471 bushels of rye, at 624 c:s a bushel? Ans. $294.37. 73. What cost 600 bls. of cider, at 662 cts. per barrel? Ans. $400. 74. What cost 300 lbs. of tea, at 75 cts. per pound? Ans. $225.?5. What cost 400 yds. of satinet, at 871 cts. per yard? Ans. $350. 76. What cost 150 acres of land, at $8,25 per acre? Ans. $1237.50. Suggestion.-$8.25 = $8; therefore multiplying 150 by 8-, it s evident we shall have our answer in dollars. HIenceWhen the price of an article is in dollars and cents, Obs. 6. Multiply the number of articles, by the numbeo of dollars, take aliquotparts for the cents, and add the several results to ether. 'Explain the aolution'of Ex. 69. When the price of an ttUiat ii in dollarO i d centi how do we proceed? - ' Art 2. ANALYSIS. 209 77. What cost 250 bushels of wheat, at $1.1 2 per bushel? Ans. $281.25. 78. What cost 360 yds. of cloth, at $2.162 per yard? Ans. $780. 79. What cost 480 silver pencils, at $2.25 apiece? Ans. $1080. 80. What cost 260 gallons of wine, at $2.33- per gallon? Ans. $606.66|. 81. What cost 80 yds of cloth, at $4. 37' per yard? Ans. $350. 82. What cost 160 acres of land, at $12.50 per acre? Ans $2000. 83. What cost 40 sacks of coffee, at $12.62y per sack? Ans. $505. 84. What cost 72 bls. of flour, at $7.66] per barrel? Ans. $552. 85. What cost 120 acres of land, at $5.75 per acre? Ans. $690. 86. What cost 24 stoves, at $15.87' apiece? Ans. $381. REMARK.-The preceding examples in practice, are confined to Federal Money; we now design to show its application to compound numbers. 87. What cost 12 acres, 40 sq. rds. of land at $20 per acre? Ans. $245. 40 sq, rds. = of an acre. Then $20 X 12 = $225. HenceOperation. To find the value of a quantity 4 $20 consisting of several denomina12 tions: $240=cost of 12 acres. 5= cost of 40 sq. rds. Ans. $245=costof the whole. Obs. 7. Multiply the price by the number of the denomination at which the price is rated, and take aliquot parts, to find the value of the lower denominations. The sum of the several values will be the value of the entire quantity. A few of the principal aliquot parts in compound numbers, are shown in the following How do we find the value of a quantity consisting of several denominations? COMMON ARITHMETIC. Sect. XI TABLE, LONG MEASURE. 2 furlongs are, of a mile. 4 'e ' I (( 6 "' * 3 -_ 40 rds. " - of a mile. 80 '. ' ( 160 rds. I4 " C 240 '" "-3 +I 1 4 -2[41 20 " 1 < a furlong. 1 ft. is of a yard. 6 inches " 2 a foot. SQUARE MEASURE, 40 sq. rds. are - of an acre. 80 " i t t, 120 " " 3=4 2+ AVOIRDVPOIS WEIGHT. 5 cwt. is 4 of a ton. 10 " is 1 15 " is 3.+- ton. 4 oz. is 1 of a pound. 8 oz. is 1 " a 12 oz. is 3 =_ 1 (,,~i~ These are the parts that occur the most fi'cqucntly in common business. Others might be given. but it is thought unnecessary, as the pupil can easily discover them by the following directions: Obs. 8. Make the given number the numerator of a fraction, and that number which it takes of this denomination to make 1 of the next higher, the denominator. Thus, if it were required to find what pa:t of an hour were 30 minutes, we make 30 the numerator, and 60 (minutes in an hour) the denominator; thus, -- 88. How much would 12 bu. 2 pks. 4 qts, of wheat cost, at $1.25 per bushel? Ans. $15.78'. Oleration. 2 $ 1.25 2 pks.- a bushel. Hav12 ing found the cost of 12 bush-- -els, we find the cost of 2 pks., $15.00 c= cost of 12 bu. or 2 a bushel, by taking 1 the 4.62 = cost of 2 pks. price of 1 bushel, ($1.25.) 4.15 -= cost of 4 (its: qts. = a peck, or - of 2 pks. -Therefore, we take - of the $15.78- = cost of the whole. cost of 2 pks., ($0.62j) as the cost of 4 qts., and the sum of all these is equal to the whole.(Sect. IV. Art. 4. Obs 4. Rem. 2.) 89. How much will 15 bu. 3 pks. 6 qt-. of wheat cost, at $1.44 per bushel? Ans. $22.95. The learner will perceive that as many decimals are pointed off in the result, as there are in the given price of the single article. Name some of the aliquot parts of Long Measure? Square Measure? Avoirdupois weight? How do we find what part one number is of another? In finding the quantity of several denominations, how many decimals do we point off in the result? Art. 2] ANAL Y SIS. 211 90. If a man walk 24 miles, 80 rods per day, low far can he walk in 48 days? Ans. 1164 miles, 91. How much would 15 acres, 2 R. 20 sq. rds. of land cost, at 8. 50 per acre? Ans. $132.81. 92. How much would 18 A. 1 R. 10 sq. rds. of land cost, at $10 an acre? Ans. $183.12P. 93. How much would 32 A. 3 R. 30 sq. rds. of land cost, at $15 an acre? Ans. $494.06j. 94. How much would 6 T. 10 cwb. 20 lbs. of hay cost. at,8 a ton? Ans. $52.08. 95. How much would 15 cwt. 2 (rs. 12 Ibs. 8 oz. of cheese cost, at $12.50 per cwt.? Ans. $195.31~. 96. If a man work 12 hrs. 30 min. per day, how much time will he work in 36 days? Ans. 18d, 18 hrs. 97. If a man earn $1.50 per day, how much will he earn in 24 days, 18 hrs. 45 min.? Ans. 537.17L. 98. How much would 15768 lbs. of hay cost, at $8.40 per ton? Ans.;66.25. Solution.-A ton is 2000 Ibs. Tlerefore, the hly costs $8.40 -2 = $4.20 per thousand pounds, and its cost is found according to Sect. VIII. Art. 11. Obs. 1. to be G66.2255.I-encreTo find the cost of articles by the ton, or 2000 lb,.: Obs. 9. Multiply half the cost of a ton by the nuulcr of pounds, and point off three additional decimals from the riglat thand. 99. How much will be the storage on 27480 lbs. of goorls, at $3.60 per ton? Ans. 849. 164. 100. How much would 7240 lbs. of coal cost, at:6.90 pFe ton? Ans. 824.978. 101. How much would be the transportation of 12415 lbs. of goods at $11.80 per ton? Ans. $73.248. 102. At $7 per ton, how much would 1694 lbs. of hay cost? Aus. 85.929. 103, A farmer sold 50 bushels of wheat, at $1.25 per bushel, and took his pay in cloth, at $0.75 per yard. How many yards did it take? Ans. 83. ' Solution.-We first find the cost of the wheat to be $62.50. (Obs. 6') Next we divide this by.75, because the coth is $0.75 per yard, and he can evidently buy as many yards as.75 is contained in $62.50. (Sect. VI. Art. 1. Obs. 24. a.) How do we find the cost of articles by the ton of 2000 lbs.? Denonstrate this rule. 212 CGMMON ARITIHMETIC. Sect. Xt Questions of this kind are sometimes solved by a rule calied BARTER. Barter signifies an exchange of different commodities, at prices agreed upon by the parties. NOTE.-It is not necessary to give an additional Rule for such questions. They can all be solved by Analysisl and the process is so simple that any scholar of moderate capacity can easily understand it. 104. How many bushels of oats, at $0.31-J a bushel, must be given in exchange for 48 yards of muslin, at 12j cts. a yard? Ans. 19-. 105. How many cords of wood, at $2.75 a cord, must be given for 75 lbs of sugar, at 10- cts. per pound? Ans. 254. 106. How many pounds of butter, at $0.16 per pound, must be given for 48 yards of calico, at $0.25 p:er yard? Ans. 75. 107. If I give 45 acres of land, worth $25 an acre, for 15 horses, how much do the horses cost me apiece? Ans. $75. 108. Bought a hogshead of molasses, at 80.45 per gallon, and paid for it in butter, at $0.18 per pound. How many pounds did it take? Ans. 1571. 109. How many sheep, at $1.75 apiece, must I give for 35 cows, at $10 apiece? Ans. 200. 110. How much tobacco, at 12- cents a pound, must I give in exchange for 300 pounds of saleratus, at 5 cts. a pound? Ans. 120 lbs. 111. How many pair of shoes, at $2. 25 a pair, will it take to pay for 144 yards of satinet, at $ 1. 25 a yard? Ans. 80. 112. Bought, 720 bushels of apples. at $0.75 a bushel, and paid for them with 1620 bushels of potatoes. How much were the potatoes per bushel? Ans. $0.3.3'. 113. A man mixed 5 lbs. of tea, worth $0.50 per pound, with 6 lbs. worth $0.75 per pound, and 8 Ibs worth $0.80 per pound. How much was a pound of the mixture worth? Ans. $0.70'. Operation. $0.50 X 5 = $2.50 = the price of 5 lbs. at $0.50 per pound. 0.75 X 6 = 4.50 - " " 6 lbs. at $0.75 0.80 X 8 - 6.40 = " " 8 lbs. at $8.80 " 19 ) 13.40 (.70-0 cts. Ans. 13.3 10 The whole number of pounds mixed are 5+-6 + 8 = 19!bs. What does barter sifinify? Art. 2. ANALYSIS. 213 The cost of the whole mixture is $2.50 + $4.50 - $6.40 = $13.40. Then one pound must be worth $13.40 4- 19 $0.70~-. HenceTo find the cost of a pound, bushel, &c., of a mixture: Obs. 10. Divide the whole cost of the mixture by the whole number of simples. Obs 11. The mixing of several simples of diferemt qualities together, as in the last example, to form a compound of a mean or middle quality, is called ALLIGATION. Alligation is of two kinds-MEDIAL and ALTERNATE. Obs. 12. ALLIGATION MEDIAL is used when the prices and quantities of the simples are given, to find the price of the mixture compounded of them, as in the last example. REMARK.-Alligation Alternate is not much used in common business, and is not treated of in this work. Its chief object is to find the proportional quantity to be taken of each simple, when the price of the mixture, and ofthe several simples, is given. 114. A farmer made a mixture, containing 12 bushels of oats, worth $0 25 per bushel; 15 bushels of corn, at $0. 372 per bushel,.and 8 bushels of peas at $0.624 cents per bushel. What is a bushel of the mixture worth? Ans. $0.38',-. 115. A grocer mixed 30 gallons of water with 60 gallons of whiskey, worth $0.20 per gallon. How much is a gallon of the mixture worth? Ails. $0. 13k. 116. A grocer mixed 18 lbs. of sugar, at 8 cts. per pound, with 22 lbs., at 10 ots; 24 lbs., at 12 cts,; and 30 lbs.. at 14 cts. per pound. How much is a pound of the mixture worth? Ans. $0. 11. 117. A man mixed 40 gallons of wine, worth $2.25 per gallon, with 30 gallons of brandy, worth $3. 5 per gallon, and 20 gallons of water. How much was a gallon of the mixture worth? Ans. $2.16. 118. A goldsmith mixed 8,oz. of gold, 16 carats fine; 12 oz., 18 carats fine; 16 oz., 20 carats fine; 20 oz., 22 carats fine; and 24 oz., 24 carats fine. Required-the fineness of the mixture. Ans. 21 carats. 119. James and William wish to divide 24 cents between them, in such a manner that James may have 3 as many times as William. How many must each one have? Ans. James 18, and William 6. How do we find the oost of a pound, bushel, &c., of a mixture? What is Alligation? How is it divided? When is Alligation Medial uwdl 214 COMMON ARITHIMETIC. Sect. XI Solution.-Whilst James takes 3 cents, William takes 1; and both take 4 cents. Then James must rec ive as many times 3 cents, and William as many times 1 cent, as 4 is contained in 24. 24 -4= 6. Therefore, 3 X 6 = 18 cents for James' share, and 1 X 6 = 6 cents for Willia:n's share. PROOF.-18 +6 = 24. 18 = 6 X 3. Obs. 13. T'le process of dividinq a numbcer into two or more parts that shall bear a certain relation to cach other, as in the last example, is called lRoroPITIONA LDIVISION. To divide a number into proportional parts: Ohs. 14. Divide the nu7rnber to be divided by the sum of iheparts the result will be one part, fJ om which the other parts can be found. PRiOF -Add the sever(d rcsults together; if their sum is equal to the numitber divided, the work is correct. REFMARK 1.-The reason of the proof depends upon the selftevident principle, tllat the whole is equal to the sum of all its parts. 120. Suppose tvwo men start, one from Columbus, and the other from Wooster, Ohio, and walk towards each other, the former at the rate of 3 miles per hour, and the latter at the rate of 4 milesper hour. How far would each one walk before ihey would meet, the distance being 85 miles? Ans. The former walked 36- miles, the latter 47- miles. 121. Divide 126 into three parts, that shall be to each oth, r as the numbers 2, 3, and 4. Ans. 28, 42, 56. 122. Divide 144 into 4 parts, the numbers of which shall be to each other as the number 3, 4, 5, a' d 6. Ans. 24, 32, 40, 48. 123. Divide 120 into four parts, that shall be to each other as the fractions 12,, s and a. Ans. 48, 32, 24, 16. REMARK t2.-By multiplying each fraction by the least common multiple of their deaominators, we find they are to each other as the numbers 6, 4, 3, and 2. 124. The standard for gold and silver coins in the United States is 9 parts pure metal to 1 part alloy. How much of each is there in an eagle, which weighs 10 pwts. 18 grs.? Ans. Gold, 9 pwts. 16' grs. Alloy, 1 pwt. 1 grs. 125. Gunpowder is composed of 76 parts of nitre, 14 of char What is Proportional Division? How do we divide a number into proportional parts? What is the method of proof? Upon what does the reason of this method of proof depoed? How do we find the relation which fractions bear toch other? Art. 2. ANA LYSIS. coal, and 10 of sulphur. How much of each is there in 200 lbs.? Ans. Nitre, 152 lbs. Charcoal, 28 Ibs. Sulphur, 20 lbs. 126. If a man ride 288 miles in 8 days, riding 9 hours per day, how far, at the same rate,-can he ride in 15 days, riding 12 hours per day? Soutkon.-If he rides 288 miles in 8 day>, he will ride 288- 8 = 36 miles per day; and if he rides but 9 hours per day, he will ride 36 — 9 = 4 miles per hour. Again-if he rides 4 miles per hour, he will ride 4 X 12 = 48 miles in a day of 12 hours length; and in 15 such days he would ride 48 X 15 = 720 miles. Ans. 720 miles. Or, the operation may be performed by cancelation. Thus: p 15 The learner will perceive that we multiply 12 and divide the same as in the above solution. Ans. 720 miles. NoTE.-Questions of this kind are usually solved by Compound Proportion, which is explained in the next section. But the solution by Analysis is preferable to any other. The learner will observe that the canceling part is not the solution, but merely the operation. The operation is the mechanical, and the solution the mental, or reasoning part, of the work, aud a pupil ought never to accustom himself to the use of the former, without the aid ot the latter, as a thorough knowledge of all the reasons why he performs his operations as he does, is absolutely necessary, if he ever wishes to become a good arithmetician. 127. If a man ploughl-~acres in 10 days, how long will it take 18 men to plough 230 acres? Ans. 10-4 days. 128. If 5 men earn $75 in 12 days, how long will it take 8 men to earn $200? Ans. 20 days. 129. If 15 men dig 360 rods of ditch in 4 days, how many rods will 24 men dig in 12 days? Ans. 1728. 130. If 12 horses eat 126 bushels of oats in 14 days, how many bushels will 16 horses eat in 24 days? Ans. 288. 131. If a family of 5 persons spend $87.50 in 7 months, how much will they spend in 16 months, if 6 more persons are added to the family? Ans. $440. 132. A farmer has I of his hogs in one field, - in another, and 24 in a third field. How many has he in all? Ans. 144. Solution.-'- + -- =5 what he has in two fields; then the third field must contain - -- - =- of the whole. But the third Explain the difference between the operation and the solution of a question in mathematics? Should the former ever be used without the latter? Whv not? 216 COMMON ARITHMETIC. Sect" XI field contains 24; therefore, 24 is i of the whole, and 24 X 6= 144, the whole lot. NOTE.-Questions of this kind are sometimes solved by a rule called Position, or Trial and Error. Analytic solutions, however, are preferable. REMARK 3.-Position might, with propriety, be called the Guess Work Rule. In Arithmetic, all questions should be solved upon trnueprinciple,t without any guess work about it; and as all examples in Position can be solved by Analysis, to give an extra rule for their solution would be unnecessary. 133. In a certain school 4 of the pupils study Arithmetic, - study Philosophy, 12 study Algebra, and the remainder, who are 7 of the ' school, study various branches. How many scholars are therein the school? Ans. 84. 134. Of a certain army, 1 were killed, 4 were wounded, - were taken prisoners, and 3000 fled. How many were in the army at first? Ans. 12000. 135. A drover has sheep in 6 lots: In the first he has - of his flock; in the second,,; in the third, i; in the fourth, 8; in the fifth, ~s; and in the sixth he has 84. How many sheep has he? Ans, 480. 136. If a pole is 1 in the mud, ~ in the water, and 20 feet out of the water, what is its length? Ans. 48 feet. 137. A schoolmaster being asked how many children he had, answered: "' If I had as many more as I now have, 3 as many, - as many, x as many, and 1- as many, I should then have 270." How many had he? Ans. 75. 138. What number is that, -,, and l of which is 36? Ans. 48. 139- What number is that, 2 of which exceeds - of it by 10? Ans. 60. 140. What number is that, from which if you take -, l, and - of itself, the remainder will be 44? Ans. 96. 141. What number is that, which, if you increase it by -, I, and -1 of its if, the result will be71;. Ans. 120. 142. A.'s age is 3 times B.'s, and twice A.'s age equals C.'s; the sum of all their ages is 80 years. How old is each? Ans. A. 24, B. 8. and C. 48 years, This example might be performed by Proportional Division, B.'s part being 1, A.'s part 3, and C.'s 3 X 2 - 6; then 6 +3 + 1 = 10, the sum of all the parts. What might Position with propriety be called? I-ow shbold all questions in arithmetic be solved? What do you mean by true principles? How can all questions in Position be solved? tBv t rue prtincipl, are meant principles 6apable of demonstratlon or plftw Art... ANALYSIRn 217 143. A., B.. and C. talk of their ages. A. said his age was 1 of B.'s, and C. said his age was 7 of the sum of the ages of A, and B. both; the sum of all their ages was 121 years. Required-the age of each. Ans. A.'s age, 30 years. B.'s, 1.4 years. C.'s 77 years. 144. A m n bought a sheep, cow, and horse; the cow cost 5 times as much as the sheep, and the horse cost 5 times as much as the cow and sheep both. They all cost $72. Required-the cost of each? Ans. The sheep, $2; the cow, $10; and the horse, $60. 145. A. and- B. have the same annual income. A. saves - of his, but B., by spending - as much again as A, loses $75 in the course of the year. Required-the amount of the annual income of each. Ans. $300. 146. A. and B. have the same annual income: A. saves -' of his, but B., by spending $50 per year more than A., at the end of 5 years finds himself $75 in debt. Required-the annual income of each. Ans. $350. Solution.-B. runs in debt $75 in 5 years. This is $75-5 = $15 a year more than his income; and as he spends $50 per year more than A., A. must save $50- $15 = $35 a. year; and as A. saves -t of his income, their income must be $35 X 10 = $350. 147. A men hired 90 days on these conditions: for every day he worked, he was to receive $0..75; and for every day he was idle, he was to. pay $0.25 for his board. At the end of his time he received $40. How many days did he work, and how many days was he idle? Ans, He worked 62' days, and was idle 27Y-. 148. A. and B. commence trading with equal sums of money: A. gained a sum equal to A of his stock, and B. lost $200. B. had then as much money as A. How much had each at-first? Ans. $1200. 149. A man left his property to his 3 children, giving A. I wanting $200; to B. -, and to C. the rest, which was $200 less than the share of B. What was the value of the estate, and what was: each one's share? Ans. Estate, $3000; A.'s share, $800; B.'s share, $1200; and C.'s.. [share, $1000. 150. A person being asked the time, replied: "The time past. noon is equal to -5 of the time till midnight." What time was it? Ans. 45 min. past 3 o'clock. isA. A certain pole is composed of three pieees; The top piec is 15 feet long, the middle piece is as long as the top piece, and 4 the length of the loWer. piece; and the lower piet' i, as... By; 218 C0*10N ARITH1METIC. Sect. XI both the oth r pieces. How long is the pole and how long is each apiece? 'Ans. Length of the p ec, 100 feet; length of pieces, 15, 35, 50 ft..152. A man has a gold and a silver watch, and a chain worth $20. If he put the chain on the gold watch, its value will be 6 times that of the silver watch; but f he put the chain on the silver,watchi, its value will be but half that of the gold watch. Required -the -yalue of th.e two watches. Ains. The gold one $70; the silver one $15. Solution. —As tta chain, when put on the gold watch, makes its value 6 times that of the silver watch, the silver watch must be worth $20- 6 = $3- -+4 f the value of the gold watch. Ther — fore when the chain is put on the silver watch its value must be $20 +- $33+ - of the y'lIe of the gold watch. But by the question this is E the value of the gold watch; therefore, twice this, or $40 + - $6W + - of the value of the (rold watch must be the value of the gold watch. Now $44 ) + -6- =_,46; and from the above reasoning, this must be 3 of the value of the gold watch. If $46|- is:, $46- 2 = $23-, is - and $233 X 3 - $70, the value of the gold watch. 153. A ma'n has two horses, and a saddle worth $25. If he puts the saddle on the first horse, his value will be U'7 of that of the second horse; but if he puts the saddle on the second horse, his 0l1ue will be 7 of that of the first horse. Required-the value of each horse. Ans. The first $60; the second $45. ';54. The hou and minute hand of a clock are together at 12 o'clock. At what time are they next together? 0n11~~ WGAns. 5 min. 27i- s-c. past 1. Soluton.-The minute hand moves over 12 spaces, whilst the hotr hand moves over but 1, and therefore gains 11 spaces in an hour. If the minute hand overtook the hour hand precisely at 1,'clOck, it would gain 12 spaces in an hour. But it gains only 11,spaces in an hour; consequently to overtake the hour hand it must ga n i'r of the distance it has to run. "Ti' of 1 hour, or 60 minutes is 4 W,.r,.5 minutes, 272-~ seconds. (Sect. IX. Art 3. Case 2.) 15t5 At what time between 3 and 4 are the hands of a clock togeter! Ans. 16 min. 21 2i sec. past 3 o'clock. - 156, At Wyi. tiae between 10 and 11 are the hands of a clock,gether?9 Ans. 54 min. 32TV sec. past 10 o'clock. 157. At what tile betweejr 9 and 10 are the hands of a clock,exactly opposite each otke'? Ans. 16-1 min. past 9 o'clock. Suggestion. —The learner will perceive that to be opposite each gee'~, pe sands must be 30 min'ites apart. Now at 9 o'clock the Ar,. A. NALYS6lq 19 minute hand is 15 minutes alhead of the hour lhand; and thlerefore, it has but 15 minutes more to gain to be opposite the hour hand. 1 58. At what time between 2 and 3 are bthe bhands of a clolk exactly opposite each other? Ans. 43-7- min. past 2 o'clock. 159. Four men trade together: A. put in 7Y as much as B.; C. put in ' of as of what A. and B. both put in; and D. put in 17 of 8 of what B. and C. put in. They all put'in $9875. What waseach one's share? Ans. A.'s $1750; T.'s $2750;.'s.s $2500; and D.'s $1.875. 160. A., B. and C. tra!'e in company: A. put in $4500; - of what A. put in was equal to 5 of what B. put in; and of what B. put in, added to -5 of what A. put in, was T-r of what C. put in. They gained a certain sum, of w'hionh A.'s share was eqtual to 2 of what B. and C. put in; B.'s share was equal to - of what A.,nd C put in; and C.'s share was equal to L of what they all put in. What wa; the whole amount invested? How much did B. and C. put in; what did they gain, and 'what was each one's -share of the gain? Whole stock, $15000; B. put in $5000; C. piut in $5500. Ans. < Whole gain, $0000; A.'s share,;3J00:;.s share, ( $3333.331-; C.'s share. $3666.6G6-. SECTION XII. RATIO AND PROPORTION. ARTICLE 1. RATIO. Obs. 1. RATIO means relation, and signifies tswat relation which -one number has to another of the same kind. It is expressed by the lquotient of one divided by the other. Thus, if we inquire, What is the ratio of 4 to 2? if we make the,first number (4,) the standard of comparison, the question resolves itself into this: What part of 4 is 2? and the answer is, 2 is 1 of 4. Again: if we make a second number (2,) the standard of comparison, the question is, What p2art of 2 is 4? and the answer is, 4 is twice 2. The first of these methods is called the Frmch method, by which the ratio of 4 to 2 is 2 - 4 =- 2 Whatdoes ratio mean? What does it signify? How is it expressedt How do we find the ratio between two numbers? 320; CLMMON ARITHMETIC. Sect. XII T le second method isfcalled the English method, by which the atio of 4'to 2 is, 4'- 2 '=.-2. Each method has advlantages peculiar to itself, but we shall follow 'ie French method, as it is generally adopted in this country.. - ence-* To,find, the ratio of cte riumber to another:' Obs. 2. Divide the second'number by thefirst.. MENTAL EXERCISES, 1:. What is the ratio of 9 'to 3? of 6 to 12? Ans. -, 2. 2: What is the ratio of 16 to- 2? Of 01 to 7? Of 48 to- 12? f 63'to 9? Of 84'to 7? 3. What is the ratio of 30 to 6? Of 36 to 4? Of 120 to 10T; Of 132 to 11? 4. What is the ratio of 3 to 4q Of 54o-82 Of 7 to-9? Of 8' t1I 5? Of 23 to 37? 5. What is the ratio of 15 to 25- Of 18 to 36? Of 36 to 84? Of 90 to 100? 6. What is the ratio, of 75 to 25? Of 144 to 48? Of 132 to, 1l728? 7; What is the -ratio of 288 to 5.76? Of 132 to 96? Of W96:to 210? Obs. 3. The twonumbers thus compared, when spoken of together, as in the preceding examples, are termed at COUPLET; but: when spoken of separately, they are called the terms of the couplet. Obs. 4: The first term is called-the ANTECEDENT, because it comes before the other; the second term is called the CONSEQUENT, because iit,filtows after the other. 8. Of what numbers is ' the ratir? Or, what number, divided ry. another, will produce? It is evident that the least numbers that' will dothis are sand 6,; c-. the ratio of 5 to 6 is 5. HenceTlo find two numbers of which a given fiaction; is, the-ratio: Obs. 5. Make the numerator thie scond. number, and the denomina — tcr'the first term of the couplet. 9' Of what two- numbers are-the following fractions the ratio? 2. 7-.. 12. 12. 6 20. 288. 7 9. o1 0. 1723. 2 31 T3' 8- 7 T' 4 ' * 2 e5'~ 144T '63" 48 T T437'T 5'76 1t: Of what two numbers is 7 the ratio? When two numbers compared are spoken of togetlwr, what are they called?' When- spoken of. awprately, what are they called? What is the first term caled?' Why? What is-the second term called? Why? How do we find tw-ebulnbers,,of which la given fraction is the ratio? Art. 1. RATIO AND PROPOcTIjN N. It is evident that we wish to find twf,?ii c of vhiicih mitiplied by 7 will produce the other. T h aIssuming 1 as onelnaulber, the other must be 1 X 7 7; or, asuming 2 as one numnibr,:the other is 2 X 7 = 14; also, 3 X 7 -- l,. Icc. That is,,7 is.i}3.ratio of 1 to 7; of 2 to 14; of 3 to 2-1, &c. HenceTo find two numbers of which a given integer is the ratio.: Obs. 6. Take any number as the first term of the couplet,acnr d:multiply it by the ratio to obtain the second term of the coe!,plet. [Ilihe same rule will apply when tle ratio is a frac.tion, and other terms are desired than the numerator and deiomiinator. 11. Of what two numbers is 6 the ratio? 9; 4; 8; 12; 10; X.; 26; 38; 18? 12. Of what two numbers is 3 the ratio? 5;!11; 13; 23;,; 108; 150? REMARK 1.-As the ratio is fotrnd by dividing cue numrer by another, it caa always be expressedfractionally. Thus, the ratio of 6 to 3 is; -and.the ratio of 8 to 16 is t6. 2.-Or, the ratio may be expressed by a couplet of (lots (: ) placed betweea the terms of the couplet. Tnus the ratio of 10 te.15 may be ep, essed'thus10: 15. Hence — lObs. 7. Rato may be esp2ressed in two ways, anul it is 'imriterial ~which we use. Obs. 8. As ratio expresses the elation of nunibers, one enumber cannot have a ratio:toanother number of -a difercnt kiund. Berause:t would be absurd to c,mpare bushels with hours, or feet with-pounds as a bushel cannot be said to be part of an hour, or a foot longer or shortor than a pound. But we can compmre buslhels with bushels, 'or feet with fcetor pounds withpounds, because they express things of the same kind. REMARK. —From this we perceive that ratio refers only to the relative, andr not to tie absolute magnitude of the two quantities. For.1 Ib.:hasthe sanem relation to 2 Ibs. as 1 ton to 2 tons, or 1 ounce to 2 ounces.;13. What is the ratio of 2,pecks to 1 peck? Ans. ', or I peck is' of 2 pedks. How do we find t-wo numbers of which a given integer is the-ratio? W:ill the same rule ajpply when the ratio is a fraction.? How can ratio always be expressed? How may it,otherwise be exwpressed? What is the inferecet deduced from this? Can ratio express the relation between numb)ers of diffrrenit kitils? Cau we cointare bushels with hours? Why not? Can we rompipyae efe:t with pounds? Why not? To what can we compare bush-ls? With what can we compare fe.e? With wliat can we compare pn)uud? <Vlhy? Does ratio refer to the relative or absolute magnitlde of two quautities? Viv. an example. 1... _7 f *UZ -~, OCOMMON ARITHMETIC;, ectC XI11 14. What is the rati' bushels to 3 pecks? NoT:.-First redAce botrl t6 tite same denomination., 2 bu. = 8 pkis. Ans. 3. Obs. 9. Hence —When the terms of the couplet are of the same kindt, but of different denominations, they must both be redvoced to the same denomination- before their ratio can befozund. 1-5. What is the ratio-of 2 feet to 4 yards? 16. What is the ratio of 1I ton to 161 cwt.-? Of 3 seconds to 4 3mnutes? 17. What is the ratio of 120 sq. rds. to 2 acres? Of 7 acres to 21 acres? Of 320 acres to 1- sq. mile? Of 6 miiles to 3 miles? Of 8 miles to 4 furlongs? 1-8. What is thie ratio of 2 miles to 1-80 rods? Of t bushel to 6;qts.? Of 1 gallon to 2 qts.? Of 1 ton to 1 lbs.? RNMARK.-As the ratio of any two numbers can always be expressed fractie.nally, it is obvious, that as fiir as the value is concerned, the same general principles are equally applicable to both ratio atnd fractiots. HfenceObs 1t. I 7 wZe multiply the crnsequent, or divide the antecedent by any, number; we multiply the ratio by that number. (Sect. VIII. Art. ` Obs. 1 and 5.) TI us,: The ratio of 2 4 is 2. M-ultiplyliDg. the consequentby 2, the ratio of 2 8 is 4, or 2X.2. Dividing the antecedent by 2, I' 1: 4 is 4, or 2X2.. R:slARK.-This is evident from the fact that ws either multiply the numeraltor or divizie the drynomilatJr, either of whiclh has- the same effect upon fthe quotient. (Sect. VIII. Art. 2. Ohs. 7.) Obs. 'h. If we divide the consequent, or mvt!lely theantecpedent by any number, we dividi the 1ratio by that nuvwmber. (Sect. Vll. Art. 2. 0bs. 2' and4 4~.) Thus, The ratio of 4 16 is 4. Dividing the consequent by 2, ' 4 8 is 2, or 4 —2. Multiplying the anwtce dent by 2;. " 8: 16 is 2, or 4-2. RE.ARK. —Tiis is eviden;t from tle fact. that swe cither divide the numerator *r'muoltiply;the denoomi:otfor, either of whielh h;is-.th stame effect upon the quo.tient. (Sect. V1:I. Art. 2. Obs. 7.) I When tiHe terms of the couplet are of tlie same kind, but of different dernsinations, how do we proceed? Thli rntio of ay two numibers can always b+ expres.ed fractionalv; what.tl is he irfr!l,:e (!cd(ced Irom this fact? What eSrict does it have up on the ratio, to r::l 'ity ll;e co seq ueit, or divide lhe.autesedeitt,by aly nlun,1ierT ' Shew w,'h tlis is ((' ' clt. What eff;: (oes it have upon the ratio to divide the conseqtunt, or nnmltiIuly the antecedent, by any nwmber?. Show why this is correct? Art. 1. R MIO AND PROAORT ON. 2*3 Obs. 12. If we multiply and d'vide both the antecedent and the consequent by the same number, we do not alter the ratio. (Sect V.ILi Art. 2. Obs. 8.) T'h'.. Thne ratio o 1 2 6 is Maltiplying both terms of the coiup'et by 3, "36. 18 is. Dividing both term-: of the couplet by 3, " " 4 2is 1 REMARK.-Thiq is evident fro'l thl, fact, thitt we me!tiply or dn:ni' both dairisor and dividend by the s<rme number, wicll does not alter tne quo'ient. (Sect. VI. Art. 1. Obs. 28.) Obs. 13. If the antecedent and consequent are equal. the ratio is 1, and it is called a ratio of equality (Sect. VIIi. Art. 2. Obs. 10. a.) Thus. the ratio of 3: 3 is 1, or 3 is equal; to 3. RaEMYARK.-Tl'hi is -vid(,nlt from the fia t, tlht the d^isior a:l d dli nle:ad rre equal. (Sect. VI. Art. 1. Ohs. 26. d.) Obs. 14. If the ani e'edent is qref!e:;,:i i,,':.^?U':d fe r, to is less than 1, andu it is called a:rati) oft i'-!:;iy;i-...' i! ' t. 2. Obs. 10.) REMARK.- This is evident from the fatc, thlt the divisor is greater than the: diviieud. (Sect. V1. Art. I. Obs. 26 d.) Obs. t5. If the antecedent is less than the conseuent. the rati'o is greater than 1, and it is called a ratio of greater irnqualtyt. (Sect.. VIII. Art. 2. Obs. 10. b.) Thus, the ratio of 5: t5, or 3, is a ratio of greaesr-ieequality.. REMARK.-This is eviient fro'n thf fiact, that the dividend is greot:, than the divisor. (Sect. VI. Art. I. Obs. 26. d.) 19. Of what kind is the ratio of 5 to 7? Ans. A ratio of greater inequality. 20. Of what kind are the ratios of the fr]lowing couplets: 6: 4;; 8:8; 7: 14; 36: 24; 15: 15? 20: 40;: 60: 30; 15: 24; 23 17? Obs. 16. Ratio is of two kinds-SINMPLE and' COMPOUND. What effect does it have upon the ratio to muiltiply or divide both the antecedeint nmd coosequeint by the sane nuillber? S lhow w y this is correct. I' the antecedent and colneqluent are equail, whit is the ratio,? k hat is it called? Why is this correcti? If the ant!-cdleet is reat^ar thi., tl!e.aolneq. lent, what is the ratio? What is itcalled? Whyy is thlis corretl?' f thl ant. ced-nt is les than the consequent, what is the ratio? WlLat is it ctled4 Why iatlacoz,-. rect? How is ratio divided? 224 C'OMMTON ARITHM~ETIC. Sect. XII a. A SIMPLE RATIO is the ratio of a single expression, or couplet. b. A COMPOUND RATIO is the ratio of the products of the corresponding terms of two or more couplets. Thus, the ratio of 6: 2, or 12: 24 is a simple ratio. Again the simple ratio of 3: 6 is 2. And the simple ratio of 2: 8 is 4. And the ratio compounded of these is the same ma the simple ratio of 6: 48,which is 8. NoTE.-Compound ratio is not different from any other ratio. It is only used in some cases to denote the origin of the ratio, 21. Whi' is the simple ratio of 4: 2? Of 8 16? Of 48 96; Of 56: 28? 22. What is the compound ratio of the following ratios: 6: 12. 8:. 2 8) 12 6) 18 9) 4 245 9 2; 6 24>; 24 8; 12 36. 4 12) 16 9) 4 16). v~ ~ ARTICLE 2. PROPORTION. "Obs. 1. A PROPORTION consists of an equality of ratios. Thus, the ratio 3: 9 is equal to the ratio of 4: 12; that is, the rato of both is 3. Hence, the two couplets 3: 9 and 4: 12 form a proportion. OCs. 2. The terms of the two couplets which compose the proportion are called PROPORTIONALS. Obs. 3. Proportion is usually expressed by four dots (:: ) placed between the two couplets. Thus: 4: 8:: 6: 12 is a proportion, and is read, 4 is to 8 as 6 is to 12; or, the ratio of 4 to 8 is the same as that of 6 to 12: i.e. 2. a. Proportion may also be expressed by placing the sign of equality between the two ratios. Thus: 2: 8 - 4: 16 is a proportion, and is read, the ratio of 2 to 8 is equal to the ratio of 4 to 16. What is a Simple ratio? A Compound ratio? Is there any difference between Compound ratio, and any other ratio? Why then is it used? Of what does a proportion consist? Give an example. What are the terms of the couplets which compose the proportion called? How is proportion usually expressed? How read, when expressed in this manner? How may it otherwise oe expressed? Art. 2. RATIO AND PROPORTION. 225 Obs. 4. From these illustrations, we conclude that four numbers are in proportion, when the first has the same ratio to tie second, that the third has to the fourth, Obs. 5. The learner will perceive from what has been said, that Ratio and Propo tion are different: 1st. Because a ratio consists of but two terms, or one couplet; whilst a proportion cannot exist without four terms, or two couplers. 2nd. Because one ratio may be greater or less than another; thus, the ratio of 2: 8 is greater than that of 4: 12, and less than that of 3: 15; but in a proportion the ratios must be equal, and consequently one cannot be greater or less than another. Obs, 6. Although it takesfour terms to fo:m a proportion, still it can be formed with three numbers, as one of the numbers may be repeated. Thus: 3: 6 6 12 is a proportion, of which 6 is the consequent of the first couplet, and the antecedent of the second couplet. Obs, 7. When a number is repeated, as in this example, it is called a mean proportional between the other two numbers; and the last term is called a third proportional to the other two numbers. Thus, in the above example, 6 is a mean proportional between 3 and 12, and 12 is a third proportional to 3 and 6. Obs. 8. The first and last terms of a proportion are called the EXTREMES; the two middle terms are called the MEANS. Thus: in the proportion 15 3:: 20: 4, 15 and 4 are the txtremes, and 3 and 20 the means. Obs. 9. lit every proportion the produ.t of the extremes is equal to the product of the means; otherwise, they are not proportional. Thus. 4:2: 6 3 is proportional, because 4 X 3 = 2 X 6. But 6:: 12 3 is not proportional, because 6 X 3= 18, and 12 X 2 = 24, which is reater. Howv read, when expressed in this manner? When are four numbers in proportion? Are ratio and proportion alike? VWhy not? Of how many different numbers can proportion consist? How can this be done? What term i1 each cou plet is the number repeated? What is the number that is repeated called? What is the last terim called? Tu the proportion 3: 6:: 6: 12, which is the mean proportional, and which the third proportional? Which are the extremes in a proportion? A\ hich the means? In the proportion 15: 3::20: 4, which are the extremes, and which the means? What principle is true wrth regard to every proportion? If this is not the case, what is the concltsion? Is 4: 2:: 6: 3 proportional? Why? Is.62:: 12 3 proportional? Why not? 11A 226 COMMON ARITHMlETIC. Sect. XII Illzstration 1st.-In the proportion 4 2 6: 3, the ratio of 4: 2is -; also, that of 6 ' 3 is; hence, the ratios of the two couplets being equal, the numbers are proportional, according to Obs. 1. 2nd.-From the fact that proportion is an equality of rat'os, it is evident that the product, either of the extremes or means, must contain, as factors, the antecedents of both couplets and the ratio. But the consequent of each couplet is equal to the antecedent of this couplet multiplied by the ratio. (Art. 1. Obs. 6.) Therefore, the products of the antecedent of each couplet, multiplied by the consequent of thy other couplet, are equal. Obs. 10. As in Ratio, so in Proportion, the tuo terms of each couplet must be of the same denomination. (Art. 1. Obs. 8.) REMARK.-The learner will observe that it is not necessary for all four terms of the proportion to be of the same denomination, but only those of each couplet. For $6 has the s.ame ratio to $3, that 8 days has to 4 days; that is, the ratio of each is 2. Therefore, $6: $3:: 8 days:4 days: is a correct proportion, Obs. 11. By attentively examining Obs. 9, we notice two considerations a. 1st. If we divide the product of the means by either extreme, the quotient will be the other extreme. b. 2nd. If we divide the t product of the extremes by either mean, the quotient will be the other mean. Thus, in the proporton, 5: 10:: 6: 12; If we have given the ) two means, and 12X5=60; 60 —10-6, one mean. either mean, Or,.... 12X55=60; 60 —6 —10, the other mean. If we have given the ) two extremes, and 10X6=60; 60O-5 — 2 one extreme. either extreme, Or,.... 10X6=60; 60- -12=5, the other extreme. Obs. 12. Hence-If any three terms of a proportion are given, the other may be found by dividing the product of t1e means by the given extreme, or the product of the extremes by the given mean. Explain this point by the illustration given? Proportion is an equality of ratios; wlhat inference is deduced from this?' To what is the consequent of each couplet equal? What do we conclude from this fact? What is necessary in both ratio and proportion? s 'it iecessary for all. four of the terns to be of the same denomination? Which thet? Give an example illustrating this point. What is the first consideration we notics from examining Obs. 9? The second? If we have given three tetm;s of a proportion, how, do we find the roarth7 Art. 2. RATIO AND PROPORTTON. 227 REMARK.-This is evident from the fact, that in all such cases we have thepro. duct of two factors, and one factor given to find the other, which is always found by dividing the product by the given factor. (Sect. VI. Art. 1. Obs. 16.) MENTAL EXERCISES. 1. If the extremes are 4 and 6 anJ oae mean is 3, what is the other mean? 2. If the extremes are 8 and 15, and one mean is 5, what is the other mean? 3. If the means are 6 and 12, and one extreme is 8, what is the other extreme? 4. It the means are 7 and 8, and one extreme is 4, what is the other extre i.e? 5. If the extremes are 4 and 12, and one mean is 8, what is the other mean? 6. If the means are 4 and 9, and one extreme is 3, what is the other extreme? 7. If the extremes are 8 and 9, and one mean is 12, what is the other mean? 8. If the means are 4 and 10, and one extreme is 20, what is the other extreme? 9. Is 3: 4: 9 12 proportional? Is 7: 8:: 9: proportional? Is 12 6:: 20: 8 proportional? 1 ). Is 21: 7::24 8 proportional? Is 10: 12:: 15 18 proportional? 11. s 36: 9:: 42: 7 proportional? Is 72: 8:: 108: 9 proportional? 12. Is 40: 50:: 70: 90 proportional? Is 40: 6:: 60 90 proportional? NoT. —The learner should be required to give his reasons in the above examples, why they are proportional or not, as the case may be. Obs. 13. Proportion is of two kinds —SIMPLE and COMPOUND. Case 1.-SIMPLE PROPORTION. Obs. 14. SIMPLE PROPORTION is an equality between two simple ratios. Thus: 6: 2 9: 3 is a simple proportion. REMARK 1.-The chief use of Simple Proportion is tofind the fourth term of a proportion, when the first terms are gioen. It is in this manner that it is chiefly applied to practical business. 2. —Simple Proportion is often called the Single Rule of Three, because three Show why this is correct. How is Proporion diviled? What is Simple Proportion? For what is it chiefly used? What,is Simple Proportion. ofte called? Why? 228 COMMON ARITHMETIC. Sect. XII terms are given to find a fourth. Proportion, however, we consider to be the most appropriate name, because we have one couplet, and the antecedent of another couplet given, to find such a consequent of the latter couplet, that both shall have the same ratio, and thus constitute a proportion. (Obs. 1.) Ex. 1. If 4 yards of cloth cost $6, how much would 7 yds. cost? Ans. $10.50. Solution.-As in every couplet the terms must be of the same kind, it is evident that the numbers 4 and 7 form the terms of one couplet, because they are both yards, and $6 is one term of another couplet, and the third term of the proportion, and as this term is dollars, the fourth term must also be dollars, in order that both may be of the same kind. HenceObs. 15. In every proportion the third term must be of the same kind as the answer. We now wish to ascertain which of the other numbers is the first, and which the second term. If we make 7 the first term, and 4 the second, the proportion stands, 7 yds.: 4 yds. $6: the fourth term, or answer. Here we have the two means and one extreme of a proportion given, to find the other extreme. Then 6 X4=24; 24-7= —S, (Obs. 12,) and the proportion, when completed, is 7 yds. 4 yds. ' $6 $3. Thus, we make 7 yds. cost less than 4 yds., which is absurd. Therefore, we will make 4 the first term, and 7 the second term, and the proportion stands thus: 4 yds. ' 7 yds.:: $6: the fourth term, which we find, as above, to be $10.50.Hence Obs. 16. If the answer ought to be greater than the third term, make the greater of the remaining numbers the second term, and the less number the first term of the proportion. RgMARK.-This is evident from the fact, that if the answer should be greater than the third term, the ratio of each couplet is greater than unity; (Art. 1. Obs. 15.) and therefore each consequent is greater than its antecedent. The following form is usually adopted in simple proportion: We multiply the 6 and 7 together, yds. yds. $ and divide their product by 4, which 4: 7: 6.gives $tO0.50, as the fourth term. 6 4)42 Hence- Ans. 101 = $10.50. What is the most apprapriate name? Why? What must the answer be in Art. 2. RATIO AND PROPORTION, 229 When three terms of a proportion are given, to find the fourth: Obs. 17. Multiply together the second and third terms, and divide the product by the first term. (Obs. 12.) NOTE.-The learner must bear in mind that we cannot multiply two concrete numbers together; (Sect. IV. Art. 2. Obs. 6. Rem. 1. and Sect. VI. Art. 1. Obs 15.) thus in the above example we multiply 6 and 7 together, and divide the product by 4, because the ratio of $6 to the answer, must be the same as 4 yds. to. 7 yds. In this way we make the terms all abstract. PRooF.- 10.50X4==42; 7X6=42. (Obs. 9.) HenceTo prove Simple Proportion: Obs. 18. Multiply together the first term and the answer, orfourth term; and also the second and third terms; if the two products are equal, the work is correct. (Obs. 9.) Questions in Simple Proportion can be worked by cancelation: The above example is worked thus: 2_4 7 We place the 7 and 6 at the right for the -3 same reason that we make them the second and third terms in the proportion, and we 2 21 =10'. place the 4 at the left for the same reason that we make it the first term. Again, the second and third terms are multipliers, and the first term is a divisor. (Obs. 17., and Sect. ViI. Art. 1. Obs. 3.) Analytic Solution.-If 4 yds. cost $6, 1 yard will cost 6 —4 -$1.50, and 7 yds. will cost $1.50X7==$ 10.50. 2. If 8 lbs. of tea cost $7, what will 5 lbs. cost? Ans. $4.37'. every proportion? Why? If the answer should be greater than the third term, which of the remaining numbers do we make the first, and which the second term of the proportion? Show why this is correct. When three terms of a proportion are given, how do we find the fourth? Can we multiply concrete numbers together? How do we avoid this in Proportion? How do we prove Simple Proportion? Why is this correct? By what other method. can questions in Proportion be performed? How are the numbers arranged for canceling? Chay arranged in this manner? If the answer ought to be less than the third term, which of the remaining numbers do we make the first, and which the second term of the proportion? Show why this is correct. 230 COMMON ARITHMETIC. Sect. XII Operation. We arrange the terms as in the first lbs. Ibs. $ example, excepting that we make 5 the 8: 5:: 7 second term, because 5 lbs. will evidently 7 cost less than 8 Ibs. We then proceed according to Obs. 17. 8:35 Hence- Ans. 43 =$4.371. Obs. 19. If the answer oulht to be less than the third term, make the least of the remninygz numbers the second term, and the greater numb.r the first term of the proportion. REMxanK.-This is evident from the fact, that if the answer shiuld be less than the third term, the ratio of esch ctouplet is less than. unity; (Art. 1. Obs. 14.) and therefore, each consequent should be less than its antecedent. Analytic Solution.-If 8 lbs. cost $7, a lb. will cost $7 -8= -, and 5 lbs. will cosi -:7X5=$35=$4.37', as before. There is also nnother method by which questions in simple proportion may be performed, which may be thus explained: A proportion must consist of two couplets, (O)bs. 5, a ) and there couplets must both have the same ratio. (Obs. 1., and Obs. 5. 6.) Now, in examples in simple proportion we have given one couplet, and the antecedent of another couplet, to find the consequent of this latter couplet. (Obs. 4. Rem. 1.) But the consequent of any couplet is equal to its antecedent, multiplied by the ratio; (Art. O. O s 6.) therefore, as the first term must have the same ratio to the second, as the third has to the fourth, (Obs. 4.) Obs. 20. The fourth term of a proportion may be found by multiplying the third term by the ratio of the first to the second. Thus, in the first example, we have the proportion 4 7:: 6 the fourth term. Here the ratioof 4 7 is -, (Art. 1. Obs. 2.) and $6 X7= l10.50, as before. In the second example, we have the proportion 8: 5:: 7 the fourth term. Here the ratio of 8: 5 is -, (Art. 1. Obs. 2.) and $7X -=-$4.372, as before. 3. If 3 bu. 3 pks. 6 qts. of wheat must be given for 2 galls. 3 qts. 1 pt. of wine, how much wine must be given for 6 bu. 2 pks.: of wheat? Ans. 4 galls. 2 qts. l"n. pts. Of how many couplets must a proportion consist? What relation must the ratios of these c3uplats have to each other? What have we given in examnpies in Simple Proportion? What have we to find? To what is the consequent of any co iplet equal? What relation have the four terms of a proportion to each othbir? By what other method then can the foirth term of a proportion be found? Art. 2. RATIO AND PROPORTION. 231 Operation. bu. pks. qts. bu. pks. galls. qts. pt. 3 __ 3 6 6 2: 2.. 3 1 4 4 4 15 pks. 26 pks. 11 qts. 8 8 2 126 qts. 208 qts.:: 23 pts. 23 624 416 126)4784(378- pts.=-4 galls. 2 qts. 1 ps. Ans. 378 1004 882 19122 f 1 _o I-2 W 3l In this example, it is first necessary to reduce the first and second terms to one, and the same denomination, and also to reduce the third term to the lowest denomination mentioned in it, before we can proceed further. We then work as usual, and the fourth term of course is pints, which is the denomination to which the third term was reduced. (Obs. 10.) Hence — Obs. 20. If the first and second terms contain different denominations, reduce both to one and the same denomination; and if the third term is a compound number, reduce it to the lowest denomination mentioned in it;. after t/is, proceed according to Obs. 17; the result will be of the denomination as the third term, when reduced. REMARK.-The arrangement of the terms, as in the preceding examples, is called stating the question. Analytic Solution.-If 126 qts. pay for 23 pts., 1 qt. will pay for T of 23 pts., or 6 pts. Again, if 1 qt. pays for -, pts. 208 qts. will pay for 208 times as much, and 2^ X 208 = 23 2ts. 4 galls, 2 qts. 1 - pts., as before. From the preceding examples, we perceive that all questions i, If the first and second terms contain different denominations, how do we proceed? If the third tenn is a compound number, how do we proceed? Of what denomination is the result? 232 COMMON ARITHMETIC. Sect. XII Simple Proportion are solved on precisely the same principles. HenceObs. 22. All questions in Simple Proportion can be solved by Analysis. From the preceding remarks and illustrations, we derive the following GENERAL RULE FOR SIMPLE PROPORTION. I. Make that number the third term which is of the same kind as the answer requirel. (Obs. 15.) II. Consider from the nature of the question whether the answer should be greater or less than this term. III. If the answer ought to be greater than the third term, make the greater of the two remaining numbers the second term, and the less number the first term. (Obs. 16.) IV. If the answer ought to be less than the third term, make the least of the two remaining numbers the second term, and the greter number the first term. (Obs. 19.) V. Then multiply the second and third terms together, and divide the product by the first term; the quotient will be the fourth term, or answer, (Obs. 17) which is always of the same denomination as the third term. (Obs. 21.) PROOF.-Multiply the first and fourth terms together, and also the second and third terms; if the two products are equal, the work is correct. (Obs. 18.) REMARK 1.-If either, or all the terms, are compound numbers, the first and second terms must be reduced to the same denominations, and the third term must be reduced to the lowest denomination mentioned in it, before the multiplication or division can be performed (Ohs. 21.) 2. —If the le;rner choose, he may reduce the lower denoninations to fractions or decimals of a hi'her, instead of reducing the highcr to the lower denominations. (Sect. IX. Art. 3. Cases 2, 3, and 4. Rules.) 3.-We can often shorten the operation by multiplying the third term by the ratio of the first term to the second. (Obs. 20.) When the ratio is an integer, or smUll fraction, this method it preferable. 4.-If there is a remainder after the division has been performed, reduce it to the next lower denomination, and divide as before. 5.-This rule is equally applicable, whether the numbers are Integral, Fractional, or Decimal. What is the arrangement of the terms as in the preceding examples called? How can all questions in Simple Proportion be solved? What is the rule in Sitmple Proportion? What id the proof? If either or all the terms are compound numbers, how do we proceed? By what other method can we proceed with compound numbers? How can we often shorten the operation? When is this method preferable? If there is a remainder after the division has been performed, how do we proceed? Art. 2. RATIO AND PROPORTION. 233 RULE —By Cancelation: Consider from the nature of the question which are the di.ierent terms; then after placing the second and third terms at the rigit, and the first term at the left, cancel as usual. NOTE.-The above Remarks, (the third exce pted,) are equally applicable to Cancelatiou or Proportion. EXERCISES FOR THE SLATE. 1. If 8 bushels of apples cost $4, how much will 14 bushels cost? Ans. $7. 2. If 9 bushels of peaches cost $6.75, how much will 12 bushels cost? Ans. $9, 3. If 14 yds. of cloth cost $21, how much will 11 yds. cost? Ans. $16. 50. 4. If 13 pencils cost $22.75, how much will 9 pencils cost? Ans. $15.75. 5. If $9 buy 12 bushels of apples, how many bushels will $15 buy? Ans. 20. 6, If $24.50 will pay for 14 saws, how many saws can be bought for $14? Ans. 8. 7. If 6 men can do a piece of work in 18 days, in what time can 9 men do it? Ans. 12 days. 8. If 12 men can do a piece of work in 21 days, in what time can 7 men do it? Ans. 36 days. 9. If 10 lbs of sugar cost $1-25, bow much will 1 cwt. cost? Ans. $12.50. 10. If a man can travel 75 miles ii 3 days, how far will he travel in 3 weeks? Ans. 450 miles. NoTE.-The learner will observe that he travels 6 days in the week. 11. If 2 cwt. 3 qrs. 15 lbs. of coffee cost $30.45, how much will 2 qrs. 22 lbs cost? Ans. $7.56. 12. If 2 yds. 3 qrs. of cloth cost $5. 50, how much will 3 qrs. 2 na.cost? Ans. $1.75. 13. If the moon moves 105~ 24' 40" in 8 days, how long will it take it to perform one revolution, or 360~? Ans. 27 d. 7 h. 43 m.+. 14. If a pole 5 feet high east a shadow 3 ft. 6 in. in length, how high must a pole be to cast a shadow 130 feet? Ans.. 185- feet. 15. If 7 yards of cloth cost $10.50, how much will 15 bushels of apples cost, if 9 bushels of apples are worth 3 yards of cloth? Ans. $7.50. Is this rule applicable to either integral, fractional, or decimal njumbers? What is the rule by cancelation? COMMON ARITHMETIC.e Sect. XII 16. If from a: staff that's three feet long, A shadow five is made, What is a steeple's heighth in yards, That's ninety feet in shad.t Ans. 18 yards. 17. If I gv'e $r5 for the use of $200 a certain time, how much must I give for the usy of $700 for the same time? Ans. $52.50. 18. If a family of 6 persons consume 50 lbs. of flour in a certain time, how much will they consume in the same time, if 5 persons more are added to the same family? Ans. 912 lbs. 19. If a man can perform a piece of work in 16 days, working 10 hours per day, how long will it take him to perform it if he works 12 hours per day? Ans. 1:3l days. 20. If 84 bushels of oats last 30 horses 7 days, how long will 215 bushels last them? Ans. 17' days. Upon carefully ex.amninng the question, it will i e perceived that the number of horses is not one of the terms of the proportion. 21. If the interest of $600 for 12 months is $36, what is the inte:est of the same sum for 9 months? Ans. $27. Fractions.-22. If 3 of a yard of cloth cost 7 of a dollar, how much will TM of a yard cost? 3 yd,.: yd.:: 7$: - $1. Ans. 23. If V of a barrel of flour cost $8, what will ' of a barrel cost? $4'2 24. If; of a pound ef tea cost $70, how much will 7 of a pound cost? Ans. $0.75. 25. If 2 of 5 of a gallon of wvine cost u of a dollar, how much would s of - of a garllon cost? Ans. $0.924. 26. If $187- buy 15 bushels of wheat, how many bushels can be bought for 828o-? Ans. 221. 27. If 4- yds. of cloth cost $5-, how rmach will 7' qrs. cost? Ans. 2'!. 28. If 215 brtshels of oats last 30 horses 17- days, how long will 84 bushels last them? Ans. 7 days. 29. If of' an acre of land cost $2, how much would. of an acre cost? Ans. $8 -7 30. If a railroad car-goes 162.75 miles in 7.75 hours, how far, at this rate, would it travel in 2. 5 days? Ans. 1P.60 miles. 31. If 22- Ibs. of butter cost $4.74, how much would 1 cwt. cost? Ans, $39.37-. 32. If 5.25 bushels of wheat cost $4..721, how much would 8.37 bushels cost? Ans. $7.533. 33. If 12.45 pounds of tea cost,$9.3375, how much would 9.875 bs. cost? Ans. $7.4Cl. Art. 2. ItATIO AND PROPORTION. 335 34. If 20.125 acres of land cost $32-2, bow mnch woiuld 15.5 acres cost? Ans. $248. 35. If 12 horses eat 62.75 bushels of oaits in 7 days, how much will they cat in 15 days? Ane..!34.46+ bushels. 36., ftield was measurdt;wmdd found to be f48 rodi in length; but afterwards the line witih vwich t was;i easured was found to be but 31 - feet long; instead of 33 feet as was supposedt. Requiredthe true lenugtlh of the field? Ans: 142|- rds, Case 2.-COMNPOUND PROPORTIONS Obs. 23. A COMPOUND PRmOoTIONx i.' an equality between a com)poun2 d t(d a simple ratio. Thus: 43 t:2: 5: 100 is a compound proportion, and is read —" 3 into 4 is to, 12 into 23 as 5 is to 100;" or, ' the ratio of 3 X 4: 12 X 20 is equal to tle ratio of 5: 100." RET.anTI 1. —The learner will perceive that it is not the. ratio of 3: 12, or that of 4: 20 alo!ne, whichl is equa;il to tl;ht of 5: lOt); ilIt it is t/ e rati ecso - pounded of liese which is equal to that of 5: i00. Thus: 3X4: 12X20:: 5:100; because 3X4X 100 12X20X5.(Obs. 9.) 2.-Compoanl Proportion is clii/fly used wthen tnore than one statement is necessary in Simple Proportion. It is often called 7'he Double Rule of Three, but we consider tle term Compound Proportion preferable, as it expresses what it really is. Ex. 1. If a man ride 246 miles in 6 days, riding 8 hours per day, how far can he ride in 14 day s, riding 12 hours per day? Ans. 8661 miles. Solution. —in this example two things are to be considered: Ist. The number of days he rode. From this we deduce this question: if a man rides 246 miles in 6 days, how far will he ride in 14 days. And the proportion and result stand thus: 6: 14 days: 246 miles: 574 miles. 2nd. The 7nEmber of /hours he rode per day. From this and the last prmoprtion we obtain this question: If a man rides 574 miles in a certlain number of Idys, riding 8 hours per day, how Jar will he ride in the same time, riding 12 hours per day? And the proportion and result stand thus: 8 hrs.: 12 hrs.: 574 miles. 861 miles. W~hat is omnpoundi Proportion? H:ow do you read the expression given? Ts it the ratio of " 1'2, or of 4: 20 alone which is equal to that of 5: 100? What ratio ll.on is equal lo that of 6: 100? 7Why? WhPen is Compound ProporLion chiefly used? What is it often! called? What iisa. mre appropriale uaxne? Why? 236 COMMON ARITHMETIC'1 Sect. XII All questions in Compound Proportion involve the same principles. Henc.eObs. 24. Al questions in Compound Proportion can be performed by two or more statements in Simple Proportion. This example, however, can be worked by one statement, thus: Operation. 6 days: 14 days ) hours. 1 hours:: 246 miles: the answer. 8 hours. 12 hours; 48: 168: 246 miles: the answer. '246 1008 672 336 68)41328(861 miles. Answer. 384 292 288 48 48,00 As we desire miles for our answer, we make 246 miles the third term. (Obs. 15.) Then, if he rides 246 miles in 6 days, he will *evidently ride farther in 14 days; therefore, we make 14 the second term, and 6 the first term. (Obs. 16.) Again, if he rides 12 hours per d iy, he will evidently ride farther than if he rides but 8; tierefore we make 12 another second term, and 8 the first. (Obs. 16.) But we perceive this to be a compound proportion, (Obs. 23.) and is equivalent to the simple proportion 48 (6X8): 168 (14X12):: ' 46: the answer. (Obs. 23. Rem. 1.) which is found according to Obs. 17. Hence To perform operations by one statement in Compound Proportion. Obs. 25. _Arrange the third term,,and each couplet as in Simple How can all questions in Compound Proportion be performed? When we make but one statement, how do we arrange the terrns? Art. 2. RATIO AND PROPORTION. 237 Proportion, (Obs. 15, 16, and 17.) Then multiply the continued product of all the second terms by the third term, and divide the result by the contined product of all the first terms. The above example may be proved thus: 6X8 X 861=14 X 12 X246. (Obs. 9.) HenceTo prove Compound Proportion: Obs. 26. Multiply the continued products of the first terms by the fourth term; also, the continued product of the second terms by the third term; if the two results are alike, the work is correct. (Obs. 9.) REMARK.-In all questions in Compound Proportion, there is one number of a different kind from any of the rest, and this number is always made the third term. The fourth term of the above proportion may also be found thus: The ratio compounded of two couplets 6: 14 and 8: 12 is 3~. (Art. 1. Obs. 16. b.) Now the third term (246) must have t1 same ratio to the fourth term, or answer, (Obs. 23, Rem, 1.) and 246X3$=861. (Obs. 20.) Hence-we can obtain the fourth term of a Compound Proportion: Obs. 27. By multiplying the third term by the ratio compounded of the first and second terms, Analytic Solution. —If he rides 246 miles in 6 days, he will ride - of 246 miles, or 41 miles, in 1 day; then if he rides but 8 hours per day, he will ride - of 41 miles, or 4 miles per hour; and in 12 hours he will ride 12 times as far, or ' |3 miles; and in 1 days, of 12 hours each, he will ride 14 times 13 miles, which is 861 miles, as before. All questions in Compound Proportion are solved on the same principles. HenceObs. 28. All questions in Compound Proportion can be solved by Analysis. Operations in Compound Proportion can frequently be shortened by canceling factors common to the first, and the second or third terms, as in the following How do we next proceed? How do we prove the operation? What occurs in all questions in Compound Proportion? What term is this number made? By what other method can the fourth term of a compound proportion be found? How may all questions in Compound Proportion be solved? How can we often shorten the operation? 'r(-ON 11TN ARTITH1METIC. Sect. XI[ Operationb j.t1:'4- Our first tarmns are all factors of some -- 4. -8,Y.._ nunlber wlilch is ua dlivisor, and our second —._ 123 and third terms are all factors of some 123X7 86 1 lnumbtr wllich, is 1a dividend; (Obs 25.) therefore we place all our first terms at the left of thl line, iand all (he otlier numbers at the right of tie line, (Sect. VII Art. 1. Obs. 3.) and cancel as usuail. RE.MARK 1.-We have presented flt:ee dift renl mnelihods of obttalinng tire resull, both in Simple and Collmpol nd Proiortioni, for tlih purpose of showing the relation which numrbers loer.o each olher, and also to show how the satme result may be obtained by ditfflrent processes, thus provinlg clearly the correctness of the princ.pies by which we aperate. Anld tlhis we iconsider to be one of the principal beauties of mathermatics-lhat uwe are obliged to take rro principle, or believe no theory, unless it is clp,Able of dom.sonstration, or proof. And one indubitable proof of the correctness of atllemraticil principles, is, that we alway/s obtain the same result to a question or pronlemn, no matler what princi ples we apply, or what process we use, provided we snake nno mistake in our rrasoninrg, or mode of application. The case would he otherwise, were our principles incorrect, because then the aplicatiou of diflerent principles to the same question, would often produee diffetrent results; but as this never happens, we coticlnde that tlhey are absolutely trae. 2.-Of the ditferent netlhods of obtaining the result in proportion, that by Analysis is decidedly preferable, as it is better adapted to the discipline of the mind, and strengthens the mental faculties. It is recormmeTied to the pupil to solve all the questions, both in Siinple and Compound Proportion, by Analysis, after he has obtained the result by Proportion. He will by this means gain sufficient mental culture to arnply crsMpensate him for his trouble. From the foregoing remarks and illustrations, we derive the folfollowing.* GENERAL RULE FOR COMPOUND PROPORTION. I. Make that number the third term!which is of the same kind as the answer. (Obs. 15.) It. Take two of the remaining numbers of the same kind, and arrange them in the same manner as in Simple Proportion. (Ohs. 16 and 17.) Proceed in the same manner with all the other numbers. III. Multiply the continued product of the second terms by the third term, and divide t/e result by the continued product of the first terms. (Obs. 25.) How do we arrange the terms for canceling? Why do wearrange them in this manner? What do these different methods in obtaining the result in Silnple and Compound Proportion show? What does this prove? What is one of the principal beauties in mathematics? What is one indubitable proof of the correctness of mathematical principles? Would this be so were our principlesincorrect? Why not? As this never happens, what do we colnlude? Of the different methods of obtaining the result in Proportion, which is preferabls? Why so? What is the rule in Compound Proportiln? Art. 2. RATIO AND PROPORTION. 239 PRooF.-Multiply the fourth term by the continued product of the first terms; also, multiply the third term by the continued product of the second terms; and if the two results are -equal, the work is correct. (Obs. 26.) NOTE.-The'rerharks und' ifab rile in Simple Propbtrtion are equally applictbhi to'Compound Proportion, except the third, of which it will be observed in Compound Proportion that we multiply the third term by the ratio compounded of the first and second terms, Instead of the simple ratio of either couplet. (Obs. 27.) RULE-By Cancelation.-Consider whic' are the diferent ternms then place the third term together with all the second terms at the right, and all the irst terms at the tlft. EXERCISES FOR TIHE SLATE. 1. If 15 men earn $225 in 12 days, how much will 9 men earn in 24 days, at the same rate? Ans. $270. 2. If 8 men earn $'84 in 7 days, werking 10 hours per day, how long will it take 12 men to earn 6300, working, 12 hours per day? Ans. 13' days. 3. If 7 men cut 49 acres of grass in 4 days, how many days will it take 11 men to cut 1 2 acres of grass? Ans. 5,-i. 4. If the interest on 8200 for 2 years is $24, what is the interest on $159 for 1 year and 9 months? Ans. $15.75. 5. If 12 students spend $100 in 3 weeks, how much will 7 students spendin 5 weeks? Ans. $97.22-. 6. If 10 men dig a trench 240 yards long, 6 feet wide, and 5 feet deep, in 8 days, how long will it take 18 men to dig a trench 400 yards long, 8 feet wide, and 4 feet deep? Ans. 7-7 days. 7. If 6 men, working 12-l0 hours per day, dig a cellar 226 ft. long, 17- ft. w, ad ft. wide, and 4deep, in 2- days, how long will it take 9 men to dig a cellar 45 ft. long, 353 ft. wide, and 5,7l ft. deep, working 8- hours per day? Ans. 12 days. 8. If 80 men dig 20 cellars, each 45 ft. long, 28 ft. wide, and 3 ft. deep. in 10 days, working 12 hours per day, how many men will it require to dig 30 cellars, each 24 ft. long, 21 ft. wide, and 44 ft. deep, in 18 days, working but 10 hours per day, supposing the strength of the men in the former case to be 1- that of those in the latter, and the hardness of the ground in the latter case to be only 3 of that in the former? Ans. 56. What is said respecting the remarks under the rule in Simple Proportion? What must be observed with respect to remark third? Wliat is the rule by cancelation? 240 OOM~MON ARIIHMETiC. Sect. XII ARTICLE 3. PARTNERSHIP. Obs. 1. PARTNERSHIP is the associating together of two or more persons in joint trade, with a mutual agreement to share the respective gains or losses in proportion to the capital each one has invested, and the time it has been employed; this association is called a COMPANY, or FIRM, and each individual is called a PARTNER. Obs. 2. STOCK is the name given to the money, or value of the articles employed in trade. It is sometimes called CAPITAL, The gain or loss divided is called the DIVIDEND. There are two cases of Partnership. CASE 1.-This case is used when no consideration is made with regard to time. Ex. 1. A. and B. enter into partnership. A. furnished $300, and B. $500; they gained $250. What was each man's share of the gain? Solution. —A. put in $3 0, and B. $500; hence, $300+-$500= $800, the whole stock invested. Now it is evident that each one's share, of the gain must have the same relat on to the whole gain, as each one's share has to the whole stock. Hence, the following Operation. Whole A.'s Whole A.'s share $300 A.'s stock. stock. stock. gain. of gain. $500 B.'s stock. Thcn,-$800 300:: $250: $93.75. - B.'s stock. B.'s'share of g'n $800 Whole stock. And, $800: $500:: $250: $156.25. f$ 93.75 A.'s share of the gain. PROOF: 8156.25 B.'s share of the gain. l$250.00 Whole gain. Solution by Analysis.-As A. put in $300, he owns - = of the stock; therefore he must have 3 of the gain, and the 3 of $250 is $93.75, A.'s gain. Also, as B. put in $500, he owns 58 of the stock, and therefore must have 5 of the gain; 5 of $250 is $156.25, B.'s gain 2. Suppose, in the above case, they had lost $200; what would have been each one's share of the loss? What is Partnership? What is the association called? What is each individual called? What is stock? What is it sometimes called? What is the gain or loss to be divided called?.Of how many cases does Partnership consist? When is Case 1 used? Art. 3. RATIO AND PROPORTION. 241 Operation. $300 $500 Then, $800 $300: $200: $ 75, A.'s loss. -- And $800: $500:: $200: $125. B.'s loss. $800 r $ 75, A.'s loss. PROOF: q $125, B.'s loss.. $200 whole loss. Analytic Solution.-A. put in 8 of the stock, and therefore must sustain of $200 is $75 A.'s share of the loss. o $ i 5. A s o th l B. put in - of the stock, and therefore must sustain 6 of the loss: - of 200 is $125, B.'s share of the loss. From these illustrations we derive the following RULE FOR PARTNERSHIP. As the whole stock is to each man's share of the stock, so is the whole gain or loss to each man's share of the gain or loss. PRooF.-Add together the several shares of the gain or loss; if the sum is equal to the whole gain or loss, the work is correct. REMARK.-This case is often called Single Fellowship; but as a partnership cannot exist without at least two individuals, the propriety of calling it single is somewhat doubtful. 3. A., B., and C. trade together: A. puts in $500, B. $800, and C. $1000; they gain $690. What is each one's share. Ans. A.'s, $150; B.'s, $240; and C.'s, $300. 4, A,, B., and C. speculate together. A. furnishes $3000, B. $5J00, and C. $4000; they lose $3000. How much was each one's loss? Ans. A's $750; B's $1250; and C's $1000." 5. P., Y., and R. own a vessel worth $10000. P's share is worth $3500; Y.'s share is w(rth $4250, and R. owns the rest, In a certain trip they -ain $1250. How much is each one's share of the gain? Ans. P.'s $437.50; Y.'s $531.25; and R.'s $281.2.6. U. V. and W. own a store. U, owns, V. owns i, and W. owns the rest; they lose -$480. How much is each one,' share of the loss? Ans. U.'s $137.142; V. and W., each $171.428. 7. A., B. and C. loaded a ship with flour. A. put on 460 bar. What is the method, fproof? What is this case often eUlledT? t.tlit'm aprprop^iat6 Why not? 19~~~~~~~~~~~~:J2.. i:.. 242 COMMON ARITHMETIC. Sect. X11 rels, B. 350, and C. 250; in a gale the Captain threw overboard 200 barrels. What loos did each one sustain? A.'s loss 89 bbls.; B.'s 70 bbl.s; and C.'s 50 bbls. 8. Four persons trade together. A. put in $200; B. $4C0; C. $350; and D. $550; they gain $600. What is the share of each? Ans. A.'s, $80; B,'s, $160; C.'s $140; and D.'s i220. Obs. 4. BANKRUPTCY.-A Bankrupt is a person who is uni.b'e to pay his debts. Questions in Bankrultcy are usually performed by the rule of Partnership. 9. A bankrupt owes $4000; his property is worth but $3000. How much can he pay on the dollar? Ans. $0.75. 10. A man's property is worth $5000. IHe owes A. $1000, B. $3500, C. $850, D. $1500, and E. $2150. How much can he pay each in proportion to their debts? Ans. A. $555.5554; B. $1944.4444; C. $472.222'-; D. $833.333-; and E. $1194.444~. 11. A man died, leaving property worth 810000. He owed F. $2500; G. $3050; H. $4500; and.,$2450. How much can (-ach creditor receive, and how much can the estate pay on the dollar? Ans. P. receives $2000; G. $2440; H. $3600; and I. $1960. And the estate pays $0.80 cents on the dollar. Obs. 5. GENERAL AVERAGE.-When the master of a vessel, in consequence of a storm, or other casualties at sea, is obliged to throw overboard a part of the cargo, in order to save the ship and crew, the law requires that the loss shall be sustained by the owners of the vessel and cargo, ia proportion to the value of each individual's property at stake. The property sacrificed is called the Jettison. The process of finding each one's loss in such cases is called GENERAL AVERAGE. The rule is the same as in Partnership. 12. A vessel being in distress, the Captain threw overboard goods to the value of $15000 The cargo was owned by thr e Ijerons, L., M. and N. L. owned $20000; M. $35000; and N.,25000: the vessel was worth $20000. How much was each man's loss? Ans. L.'s $3000; M.'s $5250; N.'s $3750; and the owner of the vessel $3000. What is a bankrupt? How are questions in bankruptcy usually performed? When the master of a vessel is obliged to throw overboard a part of his cargo, In order to savehis ship or crew, how is the loss sustained? What is the pro,.es. of fading eaoh one's loe ia such cases called? What is the rule? rt. 3 PARTNERSHTP. 243 13. In consequence of a storm at sea, a vessel sustained the following loss: For a part of her cargo thrown overboard, valued at $4850 For repairing damages of the ship, &c., -..-..-.. 450 Other expenses,. ----. -.... 30 300 Tile loss was sustained by a general average, the value of the ship and cargo being as follows: Value of the vessel, ------------ - $25000. Part elf the cargo owned by Wm. Trader, & Co. valued at 48000. Part " " ". Thos Dealer, & Co., -_ " 15000. Part " " " James Loverain, 27000. Part " (" " Captain of the vessel,." 10000. What share of the loss must each sustain? Ans. The owner of the vessel $1120; Tra 'er & Co. $2159.40; of the vessel $448. A number can also be divided into proportional parts by the ru!e of Partnership. 14. Divide the number 576 into three parts that shall be to each other as the numbers 7, 8, and 9. 7+-8 +9 =24; 24: 7:576: 168, &c. Ans. 168, 192, and 216. 15. Divide the number 1728 into 6 parts, that will be to each other as the numbers 3, 4, 5, 6, 8, and 10. Ans. 144, 192, 240, 288, 384, and 480. CASE 2.-Obs. 6. This case is used when time is considered. Ex. 1. A. and B. hire a pasture for $10. A. put in two cows 4 months, and B. put in 4 cows 3 months. How much ought each to pay? Solution.-A. put in 2 cows 4 months; this is the same as 8 cows 1 month. B. put in 4 cows 3 months; this is the same as 12 cows 1 month. Hence, they must lpay in the proportion of 8 to 12. Operation. 2X4= 8. 4 X 3 = 12 Then 20: 8: $10: $4 A.'s share. And 20:12:: $10: $6 B.'s share. 20 r$ 4 A.'s share. PROOF. 6 B.'s share. PROOF. _ $10. Whole cost. When is Case 2 used? 244 CONLIMON ARITHMETIC.C Sect. XIII Analytic Solution.-A. has in the same as 8 cows 1 month, and B. has in the same as 12 cows 1 month; and both have in the same as 8 + 12 = 20 cows 1 month. Then A. must pay, =- -, and B. 1 32 of the cost. of $10 is $4, A.'s share, and 3 of $10 is $6, B.'s share, as before. HenceTo solve questions in Partnership, when time is considered: Obs. 7. Multiply each man's stock by the time he continues it in trade, and use the productfor his share. 2. A., B. and C. trade together. A. put in $400 for 6 months; B. put in $500 for 4 months; and C. put in $350 for 8 months; they gained $300. How much was each one's share of the gain? Ans. A.'s $100; B.'s $83.333-; and C.'s $116.666~. 3. Suppose in the last example they had lost $200. How much would have been each one's share of the loss? Ans. A.'s $66.6662-; B.'s $55.5551; C.'s $77.777-. 4. W., X,, Y. and Z. trade together for one year. W. put in $500 for 6 months, and then put in $700 more; X. put in $1200 for 4 months, and then took out $500; Y. put in $800 for 5 months, and then took out $200, but put it back again at the end of 9 months, with $400 more; Z. put in $700 for 7 months, and then put in $800 more; but he took out the whole of his stock at the end of 11 months; at the end of the year they have gained $2490, How much is each one's share? Ans. W.'s $612; X.'s $624; Y.'s $600; and Z.'s $654. SECTION XIII. PERCENTAGE. ARTICLE 1. DEFINITIONS, MENTAL EXERCISES, &C. Obs. 1. The terms percentage and per cent are derived from two Latin words per and centum, which signify biy the hundred. Therefore, when we speak of a certain per cent of any number, we evidently mean such a hundredth part of that number. Thus, 5 per cent. of any number is, ' of that number; 12 per cent of any number is '-, of that number, &c. How do we solve questions in Partnership when time is considered? From what are the terms per cent and percentage derived? What do these words signify? What do we mean when we speak of a certain per cent of any number? Art. 1. PERCENTAGF. 245 MENTAL EXERCISES. 1. What is 2 percent of 100? 2 per cent of 100 is 2 of 100. Now -0 of 100 is 1, and 2 of 100 is 1 X 2= 2. Ans. 2. 2. How much is 4 per cent of 100 cents? Ans. 4 cents. 4 per cent is y-. - 4-o of 100 cents is 4 cents. 3. A man borrowed $100, paying 6 per cent for the use of it. How much did he pay? Ans. $6. 6 per cent is yo. -oB of $100 is $6. 4. A man having 100 sheep, lost 10 per cent of them by disease. How many did he lose? 10 per cent is - fo-. -, of 100 is 10. HenceObs. 2. Per cent implies so many units for each 100 units, so many cents for each 100 cents, so many dollars for each 100 dollars, so many articles for each 100 articles, &c.: and theper centage of any number is as many times the per cent as there are hundreds in this number. 5. How much is 6 per cent of 300?3 Ans. 6 X 3 = 18. 6. How much is 3 per cent of 200? 400; 700; 500; 800; 900; 290? 7. How much is 8 per cent of 100? 200; 500; 700; 900; 400; 800; 1200? 8- How much is 10 per cent of 200? 1200f 1500; 900; 700; 1000; 2000? 9. A man having found $6, the owner give him 4 per cent for finding it. How much did he receive? Ans. 24 cents. Solution. I- e received 4 cents for each 100 cents; in $6 are 600 cents. Therefore he received 4 X 6 = 24 cents. 10. A merchant sold a coat tor $12; his gain was 10 per cent. of what he sold it for. What was his gain? 11. What is 10 per cent of $5? of $8; of $10; o $11; of $6; of $9; of $20? 12. A constable collected $8, and received 5 per cent for his services. How much did he receive? 13. How much is 5 per cent of $2? Of $4? Of $10? Of 7? Of 5? Of $12? Of $6? What is 5 percent of any tumber? 12 per cent? 6 per cent? 3 per cent? 9 percent? 1 per cent? WhVut is 2 per ce.nt of 100? 6 per cent of 100 cents? 9 per cent of 100 cents? per cent of 100? 2 per cent of $100? 12 per cent of $10)? What does per cent Imply? What is the percentage of any number? 246 CO0,1MON ARITHMETIC. Sect. XIII 14. A man had 200 sheep; in two years they increased 50 per cent. Iow many had they increased? 15. An auctioneer sold $1200 worth of gco.s, and received 4 per cent for selling. How mnuch did lie receive? 16. What is 4 per ctnt of $200? Of $800? Of $600? Of $1000'? Of $300? 17. A man borrowed $460, and gave 6 per cent for the use of it? -low much did it cost him? 18. How much is 6 per cent of $200? Of $800? Of $300? Of $500? Of 1200? 19. A man bought a farm for $1000, and sold it so as to gain 12 per cent? How much did he gain? 20. What is 12 per centof $800? Of $1200? Of $700? Of $1000? Of $500? Of $900? Of $300? Of 600? Of $200? Of $1500? Of $2000? Obs. 3, Since I per cent sigqnfies yl-,, 2 per cent &1, &Jc., it is evident tha t t moy be expressed decimally. Thus: 1 per cent may be written... --- —.._.01. 2 per cent i" ------—..02. 3 per'cent {".. --- — - -..__.03. 4 per cent." '-. --- —--- -. ---_.04. 7 per cent " —. — ___ ----.07. 9 per cent " "- - -.09. a. When the given per cent is 10 or more, it is merely written with the decimal point before it. Thus: 10 per cent is written -..10. 12 per cent " -.-.12. 15 per cent " ----.....15. 25 per cent ". -.25. 50 per cent " - _.. ---_.. —.. ---. —.5). 75 per cent "._... __.75. 99 per cent:'..........-.99. b. When the given per cent is 100 or more, it is evidently a mixed number, and must be written accordingly. Thus 100 per cent is written..__._. --- - 1.00, or i. 104 per cent ",-..._.-... 1.04. 125 per cent ".... 1.25. 274 p:r cent "........- 2.74. HIow c 1 i ny per cent,1): exi)rjs l? S ' ) v whv thli is (orrte.t? When tle given per cent is 10 or nmore, Ihow is It writl,1? \'Whenl the given per cent is 10J or more, what kind of an cxri.ss.:oni is it? fIow written? Art. 1. PERCENTAGE. 247 c. When the riven per cent is less than 1, it is either written racionally. or t occupies three or more decimal places. Thus:, per cent, that is 2 of 1 per cent is written - -.001, or.005. per cent. that is of 1 per cent is written -.00-, or.0025. \ per cent. that is I of 1 per cent is written.003, or.0075. 2- per cent, is written -.02, or.025&C 21. Write 5 per cent; 6 per cent; 8 per cent; 11 per cent; 14 per cent; 18 per cent; 22 per cent; 39 per cent; 50 per cent; an-l 18 per cent. 22. Write 120 per cent; 150 per cent; 130 per cent; ' per cent and 1 per cent i. decimals. 23. Write 35 per cent; - per cent;,, er cent; and 2- per cent in decimals. 24. Write 2'- per cent; 4' per cent; 6' per cent; 8- per cent; an] 25,1 per cent in decimals. 25. Write 842 per cent; 3261 per cent; and 5123 per cent in decimals. EXERCISES FOR THE SLATE. 1. A constable collected $600, for which he was to receive 4 per cent for collecting. How much did he receive?' Operation. 4 per cent is expressed.04. Therefo'e. if we $600 iulLiply $600 by.04, the product \vill be what the.04 constable must receive. (Obs. 2.) Hence- -- Ans. $24.00 To find the percentage of any numberr: Obs. 4. Miultiply the given number by tie given per cent expressed decimally, and point qf in the product as in multiplication of decimals. (Sect. VIII. Art 11. Rule.) 2. What is 6 per cent of $80? Of,$100? Of$150? Of $300? Of $1000? 3. What is 5 per cent of $30.26? Ans. $1.513. 4. What is 10 per cent of $75.84? Ans. $7.584. 5. What is 15 per cent of $125.75? Ans. $18.86^. 6. A broker exchanged $1000, for which he was to reoeive ' per cent. How much did he receive? Ans. $5. 7. What is ' per cen. of $2000? Of $800? Of $4000? Of $3000? Ans. 85; $2;.S10; $7.50. 8. What is ' per cent of:*45). 80? Ans. $1.127. When the given per cent is less than 1, how is it written? How do wo find the percentat.e nof -.r -"-.lwer? 248 COMMON ARITHMETIC. Sect. XIII 9. What is.- per cent of $400? Of $800? Of $1200? Of $1500? Ans. $.50; $1; $1.50; $1.871-. 10. What is 5 per cent of $1000? Ans. $6.25. 11. What is - per cent of $600? Ans. $2. REMARK.-When the given per cent is a fraction, we may first find the per centage at 1 per cent, and then takefractional parts of this for the given per cent. This method is generally preferable. 12. What is ' per cent of $1200? Of $1500? Of $1800? Ans. $2; $2.50; $3. 13. What is 41 per cent of $200? Ans. $9. 14. What is 3} per cent of $60? Of $80? Of $100? Of $200? Ans. $1.92; $2.56; $3.20; $6.40. 15. An auctioneer sold $1000 worth of goods, for which he was to receive 12 per cent. How much did he receive? Ans. $120. NOTE.-The learner will bear in mind that he received 12 per cent on what he sold, and not on what was paid over, as is often supposed. In the latter case he would only receive-j-,a* instead of -lo2o of the value of the goods sold. The same remark is applicable to moneys collected by constables, &c. 16. A merchant bought $15000 worth of goods, and sold them so as to gain 15 per cent. How much did he gain? Ans. $2250. 17.: clock dealer shipped 800 clocks for New Orleans, but on the passage 12 per cent of them were washed overboard. How many were lost? Ans. 96. 18. Two men engage in trade with $1500 apiece. One gains 16 p( r cent, and the other gained 20 per cent. How much did one gain rwore than the other? Ans. $60. 19. A gentlemen deposited $81200 in a bank, and afterwards drew out 19i per cent of it. How much did he take out, and how mucd had he left? Ans. He took out $234, and had left $966. 20. What is the difference between 18 per cent of $300, and 24 per cent of $400? Ans. $42. 21. What is the sum of 14 per cent of $200, and 9 per cent of $700? Ans. $91. 22. What is 112 per cent of 200; 800; 1200; 1500; 1824. 4276; 5000? Ans. inorder. 224; 896; 1344; 1680; 2042.88; 4789.12; 5600. 23. What is 103 per cent of 400? Ans. 412. 24. What is 250 per cent of 100? 400; 600; 900; 1400; 1846; 8120? Ans. in order. 250; 1000; 1500; 2250; 3500; 4615; 20300. When the given per cent is a fraction, how do we proceed? Is the percentage received by constables, auctioneers, &c., calculated on the value of the property they sell or collect, or on what they pay over? Why so? Art. 2. COMMISSION AND INSURANCE. 249 25. What is 6- per cent of $240? $273.80; $320; $480.40; $512.96? Ans. in order. $15; $17.11-; $20; $30.02j; $32.06. 26. What is 83 per cent of $240? $345; $462.36; $512.34; $680.64? Ans, in order. $20; $28.75; $38.53; $42.691; $56.72. 27. What is 371 per cent of $250? $300; $412.25; $560.59; $618.90? Ans. in order. $93.75; $112.50; $154.593+; $210.181; $232.0834. 28. Which is the most-8 per cent of $500, or 6 per cent of $700? How much is the difference? Ans. 6 per cent of $700 is the greatest by $2. 29. What is the difference between 4- per cent and 6- per cent of $1500? Ans. $32.50. 30. What is I per cent of $200? $240.75; $300; $312.25; $473. 95? Ans. in order. $0.40; $0.48+; $0.60; $0.624+; $0.947+. 31. What is -, per cent of $120? $160; $190; $210; $235; $375. 50? Ans. in order. $0.84; $1.12; $1.33; $1.47; $1.645; I $2. 628+. 32. What is-2 per cent of $400? $800; $1000; $1200; $1500; $2400? Ans. in order. $1.80; $3.60; $4.50; $5.40; $6.75; $10.80. 33. What is,5 per cent of $48? $72; $84; $120; $288; $324; $1728. Ans. in order. $0.20; $0.30; $0.35; $0.50; $1.20; $1.35; ~$7.20. 34. A drover having.300 cattle, sold 40 per cent of them. How many did he sell? Ans. 120. 35. Two men had each $1500. One gained 20 per cent of his, and the other spent 20 per cent of his. How much ihen had one more than the other? Ans. $600. Ohs. 5. Percentage is applied to various calculations in practical business. The most important of these are Commission, Insurance, Profit and Loss, Stocks, Brokage, Duties, Taxres, and Interest. ARTICLE 2. COMMISSION AND INSURANCE. Obs. 1. COMMISSION.- Commission is theper cent, or sum charged by an agent for transacting business for dhs employer. It is generally applied to buying aad selling goods. To what is Percentage applied? Which are the most important of the.e? What is Commission? To what is it generally applied? 12i 250 COMMON ARITHMETIC. Sect. XIII REMARK 1.-The person who buys or sells goods for another is called d Commission Merchant, Correspondent, or Factor. 2.-As ccinmission is reckoned at so much per cent on the money employed the rule is the same as in percentage. (Art. 1. Obs. 4.) EXERCISES FOR I HE SLATE. 1. A. sold $200 worth of goods for B., at 6 per cent commission How much did he receive? Ans. $12. 2. A.in sold 100 bushels of wheat at $1.25 per bushel, on commission, for which he received 5 per cent. How much was his commission? Ans. $6.25. 3. How much is the commission for selling $500 worth of goods, at 8 per cent? Ans. $40. 8. How much is the commision for buying $1200 worth of goods, at 4 per cent? Ans. $48. 5. B. sold on commission for C. $427.80 worth of goods, at 8.per cent; $512.40 worth of goods at 10 per cent; and $350.50 worth of goods at 122 per cent. How much did his commission amount to? Ans. $132.914. 6. D. sold on commission for E. 120 yards of cloth, at $4.50 per yard, for which he was to receive 4' per cent. How much was his commission, and how much must he return to E.? Ans. The commission was $23.40, and the sum returned was $516.60. 7. A Southern merchant consigned to a merchant in Boston the following art cles, to be sold on commision, at 73- per cent: 4200 lbs. of cotton, at 71 cs per pound; 7600 lbs. of sugar, at 8; cts. per pound; 19 bags of c,ffee, each contaiinng 90 lbs., at 7 cents per pound; 2100 lbs. of rice, at 43 cents per pound; and 25 barrels of molasses, at $15 a barrel. How much was the commission, and what sum must be returned to the owner? Ans. The commission was $1 16.77, and the sum returned was l $1419.68. INSURANCE. Obs. 2. By INSURANCE is meant security from loss or damage occasioned by fires, storms, shipwrecks, &c. Obs. 3. This security is usually effected with imdividuals, or insurance companies, by the payment of a stipulated sum, which is generally a certain per cent of the value of the property insured. What is the person who buys or sells goods for others called? How is commission reckoned? What then is the rule? What is rnm-ant by iu.urnnce? How is the security effected? Wheu an insurlnce is eff-cted by u contract with individuals, what is it termed? What is Marine Insurance? How offeoW? Art. '2. COMMISSION AND INSURANCE. REMARK 1.-When insurance is effected by a contract with individuals, it is termed out door insurance. Insurance at sea is called Marine Insurance. It is usually effected for a certain voyage. 2.-Thie insurers, whether individuals or an incorporated company, are often called underwriters. Obs. 4. The sum paid for the insurance is called the PREMIUM; and it is paid when the insurance is effected. The written instrument that binds the contracting parties is called the POLICY. T'he amount of pecuniary responsibility taken by the insurers is called the RISK. It is sometimes equal to the whole of the estimated value of the property insured, and sometimes it is equal to only a part of the estimated value. Case 1.- Whlen the premium. is, a certain per ecnt of the value of the property insured. REMARKK.-The v-ry nature of this case shows that the rule is the same as in Percentage. We must add tlhe cost of the policy to the percentage, however, to find the actual cost of thia insurance. EXERCISES FOI. THE SLATE. 1. How much must be paid for insuring a house valued at $1200, at 4- per cent, and the cost of the policy being $1. Ans. $55. 2. How much must be paid for insurance on a store valued at $4000, and on the goods valued at, 4$15)00, at 3^ per cent; the policy costing $1.50? Ans. $647.50. 3. How much must be paid for insuring a steamboat valued at $10000, the policy costing 1, and.he pr mium being 5 per cent on W of the value of the boat, that being the amount of risk taken by the company? Ans. $401, 4. How much must be paid f r insuring a ship valued at $20000, and the cargo valued at $50000, the premium on the ship being 6 -per cent on 7 of its estimated value, and the premium on the cargo beirg 4,o per cent on 3- of its estimated value, and two policies being required, each costing $1? Ans $2858,25. REMARK.-In mutual insurance companies each one gives a premium note of so much per cent on the property which he wishes to insure, the rate being determined by the amount of Risk. A certain per cent is paid down on these notes for immediate use, and any losses that occur more than this per cent are What are the insurers often called? What is the Premium? When paid? What is the Policy? The Risk? To what is the Risk equal? What is the rule when the premium is a certain per cent of the value of the property ins sured? 252 COMMON ARITHMET[C. Sect. XIII averaged on the premii um notes, and thus eacll one helps to supply the deficiency in proportion to the amount of property he lias insured. The uinlouat of the premium notes forms tie capital of the company. 5. How much must be paid for insuring a store valued $15420, the premium note being 8 per cent, and the assessments 2, 2-, 4, 3j. and 1 per cent? Oleration. $415420.08 81233.60 premium note.. 131.- Yumu of assessments. 6 GPI, 370080 123360 Ans. 4 166. 5360 By taking 8 per cent of S15420, we find our premium note to he $1233.60. The sum of the several assessments is 131. Then 13~ per cent of;1i233.60 gives 166(. 5:3t.;s our answer. 6. How much must be paid for insuring- $8540 worth of property, at 7 per cent, the assessments being 1-, 1,, 3-, nd 7 per cent? Ans. $38.109. 7. How much must be paid for insuring 12.500 worth of property, at 12~ pec cent, the assessments beingr 2, 2 -, 3I-, 4, 33 and, per cent? Ans. $263.02. 8. A gentleman paid $50 annnualy for insurance on his property, at 4 per cent. How much was tlhe valuel of the property covered by the policy? Solution.-As the value of the property insured multiplied by the rate per cent of insurance, give\s the premium, it is plain that t e premium divided by the per cent will give the value of the property insured. Therefore, 53J --.04 = $1250. Ans. $1250. PRooF. —$1259 X.0.1 = $59 the premium. HenceWhen we have given the premium and the ratd per cent of insurance, to find the value of the property insured: Obs. 5. Divide the premium by the rateper cent of insurance What is the method pursued by mutual insurance co npanies? How do we find the value of the property insured, when we-have given the premium and he rate per cent of insurance? Art. 2. COMMISSION AND IN.URANCE. 253 expressed decimally, and point ojf in 'te quotient as in Division of. ecimals. (Sect. VII. Art. 12. Rule.) 9. If I pay P270 premium, at 3 per cent, what is the value of the property insured? Ans. $99000. 16. A merchant paid $1000 premium on his property, at 5 per cent. What was the value of his property? Ans. )200t)0. 11. A gentleman paid 81500 lJremium, at 3 per cent, on a ship and cargo. Itow much were they worth? Ans. fi500000. 12. A man paid $50 premium on his property, valued at $1250. What was the rate per cent of insurance? Ans. 4. Solution.-As the value of the property, multiplied by the rate per cent of insurance, gives the premium, it is evident that the premium divided by the value of the property will give the per cent. Therefore, 50. 1250 = - ==.04. HenceWhen the premium and value of the property are given, to find the per cent of insu.ance: Obs. 6. Mcke the premium the numerator, and the value of the property the denominator of a common fraction; then reduce it to a decimal. (Sect. VIII. Art. 9. Obs. 4,) NOTE.-The pupil will perceive that these three cases of insurance reciprocally prove each otler.l 13. A gentleman paid $400 premium on his property, which was valued at 616000. What was the per cent of insurance? Ans. 2~. 14. A ship captain paid $1250 premium on his vessel, valued at $25000. What was the per cent of insurance? Ans. 5. 15. A merchant paid $1800 premium on his store and goods, valued at $60000. What was the per cent of insurance? Ans. 3. Case 2.- TVhen a person wishes to insure such a sum that in case of a loss, le may receive both the value of his property, and the price for insuring Ex. 1. A gentleman wishes to insure his property, worth $1140, at 5 per cent, so that if it is destroyed, his policy will cover both the premium and value of the property insured. What sum must his policy cover? Solution.-The rate of insurance is 5 per cent; therefore on $100 How do we find the rate per cent of insurance, when we have given the premium and the value of the property insured? What relation have these three cases of insurance to each other? COMMON ARITHMETIC. Sect XIII lie receives but $.95, as he pays $5 for insuring. Therefore, he recelves 9o-5 of the property he insures. Then $1140 is oL — of the sum necessary to be insured, and $1140 is 'L5- of $1200. Ans. $1200. PRooF.-$ 1200 X.05 = $60, the premium he would pay. $1200 - 60 _= $1140, the value of his property. HenceTo solve all such questions: Obs. 7. Multiply the sum necessary to be insured by 100, and divide the product by 100 diminished by the rate per cent of insuring. 2. What sum must I have insured, at 8 per cent, to cover $2760? Ans. $3000. 3. A merchant sent a cargo to Liverpool, valued at $12500. What sum must he have insured at 8 per cent, th t in case of a wreck he may sustain no loss by the operation? Ans. $13586. 951. 4. What sum must be insured at 10 per cent, in order to cover both the premium, and $10800 worth of property? Ans. $12000. ARTICIE 3. PROFIT AND Loss. Obs. 1. PROFIT and Loss in trade signify the sum gained or lost in common business transactions. They are calculated at a certain per cent on the purchase price, or sunm paid for the articles under consideration. MENTAL EXERCISES. 1. A man bought a sack of wheat for $5, and sold it at a profit of 10 per cent. For how much did he sell it? Solution.-As he sold it at 10 per cent profit, he sold it for $5, together with 10 per cent of $5. Now 10 per cent of $5 is $0.50, (Art. 1. Obs. 4.) and $5 + $0.50 = $5. 50. Ans. $5.50. 2. Suppose in the last case he had sold the wheat at 10 per cent loss. How much would he have received for it? Solution.-10 per cent of $5 is $0.50. Then if he sold it at 10 per cent loss he sold for 50 cents less than it cost him, and $5 - $0.50 = $4.50. Ans. $4.50. 3. Bought a stack of hay for $12, and wish to sell it at 8 per cent profit. For how much must I sell it? What is the rule, when a person wishes to insure such a sum that in case of a loss he may receive both the value of his property, and the price paid for insuring? What do Profit and Lss signify? How culculatitd Art. 3. PROF;T AND LOSS 4. C. bou hlt a piece of cloth for 810; but it being damgtlmil!, he is willinr to sel i at 4 per cent loss. For how much must he sell it? 6. A farmer boughlt sonm * ogs for $1 1, and wishes to sell them at 12 percent profit. For how much must he sell them? 6. A man bought a coat for $8, and was offered 12' per cent for his bargain, flow much was he offered for it? 7. A merchant bought a barrel of flour for $5, and sold it at 5 per cent loss. For how much did he sell it? 8. A man bought some tea for $6, and sold it at 20 per cent profit. For how much did he sell it? 9. A man bought some paper for $12, and sold it at 8- per cent profit. For how much did he sell it? 10. A boy bought a cap for $3, and sold it at 3- per cent loss. For how much did he sell it? 11. A man bough: some land for $50, and sold it at 6 per cent loss. For how much did he sell it? 12. A drover having 400 sheep, lost 9 per cent of them by disease. Hlow many had he left? EXERCISES FOR TIHE SLATE. Case 1.-To fined at uchat price to sell an article in order to gaia or lose a certain per cent, the parchase price being given. 1. A man bought a wagon for $60, and sold it at 5 per cent profit. For how much did he sell it, and how much did he gain? Operation. $ 60 By multiplying the purchase price.05 by the per cent, we find his profit to be $3. Then if he gained $3, he $ 3.00 gained. must have sold it for,$60 + $3 = 60.00 $63. Ans. $63.00 selling price. 2. A man bought a farm for $1200, and sold it at 8 per cent lossHow much did he lose, and for what price:did he sell it? Operation. $1200 $1200 purchase price..08 96 lost. Ans. $96.00 $1104 selling price. 8 per cent of $1200 is $96. Then as he sold it at 8 per cent loss, he evidently sold it for $96 less than he gave for it; and $1200 - $96 = $1104. Hence 256 COMMON ARITHMETIC. Sect. XIII To finl at what price to sell an article in order to gain or lose a certain per cent, the purchase price being given: Obs. 2. Firs!, nulltiply the purchaase 2price by the peer cent; this will give the gain or loss. Then add the gain to, or subtract the loss from, the purchase price, as the case may require; the result in either case will be the selling price. REMARK.-As 100 per cent is 1, or unity, it is evident that any per cent profit is equal to 100 increased by this per cent, and any percent loss is equal to 100 diminished by this per cent. Hence, the selling price may be found thus: Obs, 3. Add the per cent gained to, or subtract the per cent lost from 100, and multiply thepurchase price by the sum or remainder, as the case may be, and point off two decimals; the result in either case will be the selling price, NOTE.-The learner will bear in mind that it is only the selling price that we obtain by this method, and not the sum gained or lost. The operations of Ex. 1 and 2, by this method,' may be seen, as follows: Ex. 1. Ex. 2, $60 $1200 100+5=105 100-8= 92 300 2400 60 10800 $63.00 selling price. $1104.00 selling price. 3. A merchant bought tea for $0.50 per p(und, and sold it at 20 per cent profit. For how much did he sell it per pound? Ans. $0.60. 4. A merchant bought a lot of goods for $450; but getting them damaged, he sold them at 10 per cent loss. How much did he lose, and for how much did he sell them? Ans. He lost $45, and sold them for $405. 4. A droverbought a lot of sheep for $500, and sold them at 25 per cent profit. How much did he gain, and for how much did he sell them? Ans.. He gained $125, and sold them for $625. 6. Bought a span of horses for $250, and sold them at 15 per cent profit. Required-the profit and the selling price. Ans. Profit, $37.50; selling price, $287.50. How do we find at what price to sell an article in order to gain or lose a certain per cent, the purchase price being given? By what other rule can the selling price be found? Demonstrate this rule. Art. 3. PROFIT AND LOSS. 257 7. A man having a flock of 200 sheep, lost 40 per cent of them by disease. How many did he lose, and how many had he left? Ans. He lost 80, and had left 120. 8. A merchant bought cloth for $6 per yard, but getting it damaged he is willing to sell it at 10 per cent loss. For how much must he sell it per yard? Ans. $5.40. 9. Suppose he wished to gain 5 per cent on the above cloth. For how much must he sell it per yard? Ans. $6.30. 10. If he would gain 15 per cent, at what price per yard must he sell it? Ans. $6.90. 11. A merchant bought 500 bales of cotton, each bale containing 225 lbs., at 93 cents per pound; but the price declining, he sold it at 4- per cent loss. How much did he lose, and at what price did he sell it? Ans. Lost $475.311; sold it for $10'93.43. 13. A dealer in grain bought 48 leads of wheat, each containing 44 bushels, at $1.12~ per bushel;:'0 loads of rye, each containing 52 bushels, at $0.62' per bushel; 115 sacks of oats, each containing 15 bushels, at $0.20 per bushel; and 230 loads of corn, each containing 35 bushels, at $0.28 per bushel: he sold the entire quantity so as to gain 16| per cent on the cost. Required-his gain, and the price at which he sold. Ans. Gain, $99 1.662; selling price, $6941.66-. Case 2.-Two numbers being given, one of which is regarded as a certain per cent of the other, to find the rate per cent. 1. What per cent of $20 is $5? Solution.-From the nature of the question, $5 is some per cent of $20; that is, $:0 multiplied by some number, will produce $5. Therefore, we have $5 as the product, and $20 as one factor, and the other factor must be 85 $- $20 - - =.25. (Sect. VI. Art. 1. Obs. 16.) PRooF.-$20 X.25 = $5. HenceTo fnd the per cent in such cases: Obs. 4. Mlake the number which is regarded as the percentage the numnerator, and the number of which it is so much per cent the denominator of a commonfraction, and reduce this to a decimal. (Sect. VIII. Art. 9. Obs. 4.) 2. What per cent of $40 is $11? Ans..275- =271. 3. What per cent of f90 is $30? Ans. 33-. 4. What per cent of $80 is $40? Ans. 50. When two numbers are given, one of which is regarded as a certain per cent of the other, how do we find the per cent? 258 COMMOON ARITlIMETIC. Sect. XIII 5. What per cent of 120 is 6? 6. What per cent of 1 is 2? 7. What per cent of 60 is 75? 8 What per cent of $5 is 5 cent? 9. What per cent of 680 is $120? 10. Whalt per cent of $40 is 5 cents? 11. What per cent of t90 is 6 cents? 12. What per cent of $450 is $0.90? Ans. 5. Ans. 200. Ans. 125. Ans. 1. Ans. 150. Ans..00125 -= ". Ans. 1%. Ans.. A. The principles of this case are applied to the transactions of business as follows: 13. A merchant bourht some cloth at $5 per yard, and sold it for 66.25 per yard. What per cent did he gaiin? Operation. As he boulght for $5, and sold for $6.25, he evidently gained; 6.25 -$5 = 1.25 on each yard. The question then is, Wl/f't per cent of $5 is $1.2'5? By Obs. 4 we find this to be 25. 66.25 selling price. 5.00 purchase price.:1.25 gain per yard. o -= 25 pter cent. rOo,.- 5 X.25 = 81. 25; 85 + 81. 5- = 6. 25. HenceWheln we have given both the purclhase ln tIlhe selling price, to fiid l:e gain or loss pecr cent on the cost or ptuchase p)ice: Obs. 5. ]'ind t/ie g/fin or loss ol thie arti cle 5y sbtLractc tg one pr;ic Jforo tIhe o/hcr;,1 m(ke this the rnunerea!or,!n th-e cosI or purctase price the denomin(aor o' a comnion f3racticn; thelz reduce this to a decimal. (Sect. V1II. Art. 9. Obs. 4.) RlEMARK. —T'l1 Ilarner will perceive that tihe percnltau? is eale.lated on the co.t, ol sum paid, for the article, anid iot onl tilet' cl'ing! rice, as is often sup14. B3l"illt a wagon for 860, and "sold it for $70. lIequiredthe per cenit gained. Ans. 16.. 15. A man bought a horse for t l0d, and sold him for 875. Required -the gain per cent. Ans. 25. 16. A tailor bought some cloth for,120, and sold it for 1.35. tRequired-the gain per cent. Ans. 12'. 17. A nman bought a carriage for,$00, and sold it for $180. Ieuquire:!'- thi loss pe cent. Ans. 10. When we have given both thle turci-ase a',)' s-lling pric,, how (10 we fin:l the gain or loss per cent? In the percentage calculated on the buying or sell Art. 3. lPROIT AND LOSH. 18. A grocer b;ilght butter for 18 cts. per pound, and sold it for 24 cts. per pound. Required-the per cent profit. Ans. 33-. 19. A gentleman bought a house and lot for 86'2500, and sold them for $2000. What per cent was his loss? Ans. 20. 20. A merchant bought 60 yards of clo'h at $6 per yard, and sold the entire quantity for $400. What per cent was his profit? Ans. 11. 21. A speculator laid out $2000 for land, and afterwards sold his land for;2800. What per cent did he gain? Ans. 40. 22. Bought cloth for 24 cents per yard, and sold it for 21 cents per yard. What per cent was lost? Ans. 122. 23. A merchant bought cloth at $0. 50 per yard, and sold it for $0.60 per yard? Required-the per cent gained. Ans. 20. 24. A man bought a pair of tine boots for $5, and sold them for '5. 25. What per cent was gained? A ns. 5. 25. A grocer bought 6 sacks of cofl'(e, each containing 200 lbs., at 9' cts. per lb.; and 3 cwt. of sular, at 8 cts. per lb.: he sold the whole for $161, What per cent did he gain? Ans. 16. Case 3.-The price at which an arl'iwle is sold, and the rate per cent of gain or loss being given, to find the originri1 cosf. 1. A man sold a horse for 890, andl by so doing igained 20 per cen-. How much did the horse cost him? Ans. $75. Solution.-From the nature of the (question, tl.e sellinFtr price is 120 per cent of the purchase price. 120 per cCnt is '0. '1Then 690 is 2 of what number? PROOF.-$75 X.20 = 815; '75 + 1 = - 90, the sellIng price. 2. A man sold cloth for $4 per yard, and by so do n: lost 20 per cent. Required-the original cost. Ans. n 5. Solution.-The selling price in this example is evidently equal to 80 per cent of the original tost. 80 per cent is _8o. Then $4 is 80 of what numbelr? PROoF.-$5 X. 20= 1; 85-.1 = $4, the selling price. Hence-To find the original cost, when the selling price and the rate per cent of gainl or loss lare given Obs. 6. Add the gain to, or suhtract the loss from 100; then multiply the selling price by 1 00, and dic'ide the product Iy thle sum or remainder, as the case myn be. When the selling price, and the per cent of gain or los are given, how do wo find the origiual cost? 260 COMMON ARITHMETIC. Sect. XIII REMARK.-Pupils sometimes think that if we find the percentage on the selling price at the given rate, and add it to, or subtract it from, the selling price, the sum or remainder, as the case may be, will be the original cost. This error may be avoided by recollecting that the cost or purchase price is always made the basis upan which the gain or loss is calculated. (Obs. 1.) 3. By selling cloth at $5 per yard, 25 per cent was gained. Required-the original cost. Ans. $4. 4. By selling sugar at 11 cents per pound, 10 per cent was gained. How much did it cost? Ans. 10 cts. per lb. 5. A grocer, by selling tea at $1 per lb., gained 11 per cent, How much did it cost him? Ans. $0. 90 per lb. 6. By selling cloth at $1.50 per yard, 25 per cent was lost. Required-the original cost. Ans. $2 per yd. 7. A bookseller sold a lot of books for $400, and by so doing gained 33x per cent. How much did they cost him? Ans. $300. 8. A merchant bought a lot of damaged goods for $1000, which was 162 per cent less than cost. Required-the cost. Ans. $1200. 9. A man bought two horses, paying the same sum for each. He afterwards sold one of them for e180, which w.is 10 per cent loss, and the other at 15 per cent profit. For how much did he sell the latter horse? Ans. $230. 10. A man sold a piece of land for $2000, by which means he lost 16- per cent. He tried at first to sell it at 12- per cent profit. For how much did he try to sell it? Ans. $2700. ARTICLE 4. STOCKS, BROKAGE, DUTIES, AND TAXES. STOCKS. Obs. 1. By STOCK is meant the capital of Trading Companies, Banks, Rail-road and Insurance Companies, Manufactories, dcc.; also, the funds of governmnnt, state bonds, &c. It is generally divided into a certain number of parts, called Shares. A share is generally valued at $100, and owners of shares are called Stock-holders. Obs. 2. Stock has two values: a nominal, and a real value. The nominal value is the original cost, or price paid for a share. The real value is the price at which a share can be sold, which varies at different times. Obs. 3. When stock sells for its nominal value, it is said to be What error do pupils sometimes fall into with respect to questions of this kind? How may this error he avoided? What is meant by Stock? How is it generally divided? What sum usually constitutes a share? What are the owners of a share called? How many values has stock? What are they? What is the nominal value? The real value? Art. 4. STOCKS. at par. When it sells for more than its nominal value, it is said to be above par, or at an advance; and when it sells for less than its nominal value, it is said to be below par, or at a discount. Par is a Latin word, signifying equal. REMARK.-Persons who deal in Stock are usually called Stock-brokers, or Stoak-jobbers. Obs, 4. The variation of the real value of Stock is called its rise or fall. It is reckoned at a certain per cent on its nominal or par value, EXERCISES FOR THE SLATE. 1. What is the real value of $1000 of bank stock at 110 per cent; that is, 10 percent above par? Ans. 1100. Operation. —$1000 X 1.10 = $1100. Or, $1000 X.10 = $100; $1000 +$100 = 1100, as before. 2. What is the value of $1500 in bank stock, at 94 per cent; that is, ^per cent below par? Ans. $1410, Operation.-$1500 X.94 =$1410. Or, $1500 X.06 $90; $1500 -90 =-$1410, as before. REMARK.-From these operations we perceive that questions in Stocks are performed in the same manner as in Percentage; (Art. 1. Obs. 4.) or as questions in Profit and Loss, Case 1. (Art. 3. Obs. 2.) 3. Bought 8 shares of railroad stock at 12 per cent advance. How much did ii cost? Ans. $896. NoTE. —When the value of the share is not given in these examples, $100 is understood as its value. 4. A gentleman bought 6 shares of stock in a manufacturing establishment, at 4 per cent discount. How much did it cost? Ans. $576. 5. A gentleman bought 12 shares of stock in a cotton factory, at 4 per cent discount, and sold it at 6- per cent advance. How iAuch did he gain, the par value of each share being $75? Ans. $92.25. 6. A stock-broker bought 20 shares of canal stock, at 2' per cent discount, and sold it at 3- per cent advance. Required-his gain, the par value of a share being $100? Ans. 120. When is stock said to be at par? When above par? When below par? What doeq par mean? What are persons who deal in stock usually called? 'What is the variationl of the real value of stock called? How is it reckoned? How are questions in stock performed? 262 COM7MON ARITHIMETIC. Sect. XIII 7. Boughlt 6 shares of railroad stock at 94 per cent, and sold it at 112 p:r cent. Required —the gain? Ans. 108. 8. Boutght 9 shares of railroad stock at 106 per cent, and sold it at 96 per cent. IIow much was lo.st by the transaction? Ans. $90. B:OKA.F. Ohs. 5. BIROKA:GE ii the operation of finding the percentage on Bills of Erchange, lcBak notes,,&c. Persons who buy and sell b1ank notes are cliled Brokers. Obs. 6. Bank notes have two values: a nominarl, or par value, and a real value. The nominal, or par value. is the sum named on the face of the note. The real value is the price at which the note will sell. The remarks of Obs. 3 apply to bank notes as well as to Stock. The rule is the same as in stocks. 1. What is the value of,$1000 in bank notes, at 1- per cent advance? Ans. $1012.50. 2. What i. the value of $1590 in bank notes, at 2 per cent discount? Ans. $1470. 3, A merchant negotiated a bill of exchange of $2500 with a broker, for wlich hle was to give him 425 per cent. How mufch (lid the broker receive, and what sum (lid he return to the own r? Ans. The broker received $110, and returned.: 2390. 4. What is the value of $5000 in bank notes, at 3 per cent premium? Ans. $5018.75. 5. What is the value of $10000 in bank notes at 5 per cent discount? Ans.,b9937. 50. 6. A gentleman in St. Louis exchanged $1500 in specie for Cincinnati bills, at 5 per cent advance; he then started for Cincinnati, an l paid for his passage 15; at Cincinnati, he exchanged his bills for notes on a bank in Albany, N. Y., at 2- per cent discount; he then started for Albany, and paid for h s expenses $45; in Albany he sold his notes for Bosion bills, at - per cent dis ount: his expenses from Albany to Itoslon were 810. How much had he when he arrived at Boston; and did lie gain or lose by exchanging, and how much, supposing the specie to be every where current i Ans. I He lad when he arrived 81458 62. Gained by exchanging,.,28.62. DUTIES. Obs. 7, DUTIES are sums of money required Jby government, to What is Brokage? What are persons who buy and sell bank notes called? How mtnyvaltes have bank notes? What is the nominal, orpar value? The real value? What isthe rule iR Brokage? Art. 4. DU tIES. be paid on thc importation of goods. They arc of two kinds: SSpeci'ic awn Ad Valoram. The inco:me arising from duties, &c., is called RPEVENUE; and a table of the dutlies i; called a TARIFF. Case 1.-SPECIFIC DUTIES. Obs. 8. A Specfic Dauty is a certain sum imposed upon a ton, pound, yard, gallon, hogshead, &c., without regard to the value of the article. REMIARK 1. —Accordiwg to law, c rt;:iu ded uctioos are made, called tire, draft, lekailre, &c., before specific (duties are impoled. 2.-Tare i.. the allowance made m i:r the weight'of the box, cask, or whatever contains the article on which the duty is paid. Draft, or Tret, is the allowance made on the weight of the goods for waste or refuse rmitter. Leakage is an allowance for waste of liquors in casks, &c. It is usualiy about 2 per cent. 3.-For breakage, 10 per cent is allowed on,II a!e, beer, and porter, in bottles, and 5 per cent on all other liqluors in bottles; or tle i:iporter may have the duties computed on the actual quantity by tale, if he so chooses at the tinie of entry. 4.-At the Custom [louses, common sized bottles are estimated to contain 223 gallons per doz0en. 5.- he ertire weight of any lot of goods, together with tIe box, or whatever contains the gfods, is called the i ros., wei:,~ht. T'lhe wIg}ht Cof t' g",ods alone, afttr l l deducltins for tare, draft, &c., is called the meat weitht. (Sect. IX. Art. 2. Obs. 21. Reni. 2.) 1. What is the specific duty on 30 hhds. of sugar, at 21 cents per pound; the gross weight being 9 cwt. 2 qrs. each; the draft 4 lbs. per hhd., and the tare 12 per cent? Oierat:on. 9 cwt. 2 qts. - 950 lbs. in each hhd. 30 28500 lbs. in 30 hhds. 30 X 4- 120 Ibs. draft. 28380 283P80 X. 12 34' 6 lbs. tare. 2)24974 lbs. neat weight..02' duty per lb. 49948 12487 Ans. $624.36 264 COIMM'N ARITHMETIC. Sect. XIII In this operation we notice two considerations: 1st. The tare is oomputed on the remainder, after the draft is deducted. 2nd. The duty is computed on the neat weight, or the remainder after all deductions are made. NoTE.-In casting tare, any remainder which does not exceed half a pound, is not reckoned; but if it exceeds half a pound, it is reckoned a pound. By remainder, in this case, is meant the decimals. HenceTo find the specific duty on any article of merchandise: Obs. 9. First deduct the draft, tare, leakage, &c., from the given article, and then multiply the remainder by the duty per pound, gallon, yard, dcc. 2. What is the specific duty on 10 hogsheads of wine, at 12 cents per gallon, deducting 2 per cent for leakage? Ans. $74.088. 3. What is the duty on 5 gross of porter, at 15 cents per gallon, 10 per cent being allowed for breakage? Ans. $300. 4. What is the duty on 30 casks of nails; weight, 140 lbs. per cask, at 4' cts. per pound, allowing 1 lb. per cask for draft, and 3 per cent for tare? Ans. $182.02. 5. What is the specific duty on 6 hhds. of sugar, at 3 cents per pound, gross weight as follows: the first. 6 cwt. 3 qrs.; 2nd, 7 cwt. 1 qr.; 3d, 4 cwt.; 4th, 5 cwt. 3 qrs.; 5th, 6 cwt. 1 qr. 20 Ibs; 6th, 8 cwt. 15 lbs.: allowance for draft, 4 lbs. per hhd., and for tare 12 per cent? Ans. $100.62. Case 2.-AD VALORAM DUTIES. Obs. 10. AD VALORAM D-UTIES consist of a certain per cent of the actual cbst of the goods in the couzntry from which they were brought. Ad Valoram is a Latin term, signifying according to value. 1. What is the ad valoram duty on a quantity of broadcloth, invoiced at $2000 in London, at 12 per cent? Ans. $240. Suggestion.-When the duty is imposed on the actual cost of the goods, of course no deductions are to be made. Then 12 per cent What are Duties? How many kinds of duty are there?- What is Revenue? What is Tariff? What are Specific Duties? What deductions are made before Specific duties are imposed? What is Tare? Draft? Leakage? What per cent is allowed on liquors in bottles, for breakage? How much are common sized bottles estimated to contain? What is gross weight? Net weight? Upon what is tare computed? The duty? How do we find the specific duty on an article of merchandise? What are Ad Valoram duties? What does the term ad valoram mean? Are any deductions to be made on ad valoram duties? Why not? Art. 4. TAXES. 265 of $2000, is $2000 X. 12 = $240. It will be perceived that the operation is the same as in percentage. (Art. 1. Obs. 4.) REMARK. —An Invoice is a written statement of the goods in question, with the prices of the articles annexed. 2. What is the duty on a quantity of books, invoiced at $800, at 10 per cent, ad valoram? Ans. $80. 3. What is the duty on a quantity of wine, invoiced at $1000, at 16T per cent, ad valoram? Ans. $162.50. 4. What is the duty on a quantity of jewelry, invoiced at $500, at 18 per cent, ad valoram? Ans. $90. 5. A merchant bought a quantity of silks in Canton for 1200 talcs; upon importing them the duty was 15 per cent ad valoram. Required-the duty, the tale being estimated at $1. 48. Ans. $266. 40. TAXES. Obs. 11. TAXES are sums paid by the people for the support or benefit of the Government, County, District, &c. Obs. 12. Taxes are levied either on the person or property of the citizens. When the tax is levied on the person, it is called a poll tax, and is usually a specified sum.t When the tax is levied on the property, it consists of a certain per cent of the value of the entire property possessed by each taxable person. REMARK.-Property is of two kinds: Real Estate and Personal Property Real Estate denotes property that is fixed: as, lands, houses, &c. Personal Property denotes property that is sloveable: as, horses, cattle, carriages, notes, tools, &c. Obs. 13. The first thing necessary to be obtained in levying a tax is an exact inventory, or written account, of all the taxable property, both real and personal, in the State, County, or District &c., where the tax is to be levied, together with the value of the entire property owned by each individual who is to be taxed, and the number of polls. An Inventory is a list of articles. 1. A certain district was taxed $1850. The district contained 50 polls, which are assessed at $1 each; and the whole amount of taxHow do we find the ad valoram duty on an article? Whit is an inrvice? What are taxes? How levied? What is a poll tax? When a tax is leviedon property, of wlat does it consist? Of how many kinds is property? What is Real Estate? Personal Properiy? What is the first thi ng necessary to be obtained in levying a tax? tin 5At* States tiret isr nn pol tax.. t$ '. v 266 COMMONf ARITHMETIC. Sect. XIII able property, both personal and real, in the district, was valued at $120000. Required-the tax on a dollar, and also the tax of an individual whose real estate is worth $2500, and whose personal property is worth $800, and who pays for 1 poll? Ans. Tax on $1 is 1l cts.; on the individual $50.50.;"~ Operation. $ 1850 120000)1800.00(. 01 "- =1I cts.on 50 poll tax. 1200 00 [the doll. $1800 tax on property. 600000 $2500 real estate. $3300 800 personal property..01 $33O0 taxable property. $3300 1650 $49.50 -+$1 = $50.50. As the poll tax is a specified sum on each individual, it is evident that this must be deducted from the tax to be raised before calculating the tax on the property. By this means we find the tax on the property to be $1800. Then if $120000 pays $1800 tax, $1 will pay $1800 -$120000= I' cts. Also, if a man's real estate is worth $2500, and his personal property worth $800, his entire taxable property is worth $2500 + $800 = $3300; and $3300 X. 1- =- $49. 50, and $1 for his poll tax makes his entire tax to be $50.50. HenceTo assess a Statt, County, or other tax, we have this RULE. I, F?;d the tax on all the polls, (if any) at the given rate, and subtract (tis from the whole tax to be raised. Divide the remainder by the whole amount of taxable property in the State, County, District, &c., and the quotient will be the tax on one dollar. II. Multiply the value of each man's property by the tax on $1, and the product will be the tax on his property, to which add his poll tar, and the suh will be his total tax. PRoor. —Add together the taxes of all the individuals taxed; if the pnm s equal to the whole tax assessed, the work is correct. 2. A town was taxed $600. It contained 40 polls, valued at $0. 75 each, and the value of the taxable property was $23500. What murt be done before the tax is levied on the property? Why so? What is the rule for nsing taxem? The prof? Art. 4. TAXE'. 267 What per cent was the tax, and what is the tax of an individual who pays for 3 polls, and whose property is worth.$1575? A ax on $1 is 2 cents. ns. Individual's tax, $33.75. 3. A company was taxed $313&. The value of the taxable property was $125000, and there were 20 polls valued at $0. 50 each. Required-the tax on $1. Ans. 2' cts. REMARK.-Assessors generally make a table containing the tax on $1, $2, &c., up to $1000; and thus, by knowing the value of any one's property, his tax can easily be found. Thus, in the last example, the tax on $1 being 2& cts., on $2 it will be twice as much, or 5 cts., &c. Hence-the following TABLE. $1 pays $0.02'. $20 pays $0.50. $200 pays $5.00. 2 ".05. 30 ".75. 300 " 7.50. 3 ".07(. 40 " 1.00. 400 " 10.00. 4 ".10. 50 " 1.25. 500 12.50. 5 ".122' 60 " 1.50. 600 " 15.00. 6 ".15. 70 " 1.75. 700 " 17.50. 7 ".17'. 80 " 2.00. 800 " 20.00. 8 ".20. 90 " 2.25. 900 " 22.50. 9 ".22'. 100.50. 1000 " 25.00. 10 ".25. 4. Find, by the above table, the tax of an individual whose property is valued at $1426, and who pays for 2 polls, at $0.50 each. Ans. $36.65. Operation. The tax on $1000 is $25.00 1426-=1000+400+20+6. The " 400 is 10.00 Therefore, by finding the tax The " 20 is.50 on each of these in the table; The " 6 is.15 and adding them together we find the tax on the whole. And the tax on $1426 is $35.65 (Sect. 4. Art. 4. Obs. 4. Two polls at $0.50 each 1.00 Rein. 2.) Total tax, $36.65 5. What would be A's tax in the above company, who pays for 1 poll, and whose property is worth $1594? Ans. $40.35. How do assessors proceed in levying a tax? How do we find a.ax by the Table? 268 COMMON ARITHIMETIC. Sect. XIII 6. C pays for 3 polls, and his property is worth $2875. Required his tax? Ans. $73.37-. 7. D pays for 1 poll, and his property is worth $882. Required his tax? Ans. $22.55. 8. E pays for 2 polls, and for property worth $27740. Required his tax? Ans. $694.50. 9. F pays for 4 polls, and for property worth $48450. Required his tax? Ans. $1213.25. 10. G pays for 3 polls, and for property worth $30012. Required his tax? Ans, $751.80. 11. H. pays for 4-polls, and for property valued at $12021. Required his tax? Ans. $302.523. 12. Prove by the 8 preceding examples, that the result in exampli a i., cs;;c 1 *v. ARTICLE 5. INTEREST. Obs. 1. INTEREST is the sum paid for the use of money, It is computed byper centage: that is, so many dollars are paid for the use of $100 for 1 year; and in the same proportion, for a sum greater or less than $100, or for a longer or shor cr time than a year. Obs, 2. The sum on which the interest is paid is called the PRINCIPAL, The interest is generally a certain per cent. of the principal. The per cent. on $100 for 1 year is called the RATE. The sum of the principal and interest is called the AMOUNT, Illustration.-If I borrow $100 for 1 year, and agree to pay 6 per cent, for the use of it, at the end of the year, I must pay back, not only the $100 which I borrowed, but 6 per cent. of it, or $6 also, making in all $106. In this case $100 is the principal; $6 the interest, 6 per cent. the rate; and $106 the amount. Obs. 3. Per annum signifies by the year. Thus, 6 per cent. per annum, signifies that $6 is paid for the use of $100 for 1 year. REMARK.-Whpn the tine is not mentioned with the rate per cent., a year is always understood. Obs. 4. Legal interest is the rate per cent. established by law. What is interest? How is it colmp-uted? What is the principal? The interest? The rate? The amount? In lthe illustration given, which is the principal? The interest? The rate? The amount? What does per annum signify? Give an example. When the time is not mentioned, with the rate per tent.3 what is haidersttai7 What is legal interest? Art. 5. INTEREST. 269 It varies in different countries, and also in the different States. Thus, it is 6 per cent. in all the New England States, in New Jersey, Pennsylvania, Delaware, Maryland, Vir'inia, North Carolina, Tennessee, Kentucky,Ohio, Indiana, Illinois, Missouri, Arkansas, District of Columbia, and on debts or judgments in favor of the United States. 7 per cent. in New York, South Carolina, Michigan, Wisconsin, and Iowa, 8 per cent. in Georgia, Alabama, Mississippi, Texas, an d Florida. 5 per cent, in Louisiana, REMARK. 1. When the rate per cent. is not mentioned in transacting business, the legal rate is always un derstood. 2. Any rate per cent. higher than tile leg:l rate is called USURY, and the person who exacts it is liable to puuililmenat. If the parties agree, however, any rate less than the legal rate ilay be taken. Obs. 5. The learner will observe this diTerence between per centage and interest: in interest, time is consdetred; in per centaye it ts not. Interest is of two kinds: SIMPLE and COMPOUND. SIMPLE INTEREST. Obs. 6. SIMPLE INTEREST is the interest on the principal only. CASE 1. 2o calculate interest for any number of years, at any rate per cent. Ex. 1. What is the interest of $450 for 3 yrs. at 5 per cent? The amount? The interest is 5 cents for 100 cents, or Operation. $1 for 1 year. Therefore, the interest of $459 $450 for 1 year is $450 X.05 = $22.50;.05 and for 3 years it is 3 times as much or $67.50. The sum of the interest ($67.50) $22.50 int. for 1 yr. and the principal ($450) is the amount. 3 (Obs. 2.) Ans. $67.50 int. for 3 yrs. 450.00 principal. Ans. $517.50 amount. Hence-to find the interest of any sum of money, for any number of years, at any rate per cent. What is the legal rate i n most of the United States? When the rate per cent. is not mentioned in transacting business, what is understood? What is Usury? What is the difference between per centage and interest? How is Interestdivided? What issimple interest? How do we find the interest of any sum, for any number of years, at any rate per cent? 270 COMMON ARITHMETIC. Sect. XIII Obs. 7. Malt;ply theprincipal by the rate per cent. expressed decimally, this will be the interest for I gear. Mlultiply the interestfor 1 year by the number of years, if the time is longer than 1 year. REMARK. To finJ the amount; Aid the principal to the interest. EXERCISES FOR THE SLATE 1. What is the interest of $120.50 for two years at 6 per cent? At 8 per cent.? Al 7 per cent.? Ans. in order. $14.46; $19.28, $16.87. 2. What is the interest of $640.20 for 5 years at 4 per cent.? At 5 per cent.? At 7- per cent.? Ans. in order. $128.04; $160.05; $240.075. 3. What is the interest of $300 for 7 years, at 12- per cent.? Ans. $262.50. 4. What is the interest and amount of $800 for 12 years, at 83 per cent.? Ans. Int. $840; Amt. $1640. 5. What is the interest and amount of $1000 for 5 years, at 47 per cent.? Ans. Int. $243.75; Amt. $1243.75. 6. What is the interest and amount of $575 for 6 years, at 4 per cent.? Ans. Int. $138; Amt. $713. 7. What is the interest and amount of $860 for 10 years. at 7, per cent.? Ans. Int. $645; Amt. $1505. CASE 2. To calculate interest for years and months, at 6 per cent. 1. What is the interest of $400 for 1 year and 6 months, at 6 per cent? At 6 per cent., the interest of $1 for I year is Operation. 6 cents. Then as there are 12 months in a year $400 the interest {or I month is 6 - 12 == 6 = 1 a.09 cent or 1 cent for every 2 months. Then if it is 1 cent for ( very 2 months, for 8 months it will Ans. $36.00 be 4 cents; for 10 months, 5 cents; for 6 months, 3 cents, &c.; that is,-the interest of I for any number of months, is half as many cents as there are months in the given time. Therefore, in the above example, 1 yr. 6 mo. -18 mo, anl the interest (f I.I for 18 mo. is 9 cts.; and the interest of $400, is.09 X 400 = $36. Hence How do we find the amount? To what ix the interesl of $1 for any number of months equal? Show why tlis is correct. Art. 5. INTERBSr. 271 To find the interest of any sum for years and months, at 6 per cent. Obs. 8. Call half thenumber of months cents; and multiply the principal by this, expressed decimally, 2. What is the interest of $350 for 3 yrs. 8 mo., at 6 per cent.? Ans. $77. 3 What is the interest and amount of $1000 for 5 yrs.? Ans. Int. $300; Amt. $1300. When the rate per cent. is not mentioned in these examples, 6 per cent is understood. 4. What is the interest of $1200 for 4 yrs. 7 mo.? Ans. $330. 5. What is the interest of $750 for 15 yrs. Ans. $675. 6. What is the interest of $225.50 for 2 yrs. 4 mo.? Ans. $31.57. 7. Required the interest of $120 for 7 mo.? Ans. $4.20. 8. Required the interest of $284.66 for 3 yrs. 9 mo.? Ans. $64.048 9. What is the amount of $427.86 for 7 yrs. 5 mo.? Ans. $618.257_ 10. What is the amount of $368.27 for 3 yrs. 3 mo. Ans. $440.082.+ 11. What is the amount of $218.75 for 10 yrs. 2 mo.? Ans. $352.187.+ 12. What is the amount of $768,25 for 4 yrs. 10 mo.? Ans. $991.042.+ CASE. S. To calculate the interest on any sum, for any number of days at 6 per cent. 1. What is the interest of $80 for 24 days? Operation. At 6 per cent the interest on $1 for 1 $80 month is I a cent; (solution of Ex. 1. Case 2.).004 In interest 30 days are reckoned as a month; ---- therefore the in erest of $1 for 1 day is $0.005 Ans, $0.320 - 30 = -5 =- of a mill; or it is 1 mill for every 6 da s. Then as it is I mill for every 6 days, for 21 days it would be 33 mills; for 12 days, 2 mills; for 15 days, 2- mills, &c.; that is, the interest of $1 for any number of days, is one sixth as many mills as there are days given. How do we find the interest of any sum for years and months, at 6 per cent.? To what is the interest of $1 for any number of days equal? Show why this is correct. 272 COMMON ARITHMETIC. Sect. XIII Therefore, in the above example the interest of $1 for 24 days is 4 mills, and the interest of $80 is.004 X80 = $0.32. HenceTo find the interest on any sum for any number of days: Obs. 9. Call one sixth of the number of days mills; and multiply the principal by this, expressed decimally. 2. What is the interest of $50 for 28 days? Ans. $0.233.4 -3. What is the interest of $120 for 20 days? Ans. $0.40. 4. What is the interest of $145 for 18 days? Ans. $0.435. 5. What is the interes of $180 for 17 days? Ans. $0.51. 6. What is the interest of $200 for 15 days? Ans. 80.50. 7. Required the interest of $250 for 12 days? Ans $00.50, 8. Required the interest of $300 for 7 days?.Ans. $0.35. 9. Required the interest of $900 for 5 days? Ans. $0.75. I0. Required the interest of $100C for 1 day? Ans. $0.16-. CASE 4. To find the interest of any number, for any time, at 6 per cent. NOTE.-This case is simply the three preceding cases united into one. 1. What is the interest of $1200 for 2 yrs. 5 mo. 18 da. at 6 per cent. Operation. 2 yrs, 5 mo. = 29 mo.; 29- 2 = $0.14 - $1200 $0.145; 18 -6 = 3 mills. $0.145 — $0.003.148 = $0.148, the interest of $1 for the given time, and the interest of $1200 is.148 X 1206 -9600 $177.60. 4800 1200 Ans. $177.600 NOTE.-We generally use the interest of $1 as the multiplier, as it is usually the smallest number and it makes no difference which factor we use as the multiplier. (Sect. IV. Art. 2, Obs. 5: Rem.) Hence-To calculate interest at 6 per cent. we have the following GENERAL RULE. I. Reduce the given number to months and days: then take half the number of months, and call it cents; (Obs. 8.) end if there is an odd month call it 6 mills, and count another millfor every 6 days; (Obs. 9.) this will gzve the interest of $1 for the given time. How do we find the interest of any rum for any number of days? Which number do we generally use as the multiplier? W'y? What isthe General Rule for calculating interest at 6 per cent.? Art. 5. INTER EST. 273 II. Multiply the given nrincipal by the interest of $1, already found, eind point off as in multiplication of' decimals. (Sect. VIII, Art. 11. Rule. NOTE.-If there is no o Id month, anl tile nunnb3r of days bo l3es than 6, a cilphr must be put in the place of mnlls. For the convenience of the learner rwe subjoin the following TABLE By which to find the number of days from any given date to any required date; or to find any re quirel date from 'a;ivel date. Ild'x- Jan. Feb. Mar.' Apr. I-ay. June. July lAug. 1 1 | 2 60 91 121 152 182 213 2 2 33 61 92 122 153 183 214 3 3 34 62 93 123 154 184 215 4 4 35 63 94 124 155 185 216 5 5 36 64 95 125 156 1 186 217 6 6 37! 65 96 126 157 i 187 218 7 7 38 66 97 127 158 188 219 8 8 39 67 98 128 159 189 220) I i LAn 1 Q a ~ e1a a 10 10 11 11 12 12 13 13 14 14 15 1 15 16 16 17 17 18 118 19 119 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 2S 29 29 30 30 31 31 4u I 00 41 1 69 42 1 70 43 I 71 44! 72 45 73 46 74 47 75 48 t76 49 77 5) 78 51 79 52 1 80 53 81 54 82 55 83 56 84 57 85 58 86 59 87 60 88 *; 89 ** 90 YY 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 ~*~+ 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 I 151 ItU fl) Y 161 191 162 192 163 193 164 194 165 195 166 196 167 197 168 198 169 199 170 200 171 201 172 292 173 203 174 1204 175 205 176 206 177 207 178 208 179 009 180 210 181 211 * 212 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 Sept./ Oct. INov. Dec, 244 274 305 335! 245,275 306 336 246 276 307 337 247 277 308 338 248 278 309 339 249 279 310 340 250 280 311 341 251 281 312 342 252 282 313 343 253 28 314 344 354 284 315 345 255 2z5 316 346 256 286 317 347 257 287 318 348 258 288 319 349 259 289 320 350 260 290 321 351 261 291 322 352 262 292 323 353 263 293 324 354 264 294 325 355 265 295 326 356 266 296 327 357 267 297 8128 358 268 298 329 359 269 299 33,360 270 300 331 361 271 301 332 362 272 302 333 363 273 303 334 364 *** 304 *** 3!65 This tablt may be applied to the solution of two problems. Problem 1. To find the time between any two dates: Hlow do we fiad the time betweea two dates by the tablet 13A 274 COMMON ARITHMETIC. Sect. XIII Obs. 10. Find the day of the month in the index, and opposite this in the column of the given months will be found the day of the year upon which the diferent dates fall. The diference of these days, less 1, wdll express the time between the two dates. Required-the time between March 12th and June 26th. Solution.-Opposite 12, in the column for March is found 71; opposite 26, and in the co'umn for June is found 177. 177-71 = 106; 106 less I = 105. Ans, 105 days. REMARK 1. In leap years when the 29th of Feb. comes between the dates the difference between the (ays is the time. 2. When the time overruns the end of the year, take the earlier date from 365, and to the difference add the n tumber found against the latter date, less 1. Required-the time between Jan 17th, and March 8th, 1848, Ans. 50 days. Required-the'time between Dec. 14th, and Feb. 9tl. Ans. 56 days. Required-the time between Aug. 23d, and Oct. 17th. Ans. 54 days. Problem 2. To find any required date: Obs. 11. Find the day of the year as before; to this add the time between the given and required dctes, and find the sum in the table. The column in which the sum is fond will express the month, and opposite the sum in the index will be found the day of the month, less 1. A note was given May 7th, to run 90 days; when does it become due? Ans. Aug. 6lh, Solution.-Opposite 7, in the column for May is found 127; 127 - 90 - 217; 217 is found in the column for Aug. and opposite 5. 5+1 =6. REMARK 1. In leap years, when the 29th of Feb. comes between the dates, the result found in the table is tlhe correct date. 2. When the time overruns the end of the year, take 365 fromn the sum. and the difference will be the number to find in the table, for the required date. A note given Jan, )6th, 1348, fell due in 2 months, at what time did it fall due? Ans. March 17th. Solution.-2 mo. = 60 d, 16 +60 = 76; 76 is in the column for March, and opposite 17. What must be observed in Leap Years when the 29th of February comes between the dates? Wh}en the tlins tuverruln tihe end of tlh.* ye.r how do we procet i? How do we, fil(n a relitirc.d date bvy tli, t til'-? Wh'l-, tlhe 29h, of Feb. comiies between the dates how do we proceed? Wheu the tile overruns the end of the years how do we proceed?... Art. 5. INTEREST. 275 What date is 54 days later than the 22nd of Dec.? Ans. Feb. 15th. A note was given Aug. 29th, payable in 6 months; when did it fall due? Ans. Feb. 26th. A note was given Oct. 11th, 1847, payable in 5 months; when did it fall due? Ans. March 10th, 1848. A note was given Jan. 12th, 1848, payable in 9 mo.; when did it fall due? Ans. Oct. 9th, 1848. NoTI.-In these examples e have calculated 30 days to the month. When the exact time is required, the surplus time must be added or the result be diminished for Feb. if required. The exact answer for the last example is Oct. 13th 184P. Find the interest and amount of the following sums at 6 per cent. 2. $140 for 1 yr. 3 mo. 12 da. Ans. Amt. $150.78. In the answers to the following questions the amount only is given. 3. $84.23 for 2 yrs. 4 mo. 18 da, 4. $300 for 1 yr, 8 mo. 5. $24 for from June 12th, 1847 to April 24 6. $650 for 3 yrs. 1 mo. 6 da. 7. $0.96 for 2. days. 8. $175 for 1 yr. 1 mo. 1 da. 9. $280 for 2 yrs. 4 mo. 15 da. 10. $315.25 for 3 yrs. 24 da. 11. $420.50 for 4 yrs. 8 mo. 10 da. 12. $742.80 for 5 yrs. 2 mo. 20 da. 13. $896.96 for 6 yrs. 8 mo. 14. $1457.12 for 6 yrs. 6 mo. 6 da. 15. $212.25 for 4 yrs. 1 mo. 2 da. 16. $320.75 for 6 yrs. 3 mo. 9 da. 17. $478.60 for 1 yr. 1 1 mo. 18. $317 for 7 yrs. 9 mo. 16 da. 19. $400 for from Jan 6th to May 17th. 20. $700 for 7 yrs. 7 mo. 7 da. Ans. $96.274.4 Ans. $330. 6 1848. Ans. $25.272. Ans. $770.90. Ans. $0.9632. Ans. $186.404. Ans. $319.90. Ans. $373.256. Ans. $538.94. Ans. $975.544. Ans. $1255.744. Ans. $ 2026.853.Ans. $264.322. Ans. $441.512.Ans. $533.639. Ans. $465.249. Ans. $408.666. Ans. $1019.316, Required the interest and amount of the following sums at 6 per cent: 21. $1250 for 9 yrs. 2 mo. I day, Ans. $1937.708. a. When the interest is required for a large lumber of years, it is generally more comnenient, to find the interest for the years separte m h at fortmtbsm antd y ad tha add s C twe Pb 276 COMMON ARITHMETIC. Sect. XIII sults together. Thus, the interest of $1260 for 9 yrs. is $1250 X,06 X 9 $675; (Obs. 7.) and for 2 mo. 1 da. the interest is $1250 X.010 = $12:708; and $65 + $12,708 = $687.708 as the interest of $1250 for 9 yrs. 2 mo. 1 da.; and the amount is $687.708 + $1250 = $1937.708. 22, $590 for 9 yrs. 5 mo; 28 da. Ans. $926.002. 23. $440 for 10 yrs. 7 mo. 29 do. Ans. $721.526. 24. $111.11 for 11 yrs. 11 mo. 11 da Ans. $189 647. 25. $1200 for 11 yrs. 10 mo. 23 da. * Ans. $2056.60, 26. $1500 for 14 yrs. 7 mo. 17 da. Ans. $2816.75. 27. $319.90 for 15 yrs. 8 mo. 26 da. Ans. $t621.992.-4 -28. $975.50 for 23 yrs. 10 mo. 14 da. Ans. $2197.151.29. $500 for 40 yrs. 6 mo. 24 da. Ans. $1717. 30. 1000 for 60 yrs. 12 da Ans. $4602. CASE 5. To find the interest of any sum,for any time, at any rate per cent. FIRST METHOD. REMARK. We have now considered how interest is computed at 6 per cent., but it often happens that we wish to find the interest on sums of money, at other rates per cent. than 6. The interest of $1 for 1 year at 6 per cent. is 6 cents; At 1 per cent. it is 1 cent, or ^ of 6 per cent.; At 7 per cent., it is 7 cents, or 7 times 1 per cent.; At 3 per cent. it is 3 cents, or 3 times 1 per cent; &c. Hence — To find the interest of any sum, for any time, at any rate per cent, Obs. 10, Find the interest at 6per cent. as usual; divide this by 6 and the result will be the interest at 1 per cent. then mnultiply the interest at 1 per cent. by the given rate.* REMARK. - If the learner chooses, he can multiply the interest at 6 per cent. by the given rate, and divide the product by 6, as it will always produce the When the interest is required for a large number of years, how do we proceed? Howdo we compute intertst at other rates per cent. than 6? Show why'this ureiscorrect. By what otherr-.ethod can the interest be found? Demronstrate this rule. *When the rate per cent. is 3, we may multiply the principal by the numler of months -when the rate per cent. is 4, we may multiply the principal by ' the nanumer of months wjen the rate per cent. is 8 we may multipl. by * the lunihler of 1 onths; and when t he rate per cent. is 9 we may multipfl..']' t'p niunler of nrlltlhF;,a;li in ea.h a-e -noint o0T Itwo decimals. These rules wiil only api;;t however, wvinti Ithe tin!t' ca;i be ci(l 'ieti to riomlthl -DEMOYFSTnATIp -.3.1 e cellt. i.- 3 " I tskfor iCO cents tol; I. ilo!l).,;; 1i: s F 3 divided ly 12 = 'of a cent per innolltt; pr:pe c.rit is ' ofare'tilt in'r iMltti, pi er re.t-.' 1s " -tr' ei' t' per month; and 9 per cent. ii of a cent per month. Hence, the rules are evidcnt, Art. 5. INTEREST. '277 same result and by this means fractional numbers may often be! avoided in the operation. 1. What is the interest of $400 for 2 yrs. 6 mo. 18 da., at 4 per cent? Ans. 40.80. So'ution.-$400 X.153 = $61.20, the interest at 6 per cent: 61. 20 6 = $10.20, the interest atl 1 per cent.; and 10. 20 X 4.= 40. 80 the interest at 4 per cent. Find the interest and amount of the following sums: 2. $15.30 for 9 mo., at 7 per cent. Ans. Amt. $16.103. 3. 840 for 1 yr. 6 mo., at 8 per cent. Ans. $44.80. 4. $120.60 for 18 da., at 9 per cent. Ans. $121.142. 5 $2(00 for 2 yrs. 4 mo. 18 da. at 3 per cent. Ans.,214.30, 6..250.75 for 1 yr. 8 mo. 24 da,, at 4n- per ce. t. Ans. 8270.308. 7. $300 for 3 yrs. 2 mo, 12 da., at 2- per cent. Ans. $326.40. 8. $325 for 2 yrs. 6 mo. 16 da., at 5 per cent. Ans. $366.347, 9. $400.50 for 4yrs. 11 mo. 29 da., at 7' per cent. Ans, $550.604. 10. $525.75 for 5 yrs. 9 mo. 21 da., at 9 per cent. Ans. $800. 585. 11. $600 for 6 yrs. 7 mo. 11 da., at 11 per cent. Ans. 10u36.516. 12. $1000 for 8 yrs. 5 mo. 15 da., at 12 per cent, Ans. $2015. SECOND METHOD, Since 1 month is -i, of a year; 2 months,, or, of a year. &c., it follows that the interest of any sum for I month, is ', or the interest of the same sum for 1 year; for 2 months.,2 or l of the interest for 1 year; for 3 months, _- or x of the interest for 1 year, &c. Also, since 1 day is J of a month; 2 days, -2 or 2- of a month; 3 days, '3 or - of a month, &c., it is evident that the interest of any sum for I day is 2- of the interest of the same sum for 1 month; for 2 days, -i or 2- of the interest for 1 month; for 3 days, -L or ', of the interest for 1 month; &c. HenceTo find the interest of any sum, for any time, at any rate per cent: a. iFind the interestfor tide years according to Case 1, Obs. 7; then find the interest for thi months and days by taking aciquot parts. COMMON ARITHMETIC. Sect. XIII 13, What is the interest of $360 for 3 yrs. 11 mo. 18 da., at 5 per cent? Ans, $71.40. Operation. * 360.05 $18.00 = Int. for yr. 3 $54.00 - Int. for 3 yr. m 6 mo. =- a year. 2 9.00 =Int. for 6 mo. I. mo. 4mo. - of a year. 3 6.00 =Int. for 4mo. 1 mo.= of 4 mo. 4 1.50 =TInt. for 1 mo. 15 da,=' of amo. 2 75 =Int. for 15da. 3 da.= of 15da. 5 15 = Int. for 3 da. o =5 18d. I _ —I Ans. $71.40 Find the interest and amount of the following sums: 14. $360.80 for 2 yrs. 6 mo. 15 da., at 6 per cent. Ans. $415.822. 15. $212.59 for 3 yrs. 8 mo. 10 da., at 7 per cent. Ans. $267.454. 16. $400 for 5yrs. 10 mo. 12 da., at 4 per cent. Ans. $493. 65, 17. $525 for 4 yrs. 29 da., at 8 per cent Ans. $696.383. 18. $612.75 for 7 yrs. 1 mo. 20 da. at 9 per cent. Ans. $1006.441. 19. $700 for 5 yrs. 10 mo., at 5 per cent. Ans. $904. 166. 20. $1249.60 for 12 yrs. 11 mo, 29 da., at 7- per cent. Ans. $2449.926. CASE 6. To find the interest on Jotes, Bonds, &c., on which partialpayments have been made. Obs. 11. A TOTE OR BOND is an instrument of writing, in which the debtor promises ts pay the creditor his due, in such a manner, and at such a time as may be agreed upon. REMARK 1. The debtor is the one who borrows the money, or in any way owe. ttie other; the creditor is the one who lends the money, or to whom any. thing is due. Wht is j ot, or onwud Wich poa is t bJt;F? *b;i t io tor? Art. 5. INTER EST. 279 2. It often harvp3m that the debtor plys only a part of his note, or obligation, at a tinim. Waetii a p)tyinm: t is thui an111 it is cll. d an ea.lrsernent, and is written oa thl back of th3 note. Obs. 12. When the debtor pays th- crelitor a parL of his note, or obligation, it is evident thait the su'n pail, should be deducted from the su:n dec; the sum tlue b, in/l tlA facer of the note, together with the ineteiest of the s.an u:?zil the ti tn of paymf!nnt, Hence, the following 1. RULE. I. Find the amount of the priznipal to the first time when a payment was mide. which either alone, or together with the preceding payments, (if any,) exceeds the interest then due. I [. Fromn this a.?tu.nt suMtra tf th/l p o /)symt, 0or stCn of the payments madle within the tin for which the interest was computed, and the re mainder will be a new principal, with which pro:eed as before. NoT:.-The abh)vi rule, with so nli slirzht alt.-raution;of phrase)lIogy, is adopteJ by th., Supre:ne Court of the United Sltaes, and also by the differestl States of the Union, with but few exc~ptions. In the following examples, 6 per cent. is understood unless otherwise stated: (1.) $800 COLUMABUS, Ohio, Jan. 1st, 1836. For value received I promise to pay Thomas Trueman, or order, eight hindcired dollars, on demand, with interest. CHARLES PAYWELL. On this note were the following endorsements: Sept. 15th, 186, $300.) July 9th, 1837. $20. Dec. 12th, 1837 $400.) What remained due Aug. 7th. 1838? Ans. $159.655. What is a cendorsement? Wlt.re written? When part of a note or *blig;ition iFs ptidl what fact et viilullnt Wi,at is lhe urnn *iue? Vhat then is the rule fjr comiiutiug iulrc9lC o uotsa, lbosud, c., wlho. partiil pWla eut hilve been madw art COMMON ARrEIMETIC. Sect. XIII Operation. Principal $8C0. Time from Jan. 1 t, 1 336 to Sept. 15tll, 1 836, 8 mo. 14 da.; (Sect. IX, Art. 5, Obs. 1,) interest of $800 for this time, 33.866. Amount, $833. 866. Payment, Sept. 15th, 1836, exceeds the interest, 300 Remainder for a new principal, 8533.866. Interest from Sept. 15lh, 1836 to July 9th, 1837, (9 mo. 24 da.) $26.159. Payment, July 9th, 1837, less than the interest, 20. Interest firm Sept. 15th, 1836 to Dec. 12th, 1837, (1 yr. 2 mo. 27 da.) 39.772, Amount,.$a73.638 Payment, July 9th, 1837, $20. Payment, Dec. 12th, 1837, $400. Sum of these payments, 420.000 Remainder for a new principal, $153.638 Interest from Dec, 12th, 1837 to Aug, 7th, 1838, (7mo. 25 da.) 6.017 Amount due Aug. 7th, 1838, $159.655 (2.) $1200 CLEVELAND, June 29th, 1840. For value received, I promise to pay Timothy Just, or bearer, twelve hundred dollars, with interest. JAMES GCODMAN. Endorsed, Dec. 18th, 1840, $100. Aug. 1st, 1841, $250.75. Jan, 7th, 1842, $10.25. Nov. 3d, 1842, $600. July 1st, 1843, $200 What was due Jan. 1st, 1844. Ans. $205.889 (3.) $623.75. CINCINNATI, May 2nd, 1843. For Value received, six months after date, I promise to pay Rob Art. 5. INTEREST. ert Brown, or bearer, six hundred and twenty-three dollars, and seventy-five cents. JOHN SMITH. Endorsed, Jan. 1st, 1844, $75.25, Dec. 13th, 1844, $200. Aug. 7th, 1845, $10. 75. Jan. 1st, 1846, $350. J And April 4th 1846, he paid the balance How much was this balance? Ans. $70.942. REMARK 1. When a note is given without mnntion of interest, as the above, it is not customary to charge interest. 2. After a note becomes due, however, if pavment is delayed, it will draw legal interest, although no mention be made of iaturest. 11. VERiMONT RULE. "Find the amount of the principal from the time the note was given, till the time of settlement: next, find t/le amount of each paynment from the tie t hepayment was made till the time of settlement; finally, add together the amounts of the several payments and subtract their szumfrom the amount of theprincipal already found; the remainder will be the sum due." (4.) $1000. PITTSBURGH Feb. 2nd, 1840. For value received, I promise to pay Amos Bush, or bearer, one thousand dollars, on demand, with interest. CHESTER DALE. Endorsed, Nov. 1st, 1840, $300.July 5th, 1841, $275, Feb. 19th 1842, $325. Oct. 12th 1842. $100. What was due Jan. 1st., 1843? $93.051. Operation. Principal, $1000. Interest of principal from Feb. 2nd, 1840 to Jan. 1st, 1843; 174.833 Amount of same; $1174.833 What is said respecting interest on notes in which no mention is made of interest? What is said respecting interest on notes which are not paid when they become due? What is the Vermont rule for computing interest e* notes? 282 COMMON ARTHMETIC. Sect. XIII First payment, made Nov. 1st, 1840. $300.000 Interest of same till Jan. 1st, 1843. 39.000 Second payment made July 5th, 1841. 275.000 Interest of same till Jan. 1st, 1843. 24.566 Third payment, made Feb. 19th, 1842. 325.000 Interest of same till Jan. 1st, 1843. 16.900 Fourth payment made Oct. 12, 1842. 100.000 Interest of same till Jan, 1st. 1843. 1.316 Total amount of payments, and interest; $1081-782 Amount due; $93.051 (5.) $710.40. BURLINGTON. July 4th, 1841. For value received, on demand I promise to pay Edwin Farr, or order, seven hundred and ten dollars, and forty cents, with interest. GEORGE HUNTER. Endorsed, Jan. 1st, 1842, $400..March 3d, 1843, $2500.0. Feb. 19th. 1844, $100. The balance, July 4th, 1844. Required the balance? Ans. $5.241. (6.) $400. BUFFALO, March 1st, 1842. For value received, on demand I promise to pay Ira Jones, or bearer, four hundred dollars, with interest at 7 per cent. LEWIS MAKER. Endorsed, Jan. 1st, 1844, $215.871. April 19th, 1845, $212.371. The balance, Sept. 1st, 1845. Required the balance? Ans. $39.115. III. CONNECTICUT RULE. I. "FinJ the amount of the principal to the time of the first payment, if it be a year or more from the time the interest commenced, and Irom this amount subtract the payment. II. The remainder will be a new principal; find the amount of this to the next payment, from which subtract the payment as above. So continue to do from paymiut to payment, until all are employed, provided a year or more intervenes between each two payments. III. Bult if the time between any two payments be less than a year, find the amount of the last principal for a year, and of the payments up to the What is the Connecticut rule for computing interest on notes? Art. 5. INTEREST. same time. and subtract ths latter from the former. If, however, a year overruns the time of settlement, find the amounts up to that time instead of for a year. IV. If any remainder, after sbhtraction, be greater thait thn preceding principal, the preceding principal is still to be continued as the principal for the succeding time, instead of the remainder; and the difference is to be regarded as so much unpaid interest which is to be added to the principal at the time of the next payment." (7.) $1200. HARTFORD, Oct. 7th. 1836. For value received, on demand I promise to pay Henry Inman, or bearer, twelve hundred dollars, with interest. PETER Russ. Endorsed, April 3d, 183t, $600. Jan. 1st, 1839, $300. Sept. 2d, 1839, $250. What was due Jan, 12th, 1840? Operation. Principal, Interest of the same to April 3d, 1838, (17 mo. 26 da.) Amount, 1st payment, deduct, Remainder for a new principal, Interest April 3d, 1839, (12 mo.) payment being made before a year has elapsed, Amount, 2nd payment made Jan. 1st, 1839, $300 ) Interest of same to April 3d, 1839, (3 mo. 2 da.) 4.60 Amount of payment, deduct, Remainder for a new principal, Interest from April 3d, 1839 to Jan. 12th, 1840, (9 mo. 9 da.) settlement being made within the year, Amount, 3d payment made Sept. 2d, 1839, $250 ) Interest of same to Jan. 12th, 1840, (4 mo. 10da.) 5.416 Amount of payment, deduct, Ans. $210.309. $1200. $107.20. $1307.20. 600. $707.20 42.432 $749.632 304.60 $445.032 20.693 $465.725 255.416 $210.309 Remainder due Jan 12th, 1840, 284 COMMON ARITHMETIC. Sect. XIII (8.) $900. NEW HAVEN, May 3d, 1842. For value received, on demand, I promise to pay Silas Weeks, or bearer, nine hundred dollars, with interest. JOHN LEAK. Endorsed. June 1st, 1843, $300. Feb. 2nd, 1844, $259. Jan. 1st, 1845, $300. Nov. 9th, 1846, $100. The balance Jan. 1st. 1847. Required the balance. Ans. $76.422. By each of the preceding rules let the learner find the balance due on the followings notes: (9.) $1500. SANDUSKY, June 1st, 1830. For value received, on demand, I promise to pay Jonas Trusty, or bearer, fifteen hundred dollars, with interest. THEODORE HAND. Endorsed. Jan. 1st, 1831, $500 ) July 4th, 1832, $700. I Feb. 2d, 1833, $30u. What remained due April 1st, 1833? ( By the Mass. and N. Y. Rule, $164.878. Ans. By the Vermont Rule, $153.40. By the Conn. Rule, $165.065. (10.) $12500. LOUISVILLE, Sept. 1st, 1841. For value received, I promise to pay Henry Loan, or bearer, twelve thousand, five hundred dollars, with interest. JOSEPH FAIRMAN. Endorsed. Jan. 1st, 1842, $1800.' Oct. 9th, 1843, $6000. Dec. 15th, 1843, $ 50. Nov. 25th, 1844, $3000. I Sept. 30th, 1845, $1500. What remained due Jan. 1st, 1846? Ans. Bythe Vermont Rule, $1939.118. By the Conn. Rule, $2202.525. Art. 5. INT E 1r ES'r 285 CASE 7. Problems in Interest. Obs. 13. A PROBLEM is a question proposed requiring a solution. This question may apply either to the investigation of some truth, or principle, or to the operation of practical questions. REMARK.-In the preceding examples we notice five different terms, or parts connected with Interest; viz: the Principal, the Time, the Rate, the Interest, and the Amount. The relation of these terms to each other, is such, that if any three are given, the other two can be found. PROBLEM I. The Principal, Rateper cent. and time being given, tofind the Interest, and Amount. This is deemed the most important problem. It has already been exemplified in the preceding Cases of this Article. PROBLEM II. The Principal, Interest, and Time being given, to find the rate per cent. 1. A man borrowed $250 for 3 years, and paid $45 interest.What was the rate per cent.? Ans. 6. Solution.-The interest of $250 for 3 years at 1 per cent is $7.50. (Obs. 10.) Therefore, as many times as $7.50 is contained in $45, so many times more than 1 per cent. was the per cent. paid; and $45 $7.50 = 6. PROOF. —3 yrs 36 mo. 36 2 = $0.18; $250 X.18= $45 interest. (General Rule, Case 4.) HcnceTo find the rate per cent, in such cases. Obs. 14. Divide the given interest by the interest of the given principal,for thegiven time, at 1 per cent. REMARK.-The amount can be found by adding together the principal and interest; (Ohs. 7, Remark) also, the principal can always be found by ubtracting the interest from the amount and the interest can always be found by subtracting the principalfrom the anto'nt. (Sect. VI, Art. 1, Obs. 11.) 2. If I borrow $420 for 2 yrs. 6 mo., and pay $73.50 interest, what is the rate per cent.? Ans. 7. What is a problem? To what may this question apply? How many terms are connected with interPst? Name thelrm. What relation have these terms to each other? Which is the most important probllem? Why do you suppose it to be 1he mot important? How do we tind the rate per cent. when the principal, interest, and timP art. iven? How can tile amount be found in suohcases? The principal? The interesi? 286 COIMMON ARITHMETIC. Sect. XIII 3. If $117.60 is pai.l for the use of $700 3 yrs. 8 mo, 24 da., what is the rate per cent.? Ans. 4'. 4. If a person pays $151.20 for the use of $810 2 yrs. 4 mo., what rate per cent does he pay? Ans. 8. 5. A man lent $1000, and at the end of 3 yrs. 6 mo. received for his due $1236.25. R equi red the rate per cent.? Ans. 6~. 6. Lent $750 for 6 mo., and receive I for principal and interest $765. Required the per cent.? Ans. 4. 7. At what rate per cent. must 81250 be loaned, in order to amount to $1343.75 in 1 yr. 6 mo.? Ans. 5. PROBLEM III. The Principal, Rate per cent., and Interest being given to find the Time. 1. A person borrowed $400 for a certain time at 7 per cent. interest: at the end of the time the interest was $70. How long did he have the money? Ans. 2 yrs. 6 mo. Solution. —The interest of $400 for 1 year, at 7 per cent.,is $28, (Obs. 7.) Then if the interest of $400 for 1 year is $28, it will take it $70- $28 -=. yrs to produce $70 interest. PROOF.- 2 yrs. =30 mo. 30 2 = $0.15; $400 X.15 = $60 interest at 6 per cent. (Obs. 8.) $60 6 = $10 interest at 1 per cent. $10 X 7 == $70 interest at 7 per cent. (Obs. 10.) HenceTo find the time in such cases: Obs. 15. Divide the given interest by the interest of the principal for 1 year at the given rate. REMARK.-If there is a remainder after dividing, it may either be considered as the fractional part of a year, or continued to decimals. In either case the value can easily be found in months and days. (Sect. 1X, Art. 3, Case 2, Ex. 7, andCase3. Rule.) 2. In what time will $375 gain $90 interest, at 6 per cent.? Ans. 4 yrs. 3. In what time will $700 gain $210 interest at 9 per cent? Ans. 3 yrs. 4 mo. 4. In what time will $560 gain $45.36 interest, at 4- per cent.? Ans. 1 yr. 9 mo. 18 da. 5. In what time will $840 amount to $1015.08- at 5- per cent.? Ans. 3 yrs. 7 mo. 15 da. How do we find the time, when the principal, rate per cent., and interes are given? If there is a remaindet after dividing, what is done with it? Art. 5. INTER EST. 287 6. In what time will $900 amount to $1710, at 10 per cent.? Ans. 9 vrs. 7. In what time will $1200 amount to $2200, at 8- per cent.? Ans. l0 yrs. 8. In what time will any sum of money double itself, at 5 Ier cent.? Suggestion.-Assume $100 as the given sum. Then, in what time will $100 gain $100 interest, at 5 per cent.? Ans. 20 yrs. 9. In what time will any sum of money double itself, at 8- per cent.? Ans. 12 years. 10. In what time will any sum double itself, at 6 per cent.? Ans. 16 vrs. 8 mo. 11. In what time will aity sum double itself,; t the following rates per cent: 1; 4; 7; 3; 9; 2; 8; 10; 12; 15; 20; 24; 25; 30; 40; and 50per cent.? At l-; 2'; 3}; 41; 51; 7'; 81; and 12 per cent.? Ans. in order. 100 ys.; 25 yrs.; 14~ yrs. 33 yrs. 4 mo.; 11I yrs.; 50 yrs.; 12 yrs. 6 mo.; 10 yrs; 8 yrs. 4 mo.; 6 yrs. 8 mo.; 5 yrs.; 4 yrs. 2 mo.; 4 yrs.; 3 yrs. 4 mo.; 2 yrs. 6 mo.; 2 yrs.; 66 yrs. 8 mo.; 40 yrs.; 284 rs. 22' yrs.; 18-2- yrs.; 13 yrs. 4 mo.; 11 — yrs.; 8 yrs. 12. In what time will any sum treble, aud quadruple itself, at 121 per cent. Ans. S Treble itself in 16 yrs. Ans Quadruple itself in 24 years. PROBLEM IV. The Interest, Time, and Rateper cent. being given to find the Principal. 1. The interest of a certain note is $30, it having been at interest 2 yrs. at 5 per cent. Required the face of the note, or principal? Ans. $300. Solution.-The interest of $1 for two years, at 5 per cent. is $0.10. (Obs. 7.) Then the principal is as many times $1, as $0.10 is contained in $30; and $30 —. 10= $300. HenceTo find the principal in such cases: Obs. 16. Divide the given interest by the interest of $1 for the given time, and at the given rate per cent. How do we find the principal, when the interest, time and rate ptr cent. are given. 288 CjilMON ARITHMETIC. Sect. XIII 2. What principal at 6 per cent. will gain $27.45 in 1 yr. 6 mo.? Ans. i305. 3. What principal at 4' per cent. will gain $76.50 in 3 yrs. 4 mo. 24 da.? Ans. $500. 4. What principal at 7 per cent will gain $236.25 in 4 yrs. 6 mo.? Ans. $750. 5. What principal at 9 per cent will gain $237 in 2 yrs. 7 mo. 18 da.? Ans. $1000. 6. What principal at 121 per cent will gain $1265.625 in 6 yrs. 9 mo.? Ans. $1500. PROBLEM V The Amount, Rate per cent.. and Time being given to find the Principal and interest. The amount of a certain note is $660; it has been on interest 1 yr. 8 mo. at 6 per cent. Required the face of the note, and the interest? Ans. Face of the note $600; and the interest $60. Solution.-The amount of $1 for 1 yr. 8 mo. at 6 per cent., is $1.10. Then the principal, or face of the note, must be as many times $1, as $1.10 is contained in $660; and $660 - $1.10 = $600 the principal, and $660-$600 = $60 the interest. HenceTo find the principal and interest in such cases: Obs. 17. Divide the given amount, by the amount of $1 for the given time, at the given rate per cent. To find the interest; Subtract the princilpalfrom the amount. (Obs. 14. Rem.) 2. What sum of money will amount to $77.76, in 1 yr. 4 mo. at 6 per cent? T Ans. 72. 3. What sum of money will amount to $1190 in 5 yrs. at 8 per cent? Ans. $850. 4. The amount of a certain note is $776.3 '; it has been on interest 4 yrs. 8 mo. 24 da. at 41 per cent. Required the face of the note and the interest? Ans. Face of the note $640. Interest $136.32. 5. The amount of a note, after having been on interest 3 yrs. 10 mo. 1 da. at 7 per cent was $1144.65. Required the face of the note, and the interest. Ans. Face of the note 8990. Interest $244.65. How do we find the principal, when the amount, rate per cent and time are given? How do we find the interest? Art. 5. INTEREST 289 6. If the amount of a note is $2139 after having been on interest 7 yrs. 9 mo. 27 da. at 10 per cent., what is the face of the note, and what the interest? o Ans. Face of the note $1200. Interest 93,. DISCOUNT. Obs. 18. DISCOUNT is a deduction made front a debtforpaying it before it becomes due. Thus, if I owe a man $106, due 1 year hence, and he wishes me to pay him now, it is evident he ought to make some deduction for present payment. And I ought only to pay him such a sum as would amount to $106 in 1 year. If the discount is made at 6 per cent., for instance, I ought to pay him $100, because the interest of $100 for 1 year is $6, and the amount is $100 + $6 = $106. Obs. 19. The sum, which in the given time, with its interest, would amount to the sum on which the discount is made, is callPd the PRESENT WORTH. Thus, in the example mentioned above, $106 is the sum on which the discount is made, $6 is the discount, and $ 100 the present worth. 1. What is the present worth of $440, payable in 1 yr. 8 mo., discounting at 6 per cent. What is the discount? Ans. Present Worth $400. Discount $40. Suggestion.-By examining this question attentively, we perceive that it does not differ from those under Problem V; the Debt corresponds to the Amount; the present worth to the Principal; and the Discount to the Interest for the given time, at the given rate per cent. Therefore, $440- $1.10 = $400 the present worth; and $440-$400 = $40 the discount. REMARK.-It is not unfrequently supposed thatif we find the interest on the given sum for the given time, that this interest will be the discount, which subtracted from the given sum, will give the present worth. But this error may be avoided if the pupil will recollect that the debt answers to the amount in interest, and the principal is always the sum on which the interest is calculated. 2. What is the present worth of $ tpayable in 10 mo. at 6 per cent.? Ans. $952.38X.: 3. What is the present worth of $1488, due 4 years hence- i- Ans. $1200. In these examples, 6 per cent. is understood when no other rate is mention d. What is discount? What is the present worth? In the example given which is the sum on which the discount is made? Which is the discount? Which is the present worth? What error do rmany fall into ropecting diWsount? How may this error be avoided? 14 290 COMMON ARITHMETIC. Sect. XIII 4. What is the present worth of $1590 due 1 yr. hence? Ans. $1500. 5. What is the present worth of $1360 due 6 yrs. hence? Ans. $1000. 6. What is the discount of $872 due 2 yrs. hence? Aus. $93.429.+ 7. Whaat is the discount of $1736 due 3 yrs. 6 mo. hence? Ans. $301.29.8. What is the discount of $2412.75 due 6 yrs. 1 mo. 24 da. hence.? Ans. $694.263 + 9. A merchant bought goods to the amount of $1968.75, payable in 10 mo. How much ready money should pay the debt? Ans. $1875. 10. What is the present worth of $2303.40 due 2 yrs. 4 mo. hence, discount being made at 7 per cent? The discount? Ans. Present worth $1980. Discount $323.40. 11. What is the present worth of $2156.371 due 3 yrs. 7 mo. 18 da. hence, the money being workh 9 per cent.? The discount? Ans. Present worth $1625. Discount $531.3 7. 12. A mansold a farm for $5836.80, payable in 3 yrs. 6 mo.; he could have received $4650 down for it. Did he gain or lose by selling on credit; and how much, the money being worth 8 per cent.? Ans. He lost $90. 13. What is the difference between the interest of $2352 for 4 yrs. 8 mo, 12 da., at 10 per cent., and the discount of the same sum, for the same time, and at the same rate? Ans. $353.44. 14. A merchant bought a lot of goods for $2500 ready money, and sold them the same day for $3044.25 payable in 1 yr. 9 mo. 12 da. at 6 per cent interest. Did he gain or lose by the transaction, and how much? Ans. He gained $250. When payments are to be made at different times, to find the present worth: Obs. 20. Find the present worth of each payment separately, an add together these results, for the present value of the debt. This rule is so evident that it needs no demonstration. Let the learner see if he cannot demonstrate it himself. 15. What is the present worth of $3000, one half payable in 2 yr s. 6 mo. 18 da., and the other half payable in 3 yrs. 4 mo. 24 da.? T e discount? Ans. Present worth. $2546.801+. Discount $453.199-. i6. A merchant bought goods to the amount of $2460; 3 of When payments are made at different times, how do we find the present worth? Art 5. INTEREST. 291 which wa to be paid in 4 mo.;. in 8 mo.; and 3 in 12 mo. What suml slhould be paid at the present time to balance the debt? Ans.;;i-2365.966-. 17. An a'ent has 1260 to uce in trade, and is to rcceive 5 per cent,. on what, hle lays out, What sum does he lay out, and how much does lie receive? Ans. tIe lays out $ 1200, and receives 860. Solution.-As often as he lays out $1, he receives $0.05. Therefore, 81260 — 1.05 = $11200 what he lays out, and $1260 -$1200 == 60 what he receives. Had we multiplied $12o60 by.05, (as many a c reless pupil would have done,) we should have given the agent 5 cents fcr every 95 that he laid out, instead of 100 as required by the quest on.) Pi1ooF. —S1200X-05= -_ GO60. $1200 -+ $60 = $1260 18. An agent ihas,918 to use in trade, for which he receives 8 per cent on what he lays out. How much does he lay out for his eml)loyer, and how much does h1e receive? Ans. He lays out $850, and receives $68. CASE 8.-Banking. Obs. 21. A BIAN- is an institZ1ton that deals in money. Its Capital or Stoik is divided into parts called Shares, and owned by individuals calledl Stockholders. A share is usually $100. Obs. 22. Tl1e operations of;i bank are conducted by a President and LBoa'd of Dire'-tors. It has a deposite of ipecie. and issues notes, or hills, intenlded to be used as a circulating medium instead of gold and silver. These bills are usually obtained from the bank in loans. Obs. 23. In loaning money, the banks usually deduct the legal interest off the face of the note for the given time, and pay the holder the balance. The sum which the holder of the note receives, is called the Proceeds of the note. The sum deducted is called the Bank Discount, and is the same as simple interest in advance. Obs. 24. The time from the date of a note until it becomes due, is called the Days to run. Besides the time mentioned in'the note, three additional days are allowed before it is legally due. These are called D)tys of Grace*; and as notes are not usually pa'd until the last of these dates, the banks charge interest on the days of grace. What is a bank? Hlow is the capital divided? What are stockholders? How much is a share? How are the operations of a bank conducted? What has it? What does it issue? For what? How are these bills usually obtained? What is usually done by the banks in loaning money? What are the proceeds of the note? The bank discount? Days to run? Days of grace? 'Some Banks allow 4 days of Grace. 292 COMMON ARITHMETIC. Sect. XIII Obs. 25. A note is said t, be discounted when the Bank Discount has been deducted. The difference between the Bank Discounf, and the true discotnt, is the interest of tlih true (!iscount for the given time. 'This gives an extra pi)ofit to the IBank. O small sums for a short time, however, the diitlerence is triflilng; but when the sum is large, and the limle for which it s discounted, is long, the ditierence amounts to a considerable sum. 1. What is the bank discount on a note of $200 payable in 1 yr. 6 mo. at 6 per cent.? What are the proceeds? Operation. The interest of $200 for 1 yr. 6 mo. is $18.03 The " " " 3 days of grace is.10 Bank discount, $18.10 Face of the note $200,00 Bank discount 18.10 Proceeds, $181.90 Hence-To find the bank discount of any note, draft &c.: Obs. 26. Add the days of grace to the time the note has to run and find the interest of the face of the note or this time. To find the proceedsDeduct the Bank discountfrom the face of the note. 2. What is the bank discount of a note of $500, payable 6 mo. after date? Ans. $15.25. NOTE.-In these questions, 3 days of grace, and 6 per cent. are understood, unless otherwise stated. 3. What is the bank discount on a note of $350 payable in 2 yrs. 4 mo.? Ans. $49.175. 4. What are the proceeds of a note of $400, payable 9 mo, after date? Ans. $381.80. 5. What are the proceeds of a note of $750 for 1 yr. 8 mo. 21 da.? Ans. $672. 6. What is the bank discount on a note of $900, payable 30 days after date, at 7 per cent? The proceeds? An. Discount $5.775. Ans. Proceeds $894.225. 7. What is the bank discount on a note of $1500, payable 90 When is a note said to be discounted? What is the difference between the bank discount, and the true discount of a note? How do we find the bank discount of a note, draft, &c.? The proceeds? Art. 5. INTEREST. 293 days after date, at 5 per cent. and allowing 4 days of grace? The proceeds? Ap e. s Discount $19,58-. ns. Proceeds $1480.41 -. 8. 'What is the difference b tween the bank discount of $1200 for 3 vrs. 6 mo., and the true discount of the same sum for the same time? Ans. $44.335. 9. What is the bank discount of $1000 payable 6 months after date? Ans. t30.50. 10. What are the proceeds of $1260.50, payable 90 days after date? Ans. $1240.963. 11. A man bought 1250 barrels of flour at $4.50 a barrel, cash, and sold them the same day for $5 a barrel, payable in 9 months, without interest. IIe got his note discounted at the bank, wh;n money was worth 6 per cent. Did he gain or lose by the operation, and how much? Ans. Ile gained $340.621. 12. A man got a note of $1500 payable in 90 days, without interest, discounted at the bank at 6 per cent., and put the proceeds at interest for 1 year at 7 per cent. He renewed his note at the bank each 90 days, by paying the bank discount at (aclh renewal. At the end of the year, lie received his due on what hle had lent, and paid his note at the bank. Did he gain or lose by the operation, and how much? Ans. Ile gained $6.71. In this example the money is supposed to be worth 7 per cent., although he paid but 6 per cent, discount at the bank. CASE 9. Equation of Payments. Obs. 27, EQUATION OF PAYMENTS is the method of finding the time, when two or more paymecnts, due at dfl'erenl times, may be made at once, without injupry to either party. This time is usually called the mean, or equated time. 1. A. owes B. $60, to be paid as follows; $10 in 1 month; $20 in 2 months, and $30 in 3 months. lHe wishes to pay it all at once, in what time ought he to pay it? Ans. 2 mo. 10 da. Solution.-The interest of $100 for 1 year is 86; for 2 years, the interest is $12, 1which is the same as the interest of $200 for 1 year. Also the interest of $100 for 3 yrs. is $18, which is the same as the interest of $300 for 1 year. H-ence, universallyOh)'. 28. The. in!tcrest!f any sum of moncy for any time, is equal to t/e a.'le'ctl (f twice s wiaciJu'r /j '/ li / tJ o;me, tihree it:nes as v 'c/i, What is equation of payments? What is the time usually called? 294 COMMON ARITHMETIC. Sect. X 11 for one 1th7rd of thi time, &c. And conversely; The interest of any svm of money for any time is e terest of one-I lf as much bor twice this time; one-third as much for three tiehis time, dc. Therefore, The int. of $10 for I mo., is the same as the int. of i81 for 10 mo. The " 820 (2 (C. c 1 for 40 mo. The " 30 " 3..' 51 for 90 mi. And the int, of t60 for the mean time, i. tle ssame as the int. of 2:31 ifo 14I mo. JBut:G36 is 60 times 1, turefoe, it il take it w but ' plart of tlhis time to g'ain the samen interest and 1- of 140 moi. is 2 mo. 10 dta. PiOOF.-The interest of,60 for 2 mo. 10 (a. is 60 X.01 12 $0.70. (Case4, IRule.) Tel interest of $1 tfor 1.0 imo. is 1 X.70 0= i0.70. (Case 4., u:ee). Hence — To find the mean or equa ted time of several p);,1vnelts: Obs. 29. Ji(ulti1pl each paymeznt bl the thime b'/ore it becom;,es d(.e, an ddivide the sum of the p yroducts by the suim tle /:,(i cmet s. REMARK 1. This rule is founded on the sulpposition ltiat tie interest of any sum of nmoney for a certain time, is equa;l to the d(iscoinlt of tile saidno sumI for the san(e tinle, This is not tle case, as the discount is dtl!'(,y IC3. e than the interest; the difFereice heinig eqiul to tlie interest of t'i ti r,.-: (liscointl fir tthe given ti!ne. T1he difFerence in s;nall sums, for a short tliii, however, is too slight to be noticed. 2. A merchant lias owing to him li 1000, to b)e p!li I 1as ollows: $200 in 3 months $; 300 in 5 months; $250 in 6 monthls, and ~259 in 8 montls. It is agreed to make but one payment, In what time must this payment be made? Ans. 5 mo. 1 da. 3. A. owes B. $120 to be paid in 5 montlhs; 884 to be paid in 8 months; $132 in 9 montls; and 8160 in 10 nontlhs. Required the mean time for the payment of the whole? Ans. 8 mo. 5 da.-+ 4. A. owes B. 12030 to be paid as follows: ' in 6 months; in 8 mo., and the balance in 10 mo. IRequired the equated time for the payment of the whole. Ans. 7 mo. 10 da. 5. A merchant in Columbus, Olio, orders g(oods from New York as follows: To what is tihe interest of any s)um of money for anin tif!fo eqnall? How do we find the inean or equated time of several payments? Upon what is this rule founded? Is this correct? Why not? Why then is the rule used? What is the correct rule? Art. 5. INTEREST. 295 Oct. 12th, 1849, to the value of $100, payable in 6 mo. Oct. 25th I. " 150. " 4 mo. Nov. 10th " t t. 300, " 3mo. After ordering the latter quantity he wishes to give his note for the entire lot. Required the date to which the note should run that neither party may lose by the operation. Ans. Feb. 25th, 1850. Solution. —Ic evidently owes the whole amount from Nov. 10th. From Oct. 12ih, to Nov, 10th, is 28 days; from Oct 25th to Nov. 10th is 14days. The accounts then stand $100 for 152 da. (180 -28); $:1 50 for 106 da. (120-14); and $300 for 90 da. Theequated time is then found as usual. The date is found by the table on page 273. As we have before said, the rule given is not strictly correct. The following will give the correct mean time if no mistake is made in the operation. Obs. 30. In'del the present worths of the several debts at the given rate. Tien find in what time at the same rate the sum of the present worths would (olnunt to the sum of the debts. The result will be the equated time. REMARK. When the date of the several transactions is different, the date from which to maike the calculations must be found as in Ex. 5. The date from which we calculate in this example is Feb. 10th. Find the answers to the following questions by both Rules. 6. Required the date for the payment of the following sums at one time. $300, payable in 8 mo.; date Jan. 6th, 1849. 459, " 6 mo.; " Jan. 20th, " 500, " 4mo.; " Feb. 1st, " 1000, " 3mo.; " Mar. 3d, " Ans. By Obs. 29. June 24th, 1849. By Obs. 30, June 25th 1849. 7. Required the date for the payment of the following sums, at one time. $800, payable in 1 yr.; date Aug. 12th, 1849. 1000, " 9 mo.; " Sept. 1st, " 1500, " 8 mo.; " Sept. 15th, " 1800, " 6 mo.; " Oct. 1st., " Ans. By Obs. 29. May 16th, 1850. By Obs. 30, May 17th, 1850, nearly. 8. Required the date for the payment of the following sums at one time. 296 COMMON ARITHMETIC. Sect. XIII $1000, payable in 11 mo.; date Jan. 1st, 1849. 1200, " 10 mo. " Jan. 15th " 1500, " 9 mo..: Feb. 1st " 1800, " 8 mo.; " Feb. 22d, " 2000, " 6 mo.; " March 18th" 2500, " 3mo.; " April 9th, ' Ans. By Obs. 29. Sept. 27th 1849; by Obs. 30. Sept. 28th, 1849. COMPOUND INTEREST. Obs. 31. COMPOUND INTEREST is the interest, on both theprincipal and interest, when the latter is not paid at the time it becomes due. It is calculated by adding the interest to the principal at the end of each year, or other stated time, and making their sum the principal for the next succeeding year o: stated time. The compound interest is found by subtracting the first principal from the last amount. 1. What is the compound interest of $400 for 4 years. at 6 per cent? Ans. $104.99. Operalion. $400 - Principal. $400X.06 = 24 = Interest for 1 year. (Obs. 7.) $424 =- Amount for I year. $424 X.06 = 25.44 - Interest for 2d year. $449.44 - Amount for 2d vear. $449.44 X.06 = 26.966 = Interest for 3d year. $476.406 = Amount for 3d year. $476.406 X.06= 28.584 = Interest for 4th year. $504.930 = Amount for 4th year. Deduct $400.0C0 = First principal. Ans. $104.99 = Compound interest for 4 yrs. 2. What is the compound interest of $300 for 3 yrs. at 7 percent? Ans. $67.5129. 3. What is the compound interest, and amount of $500 for 4 yrs. at 5 per cent? Ans. Amount. $607.647.. C. Int. $107.647: What is compound interest? How is it calculated? When months and days are given, hiuw do we proceed? Art. 5. CO'MPOUND INTERE'ST. 2967 4. What is the con-poand interest, of $700 for 5 yrs., at 6 per cent.? The amnounlt. Ans. Amt.,936.762. Comp. Int. $:236.7G2. When months and days are given: Obs. 32. First find the amount for Itec required num2bcr of years, twz ont lhe /icst amoa(nt/jr tle mJonIth and (days. What is the compound interest of '$800 for 5 yis. 6 mo. 24 da., at 6 per cent.? Th'e amount? Ans. Amt. `$1106.98. Compt. int. 8306.98. 6. What is the compound interest of,500 fir 3 yrs. 8 mo. 18 da., at 6 per cent.? The amount? Ans. Amt. $(621.114. Comp. int. 1 21.114. 7. What is the interest of $200 for 3 y'rs., compounded every 6 months, at 6 per cent? Ans.;'38.79. NOTE.-The learner will observe to add the interest to the plincipal, at the end of each 6 nmoths. (Obs. 30.) 8. What is the amount and interest of $600 for 5 yrs., compounded every 6 months, at 6 per cent? Ans. Amount. $806.34. Comp. Int..206.34. TABLE Showing tlhe amount of $1, at 5, 6, and 7 per cent, for any number of years from 1 to 30. Yrs. 5 percen 3 per centec -- - Yrs - pcni per cent Yrs. 5 prcent6 per cent ) per cent 1 $1.050(.C600G,1.07001 16 $42.18 28.,2.5403$2.9521 2 1.1'025 1.1236 1.1449 17 2.29201 2.6927' 3.1588 3 1.1576 1.1910 1.2250! 18 2.40C66 2.8543 3.3729 4 1.2155 I.2624] 1.3107' 19 2.5269 3.0256 3.6165 5 1.2762 1.3382 1.40251 20 2.65321 3.2071I 3.8696 6 1.3400 1.4185 1.5007 21 2.78591 3.3995 4.1405 7 1.4071 1.50361 1.60,- 7 22 2. 9252 3.6035i 4.4304 8 i 1.4774 15938 1.71811 23 3.07151 3.8197 4.7405 9 1.5513 1.6894 1.8384 24 3.22511 4.0489! 5.0723 10 1.6288 1.7908 1.9671 25 i 3.38631 4.2918 5.4274 11 1.7103 1.8982 2.1048 26 3.5556 4.5493 5.8073 12 I 1.7958 2.0121 2.2521' 27 3.7334! 4. 8223 6.2138 13 I 1.8856 2.1329 2.40981 28 3.9201 5.1116 6.6488 14 1.979 2.209 609 2.5785i 29 4.11611 5.4183 7. 1142 15 2.07891 2.3965 2.7590j 30 4.3219! 5.7434 7.6122 14A 298 COMMON ARITHMETIC, Sect. Xlll The amount of $2 (whether at simple or compound interest,) is twice as much as the amount of ^1 for the same time; the amount of 53 is three times as much as $'1; &c. HenceTo find the amount of any sum by the table: Obs. 33. ilft'17fily i/te amoaanz of 1I fir the require number of years, by the given stum To find tile interest: See Ohs. 30, 9. What is the amount of $2000 for 15 yrs., at 5 per cent., compound interest? Solution. By the table, the amount of 81 for 15 yrs. is $2.0789 and $2.0789 X 2000 =- 4157.80. Ans. $4157.80. 10. What is the amount of $2530 for 23 yrs., at 6 per cent, c mpound interest? Ans. $8017.75. 11. What is the amount of 81590 for 33 yrs., at 7 per cent. compound interest? Ans. $11418.30. 12. What is the amount of $50000 for 18 yrs. 6 mo. 12 da., at 6 per cent., compound interest? (See Obs. 31.) Ans. $147281.88. 13: What is the amount of $10000 for 8 yrs. 8 mo. at 6 per cent compound interest'? Ans. $16575.52. 14. What is the amount of 3000 for 40 yrs., at 6 per cent., compound interest? Ans. $30855.842. NOTE.-First find the amount for 30 yrs., and then on this amount for 10 yrs. 15. Required the amount of $5000 for 37 yrs., at 1 per cent., compound interest? Ans. 6G1114.5477. REMARK.-In several of the preceding exanples the principal has doubled itself. Any sum at 5 per cent., compound interest, will double itself in 14 vrs 2 mo. 13 da.: at 6 per cent. in 11 yrs. 10 mo. 22 da.: and at 7 per cent. in 10 yrs. 2 mo. 27 da. How do we find the amount of any sum by the table? bUODECnMALS. 299 SECTION XIV. DUODECIMALS, Obs. 1. DUODECIMALS are fractions of a foot; or a species oJ numbers, of which, the ratio of increase and decrease, is twelve. The term is derived from the Latin numeral duodecim, which signifies twelve. The denominations are FEET, INCHES or PRIMES, SECONDS, THIRDS, FOURTHIS, & c. TABLE. 12 fourths ("") make 1 third, marked "' 12 thirds, " 1 second, " " 12 seconds, " 1 inch or prime, " in.or '. 12 inches, " 1 foot, " ft. REMAIK 1. The marks ', ", "',.", &c., which distinguish the different denominations, are called indices. The foot has no index, being considered as the unit. 2. Duodecimals may be added or subtracted in the same manner as other compound numbers. (Sect. IX, Arts. 4 and 5.) MULTIPLICATION OF DUODECIMALS. Obs. 2. Duodecimals are used principally in measuring surfaces or infinding the solidity of bodies. The former is ascertained by multiplying together the length and breadth. (Sect. IX, Art. 2, Obs. 15.) And the latter is found by multiplying together the length, breadth and thickness. (Sect. IX, Art. 2 Obs. 18.) REMARK 1. As 12 inches make 1 foot, 12" make 1 ', &c., it folows that 1 ' is - - - - -, of a foot. 1 is yS of an inch, or of 1= - of a foot. 12 144 1 i"' is -1 of 1 ",or jy2 o f I r o f 1 ',or of o = - of a ft. 1 "" is of, r - of- I r of 1 ', or of I of 1, or of of of o T- of ~ of ^ of T2 - 20736 of a foot, &c. The foot being regarded as the unit, and the other denominations as the fractional parts of this unit, or foot, it follows, that if we What are duodecimals? From what is the term derived? What are the denominations? Repeat the table. Whatare the marks ', ", "', "", &c., called Why has the foot no index? How may duodecimals be added and subtracted? For what are duodecimals principally used? How is the former ascertaiued? How is the latter found? What part of a foot is 1 '? What part of a foot is 1 "? Is 1 'k'? Is ""'? If we multiply feet by feet, what shall we obtain as the resiult? WhyS 3oo COMMON ARITHMETIC. Sect. XIV multiply feet by feet, we shall obtain feet as the result; as 1 X 1 = 1. But if we multiply feet by inches (') or inches (') by feet, the result will be inches (') because 1 inch is y ot a foot, ard 1 X - -1 12 12 Also, inches (') multiplied by inches (') produce seconds, (") as T Y X 4 — 1 " Again, feet multiplied by seconds (") or seconds (") by feet, produce seconds, (") as 1 X T == -1 ". And inches (') multiplied by seconds, (") or seconds (") by inches (') produce thirds, ("') as l,- X,, T -- = 1 '"' And seconds (") multiplied by seconds (") produce fourths, ('"). as T T X 1 =- 3 = 1 "".-Hence144 20736 - Obs. 3. The product of any tzo denominations is always of that enomination, denoted by the sum of their indices. Ex. 1. How many square feet in a room 14 ft. 8' 10" in length, and 12 ft. 7' 9" in width? Ans. 186 sq. ft. 4' 2" 6 "' 6"". 1 st Operation. 14 ft. 8' 10" 12 ft. 7' 9" 11' " 7'" 6."" 8 ft. 7' 1" 1"' 176 ft. 10' 0" Ans. 186 ft. 4' 2" 5"' 6"' 2d Operation. 14 ft. 8' 10" 12 ft. 7' 9" 176 ft. 10' O" 8 ft. 7' 1" 10"' 11' 0" 7// 6"" Ans. 186 ft. 4' 2" 5"' 6"" In the 1st operation, we multiply first by 9". 10" X 9" - 90"" (Obs. 3.) 90"" -12 = 7"' 6"". We set down 6"" and carry 7"' Feet and inches multiplied together, produce what? Why? Inches into nches produce what? Why? Feet into seconds produce what? Why?i/iahes into seconds produce what? Why? Seconds intoseconds produce what? Why? Of what denomination is the product of any two denominations? In the operations of Ex. 1, how are the partial products written? DUOrD1C TMA LS. 301 9' X 9" 72 "' and 7"' to carry m-ke 79"'. 79"' * 12= 6" 7"', We set down 7"' and carry 6". 11 ft. X 9" = 126", and 6 to carry make 132". 132"- 2 11' 0", which we set down. We multiply by 7' and 12 ft., setting down and carrying in the same man:ier. Finally, we add the several proJucts together, observing to carry for every 12, as in multiplying. In the 2nd operaion, we multiply first by 12 ft., anl afterwards by 7', and 9" setti g down and carrying as in the first operation. RENtMRK. —In both th'lle operations the learner wi!l perceive that we place those numbers of the same denomination under eacha other. From the preceding remarks we derive the following RULE. I. T'Write th/e 7mdltiplier under th e multiplica2nl. II. Mlltiply each denomination of the multiplicand (commencing at the rigid,) bly eac denomination of the multiplier and write each p;roduct under i s corresponding denomination. III. F'inally, add together the several partial products, as they stand, carrying for every 12 both in multiplyingr and adding. REMARK 1. The learner will observe that feet multitldied by feet produce square feet, and the same remark is observed of all the denominations of linear measure. (Sect. IX, Art. 2, Obs. 15. Rem). Also, square measure multiplied by linear measure produces solid, or cublic measure. 2. The inches in dno;lecirnals are usually called Carpenters' inches. They are neither linear, square, nor cubic. In measuring length, an inch is one-twelfth of a foot. In measuring surJfices, an inch is one-twelfth of a square foot; and measures 1 foot in length, and I inch in wilth. In measuring solids, an inch is one-twelfth of a solid, or cubicfoot; and measures 1 foot in length, 1 foot in thickness, and 1 inch in width. 2. How many square feet in 12 boards, each 18 ft. 7' 6" long, and 1 ft. 4' wide. Ans. 298. 3. How many square feet in a room 24 ft.6' 9" in length, and 18 ft. 9' 10" in width? Ans. 462 sq. ft. 3' 0" 4'" 6 "". 4. How many solid feet in a vat 9 ft. 6' long, 7 ft. 8' wide, and 6 ft. 9' deep? Ans. 491 sq. ft. 7' 6". 5. How many solid feet in a box, 17 ft. 8' 4" in length, 6 ft. 7' 3" in width, and 4 ft. 6' 6" in depth? Ans. 530 sq. ft. 8' 8" 6"' 2"" 6""'. Obs. 4. Painters', Pavers' or Plasterers' work is generally computed by the square yard. Square feet are reduced to square yards by dividing by 9. (See table, square measure.) What is the rule? What does a linear measure multiplied by linear measure produce? What does square measure multipllied by linear measure produce? What are the inches in dtiodecimals usually called? What is an inch, and how much does it measure in measuring lengths, surfaces and solids? How is painters' pavers and plasterers' work computed? CG2 COMMON AlITHMETIC. Sect. XIV 6. I-ow many square yards in the walls of a room which measures C 1 it. 4' in compass, and 11 ft. 3' in height? And what will it cost to paintt the walls at 12-\ cents per square yard? Ans. 76J- sq. yds.; cost $9.58 —. 7. ilow muc- r wil it cost to paint a building measuring 96 ft. 9' il co mpl;ass, and 1t 5 ft. 8' in height, at 10 cents per square yard? Ans. 1 (i.841-. O. I~ow nliu:: will it cost to palve a side walk 19 24 ft. 6' in length, and 7 ft. 1' i:l;vwi-lh, at 25 cCell:s per s(uarfe y rd' An. 27.0(9. 9. lio,v miuci: will it co-t, to plave;a -';1rd 35 ft. 6' in length, and 25 ft. 8' in widtlt, at 30 cents per square y1ard' \Ais..30.37-. 10. iHow rmu.'h,! xill il cost to plaster a ceilin-, at 18 cents per square y'lard, iL meaI]srin' '-21 ft. 8' in lengtll, and 12 f t. 9' in width?! Ans.,5.52~-. 11. -Iow muctl will it co-t to p1)astel A ceiiing 18 ft. 6' in length, and 10 ft. 2' in width, at 20 cents per square yard? Ans. 84.18 nearly. 12. A certaint room measures 25 ft. 3' in leng'th, 18 ft. 4' in wildth, and 9 ft. in lheighth. Required the cost of plastering it, at 25 cents per s(quae yard, deducting 3 doors, eac'i 6 ft. 6' by 2 ft. 8', a fire place.5 ft, 6' by 4 ft. 4', and 4 windows, each 5 ft. 8' by 3 ft. 3'"1 Ans. $32.032, 13. A certain building is three stories in helght; each story has 12 windows, each 6 ft. 10' by 3 ft. 6'. Required the cost of the glazing at 15 cents per square foot. Ans. $129.15. Obs. 5. Some parts of Carpenters' and Joiners' work are computed by the sq. yd.; other parts such as flooring, &c., are computed by the square. A square consists of 100 square feet. 14. How much will it cost to lay a floor 22 ft. 6' in length, and 18 ft. 6' in width, at $3.50 per square? Ans. $14.567. 15. How much will it cost to lay a floor 30 ft. 8' in length, and 26 ft. 6' in width. at $4.50per square, deducting a place for the stairs 9 ft. 10' by 4 ft. 6'; and a fire place 5 ft. 10' by 4 ft. 6'. ~~~~~~~~, ~Ans. $33.391. Obs. 6. Masons' work is sometimes computed by the solid foot and at other times by the perch. A perch measures 16- feet in How are square feet reduced to square yards? How is flooring, roofing, &c., computed? Of how much does asquaie consist? How is masons' work sometimes conTpu ted *ft OtWErtS AMN) ROOTi letngth, 1 ft. inbreadth, and 1 foot in depthi, and contains l16 X 1- X 1 = 24a solid feet. (Sect. IX, Art. 2, Obs. 13.) Ilcnce1To find the number of perches in any wall or solid body: Obs. 7. Find the contents in solid felt, and dividc this by 24. 16. HIow many perches in a wall 72 ft. 3' lon:, 12 fi, 8' -igh, and 2 ft. thickv? Ans. 73.952-t-. 17. How many perches in l wall 43 ft. 10' loni, 15 ft. 6' Jiigh, and 2 ft. 3' thick? 1 What will it cost to build it ai.' 2.5',)I, 'r pc rclih' Ans. 68.81 -{-p(lrceis. ('ost 1 72. 02'5. 18. How manv bricks in a wall 100 ft. 4 long, 10 ft. 6' ligh, and 1 f. 8' tliick, allowinr 620 bricks to the solid liot, and how much w ill it cost to lay this wall, at $6 per thou sandl brliclks Ans. 35116 brieks. Cosst 2 i0.70, 19. HIow manv bricks, each 8 inches longl, 4 inches w i:le,;and 2 inches thick will it take to btlild a wall 40 ft. 6' long, 14 fIt. 4' high, and 2 ft. 8' thick? Ans. 41796. 20. A certain brick building is 28 ft. 6' in length, 24 ft. 6' in width, and 15 ft. 9' in heighth, and the walls are 1 Ioot thlick. Now deducting 4 doors, each 6 ft. 9' by 2 ft. 8' and 10 windows, each 6 ft. 6' bv 3 ft. 6', bow many bricks, each 8' long, 4' wilde, nnd 2' thick, did it require for this building. and how much did they cost at $3.25 per thcusand? Ans. It took 36990 bricks, and they cost $ 120.21 SECTION XV. POWIERS AND ROOTS, Def, 1. A POWER of any number, is the product arising from mutiplying it into itself. Thus; 4 is a power of 2, because 2 X 2 -4; and 27 is a power of 3, because 3 X 3 X 3 = 27, &c. Def. 2. A RoOT of any number, is a number which multiplied into itself a certain number of times will prcduce the given number. Thus; 2 is a root of 4, because 2X 2 = 4; 3 is a root of 27, because 3X 3X3 = 27; and 4 is a root of 256, because 4 X 4 X 4 X 4 256. What does a perch measure? How mnany solid feet docs it contain? How do we find the uumber of perches in a wall or solid body? What is a power? Give examples. What i a root? Give examples? 304 COMMON ARITHMETIC. Sect, XV ARTICLE 1. INVOLUTION. Obs. 1. INVOLUTION is the method of findicng the powCers of numbers. Obs. 2. Powers are divided into several orders, called the first, second, third, fourth power, &e. Obs. 3. The numnber which is to be involved is call d the root, orfirst power. The (ther powers derive tleir name from the number of' times the roet is ce:ployed as a factor, izn p)roducicn this p)ower. Thus 2 is the first power of 2. 2X2= 4 is the second power of 2. 2X2Xo= 8 is the third power of 2. 2X2X2X —16 I is the fourth power of 2. REMARK.-The second power of a number is usually called the siare The third power is lusally called the cubt.: The fourth power is usually called the hiquadlrale: The fifth power is usually called the sursolid: The other powers generally receive no other name tlhan their numeral distinction, as the sixthpower, the seventhpower, &c. of the root, or first power. MENTAL EXERCISES, 1. What is the square of 3? Of 4; 5; 6; 8; 11; 7; 10;-9; 12? 2. What 's the cube of 3? Of 4; 7; 6; 5; 12. 9; 11; 10; 8? 3. What is the fourth power, or biquadrate, of 3? Of 5; 4; 6? 4. What is the fifth power, or sursolid, of 2? Of 3; 4; 5f 7? 5. What is the sixth power of 2? Of 3? 6. What is the seventh power of 2? Of 3? Obs. 4. The number that denotes the power to which the root is to be raised is called the index, or exponent of the power, and is usually written at the right, and a little above the root. Thus: 31 signifies 3, or the first power of 3. 32 signifies 3X 3 9, or the square of 3. 33 signifies 3 X 3 X 3=27, or the cube of 3. 34 signifies 3X3X3X3=81, or the biquadrate of 3, &c., &c. REMARK 1.-To find the second power, or square of 3, we use 3 as a facto r twice; that is, we multiply it into itself once. Thus: 3 X 3= 9. What is Involution? How are powers divided? What is the number to he involved called? From what do the other powers receive their name? Give examples. What is the second power of a number usually called? The third power? The fourth power? The fifth power? What name do the other powers receive? What is the number denoting the power to which the root is to be involved called? Art. 1. INVOLUTION, 305 2.-To find the cube, or third power of 3, we use 3 as a factor three times; that is, we multiply it into itself twice. Thus: 3 X 3 X 3 = 27. 3.-To find the biquadrate, or fourth power of 3, we use 3 as a factorfour times; that is, we multiply it into itself three times. Thus: 3 X 3 X 3 X 3 =81. HenceTo involve any number to any required power: Obs- 5..fultiply the given number into itself, until it has been employed as a factor as many tines as there are units in the index, denoting the power to which it ts to be raised. REAARnI.-The number of multiplications in involving any number to a required power, is always 1 less than the index. Thus, 33 signifies that 3 is to be employed as a factor tlree times, but to be multiplied into itself but twice. Thus: 3X3 X 3 = 27,&c. Obs. 6. All the powers of 1 are the same-that is, 1. For 12 =1; 14= 1, &c. EXERCISES FOR THE SLATE. 1. What is the square of 14? Ans. 196. 2. What is the cube of 13? Ans. 2197. 3. What is the fourth power of 15? Ans. 59625. 4. How much is 132? 143? 182? 204? 305? 406? Ans. in order. 169; 2744; 324; 160000; 24300000; Ans. in order. t 4096000000, 9 is the square of 3. To multiply 9 by 3, and that product by 3, we obtain 81 = 34, which is the same as 9 X 9, or 3" X 32. 33 = 27; 27 X 3 X 3 = 243 = 3, which is the same as 27 X 9, or 33 X 32. 33 = 27; 27 X 3 X 3 X 3 = 729 = 36, which is the same as 27 X 27, or 33 X 33, &c. Obs. 7. Hence-The fourth power is equal to the square multiplied by the square. The fifth power is equal to the cube multiplied by the square. The sixth power is equal to the cube multiplied by the cube. The seventh power is equal to the fourth power multiplied by the cube, d'c. That is-The product of any two powers of a number, is equal How do we involve iunl bers to any required power? How many tites is a number used as a multiplier i involving it to any required power? What are all the powers of I? To what is the fourth power of any unuber equal? 'Ihe fifth power? The sixth power? The seventh power? To what is the product of any two powers of a number equal? 306 COMMON AR1THMETIC. Sect. XV to that power of the number denoted by the sum of their exponents. From the above illustrations, the following proposition is also evident: Obs. 8. The quotient arising from dividing any power of a numEber by another power of the same 7number, is equal to that pouer of the number delenoled by the diference of their exponents. Let the learner find the results of the following questions by the wo preceding obselvations: 5. lRcquire:l-teli fiurlt power of 16? 18? 22? 25? Ans. in o der. 65536; 104976; 234256; 390625. 6. Required-the fifth power of 17? 19? '21? 24? 309? Avnsf. i, ( i. 3 1419857; 2476099. 40841(1; 7962624; f (T 241301000. 7. lRequired —the sixtlh power of 16? 18? 22? 25? Ans. in loder. 16777216; 34012224; 113379904 * oe244140625. 8. Required —the seventh power of 17? 19? 21? 24? 30? Ans. in order. 410338673; 893871739; 1801088541; 4586471424; 21870000000. 7. Divide the ninth power of 12 by the seventh power of 12. Ans. 144. 10. Divide 14- by 143. Ans. 196 = 142. 11. Divid e I Aby 1 8'. Ans. 5832 183 12. Lividel 36' by 3625. Ans. 60466176. Obs. 9. Fractions nayv be involved as well as whole numbers, by involeving both the nuanrator and denominator to the power i2ndicat d by the ex.ronen.t. REMARK 1.-A fraction to be involved is generally written within a parentlesis; thus (G)2. This signifies that - is to be multiplied into itself, or squared. Thus: - X - 4 2; and (.)3= X3 2 7 Ut 2s3 - ~ _ X2 - ad n3 d X & C. 27; but '3 2: and -3 - &c. 2. —Any pouer of a proper fraction is less than the fraction itself. This is evident from the fact that the numerator is less than the denominator; (Sect. VllI. Art. 1. Obs. 11.) and therefore, when the fraction is involved, the multiplier of the denominator is greater than the multiplier of the numerator, and this being the case, the ratio To what is the quotient arisin?.' from dividing any power of a number, by another power of the saine nulmber, equal? How inny fractions be involved? How is a fraction to be invoived generally written? What is the value of the power of a proper fraction, compared with the value of tihe fraction itself? Show why thli. is tlhe c:as'? Art. 1. INVOLUTION 307 between the terms of the fiaction is diminisihed; hence, as the ratio is the quotient of one number divided by another, (Sect. XII. Art. 1. Obs. 1.) and the vtalue of the fraction answers to the quotient in division, (Sect. VII.1. Art. 1. Obs. (9.) it is e\vi lent thlt if the quotient, or ratio, is dimini.sllhe, theic voalte of ti:,' tfa 'tion is also dimir,ishced. 3.-I1t is ge'ncrall st l: cony tnil'l to reciCe 1n.Ce 1 t7mrbers t) iproperfra: ' tionizs, or Lthe fr'l:t ionill p rt to a. de' n;d. btfore itlvo'vill it to the pow)Ver ijdliAlt(d by the in lex. Likewise, com/2,:n,..:l and complex. f' ( ti.>~ must be e Ineedl t(o;simp1e ov1e.s bc-Y ilc illvolvhiL, them. 3 59 13, W hIIat is tell sq, ire Of I I -? -,3? \ ], 7-? 4 of 6 14. W\lat is t l), cll) o yu- -.t' -? ' 4 ' - of 7,'.(:1ie22,,2 I )3 15. How muc] is - -[ '-!( i:;:J, (, 5'.I of ) of 9 16. What is tlhe s luare of 25. Ains. 625. Common OP)ertKio.i..7nai/lwti Operalti 'm. 25=2U?+5 "25;, -2O-+ 5 125 1'0-f-'25 _= -I'!ur t,,f G2 )+-'5!). 50 4'10- { — 10 = Pro u1', ol 2-.-5 }y 20. Ans. 625 Ans. 400-+200-J1-25= t625=: i s u!:i 8-r. Thus, we 1)lrcive the result to be equnl to t1 sljaIl e oft i)e first part, (20,)2 plus twice the product of tlh t-wo p rt; (2) X 5 X 2.) plus the square of thie second part, (5)2. 'Ilte opelralion may be contracted, thus: 25 5X 5, or 52 =C25; set down 5 and carrly. 2 5- 10; 25 10X2=20, and 2 to carry =22; set down 21, and carry -- 2. 2X'2, or 22-4, and 2 to carry = 6. (e'. VI. Art. 625 2. Ex. 13. Hence — To find tlle squ(lre of any number consisting of but two figures: Obs. 10. SqirCl te trigh ha/n2i digit for the unit iyure of the Hlow do we proceed with mixed numbers, when involvilng tihlm? With compound auId comlpl.-x fractions? In the analytic operation of;Ex. 16, te what is thle result equal tHow tlheu may we find the square of a nnunbr ccusisting of but two figures? 308 COMMON ARITHMETIC. Sect. XV result, double the product of the two digits for the ten's figure of the result, and square the left hand digit for the remaining figures of the result, observing to carry in each case as in Multiplication of Simple Nlumbers. NOTE.-By this method the learner can often find the square of numbers mentally, which would otherwise require the use of the slate. 17. How much is 993? 272? 462? 752? Ans. 9801; 729; 2116; 5625. 18. How much is 922? 862? 502? 1002? 19. IHow much is 313? 443? 374? 304? 605? 109? 207? 20. Iow much is 1203? 1442? 3006? 17282? 2564? 21. -ow nmuch is 146? 910? 153? (4)4? 2? 7? (41)3? ~~~~~1 1?4' 7 32 4 9 l2J 22. How much is ()c? (-)9?.52?.073?.00110? 4.753? 1.15? 23. How much is.42?.026? 1.00022? 3.07022? (10- )3? 24. How much is 7.92? 1.163? 2114? 200.0023?.00016? 25. How much is(10-)4? 17.93? 216.012? 100.015? ARTICLE 2. EVOLUTION. Obs. 1. EVOLUTION is the method of finding the roots of numbers. REMARK 1.-Evolution is the opposite of Involution, and each reciprocally proves tie other. By the one, we multiply a nurmber into itself to find the power: by the other, we resolve the power into cqual factors to find the root. 2.-Different roots have different names, corresponding to those of the different powers. Thus: 2 is the second root of 4, because 2 2X = 4; and 3 is the -third root of 27, because 3 X 3 X 3 = 27. Hence — Obs. 2. A root takes its name fom the number of times it is employed as a factor to produce the given power. REMI.IRK 1.-The second root is usually called the square root. The third root is usually called tlie cube root. T''e other roots generally receive no other name than their numeral distinction-as the fourth root, the fifth root, &c. la! i:s iij;:itv: ntage of thiis Irle? W Vht;. voi.i,: ' W at rel-Ation has Evtl'!i,)u 10o Ivol itlion'? iiow i - lI i '? Fri ' oi weilt doi a root tlke its name? What is the eecond root usually called? The third root? What name do the other roots generally receive? Art. 2. EVOLUTION. 309 2.-Powers and roots are correlative terms; because, if one number is a power of another, the latter is a root of the former. Thus, 8 is the cub3 of 2, and 2 is the cube of 8. MENTAL EXERCISES. 1. What is the square root of 9? Of 16; -s; 49; 36; 81; 100? 2. What is the cube root of 8? Of 27; 64; 125? 3. What is the fourth root of 16? Of 81? 4. What is the square root of 144? The cube root of 729? Obs. 3. This character (V/) is the sign of the square rcol. Other roots are indicated by the same character, with the index of the root placed above it. Thus: f indicates the cube root; J indicates the fourth root, &c, REMARK 1.- Roots are also iunlicated by a commonn fraction, the numerator of which is 1, and the denominator the index of the root. This is written in the same manner as the exponents of powers. 1 1 1 Thus: 4' is the same as l/4; 83 is the same 388; 164 is the same as f 16, &c. 2.-As tile square multiplied by the square produces the fourth power, (Art. 1. Obs. 7.) it is evident that the square root of the square root is the fourth root. Thus: V16 =4; /4 =2= 1 6. Likewise, the sixth root is equal to the square root of the cube root, or the cube root of the square root. Thus: V64=-4; d,4=2 = 64. And 764=-8; /8= 2 = /64. HenceObs. 4. When the index of the root can be resolved into factors, these factors denote the roots, which being successively found, will give the required roof. Obs. 5. The process of finding the roots of numbers is called the Extraction of Roots. It consists in finding such a number, as multiplied into itself a certain number of times w 11 produce the power. (Definition 2.) 5. How much is 1144? /81; 3V216; 3344;,f81? What kind of terms are powers and roots? Why? What is the sign of the square root? How are other roots ind;cated? By what other method are roots indicated? How is this fraction written? To what is the square multiplied by the square equal? To what is the square rootFOf the square root equal? To what is the sixth root equal? When the index of.root can be resolved into factors, what do these factors denote? What is the process of finding the roots of numbers called? In what does it consist? ",![O C 0 oM.;;A. I'a S t, F t. Sect XV o. tow 1(l i.. Jl21.",1 /0:: 5 ')12; /729; J2243? 7. ltow illnt- is v G i 3; t125; V16; /64; /729? (O)1. 7. '/,,(fs /, ' ',rt.'',. 1: may be ext racte, by finclino' the r/:;(/)s o/ l&olh t/'e nu/eat',or i'nd tle den2ominolo:r. ^4,5 'I'.-_;l l,=7-; 'L =," / r,. Ths r4= ^7::',ti -=t and 5 -- - 21 c. '7 3. 8 7,a. MAi.ie itiz26hers must be rieduced to 'in pro;er fractions before ( xitr;icltiflr thleir i ro )ts. 8. hIW m I.U/I i -f ' I/ 4; ~ / 4? 1 9. 1(ow mucl ish 1j- I ^/; 2/ 7 -^/64 10. Flow mituch is?/l 37 3 /3:; \/21 Ob: 8.. ln:i; inl's (xxi i;plels vwe notice this consideration: 4f the <i':,?Ce. a'o;;/;/(,,.s' fraction, the root also contains ( p'oper) i (ativon. And coli1ne::' (. y: ct. If the/ roit ('o.aii.s' pro'l.: er /'raction, ay/ ])power of this root nmt.sl also contain a proper fractionh. This is evident from Art. 1. Obs. 9. Rn.em. A. lso Ob.s. 9. /f thi i:'c't ' ('c'Ti/l.!' a decimal, its root m2ust also contain a decimi.al. And colverselya. I' the root conlt(inis a deCim7l, any 2pou-er of tids root must contain a ldecimall. Thlis is evident from the fact, that if we multiply by a decimal, the product must contain a de imal. (Sect. VIII. Art. 11.) RE,,IAnK.-From the f. ct that the product must contain as many decimals as botil faclors, it is e.:id.clt lhat the root must contain one half as many decim,,nls as its squllrc, onri thirl as miny decimals as its cube, -Cc. 11. How muclh is v/.25? /.64? V/.027? 12. How much is -3/.343t /.729? V/. 16X9? /30-3? Obs. 10. A number whose root can be exactly found, is called a perfect power, and its root is called a raiional?um.ber. Thus: 9, 16, 25, 27, &c., are perfect powers, and their roots, 3, 4, 6, and 3, (the cube root of 27,) are rational numbers. How may roots of fractions be extracted? How do we proceed with mixed numbers when they occur? W\ hat consideration is noticed from Ex. 8, 9, and 10? Show why thiisisorrect? What is said respectingdecimalsin the power or root? Show why this is correct. Hzw many decimals must the root contain, compared with tlose of the square? Of its cube? Show why this is correct, What is a perfect power? A rational number? Art. 2. EVOLUTION, 311 Obs. 11. A number whose root cannot be exactly ascertained, is called an impefect power, and its root is called a surd, or irratlioldl inzmber. Thus: 11, 17, 45, &c., are imperfect powers, and their roots 3.3 +, 4.1+-,. 7+ are surds. REMAKK.-A number may often be a perfect power of one order, and an imperfect power of another order. Thus, 25 is a perfect power of the second order, and an imperfect power of the third order. Also, 64 is a perfect power of the second, third, and fifth orders, and an imperfect power of the fo'urth order. Obs. 12. Every root and power of 1 is the same-that is, 1. Thus, 12, 1I, 4, /1, 1, 51I, 11, &c., arc all equal. NOTrE.-Tlhe preceding exercises in Evo!ution arc designed to be performed mentally; but it frequently happens that the numbers are too large to ascertain their roots in this manner, and we have to use the slate. In such cases there is a particular process by which the roots are found. lThe miethods of extracting the square we will now explain. As the other roots seldom occur in practical business, they are not treated of in thi; \work. Case 1.-Extraction of the Square Root. Obs. 13. To extract the square root of a ntmler', is to find a number, which being multiplied into itself, will prodtce the given number. (Art. 1. Obs. 4. Rem.) REMARK.-Tlle terms S.quare and Square Root are derived fromn Geometry. The [sqnare may be considered as a square figure, made up of several smaller squarer, and the square root as the number of smaller squares on one side of this figure. Thus, in the diagram, the large square is made up of 16 smaller ones, and there are 4 of these on a side; therefore, 16 is the square, and 4 is the square root. HenceObs. 14. T'he extraction of the Square Root may be defined, ( finding one side of a square, when the square contents are given. Obs. 15. Take the following numbers: 1, 2, 3, 4, 5, 6, 8, 9, 10, 99, 100. Their factors are 1, 4, 9, 16, 25, 36, 64, 81, 100, 9801, 10000. The numbers in the first line may also be regarded as the square roots of those in the second line. (Obs. 2. Rem. 2.) What is an imperfect power? A surd or irrational number? Can a number be a perfect power of one order, and an imperfect power of another order? Give examples. What is every root and power of 1? What is it to extract the square root of a number? From what are the terms square and square root derived? What may the square be considered? The quare root? How then may the extraction of the square root be definedl 312 COMMON ARITHMETIC. Sect. XV From this we notice the following considerations: a. 1st. The square of no digit exceeds two figures. b, 2nd. When a number contains but two figures, its square root contains but one. Also: 1When the number co. tains three or four places of figures, its square root contains 2. c. 3rd. The square of no number contains more than double the number of figures in the number squared, and but I less. Obs. 16. Hence- 'When we extract the square root of a inumber, we first point it of into periods of two figures each, commencing at the right hand. By this means we ascertaina. ist. Hfow many figures the left hand period contains, or how many figures must be taken to find the first fi2ure of the root. This is the most important object gained. b. 2nd. Of how many figures the root will consist. REMARK.-From Obs. 15, c., it is evident that the square root must contain as many figures as there are periods in the square. If the sqiare contains an odd number of figures, the root will contain 1 more than half as many; if the square contains an even number of figures, the root will contain just half as many. Thus, the square root of 144 is 12; and of 6400, the squa'e root is 80. These numbers when pointed off into periods stand thus: i44; 6406. Ex. 1. A man has a square field containing 625 square rods. How many rods does it measure or each side? Ans. 25. Operation. 625 (25 1st step.-In this example we wish to find one 4 side of a square, having given its contents; that -is, we wish to extract the square root of 625. 45)225 (Obs. 14.) We first point off the given number 225 into periods, and then ascertain that the left hand - -- period contains one figure, and that the root will 000 contain two figures. (Obs. 16.) '1 hese two figures (of the root) of course are a ten and a unit. 2nd step.-We now seek the greatest square in the left hand period, (6) and find it to be 4, the square root of which is 2. This we place at the right, with a liiie to separate it from the Fower. As this is the root of the left hand period, (6 hundreds) it is in reality 2 tens, (because the square of 2 tens, or 20, is 4 hundreds, or 400,) and is equa! to one side of a square measuring 2 tens (20) rods on a side. This may be shown by a diagram: What is the first consideration that we notice from examining the given numbers and their squares? The second? The third? What is the first thing we do in extracting the square root ot a number? What do we ascertain by this means? How many figures must every root contain? If the square contains an odd number of figures, how many figures will the root contain? If the square oentains an even number of figures, how many will the root centin? Art. 2, EVOLUTION. 313 A Fig. 1. B 400 sq. rds. D20 rods.C D 20 rods. C The contents of this figure we find to be 20 < 20 -400 sq. rds., now disposed of, aid which we will deduct from the whole number of rods!a the field, (.625) showing 225 sq. rds. remaining. 'his deduction is readily eff -ted by subtracting the 4 /hili^e nds) the square of the 2 (tens,) from the left hand period, 6, (hundreds,) leaving 2 (hundreds,) and bringing down the next period, ('65,) making 225 as seen in the operation. 3d step.-We now wish to enlarge our figu e by the addition of 225 sq. rds., and it is evident that this addition must be made on two sides, in order to preserve the square. This is shown in the following diagram: FIGURE 2. 20 rds. 5 rds, C 20 5 5 B I '5 100 sq. rds. 25 sq. rds A D 20 20 20 5 400 sq. rds. 100 sq. yds Now as the addition is made g on two sides of the square, if we divide the number of rods to be added, (225,) by the length of the two sides, (20 +20=40,) the - quotient w ill $ evidently b e the width of the additions made to the two sides AB, B C, Fig. 1. Accordingly, we doubTe 20 rds. 5 rds. In the operation of Ex. 1,. what is the first step? What do we ascertain by this? Whatis the second step? What is this root? WhyT To^-,wft is it equal? What is done with the contents of this figure? 15 314 COMDMON ARITHMETIC. Sect. XV the root (2 tens,) already found, making 4 (tens) for a divisor, 4th step. —We now seek how many times our divisor (40) i. contained in our dividend, (225) and find it to be contained 5 times; therefore, 5 rds. is the width of the additions, and 40 X5=200 additional rods disposed of, and 25 rds. remaining. 5th step.-But looking at Fig. 2 we perceive that there is still wanting in the corner, C, a small square, equal in size to the width of the additions at the sides, (5 rds.) As this is a square, its contents are 5X5==25 sq. rds., which completes the square, and gives 25 rds. as the lengthl of one side. If the learner examines Fig. 2 attentively, he will perceive the additions measure 45 rds. in length, (20 rds. on each side, and 5 rds. at the corner,) and 5 rds. in width; and therefore, if we multiply 45 by 5, we shall obtain the same result (225.) This is effected, in our operation, by rejecting the cipher at the right of our divisor, (40,) and also the right hand figure (5) of our dividend, (225,) and dividing the remaining figure of our dividend (22) by the remaining figure (4) of the divisor, and writing the quotient figure both in the root, and also at the right of our divisor (4.) Next, we complete our dividend, (225,) and multiply our divisor (45) by the last figure of the root, (5.) and subtract the product from our dividend, which in this case leaves no remainder. Rejecting a figure at the right of our divisor and dividend, is in effect dividing both by 10, (Sect. V. Art. 4. Obs. 1.) and this cannot affect the result. (Sect. VI. Art. 1. Obs. 28.) PRooF.-The operation may be proved by adding together the several parts of the last diagram. Thus: The square A contains 400 sq. rds. The additions B and D contain 200 sq. rds.; that is, 100 sq. rds. The corner C contains 25 sq. rds. [each. And the entire figure contains 625 sq. rds. Or, by Involution: 25 X 25 = 625, as before. (Art. 1. Obs. 5.) REMARK 2.-Perhaps the learner has inquired, before this time-Why do we commence at the right to point off the given number into periods? The anHow is this deduction effected? What is the third step? On how many sides must this addition be made? Why? How do we find the width of these additions? How do we obtain our divisor? What is the fourth step? Howy are the contents of the corner C ascertained? What is the length of all the additions? How is this found? What is their width? How then may we find the contents of all the additions7 In the operation, how is this effected? What effect does it liave upon the divisor and dividend to reject a figure ftis the right of each? Does this effect the result? Why not? How do we prove the operatiot? By what other method? Why do we commence at the right to point off the given sum into periods? Art. 2. EVOLUTION. 315 is-Because, after the first figure of the root is obtained, there must be two figures in the given sum for every additional figure in the root. [Rem. 1 ] Hence, we must point off from the right, as the first figure of the root is found from the left hand period. We colmmence our operations at the left on account of the remainders which occur. [Sect. V. Art. 2. Rule for Short Division. Rem. 1.] From these remarks we derive the following RULE FOR EXTRACTING TIHE SQUARE ROOT. I. Point qof the given sum into periods of two figures each, placing a dot over the unit figure, another over the hundreds, and so on. (Obs. 16.) II. Find the greatest square in the left hand period, and place its root at the right, like a quotient in Division. Subtract the square of the root from the left hand period, and bring down the first figure of the next period for a partial dividend. III, Double the root already found for a defective divisor; seek how many times this defective divisor is contained in the partial dividend, and place the result in the root, and also at the right of the defective divisor to complete it. Likewise, bring down the remainingfigure of the period at the right of the partial dividend, for a true dividend. IV. Multiply the true divisor by the last figure of the root, subtract the product from the dividend, and to the right of the remainder bring down the first figure of the next period for another partial dividend. V. Double the root rdready found for another defective divisor, with wuhch proceed as before, until all the periods have been brought down and divided. VI. If at any time the defective divisor is not contained in the partial dividend, write a cipher in the root, and bring down the remaining figure of the period, together with the first figure of the next period, for another partial dividend, with which proceed as before. PRooF.-Multiply the root into itself, and if the product is equal to the given sum, the work is correct. 2. What is the square root of 41616? Ans. 204. Operation. Proof. 41616(204 204 4 204 404)1616 816 1616 408 0000 41616 (Art. 1. Obs, 5.) 316 COMMON ARITHMETIC. Sect. XV The operation may also be demonstrated in the following manner: 3. What is the square root of 1296? Ans. 36. Operation. Proof. 1296(36 36=30+ 6 9 30+ 6 66)396 180+36 396 900+180 000 900+360+36=1296. (Art. 1. Obs. 10.) 1st step.-We find the greatest square in the left hand period to be 9, and write its root (3) in the quotient, subtract the square (9) from the period (12), and bring down the next period for a dividend. 2n step.-As the root consists of but two figures, the second figure must be such, that twice the product of the first and sec ond terms, together with the square of the second, must complete the square [Art. 1. Obs. 10.]; therefore we divide our dividend [396] by twice the figure of the root already found, viz.: 3 [tens]X2= 6 [tens], or 60, and find the second figure of the root to be 6. 3d step.-We then add the last figure of the root [6] to our divisor [60], making 60+6=66, (or merely write the 6 in the place of the cipher,) and multiply this by the last figure of the root [6], and subtract the product from the dividend. By this means we obtain twice the product of the two terms, and the square of the second term; because 60 is twice the first term, which being multiplied by 6 gives twice the product of the two terms, and 6, multiplied by 6, gives the square of the second term. It produces the same result to multiply 66 by 6, because 66=60+ 6, and it is an established principle in mathematics that if equals are multiplied by equals, their products will be equal. Obs. 17. From examining the first nine digits and their powers, we perceive that every perfect square must end with 0, 1, 4, 5, 6, or 9; because the right hand figure of any root must be one of the nine Why do Are commence at the left to extract the root? What is the first step in extracting the square root? The second? Third? Fourth? Fifth? If at any time the defective divisor is not contained in the partial dividend, how do we proceed? How do we prove the operation? Wlat is the first step in the operation of Ex. 3? The second? The third? What do we obtain by this method? Explain why this is correct? By what other method can we obtain the same result? Why is this correct? What consideration do we notice from examining the first nine digits and their squares? Why is this correct? Art. 2. EVOLUT;ON. 317 digits, or a cipher; and the square of no digit ends with any other figure than one of the above. Again-as no digit, multiplied by itself, will produce 0 as the unit figure of the result, if the right hand place of the root is a cipher, the last period must be ciphers; that is, if a perfect power must end with 1, 4, 5, 6, 9, or 00; likewise, if a perfect power ends with 00, the remaining figures must be a perfect power. This is evident without further demonstration. Also-since every defective divisor is an even number, (it being twice the root,) the product of this defective divisor, multiplied by 5, must end with a cipher; therefore, if the root ends with 5, the square must end with 25; because the square of 5 is 25, and the 2 occupies the place of the cipher above mentioned. Hence, every perfect square must end with 00, 1, 4, 25, 6, or 9, and if a perfect square ends with 00, the remaining figures must also be a perfect square. EXERCISES FOR lTHE SLATE, 1. What is the square root of 502681? Ans. 709. 2. What is the square rout of 15876? Ans. 126. 3, What is the square root of 75076? Ans. 274. 4. What is the square root of 18541636? Ans. 4306. 5. What is the square root of 81072016? Ans. 9004. 6. What is the square root of 73960000? Ans. 8600. 7. What is the square root of 174740720400? Ans. 418020. Obs. 18. If there "is a remainder after the last period has been divided, we may form new periods, by annexing ciphers, and continue the root to decimals, to any degree of exactness. The learner must not expect to find an exact root in such cases, however, as the square of no digit ends with a cipher; but we can approximate sufficiently near by the aid of decimals. To prove the operation, he must add the remainder to the square of the root. 8. Required, the square root of 8? Ans. 2.828+. 9. Required, the square root of 590? Ans. 24. 29+. 10. Required, the square root of 46780? Ans. 216.286-. Obs. 19. To extract the square root of decimalsCommence at the hundredth's place, and point towards the right. The square of no digit ends with a cipher; what is the inference deduced from this fact? If a perfect square ends with 00, what must the remaining figures be? Why? What is every defective divisor? Why? What inference is deduced from this fact? If the root ends with 5, with what must the square end? Why? With what then must every square end? If there is a remainder after the last period has been divided, how do we proceed? Can we find an exact root in such cases? Why not? How can we approximate sufficiently near? How prove the operation? How do we extraet the square root of decimals? 318 COMMON ARITHMETIC. Sect. XV When the power consists of whole numbers and of decimals: Point both ways from the unit's place, REMARK.-The reason why we point decimals from the left towards the right, is evident from the fact, that every perfect square must contain twice as many decimals as its root. (Obs. 9. Rem.) Therefore, for every decimal in the root, there must be two decimals, or a full period, in the square; and if the periods are not complete, a cipher must be annexed to the decimal. Hence-we must commence at the left to point off decimals; because if we prefix a cipher to the left to complete the period, the value of the decimal is altered, whilst if a cipher is annexed to the right it is not. (Sect. VIII. Art. 7. Obs. 6 and 7.) 11, What is the square root of 356.164? Ans. 18.872 —. 12. What is the square root of.0225? Ans.. 15. 13. What is the square root of 225.6004? Ans. 15.02. 14. What is the square root of.0000163216? Ans..00404. 15. What is the square root of 18.211? Ans. 4.26731+. 16. What is the square root of 1.004030076? Ans. 1.002012+. b s. 20. To extract the square root of a common fraction: First reduce the fraction to its simplest form; then if both terms are perfect squares, extract the square root of each (Obs. 7.); but if both terms are not perfect squares, reduce it to a decimal, and then extract the square root, according to Obs. 19. 17. What is the square root of s56? Ans. 2. 18. What is the square root of "2? Ans.. 6856&' 242' 19. What is the square root of 1 — '! Ans..3535 —. 20, What is the square root of y3- '? Ans.. 16. 21. What is the square root o' l10o? (Obs. 7. a.) Ans. 1 3. 22. What is the square root of 184? (18- = 18.8.) Ans. 4. 3354-. 23. What is the square root of 20024~9s4? Ans. 14. 24. If an army of 186624 men were drawn up in a perfect square, how many men would there be on a side? Ans. 432. 25. A certain square field contains 86436 hills of corn. How many are there on a side? Ans, 294. 26. A company of men paid $2304 to a charitable institution, each man paying as many dollars as there were men in the company. How many men were there in the company? Ans. 48. 27. A. has two lots, one measuring 55 rods in length, and 40 rods in width, the other measuring 70 rods in length, and 20 rods When the power consists of Loth w-holb numbers and decimals, how do we proceed? Why do we point decimals from the left hand towards the right? How we extract the square root of a conmmon fraction? Art. 2. EVOLUTION. 319 in width. B. offers him for these, a square lot containing as many rods as both. How many rods square is B.'s lot? Ans, 60. 28. What is the difference between a square half foot, and half a square foot? Ans. 36 sq. in. 29. If it takes 4356 square tiles to pave a square floor. how many are there on a side? Ans. 66, 30. A nurseryman wishes to set out 1521 fruit trees, so as to form a perfect square. How many must he put in a row? Ans. 39. Case 2.-APPLICATION OF THE SQUARE ROOT. TRIANGLES. Obs. 21. A TRIANGLE is a figure bounded by three straight lines. Obs. 22. When one line meets another so as to form a square corner, or right angle, it is called a Right Angled Triangle. A Thus: A B C is a right angled trian- gle, of which the right angle is B. The side A B is called the base; ihe side B C I is called the perpendicular; and the side A C is called the hypothenuse. The hy- j pothenmse is always the side opposite the A ase. C right angle; the hypothenuse A C is opposite the right angle B. Obs, 23. In every right angled triangle, the square describe ' on the hypothenuse is equal to the sum of the squares of the other two sides, Thus: if the base is 4 feet, and the perpendicular 3 feet, the hypothenuse is /42 2 - o = 5 feet. This point is also illustrated by the adjoining diagram. What is a triangle? A right angled triangle? Which side is th% hypothenuse? To what is the square of the hypothenuse equal? FThis deiniition has reference only to rectilinear triangles; a curvilinca triang'e is bounded b y curved lilies. S20 OMMON ARITHMETIC. Sect. XV From this we derive the following considerations: a. 1st. If we add the square of the base to the square of the perpendicular, and extract the square root of their sum, the result will be the hypothenuse. b. 2nd. If we subtract the square of the base from the square of the hypothenuse, and extract the square root of the remainder, the result will be the perpendicular. c. 3d. If we subtract the square of the perpendicular from the square of the hypothenuse, and extract the square root of the remainder, the result will be,the base. REMARK 1.-When the base and perpendicular are equal, the hypothenuse may be found by multiplying the base (or perpendicular) by 1.4142. This is the hypothenuse of such a triangle, the equal sides of which are unity, or 1. 2 -The base and perpendicular are sometimes called the legs of the triangle EXERCISES FOR THE SLATE. 1. A room measures 16 feet in length, and 12 feet in width. How long is a line that will just reach from corner to corner? Operation. 162 =256 122 =144 400(20 feet. Ans. 4 00 2. A ladder 30 feet long will just reach the top of a wall by placing its foot 24 leet distant. Required-the heighth of the wall. Ans. 18 feet. 3. A certain pole is 60 feet high. How far from the foot of it will a line 100 feet long reach the glound, by being fastened at the top? Ans, 80 feet. 4, The distance between the foot of two rafters is 32 feet, and the heighth of the ridge above the plate on which the rafters rest is 12 feet. Required-the length of the rafters. Ans. 20 feet. 5. If a square field contains 35 acres, 25 sq. rds., how many rods d es it measure on a side? What is the distance from the centre to: ch corner? Ans. to the last. 53.033+ rds. 6. The walls of a certain building are 32 feet in length, 30 feet How then do we find the hypothenuse w hen the other two sides are given? How do we find the perpendicular when the other two sides are given? How do we find the base when the other sides are given? Art 2. EVOLUTION. 321 in width, and 24 feet in heighth. Required-the length of a line that will connect the two corners farthest distant fiom each other. Ans. 50 feet. 7. Suppose a pole 60 feet long to be so planted between two straight trees, as to reach a limb of one 48 feet high, and without moving the foot, to reach a limb of the other 36 feet high. How far apart are the trees? Ans. 84 feet. Obs. 24. A mean proportional between two numbers is found by extracting the square root of their product. Thus, a mean proportional between 2 and 8 is 4, because 8X2= 16; and /16=4. PROOF.-2: 4:: 4: 8; 2 X 8= 4X4. REMARK.-This rule is evident from the fact that a mean proportional forms the two means of a proportion. (Sect. XII. Art. 2. Obs. 7.) But the product of the extremes is equal to the product of the means. (Sect. XII. Art. 2. Obs: 9.) Therefore, the square root of the product of the extremes is the mean proportional. 8. What is a mean proportional between 16 and 64? Ans. 32. 9. What is a mean proportional between 68 and 612? Ans. 204. To find the side of a square equal in area to a given surface. 6 ft. lon r. IOTT O1.... 9tl, ft. ]ong. I I * *-J 71, en I-? HE= ________; ^^ __ __ __ __ ' Ct -'-1,! -- I 4. CD 1L 111 1 9X4=36 sq. ft. 6X6=-36 sq. It. We perceive from the diagrams that a square 6 ft. long, and 6 ft. wide, contains 6X6= 36 sq. ft. We also perceive that a rectang!e 9 ft. long, and 4 ft. wide, contains 9X4=36 sq. ft.; (Sect IX. Art. 2. Obs. 15.) or a square 6 ft. long and 6 ft. wide, is equal in area to a rectangle 9 ft. long, and 4 ft. wide. But the side of a square is formed by extracting the square root of its contents, or superficial area. (Obs. 14.),-How do we find a mean proportional between two numbers? Show why this rule is correct. How do we find the side of a square equal in area to a given surface? Show why this is correct. 15A 322 COMMON ARITHMETIC, Sect. XV Therefore, if we extract the square root of the area of a rectangle, we shall have one side of a square 'of equal area, as the result. It is ilso evident, from Obs. 14, that the square root of t/ie area of any Furface is one side of a square surface of equal area. HIenceTo find the side of a square equal in area to a given surface: Obs. 25. Extract the square root of the given area. 10. A certain field is 48 rods long, and 12 rods wide. Required-the length of one side of a sqluare field containing the same quantity of land? Ans. 24 rds. 11. A parallelogram is 16 ft. long, and 9 ft. wide. What is one side of a square equal in area? Ans. 12 ft. 12. A circular field contains 2025 sq. rds. How many rods on one side of a square field containing the same quantity? Ans. 45. 13. A certain triangular field contains 10 acres, 128 sq. rds. Required-one side of a square field containing the same quantity. Ans. 41.569+ rds. 14. If a certain square field measures 30 rds. on a side, what will be the length of one side of a square field containing 4 times as much? Ans. 60 rods. Operation. 30 30 900 sq. rds. = contents of the given field. (Sect. ]X. Art. 2. 4. [Obs. 15.) 3600 sq. rds. = contents of the required field. o/3600'=o60 rds., one side of the required field. (Obs. 14.) 15. If the above field measured 28 rods on a side, what is the length of one side of a square field containing 9 times as much? 16 times as much? 25 times as much? 64 times as much? Ans. in order. 84 rds.; 112 rds.; 140 rds.; 224 rds. 16. If a hall measures 40 feet in length, and 10 feet in width, what is the length of one side of a square room 9 times as large? as large?, as large? Ans. in order. 60 ft.; 0 ft.; 15 ft. 17. I have 512 sq. rds. of l:nd in afield which is twice as long as it is wide. Required its leng(th and width? Suggestion. —If tlhe field is divided at the middle of the sides, it Art. 2. EVOLUTION. 323 will form two equal squares, each containing 512~ 2 = 256 square rods. Ans. Length 32 rds.; width 16 rds. 18. A nurseryman wishes to set out 1280 frut trees, having the length of his orchard 5 times its width. How many rows must he have, and how many trees in a row? Ans. 16 rows, and 80 trees in a row. 19. A certain room contains 300 sq. ft., and its width is 3 of its length. Required its length and width? Ans. Length 20 ft., Width 15 ft. CIRCLES. Obs. 26. A CIRCLE is a portion of space enclosed by a curved line, called the CIRCUMFERENCE, every part of which is equally distant from a point within, called the CEN- B TER.A CA A straight line passing through the center from one part of the circumference to another is called the DIAMETER. E A straight line from the center to any part of the circumference is called a RADIUS. Thus, in the above figure, the point A is the Center; the line B E C D is the circumference; the line B A C is the Diameter; and the line A D is the Radius. REMARK 1. It will be perceived that A B, and A C are radii, and that the radius is half the diameter. For this reason all radii of the same circle are equal. 2. The learner must not confound the circle with the circumference. The circle is the surface enclosed, and the circumference is the line which enclosesf it. Obs. 27, The areas of circles, (and also of all othF( urfaces,) are to each other as the squares of their like dimensions.. NOTE.-This proposition cannot be demonstrated without a knowledge o geometry. To find the dimensions of a circle which shall contain 2, 3, 4, 1-, 1, &c. times as much as a given circle: Obs. 28. Square the diameter, or circumference (whichever is given,) of the given circle, multiply the square by the given proportion. What is a circle? What is the circumference? The diameter? The radius? What is the ratio of the radius to the diameter? What is the differenee between the circle and the circumference? What relation have the areas of circles to each other? How do we find the dimensions of a circle which shall contain 2, 3, 4, i, k, i, &c., times as much as a given circle? 324 COMMON ARITHMETIC. Sect. XVI and extract the square root of the product; the result will be the similar dimensions of the required circle. 20. A circle measures 6 inches in diameter. Required the diameter of one 4 times as large? Ans. 12 inches. 21. The circumference of a circle is 8 feet. Required the circumference of one 9 times hs large? Ans. 24 feet. 22. The circumference of a circle is 15 feet. Required the circumference of one 27 times as large? Ans 75 feet. 23. A gentleman has two circular ponds on his farm; one of which is 16 yards in diameter, and the other is 16 times as large. Required the diameter of the larger? Ans. 64 yds. 24. The diameter of a circle is 16 feet. Required the diameter of a circle but - as large? Ans. 8 feet. 25. The circumference of a circle is 64 feet. Required the circumference of a circle but y as large? Ans. 16 feet. SECTION XVI. MIENSURATION. Def. MENSURATION teaches the art of finding the area of surfaces or the contents of solids. The area of a surface is the space enclosed by its boundaries. (Sect. IX, Art. 2, Obs. 14.) ARTICLE 1. MENSURATION OF SURFACES, Obs. I. When we speak of a surface we mean the face of any thing. A SURFACE, therefore has length and breadth, but not thickness. ^ " CaSE 1. To find the area of a Square, Rectangle, &c. Obs. 2. Multiply together its length and breadth, (Sect. IX, Art, 2, Obs. 15.) IREMARK. The area of a surface is expressed by Square Measure; as, square fee,- square rods, square miles, &c,.. Ex. 1. How many square feet in the floor of a room 22 ft. long ^id 16tt. wide? Ans. 352. What is Mens. lration? What is the area of a surface? When we speak of a surface,wht do ve mean? What then has a surfaea? How do we find the area of a square, recta ngle, &c, Art. 1. MENSURATION. 325 2 A square field measures 34 rds. on each side. Required, its contents. Ans. 1156 sq. rds, = 7 A. 36 sq. rds. 3. A certain field is 125 rods long and 65 rds. wide. Required, its contents, it being in the form of a rectangle. Ans. 50 A. 125 sq. rds. CASE 2. To find the area of a Triangle. 1. How many square feet in a board 12 ft. long, and 4 ft. wide at one end, and tapering to a point at the other? Ans. 24. A 12 ft. D The surface of the board would ^'^^pv~~ Ievidently describe a triangle (which we will suppose to be right-angled,) like the triangle A B C in the diaB C gram. Now by drawing the lines A D and D C, parallel to B C and A B, we form another equal and similar tiiangel A D C, and also the rectangle A B C D. The length and width of this rectangle are the same as the base and perpendicular of the triangle A B C; that is 12 feet long, and 4 feet wide; and its area is 12 X 4 = 48 sq. ft. (Obs 2.) Therefore, as the rectangle is composed of two equal triangles, the area of each must be 48- 2 = 24 sq. ft. HenceTo find the area of a right angled triangle: Obs. 3. Multiply together the base and perpendicular, and take half their product. " REMARK 1. Parallel lines are those which are every where equally distant from each other. 2. From the above demonstration we perceive that every Triangle is half a Rectangle havingthe same base and altitude. 2. What is the areaof a triangle the base of which is 40 feet, and the perpendicular 24 feet? Ans. 480 sq. ft. 3. How many acres in a triangular field the side or base of which is 124 rods, and the end or perpendicular, 40 rods? Ans. 15 A. 80 sq. rds. 4. How many acres in a triangular field, the base of which is 150 rods, and the perpendicular 50 rds? Ans. 28 A. 20 squa;e rods. When the triangle is not a right angled triangle, and we have the three sides given we proceed as follows: How is the area of a surface expressed? How do we find the area of a right angled triangle? What are parallel lines? What is every triangle? 326 OMMION ARITHMETIC. Sect. XVI Obs. 4. Add together the three sides awd take half tieir sum. From the half sum take the three sides severally: Finally, multiply together the half sum and the three remainders, and extract the square root of the product. RIMARK. When the perpendicular height is given we work by Obs. 3. NOTE.-The demonstration of this rule cannot be understood without a knowledge of geometry. 5. Required the area of a triangle the sides of which are 6, 7 and 8 feet, respectively? Ans. 20.333 + sq. ft. 6. Required the area of a triangle the sides of which are 24, 30, and 36 feet respectively? Ans. 357.176 + sq. ft. 7. Required the area of a triangle the sides of which are 26, 28, and 30 feet, respectively? Ans. 336 sq, ft. 8. How many sq. yds. of plastering are there in a triangle whose sides are 30, 40, and 50 feet, respectively? Ans. 66i. CASE 3. To find the area of a Trapezoid. / --.... \ O hObs. 5. A figure having two sides parallel, and the other two not is \. called a Trapiezoid. 1. A man has a board in the form of a trapezoid; the length of one side is 60 inches, and of the other side 48 inches, and its width is 24 inches. ' Required its area? Ans. 9 sq. ft. f 60 X 48 108 inches=the length of both sides. 108 -2.-54 inches=the mean, or average length. (Sect. IX, Art. 2, Obs. 24.) Operaionn 54X 24=1296 sq. in.=the area of the surface of the board. (Obs. 2.) 1 296- t44=9 sq. ft. HenceTo find the area of a trapezoid: Obs. 6. iubltiply half the sum of the parallel sides by the width. 2. Required the area of a trapezoid whose parallel sides are 64 and 57 feet, and the width, 39 feet. Ans. 2359' sq. ft. 3. A farmer has a field in the form of a trapezoid: the parallel sides measure 40 and 50 rods, and the width measures 20 rods.How many acres does it contain? Ans. 5 A. 100 sq. rds. CASE 4. Circles and Polygons. When the triangle is not right angled how do we find i' area? What is a Trapezoid? How do we find the area of a Trapezoid? Art. 1. IMENSURATION OF SURFACES. 327 Obs. 7. A iOLYGOoN is a fiqure bounded by straiqht lines. A Triangle is a polygon; also, a square. PROBLEM 1. To find the area of a regular polygon: Obs. 8. Multiply the perimeter by half the perpendicular, let fall from the centre to the middle of one side. A Demonstration.-In the diagram we draw the right angled triangle 0 S A within the E B polygon A B C D E. In the same manner the entire polygonff imy be divided into \ 0 / right angled triangles. But the area of the triangle A S 0 is found by multiplying the base A S by half the the perpendicular S D C 0; (Obs. 3.) and the area of the othertriangles is found by multiplping their bases by half of an equal perpendicular. Therefore, since the sum of their bases forms the perimeter of the polygon, the rule is evident. REMARK 1. A Regular Polygon is one in which each corner is equally distant from the center. The middle point of each side is also equally distant from the center, and the sides are all of equal length. 2. A polygon of five sides is called a Pentagon; of six sides, a Hexagon; of seven sides, a Heptagon; of eight sides, an Octagon; of nine sides, a Nonagon; and of ten sides a Decagon. 1. I-ow many square feet in the floor of a school house in the form of an octagon, each side measuring 10 feet, and the line from the center to the middle of one side measuring 12 feet? Ans. 480 sq. ft. 2. A man has a field in the form of a hexagon, each side of which is 60 rods, and the distance from the center to each corner is 60 rods. Required. its area? Ans. 68 A. 8 sq. rds., nearly. PROBLEM 2. To find the area of irregular Polygons: Obs. 9. Draw diagonals dividingr tlhe;olygon into triangles, rectangles, c&c. Then find the area of each of these, separately, andz add these several areas together. REMARK. A DIAGONAL is a line joining two angles not adjacent to each other. What is a Polygon? Give examples. How do we find the area of regular polygon? Demnostrate this rule? Whut is a iegular polygon? What is a polygon of five sides called? Of six sides? Of seven sides? Of eight sides? Of nine sides? Of ten sides? How do we find the area of an irregular polygon? 328 COMMON ARITHMETIC. Sect. XVI B _ 'C E X D 3: There is a certain field in the form of the adjoining diagram. The side A B measures 26 rds.; the side B C measures 28 rds.; the side C D measures 12 rds.; the side D E measures 20 rds.; and the side E A mea15 rds.; diagonal A C measures 30 rds.;-and the line F G measures 10 rds. How many acres in the field? Ans. 3 A, 106 sq. rds. PROBLEM 3. Circumscribed and Inscribed Polygons. A Obs. 10. When a Polygon is drawn about a circle so that the middle point of F a each side touches the circumference, it is / called a Circumscribed Polygon. When a polygon is drawn within a circle so that each corner shall touch the circum- ference, it is called an Inscribed Polygon. E Thus, in the diagram, A B C D E F is a circumscribed polygon, and a b c d ef is d an inscribed polygon. Also, in the same diagram, the circle is said to be inscribed within the polygon A B C D E F, and to be circumscribed about the polygon a bc d ef. REMARK 1. The term Circumscribed is derived from two Latin words, Circum, about, and Scriho, to write, or draw, The term Inscribed is derived from the same word Scribo, and In, within. 2. The lines which bound the polygon are termed the Perimeter of thec polygon. If in the above diagram we bisect, or divide into two equal parts each side of each polygon it is evident that each polygon will appt oach nearer the circumference, and consequently nearer each other. Therefore, if the sides of each polygon be bisected indefinateWhat is a diagonal? What is a circumscribed Polygon? An Inscribed Polygon? What is said respecting the circle in the diagram? From what is the term circumscribed derived? What does it mean? From what is the term inirribedderived? Whatdoes it mean? What is the Perimeter of a Polygon? What effect does it have upon the perimeter of a circumscribed and inscribed olygon, to bisect each side indefinately? Where will the perimeters meet? V;hy? Art. 1. MENSURATION OF SURFACES. 829 ly, the perimeters of the two polygons will meet. But as the inscribedpolygon cannot fall without the circumference, (for then it would be a circumscribed one,) and as the circumscribed polygon cannot fall within the circumference, (for then it would be an inscribed one,) they must meet upon the circumference. HenceObs. 11. A Circle may be defined as a polygon uith an indefinite number of sides. REMARK. This definition is not strictly correct, as the side of a polygon, and the circumference of a circle cal never coincide so as to form one and the same line. It is sufficiently correct, however, for all practical purposes. PROBLEM 4. The Quadrature of the Circle. The problem is not only difficult, but impossible to solve exactly, although we can approximate very nearly. What is meant by it, is, to determine the area of a circle, the diameter of which is equal to the sfde of a given square; that is, to find the area of a circle inscribed in a square. The first step is to find the ratio of the circumferance to the diameter. The process used to obtain this is too complicated for a treatise on Arithmetic; therefore we will merely tell how it is done, and take the result as correct without further trouble. The learner must, however, bear in mind, that the exact ratio can never be ascertained, neither the area of the circle, and all good mathematicians have long since abandoned both as impossible. The process used, is, to inscribe and circumscribe a circle with regular polygons, and then to bisect these until the sides of the two polygons and the circumference of the circle all seem to unite in one common line. They do not exactly coincide, (Obs. 11. Rem.) but come sufficiently near to render accuracy almost absolutely certain. The circumference of the circle must always be between the perimeters of the two polygons, and since the ratio of the diameters to the perimeters of the polygons can be ascertained, the ratio of the diameter to the circumference of the circle must lie between these two ratios. ARCHIMEDES a Grecian, first commenced the investigation of this problem, and by increasing the number of sides of the polygon to 32768, he found the ratio to be between ')4 and 3l 1. 3 22, or the ratio of the circumference to the diameter is as 22 to 7. METIUS How then may a circle be defined? Is this strictly correct? Why not? What is meant by the quadrature of the circle? Is this problem capable of being solved exactly? What is the first step in solving it approximately?What is the process used to find this? Can the perimeters of the polygons, and the circumference of the circle ever coincide? Where does the circumference of the circle lie? Where then must the ratio of the diameter to the circumference bet Why? 330 COMMON ARITHMETIC. Sect. XVI a German, afterwards found the ratio to be as 355 to 113. Van Ceulen, a Dutch mathematician, carried the process much farther, and found that if the diameter was 1, the circumference would be less than 3.14159265358979323846264338327950289, and greater than 3.14159265358979323846264338327950288. Later mathematicians have carried the process still farther, and have ascertained the ratio of the diameter to the circumference to be as 1 to 3.141592653589793238462643383279502884197169399375705820 -974944592307816406286208998628034825342117067982148086 -5132823066470938446460955051822317253594081284802. The ratio then may safely be put down as 3.141592. This gives the exact ratio to 5 decimal places, and making the error as small as Tlfo of- O. For all practical purposes, 3.1416 may be us: d, changing the 9 in the fifth order, into 6 in the fourth order of decimals. HenceTo find the circumference of a circle when the diameter is given: Obs. 12. Mlultiply the diameter by 3.1416; or, forgreater accuracy, by 3.141592. 1. A certain wheel is 36 in. in diameter. Required its circumference. Ans. 9 ft. 5.0976 in. 2. If the diameter of a circle is 15 ft., what is the circumference? Ans. 47.124 ft. 3. If the earth is 7912 miles in diameter, what is its circumference? Ans. 24856.3392. To find the diameter of a circle when the circumference is given. Obs. 13. Divide the circumfereence by 3.1416. Or, Multiply the circumference by.31831. REMARK. Since the circumference is I multiplied by 3.141f, the diameter is 1 divided by 3.1416. 1 divided by 3.1416 equal.31831. 4. If the circumference of a circle is 37.6992 ft., Gwlat is the diameter? Ans. 12 ft. 5. If the circumference of a circle is 60 ft., whlat is its diameter? Ans. 19.0985 ft. 6. If the circumference of a circle is 42 ft., 6 in., what is its diameter? Ans. 13 ft. 6.337, in. 7. If a tree measures 12 ft. 9 in. around, wliat is tle distance though it. Ans. 4.0584 ft. To what may this ratio be safely put down? IHow sinill iQ the error in this case? What number is used for practical purpos.s. Howv then do we find the circumference of a circle when the diameter is given? flow (o we find the diameter when the cireuinference is given? Art. 1. MENSURATION OF SURFACES. 331 PROBLEM 5. The area of the Circle, In the diagram (Obs. 8.) we have shown the right-angled triangle a. S 0, the area of which is equal to one-half the perpendicular (O S) multiplied by the base a S. (Obs. 3). The whole polygon may be separated in triangles in the same manner, and the area of each is found by the same process. Now if the sides of the polygon are bisected until they are indefinitely increased, its perimeter will be the circumference of the circle, and the perpendicular O S will be the radius of the circle. (Sect. XV, Art. 2, Obs. 26.) HenceObs. 14. The area of a circle is found by multiplyiug the circumference by half the radius.* 1. If a circle is 15 ft. in diameter, and 47.124 feet in circumference how many square feet does it contain? Ans. 176.715. The learner will remember that the radius is half the diameter Sect. XV, Art. 2, Obs. 26, Rem. 1.) 2. If the diameter of a circle is 25 ft., and the circumference 78.54 ft., what is its area? Ans. 490 sq. ft., 126 sq. in. As the radius is half the diameter, half the radius is - of the diameter, the area of a circle is equal to the product of the circumference into 4 the diameter; or which is the same thing - of the product of the circumference and diameter. But the circumference is equal to the diameter multiplied by 3-1416; therefore, (Obs. 12.) the circumference multiplied by the diametcr is the same as 3.1416 times the diameter into itself, or 3.1416 times the square of the diameter; and the area of the circle is I of this. or 3.1416 4 =.7854 times the square of the diameter. Hencea. The area of a circle may be fcuzid ly multijlying the square of the diameter by.7854. REMARK. The learner will observe from this that when the diameter of the circle is 1, the area is.7T54. How do we find the area of the circle when tile circumrference and diameterare both given? Demonstrate this rule. When the diameter only is given how do we find the area? Demonstrate this rule. WNhen the circumference only is given how do we find the area? Demonstrate thia rule? When we have given the area howv do we find the dimenlsioiis? *In this caFe as the circuiiference is the p;erimeter c-f a Tol; ' n l w-ilh a;: indpfini'e rn!iler of sides. the circle is conposed of al illndeliite lnltil ec of tlianlzI(', ot' \w h ti he ar a of each is fouirn by 019. 3, and the sum of all these areas is e u!al! to tl!( a-ta of the cir lel. Secrt. IV, Art. 4, Obs. 4, Rein. 2). Or, as tie perpendieilla!s to all these trlanlles are equal, the suiti of their bases, (or circumferences) multiplied iy ha'f tile iertlendicular (or radius) ltust give the area. 332 COMMON ARITHMETIC. Sect. XVI 3. Required-the area of a circle the diameter of which is 60 ft. Ans. 1963- sq. ft. When the circumference is 1, the diameter is.31831, (Obs. 13, Rem.) and the area.318312 X.7854. But.318312 X.7854 =.08 nearly. Henceb. Ihe area of a circle may be found by multiplying the square of the circumference by.08; or, for greater acuracy by.0795, 4. If the circumference of a circle is 15 ft. what is its area? Ans. 18 sq. ft. 5. If the circumferance is 28 ft. what is the area? Ans. 62.72 sq. ft. 6. The area of a circle is 176.715 sq. ft., what is its diameter and circumferance? Ans. Dia. 15 ft., Cir. 47.124 ft. Oe. n 176.715 -.7854 = 225; /2255, = 1 ft. diameter. 1n 5 X 3.1416 - 47.124 ft. circumference. REMARK. This question is exactly the reverse of those under Obs. 14, a. Therefore, the rule is evident. HenceWhen we have the area of a circle given to find the dimensions: Obs, 15. Divide the area by.7854, and extract the square root of the quotient for the diameter. The circumference is thenfound by Obs. 12. 7. A square field contains 490.875 sq. rds. Required-the diameter of a circular field containing the same quantity. The circumference. Ans. Dia. 25 rds., Circle 78.54 rds. 8. How long must a halter be, which being fastened to a post at tle center, will just allow a horse to feed on lialf ap acre of ground? Ans. 83 ft. 3.1476 in. PROBLEM 6. To find the Area of a Square inscribed in a Circle. The rule is deduced from the following proposition: 1. The area of a square inscribed within a circle is one-half the area of a square circumscribed about the same circle. From what principle is the Rule for finding the area of a square inscribed in a circle deduced? Demonstrate thisprinciple. To what is thearea of an inscribed square equal? Art. 1. MENSURATION OF SURFACES. 333 G B F Demonstration.-Let A B C D be. -- the inscribed, and E F G H tle circumscribed square. It will be per- ceived that the angles of the inscri- bed square are located at the middle C _ of the sides of the. circumscribed \ square. Draw the diagonals B 0 D and C O A. These divide the circumscribed square into four equal smaller squares. The sides of the H D E inscribed square divide each of these smaller squares into two equal and right angled triangles. (Obs. 3, Rem.) Therefore, since both of these triangles are included in the circumscribed square, and but one of them in the inscribed square, the circumscribed square contains eight triangles, and the inscribed square but four-hence, the proposition is correct. Now the side of the circumscribed square is equal to the diameter of the circle, and its area is the square of this diameter.(Obs. 2.) HenceObs. 16. The area of a square inscribed within a circle is one half the square of the diameter of the same circle. a. The area of an inscribed square is also equal to the diameter of the circumscribed circle multiplied into one half of itself. This is evident from an examination of the above diagram. We perceive that the inscribed square is composed of the two equal triangles, A B C and A D C. (Obs. 3, Rem. 2.) Now the area of each of these triangles is equal to the hypothenuse A C, multiplied by half the perpendicular height B 0, or O D, (Obs. 3.) But 2 of B + - of O D is equal to either B O or O ), since B O and O D are equal. Hence-the area of the inscribed square is equal to A C (the diameter of the circumscribed circle) multiplied by B 0, or 0 D, (half. this diameter.) 1. Required- the area of a square inscribed in a circle 12 feet in diameter: Ans. 72 sq. ft. Solution.-12 X 12 = 144; 144 -- 2 - 72. Or, 12 X 6 72. 2. Required-the cost of a log when squared, at $0.40 per To whatelse is it equal? Why? How do we find the side of an inscribed square? Why is this correct? When the diameter of the circle only is givei how do we find the side of the inscribed square? Demonstrrate this rule. When the circumference only is given how do we find the area? Demonstrate this rule? To what may this principle be applied? 334 COMMON ARITHMETIC. Sect. XVI square foot, which measures 3 ft. in diameter, and 22 ft. in length. Ans. $39.60. Suggestion.-Find the size which the log will square, and multiply this by the length for the number of sq. ft. 3. How much would a log cost at $4.50 per 100 ft., which is 4! ft. in diameter, and 43 ft. inklngth? Ans. $21.87. 2. How much would a log cost which is 4- ft. in diameter and 64 ft. long, at 650 per M.? Ans. $32.40. 5. How much would a log cost which is 5- ft. in dia., and 80 ft. long at $75 per M.? Ans. $82.683. PROBLEM-TO find the side of a Square inscribed in a Circle, The side of a square is equal to the square root of its are a (Sect. XV, Art. 2, Obs. 25). HenceTo find the side of an inscribed square: Obs. 17. Extract the square root of its area found according to Obs. 16. From examining the diagram for illustrating the preceding case, we perceive that the inscribed square consists of four equal rightangled triangles, and that eacii of these triangles has two of its sides equal, since these sides are radii of the same circle (Sect. XV, Art. 2, Obs. 26, Rem. 1). The hypothenuse of these triangles then is equal to one of these sides (or radii) multiplied by 1.4142. (Sect. XV, Art. 2, Obs. 23, Rem 1). But this hypothenuse is the side of the inscribed square, and twice the other side of the triangle, (radius) is the diameter of the circumscribed circle. (Sect. XV, Art. 2, Obs. 26, Rem. 1). Therefore, this diameter multiplied by.7071 (1,4142 2) must be the side of the inscribed square.(Sect. VI, Art 1, Obs. 30.) Again, when the circumference is 1 the diameter is.31831, (Obs, 13, Rem ) and when th;: diameter is I the side of the inscribed square is.7071; therefore when the circumferance is 1, the side of the inscribed square is,31831 X.7071 =-=.2251 nearly, HenceTo find the side of an inscribed square: a. Multiply the diameter by.7071, or the circumference by.2251. 1. How large a square can be hewn from a log 2 ft. in diameter. How many dimensions are used in measuring lumber? What are they?What is the thickness considered? How is lumber more than an inch in thickness counted? When the waste by sawing is considered how do we proceed? When the lumber ialess than an inch in thickness how do we proceed? How do we find the dimeusions of a circumscribed circle when the side of the inscribed square is given? Demonstrate this rule. Art. 1. MENSURATION ( F SURFACES. 335 Solution.-22=4; 4 2 = 2; /2 = 1.4142. Or,.7071 X 2= 1.4142 as before. Ans. 1.4142 ft. 2. How many inches will a log square that is 120 inches circumference? Ans. 27.012 in. Solution.-120 X.2251 = 27.0120. 3. I wish a sill 18 inches square, and find a tree 25 in. in diamcter. Will it answer my purpose or not? Ans. It will not as it will square but 17.67 in. 4. I find another tree measuring 80 in. in circumference. Will this answer my purpose? Ans. It will, as it squares 18.008 in. This principle may be applied to finding the quantity of lumber a log will make. 5. A log measures 24 ft. in length, and 15 inches in diameter how many feet of inch lumber will it make, allowing no waste for sawing? Solution.-15 X.7071 = 10.6065 inches that the log will square. Then 24 ft. = 288 in.; 10.60652= area of the end, and 10.60652 X 288 =- square contents in inches, which divided by 144 = 214.389 + sq. ft. Ans. The operation may be shortened somewhat by canceling. REMARK 1. In measuring lumber but two dimensions are generally used, viz: length and width. The thickness is considered a unit 1 inch being the standard. Lumber more than an inch in thickness, is counted an additional thickness for every additional inch. When the waste by sawing is considered, it should be deducted from the result. 6. A log measures 25 feet in length, and 12 inches in diameter. How much two inch lumber will it make allowing it to lose - of an inch in sawing? Ans. 132.3225 sq. ft. Solution.-We find the contents of the log, as in the last example, to be 150 sq. ft. nearly. The log squares 8.4852 in., and as the lumber is 2 inches thick, the saw passes through four times, which makes 1 inch waste. This 1 inch measures 8.4852 in. in width, and 300 in, (25 ft.) in length which makes 17.6775 sq. ft. loss by sawing, and 150-17.6775 = 132.3225 sq. ft of lumber. REMARK I. Had the width of the lumber been given, and the log been sawed both ways across the end, an additional waste would have accrued from sawing which would have been deducted. When the lumber is less than an inch in thickness the calculations for the quantity are made by the inch as usual, although the allowance for waste in sawing must be made according to the thickness of the lumber sawed. 336 COMMON ARITHMETIO. Sect. XVI 7. A log measuring 2 ft. 8 in. in diameter and 24 ft. in length, is to be sawed into lumber, 6 inches wide, and 1 inch thick. Required the cost of this lumber at $1.75 per C., the waste being -5 of an inch for sawing. Ans, $5.17, nearly. 8. Twelve logs, the average diameter of which is 2 ft. 6 in., and the lengths, 12, 15, 10, 22. 14, 16, 20 18, 15, 14, 24 and 20 ft. respectively are to be sawed into lumber 1 inches in thickness, and 6 inches in width. Required the cost of the lumber at $22.50 per M, allowing 3e of an inch waste for sawing. Ans. $142.392+-. 9. How much half inch lumber, 4 itlches in width, can be sawed from a log 3 ft. in diameter, and 25 ft. long, allowing - of an inch waste for sawing? Ans. 589.488+sq. ft. PROBLEM 8. To find the Circumference or Diameter of the Circumscribed Circle when the side of the Inscribed Square is given. By examining the diagram for finding the area of the inscribed square, we perceive that the diagonal of this square, is the hypothenuse of a right-angled and equal sided triangle, the legs of which are the sides of the inscribed square. But this diagonal is the diameter of the circumscribed circle,and is equal to either of the sides of the inscribed square multiplied by 1.4142. (Sect. XV. Art. 2, Obs. 23, Rem. 1). Again, the circumference is equal to the diameter multiplied by 3.1416. (Obs. 12.) But the diameter is equal to the side of the inscribed square multiplied by 1.4142. therefore the circ. is equal to the side of the inscribed square X 1.4142 X 3.1416. 1.4142 X 3.1416 = 4.443 nearly; therefore the circumference is equal to the side of the inscribed square multiplied by 4.443. HenceTo find the dimensions of circle when the side of an inscribed square is given: Obs, 18. Multiply the side of the inscribed square by 1.4142 for the diameter, or, by 4.443for the circumference. 1. How large must a tree be in diameter and circumference to square 10 inches? Solution.-10 X 1.4142 = 14.142 in. dia, and 10 X 4.443-= 44. 43 in. Circ. Ans. How do we find the side of asquare of equal area to a given circle? Demonstrate this rule. How do we find the dimensions of a circle of equal area to a given square? Demonstrate this rule. Art. 2. MENSURATION OF SOLIDS. 337 2. I wish a stick of timber 12 inches square, and find a tree 16 inches in diameter. Will it answer my purpose or not. Ans. It will not, as to square 12 in. it must be 16.9704 inches in diameter. 3. How large must a tree be in circumference to square 12 inches? Ans. 53.316 inches. 4. How large must a tree be in diameter to square 15 inches? Ans. 21.213 inches. We learn (Obs. 14. a. Rem.) that if the diameter of a circle is 1, the area is.7854; therefore, the side of a square of equal area is /.7854 =.8862; and.8862- 3.1416 =.2821 nearly. HenceTo find the side of a square equal in area to a given circle: Obs. 19. Multiply the diameter of the circle by.8862; or the circumference by.2821. 5. A circle is 12 ft. in diameter. Required the side of a square of equal area. Ans. 10.6344 ft. 6. A circle is 40 ft. in circumference. Required-the side of a square of equal area. Ans. 11.284ft. Since the diameter of a circle multiplied by.8862, or the circumference multiplied by.2821, will give the side of a square of equal area, it is evident that the side of a square divided by.8862 will give the diameter,or by.2821 will give the circumference of a circle of equal area. Then assuming unity, or 1 as the standard, 1 -.8862 = 1.128 as the diameter, and 1 -.2821 = 3.545 nearly, as the circumference of a circle, the area of which is 1, or equal to the area of a square the side of which is 1. Hence'io find the dimensions of a circle, the area of which shall be equal to the area of a given square. Obs. 20. Multiply the side of a given square by 1.128 for the diameter; or, by 3.545for the circumference. 7. The side of a square measures 150 feet. Required-the diameter and circumferenc of a circle of equal area. Ans. Dia. 169.2 ft. Circ. 531.75 ft. ARTICLE 2. Mfensuration of Solids, Obs. 1. In Mensuration of solids, two things are to be considered: 1st, The Mensuration of their surfaces; and Wow many things are to be considered in the menguratiom of lIids? 16 338 COMMON ARITHMETIC. Sect. XVI 2nd. The mensuration of their Solidities. A SOLID has length, breadth, and thickness. REMARK. It is aa established theorem in G.omrntry, that all solid bo lies are to each other as the cubes of hieir like dimtnn;ons. 1. If a ball weighing 6 Ibs., be 4 inches in diimeter, what is the diameter of a bali of tile same metal weighinll 48 Ibs. Ans. 8 inches. Operation,.-6 lbs.:48 lbs.: 64 (43): 512 -1 512 = 8. 2. If a ball 8 inches in diameter weighs 48 lbs., what is the weight of a ball 4 inches in d(iameter? Ans. 6 Ibs. 3. If a cubical vessel is 8 inches in length, what is the side of a cubical vessel that shall contain 27 times as much? Ans. 24 inches. CASE 1. Prisms. Obs. 2. A PRIsM is a solil having its ends, or bases equal and parallel. It is sail to be lrianyalar, Q tad- I rangular, &c., according as its sides are triangles, squares, &c. To find the area of the surface of a prism: Obs. 3. Multiply theperimeter of th.e ase by its altitude, or heighth, and to the product add the area of the bases when the entire shrface is required. Demonstration.-The sides of the prism are parallelograms, (Obs. 2.) and the area of each of these parallelograms is found by multiplying together its length and width; (Art. 1, Obs. 2). But as their lengths are equ 1, their area is equal to the uum of their width, or bases multiplied by their length, or altitnde; hence the above r.lle is evident Ex. 1. What is the entire surface of a triangular Irism, whose sides are 3, 4, and 5 leet, and whose altitude is 6 f et: Ans. 84 sq. ft. What are they? What has a solid? What relations have solid bodies to each other? What is a prism? WVhen is a prism triangular, oua rangular, &c.? How do we find the surface of a prism? Demonstrate this rule? Art. 2. MENSURATION OF SOLIDS. 339 2. What is the entire surface of a triangular prism the sides of which are 8, 10, and 12 feet, and its altitude 9 feet? Ans. 349.36-sq. ft. 3. What is the convex surface of a quadrangular prism, each side of which is 5 feet, and its altitude 6 feet? Ans. 120 sq. ft. To find the solid contents of a prism: Obs. 4. Mfultiply the area of its base by its altitude, or heighth. Demonstration.-The area of the base comprises the two dimensions, breadth and thickness, and is expressed by square measure. (Art 1. Obs. 2. Rem.) But the altitude is the length of the prism; and square measure multiplied by linear measure gives solid or cubic measure. (Sect. XIV. Rem. 1., under the Rule.) Hence, the rule is evident. 4. Required, the solid contents of a triangular prism, whose length is 12 inches, and each side of its base 3 inches? 3-+3 —3 =9; 9-2=4.5= the half sum of the sides of the base. 4.5 —3=1.5; the three remainders are equal; since the sides are equal. Operation. 4.5X.5X1.5X11.5=15. 1875; 115.1875= area of base. (Art. 1. Obs.) ^15.1875 X 12 = contents of the prism. 12 -^144. /15.1875 X 1/144 = /2187 = 46.7653 + sq. in. Ans. 5. Required, the solid contents of a quadrangular prism 36 inches in length, and each side of its base 8 inches? Ans. 1 s. ft. 576 s. in. 6. Required, the solid contents of a triangular prism, whose sides are 4, 6, and 8 feet, and its length 15 feet? Ans. 174.2812+ s. ft. Case 2.-CYLINDERS. Obs. 5. A round pr'ism hainq equal circlesfor its ends, is called a CYLINDER. To find the surface of a Cylinder: Obs. 6. JLultiply the circumference of its. end by its leng'h. How do we find the solidity of a prism? D; riionstt ait tlhis Rule. What is a Cylindcr? How do we find the surface of a Cylinder? 340 COMMON ARITHMETIC. Sect XVI1.Demonstration.-The surface of any prism whose base is a polygon, is found by multiplying the perimeter of its base by its altitude. (Obs. 3.) Now if the sides of a base are bi ected indefinitely, the perimeter will become a circle, and therefore the prism will become a cylinder. Hence, the rule is evident from Art. 1. Obs. 9. REMARK.-The above rule only gives the convex surface. When the entire surface is required, the area of the bases must be added to the result. 1. A cylinder measures 6 inches in circumference, and 18 inches in length. Required, its convex surface. Ans. 108 sq. in. 2, A cylinder is 60 inches in diameter, and 15 feet in length. Required, its entire surface. Ans. 274.89 sq. ft. To find the solidity of a cylinder: Obs. 7. Multiply the area of its base by its length. Demonstration.-The solidity of any prism whose base is a polygon, is found by multiplying the area of its base by its altitude. (Obs. 4.) Now if the sides of the prism are bisected indefinitely, the prism will become a cylinder. (Art. 1. Obs. 9.) Hence, the rule is evident. 3. Required, the solid contents of a cyl;nder 12 feet in length, and 5 feet in diameter? Ans. 235.62 s. ft. 4. How many solid inches in a common, or Winchester bushel, the diameter of which is is 18- inches, and the depth 8 inches. Ans. 2150.4252 s. in. Case 3. —CONTENTS OF BOILERS. Obs. 8. A BOILER may be regarded as a cylinder, with several smaller cylinders (flues) within it. A cubic foot contains 1728 solid inches; a wine gallon contains 231 cubic inches, and a beer gallon contains 282 cubic inches. Therefore a solid foot contains 1728- 231 = 7.48 wine gallons; and 1728- 282- =6.127 beer gallons. 1728X.7854=1357.1712 cubic inches in a cylindric foot; and 1357. 1712 - 231 = 5.875 wine gallons, and 1357. 1712 282 = 4. 812 beer gallons in a cylindric foot. 5. 875 = 57 =, 4; and 4.812 4= = HenceTo find the contents of Boilers: Obs. 9. Mutliply the square of the diameter by the length and Demonstrate this rule. What surface does this rule give? When the entire surface is required. how do we proceedl How do we find the solitity of a cylinder? Demonstrate this rule. As what may a boiler be regarded? How d wBe find the c'dntents of a bbalert Demongtrate this rule. Art. 2. MENSURATION OF SOLIDS. 47, and divide the result by 8. The result will be in 'wine gallons. For beer gallons use 24 and 5, ins'tead o' 47 ai,4i 8. Find tie contents of the Jlues in the same way, which subtract from the entire contents. REMARK 1.-The dimensions must be taken in feet by this Rule 1. How many wine gallons in a boiler 30 inches in diameter, and 25 feet long, having two flues, each 9 inches in diameter? Operation. 30 in.=2j=5 'ft.; (5)2 X25X47 -8=917.96875 entire contents. 9 in.= — ft.; (3)2 X25X47X 28 =165.234375 flues. Ans. 752.734375 gallons. We multiply by 2 because there are are 2 flues. 2. Required, the contents, in wine and beer gallons, of a boiler 36 in. in diameter, and 45 ft. long, having 3 flues, each 8 inches in diameter? Ans. 2026.875 wine gallons; 1656 beer gallons. REMARK 2. ---The contents of circular cisterns may also be found by the above rule. hi hen the cistern is square, rectangular, &c., multiply its solid contents by 7.48 for wine gallons, and 6.127 for beer gallons. In this case we do not use the numbers 47 and 8, or 24 and 5. Notice that the dimensions must all be taken in feet. NOTE.-The wine measure of 231 cubic inches to the gallon, is generally used as the standard in the United States, and is so understood in the following examples, unless otherwise mentioned. 3. Requ'red, the contents, in wine and beer gallons, of a circular cistern, 6 It. 8 in. in diameter, and 8 ft. deep. Ans. 20884 win gallons; 1706| beer gallons. 4. Required, the cont nts, in wine and beer gallons, of a rectangular cistern that contains 300 solid feet? Ans. 2244 wine gallons; 1838.1 beer gallons. 5. A rectangular cistern is to be made containing 1400 gallons; its length and breadth are 8 feet each. Required, its depth? Ans, 3 ft. nearly. Suggestion.-Find the contents of the cistern by multiplying 1400 by 231. We then have the contents, and two of the dimensions given. 6. A circular cistern measures 5 ft. in diameter, and contains 1000 gallons. Required, its depth? Ans. 6-4 ft. What is the standard for Liquid measure in the United States? 342 COMMON ARITHMETIC. Sect. XVI 7. A circular cistern is to be made to contain 1692 gallons, and to be 8 ft. deep. Required, its diameter? Ans. 6 ft. Suggestion.-By putting some character, as x, in the place of the dimension wanted in the last two examples, the method of working will be at once perceived. Case 4. PARALLELOPIPEDS. Obs. 10. A PARALLELOPIPED is a solid or prism having six faces, of 1 Iill which those opposite each other are i iiiii i equal and parallel. I J Ifii i' REMARKC.-A CUBE is a paraleleopiped, each face of which is a square. (Sect IX, Art. 1, Obs. 17.) A common brick, or chest, is a parallelopiped, each side of which is a rectangle. To find the surface of a parallelopiped: Obs. 11. Find the surface of each side separately, and add the several results together. REMARK.-The surface of any solid bounded by plane, or straight edged surfaces, is found by the same Rule. 1. Required, the surface of a parallelopiped 6 ft. long, 4 ft. wide, and 4 ft. deep? Ans. 128 sq. ft. 2. Required, the surface of a solid 4 ft. 9 in. long, and 3 ft. 10 in. wide, and 2 ft. 8 in. thick? Ans. 68 sq. ft. 104 sq. in. To find the solid contents of a parallelopiped: Obs. 12. Multiply together its length, breadth, and thickness. (Sect. IX. Art. 2. Obs. 18.) 3. What is the solidity of a parallelopiped 12 ft. long, 6 ft. wide, and 4 ft. deep? Ans. 288 s. ft. 4. Required, the contents of a solid 24 ft. long, 20 ft. wide, and 16 ft. thick? Ans. 7680 s. ft. Case 5. PYRAMIDS. Obs. 13. A PYRAMID is a solid which decreass gradually from its base, until it comes to a point at the top. This point is called the VERTEX. What is a Parallelopiped? Give examples. How do we find the surface of a Parallelopiped? What olh'r soliiJscn have their surfaces found by the same rule? How do we find the solidity of a Parailelopi,.ed? What is a Pyramid? What is the poiut at the top called? Art. 2. MENSURATION OF 190L1PI. 343 Pyramids are triangular, guadragvlar, circular, &c., according as their bases are trianoles, squaresl, circles. &c. A circular pyramid is called a CONE. Obs. 14. A FRUSTRFM is what 'rew. ips (:f/er the top of the pyramid or -one has been taken off. The;o an I bo.t)m of the frustrum are called its bases. 0 0,i I j/11;!, ^ mfi~nA B Sq. Pyramid. Triangular ilyrtiind. Cone. If in the above diagrams, the par' O-D E F he taken from the triangulaur pyramid, the remaining part ABC-DEF will be the frustrum. Also, if the part O-CD be taken from the cone, the remaining part AB-CD will be ihe frustrum. Obs. 15. The Altitude of a pyramid, or cone, is a line passing through the centre from the vertex to the base. A line drawn from the vertex perpendicularly to one si.e, is aliled its Slant Heighth. ''hus, OS is the slant heig.th of the pyramid O-ABC. To find the surface of a pyramid, or cone: Obs. 16. Afultiply the perimeter of its base by half its slant he y//th. Demonstration,-'lhe sides of pyramids are triangles, which the slant ci-hilth divides into right angled trianglcs. of which the area is found by multiplying he base by half the iperpendicular. (Art. 1. Obs. 3.) But the sum of all the bases of these triangles is the perimeter of the pyramid, which multiplied by half the common perpendicular, or slant heighth of the p. ramid gives the surface. When are pyramids said to he triangular, quadrgngular, &c.? What is a Cone? A Frustrum? What are the top and bottom of the frustrum called? What is the altitude of a Pyramid, or Cone? Its slant heighth? How do we find the surface of a Pyramid, or Cone? Demonstrate this rule? 344 COMMON ARITHMETIC. Sect. XVI Again: if the bases of these triangles are bisected indefinitely, they will become a circle, and the pyramid will become a cone. Hence — -'le surface of a cone is found by the same process. (Art. 1. Ob:,. 9.) 1. Required, the area of the surface of a triangular pyramid, the slant heithth of which is 30 ft, and each side of its base 12 ft.? Ans. 540 sq. ft. 2. Required, the surface of a quadrangular pyramid, each side of which is 9 ft., and the slant heighth 15 ft.? Ans. 270 sq. ft. 3. Required, the area of the surface of a cone 3 ft. in diameter, at the base, and its slant heigth 12 ft.? Ans. 56.5488 sq. ft. 4. Required, the area of the curved surface and base of a cone, the slant heighth of which is 12 ft. 11 in., and the diameter of its base 4 ft. 7 in.? Ans. 109 sq. ft. 70.905 sq. in. To find the surface of the frustrum of a pyramid or cone: Obs. 17. Add together the perimeters of the upper and lower bases, and multiply half their sum by the slant heighth. Demonstration.-From the diagram under Obs. 12, we perceive the sides of the frustrum of a pyramid are trapezoids. (Art. 1. Obs. 5.) But the area of these is found by multiplying half the sum of the parallel sides (or bases,) by their altitude, (or slant heighth.) (Art. 1. Obs. 6.) Hence, as the sum of the several bases forms the perimeters of the two bases of the pyramid, and their altitude, or slant heighth is the same in all, the rule is evident. Again: If the encs, or bases of these trapezoids are bisected indefinitely, they will become circles, and the frustrum of a pyramid will become the frustrum of a cone. Hence, the surface of the frustrum of a cone is found by the same process. (Art. 1. Obs. 9.) 5. Required, the convex surface of the frustrum of a triangular pyramid, of wlich the upper base measures 6 ft., and the lower base 9 ft. on a side, and the slant heightb is 12 ft.? Ans. 270 sq. ft. 6. Required, the convex surface of the frustrum of a quadrangular pyramid, of which the slant heighth is 36 inches, and the upper and lower bases 14 and 26 inches on a side? Ans. 20 sq. ft. 7. Required, the convex surface of the frustrum of a cone of which the slant heighth is 30 ft., the diameter of the upper base 12 ft., and that of the lower base 18 ft.? Ans. 1263 sq. ft. 103.68 sq. in, To find the solidity of a pyramid or cone: Obs. 18. Mulltpilg the area of its base by one-third of its altitude. How do we find the surface of the frustrum of a Pyramid, or cone? Demonstrate this rule. How do we find the solidity of a pyramid, or cone? Art. 2. MENSURATION OF SOLIDS. 345 The demonstration of this and the next rule in this case, and also of the rules of the next case, is too complex to be understood without a knowledge of Geometry. 8. Required, the solid contents of a triangular pyramid, whose base is 6 ft. on each side, and altitude 25 ft.? Ans. 130 s. ft. nearly. 9. One of the Egyp ian pyramids is square at the base, measuring 720 ft. on a side, and its heighth is 477 ft. Required, ifs contents? Ans. 82425600 s. ft. 10. Required, the solid contents of a cone 20 ft. in heighth, and 5 ft. in diameter at the base? Ans. 130.9 s. ft. 11. How many times can a conical vessel 5 inches in diameter, and 8 inches deep, be filled from a hogshead of cider? NOTE.-The learner will recollect that a gallon of cider contains 231 cubic inches. Ans. 277'4 times. To find the solidity of the frustrum of a pyramid or cone: Obs. 19. Find the area of each end, or base; Multiply the two areas together, and extract the square root of their product. Finally, add together the areas of the upper base, the lower base, and the root just found, and multiply their sum by one-third of the altitude. 12. Required, the solidity of a block whose ends are squares, each side of the lower base being 4 ft., each side of the upper base 3 ft., and its length 15 ft.? Ans. 185 s. ft. 13. Required, the solidity of a block whose ends are triangles, the sides of the top end being 2, 23, and 3' feet, the sides of the lower end being 31, 4, and 5' ft., and its length being 12 ft.? Ans. 56.5+ s. ft. 14. Required, the solid contents of a glass in the form of a frustrum of a cone, the diameter of the top being 4 inches, and of the bottom 5 inches, and its depth 9 inches? Ans. 143.72823 in. 15. Required, the solid contents of a mound in the shape of a frustrum of a cone, the diameter of the upper base being 15 ft., of the lower base 20 ft., and its heighth 30 ft.?!~i~~~~~ ~~Ans. 7264.95 s. ft. How do we find the solidity of the frustrum of a pyramid or cone? S46 COMMON ARITHMETIC. Sect XVI Case 6. SPHERES, OR GLOBnUS. Obs. 20. A GLOBE, or SPHERE, is a solid, t in which all parts of the surface are equally dis- f!'ii ''. tant from a point within called the centre; as a a. /,J cannon ball, a ball of yarn, &c. 't To find the area of the surface of a Globe, or Sphere: Obs. 21. Multiply the circumference by the diameter; or Multiply the square of the diameter by 3.1416; or, Multiply the square of the circumference by.3183. 1. Required, the area of the surface of a globe 14 in. in diameter? Ans. 4 sq. ft. 39.7536 sq. in. 2. What is the area of the surface of a sphere 47.124 ft. in circumference? Ans. 706.86 sq. ft. To find the solidity of a Globe, or Sphere: Obs. 22. Multiply the cube, or diameter, by.6236: or,.Multiply the surface by one-sixth of its diameter. 3. Required, the solidity of three globes, whose diameters are 12, 16, and 20 inches respectively? Ans 904. 808; 2144.6656; and 4188 8 s. in. respectively. 4. If the diameter of the earth is 7912 miles, what is its solidity, supposing it to be a perfect sphere? Ans. 259333411782.8608 s. miles. Case 7. TONNAGE OF VESSELS. Obs. 23. The first thing necessary in ascertaining tonnage, is to find the number of cubic feet of water displaced by the vessel. This is done by finding the s lid contents of the hull, which is equal to the product of the length, width, and depth of the vessel. A body floating on a fluid will displace as much of the fluid as is equal to its own weight; and a cubic foot of water weighs 1000 oz. Avor.; therefore, 95 cubic feet, (which is allowed for a ton of 2240 lbs.), will weigh 5937- lbs.; that is, nearly three times the weight of th freight in water is allowed to the ton. *\V;i; i G^. I * r hr.? flow;'o wve ti.d thl surface of a globp, or splr~f? HIow do we finid its so;idity? \V ai;t is lite fist tij;ilt nc(:i.-s. rv in ascertainingtonn;age? How is this folnd? What quaitity of fluid will a bold floating on its surface displace? How much does a cubic foot of water weigh? How many cubic feet are allowed to the ton? How mush will a ton weighl Art. 2. MENKSURATION OF SOLIDS. 347 There are two rules for calculating tonnage-the Government Rule, and the Caipenter's Rule. GOVERNMENT RULE: For DI uble-d?(ked Vessels. " f the vessel be double-decked, take the length thereof, from the fore part of the main stein to the cafter part of the stern part, above the upeper deck; the breadth thereof at the broadest )pa't above the main wales, half of which breadth shall be accounted the depth of such vessel; and then deduct from the length three-fJfths of the breadth, multiply the remainder by the breadth, and the product by the depth, and divide this product by 95, the quotient whereof shall be deemed the true contents, or tonnage of such ship or vessel." 1or Single-decked Vessels. "If the ship or vessel be single-decked, take the length and breadth, as above directed, deduct from the said length th.ee-fifths of the breadth, and take the depth from the under side of the deck plank, to the ceiling of the holJ; then multiply and divide as aforesaid and the quotient shall be deemed the tonnage." REMARK 1.-The Carpenter's Ru'e is the sarne as the Government Rule, except that the 3 of the breadth is not deducted from the length, but all is calcu lated. 2.-The Keel of a vessel is its main bottom in length; the beam is its greatest width from side to side of the hull; and the hold, the depth fro... the main deck to t.e bottom of the hull. 2. Required, the tonnage of a single-decked vessel, by both rules, 125 ft. keel, 38 ft. beam, and 10 ft. hold? Ans. By Gov. Rule, 2085 T.; By Car. Rule, 500 T. Operation. — of 3 - 22; 125 - 224 1022; (102. X 38 X 10)- 95= 208 —. By Carpenter's rule: (125 X38X 10) 95 = 500. 2. Required, the tonnage, by the Government rule, of a vessel 228 ft. long, 35 ft. wide, and 141 ft. deep? Ans. 10863 T. 3. Required, the Government tonnage of a double-decked vessel 300 ft. long, and 35 ft. wide? Ans. 1798l T. 4. Required, the Government tonnage of a double-decked vessel 379 ft. in length, and 35 ft. in width? Ans. 1710 T. How manly rules are there for calculating tonnage? What is the Government Rule for double-decked vessels? For single-decked vessels? What is the Carpenter's rule? Wnat is the the keel of a vsel? The beamt T4h Iheolb 348 COMMON AITHMETIC, Sect. XVI Case 8. IRREGULAR BODIES. To find the solidity of any irregular body which cannot be reduced to a regular form: Obs. 22. Immerse it in a vessel partly filled with water; then the solid contents of Itat pa(rt of/ the vessel filled by the rising of the water, will be the solidity of the body immersed. This rule is so evident that it needs no demonstration, I. A stone of irregular form being put into a tub partly filled with water, raised the water 4- inches. Required, the solidity of the stone, the diameter of the tub being 18 inches? Ans. 1145.1132 s. in. 2. A lobster being put into a bucket partly filled with water, raised the water 3 inches, The diameter of the bucket at the surface of the water, before the lobster was put in, was 9 inches, and the diameter at the surface after the lobster was put in was 10 inches. Required, the solidity of the lobster? Ans. 212.8434 s. in. 3. Required, the solidity of a brush-heap, which being put into a conical cistern, raised the water 25 inches, the diameters being 5 ft. 10 in. and 6 ft. 3 in. respectively? Ans. 59 s. ft. 12953 s. in. ARTICLE 3. GAUGING. Obs. 1. The mezJod of finding the contents of any vessel in gallons, bushels, &c., is cafled (GAu( TNG. To find the number of bushels which any vessel will contain: Obs. 2. Find thle solid corce.lts of the vessel in feet; then multiply this by 45, and divide the pi rodlct bg 5&. Demonstration.-In a solizI foot are 1728 solid inches; (see table Cubic Measure;) in a bush(l Il (rc are 2150.4 solid inches; (see table Dry Measure;) therefore a bushel contains 2 50 0 5= solid feet. Hence, the vessel will contain as many bushels as 4 is contained in its solid contents. That is, we divide by 54. a. For heaped measure use 9 and 14 instead of 16 and 56, because 68 = 4 Ex. 1. How many bushels will a box hold that is 6 feet long, 4 ft. wide, and 6 ft. deep? Ans. 1155. 2. How many bushels will a gum contain which is 12 ft. deep, and 4 ft. in diameter? Ans. 121.176. How do we find the solidity of irregular bodies? What is guaging? How do we find the number of bushels any vessel will contain? How do we proceed with heaped measure? Why? Art. 3. MENSURATION OF SOLIDS. 349 3. How many bushe's would a gum contain, which is 8 ft. deep, and measures 6 ft. in diameter at the top, and 5 ft. in diameter at the bottom? Ans. 153.155. 4. How many bushels of apples can be put in a bin 35 ft. long, 6 ft. wide, and 5 ft. deep? Ans. 6'5. NoTE.-The learner will observe that this is heaped measure. 5. How many bushels of potatoes can be put in a bin 25 ft. long, 7 ft. wide, and 6 ft. deep? Ans. 675. 6. A bin is 6 ft. wide, and 6 ft. deep; how long must it be to contain 810 bushels of apples? To contain 810 bushels of wheat? Ans. 35 ft. for apples; and 28 ft. for wheat. REMARK.-When it is desired to obtain the contents in barrels, we divide the r sult obtained by 5. If corn is the thing measured, and it is in the ear, we divide by 10; if in the husk, by 20. 7. How many gallons of water will a conical cistern contain, that is 7 ft. 9 in. deep, and the diameter 5 ft. 6 in. at the bottom, and 4 ft. 9 in. at the top? Ans. 1217.0636. NOTE.-We find the solid contents in inches, and divide this by 231, because 231 cubic inches make a gallon. (See table Wine measure.) To find the numb r of wine and beer gallons contained in a cask, or barrel: Obs. 3. Take the dimensions of the cask in inches: that is, the diameter at the bung and head, and the length of the cask. If the staves are MUCH curved, multiply the difference between the bung and head diameter by.7,; if but LITTLE curved, by.6; if they are of a MEDIUM curve, by.65; or. if they are ALMOST STRAIGHT by.55; in either case, add the product to the head diameter, and the sum will be the mean diameter, and the cask will be reduced to a cylinder. Multiply the square of the mean diameter by the length of the cask, and that product by 34 for wine measure, and 28 for beer measure, and point off four decimals in the product; the result will be in gallons, and decimals of a gallon. Demonstration.-The demonstration of this rule, until the cask is reduced to a cylinder, cannot be understood, without a knowledge of the higher departments of mathematics. After it is reduced to a cylinder, we find its solidity by multiplying the square of the mean diameter by. 7854, and its length; (Alt, 1. Obs. 11. and Art. 2. How do we find the contents in barrels? How do we find the number of wine and beer gallons contained in a cask or barrel? Demonstrate this rule. 350 COMMON ARITHMETIr. Sect. XVII Obs. 7.) and this is reduced to gallons by dividing by 231 for wine measure, and 282 for beer measure. (Obs. 3.) Thus, in wine measure,. 7854 is a multiplier, and 2 31 is a divisor; that is, we multiply by '*75 4 but ' 4.0, s 0.34. Also, in beer measure, we multiply by '851 t, and -'-7 _8.0028 nearly. Hence, this part of the rule is evident. REzMARKu.-Guaging of casks by computation is of but little u!ility, as it is now mostly done by a rod, with calculations already made, in tabular form. 8. Required, the contents, in wine and beer gallons, of a cask, of which the staves are much curved, the diameter at the head being 20 inches, and at the bung 30 inches, and its length 40 inches? Ans. 99.144 wine gallons 81.648 beer gallons. 9. How many wine and beer gallons in a cask 45 inches in length, the bung diameter 30 inches, and the head diameter 25 inches, the staves being a medium curve? Ans. 122.1035+ wine gallons; 1 0.5558 beer gallons. 10. Required, the number of wine and beer gallons in a cask of which the staves are nearly straight, and the dimensions as follows: bung' diameter, 20 inches; head diameter, 16 inches; length, 50 inches? Ans. 56.3108 wine gallons; 46.3736 beer gallons. 11. How many wine and beer gallons will a hollow cylinder contain, which is 4 feet deep, and 1 ft. 6 in. in diameter? Ans. 52,8768 wine gallons; 43.6456 beer gallons. SECTION XVII. PHILOSOPIIICAL AND MISCELLANEOUS QUESTIONS. ARTICLE ]. THE MECHANICAL PowERS. Obs. 1. T1E MECHANICAL POWERS consist of six simple instrurmetds, vi'.: the Levei, the Wheel andl Axle. the Pulley, the Inclined P-lane, the Wedgye, and the Screw. Obs. 2 A MACHINE is an instrument combining one or more of t/ie mechanical o)owers. It may be very simple, as a pin; or very complicated, as a steam-engine. 0Obs. 3. TIE POwER1 is the body whlich ap]plies the force; a.d the WTEIGHT is tl;e body whlich resixsts the folce. What are the Mebhanical Powers? What is a Machine? The powerrT Thb woight? Art. 1. TiE WMEOIJANICAL POWERS. Case 1. THE LEVER. Obs. 4. THE LEVER is simply a bar, used in raising weight.s It is moved about a fixed point, called its Fulcrum, or Prop.: REMARK.-There are three kinds of levers. Force exerted by the first kind is called prying. In this the weight and power are at the ends of the lever, end the fulcruin between them. Force exerted by the second kind is called lifting. In this the power and fulcrum are at the ends of the lever, and the weight between them. Force exerted by the third power is called lifting, or raising. In this the weight and fulcrum are at the ends of the lever, and the power between them. Railing a ladder to the roof of a house is an example of this kind of lever. 2.-The arms of the lever are the parts between the fulcrum and its ends. Obs. 5. A power and weight acting upon the arms of a lever will balance each other, when the product arising from multiplying thIe power by its distancefrom the fulcrum. is equal to the product arising from multiplying the weight by its distance from tte fulcrum. By keeping this principle in view, we can always find any one of these four things: the power, the weight, the distance of the weight from the fulcrum, and the distance of the power from the fulcrum, if the other three are given. NOTE —By this rule, the power and weight exactly balance; in order to raise the weight, there must be a snall increase of power. It is also customary in practice, when estimating the mechanical powers, to deduct onethird for friction. Ex. 1. Suppose a man weighing 140 lbs. rests on a lever 8 feet long, what weigllt will he balance on the other end, supposing the fulcrum to be 6 inches from the weight? Ans. 2100 lbs. 2. Given, the weight, d stance of the weight fiom the fulcrum, and the distance of the power fromn the fulcrum, in the last example, to find the power? Ans. 140 lbs. 3. Suppose a power of 150 lbs. be applied to a lever, 16 ft. fr: m the fulcrumn, what weight will it balance at the other end of the lever, which is 4 inches from the fulcrum? Ans. 6750 lbs. What is a lever? How many kinds of levers are there? r What is force exerted by the first kind called? How are the power, weight,::(nd fulcruln arranged with ruespect to each other? WVlat is force exerteil iy tre Fecoiii kind of lever cilled? How are the power, wieight, and fllcrun i irranlgel \ilth revpect to rachi olti: r? Wliat is force exeorted by the 'li id lti; ( i,f' {lever c:!!ed? How are the weight, power, aid frllcruin arrang,'ed \vi!li respl"ct to e;chl nthl. r? Give an exapl:rle of Ihis kirl (iof It-ver. WlVhat are lie arls o( a lever? Wrhen will the power and weight acting tipon the arlis of a lever balance each other? What is necessary in order to raise the weight? What deduction is made for friction in the practica lapplication of the mechanical powers? COMMON ARITHMETIC. Sect. XVII 4. By a lever 30 ft. long, what weight will 400 lbs. sustain, the weight being 6 inches from the fulcrum? Ans. 23600 lbs. 5. By a lever 12 ft. long, what weight will 200 lbs. balance, the power being applied 18 inches from the fulcrum? Ans. 28- lbs. Case 2. TIIE WHEEL AND AXLE, Obs, 6. THE WHEEL AND AXLE consists of two wheels, one of which is larger than the other, but the smaller one passes through the larger, and thus both have a common centre on which they turn. The power is applied to the circumference of the larger wheel, and the weight tJ that of the smaller wheel (or axle), by means of cords. Obs. 7. The power, when applied to this machine, will exactly balance the weight, when the product arising from multiplying the power by the diameter of the wheel, is equal to the product of the weight by the diameter of the axle. Hence, if any three of these are given, the other can be easily found. 1. If the diameter of the wheel is 4 ft., and the diameter of the axle 3 inches, what weight on the axle will balance 125 Ibs, applied to the wheel? Ans. 2000 Ibs. 2. If the diameter of the wheel is 6 ft. 8 in., and the diameter of the axle 10 inches, what power must be applied to the wheel to balance 2600 lbs. at the axle? Ans. 325 lbs. 3. If the diameter of the wheel is 5 ft., the power 224 lbs., and the weight 2688 lbs., what is the diameter of the axle? Ans. 5 inches. 4. If the diameter of the axle is 8 inches, the power 340 lbs., and the weight 2975 lbs., what is the diameter of the wheel? Ans. 5 ft. 10 in. 5. Suppose there are two wheels: one 6 ft. 9 in. in diameter, with an axle 9 inches in diameter, and the other 5 ft. 5 in. in diameter, with an axle 6 inches in diameter; if the power cord of the smaller be attached to the axle of the larger, what weight at the axle of the smaller wheel can be supported by 275 lbs. at the power cord of the larger? Ans. 26812- lbs. Case 3. THE PULLEY. Obs. 8. A PULLEY is a grooved wheel, which turns about its axis by means of a cord passing over it. It may be either simple or compound, fixed or movable. What is a Wheel and Axle? How is the power applied? The weight? When will the weight and power exactly balance? What is a Pulley? Art. 1. THE MECHANICAI, POWERS. 353 Obs. 9. The simple pulley consists of a single wheel and its cord, to one end of which the uweight is attached, and to the other end the power. No advantage is gained from this machine, as the power and weiglht must always be equal. When a number of,ulljtys are used, one half of them Lre niova.le, and the whole is called a system of pulleys. It is evident that the weight is divided between the number of strings, as there are always double the number of movable pulleys: HenceObs. 10. The weight divided by twice the number oJ movable pullies, will give the power; and the power multiplied by twice the number of movable pulleys, will give the weight. 1. What weight will be balanced by a power of 200 lbs. attached to a cord, which passes over 4 movable pulleys? Ans. 1600 lbs. 2. What power must be applied to a cord passing over 6 movable pullies to balance 3600 lbs.? Ans. 300 lbs. 3. What number of movable pulleys would be necessary to raise a weight of 1000 lbs. by a power of 125 lbs.? Ans. 4. 4. If a cord that passes over 5 movable pullies be attached to an axle 4 inches in diameter, and if the wheel is 6 feet in diameter, what weight can be raised by the pulley, by applying 200 lbs. to the wheel? Ans. 36000 lbs. Case 4. THE INCLINED PLANE. Obs. 11. AN INCLNI:D PLANE is a plane surface?which inclines to or frm the earth. Thus, a board with one end on the ground, and the other end on a block of wood, forms an inclined plane. Obs. 12. The power when applied to this machine will exactly balance the weight, when the weight multiplied by the heighth of the plane, is equal to the power multiplied by its length. 1. If an inclined plane is 150 ft. long, and 15 ft. high, what weight will 200 lbs. sustain? Ans. 2000 lbs. 2. If an inclined plane is 80 ft. long, and 12 ft. high, what power will sustain 2880 Jbs.? Ans. 432 lbs. 3. What power would be required to draw a train of cars weigh ing 50000 lbs. up an inclined plane 3 miles in length, and 100 ft. n heighth? Ans. 315k lbs. nearly. What is a Simple Pulley? What is a system of pulleys? How is the weight divided? To what are these equal? How do we find the power? The weight? What is an inclined plane? Give an example. When will the power balance the weight in this machine? 354 COMMON ARITHMETIC, Sect. XVII Case 5. THE WEDGF Obs. 13. THE WEDGE may be regarded as two inclined planes placed base lo base. Obs. 14. From Obs. 12 we conclude the force will balance the power when the product a ising from multiplyinr the jfrce produced at the side, by the breadth of the head of the wedqe, is equal to that of multiplying the power acting against the head by the length of its side. NOTE 1.-This force only respects one side of the wedge; when the force against both sides are required, only half the breadth of the head must be taken into consideration. 2. —n applying the wedge, the friction is at least equal to the force to he overcome; and therefore not less than one-half of the power is lost, for which no allowance is made in the above rule. The advautage ef the wedge arises from the force of percussion or blow with which it is struck, which is nlmch greater than that of any dead weight or pressure, such as is usually employed on the other mechanical powers. 1. What weirght of force would be effected on either side of a wedge, the head of which is 3 inches broad, and the side 15 inches long, by a power of 85 pounds? Ans. 425 lbs 2. What power must be applied to a wedge 8 inches long and 6 inches broad at the head, to effect a force of 1575 lbs., allowing no loss for fiiction? If the friction is equal to the force to be overcome, what power must be employed?. 525 lbs. allowing no loss for friction. * 1050 lbs. allowing loss for friction. Case 6. THE SCREW. Obs. 15. 'IIE SCREW may be considered as a thread, or groove, running spirally around a cylinder. REMARK.-The principle of the screw is the same as lh;t of the inclined plane; the distance belween the threads being the leiihiih, and the circurifereece of the cylinder the base of the plane. Obs, 16. The power when applied to tl-is machine will exactly balance the weigllt, twhen the power nmultiplied by tI/e circu'f(erence of the circle described by the power, is eqpcl to the rwe;?dt multiplied by the distance betwe en the threads. What is a wedre? Whlin will the force 1;!nicl 1e 1: power? What has this force reference to? When l le force afgainst loth si/its is req 1ired, how do we proceed? XVhat part of the power is lost in alpp)ling the wed ge to p'actical purposes? How so? What advantage tlhen uas the wedfir? What is a screw? What is the principle of the screw? Show the comp,rison. When will the power balance the weight when apllied to this nmachiine? Art. 1. THE MECHANICAL POWERS. 355 Hence, if any of these four quantities are given, the other can be easily found. 1. What weight may be supported by a screw, the lever of which is 6 feet long, and the distance between the threads of a screw ' of an inch, by a power of 125 lbs.? Ans. 113097.6 lbs, The learner will observe that the length of the lever is but half the diameter of the circle. 2. Whatpower would be required to balance 12000 lbs., if the distance between the threads of the screw was, of an inch, and the length of the lever 8 feet? Ans. 12.43 lbs: Case 7. MACHINERY. Obs. 17. By MACHINERY is understood the connection of two or more of the mechanical powers, by means of belts, bands, cogs, dsc. REMARK.-The velocities of wheels are to each other as their like dimensions. 1. A cog wheel 150 inches in diameter, runs into another but 12 inches in diameter. Required, the number of revolutions the smaller wheel makes per minute, the larger performing 8? Ans. 100. Operation.-12: 150:: 8: 100. Or, whilst the larger wheel makes 1 revolution, the smaller one makes 150. 12 = 12l-; and 12' X 8 = 100, as before. 2. A wheel 56 inches in diameter is connected with another 6 inches in diameter, and this is attached to a shaft, on the end of which a wheel 42 inches in diameter is joined to another 4 inch s in diameter. Required, the velocity of the latter wheel per minute the larger one making 50 revolutions. Ans. 4900 revolutions. 3. How many revolutions will a spindle 2 inches in diameter make per minute, which is connected with a drum 4 ft. in diameter, that performs 40 revolutions per minute? Ans. 960. 4. A belt connects a drum 5 feet in diameter, makinul 50 revolutions per minute, with a spindle 3 inches in diameter. What is the velocity of the spindle per minute? Ans, 1000 revolutions. 5. If a drum 5 feet in diameter performs 40 revolutions per minute, what is the diameter of a cylinder connected with it that performs 600 revolutions per minute? Ans. 4 inches. 6. If the spindle of a comrn on spinning wheel is - of an inch in diameter, and the rim is 4 ft. 9 in. in diameter, how many revolutions will the spindle perform whilst the rim performs 15? Ans. 3420. What is understood by Machinery? What is the ratio of the velocities of wheels to each other? 356 COMMON ARITHMETIO. Sect. XVII ARTICLE 2. METHOD OF KEEPING BooKs, FORMS OF NOTES, &C, REMARK.-It is not our design to present here an elaborate treatise on book-keeping. We shall only give a form or two for the benefit of those whose entries are few, and wlo have but little business to transact. Those who desire further information are referred to treatises on book-keeping, where they will find the subject treated upon more fully. Obs. 1. It is necessary that every person should have some regular, systematic method of keeping his accounts, because the law requires in all cases of dispute, that the book in which the charges were originally made, be produced before the court, and legal evidence given of their correctness. The following form is recommended for farmers and mechanics, whose entries are but few, and generally made at the close of the day or week. It consists in writing the name of the person on the left hand page, Dr. (debtor) for the sum he is owing you, and on the right hand page, Cr. (creditor) by the sums you owe him. The difference between the Dr. and Cr. sides will at once show how much is due from one to the other. HENRY WORTHY, Dr. HENRY WORTHY, Cr. lt4d ~1.~ cts. l14rt $ cts. Jan. 1. For 8 bushels of Wheat, Jan. 3. By Cash.......... 5 00 at $1 per bushel,.... 8 00 " 4 For one day's work, 6 By Goods......... 475 Chopping,......... s 75 " 7 For 12 lbs. butter, at " 7 By Cash......... 1 00 16% cts. per lb......2 00 " 8 For 2 doz. Eggs, at 12 i " 8 By balance, being the cts. per doz........ 25 amouut now due me. 25 i___ ll 100 11 00 184S Jan. 9. For bal'nce due me from 25 old account........ NOTE.-It is immaterial whether we place the debits and credits both on the same page as above, or on opposite pages. The same rules are observed in both cases; that is, write the name of the person on the left hand page, Dr., and on the right hand page, Cr. The pages should be ruled as above. Obs. 5. The following remarks may also be of use to the learner: a. The person who receives any thing from you is Dr. to you for the amount he receives from you; and the person from whom you receive any property is Cr. by the amount you receive from him. b. Places of residence should be named when they are not the same as that where the book is kept. Also, if different persons bea r the same name, it is best to designate the occupation or particular Art. 2. BOOK-KEEPING, &C. 357 place of residence. The name of the person who owns the book, and his place of residence should be written on the first page. c. The date of each transaction should be written against the entry of it in the column for that purpose. d. When the account of any person is closed To (or For) Balance, you are debtor to him, and when it is closed By Balance, he is debtor to you for the amount of balance; and in opening a new account, he must be credited By Balance, or made Dr. To (or For, Balance from one account to another, the balance always crossing from the Dr. to the Cr. side, or from the Cr. to the Dr. side, in passing from the old account to the new. e. When a person has dealings with several individuals, he should have an index to his book, in which the name of every person with whom he transacts business should be written under its nitial, and the page where the account is kept should be noted down. f, Care should be taken to preserve a book free whom blots, and mistakes, and every entry should be made in a full, bold and legible hand. REMARKS.-For drovers, and farmers who raise considerable stock, or grain, and are in the habit of receiving and paying out money, frequently, the following form is recommended: CASH ACCOUNT, Dr. CASH ACCOUNT, Cr. 148 $ icts. lEt4 $ Ics. Jan. 1 To Cash on hand... 0 00. 7 Ja 7By Cash paid for " 4 " Sales of Cattle.. 75 50 Goods....... 300 0 " 9 " Sales of Wheat. 50 00 oo" 1 Cash paid for work 15 25 " 12 " Cashrec,dofTho- " 25 " Cash p'd forCattle 66 75 mas True on his " 28 " Cash paid for note...15' 75 wheat........ 20 00 " 30 " Cash rec'd for hor- 30 "Balance, being the ses............ 495 75 cash on hand... 365 00 767' 00 ____ 7671 00 1848 I, Feb. I fo hal. from olh acc't. 135 00 Obs. 3. The learner will now notice the following suggestions, with reference to th, C(ash-Book: a. The CASH-BOOK is kept by making Cash Dr. to what is on l:and, and what is received, and Cr. by whatever is paid out. 6. The excess of the Dr. side over the Cr. side ought always be equal to the cash on hand. c. When the cash on hand is counted, it should be entered on 358 COMMON ARITHMETIC. Sect. XVII the Cr. side; but when a balance is struck, the cash on hand should be entered on the Dr. side. NOTE.-Ornament and perspicuity being the object of many accoun tants, they often use capital letters, notwithstanding the rules they frequently see to the contrary. 'Tey likevise, sellomn use tile preposition of, writing 12 lbs. Butter, 6 yds. Sheeting, &c., instead of 12 Ibs. of Butter, 6 yds. of Sheeting, &c.But the learner need observe but one particular viz: to have a good pen, and clean fingers, and to preserve his book from blots, mistakes, and every thing else but his entries. FORMS OF ORDERS, RECEIPTS, AND NOTES. ORDERS. COLUMBUS, June 7th, 1848. Mr. Silas Brown-Please pay the bearer, Mr. Samuel Jones, six dollars and twenty cents, and charge the same to my account. THOMAS HENRY. NEWARK, Sept. 1st, 1848. Mr. Simon Dealer-Please pay Mr. Chares Lane such goods as he may call for, not excee ling the sum of fifteen dollars, and charge the same to your humble servant. THOMAS THRIFTY. REMARK 1.-If the persons on whom the order is made resides in another town from the one in which the order is dated, the name of the town in which they reside should be written with their names. RECEIPTS. GENOA, Oct. 3d, 1848. Received of Henry Martln two dollars in full of all accounts. THOMAS STONE. Received of Moses Hale five dollars in full of all demands. Dublin, Dec. 4th, 1848. PETER GOODRICH, REMARK 2.-When a receipt is given "in full of all accounts," it cuts off only the claims to accounts; but when it is given " in full of all demands," it cuts off all claims of any kind. Received of Thomas AMosely, five dollars and fifty cents on his note for ten dollars, dated Sept. 2nd, 1847. CHARLES WAKEFIELD. CHILLICOTHE, June 6th, 1848. Received of William Jones twelve dollars, to pay on the account of John Smiths IIENRY MARKLEY. Art. 2. BOOK-KEEPING, &C. 359 DUE-BILLS. Obs. 4. A DUE-BILL is a mere pledge to pay a certain amount of money or other property therein specified, in consideration of an equivalent specified. Form of a Bill. Due Isaac Dealer, or bearer, twelve dollars for value received. Columbus, Dec. 4th, 1847. JACOB FAITIFUL. NOTE.-If payment is to be made in any thing besides money, it should be so specified in the due-bill. NOTES. No. 1. Payable to Order. $200. COLUMBUS, Feb. 2nd, 1848. For value received, on demand, I promise to pay AbraJham Driveall, or order, two hundred dollars. JAMES CONSTANCE. No. 2. Note payable to Bearer. 8500. COLUMBUS, March 4th, 1848. For value received, on or before the first day of February next, I promise to pay Joseph Goodwill, or bearer, five hundred dollars, with interest from date. EZRA FAIRFACE. No. 3. Note by two Persons. $75. CLEVELAND, April 3d, 1847. One year after date, for value received, we, jointly and severally promise to pay Charles Good, or bearer, seventy-five dollars, with i nterest after thirty days. PETER TRUEMAN. Attest: Henry Jones, Samuel Albright. REMARKS ON NOTES. 1. The drawer or maker of the note, is the one who signs it; the holder of the note is the one who has possession of it. 2. No note is negotiable or transferable, unless the words "or order," or the words "or bearer" are inserted in it. 3. When the holder ol a negotiable note payable to order, (Note 1) wishes to transfer it, he must endorse it; that is write his name on the back of it. The holder of the note is then authorized 360 COMMON ARITHMETIC. Sect. XVI1 to collect it of the drawer, but if the drawer refuses to pay it, he can collect it of the endorser. 4. When a note is given with surety, the surety is not responsible only through the inability of the principal, or drawer. 5. When a note is made payable to bearer, (Note 2.) any person who has thenote can collect it of the drawer. 6. Tho words "'For value received" should be written in every note, and the amount of the note should always be written in words. 7. Every note should be made payable on demand, or at some specified time. 8. When the time expires for which the note, was given, the note will draw interest, although no mention be made of interest.Also: A note payable on demand, will draw interest after a demand of payment has been made, for it is then due. 9. When the time in which a note becomes due is not specified, it becomes due as oon as a demand of payment;s made. 10. Ail notes have three days of grace after they are nominally due, before they are!egally due. 11. The rate of interest is always understood to be the legal rate, unless otherwise specified in the note. 12. When a note is given by two persons, jointly and severally, (Note 3,) it may be collected of either of them, but not of both. 13. If a note is payable by installments, the amount of each installment, and the time when it is to be paid should be specified in the note. 14. If a note is given for specific articles, as wheat, corn, &c., payable at some fixed time, and paym' nt is not made at this time, the holder can claim and recover the value of the note in money. ARTICLE 3. MISCELLANEOUS PROBLEMS AND RULES. Ex. 1. A vessel 6 feet square is filled with water to the depth of 10 feet, what pressure does the bottom sustain, a cubic foot of water weighing 1000 ounces Avoirdupois? 6 X 6 =- 36 sq. ft. area of the bottom of the vessel. Solution. 36 X 10 = 360 cubic feet of pressure. [Ans. 360 X 1000 = 360000 oz.; 360000 16=22500 lbs. Obs. 1. The sides of a vessel sustain a pressure equal to the area of the sides multiplied by half the depth of the water. 2. What would be the pressure of water against the gates of a sluice 18 ft. deep, and 24 ft. wide, when filled? A _ _ to A a m 1~~~~'I' Ans. %4aUW lDS. Art. 3, MISCELLANEOUS PROBLEMS AND RULES. 1 861 3. What would be the pressure of water against the gates of a sluice 18 ft. wide, and 12 ft. deep, when full? Ans. 81000lbs. To find ile height of an object by knowing its distance:? Obs. 2. Take the distance in miles: two thirds of the square-of this will be the height of the object, in feet. NOTE.-In this calculation, of course there must be no obstruction to the range of vision more than the natural convexity of the earth. 4. Looking across a plain, I saw the top of a pole exactly 10 miles distaht. Required the height of the pole? Ans. 66 ft. 8 in. 5. I saw the top of a mast which I knew to be just 12 miles distant. Required the height of the mast? Ans. 96 feet. To find the distance of an object by knowing its height: Obs. 3. Take the height infeet; increase this by one-half of itself and extract the square root of the sum. The result wll be the distance in miles. 6. I saw the top of a mast, which I knew to be 150 ft. in height. Required its distance? Ans. 15 miles. 7. To what distance on a level plain could a pole be seen which measured 80 ft. 8 in, in height? Ans. 11 miles. To find the time a body has been falling: Obs, 4. Divide the velocity by 32-. Or, Divide the space reduced to feet by 16-, and extract the square root of the quotient. 8. How long will a body be in falling, to acquire a velocity of 193 feet per second? Ans. 6 seconds. 9. I dropped a stone into a pit 400 ft. deep. How long was it in falling? Ans. 4.98 + sec. 10. How long. would it take a body to fall 2 miles? Ans. 25.62 + sec. To find the velocity, which a body has acquired in falling: Obs. 5. Multiply the space reduced tofeet by 161, and double the square root ofthe product. Or, Multiply the time by 32. 11. A stone was dropped from the top of a ledge of rocks 228 ft. in height. Required its velocity per second when it had reached the bottom? Ans. 121.11 ft. 12. What velocity per second, would a stone acquire in falling 1 mile? Ans. 582.82 ft, persec. 17 -362 COMCIMON ARITHMETIC, Sect. XVtI 13. A body has been falling 12 seconds, Required the velocity it has acquired? Ans. 386 ft. per sec. To find the space a body has fallen through: Obs. 6. Multiply the square of the time by 161-. Or, Divide the square of the velocity by 64-, 14. A body has been falling 9 seconds: Required the space it has passed through? Ans. 13023 ft. 15. Wishing to ascertain the depth of a chasm, I dropped a stone from the top and saw it strike the bottom in 4 seconds. Required the depth of the chasm? Ans. 257 ft. 4 in. 16. A body has fallen so far as to have acquired a velocity of 579 ft. per second. Through what space has it passed? Ans. 5211 ft. NOTE.-In the above rules, no allowance is made for the resistance of the air. Given the base (or perpendicular) of a right-angled triangle, together with the sum (or difference) of the other two sides, to find those sides: Obs. 7. To the square of the given side, add the square of ehe sum (or diference) of the other two sides, and divide the result by twice this sum (or difference;) the quotient will be the length of the longest side. 17. The base of a right angled triangle measures 16 ft., and the sum of the other two sides is 32 ft. Required the length of those sides? Ans. Hypoth. 20 ft. Per. 12 ft. 18. The foot of a ladder was moved out 12 ft., and the top fell 6 ft. Required the length of the ladder, supposing it to have stood perpendicularly against a wall at first? Ans. 15 ft. 19. A pole 90 ft. high was broken off by the wind, and the two parts held together. The top fell 30 ft. from the foot of the pole,Required the length of each part? Ans. 40 ft. and 50 ft. To measure standing timber: Obs. 8. Place two stakes so that they will range with the top of the timber; then multiply the distance from the short stake to the tree, by the difference between the lenyth of the stakes, and divide the product by the distance between the stakes. The result will be the length of the timber above the shorter stake. 20. I wish to find a stick of timber 36 ft. long, and find one that has a knot that will spoil it above a certain height. I therefore place a stake 10 ft. long, 30 feet from the tree, and 6 ft. from this I place Art.. 3 MtSCELLANBOUS PROBLEMS AND RULES. another stake 4- ft. long, so that the top of the two stakes wiil range with the knot. Will the stick answer my purpose by allowing 2- ft. for the stump? Ans. It will not answer, it being but 35 ft. long. To find the mean temperature of any day, reckoning from sunrise to sunrise: Obs. 9. Add together the morning observation, twice the afternoon observation, twice the evening observation, and the next morning observation, and divide the sum by 6. REMARK 1. This is called DeWitt'sRule. It gives the correct mean temperature, on the supposition that the thermometer has risen and fallen regularly between the observations. The observations are taken atsunrise, at 2 P. M., at 9 P. M., and at sunrise the next morning. 2. When any of the observations are below zero, that quantity, or its double, (as the case may me,) must be subtracted, instead of added. 3. The mean temperature for any number of days is found by dividing the sum of the mean temperature of all the days by their number. 21. On a certain day the thermometer stood, at sunrise 38~ Fahrenheit, at 2 P. M. 50~, at 9 P. M. 42~, and at sunrise the next morning 30~. Required the mean temperature of that day? Ans. 42~. 22. On a certain day the thermometer stood at sunrise -6~, at 2 P. M, 100, at 9 P. M., 40, and at sunrise the next morning at -10~; what was the mean temperature of that day? Ans. 2~. This sign ( — signifies that the observation was below zero. 23. A square field is to be enclosed by a fence 8 rails high, and having two lengths to the rod, How large must this field be that it may contain as many acres as there are rails around it, or that each rail may fence an acre? Obs. 10. Multiply 4 times the number of rails on one rod by 160; this uill give the number of rods on one side of the field, Ans. 665360 acres = 1024 sq. miles = 32 miles sq. The sum and product of two numbers being given to find the numbers: Obs. 11. From the square of half their sum subtract their product, and extract the square root of the remainder; to the root add the half sum, and the result will be the greater number. 24. The sum of two numbers is 56, and their product 768; required the numbers? Ans. 32 and 24. 25. A man bought a certain number of acres of land for $4375; if the number of dollars he paid per acre were added to the num 364 lOMMON ARITHMETIC, Sect. XVII ber of acres bought, the sum would be 200. How many acres did he buy, and what did he pay per acre? Ans. He bought 175 acres, at $25 per acre. 26. Two men, A and B bought 150 acres of land for $375, of which A paid $208, and B $170; on account of a difference in the quality of the land,they divided it so that A paid $1.20 per acre more than B. How many acres did each get, and what did he pay per acre? Ans. A got 65 acres at $3.20 per acre. s B got 85 acres at $2.00 per acre. To find the number of acres obtained by him who paid least'per acre: Obs. 12. Multiply the whole nmber of acres by the diference between the prices per acre, subtract the product from the price paid for the whole land, and divide the remainder by twice the difference between the prices per acre. Square the quotient. and to this square add the product of the whole number of acres multiplied by the sum paid by him who paid the least per acre, and divided by the difference between the prices per acre; extract the square root of the sum, and from this root subtract the number that was squared; the rerainder will be the number of acres obtained by him who paid the leastper acre. \1f '~~ ~ Solution of the last example.] [whole. 150= No. acres. $378 = Amt.pd. for the $1.20= Diff. in the price per acre. 180 = Prod. to be sub-.I -— ~~ --- -[tracted. 180.00 Diff. 1.20 X 2 =2.40)198.00(825 = quotient to 1920 [be squared. 600 480 150 = No. acres. 1200 $170 = Sum paid by him who 1200 - paid least per acre. - 105 0000 15 1.210)25500.010 21250 - Number to be added to the square of 82.5. [Solution continued on next page.] ' Art. 3.,MISCELLANEOUS PROBLEMS AND RULES. 365 82.5 82.5 4125 1650 6600 6806.25 21250 28056.25(167.5 150-85 =65 =Acres obtained by A. 1 82.5=No.sqrd. $208-65==$3.20=price per acre - paid by A 26)180 85= Ares ob156 tained by B. 327)2456 $170-.85 $2.00= price per I 2289 [acre paid by B. 3345)167 25 167 25 000 00 27. A and B bought 280 a-res of land for $738. of which A paid $326.25, and B paid $411.75. The land was divided so that B paid $0.80 per acre more than A- How much land did each get and what did he pay per acre? A i A got 145 acres at $2.25 per acre. ns B got 135 acres at $3.05 per acre. The following questions are solved chiefly by analysis, without reference to any particular rules. They are easy enough to be understood if the pupil thoroughly understands the principles previously explained. 28, A, B, and C agreed to divide $50 between them giving A 1, B I, and C -. What was the share of each in this proposition? Solution. — + -— 1 =-H I; 50 - 2 =25; 50 -3=_16~; 5 4 -12-2. Then 13: 12: $25: $23- A's share.13: 12: $162: $15 B's share.13: 12:: $12: $11 7 C's share. 29. $335 is required to be divided between A, B and C, in such a manner, that as often as A gets $3 B gets $5, and C gets $7 as often as B gets 44. Solution.-Since C gets $7 as often as B gets $4, it is evident that as often as B gets b5 C gets 7 of $5 or 8. Hence, their shares are in the proportion of 3, 5 and 8-; or 12, 20, and 35. Ans. $60 A's share. $100 B's share, $475 C's share. 366 OOMMON ARITHMETIC. Sect. XVII 30. Suppose A, B, and C start at the same point and travel in the same direction about an island 53 miles in compass, A traveling at the rate of 3, B 5, and C 7 miles an hour, in what time will they next come together? Solution.-5-3= 2 miles B's gain per hour on A. 7-3= 4 " C's " " on A. Dividing the distance about the island (53 miles) by the greatest common divisor of 2 and 4, the gains of B and C per hour on A, we find that they will all meet again in 53 -- 2 262 hours. Ans. 26' hours. 31. A hare is 60 leaps before a greyhound, and takes 6 leaps while the greyhound takes 5, but 3 leaps of the hound are equal to 4 of the hare, How many leaps must the greyhound take before he catches the hare? Solution.-As 3 leaps of the greyhound are equal to 4 of the hare, 5 leaps of the greyhound are equal to - of 4 = 6 — leaps of the hare; therefore he gains 2- of a leap on the hare for every 5 leaps he takes, and gains 1 leap for every (5- - =) 7- leaps he takes: but he must gain 60 leaps to overtake the hare, therefore he must take 60 X 7 == 450 leaps to overtake her. Ans. 450. 32. A person bought a certain number of apples at the rate of two for a cent, and afterwards bought as many more at the rate of 3 for * cent: he sold them all at the rate of 5 for 2 cents, and lost 2'^ents, How many apples did he buy in all? Sohltion.-2 apples for 1 cent is 2 a cent apiece; also: 3 apples for 1 cent is 3 of a cent apiece. Then 2 apples (that is 1 of each lot,) cost I +- 5 of a cent, and I apple cost - - 2= — 5 of a cent. Again, as he sold 5 apples for 2 cents, he sold 1 apple for 2 - 5 ==- of a cent. Therefore he lost 52 = of a cent on each apple he sold; and 1 cent on 60 apples; and since he lost 2i cents, he must have sold 60X22-==150 apples. Ans. 150. 33. A, B, and C can do a piece of work in 5 days; B, C, and D in 6 days; C, D and A, in 7 days; and D, A and B, in 8 days; in what time would they all do it, working together, and in what time would each one do it alone? Solution.-A, B, and C, can do 5 of the work in 1 day. B, C, and D can do " " C, D, and A can do ~ " " D, A, and B can do - " " Therefore they all can do 5 + + - + - 30 of the work in 3 days, or - -3 3 = 533 of it in 1 day. Then 1-.- 53 3 4A3 days in which they all would it. * YM I3iW Art, 8. * MISCELLANEOUS PROBLEMS AND RULES. 867 Also: -5o 1 - - of the work done by D in one day, and 1 - 9 —= 862 days, in which D would do it. In the same manner it can be found that A can do it in 22- 3 -days; B can do it in 14-8 days; and C can do it in 11 6-, days. 34. A certain piece of work is to be performed, requiring a framer, a carpenter, and a mason; and no two can work at the same time. The mason receives $1.25 per day; the carpenter receives $2 per day; and the framer receives $2.50 per day; the work is completed in 100 days, and each receives the same amount of money. How many days did each work, and how much did each receive? Solution,-As the framer receives $2.50 per day, whilst the mason receives but $1.25 per day. it would take a:5-S = 2 days of the mason to amount to 1 of the framcr's; likewise, it would take 22 5 ==1 days of the carpenter to amount to 1 of the framers. Therefore, the days they worked are to each other as 1, ', and 2. Then 1+1 —+2=4-; and 4-: 1: 100: 23-p' days worked by the framer. 4-: 1:: 100: 29-17 days worked by the carpenter.And, 4'-: 2:: 100: 47'1 days worked by the mason. And 23-7' X 2.50- =$58'7. 29,7- X 2 = $58-4-. the sum received by each. 47 X 1.25=$58-4. ) And 100 = the number of days worked by all, 35. A gentleman died leaving a son and daughter in foreign countries, and directed in his will, that if the son only returned, he should receive 2 of the estate and the widow the balance; but if the daughter only should return, the widow should receive -, and the daughter the balance: as it happened both returned, and thus the widow lost in equity $1600 dollars more than she would, had the daughter only returned. How much was the share of each, and how much would the widow have received had the son only returned? Solution.-By the question, the son receives twice as much as the widow, and the widow twice as much as the daughter; therefore the daughter receives 1 part, the widow 2 parts, and the son 4 parts, and 4 + 2 +1= 7 parts into which the estate is to be divided. Had the daughter only returned, the widow would have received a of these 7 parts, or 4|- parts of the whole estate; but both returning, she receives 42 —2 - 2V- parts less than she would, had the daughter only returned. But these 2- parts = $1600; 1 1600 part therefore, must be -- =$600; hence, the estate is $600X7; nd - 3 400, the widow's share hadthe son %4200; and $4200 * 3 = $1400, the widow's share had the son 368 COMMON ARITHMETIC. Sect. XVIII only returned. But as it is their shares are to eachlother as the numbers 1, 2, and 4; then 7: 1:: $4200; $600 the daughter's share. 7: 2:: $4200: $1200 the widow's share. And 7: 4:: $4200: $2400 the son's sh'are. SECTION XVIII. MISCELLANEOUS EXERCISES FOR TIE.SLATE. 1. A sum of money is to be divided between five men; A receives $130; B $170; C as much as A and B; D as much as B and C; and E as much as all the rest. Required the sum divided? Ans. $2080. 2. A mal was 21 years old when he left college; he studied 3 years for a profession; then traveled 15 years; after which he married and lived with his wife 30 years, when she died, leaving him a son 15 years of age; the son was 26 years old when the lather died. Iow old was the father at his death? Ans. 80 years. 3. Add together 4 quadrillions, 127 trillions, 9 billions, 18 millions, 102 thousand 407; 39 trillions, 7 millions, and 1; 18 trillions, and 17; 204 billions, 10 thousand; 7 millions, 4 thousands, 303; and 199. Ans. 4184213032116927. 4. A man's property was worth $75000: he lost a house worth $3000 by fire, and a vessel worth $15000 in a storm. Required the value of the remaining property. Ans. $57000. 5. From 100 quadrillions, 100 billions, and 1, take 101 billions,! 101. Ans. 99999998999999900. 6. A farmer sold as follows: a quantity of wheat for $125; corn for $200; hay for $75; oats for $50; butter and cheese $40; and some other things for $25; he received in return a piece of broadcloth worth $30; $75 worth of satinet; $90 worth of groceries; $15 worth of muslin; $40 worth of books, and the balance in money. How much money did he receive? Ans. $265. 7. Multiply 146 millions, 201 thousand, 111, by 210 millions, 12 thousand, 222. Ans. 30704020179978642. 8. There two numbers; the lesser number is 1476 times 28921, and their differerence is 1492 times 1728. Required their product? Ans. 1932269397130512. 9. Divide 1 sextillion by 333. Ans. 30030030030030030033v3. 10. Take 1 from 1 sextillion and divide the remainder by 999. Ans. 100100 10000100001001 MISCELLANEOUS EXF.REISES FOR TIlE SLATE. 369 11. How long wo illl it tlke to count a trill;on by counting 576(0 per day? Ans. 1736111 1 days = 47532 yrs. 48 dlas. 1~. ihe bible c( Itiin.s 3 1 73 vt:Ises: by reading 74 verses 1cr day, how long will it take to read it through? Ans. 421' days. 13. The sum of two numbers is 18261, and one of the numbers is 4927; what is the otler number! Ans. 13334. 14. Two men, A and B have together $15900; A has,;8976; hobw much has B? Ans. $6024. 15. (3 has t1000, which is $496 dollars more than D has; how muclh lhas D? Ans. 5-504. 16. Two men together h ve 150 cattle, and one has 8 more than the other; how many have each? Ans. One 71; the other 79, 17. What number multiplied by 144 wil. produce 34272? Ans. 238. 18. What number divided by 216 will produce 383? Ans. 82728. 19. If the dividend is 23188, and the quotient is 124, what is the divisor? Ans. 187 20. There is a certain number, to which if 460 b added, and from the sum 3u0 be subtracted. and the remainder divided by-.8, the qnotient will be equal to 9900 divided by 132. Wha is the number? Ans. 440. 21. A lad being asked low much he gave for a book, replied; if you livide the lprice by 2, to the quotient add 11, multiply the sum by 6 and from llte prol!uct subtrl;act 52, tle remalinder ill be / 20 Required the cost of the book? Ans.,$2. 22. A peddler made three trips; the seconil trip le lost 25; but the third trip he gained i',4U; in tlh tlhree trips togetlher he gained $25. Did hle gain or lose the first tlip, and liow mucll? Ans. Ie gained $10. | 23. A travels 30 miles per dlay, and B 40; if both travel the same road, and A has 5 days the start of B, how many days will it take B to overtake A? Ans 15. 24. Multiply 288 by 144, 216, 137. 2625, 1728 and 256, and divide the product by 576, 72, 274, 864, 1644, 135, and 512. Ans. 1 -f. 25, A owns 580 rods of land; B owns 796 rods; C owns 848 and D ovtns 1 232 rolls they ag(rec to divide it into equal lots fixing on the greatest number of rods for a lot, that will allow (nach owner to lay cut all his land, How many rods in each lot? Ans. 4. 26. A.pnthlrnn s;;;,. I?.,-: 9. r; respectively'; what is tilt s tsi si ii t;(t, ic r L f i lgallons ti;ht;i ust fill same number of casks of either kind? Ans. 252. 17A 370 COMMON ARITHM"ETO. Sect. XVIII 27. What fraction is that, to which if you add 4 the sum will be I? Ans.. 28. What fraction is that, from which if you subtract 4, the remainder will be 4? Ans. 4. 29. What number is that which being multiplied by 12, produces -? Ans. 'T30. What number is that which being multiplied by 9, produces 3 Ans. 4. 31. What number is that which being divided by 6 produces 4? Ans, 4. 32. What number is that which being multiplied by 3, gives 3? Ans. 181 33. What number is that which being divided by 4 gives 2 Ans. 2. 34. What number is that, from which, if you take 7 of itself, the remainder will be 12? Ans. 21. 35. What number is that to which if you add 4 of 3 of itself, the sum will be 30? Ans. 24, 36. What number is that to which if you add 4 and - of itself the sum will be 93? Ans. 36. 37. What fraction is that to which if you add I and 3, the sum will be 1? Ans, ^. 38,.of a pole is in the air,, of it in the water, and 6 feet is in the mud; required the length of the pole? Ans. 28 feet. 39. 3 of a certain number exceeds 23 ot it by 4; what is the number? Ans. 48. 40. What number is that from which if you take'- of 4 of itself and to of itself, the remainder and f sum will be 93-? Ans. 120. 41. What number is that to which if 12 be added, - of the sum will be 16? Ans. 24, 42. A young man spent 4 of his property in 18 months, and of the remainder in 12 months after; he then: had $3000 left. How much was his fortune at first? Ans, $18000. 43. A grocer bought a number of boxes of tea, each containing 566 pounds, paying at the rate of $5 for 8 pounds, and sold it at the rate of I pounds for $8, and gained $300. How many boxes were there? Ans. 52. 1 ft 23 44, If 8 - English Guineas are worth $44 - and 16 - Eng4l 163 5 413 lish Sovereigns are worth $803, how many Sovereigns are worth 1 19 Guineas? Ans. 20. 46. If A does as much work in 1 I days as B does in 2. days, I MrSeELrLAN80U9 LO EXCISES FOR TIiM ISLATE. S71 aid C does as'much in 3- days, as A does in 27 days, how many days work of B are equal to 7- days work of C Ans. 11. 46. A younger brother received $1560, which was just q2 of his elder brother's fortune, and 5- times the elder brother's fortune was 3 v as much again as the father was worth. Required the value of the estate? Ans. $19165-. 47. A person spent all his income and ' as much more one year, but after that he saved o of his income, and in 5 years made good his deficiency and had $50 left. How much was his income? Ans. $1000. 48. What number multipliedby 16 will produce.96? Ans..06. 49. What number divided by.15 will produce 18? Ans. 2.7. 50. To whit decimal is 4J equivalent? Ans..0222+. 61. What is the value of the decimal.3625? Ans. 2, 52. A farmer sold In return he received 150 bu. wheatat $1.12- perbu. 150 lbs. sugar at $0.10 per lb. 300 bu. corn at.37- perbu. 201 lbs. coffee at.08- " 250 bu, oats at.31- perbu. 30 galls molasses at.44 per gal. 175 lbs. butter at.12- per lb. 3 barrels salt at 1.75 per bl. 412 lbs. cheese at.10 " 2 quintals fish at.06[ per lb. 1200lbs pork at,061 ( 30 yds. cloth at 5,75peryd. 24 yds. satinet at 1.50 per yd. And the balance in money. How much money did he receive? Ans. $224.75. 53. If 18 grains of silver will make a thimble, and 12 pwts. a teaspoon, how many timbles and teaspoons, of each an equal number can be made from 15 oz. 6 pwts. of silver? Ans. 24. 54. Sound moves at the rate of 1142 feet in a second of time: now I saw the flash of a cannon, fired from an eminence, just 1 minute before I heard the report. How far distant was the cannon? Ans. It M. 7 fur. 32 rds. 4 yds. 65. I saw several men upon an eminence shooting at the rate of 20 shots per minute; I saw the flash of three rifles before I heard the first report. How far distant were they? Ans. 1 M. 2 fur. 15 rds. 4 ft. 6in. 66. The atmosphere presses upon all surfaces at the rate of about 15 lbs. to the square inch: how much pressure then is sustained by a surface measuring 15 sq. ft.? Ans. 16 T. 4 cwt. 67. I wish to cut off just I acre of land from the end of a lot 22 rods wide. How long a piece must I cut off Ani. 7?f t dLs. 372 COMMON ARITHMETIC. Sect. XVIII 58. Some very important news having arrived at New Yoik 7 ~ ' 8" W. Ion,, in 4 minutes it was communicated to St. Louis, Mo., 90~ 15' 16" by magnetic telegraph. At what time was it known in St. Louis, it being known in N. Y. at half past 2 o'clock, P. M.? Ans. 29 min. 4 sec. past 1 o'clock, P. M. 59. A'traveler who had set a perfectly accurate watch by the sun in Augusta, Me., 69~ 50' W. Ion., being in Cincinnati, 84~ 27', a few days after, was surprised to find it wrong when compared with the sun. Was it too fast or too slow? How much, and why? Ans. It was 58 min. 28 sec. too fast, because the time is earlier at Allluusta than at Cincinnati. 60. If a man travel 376 miles in 18 days, how far can he travel in 27 days? Ans. 564 miles. NOTE. This, and the 47 following examples are to be solved by Analysis. 61. If 838.25 pay for 17 barrels of salt, how many barrels may be bouht f,r $24.75? Ans. 11. C2. ii a b)arrel of flour lasts 17 persons 5 wveeks, how long will i tlast 23 i:rPns? Ans. 4-3 weeks. 63. If 166- acres of land coss $3253, what will 26 acres cost? Ans. $523.806. 64. A lias 120 acres of land; of A's is equal to -' of B's. How many acres has B? Ans. 165. 65. A mnn owning ' of a store, sold ' of his share for $2928; requiril the worth of the store? Ans. $5807.20. 66. Two (cxpresses starn t the same time from two places 250 miles apart, and ride towards each other; A rides 8' miles perhour, and B rides 1 l miT'es erl hour. How far does each ride before they meet? Ans. A rides 10(35-5 miles: B rides 144-,4 miles. 67. A man-of-war in pursuit of a smuggler sails 8, miles whilst the smuggler sails 5; i;e ship passes a certain island when the smuggler is 24 miles beyond? lrow far must the man-of-war sail to overtake the other? Ans. 64 miles. 68. "As I was hunting on the forest grounds, Up starts a hare before my two greyhounds; The dogs being ligl of foot did fairly run, Unto h(er fifteen rod, just twenty-one; The distance that she starcld up1 before, Was four score sixteen ro l:, just, and no more. Now this I'd have you unto rre declare. How far they ran before t1,hey can ht the hare." Ans. 336 rds. 69. A set out from Boston for Ilarlford precisely at the time when B at Hartford set out for Boston, distant 100 miles; after seven MISCELLANEOUS EXERCISES FOR THE SLATE. hours they met when it appeared that A had traveled 1 - miles per hour more than B. At what rate per hour did each travel? Ans. A, 75- miles, and B 6 nmiles per hour. 70. A cistern has two pipes, one of wlich will fill it in 24 hours, agd the other in 30 hours; it has also a d(ischalging pipe which will empty in 18 hours. If all are left open, how long will it take the cistern to fill? Ans. 51' hours. 71. A lion of bronze, placed upon the basin of a fountain, can spoat water into the basin through his throat, his eyes. and his right f ot. If he spouts thlrough his throat only, lie will fill the basin in 6 hours; if throull his right eye only, he will fill it in 2 days; if throlugh his left eye only, in 3 days; and if through his foot only, he will fill it in 4 hour. In whiat time will the basin be filled if the water flow through all the apcrtures at once? Ans. 2| 4 hours. 72. A can mow 10 acres in 6 days; B can mow 13 acres in 9 days; and C can mow 17 acre: in 12 days: in what time can they all working together mow 1082' acres? Ans. 24 days. 73. A man left 813000 to be diviled between his wife, son, and daughter; he left 8$1500 more to the son than to the daughter. ari $2500 more to his wife than to his son. Required the share of each? Ans. Wife's share $6500; Son's share $4000; Daiughter's share $2500. 74. Thomas sold 150 pine apples at $0.33S apiece, and received no more morn than hlarry (lid for a certain:numbner of watereellons at $0.25 apiece. How much money did each receive, and how many melons had Harry? Ans. Each receives $50, and Harry had 200 melons. 75. A farmer sold 59 bushels of wheat, at $1.06'- per bushel, and took his pay in cloth at $2.371 per yard. How many yards did he receive? Ans. 22T7-1. 76. A goldsmith mixed 3 oz. of silver 14 carats fine, with 4 o7 -17 carats fine, 5 oz. 2;) carats fine, and 6 oz. 23 carats fine. Required the fineness of the mixture? Ans. 19- carats. 77. A's age is double B's, and C's is 3 times A's; the s im of all their ages is 135 years. Required the age of each? Ans. A's age 30 yrs.; B's 15 yrs.; C's 90 yrs. 78. A man bought a chaise, horse and harness for $300; the horse cost twice as much as the harness, and the chaise cost 3 times as the horse and harness torether. Required the cost of each? Ans. Harness $25; horse $50; chaise $225. 79. A man has $24 to pay 20 laborers; he pays each lby 3 cents, as often as he pays each woman 5 cents, and each man 7 cents; for COMMON ARITHMETIC, Sect, XVIII every boy there were 3 women, and for every woman 2 men. How many were there of each, anl how much did each receive? ( There were 2 boys, 6 women, and 12 men. Ans. The boys received 80.60 each; the women $1 each; and the men 81.40 each. 80. If 18 workmen do a piece of work in 20 days, working 10 hours per day, how many workmen will it take to do the same work in 15 days working 12 hours per day? Ans, 20. 81. In an orchard, - of the trees bear apples, I bear peaches, bear plums, and 30 bear clhrries. How many trees in the orchard? Ans. 120. 82. A daurghter's portion is 4 of a son's portion; both their portions amount to $7200; what is the portion of each? Ans. Son's portion $4000; Daughter's portion $3200. 83. What sum of money is that whose 3, - and - parts together amount to $141? Ans. $180. 84. Wh t number is that to which if 25 be added, 1- of the sum will be 21? Ans. 46. 88. What number is that which increased by - of itself, is equal t - of itself increased by 9? Ans. 24. 86. Out of a cask of wine, from which I part had leaked 21 gallons were drawn, when the cask was found to be half full. How much did the cask hold? Ans. 126 gallons. 87. A person having spent 3 of his income, found he wanted $25 of having - of it remaining. Required his income? Ans. $600. 88. A man bought a quantity of beef for $7; he used 25 lbs., and then sold i of the remainder for $1.20 which was just what it cost. How many pounds did he buy at first? Ans. 175. 89. A father divided his estate between his children, giving the first 2 wanting $300, the second - wanting $180. and to the third the balance, which was - wanting $50. How much was the estate? Ans. $6000. 90. A general after a battle found that he had left 500 men more than ' of the whole army fit for action; 260 more thanof the whole being wounded; 125 more than.1 of the whole being klled; and the remainder, which was 1- of the whole, missing. Of how many men did the army consist at first? Ans. 15000. 91. A hare starts 12 rods before a greyhound, but is not perceived by him until she has been up 1l minutes; she scuds away at the rate of 36 rods'a minute, and the dog in view makes after at the rate of 40 rods a minute. How long will the course last, and how far will the dog run? Ans. The course lasts 141 minutes, and the dog runs 670 rods. 92. A and B have the same income; A contracts an annual debt equal I of his, but B lives on - of his, and in 8 years lends A I MISCELLANEOUS EXERCISES FOR THE SLATFI enough to pay his debts, and has $200 to spare. How much is the income of each? Ans. $300. 93. A man being asked the time of day, answered that the time past noon was equal to 3 of the time till midnight; what time was it? Ans. 30 minutes past 4 o'clock.. 94. A man performing a journey found that the distance he had traveled was equal to 4 of the distance he yet had to go; how far had he traveled, and how long was his journey in all? Ans. His journey was 165 miles, and he had traveled 60 miles. 95. Three men have a certain sum of money; A has $200; B has as much as A, and 7 as much as C; and.C has as much as A and B both. How much have they all together? Ans. $1400. 96. A lady has two silver cups and only one cover, which weighs 5 ounces: if the cover be put on the first cup its weight will be double that of the second cup; but if the cover be put on the second cup; its weight will be 4 of the first cup, Required-the weight of each cup? Ans. Weight of the first cup 25 oz.; of the second cup 15 oz. 97. At what time between 5 and 6 are the hands of a clock together. Ans. 27 min. 16, sec. past 5. 98. A harmless dove was soaring high, To stretch her wings in space; At length a hawk did her espy, And gave the dove a chase. Just forty rods was there between These birds, as we could view — 1It ~ And whilst tl:e dove flew seventeen, The hawk flew twenty-two. The hawk pursued with all his strength, As those who saw did say, Then tell the rods he flew in length, Before he caught his prey. Ans. 176 rds. 99. At what time between 8 and 9 are the hands of a clock exactly opposite each other? Ans. 10 min. 546,- sec. past 8. 100. A man bought a lot of wheat, and gave - of it to some poor families, and then sold the remainder for $1.33- per bushel, and received what the whole cost him. How much did it cost per bushel? Ans. $1.00. 101. What number is that from which if you take 4 of 7 of itself, and to the remainder add 7- of -3- of the number, the sum will be 50? Anls. 120. 102. A man when be married was 3 times as old as his wife; 15 years after he was but twice as old as his wife. Required-the age of each when married? Ans. Man's age, 45 yrs; Wife's are, 15 yrs. 376 COMMON ARITHMETIC, Sect. XVIII 103. Divide $450 between A, B, and C, giving A $50 more, and B $50 less than C. Ans. A $200; B $100; and C $153. 104. There are two numbers whose difference is 8; of the less is equal to 2 of the greater. Required-the numbers? Ans. 40 and 48. 105. If one ship containing 150 hogsheadg of wine, pays for toll, at the Sound, the value of two hogsheads, wanting $6, and another ship containing 240 hogsheals pays at the same rate the value of two hogsheads, and $18 besides, what is the value of the wine per hogshead? Ans. $23. 106, The smaller of two numbers is 15, and if we add to this 4 of both numbers, the sum will be the greater number. Requiredthe greater n imbcr? Ans. 25. 107, A man having $50, spent a part of it; he afterwarls received 6 times as much as he had spent, and then had twice as much as he had at first. How much did he spend? Ans. $10. 108. If 2 men can do a piece of work in 25 days, and it takes 3 women the same time to do it, in what time will one man and one woman together perform it? Ans. 30 days. 109. A man left his daughter,4 of his property, to his wife,ir of the remainder, and to his son what was left. Required-the share of each, the shares of the son' and daughter together a mounting to $4500. Ans. Daughter's share $2000, Wife's share $3000; Son's s share $2500. 110. If cloth worth $0.25 cash, is worth $0.31 in trade, what is wheat worth in trade that is worth $1. 06- cash? Ans. 81.3' -. 111. At the rate of $72 for 64 days' work, how many days must a man work to earn $99? Ans. 88. 112. If the hind wheel of a wagon is 14 fl. 8 in circmufcrence, and the forward wheel 11 ft. 9 in, in circumference, lhow many times will the forward wheel turn over more than the hind wheel, in going 470 miles? Ans. 42000 times. 113. If - of a farm is worth $2100, what is - of it worth? Ans. 1400. 114. If 5' acres of land are worth $34' what art 95- acres worth? Ans. $60.20. 115. If 15 men mow 250 ncres of grass in 10 days, working 12 hours per day, how many men will it take to mow 300 acres in 6 days, workin: 10 hours per dlay? Ans. 36. 116. If 33 men, vorki g 10 hlours per tday, dig a ditcl 450 yarrds lon, 5 fe:et wild,, and,1 fect t,,p, in 1.5 (lavs h,\w r;mny days will it take 45 men to dig a ditch 510 yards long, 4 feet wide, and 3 feet deep, working 1 hours per day, if the strength of the former party MISCELLANEOUS EXAMPLES FOR THI SLATE. 377 is only 8 of that of the latter, and the hardness of the ground in the latter case is 1- times that in the former? Ans. 6 days. NOTE.-Let this question also be solved by Analysis. 117. A, B, C, and D trade together: A puts in ', B, C -, and D the rest; they gain $5400. What is each one's share? Ans. A's 1800; B's$1350; C's $900; and D's $1350. 118. If the third of 6 were 3, What would the fourthof 20 be? Ans. 7k. 119. If 6 and 4 just 9 had been,j 'Pray tell how much were 7 and 10? Ans. 15-. 120. A bankrupt owes'as follows: to A $1000; to B $1200; to C. $800; and to D 61500; his property is worth $4000. How much mus each creditor receive? Ans. A,$888.888-; B, $1066.666-; C, $111.111; D, $1333.333-. 121. In a storm at sea, a vessel loaded with flour was obliged to throw overboard 150 barrels; of the cargo, A owned 1000 barrels; B owned 1400; C owned 1600; and D owned 2000, What part of the loss must each sustain? Ans. A, 25 bls.; B, 35 bls.; C, 40 bls.; and D, 50 bls. 122. A prize of $400 is to be divided between a Captain, two lieutenants, and 6 soldiers: the Captain receives a share and a half, the lieutenants a share each, and the soldiers a half share each.How much does each receive? Ans. Captain, $92; Lieut. $61-7- each; Soldiers, $30'. 123. A, B, C, and D agree to trade together with $1200 capital, of which A was to furnish -, B, C 5, and D l; D withdraws, and 0 dies. What part of the stock must A and B furnish, to have the same proportion as before? Ans. A, $6855; B, $5147. 124. A commenced trading with $1000 capital; at the end of 3 months he took in B, with $1400; two months following, they took in C, with $1200; and 4 months after, they took in D, with $1500: at the end of the year they find they have gained $800. How much is each one's share? Ans. A's, $256; B's, $268.80; C's, $179.20; and D's, $96. 125. A hired two horses and a carriage to go 30 miles and back, for $20: after proceeding 12 miles, he took in B to ride through and back again; when within 5 miles of the end of their journey, they took in C, who'also rode back with them; and when half way back, they took in D, who rode the balance of the way. How much ought each to pay? Ans. A, $7.50; B. $6.25, C, $4.371; and D, $1.87J. 126. Sold, $1500 worth of goods for a friend. How much did. my commission amount to, at 12} per cent? Ans. $187.60 378 COMMON ARITHMETIC. Sect. X ~VIII 127. What will be the cost of insuring $2500 worth of property, at 35 per cent? Ans. $90.62-, 128. A merchant bought a lot of goods for $21500, and paid $150 for transportation. For how much must he sell them to gain 18 per cent? Ans. $25547. 129. Bought a quantity of goods on credit for $75, but for cash 12 per cent was deducted. How much did they cost? Ans. $66. 130. Bought goods for $1200, and sold them for $1400. What per cent was gained? Ans. 163-. 131. Sold tea for $1.12- per pound, by which 28' p!er cent was gained. Required-its cost? Ans, $0.87'-. 132. Bought goods for $500, and sold them for $450. What per cent was lost? Ans. 10. 133. What is the interest and amount of $180 for 2 yrs. 4 mo. 23 da. at 6 per cent? Ans. Int. $25.89; Amt. $205.89. 134. What is the interest and amount of $360 for 3 yrs. 5 mo. 29 da. at 7 per cent? Ans. Int. $88.13; Amt. $448.13. 135. A man gave $300 for the use of $2500, 2 years. What was the rate per cent allowed? Ans. 6. 136. A gentleman gave $4500 for $3800 which he had used 3 years 6 mo. 15 da. What rate per cent did he allow? Ans. 5 —. 137. In what time would $300 amount to $408, at 6 per cent? Ans. 6 years. 138. What principal would amount to $221.40 in I yr. 9 mo. 12 da., at 6 per cent? Ans. $200. 139. What sum w uld gain $55.30 in 2 yrs, 7 mo. 18 da., at 6 per cent? Ans. $359. 140. 1 have a note of $540.50, due me in 2 yrs. 6 mo. without interest: my creditor wishes to pay me now. What discount ought I to make? Ans. $70.50. 141. A certain person wished to sell his property, and a gentleman offered him $2100 for it at the present time, or $2350 payable in 2 yrs. 11 mo; he chose the latter. Did he gain or lose, and how much, the money being worth 6 per cent? Ans. He lost $100. 142. What is the difference between the interest of $1000 for 3 yrs. 7 mo. 6 (la., at 7 per cent, and the discount of the same sum, for the same time, and at the same rate? Ans. 50.722. 143. What is the present worth of 6 per cent of $2400. payable, ~ in 6 mo., - in 9 mo., and the rest in 1 year. The discount? Ans. Present worth, $2296.965; Discount, $103.035. 144. A broker subscribed for 40 shares in a bank, each share being $100: le paid in 40 per cent of his stock when he subscribed, 50 per cent of the remainde: 4 months after, and the balance 3 months following. In 12 months after hle subscribed there was a dividend of 4^ per cent of the stock among the stockholders, and MISCELLANEOUS EXIRCISES FOR THE SLATB. 879 a dividend of 3 per cent accrued each 6 months afterwards. At the end of 4 yrs. 6 mo. the broker sold his stock at 8- per cent advance. Now supposing he hired his money, and paid up his notes when he sold his stock, did he gain or lose by the speculation, and how much? Ans. He gained $401.60. 145, A certain room measures 20 ft 10 in. in length, 16 ft. in width, and 7 ft. 10 in. in height: now deducting two doors, each measuring 6 ft. 7,in. by 3 ft. 6 in., and three windows, each measuring 5 ft. 4.n, by 3 ft. 4 in., what will it cost to plaster this room at $0.20 per square yard? Ans. $11.82-. 146. A certain brick building measures 64 ft. 9 in. in length, 40 ft. 8 in. ir)width, and 30 ft. 6 in. in heighth, and the walls are 1 ft, thick. It has two partition walls through the length of the building, and 6 additional walls, each 15 ft. 2 in. in length; frlf the side of the building to these partitions, all of the same heighth and thickness as the outside walls. Now deducting two outside doors, each measuring 7 ft. 8 in. by 3 ft. 6 in. and 24 inside doors, each measuring 6 ft. 9 in. by 2 ft. 8 in., and 36 windows, each measuring 6 ft. 8 in. by 3 ft. 9 in., how many bricks did it require for this building, each brick being 8 in, in length, 4 in. in width, and 2 in, in thickness, and how much did they cost at $3.59 per M.? Ans. It required 311.202 bricks, and they cost $1089.207. 147. What number added to s of 123, is equal to 3- of the square of 64? Ans. 1123. 148. What number is that, which substracted from 6 times the square root of 1849, leaves - of 552? Ans. 120. 149. A-company of men spent $51.84, each man spending 4 times as many cents as there were men in the company. How many men were there, and how much did each spend? Ans. 36 men; each spent $1.44. 150. A General, forming his army into a square, found he had 76 men over and above a square; but by increasing each side by 1 man, he wanted 161 men to complete the square. How many men had he? Ans. 14090. 151. If A travels due north 24 miles, and B due east 32 miles, how far apart are they? Ans, 40 miles. 152. A certain t.riaRgle measures 16 feet on each side. Required -the length of a perpendicular from any angle to to the opposite side? Ans. 13.85+-feet. 153. If a circle measures 9 inches in diameter, what is the diameter of one 4 times as large? Ans. 18 inches, 154. If the circumference of a circle is 36 feet, what is the circumference of one that contains but - as much? Ans. 12 ft. 155. The distance between two places is such, that if you increase its cube by 16, from the sum subtract 39, and multiply the remainder by 2, tho product will be 100. Required-the distance? Ans. 4 miles. 380 COMMON ARITHMETC. Sect. XVIII 1 156. In'measuring a certain figure like E the one in the margin, it was found that the length of the lines were such, that as often as 8 inches were measured from A to B, 15 inches were measured from A to E; and as often as 3 inches were measured from A to E, 2 inches were measured from A DA to C; and as often as 6 inches were measured from A to D, 5 inches were measured from A to C; the length of the four lines is 675 inches. Required-the length of each line, and their distance apart at their extremities? B N- f Distance from A to B 120 inches, " A to C 150 inches. " " A to D 180 inches. " A to E 225 inches. *" " B to C 90 inches. Ans,1 " " B to D 216.333+ inches. ~ " " B toE 312.129+ inches. " "6 C to D 150 inches. " " C to E 270.416+ inches. ' " D to E 135 inches. 157. There are two columns in the ruins of Persepolis left standing upright; one is 64 feet high, and the other 50 feet. In a direct line between these, stands an ancient statue, the head of which is 97 feet from the top of the higher column, and 86 feet from the top of the lower column. The distance from the foot of the lower column to the foot of the statue is 76 feet. Required-the distance between the topsqof the columns? Ais. f 169.96 ft. nearly. Or, s 157.04 ft. nearly. NOTE.-The leainer will perceive that this question admits of two diffetent answers, according Was we suppose the statue to be higher or lower than the columns. 158. How many cubical blocks, each measuring- of an inch on a side, will it take to fill a cubical box, each side of which is 2 feet? Ans, 4096. 159. In a cubical foot, how many cubes measuring 2 inches on each side? Ans. 216. 160. What is the difference between a solid half foot, and half a solid foot? Ans. 648 s. in. 161. Required-the length of one side of a solid body containing - as much as another solii body measnring 6 inches on each side? Ans, 3 inches. MISCELLANEOUS EXERCISES FOR THE SLATE 381 162. 'Required, the area of a triangle whose base is 46 feet, and perpendicular heighth 32 feet? Ans. 736 sq. ft. 163. Required-the area of a triangle whose sides are 19', 21, and 22' ft. respectively? Ans. 189 sq. ft. 164. If the wheel of a gig are 4 ft. 7 in. in diameter, how many times will they revolve in going 1 mile? Ans. 366.692+. 165. If the propelling wheels of a locomotive are 3 ft. 4 in. in diameter, and if they make 400 revolutions per minute, how far will the engine move forward in an hour? Ans. 47 miles, 192 rods. 166. A circular meadow I have on my land. Which contains just two acres and 3 tenths of ground, How long must that line be, which fixed to a pole, Will just let my horse grnze around on the whole? Ans. 10.823+ rods 178 ft., 69.54 in. 167. Required-the solid contents of a triangular prism, the sides of which are 4, 5, and 6 feet, and its heighth 12 feet? Ans. 119.0588+ s. ft. 168. Required-the solid contents of a cylinder 15 ft. long, and 3 ft. in diameter? Ans. 106.029 s. ft. 169. Required-the area of the surface and base of a triangular prism, 15 ft. in length, and measuring 4 ft. on each side? Ans. 110.851 sq. ft. 170. Required-the area of a curved surface of a cylinder 16 ft. in length, and 3 ft. in diameter? Ans. 150.7968 sq. ft. 171. What is the area of the surface of a triangular pyramid, 23 ft. in heighth, and each side of the base being 9 feet? Ans. 337 sq. ft. 72 sq. in. 172. On the fourth of July a pole was erected, Composed of three pieces all nicely connected; Five feet and 'our inches it measured around, At the place where it stood at the top of the ground. Its shape w s a cone, its surface complete, And the height of the same was twice 60 feet. How many yards of blue ribbon, procured at the shop, Will wind round this pole from bottom to top, Laying smooth and plain to be seen, By leaving a space of 6 inches between? Ans. 214.268. 173. A gentleman has a vessel in the shape of a cone, and wishes to know how many gallons of wine it will contain, it being 25 inches deep, and 5 inches in diameter at the largest end. Can you tell the the number? Ans. 708= —2 qts. 1 pt. 2] gi. 174, How many square inches of leather will it require to cover a ball 5 inches in diameter? Ans 78.54. 175. Required-the solid contents of a globe 15 inches in diametf r? Ang. 1767.16 g.d in, 382 COMMON ARITHMETIC. Seot. XVIII. 176. A stone was put into a gallon measure, which was then filled with 1 qt. of water. Required-the solid contents of the stone? Ans. 173' s. in. 177. How many bushels will a box contain that is 5 ft. long, 4 ft, wide, and 4 ft. deep? Ans. 641. 178. If a cord that passes over 4 movable pulleys, be attached to an axle 3 inches in diameter, and the wheel is 6 feet in diameter, what weight can be raised by the pulleys, by applying 250 lbs. to the wheel? Ans. 48000 lbs. 179. What weight may be supported by a screw, the threads of which are 1 inch apart, and the length of the lever 9 feet, by a power of 180 lbs., supposing of the power to be lost in consequence of friction? Ans. 81430.272 lbs. 18J. If a wheel containing 84 cogs runs into another containing 14 cogs, and at the other end of the shaft of the latter, a wheel of 64 cors juts into one of 12 cogs, which is attached to a shaft, on the other end of which a wheel of 52 cogs runs into one of 8 cogs, that is connected by a rod with a band-wheel 32 inches in diameter, and the diameter of the cylinder where the band runs is 4 inches. How many times does the cylinder revolve whilst the master-wheel revolves once? Ans. 1664 times. 181. A and B commence trading with $1000 capital, of which A furnished a part, and B. the rest. They gained 25 per cent, and shared it according to the stock each furnished, when A said to B"I have made a handsome speck." "Yes," said B, "but if I only had as many such sums as I now have as you have dollars, I should then have $375.000. What share of the capital did each furnish? Ans. A, $600, B, $400. 182. A certain pole being broken off by the wind, the top fell 27 feet from the foot of the pole: the difference between the length of the two pieces was 9 feet. Required-the length of the pole? Ans. 81 ft. - 183. Two men, A and B, are to dig 100 rods of ditch for $100, of which each receives $50. A begins at one end, and B begins at the other end. Upon settlement, owing to a difference in the hardness of the ground, A receives $0.25 per rod more than B. How many rods does each dig, and at what price per rod? Ans. A digs 43.845 rods, at $1.14 per rod; * B digs 56.155 rods, at $0.89 per rod. 184. A, B, C, and D, agree to divide $120 between them, giving 4A, B {, C ', and D 1. Required-the share of each? Ans. $A, 42r; B, $31; C, $25iJ5; and D, $21-. TW, TI, V 0, 185. A gentleman wishing to distribute some money among some children, found he wanted 8 cents to give them 3 cents apiece; he therefore gave them 2 cents apiece, and had 3 cents left. How many children were there? Ans 11. I i I t i MISCELLANBOUS EXERCISES FOR THl SLATE. 186. A sum of money is required to be divided between A, B, C, and D, in such a manner, that as often as A gets $2, B gets $3; and B gets $4 as often as D gets $5; and D gets $9 as often as C gets $8; C. gets $400. Required-the share of the others, and the sum divided? Ans. A's share, $240; D's share, $450; B's share, $360; Sum divided, $1450. 187. A farmer observed that l of his land was sowed with wheat, -,of the remainder with oats, and 5 of the remainder was p'anted with corn: likewise, he observed that his wood-land was but - of his meadow land, which together amounted to 70 acres, and this completed his farm. How many acres had he in all, and how many acres had he of each crop? Entire farm, 200 acres. Of corn, 50 acres. Ans. Of wheat, 50 " Of meadow, 50 Of oats, 30 " Of timber, 20 " 188. In turning a gig within a certain diameter, it was discovered that the outer wheel turned thrice, whilst the inner wheel turned but tw'ce: now supposinrg the axletree to be 4 ft. long, and the wheels of an e.ual size, the circumference described by each wheel is required? Ans. Outer circumference, 75.3984 ft; Inner cir. 50.2656. 189. A father left $2000 to be divided between his three so s, aged 14, 16, and 18 years, in such a manner that the share of each, being plac d at simple interest, at 6 per cent, until he arrived at the age of 21 years, should amount to the same sum. How much was the share of each? Share of the youngest, $606.851-t. Ans.( " "" second, $662.868+. ( c" " " eldest, $730.279+. 190. Four persons, A, B, C, and D, start from the same point, to travel in the same direction around an island 65 miles in compass: A goes 2 miles, B 4 miles, C 6 miles, and D 8 miles an hour. HIw long before they next come together? Ans. 1 day, 8 hours, 30 minutes. 191. In performing a journey, A has 15 days the start of B, and travels 7 days per week, whilst B, stopping Sundays, travels but'6; but B travels as far in 4 days as A does in 5. How many days before B overtakes A? Ans. 60. 192. A, B, and C can do a piece of work in 6 days; B, C, and D, in 7 days; C, D, and A, in 7' days; and D, A, and B, in 9 days. In what time can it be done by all of them together, and by tach of them singly? As. By all in 51 da.; By A in 237 da.; By B in 19 da.; ns ~ By C in 13~ da.; By D in 55"4 da. 193. A certain cellar was dug, the width of which was. of the 384 COMMON ARITHMETIC. Sect, XIX. length, and the depth was ' of the width: 160 cubic yards of earth was removed. Required-the length of the ditch? Ans. 30 ft. 194. A in a scuffle seized on - of a parcel of sugar plums; B catched 3 of them out of his hands, and C held on to -1 more; D ran off with all that A had left, except, which E afterwards secured slyly for himself: next A and C jointly set upon B, who in the conflict let fall 4 he had, which was equally picked up by D and E: B then knocked down C's hat, and to work they went anew for what it contained, of which A got 4, B, 4, D,, and C and E equal shares of what was left of that stock: D then struck 4 of what A and B last acquired out of their hands, when they with difficulty recovered 5 of it in equal shares again, but the other three carried off 1 apiece of the same. Upon this they called a truce, and agreed that the 1 of the whole left by A at first should be equally divided among them. What share of the prize after this distribution remained with each of the competitors? And supposing the least common multiple of the denominators of the fractions expressing their shares to be the number of sugar plums divided, how many did each finally obtain? Ans. tothe last. A, 2863; B, 6335; C, 2438; D, 10294; E, 4950. 195. A merchant bought some cloth at the rate of $4 per yard, and afterwards bought 4 as much more at $5 per yard: he sold it all at the rate of 3 yds. for $1t), and gained $140. How many yards did he buy in all? Ans. 150. 196. A man bought apples at 7 cents a dozen, - of which he exchlanged for lemons, at the rate of 7 apples for 3 lemons: he then sold all his apples and lemons at 1- cents apiece, and gained 36 cents. How many apples did lie buy, and what did they cost? Ans. Hie bought 7 doz., and they cost $0.49. 197. The sum of two numbers is 812-, and 4 times the less, is equal to I of the greater. Required —the numbers? i~,+~~~ - Ans. 750 and 62. 18. A man bought 6 oranges and 4 lemons for 42 cents, and at the same rate, bought 5 oranges and 8 lemons for 49 cents. Required-the price of an orange, and of a lemon? Ans. Price of an orange 5 cents, and of a lemon 3 cents. 199. Three gentlemen contribute $675 to build a church, which is situa;ed at the distance of 2 miles from the first, 2' miles from the second, and 3' miles from the third, and agreed that their shares should be reciprocally proportional to their distances from the church. Required-the amount each contributed? |Ans. The first, $289.80; the second, $201.60; 11~ ~~Ans' 7 the third, $165.60. 200. A teacher agreed to teach a certain time upon these condiinn: If he had 20 scholars, he was to receive 26&; but if he had MISCELLANEOUS EXERCISES FOR THE SLATE. 385 30 scholars, he was to receive but $30. He had 29 scholars. Required-his wages. Ans, $29,725. 201. "Between Sing-Sing and, Tarry-Town I met my worthy friend John Brown And seven daughters, riding nags, And every one had twenty bags; In every bag were thirty cats, And every cat had forty rats, Beside a brood of fifty kittens. All but the nags were wearing mittens! Mittens, kittens-rats, cats-bags. nags-Browns, Hiw many were met between the towns? Ans. 2184192," NOTE.-This question is to be taken in its plain, literal, common-sense meaning, without any quibble, and the mittens are to be counted separately and not in pairs. 202. A man spent $12 for apples and peaches, giving $1 for 4 bushels of apples, and $1 for 3 bushels of peaches: he afterwards sold ' of his apples, and ' of his peaches for $4, which was $0.50 more than they cost him. How many bushels did he buy of each? Ans. 24 bushels of apples; 18 bushels of peaches. THE END. I ERRATA Page 5, Obs., 4th line from top, for -left," read "right." 46, Rule II., last line, for "of," read "as." 57, and 58, The remarks under Obs. 20 relate to "Practical," and not o "Pure Arithmetic." 122, Second line firom bottom, for "1L-," read "I or 6 " 132, "c" should read, "'mills are reduced to cents by pointing ot one fzqure to tie right, and to dollars bypointing of three //gures," &c. 139, Thilrd and fourth lines from the bottom, for "Art. 1," read "Art. 7." " 14', Fourth line fiom the bottoin, for"Art. 1," read "Art. 7." " 155, Rem. 3, for (" lbs.," read "830 lbs." " 156, Rem., for "Liquor," read "Liquid." " 158, Note at the bottom of the page, for "691," read "69." " 161, Table, for "'160 Sq: Yds," read "160 Sq. rds." 172, Ex. 46, for '"post," read "foot." 173, Ex. 57, for 'area," read "arc." ' 182, Ex. 14, for '-Sq. ft." "Sq. in.," read 's. ft. s. in." " 185, Bottom of tlfe^paTe, for "2d," read "22d." " 186, Obs. 1, for for "Sect. XIV.," read "Sect. XIII.;" and Ex. 25, for "Area," read "Are." 223, Obs. 14, for "Ratio of Inequality," read "Ratio of Lesser equality." " 226, Obs. 11, Illustratiion. for "two Means," read "two Extremes;" and for "two Extremes," read "two Means." 227, Obs.' 14, Rem., for "tirst terms," read "first three term s." < 230, Fourth line above Obs. 20, for "Obs 4," read "Obs. 14." " 237, Top line, for "17," read "I'9." " 263, Obs. 8, Rem. 5, for "Obs. 21," read "Obs. 5." * 272, Case 4, for "any number," read "any sum;" and General Rule, for "given number," read "given time." 297, Note, for "Obs. 30," read "Obs. 31." " 2.-8, Obs. 33, for "Obs. 30." read "Obs. 31;" and Ex. 12, for "Obs. 31," read "Obc. 39," 388 ERRATA. 2 99, Obs. 2, for "Obs. 15," read "Obs. 17," and for "Obs. 18," read "Obs. 20." " 301, Rem. 1, for"Obs. 15," read "Obs. 17." 303, Second line from top, for "Obs. 18," read "Obs. 20." " 309, Third line from the top, for "cube of 8," read "cube root of 8." " 311, Obs. 15, for "their Factors are," read "their Squares are." " 321, Fourth line from bottom, for "Obs. 15." read "Obs. 17." " 322, Operation, for "Obs. 15," read "Obs. 17." 324, Definition, for "Obs. 14," read "Obs. 16;" and Cbs. 2. for "Obs. 15," read "Obs. 17." 326, Operation, for "Obs. 24," read "Obs. 23." " 331, Second line from top, for "Obs. 8," read "Obs. 10." " 338, Operation, for "^/512," read " V512." " 339, Operation, for "Art. 1, Obs. 2," read "Art. 1, Obs. 4." " 340, Obs. 2, Demonstration, for "Obs. 9," read "Obs. 11." " 342, Obs. 10,, Rem., for "Obs. 17," read "Obs. 19;" and Obs. 12, f.r "Obs. 18," read "Obs. 20." " 344, Obs. 17, Demonstration, for "Obs. 12," read "Obs. 14," " 349, Bottom line, for "Obs. 11," read "Obs. 14, a." 359, Note 3, should be signed "Peter Trueman, "Thomas Jones." These are the most prominent errors detected. Some few typographical errors have been noticed, which it is not necessary to mention here. There may perhaps be a few errors of importance that have escaped notice. All such will be corrected in future editions. U. C. BERKELEY LIBRARIES I 1111 II 11 IU l lll C061347477 I,: ~ ~~~~~~~~~~~~1-1~~~~~~~~ - iN I -." I~~~~~~~~~~~~~~~~~id~i~;ii, - -,; -4 iP t i~~~~~~ t %i - Niger -1, ~~~~~~~~~~~~~~~~~I~r*V IX".",4 ~ ~ rsuc-.rw `.~ ~ 11 I1