MCLECTiC EDUCATIONAL COURSE. IA Y'S GHER ARITHMETIC. THE PRINCIPLES OF AI I T H R E T I C, ANALSYZED AND PRACTICALLY APPL1 EAD, FOR ADVANCED STUDENTS By JOSEPH RAY, M oD. ZATE PROFESSOR OP hIATRERIATICS IN WOODWARD COLLEGE. EDITED BY CHAS. E. MATTIEWS, RM-.A R2EVISED S 2'7:iL'ror't'IE ELDfOXV CI I T N A T I SARGENT, WILSON & HINKLE. NEW YORIE: CLARIK & MIAY'NARD. TIlE BEST AND CIIEAPEST MATHEMATICAL WORKS. BY PROFESSOR JOSEPH RAY. Each Boo3 of the Arithmetical Coarse, as well as the Algebraic, iz Complete in itself, and is sold separately. PRIMARY ARITHI ETIC.-RAY'S ARITHMETIC, FIRST BOOK: simple mental Lessons and tables for little learners. INTELLECTUAL ARITHMETIC.-RAY'S ARITItIMETIC, SECOND BOOK- the most interesting Intellectual Arithmetic extant. PRACTICAL ARITHMETIC.-RAY'S ARITHAIETIC, TTILtD BOOK: for schools and academies; a full and complete treatise, on the inductive and analytical methods of instruction. KEY TO RAY'S ARITHMIETIC, containing solutions to the questions; also of test Examples for the Slate and Blackboard. RAY'S HIGHER ARITHMETIC.-The Principles of Arithmetic, analyzed and practically applied. For advanced classes. ELEMENTARY ALGEBRA.-RAY'S ALGEBRA, FIRST BOOK; for common schools and academies; a simple, progressive, and thorough elementary treatise. HIGHER ALGEBRA.-RAY'S ALGEBRA, SECOND BOOK; for advanced students in academies, and for colleges; a progressive, lucid and comprehensive work. KEY TO RAY'S ALGEBRA, FIRST AND SECOND BOOKS. In one vol.'lAtiaring for l tblitaftgllt 1.-THE ELEMENTS OF GEOM1ETRY, embracing plain and solid geometlry, with numerous practical exercises. II.-TRIGONO0IETRY AND MBENSURATION, containing logarithmic computations, pl.ne and spherical trigonormetry, with their applications, mensuriation of planes and solids, with logarithmic and other tables. III.-SURVEYING AND NAVIGATION; sur veying and leveling, navigation, baromnetl ic hights, &c. To be followed by others, forming a complete Mathematical Course for schools and colleges. CHAs. E. MALrrTTUIEWs, MA. A. for many years a pupil of, and sut:bsequently an associate Instructor with the late DR. RAY in the Mathematical department of Woodward College, will edit and superintend the publication of the unfinished parts of the series. ENTEMIEcD according to Act of Congress, in the year Eighteen lunldred and Fifty Six, by WiN'NrnaoP B. S.IIrT, in the Clerk's Officet of the District Court of the United States for the Southern District of Ohio. Stereotyped by C. F. O'Driscoll & Co. P R E F A C E. As my name appears in connection with that of the late MiL JOsEPr RAY, on the title-page of this work, it is proper to account for the circumstance by referring briefly to matters personal in their nature. As an author, Dr. Ray is known to the public at large, through the medium of his excellent mathematical publications. As an able' and faithful teacher, his merits are deeply impressed on those who have been so fortunate as to come within reach of his immediate instructions and example. Ili every line of duty he was conspicuous for unremitting industry, and in all the relations of life, his first desire was to be of service to others. To me, his memory is endeared by many acts of friendship and confidence, extending over a period of twenty years, sixteen of which were spent with him as pupil and colleague, in Woodward College and tIigh School. In his last illness, Dr. Ray expressed a wish that I should prepare for publication, his IIigher Arithmetic, then unfinished, and directed his manuscripts and materials to be handed to me for chat purpose. In performing the part thus assigned to me, I have endeavored to preserve the spirit and pursue the course indicated in the other Arithmetics of this series. With this view, the language of the autthor has been retained without material change as far as. the manuscript was complete and ready for publication. Whenever a change has been made, it has been done with great cautioni and with an anxious desire to carry out the general plan of the work. New matter has been occasionally introduced when it seemed necessary to render a demonstration more clear and con-;luJsive, or the treatment of a subject more full and satisfactory. This hals been especially the case in the Contractions in lMultiplication, Contractions in Division, Greatest Common Divisor and Least Common Multiple both of whole numbers and fractions. (iii) IV PREFACE. Pure Arithmnetic, comprising Simple Numbers, Common and Iecimal Fractions has been discussed before any of its applications. T]his arrangement is philosophical, and not open to objection in A work of this character. In questions of proportion, and generally throughout the book, the analytic method of solution has been preferred to mere formal and irrational directions; for no true development of the intellectual powers or satisfactory knowledge of any science carn be attained, unless the spirit of every operation is clearly seen through its form. Where the importance of the subject seemed to demand it, as in Insurance, Simple Interest, Compound Interest and their applications, brief methods of operation have been given, and prac. tical rules deduced for the use of accountants and men of business. Thile difficulties and obscurities attending some of the applications of Simple and Compound Interest are inseparable from the subjects themselves, and it is better to meet and overcome these obstacles in the school-room than in the counting-room. No apology is therefore offered for introducing these subjects and discussing them at large. The exercises are numerous, most of them new and interesting, and have been prepared with a view to practical utility. While they afford a full and thorough exercise in the principles of Arithmetic, at the same time, they enable the pupil to make use of his own knowledge to the best advantage. Circulating Decimals and other matters rather curious than extensively useful, may be omitted until a review, if the want of time or the character of the pupils make it necessary; and any exampies considered too difficult at first may also be postponed in the r Pa e wary THE EDITOR. STEREOTYPE EDITION. The marked approbation extended to this volume, and its wide introduction into the best Schools of the country, having led to the sale of several editions in a few months, the work has been thoroughly revised, and is. now presented in a permanent stereotype form. In the revision, by a careful abridgment of language, the ideas are presented with greater precision and perspicuity, the book condensed, and a large number of new and interesting examples added. CONTENTS. PAGE [NTROI)UCTION.. o. e. o a. - 9 SI3MPLE NU9MBERS. Numeration and Notation O -..o..... 11 Addliti on.... o.19 Subtraction. o 22 Multiplication.................26 Division............ 5 General Principles of Multiplication and Division..... 47 Summary of Principles............ 51 PROPERTIES OF NUMBERS......... 55 Factoring................. 56 Greatest Common Divisor....... 60 Least Common Multiple........ Proofs of the Rules by casting out the 9's and 11's..... 67 Cancellation............. 71 COnmItON FRACTIONS........... 2 Numeration and Notation of Fractions..... 71 Reduction of Fractions........ 77 Addition of Fractions......... 84 Subtraction of Fractions..... 85 Multiplication of Fractions........... 87 Division of Fractions. 89 To Reduce Complex Fractions to Simple ones.... 92 To find the Greatest Common Divisor of Fractions... 93 To find the Least Common Multiple of Fractions..... 94 D)ECIBIAL FRACTIONS............. 97 Numeration and Notation of Decimals..... 99 Reduction of Decimals............. 103 Addition of Decimals....107 Subtraction of Decimals.......108 MIultiplication of Decimals............. 110 Division of Decimals........... CIRCULATING DECIMIALS........ e o. 120 Reduction of Circulates....... 123 Adclition of Circulates.... 124 Subtraction of Circulates.....125 Multiplication of Circulates............ 126 Division of Circulates....... 127 ni CONTENTS. PAGE. CO1MPOUND NUMTBERS........... 128 Long or Linear -Measure -Mariners' Measure... 129 Surveyors' and Engineers' Measure....... 130 Cloth Meaesure...to). 130 Square and Surface Mleasure. B o..... 131 Land Measure.. 0. o. 132 Cubic or Solid MIeasure... a 0.. o - 133 Troy or Mint Weight.......e o.o. 134 Diamond Weight............. 135 Apothecaries Weight..... o o.. 135 Avoirdupois or Comniercial Weight.. e o. o... 136 Comparison of Weights..... o e.... 136 Wine Measure......... 137 Ale and Beer Measure... o... 0 o o.. 137 Dry Measure......... 138 Comparison of Measures.... o... 138 Apothecaries' Fluid Measure........ 139 Time........... 139 Circular or Angular Measure.......... 142 Comparison of Time and Longitude....... 142 Federal or United States Money....... 143 English or Sterling Money............145 State Currencies.........146 French Weights, Measures, and Money....... 147 Foreign Weights and Measures........ 148 Reduction of Compound Numbers,.u....... 151 Addition of Compound Numbers........166 Subtraction of Compound Numbers....... 168 Multiplication of Compound Numbers........ 171 Diivision of Compound Numbers.......... 175 ALIQUOT PARTS.......... 181 Exercises in State Currencies....... 184 Miscellaneous Examples in Compound Numbers..... 186 RATIO..... O D, e o e. e... 187 PROPORTION.......... 191 Simple Proportion.......... 194 Compound Proportion......... 199 Rule of Cause and Effect...... O O.. 200 PERCENTAGE................20 To Find any given Per Cent. of a Number....... 203 Tb Find the Rate Per Cent. one number is of another... 206 To Find a Number, when a certafin per cent. of it is known 2. 08 A Number being given, which is a certain per cent. more or less than another, to find that other...... 210 CONTENTS. vii PAGE. APIPLICATIONS OF PERCENTAGE.......... 212 Gain anid Loss....... 213 Commission and Brokerage........ 219 Stocks and Dividendsc... e 225 Par, Discount, and Premium........ 227 Insurance............. 2 Taxes...... 236 Duties or Customs.......... 239 INTERLEST............. 242 Simple Interest.............. 244 Pr-actical Rules for computing Simple Interest..... 247 Present Worth and Discount........... 253 BANKING................ 256 Pr}omissory Notes -........... 257 To Find the day a Note is legally due...... 258 DISCOUNTIN(G NOTES......... 258 To Find the Proceeds of a Note.......... 260 To Find the rate of Interest, when a Note is Discounted. 263 To Find the Face of any Note, when the proceeds, time, and rate of Discount, are given......... 264 To find the rate of Discount, corresponding to a given rate of Interest........265 Rules for Partial Payments... 266 EXCHANGE.................. 271 Table of Foreign Coins and moneys of account.... 274 Home or Inland Exchange............ 275 Foreigrn Exchange............ 276 Arbitration of Exchange.......... 278 Chain Rule..... 279 ACCOUNTS CURRENT.............. 282 To settle an Account Current whose items draw Interest. 283 To settle a Mierchant's Account Current....... 286 STOiRAGE ACCOUNrS............ 288 E](QUATION OF PAY3IENTS.......... 289 CO1MPOUND INTEREST.......... 295 To firld the Comp. Int., principal, rate and time given.. 29( To findl tlte Rate, when the principal, time, and compound interest or amount is given......... 300 To find the Principal, time, rate, and interest given.. 301 The same, timne, rate, and compound amount given.. 302 To find the Time, when the principal, rate, and compound interest or amount, are given....... 303 viii CONTENTS. PAGE. ANNUITIES............ 303 To find the Initial value of a Perpetuity...... 304 To find the Present value of a deferred Perpetuity... 305 To find the Present value of an Annuity certain.... 306 To find the Forborne or final value of an Annuity.... 307 To find an Annuity, when its present or final value, rate of Interest, and time to run, are known.... 309 To find the time a given annuity runs, when the rate of Interest an-d present or final value are known.... 309 To find the rate of Interest of a given annuity, the present value, and time to run, known......... 310 Contingent Annuities........... 311 To find the value of a Life-Estate or Widow's dower... 311 To find how large a Life Annuity can be purchased for a given sum.................. 313 To find the value of the reversion of a Life Annuity... 313 To find the Single and Annual Premiums necessary to secure a given sum at the death of the person Insured. 313 To find the value of a Policy of Life-Insurance..... 314 PROPORTIONAL PARTS............ 315 Partnership............. 317 Partnership with Time............ 319 Bankruptcy.............. 321 General average.a......... 322 Rate Bills for Schools,...o 324 Alligation........... 325 INVOLUTION.. o... 330 EVOLUTION.. 332 Square root.............. 332 Cube Root........... 338 Extraction of any Root... e a...e.. 342 SERIES........ o...... 344 Arithmetical Series.... 345 Geometrical Series. 346 PERMUTATIONS, COMBINATIONS........ 348 SYSTEMS OF NOTATION......... o o 349 DUODECIMALS.-......50 MENSURATION........... 351 Plasterers', Painters') and Pavers' Work.... 352 Mensuration of Solids...... 354 Masons' and Bricklayers' Work......... 857 Guaging............... 35 Tunnage of Vessels........... 359 MECHANICAL POWERS........ 3.....3C ARI T HMETIC. I. INTRODUCTION. ARTICLE 1. QUANTITY is any thing which can be in. creased or diminished. Thus, numbers, lilies, space, motion, time, and weight, are quantities. ART. 2. AIATHEMIATICS is the science of quantity. ART. 3. The fundamental branches of Mathematics are Arithmetic, Algebra, and Geometry. ART. z4. ARITHMETIC is the science of numbers. ART. 5. A PROBLEM is a question proposed for solution. ART. 6. A T-IEOREI is a truth to be proved. ART. 7. Problems and Theorems are both called Propositions. ART. 8. A COROLLARY is a truth deduced friom a preceding proposition. ART. 9. A DEMIONSTRATION is a process of reaso-ning by which a proposition is shown to be true. ART. 10. A DIRECT DEMONSTRATION is one which commences with known truths, and by a chain of reasoning establishes the proposition to be proved. R E V E W.-. What is quantity? Give examples. 2. What is Minthematics? 3. Whnat are its fundamental branches? 4; Define Arithmetic. 5. Whabt is a problem? 6. A theorem? 7. What common name is applied to both? S. What is a corollary? 9. A demonstratiorn? 10. A direct demonstration? (9) 10 RAY'S HIGHER ARITHMETIC. ART. 11. An NI)DTECT DEM3TONSTRATION is one which assumes the proposition to be f;/l.se, and then proves that som0e l,absurrdity will necessarily follow. This is sometimes called 2Jctdio a d adbsLuraotmL. Arr. 12. An AXIOM is a self-evident truth; that is, a proposition so evident that it can not be made plainer by any demonstration. The following are among, the most important axioms. 1. If the same or equal quantities be added to equals, the sums will be equal. 2. If the same or equal quantities be subtracted from equals, the remainders will be equal. 3. If the same or equal quantities be multiplied by the same number, theproducts will be equal. 4. If the same or equal quantities be divided by the same number, the quotients will be equal. 5. Generally, if the same identical operation be performed on two equal quantities, the results will be equal. MATHEIMATICAL SIGNS. ART. 13. For brevity, characters, called s.igns, are used in MFathematics. Those most used in Arithmetic, are -- X, x, -, =, ()or ART. 14. The sign —, read plizs, is the sign of' Adclition; it shows that the numbers between which it is placed are to be added together. Thus, 3 + 5 equals 8. ART. 15. The sign -, read mznus, is the sign of Sitbtraction; the number which follows it is to be subtracted from that which precedes it. Thus, 7 - 4 equals 3. Pleis and minus are Latin words, signifying miore and less.. ART. 16. The sign X, read times, is the sign of Maiilip)1icction. The numbers between which it is placed are to be multiplied together. Thus, 4 X 5 equals 20. ART. 17. The sign -, read (divifled by, is the sign of Div;sion. The number which precedes it is to be divided by that which follows it. Thus, 20 — 4 equals 5. lRi, v a cv.-i11. An indirect dernonstration? 12. W hat is an axiom? Naime the most important axioms. 13-19. Describe the signs most fre. quently used in Arithmetic, and give examples of their use. NUMERATION AND NOTATION. 11 AnT. 18. The sign -, read equals, or is equal to, is the sign of Lti, uality; the quantities between which it is placed are equal to each other. Thus, 5 + 3 =9 -- 1. APRT. 19. A parenthesis ( ), or vinclum, —-- shows that two or more numbers are to be considered as one, Thus, (7+4) X 3 = 33; or 8 —5 x 4 —19~ 5 2. ARITIIMETICAL DEFINITIONS. ART. 20. A unit is a single thing; a cent, a hat, &c. ART. 21. A n7zumer is a unit, or a collection of units classed under the same name, and answers to the question, 1o1o mzcany? The unit of a number is ove of the things it expresses: thus, in five cents, one cent is the unlit; in ten apples, one a21'le is the unit. Units are sometimes only 7relative in their character; thus, one jbot is a unit in regard to feet, but it is only a part of a unit in regard to yards. One diniv,, is a unit, that is, one in regard to dimnes, but it is ten in regard to cents. ART. 22. Numbers are either abstract or concrete. An ablstract iumbze? is one in which the kind of unit is not designated, as one, two, three, &c. A concrete number is one in which the kind of unit is designated, as one ccnt, tzo aplp)es, ten, bzshels, &ec. Concrete numbers are frequently called denomi[,nate nlumlner'. ART. 23. Arithmetic is founded on,NOTATION, and its operations are carried on by means of ADDITION, SUBTRACTION, MULTIPLICATION, and DIVISION. These are termed the fundamental rules of Arithmetic. II. NUMERNTION AND NOTATION. ART. 24. NUMIFERATION is the art of naming nulmbe)rs. NOTATION iS the art of representing numbers by ci;: - acters called figlres or digits. REV E w.-20. What is a unit? 21. What a numnlber? Shiow the relative character of some units. 22. What are abstract numbers? Con. crete? Give examtples. 23. On what is Arithmetic foundeel? What are its fundam ontal rules? 24. What is numeration? What is notation? RAY'S HIGHER ARLTHIMETIC.. ART. 25. The first nine numbers are each represented by a single figure, thus: 1 2 3 4 5 6 7 8 9 one. two. three. four. five. six. seven. eightt nine. All other numbers are represented by combinations o;f these and another figure, 0, called zero, naught, or Ci)h'. R EMARI K. —The cipher, 0, is used to indicate no value. The other figures are called signyficant figures, because they indicate some value. ART. 26. The number next higher than 9 is named TEN, and is written with two figures thus, 10: in which the cipher, 0, merely serves to show that the unit, 1, on its left is different from the unit, 1, standing alone, which represents a sinyle thing, while this represents a single groupn of tesn things. The numbers succeeding Ten are written and named as follows: 11 12 13 14 15 16 eleven. twelve. thirteen. fourteen. fifteen. sixteen. 17 18 19 seventeen. eighteen. nineteen. In each of which, the 1 on the left, represents a grovp oj tenl things, while the figure on the right, expresses the units or single things additional, required to make up the number. R EXrnIAR.-The words eleven and twelve are supposed to be derived from the Saxon, meaning one left after ten, and two left after ten. The words thirteen, fourteen, &c., are contractions of three and ten, four and ten, &c. The next number to nineteen, (nieve and ten), is to, ancd ten, or two groups of ten, written 20, and called tzwent(y. The next are tweenty-one, 21; twenty-two, 22; &c., up to three'ens, or thirty, 30; fortly, 40; fifty, 50; sixt7y, G0; seventy, 70; eiyght, 80; ninety, 90. REVIsW, W.-25. How are the first nine numbers represented? HIow are numbers above nine represented? Rem. For what is the cipher useld? What are the other figures called? 26. TIow is Ten written? Wh:at does the cipher show? What does the 1 represent? Write the next ninenumbers. Explain their names. Show what the figures denote in each. NUMERATION AND NOTATION. 13 The highest number that can be written with two figures is 99, called ninety-nine; that is, nine tens and nine units. The next higher number is 9 tens and ten, or ten tens, which is called a hundred, and written with three figures, 100; in which the two ciphers merely show that the unit on their left is neither a single thing, 1, nor a group of ten things, 10, but a group qf ten tens, being a unit of a higher grade than either of those already known. In like manner, 200, 300, &c., express two huzndreds, three hundreds, and so on, up to ten hundre(ds, called a thousand, and written with four figures, 1000, being a unit of a still higher order. ART. 27. From what has been said, it is clear that a figure in the 1st place, with no others to the right of it, expresses uznits or single things; but standing on the left of another figure, that is, in the 2d place, expresses groulps of telns; and standing at the left of two figures, or in the 3d place, expresses tens of tens, or hundreds; and in the 4th place, expresses tens of huncdreds or thousands. Hence, counting from the right hand, The order of Units is in the Ist place. 1 The order of Tens is in the 2d place. 10 The order of Hunudreds is in the 3d place. 100 The order of Thoutsands is in the 4th place. 1000 By this arrangement, the samile hfiye has del.rent values according to the place in which it stands. Thus, 3 in the first place is 3 unizts; in the second place 3 tens, or thirty; in the third place 3-huntdreds; and so on. AnRT. 28. The word UNITS may be used in naming all the orders as follows: Simple units are called... Ufits of the 1st order. Tens......... Units of the 2d order. hitndreds.... Units of the 3d order. Thousands.,.. U lits of the 4th order. &c. &c. REEYIE.-26. What are 20, 30, 40? What is the highest number of two figures? W\hat is the next called? Iow is it written? Explain its figures. What is a Thousand? How written? 27. What does a figure in the Ist place express? a figure in the 2d place? in the 3d place? in the 4th place? How does the value of the same figure vary? 14 RAY'S HflIGIER ARITHiMETIC. N oTE. —When units are named without reference to any particulza order, units of the first order are always intended. ART. 29. The preceding articles show the method of expressing numbers less than one thousand. For exanmpie, in the number fotlr hundrecl and twenlfy-five, there are 4 hundreds, 2 tens or twenty, and 5 units; the number is therorfe written, 4 2 5. in the number three hunedred autd nzine, there are 3 hundreds, 0 tens, and 9 units; or 3 units of the third, and none of the second, and 9 of the first order; hence, the number is represented thus, 309. ART. 30. SUIMMARY OF PRINCIPLES. 1. All nmnzbers are represeztedl by the?nine digits and zero. 2. Zero has no value; its use is to fill vacant places. 3. The base of ozlr system of notation is tent; ten uZits of any orlder making one unit of the ordler next higher. 4. The same figure has different values according to the place it occupies. ART. 31. TABLE OF ORDERS. 9th. 8th. 7th. 6th. 5th. 4th. 3d, 2d. lst. 0 2 * o e a o o o o e bers, periods of three orders each, are used. The first or-ders,'NITSI TENS, and HUNDREDS, constitute the first or UNIT ED. The second orders orm the second or TtlO.USAND PERIOD; the third 3 orders, the third or 1lILiLION PERI1OD. 0_ _ R v w. 29. Writeo the numnber four hundre and l twenty-five. Write,he number three hundred and nine. NOTATION AND NUMERATION. 15 ART. 33. List of the Periods, according to the commcn or French method of Numeration. First Period. Units. Sixth Period. Quadrillions. Second... Thousands. Seventh... Quintillions. Third... Millions. Eighth... Sextillions. Fourth... Billions. Ninth... Septillions. Fifth... Trillions. Tenth... Octillions.'t'he next tiwelve periods are, Nonillions, Decillions, Undecillions, Duodclecillions, Tredecillions, Quatuordecilliolls Quil:decillions, Sexdecilliotns, Septendecillions, Octodecil lions, Novendeeillionss, Viintillions. ART. 34. Division of the orders into periods. 6,54 21 9 8 654 c 1 egin at te ih ad oin e be ito erids of three figes eacH. Comece at the le, and read i stcessio each eiod wit its ame. RK.0 u0bers mays aby merely ning each five, or two hundreds no tens and five units.a 654 321 987 654 321 5th Period. 4th Period. 3d Period. 2d Period. Ist Period. ART. 35. RULE FOR NU3IERATION. Exprin at te wordsight, ath point the numb into eriods of three Ji08921045es each. meThe at tnu e let, adiv red into sccessio eacis 0' R ErA921'045 mbers may also be reandby merely nne ig eachiions figuzre wiUth tIhe na7m7e of the place in aoliich it stlnlds. Tlhis nletllodl, hiowsever, is rarely used except in teaclling beginnerqs. rThuls, thle nullmbers expressedl by the figures.205, maLy be reacd two huzalded aund -ve, or two hundledls no tens and five units. Express in words, the number which is represented by 608921045. The number, a4 divided into periods, is 608 921 045; and is read sis hundred and eight nlilliong nine hundred and twenty-one thousand and forty-five. R EVIE. w —30. What are the principles of Notation and Nlumeration? 31. Repeat the Table of Orders. 32. What are periods and their use? 33. Name the first ten. 34. Give an example of the use of periods in a. particular case. 35. What is the rule for Numeration? What other method may be used? 16 RAY'S HIGHER ARITHMETIC. EXAMPLES IN NUMERATION. 7 12345 1375482 29347283 40 68380 6030564 37053495 85 94025 7004025 45004024 503 70500 8025607 50340726 278 165247 9030040 60025709 1345 350304 6002007 343827544 2450 204026 4300201 904207080 3708 500050 8603004 700200408 4053 808080 2030405 502003070 7009 730003 6005010 830070320 832s045682-327825000000321 8007006005004003002001000000 60030020090080070050o060030070 504030209102800703240703250207 AIRT. 36. RULE FOR NOTATION. Write, first the number of the highest period, then, of the other periods in their proper succession, filling vacant places with cip hers. Express in figures the number four millions twenty thousand three hundred and seven. Ans. 4020307. VWrite 4 in millions period; place a dot after it to separate it from the next period: then, write 20 in thousands period; place another dot: then write 307 in units period. This gives 4' 20'307. As there are but two places in the thousands period, a cipher must be put before 20 to complete its orders, and the number correctly written, is 4020307. N O'TE.-Every period, except the highest, must have its three figures; and -if any period is not mentioned in the given number, supply its place by three ciphers. PRooF.-Apply to the number, as written, the rule for numeration, and see if it agrees with the number given. EXAMPLES IN NOTATION. 1. Seventy-five. 4. Two hundred and forty 2. Ninety. four. 3. One hundred and thirty- 5. Two hundred and forty. four. 6. Two hundred and four. NUMER ATION AND NOTATION. 17 Th. lIre'e hnre 1( andl sevenlty. 30. Twentr-three ml llin(,s 1x 8.,x }t illtt(litd 1nd( ten. i,,i nil.el nt l I f'irt — tiv(' tlheIU9 I "It hiiidretsl itl one. sadl lnine hundred and sev" V11:ty-o) l e. [0. COne thllolsanlll two hul-lndred L andl tlhOlir1tyt-fere.d 31. Ten millions one lulndredl thmt18uo l{.l. ntnd ten. 11.'1'ree thoIlsalnd four hun41Vdied and fline1ty..321. Ninety miilions ninety tholudred and l inety. sand and it limety. 12. Seven thousand and twenty- 1 it.-eilit m illions s five 3 I<'.',lt r i I Je ent y tholsaLnd and lfive. 13. Niie thousand and seven. - Sty millions seven hun34. Sixty millions Seven hun14. lotl't tllousalnd five hun- tied anld five tlhousand. drie( antl sixty-three. 35. One liunidred millions. 15. Seventyr thousand anl twen- 36. Tw'o hund-red and fifty miltvi7. lions three hlundrl(1ed(1 and [G. Eiglity-four thousand and three thousand and twentytwo. () i. 17. Nitlcty thousand and nine. 37. Eight hundred and seveln 18. Sixty thlousand six hun- nill ilins forty tholiustnd and (d1 r.ed. tlrlilrtS —,ne. LT. O)e I llnl ed Intl si.xty-for 38.,(Seven }1 111111rel ni illions sevtIle tsait1tl tl)nec hfunldided and emtty tllusai;ld an(l seven. niistv —,,luir39. T'Iw\-) illions ard twenty 20.'1\ lt( hl lllMredl and seven tHl)Mstttt11 1d )ibt lhundJi-ed and1 I liensaiid flair fiumeed 1.i 40. Thi irty trillions thlirty mil1111 1,,litis thlirty thoutsand muid 21. Fiv\e ltundred thousand andl tli ty t.. 41. Niileteen Cl1tlIdrill itns twen22..Six ttlreh1 and fortyS ttlo iou-.~ct t t ii,t i ayd five hIunI atH1 Ift. (l,(t1 h Ired blillils. 23. Ei; tit ltitil cl'and twro 4; li'I- tlv 42.'T'cn (piadrilli(ons foitr lmnred.drl atlltl tllree trillions 24. Ni i e li i idr(i e thousand nI i I t N f tIIS and six fii mithi5e f..hS(lldredC:i,- etv Illies.ns. 23. Seveti huihdr'ed thousand 43. Ei'll il ti tll tilis sfiti s e - 2G. o)lle iilliion foiupr Ilundrlred/ tvwenty-five ili itilltiois and I lIl twelttN-{me tle, usli t i tiii.Y Illl~ll Ol\'-ttit l thlttsil~ltl thirty -three ftiffiois. si X c. LI I d. and elty - six iulIr and.i lty- 44. Nine l tinhlred decillions li ye. severitv n1,tilIlins six Oc-'27. Six millions sixty thousand tilliois f;oty st'dtilittiis if-;uh sui. ty q1 1'il 1,s ii.s t( i ln28. Nite nmillions nine thou- dCreil 1ll(1 {1or' tillIIis ten S;1Ill1 anidf tn1e1i. t ili ills ortLY tllhus: iid anid'29. S;cilt Ill ill s and seventy. I sixty. iE VIE w.-36. What is the rule for notation? What is the proof? 18 RAY'S HIGHER ARITHMETIC. ENGLISH METHOD OF NUMERATION. ART. 37. Although now but little used, the learner should understand this system of Numeration. The first six orders have the same names as in the French method, and these constitute the period of Units. The next oi;Millions period, consists of six orders. Each succeeding period consists of six orders, and their names are Billions, Trillions, Quadrillions, Quintillions, and so on. The following table illustrates this method: 8,.. 0: a - HE 4 3 2 1 0 9 8 6 5 4 3 2 1 0 9 8 6 5 4 3 2 1 By this system the first twelve figures would be read, two hundred and ten thousand nine hundred and eightyseven millions six hundred and fifty-four thousand three hundred and twenty-one. By the French method they would be read, two hundred and ten billions, nine hundred and eighty-seven millions, six hundred and fifty-four thousand, three hundred and twenty-one. ROMAN NOTATION. ART. 38. In the Roman Notation, numbers are represented by letters. The letter I represents one; V, five; X, ten; L, fifty; C, one hundred; D, five hundred; and M, one thotsand. The other numbers are represented according to the following principles: lst. Every time a letter is repeated, its value is repeated. Thus, II denotes two; XX denotes twenty. hR VrE0W.-37. Explain the English method of Numeration. Give an example of its application. 38. How are numbers represented in the Roman Notation? What numbers are represented by 1, V, X, C, D, M? ADDITION. 19 2d. Where a letter of less value is placed before one of greater value, the less is taken from.the greater. If placed after it, the less is added to the greater. Thus, IV denotes four, while VI denotes six; IX denotes nine, while XI denotes eleven. 3d. A bar, -, placed over a letter increases its value a thousand times. Thus, V denotes five thousand; M denotes one million. ROMAN TABLE. I.. One. XXX.. Thirty. II... Two. XL.. Forty. III... Three. L.. Fifty. IV v.. Four. LX.. Sixty. V F.. Five. XC.. Ninety. VI.. Six. C. One hundred. IX... Nine. CCCC. Four hundred. X.. Ten. D.. Five hundred. XI. Eleven. DC. Six hundred. XIV.. Fourteen. DCC.. Seven hundred. XV... Fifteen. DCCC.. Eight hundred. XVI.. Sixteen. DCCCC. Nine hundred. XIX.. Nineteen. M.. One thousand. XX... Twenty. MM.. Two thousand. XXI.. Twenty-one. i MDCCCLVI 1 8 5 6. III. ADDITION. ART. 39. ADDITION is the process of collecting two or more numbers into one sum. Sum or Amount is the result obtained by Addition. ART. 40. Since a number is a collection of units of the same kind, two or more numbers can be unitedt into'one sum, only when their units are of the same kind. Two apples and 3 apples are 5 apples; but 2 apples and 3 peaches can not be united into one number, either of apples or of peaches. REVI EW.-38. What is the effect of repeating a letter? Of placing a letter of less value before another of greater value? Of placing one of less value after one of greater value? Of placing a bar over a letter? 39. W5hat is addition? Sum or amount? '20 ERAY'S II GH 1 ER ARITIMETIC. Neverthleless, numbers of different names may be addedl toc)etheor, if' tliey catn be brmtigliht ntider a (c)?,teiot (/ ))tt jii(:t/,'lTt. Ttlus, 2 7nel and t3 women are'5 felrsoals. Two horses, 3 shic.2p), and 4 cows, are 9 animalls. RULE FOR ADDING SIMPLE NUM3IBERS. 1. W'rife the tnm6bers to be added, so tha!t fi/grte.s of the.same orlder 7)1ay steadl int, coltnn., tnlits ut)lder' unils, tens tundtler ltes, 7I ta it (l. s l.under h i rdieds, d'c. 2. Beryinl at the right, a nld add each colamnz separately, placing (fle ulnlis of eac/~h sultm tlldle? tfhe columnl aldded, ad calrryinlq the tenls to the next coltumzn. At the last co/htltt, set dolwn1 the whole Wllhat is the sum of 639, 82, and 543? So rUT r o N.-Writing the numlers as in the margin, 3 9 say,: 111an'2 ai.e D, tl1(l 9 are 14 naits, which are I ten 8'2 nalll 4 units. Write tile 4 units btealrtllh nd carry the 54:3 I ten to tile next collmn. T']'e 1 and111 4 are -), nid 8 arle 1 13, and; 3'.Ce lt tens, vwhlich tire (3 tens to be written beneatl, iand 1 lindrlled to be carried to tile next columnn. Lastly, I and e (a, anril 6'are 12 hItudrledts, whlicl is set beneath, thiere binlg i.)no other coluitlns to carly to or add. D:osr o NS'T RAT ON.-l. Figures of the same orlder are writteii utilder each other for ( coLce'l tcc,,, siltce inoue but units of thle same riamle cn-n be added. (Art. 40.) C). Ci( mence at the riglit to ad(l, so tlhat whlen the sum of itrly c(olu1nII is ugreater than iirine, the ti,,ts Iiay be carriel to tlhe cnext column, and, there-tby, units of the s5at.ie natne lctddcd together. 3. Carry oite for every ten, since ten units of each order make oie unit of tlie order next highler. (Art. 30.) METHIODS OF PROOF, AnT. 41. 1. Add the figures dlownwatrd instead of ipE ca1 i'd; or R i v i, w -40. What is necessnry in order thlllt t1nnbers mnay he i(tided into'- oiue su ll? W-y 11? \'lat is the rile tor iddlitig simlllle inuli Iers? Explaiin the example, and give the reasons for the rule. 4L., What is the trst method of proof? ADDITION. 21 2. Separate the numbers into two or more divisions; find tile suIm of the numblers in each division, and then add thla several sums together; or 3. Commence at the left; add each column separately; place cal-h sum under that previously obtained, but extending, one figure further to the right, and then add thenz tegetlter. In all tilese methods the result should be the same as when the numbers are added upward. N o rE. —For proof by casting out the 9's, see page 68. EXAMPLES FOR PRACTICE. Find the sum, 1. Of 767; 6 75; 76 540121; 50 1; 775. Ans. 135317. 2. Of 97674; 686; 7676; 9017. Ats. 115053. 3. Of 971; 7430; 97476; 76734. Ails. 182611. 4. Of four hundred and three; 5025; sixty thousand ardT1 seven; eigilty-seven thousand; two thousand and. ninety, and 100. Ans. 154625. 5. Of 9199; 8400; 73; 47; 452; 11000; 193; 97; 9903; 42, and 5100. AiLs. 31306. 6. Of 200050; three hundred and seventy thousand two hundred; four millions and five; two millions ninety thousand sev\en hundred and eighty; one hundred thousalld and seventy' 98002; seven millions five thousand and one; and 700T 0. Ans. 13754178. 7. Of 609505; 90070; 90300420; 9890655; 789; 37599; 19962401; 5278; 2109350; 41236; 722; 876t4; 29753, and 370247. Ans. 123456789. 8. Of two hundred thousand two hundred; three hun. dcd millions six thlousand and thirty; seventy millions seventy tllousand and seventy; nine hundred and four millions nine thousand and forty; eighty thousand; ninety nillions nine thousand; six hlundred thousand and sixty; five tllousand seven hundred; four millions twenty tbousald and t\wenty. AIs. 13369000190. IRtVIsw.-'What is the second method of proof? The tUird? 22 RAY'S HIGHER ARIThTIMET1C. In each of the 7 following examples, find the sum of the n umbers from A to B, including these numbers. A. B. 9. 119... 131... Ans. 1625. 10. 987... 1001... Ans. 14910. 11. 3267... 3281... Ans. 49110. 12. 4197... 4211...Ans. 63060. 13. 5397... 5416.. Ans. 108130. 14. 7815. 7. 7831.. Ans. 132991. 15. 31989... 32028... Ans. 1280340. I16. Paid for coffee, $375; for tea, $280; for sugar, $564; for molasses, $119; and for spices, $75: whdt did the whole amount to? Ans. $1413. 17. Bought three pieces of cloth: the first cost $87; the second, $25 more than the first; and the third, $47 more than the second: what did all cost? Ans. $358. 18. Bought three bales of cotton. The first cost $325; the second, $16 more than the first; and the third, as much as both the others: what did the three bales cost? Ans. $1332. 19. A has $75; B has $19 more than A; C has as much as A and B, and $23 more; and D has as much as A, B, and 0 together: what sum do they all possess? Ans. $722. IV. SUBTRACTION. ART. 42. SUBTRACTION is the process of finding the difference between two numbers. The larger number is the Jlinuend; the less, the Subtra hend; the number left, the Difference or Renzainder. Minuend means to be diminished; subtrahend, to be os, btracted. ART. 43. Subtraction is the reverse of Addition, and since none but numbers of the same kind can be added REVIEw.-42. What is Subtraction? Minuend? Subtrahend? Ro mainder? What does minuend mean? What does subtrahend mean? SUBTRACTION. 23 togetbler, (Art. 40), it follows, therefore, that a number can be subtracted only from another of the same kind: 2 cents can not be taken from 5 apples, nor 3 cows from 8 7horses. ART. 44. RULE FOR SUBTRACTING SIMPLE NUMBERS. 1. Write the less number llnder the greater, placing units under eznits, tens under tens, &c. 2. Begin at the right, subtract each figure from the one above it, placing the remainder beneath. 3. If any figure exceeds the one above it, add ten to the tupper, subtract the lower from the sum, and carry one to the next lower figure. From 827 dollars take 534 dollars. Dollars. S o L U T I o N.-After writing figures of the same order under each other, say, 4 units from 7 units 82 7 leaves 3 units. Then, as 3 tens can not be taken 5 3 4 from 2 tens, add 10 tens to the 2 tens, which makes 2 9 8 Rem. 12 tens, and 3 tens from 12 tens leaves 9 tens. To compensate for the 10 tens added to the 2 tens, add one hundred (10 tells) to the 5 hundreds, and say, 6 hundreds from 8 hundreds leaves 2 hundreds; and the whole remainder is 2 hundreds 9 tens and 3 units, or 293. DEMONSTRarTION.-1. Since a number can be subtracted only from another of the same kind, (Art. 43), figures of the same name are placed in column, to be convenient to each other, the less number being placed below as a matter of custom. 2. Commence at the right to subtract, so that if any figure is greater than the one above it, the upper may be increased by 10, and the next lower figure by 1; both numbers being equally increased, since 10 in any place is equal to 1 in the next higher place. The difference between two numbers is the same as the difference between those numbers increased equally. Thus, the difference between 2 and 5, is the same as the difference between 2~ —10 and 5-t10, or 12 and 15. There is another method of performing the operation REtVIEw. —43. What is subtraction the reverse of? Why? What is necessary in order that one number may be subtracted from another? Why? 44. What is the rule for subtracting simple numbers? Explain the example, and give the reasons for the rule. 2'4 I.Y'S tHIGHER APRITtIBIETI(C. wllen the lower fi-ure is greater than tlhe upper, called burro,'ouIt. To explain this, take 026'rom 75. S ( r,UT 0 N.-Si nce 6 units can not. be taken fr'om 7 5 5 units, borrow 1 ten from the 7 tens, and atdd it to 2 thle 5, whtich will lma.Lke 15 units. Then, Iulnits from 4 9 15.-) units leaves 9 units; and 2 tens from 6 tenls, (the numnter left in tens' place) leaves 4 tens: and the whole remainder is 4 tels and 9 units, or 49. I' on F.-Add the remainder to the less number. The sum will be equal to the greater. NOTE.-For proof by casting out the 9's, see page 69. EXAMPLES FOR PRACTICE. 1. From 30020037 take 50009. When therle are no fitntres in the 3 0 0 2 0 0 3 7 lower ttmnihber to correspondl with those 5 0 0 0 9 in the tipper, consider the vacant places 0 2) 9 7 0 0 8 e. occupied by zeroes. FItO)I TA.E R FE.r 2. 79685. 0253 4, 94 943'.2. 3. 145-)I06.. 39876. 110600. 4. 7150490. 9 23;7. 70581 2 3. 5. 2900000. 777888. 2 1 1 1 2. 6. 7 108;540 6 1 7 730. 690 t81 0. 7. 1014: 072.. 92475(;8.!55()4. 8. 20281).;144.. 18495136. 179 1008. 9. 2504 73.1.. 71 40851 1 790651 0. 10. 101067800.. 100259063.. 08737. In each of the followin, exams ples to the I 8tll, slltract th e first nitnlher from the second, tlhen f(rola tile reUtai tldC1', anld so on, till a remainder is found. less tlh;an the first nullter. LAST I EM. 11. 7 9) i ant- 36800..... 5. 12. 753 8 - 3t 3' C ______'0_ _-_()j _110 __ _ __13 5 6 1 510OI 1911__20 __611 2 _ _24___O_2_8___Oi3_3l___8 3i_63 _ i4 t~~~~~~~i~~~~~~ O _.CS 9 I0 - I0 | 4 2 $ g p,., " O 3 28 RAY'S HIGHER ARITHMETIC. ART. 47. THEOREMS.-The product of two numbers is the same, whichever factor is the multipier." D E Mr o N S T R A T I O N.-To find the number of stars in this diagram, multiply the number of stars in a row, by the number of rows. Counting up and down, there are 4 in a row, and 3 rows; hence, there must be 3 times 4, = 12 stars. But across, there are 3 in a row, and 4 rows; hence, there must be 4 times 3, 12 stars. The two pro- ducts, 3 times 4 and 4 times 3, are equal, since each gives the whole number of stars in the diagram. The same may be shown of any two numbers. ART. 48. THEOREM.-The product is always of the same name as the miultiplicand. DE IONSTRPTATION.-The nature of a number is not changed by repeating it. Thus, 5 dollars multiplied by 2, that is, taken twice, must be 10 dollars. ART. 49. THEOREM.-The multsplier zmust always be an abstract number. D E oN T R ATI O N.-The multiplier shows the number of times the multiplicand is to be taken; hence, it can not be yards, bushels, or any concrete number. We can attach no idea to taking any thing so many dollars times, or so many inches times. It may then be asked, how explain the solution of such questions as "What will 3 yards of cloth cost at 5 dollars a yard?" Ans. By the following or a simiilar Analysis: Three yards will cost three tbnes as much as one yard; therefore, if one yard cost 5 dollars, 3 yards will cost 3 times 5 dollars, = $15. R E of A R x.-Hence, it is absurd to talk of multiplying dollars by dollars. It would be as rational to propose to find the product of 3 apples by 2 turnips. ART. 50. Multiplication is divided into two cases: 1. When the multiplier does not exceed 12. 2. When the multiplier exceeds 12. RULE WHEN THE IMULTIPLIER DOES NOT EXCEED'12. 1. Write the mrultiplicand; place the multiplier under it, and d aw a litne benleath. 2. Begin with units; multiply each figure of the multiplicand by the multiplier, setting down and carrying as in Addition. MULTIPLICATION. 29 At the rate of 53 miles an hour, how far will a railroad ear run in four hours? SOLUTTION.-Ilere say, 4 times 3 (units) are 53 miles. 12 (units); write the 2 in units' place, and carry 4 the 1 (ten); then, 4 times 5 are 20, and 1 carried 2 miles. makes 21 (tens), and the work is complete. D E A o N S T R A T I N. —The multiplier being written under the multiplicand for convenience, begin with units, so that if the product should contain tens, they may be carried to the tens; and so on for each successive order. Since every figure of the multiplicand is multiplied, therefore, the whole multiplicand is multiplied. PRooF.-Separate the multiplier into any two parts; multiply by these separately. The sum of the prodcts must be equal to the first product. EXAMPLES FOR PRACTICE. AMULTIPLICAND. MULTIPLIER. PRODUCT. 1. 195.... 3..... 585. 2. 3823 4 15292. 3. 8765. 5. 43825. 4. 98374. 6.....590244. 5. 64382. 7. 450674. 6. 58765. 8.....470120. 7. 837941. 9. 7541469. 8. 645703.. 10 6457030. 9. 407649. 11 4484139. 10. If 4 men can perform a certain piece of work in 15 days, how long will it require 1 man? S OL UTI 0 N.-One man will be four times as long as four men. 4 X 15 = 60 days. 11. How many pages in a half-dozen books, each. containing 336 pages? Ans. 2016. 12. How far can an ocean steamer travel in a week, at the rate of 245 miles a day? Ans. 1715 miles. REvIEvw.-46. Repeat the Multiplication Table. 47. Prove the theorem. 48. HIow is the denomination of the product known? Why t9. What kinld of a number is the multiplier? Why? 50. HIow is multiplication divided? What is the rule when the multiplier does not exceed 12? Give the reason. What is the proof? 30 RAY'S HIGHER ARITI-iMETIC. 13. What is the yearly expense of a cotton-mill, if $32053 are paid out every month? Ans. $384636. ART. 51. RULE WHEN THE IMULTIPLIER EXCEEDS 12. 1. WPrite the multiplier under the multiplicand, placing figures of the same order under each other. 2. AMultiply by each figure of the multiplier successively; firg by the units' figure, thene by the tens' figure, &c.; placing the right hand figure of each product under that figure of the multiplier which produces it. 3. Add the several partial products together; their sum will be the required product. Multiply 246 by 235. SoLUTIoN. —First multiply by 246 6 (units), and place the first figure 2 3 5 of the product, 1230, under the 65 (units).. Then multiply by 3 (tens), 3 0 productby b and place the first figure of the productby M 4 9 2 product by 2 30 product, 738, under the 3 (tens). 49 product by Lastly, multiply by 2 (hundreds), 5 7 8 1 0 product by 2 3 5 and place the first figure of the product, 492, under the 2 (hundreds). Then add these several products for the entire product. AN ALYSis.-The 0 of the first product, 1230, is units, Art. 50. The 8 of the 2d product, 738, is tens, because 3 (tens) times 6 = 6 times 3 (tens) =18 (tens); giving 8 (tens) to be written in the tens' column. The 2 of the 3d product, 4902, is hundreds, because 2 (hundreds) times 6 = 6 times 2 (hundreds) == 12 (hundreds), giving 2 (hundreds) to be written in the hundreds' column. The right hand figure of each product being in its proper column, the other figures will fall in their proper columns; and each line being the product of the multiplicand by a part of the multiplier, their sum will be the product by all the parts or the whole of the multiplier. METHODS OF PROOF. 1. Multiply the multiplier by the multiplicand; this product must be the same as the first product. 2. The same as when the multiplier does not exceed 12, REr AitR. —For proof by casting out the 9's, see Art. 100. R E v I w.-What ia *he rule when the multiplier exceeds 12? CONTRACTIONS IN MULTIPLICATION. 31 ART. 52. Although it is 2 46 customary to use the figures 2 3 of the multiplier in regular order beginning with units, it 7 38 productby 30 will give the same product to 4 9 2 product by 2 0 use them in any order, observ- 1 2 30 prcducb by 5 ing that the right-hand figzure 5 7 81 0 product by 2 3 5 of each partial product muvst be placed under the figure of the multiplier which produces it, EXAMPLES FOR PRACTIC'L. L. 7198X 216.... = 1554768. 2. 8862 x 189. = 1674918. 3. 7575 x 7575.. 57380625. 4. 15607 X 3094.. =48288058. 5. 93186 X 4455... = 415143630. 6. 135790 x 24680. = 3351297200. 7. 3523725 x 2583... = 9101781675. 8. 4687319 x 1987.o = 9313702853. 9. 9264397 x 9584 -.. 88789980848. 10. 9507340 x 7071 o. = 67226401140. 11. 16444005 Kx 774 9... 12742494345. 12. 1389294 X 8900... 12364716600. 13. 27785 88 x 9867.. = 27416327796. 14. 204265 x 562402. - =114879044530. 15. 39948123 x 6007.. = 239968374861. 16. 579'02468 X 5008. = 289975559744. 17. 57902468 X 5080.. = 294144537440. is. 3081025 x 2008036 ~ - 6186809116900. 19. 860389657 X 96795. = 83281416849315. CONTRACTIONS IN MULTIPLICATION. CASE I. WHEN THE MULTIPLIER IS A COMPOSITE NUMBER. ART. 53. A composite number is the product of two or more whole numbers, each greater than 1, called its factors. Thus, 10 is a composite number, whose factors are 2 and 5: and 30 is one whose factors are 2, 3 and 50 32 RAY'S HIGHER ARITHMETIC. rULE. —-Separate the mulltiplier into two or more factors. M'ub? tiply first by one of the Jt.ctors, there this product by another factor, and so on till each factor has been used as a multiplier. The last product will be the entire product required. At 7 cents a piece, what will 6 melons cost? ANArL Ysis.-Three times 2 OPERATION. fi nes are 6 times. Hence, it 7 cents, cost of 1 melon. is the same to take 2 times 7, 2 andl then take this product 3 14 cents cost of 2 melons. times, as to take 6 times 7. And the same may be shown of any other composite number. 42 cents, cost of 6 melons. EXAMPLES FOR PRACTICE. 1. At the rate of 37 miles a day, how far will a man walk in 28 days? Ans. 1036 miles. 2. Sound moves about 1130 feet per second: how far will it move in 54 seconds? Alns. 61020 feet. 3. Multiply 9765 by 35. Ans. 341775. CASE II. WHEN THE MULTIPLIER IS ONE WITH CIPHERS ANNEXED, AS 10, 100, 1000; &c. ART. 54. RULE. —Annex to the multiplicand as many ciphers as there are in the multiplier; the result will be the required product. DEnONSTRATION.-By the principles of Notation, (Art. 26), placing one cipher on the right of a number, changes the units into tens, the tens into hundreds, and so on, and therefore, multiplies the number by 10. Annexing two ciphers to a number changes the units into hundreds, the tens into thousands, and so on, and, therefore, multiplies the number by 100. Annexing three ciphers multiplies the number by 1000, &c. EXAMPLES FOR PRACTICE. 1. Multiply 375 by 100... Ans. 37500. 2. Multiply 207 by 1000. Ans. 207000. REvIErw.-52. Explain the Example. 53. What is a composite number? What are its factors? What is the rule for multiplying by a composite number? CONTRACTIONS IN MULTIPLICATION. 33 CASE III. WHIEN CIPHERS ARE AT THE RIGHT OF ONE OR BOTH FACTORS. ART. 55. RULE.-ziolltiply as if there were no ciphers at the right of the inumbers; theL annex to tle product as many ciphers as there are at tlhe right of both the fJctors. Find the product of 5400 by 130. 5 4 0 0 130 SOLITION. — Find the product of 54 by 13, 1 6 2 and then annex three ciphers, the number at the 54 right of both factors. 0'O 2 0 0 0 A N A L YsIs.-Since 13 times 54 = 702, it follows that 13 times 54 hundreds (5400) = 702 hundreds (70C200); and 130 times 5400 = 10 times 13 times 5400 = 10 times 70200 - 702000. On rTIUS: The operation is the same as that 1 3 0 in the Rule for Multiplication, (Art. 51), except that the ciphers on the right are omitted until 1 6 2 0 0 0 the sum of the partial products is found. This is 5 4 0 0 evident from the operation in the margin. 702000 EXAMPLES FOR PRACTICE. 1. - 15460 X 3o200,,. = 49472000. 2. 30700 X 5904000.. - 181252800000. 3. 67051800 X 37800.. =2534558040000. 4. 9730000 x 645100. = 6276823000000. 5. 46218000 X 757000. O4987026000000. 6. 9800100 X 6514000. = 63837851400000, CASE IV. WHEN THE 3MULTIPLIER IS A LITTLE LESS THAN 10, 100, 1000, &c. ART. 56. rruLE. —Annex to the multiplicand as many ciphers as there are figures in the multiplier, and from.the result subtract the product of the mulltiplicand by the number the multiplier wants of being 10, 100, 100 0, &c. R EVIE w.-54. What is the rule for multiplying by I with ciphers annexed? Give the reasons. 55. What is the rule for multiplying iwhen ciphers are at the right of one or both factors? Explain the example. 3o4 RAY'S IIGHER ARITHMETIC. Multiply 3046 by 997. OPERATION. ANAssLsIS. —Since 997 is the same as 3046 1000 diminished by 3, to multiply by it is 9 9 7 the same as to multiply by 1000, (that is, 3046000 to annex 3 ciphers), and by 3, and take the 9 3 8 difference of the products, which agrees with the rule; and the same can be shown in any 30368 2 similar case. EXAMPLES FOR PRACTICE. 1. 7023 X 99.... 695277. 2. 16642 x 996.... = 16575432. 3. 372051 X 9998.. =8 3719765898. NOTE.-If there are ciphers at the right of one or both the factors, the rule may still be applied, (Art. 55.) 4. 642438 X 93.. = 59746734. 5. 2814 x 995.... =28570430. 6. 99999 X 999991'.. -9999000009. 7. 525734 x 9994. -.5254185596. 8. 4127093 x 9989... = 41225531977. 9. 2634527 x 9980.. = 26292579460. 10. 37291 8 x:999600. =3729030832800. 11. 203800 x 9997000. = 2087388600000. 12. 811876 X 99988.,. 81177857488. 13. 836062583 X 9999990. =36062493937470. 14. 2055416 X 992.. = 2038972672. CASE V. ART. 57. Another rffode of contraction, of frequent use, is comprised in the following RULE.-Derive the partial products, when possible, from each other, commencing with that figure of the multbplier which is most convenient for the purTose; mcakcing use of two or more fiygures of the multiplier at once, when it can be done, and setting the righcthand.figure of each partial product under the right-hand figure of the muzltiplier in use at the time. REvvIEw. —56. What is the rule for multiplying by a number that wants but little of 100, 1000, &c.? Explain the example. If there are ciphers at the'right of one or both factors, what may be done? 57. What other method is there of contracting multiplication? DIVISION OF SIMPLE NUMBERS. 35 5Multiply 387295 by 216324. S O L U T I O N.-Comn.mence with the 3 of OPERATION. the multiplier, and obtain the first partial 3 8 7 2 9 5 product, 1161885; then multiply this pro-1 6 duct by 8, which gives the product of the 1 1 61 18 8 5 multiplicand by 24 at once, (since 8 times 92 9 5 0 8 0 8 times any number make 24 times it.) 8 3 6 5 5 72 0 Set the right-hand figure under the righthand figure 4 of the multiplier in use. Multiply the second partial product by 9, which gives the product of the multiplicand by 216, (since 9 times 24 times a number make 216 times that number.) Set the right-hand figure of this partial product under the 6 of the multiplicand; and, finally, add to obtain the total product. EXAMPLES FOR PRACTICE. 1. 38057 X 48618.... — 1850255226. 2. 267388 X 14982. =4006007016. 3. 481063 x 63721. __ =30653815423. 4. 66917 X 849612. = 56853486204. 5. 102735 x 273162... =28063298070. 6. 536712 x 729981... =391789562472. 7. 750764 x 31.5135..= 23659201140. 8. 930079 x 251255. =233686999145. VI. DIVISION. ART. 58. DIVTSION is the process of finding how many times one number is contained in another. Also, Division is the process of finding one of the factors of a given product, when the other is known. The number contained in the other is the Divisor, the other number is the Dividend, and the result obtained is the Quotienrt. The Remaninder is the number which is sometimes left after dividing. Thus, in the question, How often is 2 contained in 7? the divisor is 2, the dividend 7, the quotient 3, and the remainder 1. No TE.-Dividend signifies to be divided. Quotient is derived from the Latin word quoties, which signifies. how often. 36 RAY'S HIGHER ARITHMETIC. ART. 59. The divisor and quotient in Division, cor respond to the factors in MIultiplication, and the dividend corresponds to the product. Thus: FACTORS. PRODUCT. MULTIPLICATION.. 3 X 5 -- 1 5. DIVIDEND. DIVISOR. QUOTIENT. DIVISION... 1 5 divided by 3 = 5 or, 15 divided by 5 = 3. There are three methods of expressing division; thus, 12 -- 3, or 3)12. Each indicates that 12 is to be divided by 3. ART. 60 THIEORE.M.-DIivision is a short method of mak/in, s'eral sulittactions of the samte number. D E A c N S T R ATI O N. —To prove this, find how many times 2 cents is contained in 7 cents. Two cents fromn 7 cents leaves 5 7 cents. cents; 2 cents from.5 cents leaves 2 cents contained once. 3 cents; 2 cents from 3 Cents 5 cents. leaves 1 cent. 9 cet. nts contained 2 times. Hlere, 2 cents is taken 3 times - from (out of ) 7 cents, and 1 cent 3 cents. remna(ins; hence, 2 cents is con- 2 cents contained 3 times. tained in 7 cents 3 times, with a 1 cent left. remainder of I cent. COROLLARIES. Con. 1. The Dividend and Divisor must always be of the samze dlelominatioa. Co t. 2. The quotient is always an abstract number; merely showilg how 9many times the divisor is contained in the cdividcnd. CoR. 3. The remainder is always of the same name as the dividendc. nMultiplication is a short method of making several ad7dttions of the same number; Division is a short method of making several subtractlons of the same number: hence, itivision is the reverse of IMultiplication. Itt v,r w, w. —57. Explain the example. 55. What is Division? What ia the Divisor? Dividend? Quotient? Remainder? What does dividend mean? What does quotient mean? 59. If the dividend is the product, what are the ihctors? if the divisor and quotient are factors, what is the product?.What are the signs of Division? DIVISION OF SIMPLE N NUMBERS. 37 The next step is to learn the Division Table, which gives the quotient in all cases where the divisor and quotient are both 12 or less, and there is no remainder. It is Imade fiom the Multiplication Table. Thus, 4 in 12 is contained 8 times, because 3 times 4 are 12. ART. 61. When the divisor does not exceed 12, the operation is called Short Division. RULE FOR SHORT DIVISION. 1. TWrite the divisor on the left of the dividend with7 a curved line between them. Begin at the left, divide successively each figure of the dividend by the divisor, and set the qulolielnt beneath the figure divided. 2. TZWhenever a remantaider occurs,.prefix it to tle figure in7 the next lowoer order, and divide as before. 3. If the figlure in any order does not contain, the divisor, place a cipher beneath it, prefix it to the-figture in the next lotoer order, ard divide as before. 4. If there is a remainder after dividizng the last figure, place the divisor unlder it and annex it to the quotieint. How often is 2 cents contained in 652 cents? S 0o L U T I 0 N. —Two in 6 (hundreds) is contained 3 2 ) 6 5 2 (hundreds) times; 2 in 5 (tens) is contained 2 (tens) 3 2 6 times, with a remainder of 1 (ten); lastly, 1 (ten) prefixed to 2 makes 12, and 2 in 12 (units) is contained 6 times, making the entire quotient 326. D E o N T T 0. —Commence at the left to divide, so that if there is a remainder it may be carried to the next lower order. By the operation of the rule, the dividend is separated into parts corresponding to the different orders. Having found the number of times the divisor is contained in each of these parts, the sum of these must give the number of times the divisor is contained in the whole dividend. Analyze the preceding dividend thus: 652 = 600 + 40 + 12 2 in 6 0 0 is contained 3 0 0 times. 2 in 40 is contained 2 0 2 in 1 2 is contained 6 Hence, 2 in 652 is contained 32 6 times. IR E VIE s w.- 60. Division is a short method of performing what opera. tion? Explain the example. What is division the reverse of? why? 35 ERm.AY'S HIGHER ARITHMETIC. MIETHODS OF PROOF. 1. Multiply the quotient by the divisor, and add in the remainder, if any; the result should equal the dividend; or, 2. Subtract the remainder, if any, from the dividend; the result divided by the quotient, should give the first divisor. ART. 62. It has been shown, (Art. 60), that the divisor and dividend must be of the same denomination, and that the quotient is always an abstract number. It is now to be shown how, according to these principles, the solution of questions in Division can be explained. All questions in Division belong to two classes: I. To divide a nltumber inr a parts each containing a certain number of Wnits, acnd to find the numbler of rlCarts. II. To diuide a Lumber into a certain numnber of equal patis, and to find the number of umits in. each. EXAM1PLES OF THE FIRST CLASS. 1. At $3 each, how many hats can I buy for $15? SOL uTI o N.-Since 1 hat-costs $3, I can buy a hat as often as $8 is contained in $15; $3 in $15 is contained 5 times; therefore, I can buy 5 hats. 2. At:$5 a yard, how much cloth can I buy for $35? 3. In a school of 60 pupils, how many classes of 10 pupils each? 4. I can walk 6 miles in 1 hour: in how many hours can I walk 54 miles? EXAMPLES OF THE SECOND CLASS. 5. A man has 20 apples to divide equally among 4 boys: how many must he give to each? SOLUTION. —To give each boy 1 apple will require 4 apples; hence, each boy will receive one apple as often as 4 apples are contained in 20 apples; 4 apples are contained in 20 apples 5 times; therefore, each boy will receive 5 apples. R E V IE AV.-60. Of what denomination must the divisor be? the remain. der? WVhat kind of number is the quotient? why? 61. What is Short Division? Give the rule. Explain the example. What is the proof? 2d method? How is the remtniihder written? 62. What two classes of questions are performed by division? DIVISION OF SIMIPLIE NUTMBERS. 9 RI E l a itR.-The solution of this question evidently requires that the 20 apples should be divided into 4 equal parts, and since the answer is obtained by dividing the 20 apples by 4, it follows that to divide any number into 4 equal parts, that is, to get one fourth, (4) of a number, divide it by 4. In like manner, to divide any number into 3 equal parts, that is, to get one third of a number, divide it by 3; to divide any number into ~ equal parts, that is, to get one half of a number, divide it by 2, &c. 6. If 7 yards of cloth cost $35, what is 1 yard worth? 7. A school of 60 pupils is to be divided into 6 classes: how many will there be in each class? S. I walked 54 miles in 9 hours: how many miles per hour did I go? EXAMPLES FOR PRACTICE. 9. Divide 75191 by 3.... Atns. 25063] 10. Divide 171654 by 4.... Ans. 42913'4 11. Divide 512653 by 5.... A Is. 1025303 12. Divide 534959 by 7.... Ans. 764227 13. Divide 986028 by 8.... Ans. 1232534 14. Divide 986974 by 11.... Ans. 89724"} In the following examples, find the continued quotient of the given number, divided by 2, 4, 6, 8, 10, and 12; that is, divide the given number by 2, then the resulting quotient by 4, and so on. EXAMPLES. ANS. EXA.MPLES. ANS. 15. 138240... 3. 19. 783360... 17. 16. 230400.. 5. 20. 1336320.. 29. 17. 322560... 7. 21. 1658880.. 36. 18. 506880. ~. 11. 22. 2488320.. 5i. 23. At $6 a head, how many sheep can be bought for $222? Ans. 37. 24. At S5i a barrel, how many barrels of flour can be bought for 9$895? Anls. 179. 25. If. 8 pints make a gallon, how many gallons are there inl 2 20 pints? Alis. 315. RH EVIW iv.-t62. Give an example of the first class. Explain it. Give an example of the second class. Explain it. 1How is the half, third, &c.. of any number obtained? tIA'i' tI II i'i. A! IT II {5., 2. LAt the rate of' (9 gallons an hour w long will it ta.ke to fill a cistern of' 2115 gallons? Aws. 235 hours. 27 Wllhat number multiplied by 11 gives 638? Aes. 58.'2. I t-ihree acres of land cost V741-, what is the cost of' 1 a ce? A -. x 247. 2') if''t 1: 15 be divided into 7 equall parts, how nll y dollars wvill there be in each part? A s. a" a t-( 0fe w manly fulil weeks in 1t3 dats? flaw anyil in 67 t0 dlys? _A1s. I in 13, anlld 95 in 670 d'O".-. 31i. Hxow many tim-, call 9 be suO)t'te(l fr ol i. 5t I-:. 1 7 an d 2 le t'. Ant. 63.'ir,,rt D)ivision is perfiorired mentally, and onuly the result is written; but when all tlhe work is written, it is called Lo2ng Divi.ioTn. Short Division is generally used when the divisor does not, exceed 12; Long Division, when the divisor exceeds 12. To show the difference between them, find how often 5 is contained in 820, SHORT DIVISION. LONG DIVISION. 5 ) 8 2 0 5)820(164 Quotient. 1 6 4 Quotient. 3 2 tens. Both operations are performed on 30 the same principle. In the first, the subtraction is performed men- 2 0 uiits. tally; in the second, it is written. 2 0 ART. 64. RULE FOR LONG DIVISION. 1. Draw cervedl lites on, the right and left of the dividend, placing the divisor on the left. 2. F1'ind how often the divisor is contained in the left-hand figlure, or figures, of the dividenld, and write the ztunnber in the quotient at tile right of the dividend. 3., ultiply the divisor by this quotient figure, and write the product unlder that part of the dividend fronm which it was obtained. REVIEW.-63. Explain the difference between Short Division and Long Division. When is the latter used? DIVISION OF SIMPLE NUMBERS. 41 4. Subtract this product froel the figlure.s above it; to the re. mainder bring dowun the next flyglre of the diridend, antl dit.ide as bcf/b'e, ullitl (ll lthe fiq-ture.s (f the diviclend are broulhlt (tdonsl. 5. If at any time a/?er a.figure is brotglht do(wn, the uItz1bcr thus form) ed is too small to COntain the divisor, a cip2her tmust be placed in the quotient, and another figur1te brozught downz, after whlich divide as before. PI nooF.-The same as in Short Division. Divide $44295 equally among 13 men. So UTrs o OT. —AS 13 is not contained P, 4 in 4 (thousands), therefore, the quo-; tient has no thousands. Next, take 42 13 ) 4 2 22 ( 3 2 2 (hundrelds) as a partial dividend; 13 is 3 9 hun. contained in it 3 (hundreds) tiIncs; after multiplying and subtracting, there 3 2 G tens. are 3 hundreds left,. Then bring down 2 tens, -and 32 tens is the next partial 6 5 dividend. In this 13 is contained 2 6 5 units. (tens) times, wvithl a remrnainsder 6 tens. Lastly, bringing down the 5 units, 13 is contained in 3.5 (units) exactly (unllits) times.'Th'le entire quotient is 3 hundreds 2 tells ard 5 units, or.325. 1) it ox. qT Rtxr rto. —Tlis operation involves no principle not explaini e(t itl Solist Division, (A rt. 31 ). Thlis nmay be furtlhter shown by decomnposilng the dividenid into parts, each exactly divisible by 13, as follows: DI VISOR. DIVIDEND. QU OTIENT. 13 ) 3900 - 260 60+ 65 (3 0 0 + 0+5 3 9 00 +260 + 2 GO -+ 6t;0 +65 NOTES.-I. Tle product, must never be greater tihan thse partial:.1i'i letiI f'om whlicch it is to be suhbtlracted; if so, the quotient figire is to) large, and mIust be ciminislled. 2. Th'e remaisnder after each subtraction must be less tlsan tlhe dlivisor; if niot, the last luotient figure is too sssmall, and tllst be incressed. 3. The order of each quotient figure is the same as the lowest order in the partial dividend from which it was derived. 4 RAY'S HIGHER ARITiHMETIC. ART. 65. In the Italian method of Long Division, the multiplication and subtraction are both performed in the mind at the same time. It is illustrated thus: Divide 20877514 by 3256. So LU TON. —The first figure OPERATION. Quot. of the quotient being 6, mul- 3 5 )2 8 14(6 4 2 tiIply the divisor by it, and 1 3541 5.tohtract the product from the 39 11 dividend, saying 6 times 6 are 6 5 5 4 36, and as 36 can not be taken 4 2 Rem. from 7, add 3 tens or 30 to the figure of the dividend, making 37, and then 36 from 37 leaves 1; to. compensate for the 3 tens added to the upper figure, carry 3 to the next product when it is formed, saying 6 times 5 are 30 and 3 are 33, and 33 from 37 leaves 4, there being 3 to carry: then, 6 times 2 are 12, and 3 are 15, and 15 from 18 leaves 3, there being 1 to carry; 6 times 3 are 18, and 1 are 19, and 19 from 20 leaves 1, with 2 to carry, and 2 from 2 leaves 0. So proceed until the figures of the dividend are all used. EAXAAMPLES FOR PRACTICE. 1. Divide 139180 by 453, PROOF. 453)139180(307 307 Quotient. 1359 4 5 3 Divisor. 3280 921 3171 1535 109 1 22 13907.1 109 Rem. 1 39180 2. 1004835. 33... = 3044911 3. 5484888 67... 81864. 4. 4326422 -- 961... =4502. 5. 1457924651 - 1204. _ l2109 0 (0 I6. 65358547823 - 2789. E 23434402 7: I 9 7. 33333333333 5299. - 6290495s3 —-- 8. 245379633477 -1263.. = 194283 1616 14 9. 5555555555550 123456 45 00028 2 s7W7 10. 555555555555654321. 49056 4579 10. 555555555555 — 654321. 4=98t9056 - $64:~ DIVISION OF SIMPILE NUMBERS. 43 In the following, multiply A by itself, also B by itself: divide the difference of the products by the sum of A and B. A. B. 11. 918 1902 4.... Ans. 984. 12. 8473 9437., Ans. 964. 13. 2856 3765... Ans. 909. 14. 33698 42856.,..o An2s. 9158, 15. 47932 152604.,,.An s. 104672. In the following, multiply A by itself, also B by itself: divide tho difference of the products by the difference of A and B. A. B. 16. 4986 5369....,. Ans. 10355. 17. 397 3 4308.... Ans. 8281. 18. 23798 59635.... As. 83433. 19. 47329 65931..... Ans. 113260. 20. If 25 acres produce 1825 bushels of wheat, how much is that per acre? Ans. 73 bushels. 21. How many times 1024 in 1048576? Ans. 1024. 22. How many sacks, each containing 55 pounds, can be filled by 2035 pounds of flour? Ans. 37. 23. How many pages in a book of 7359 lines, each page containing 37 lines? Ans. 198:w 24. In what time will a vat of 10878 gallons be filled. at the rate of 37 gallons an hour? Alns. 294 hours. 25. In what time will a vat of 3354 gallons be emptied, at the rate of 43 gallons an hour? Ans. 78 hours. 26. The product of two numbers is 212492745; one is 1035; what is the other? Ans. 205307. 27. What number multiplied by 109, and 98 added to the product, will give 106700? Ans. 978. CONTRACTIONS IN DIVISION. CASE I.-WHEN THE DIVISOR IS A C03IPOSITE NUMIBER. ART. 66. RULE.-1. Divide the dividend by one of the factors /' tihe divisor, and this qzotient by anotlher Jlctor, and so on, unlid Wit tlhe factors have beez used. The last quotient will be the one required. REVIEw.-64. Give the rule; the proof. Explain the example. WVhen is the quotient figure too large? When too small? How is the order of seah quotient figure known? 44 RAY'S I-IGIIER ARITH-IMETIC. 2. Ao fild the, trule remainder: urtlltiplt each, remainder by all the preceding divisors, except that which p,odtted it, crlnd to the stnl oJ' the products, add the remainder frJom the Jilst dlivision DIvide 217 by 15. SOLUTION.-Since 15=3 X 5, divide 3)217 by 3 and then by 5. The true remainder 5) 7 2 and 1 remn is 2 X 3 + 1- 7. Hence, the quotient is 14 and 2 rem, 14, remainder 7. D E a O N S T O R T I o N.- Dividing by 3 and then by 5, is the same as dividing by 15; for, in the former case, the quotient must be multipliedl by 5 and then by 3, and, in the latter case, by 15, to produce the given dividend; and, since multiplying by 5 and then by 3, is thle satte as multiplying by 15, (Art. 53), the quotients on which these operations are performred must be alike, or they could not both produce the same dividend. To prove the rule for finding the remainder; dividing 217 by 3, the quotient is 72 threes, and 1 unit remainder. Dividing by 5, the quotient is 14 (fifteens), and a remainder of 2 threes. Hence, the whole remainder is 2 threes +- 1, or 7. EXAMPLES FOR PRACTICE. 1. 1036 28. -= 37. 3. 2332- 5'1. =43;? 2. 3640-35. =104. 4. 3800 - 56. = 67 5. 34855 168.. = 207- Gq 6. 620314 2 31.. 2685 7 931 CASE II. -W\IEN TIIE DIVISOR IS ONE WITII CIPHERS ANNEXED, AS 10, 100, 1000, &C. ART. 67. RuLE. —CU2 of as many figures fr om the right of the dliidend as there are- ciphers in. the divisor; the figures cutt q cill be the renmainder, thie other figures will be the quotient. Divide 23543 by 100. 1100)235143 2 35 Quotient, 43 Remainder. ANALvSTs.-To divide 23543 by 100, is to find how many hundreds it contains. This is done by mere inspection, the part on tle riglbt of the line, 43 (units), being the remainder, since it is less than 100: while tllhat on the left, 235, is the quotient, being the numnber of hundreds in tlhe given number. RIIvIEw. —65. \Whlat is the Ittalian method of long division? Explain the example. 66. What is the rule for dividing by a composite number? CONTRACTIONS IN DIVISION. 45 1. 4567 — 100....... 45% 832-5-1000. - 8. 3. 95043 100.. e95.0f1 4. 730015 10000.... = 73T -o0 CASE III. —-WHEN CIPHERS ARE ON THE RIGHT OF THEI DIVISOR. ART. 68. RULE.-I. CGt off the ciphers at th7e right of the divisor, and as many figlures frcnm the right of the dividend. 2. Divide the remaining part of the dividend by the remaining part of the divisor. 3. Annex the figures cut offto to the remainder, and it will give the true remnainder. Divide 3846 by 400. 4100)38146 9 Quotient, 2 0 0 + 4 6 = 2 4 6 Ren. DE.MONSTRATION.-TO divide by 400 is the same as to divide by 100 and then by 4, (Art. 66). Dividing by 100 gives 38, and 46 remainder, (Art. 67); then dividing by 4 gives 9, and 2 remainder' the true remainder is 2 X 100 + 46 =-246, (Art. 66.) EXAMPLES FOR PRACTICE. 1. 34500 - 130 1 =- 265 5 % 2. 82500. -1100. 75. 3. 60000 1700...... = 3- 5 -~o% 4. 1896000 24000...o. = 79. CASE IV.-WHEN TIlE DIVISOR WANTS BUT LITTLE OF BEING 100, 1000, 10000, &C. ART. 69. RULE. —Ct qff from the rig]ht of the dividend, by a ertical line, as manyfiqgures as the divisor contains; multiply the par't on the left of the line by what the divisor wants of being 100, 1 00V*&c., and set the product 2n0der the dividend, commencing at units' place. Multiply the part of this product on the lefJt of the lRuvEiVw.-66. Explain the rule. Ilow is the true remainder found? 67. \Wht is the rule for dividing by 1 with ciphers annexed? Explain tha examlple. 68. What is the rulo for dividing by a number that haa ciphers on its right? Explain the example. 46 RAY S HIGHER ARITHMETIC. line by the same multiplier, and set down as before. Continue so until no fiqgure falls to the left of the line; then add the several results, and for every 1 carried across the line, add to the sum already obtained on the right of the line, the number Mused as a multiplier. This will be the true remainder, and the part on the left, the quotient. N o T E. -If the remainder is larger than the divisor, carry one to tae quotient, and the excess will be the true remainder. Divide 3289662 by 95. ANALYSIS.-The reason of the rule can OPER be seen in the explanation of this example. 9 5) 3 2 8 9 6 6 2 The divisor is assumed to be 100, and the 164 quotient, on that supposition, is 32896, with 8 2 2 0 a remainder 62. If this quotient were 4 10 multiplied by the true divisor 95, which is 2 0 5 less than the one assumed, the product, 3i 6 2 7 9 after the remainder 62 was added in, would 1 5 be less than the dividend by 5 times the quotient 32896. Therefore, 32896 and 62 346 2 8 9 7 are the true quotient and remainder for oel Qulot. 9 5 a part of the dividend; the surplus remain- Rem. 2 ing to be divided is 5 times 32896 or 164480. This surplus is divided like the first dividend, the quotient and remainder going to increase those already obtained. There is a surplus also in this operation, which is treated like the first, and so on, until the quotient is nothing, when the successive divisions must cease. The sum of the remainders is 192; from which, after setting down 92, we carry 1 to the first column of the quotients. This 1 is 100, and as 1 must be carried to the quotient for every 95 in the remainder, 5 out of this 100 should continue in the remainder, and be added to the 92 already set down, making 97; but 97, being large enough to contain the divisor 95, must furnish another unit to the quotient, making it 34628, and the excess 2 is the true remainder. EXAMPLES FOR PRACTICE. 1. 75076822 - 99. = 75835199 e 24206778~-989. = 24476' 2. 24206778 - - 9 -.:: 3. 680571293 - 9996. = 680849 9 4. 8662309749~9994.. =866751fg-g% 5. 52351346504 99930 = 52388O5-5~4 6. 42104678535 - 99800 = 42189,00f 310 FUNDAMENTAL RULES 47 GENERAL PRINCIPLES OF MULTIPLICATION AND DIVISION. ART. 70. THEOREM I.-M_/]ltiplyion either factor of a prolduct multiplies the product by the stame number. I)EAONSTRATIONu.-9 multiplied by 5 gives a product 45, and 9 multiplied by 10 (2 times 5) gives a product 90, which is 2 times 45, and so in all cases; for, when a multiplier is made 2, 3, 4, &c. times as large as at first, the multiplicand must be taken 2, 3, 4, &c. times as often as before, and, therefore the product will be 2, 3, 4, &c. times as large as the first product. And since this is true of the multiplier, it is also true of the multiplicand, because it can be used as the multiplier. (Art. 47.) ART. 71. THEOREAM II.-Dividing either factor of a product divides the product by the same number. DEONSTRATION. —9 multiplied by 10 gives a product 90, and 9 multiplied by 5 ( 4 of 10) gives a product 45, which is of 90, and so in all cases; for when a multiplier is made 4,, ), &c., as large as at first, the multiplicand must be taken ~, i, 4, &c., as often as before, and, therefore, the product will be 4, -, -, &c., as large as the first product. And since this is true of the multiplier, it is also true of the multiplicand, because it can be used as the multiplier. (Art. 47.) ART. 72. THEOREM III. —Multiplying one factor of a product, and dividing the other factor by the same number, does not alter the product. D E oN S TAT.T 0 N.-Multiplying either factor by 2, 3, 4, &c., multiplies the product by 2, 3, 4, &c., while dividing the other factor by 2, 3, 4, &c., divides the product by 2, 3, 4, &c. When these opera tions are performed together, the product is both multiplied and divided by the same number, and must, therefore, remain unchanged. ART. 73. THEOREM IV. —lMultiplying the dividend, or d1iv1idig the divisor, by any number, m ultiplies the quotient by that number. DE o 0 N S aA T R 0 N.-If any divisor is contained in a given dividend a certain number of times, it must be contained in twice that dividend, twice as often; in 3 times that dividend, 3 times as often; REVEw. -69. What is the rule for dividing by a number that wants but little of being 100, 1000, &c.? If the remainder is larger than the divisor, what must be donoe? Explain example. v43 RtiAY'S IHIG-IHER ARITIiTIETIC. and so on. Thus, 3 is contained in 12 four times, and in twice 12 it is contained twice 4, = 8 tiunes. Agrain, in any given diviilend, hlal the di visor will be contained tiwice Las often as tle whole di visor; one third of tile divisor, 3 times aLs often, and so on; thus, 4 is contained in 12 three times, and one half of 4, which is 2, is contained in 12, twice 3, - 6 times. ART. 74. THEOREmI V. —Dividig tlte dividend, or mnzl tp) lying the divisor by any nuzmber, divides the LuotieCnt Iy thct number. DSEOINSTRATTrON.-If any divisor is contained in a given dividelnd a certain number of times, it must be contained in half that dividelnd, half as many times; in one third of that dividend, one third as Imany times, and so on. Thus, 3 is containe(l in 12 four times, and in one half of 12, which is six, it is contained one half of 4 -- 2 titmes. Again, in any given dividend, twice the divisor will be contained half as often as the divisor itself; three timzes the divisor, one third as often, andl so on; tthus, 3 is contained in 24 eight times, and 2 times 3, wh-lich ism6, is contained in 24 one half of 8 = 4 times. ART. 75. THIEOREMI VI.-Jlutltlplyi;ng or dividing both diri/lend anld divisor by the samne tnzmber, does?tot chcctge tte qtlutiltt. D1) l NST Rt ATTON.-This follows from the two preceding theorems; for the effects produced by multiplying or dividing both dividend and (livisor exactly balance each other; this, 6i is contained in 24 4 times; and twice 6 is contained in twice 24, juist 4 tiles; also, one ihalf of 6 is contained in one half of 24, just 4 times; and so of any other dividend and divisor. CONTRACTIONS IN MULTIPLICATION AND DIVISION. ART. 76. To multiply by any simple part of 100, 1000, &c., ItULE.-Multtiply by 100, 1000, &c., and take stmch a part *)f the result as the multiplier is of 100, 1000, &c. N Or r.. —To get two thllirds (2) of a number, divide it by 3, wllich gives one thitd ( ) of it, (Art. 62), and multiply the qtuotient by 2; or, if it is more convenient, multiply the numrlber by 2 fitst, and divide the product by 3. In like manner, to get three fourths (.-) of a number, multiply it by 3, and divide the product by 4, and so )-n. CONTRACTIONS IN MULTIPLICATION. 4. PARTS OF 100. PARTS OF 1000. 121 = Iof 100. 125 -= of 1000. 164 =. of' 100. 166 -- of 1000. 25 = 4of 100. 25 0= of 1000. 33= - of 100. 3334 = ~ of 1000. 371 of 100. 375 = 4 of 1000. 621 =- of 100. 625 =. of 1000. 662= X of 100. 666 = — of 1000. 75 = of 100. 750 =- of 1000. 834 = - of 100. 833 = - of 1000. 87 = of 100. 875 = 4 of 1000. Multiply 246 by 874. 2 4 6 0 0 SoLUTION.-Since 871 is J of 100, annex two 7 ciphers to the multiplicand, which multiplies it 8)1 72 20 0 by 100, (Art. 54), and then take g of the result. (See Note). As. 2 5 2 5 EXAMPLES FOR PRACTICE. 1. 422 x 33-... 14066] 2. 3708 X 25... 92700. 3. 6564 62... = 410250. 4. 10724x 16X = 1 787338 5. 8740 X 75... = 655500. 6. 53840 x 125... =6730000. 7. 4456 x 6663.6 = 29706663 8. 7293 x 833. =6077500. 9. 852 x 875... - 745500. 10. 64082 x 375. =24030750. ART. 77. To multiply by any number whose digits are all alike, RULE.-M- ultip7y azs if th digits were 9's, (Art. 56), and take &uch a part of the product as the digit is of 9. MIultiply 592643 by 66666. So L U T I N.-Multiply 592643 by 99999, (Art. 56), the product is 69263707357; take 3 of this product, since 6 is o- of 9; the result is 39509138238. RE V I E W.-70-75. State and prove the theorems. 5 AY'0 K t'S III(IHER ARITHMETIC. 1. 45x01 x I 3333,... - If 53 *...8 1 257 X 5 555 55a o,-)) 7;1-, 5 3. 6302 4 x4444000, 28007 154560 0. AnT. 78. To divide by a number ending in any simppl part of' 100, 1000, &eo., iUuE.-JLtfUdi:ly both di'tidde-zd and cdiisor' b? stzch/I, a( naitbe/' (3,4, 6, o wi 8,) cs,tUll cotverit the,/ial fatrces o/ the /ivisor i3to ciptiers, alnd diiet divide thie /bt, Oerl p'roduct b/y the Iltttecr. N'oTv.-i. -If there be a remainder, it should be divided by the multipil;ier, to get. the true remainder. 2. The multiplier is 83, 4, 6, or 8, according as the final portion. of the divisor is thfids, fourths, sixths, or eiyhthks of 100, 1000. 3. Several successive multiplications may sometimes be made before dividing. Divide 6903141128 by 21875. Asts. 315572 Ni.i.~ SoLUTIoa. —Multiply both by 8 and 4 successively. The divisor becomes 71)0000, and the dividend 220900O51600), while the quotient remains the same, (Art. 75). Performing tile division as in Art. 68, the quotient is 315,572, and remainder 116096. The remaindler being a part of the dividend, has been made too large by the multiplication by 8 aned 4, and is, therefore, reduced to its truie dimensions by dividing by 8 and 4. This gives 3628 for the true renMainder. EXAMPLES FOR PRACTICE. 1. 300521761 225 3. 13356522;', 2. 151 03 87" 6 4348750. = 3 5 3. 225 00712361 —l14060250. = 6000 (0 I,', 4. 62071240i -80-208333 =2 79 4, It, o. 4 -ti 5. 742 8 5 1 6 9 2 2916 254692 257 A. r. 79. To divide by any number whose digits are all alike,. iE,.- Treat the dividenl and divisor alikce by wndttl)/i:ii ti0on, and division,, if t.,ccessar?, luntil the diyits of the divisor are 9's; theaL dlivide byl Jile let Art. 69. REVIEwv.-7. What is the rule for multiplying by any simple plart of 100, 1000, &ce.? 77. For Inultiplving by any numiber whose diaitts are all alike? 78. For dividing by any number ending in any simple part of 100, 1000, &e? SUMMARY OF PRINCIPLES. 51 NOTE. —-1. If the divisor ends in ciphers, divide first as in Art. 68. 2. If a remainder occurs, reverse upon it the operations performed on the dividend and divisor, to get the true remainder. Divide 420565342 by 666. So LUTrI N. —Divide both numbers by 2, and multiply the results by 3; this malkes the divisor 999, and does not alter the quotient, (Art. 75), which is found, (by Art. 69), to be 63147'9, with a remainder 492; the latter being a part of the dividend, differs from the true remainder, by reason of the division by 2, and the multiplication by 3, and is brought to its true dimensions by multiplying by 2 and dividing by 3. EXAMPLES FOR PRACTICE. 1. 13641096 2222.... 6139)U8 2. 376802902:7770.. = 48494:$IO 3. 8811857528 33333... =2 64382 1 3 4. 5606258492 88888.. = 63071 4 SUMMI.~iARY OF GENERAL PRINCIPLES. ART. 80. Notation and Numeration, Addition, Subtraction, Multiplication, and Division, are called the fundamental rules of' Arithmetic, because all the various operations of Arithmetic are performed by them. ART. 81. NOTATION is the method of expressing num. bers in fiyures. NUMERATION is the method of expressing numbers in words. ART. 82. ADDITION is the process of finding the aggregate or sum of two or more numbers, (Art. 40). ART. 83. SUBTRACTION is the process of finding the dif/irence between two numbers, (Art. 43). ART. 84. MULTIPLICATION is the process of finding what is produced by taking one number as many times as there are units in another, (Art. 46). RaEVIEW.-79. What is the rule- for dividing by a number whose digits are all alike? How is the true remainder obtained? 80. Which are the fundamental rules of Arithmetic? Why are they so called? 81. What is Notation? Numeration? 82. Addition? 83. Subtraction? 52 RAY'S HIGHIER ARiTHMETIC. A RT. 85. DIVIsIoN is the process of finding how many times one number is contained in another, (Art. 58). GENERAL PROBLEMS. 1. When the separate cost of several things is given, how is the entire cost found? 2. When the sum of two numbers and one of them are given, how is the other found? 3. TWhen the less of two numbers and the difference between them are given, how is the greater found? 4. When the greater of two numbers and the difference between them are given, how is the less found? 5. When the cost of one article is given, how do you find the cost of any number at the same price? 6. When the divisor and quotient are given, how do you find the dividend? 7. How do you divide a number into parts each containing a certain number of units? S. How do you divide a number into a given number of equal parts? 9. If' the product of two. numbers and one of them is diven, how do you find the other? 10. If the dividend and quotient are given, how do you find the divisor? 11. If you have the product of three numbers, and two of them are given, how do find the third? 12. If the divisor, quotient, and remainder are given, how do you find the dividend? 13. If the dividend, quotient, and remainder are given, how do you find the divisor? ART. 86. PARTICULAR EXAMPLES. 1. I bought 3 horses for $165; two cows for $37; and 7 sheep for $29: what did all cost? AUs. $231. 2. The sum of two numbers is 664, and one of them is 369: what is the other? Ans. 295. 3. The difference between twxo numbers is 168, and the less number is 289: what is the greater? Ancs. 457. R E V I E w. —4. What is Multiplication'? 85. Division? SU1MMA1XRY OF PRINCIPLES. 53 4. The greater of two numbers is 753, and their difference 1 57: what is the less? Aiis. 296. 5. NAt $7 a yard, what will 5 yards of cloth cost? SOLUTION. — 5 yards are 5 times 1 yard. If 1 yard cost $7, 6 yards will cost 5 timies $7, which are $35, Ans. 6. The divisor is 753, and the quotient 245: what is the (ividend? iAns. 184485. 7. ow many classes of 9 pupils can be formed in a schlool containing 54 pupils? SoLUTION.-Since it takes 9 pupils to form one class, there will be as many classes as 9 pupils are contaiaed in 54 pupils; 9 in 54, 6 times. Ans. 6. 8. If you divide 28 apples equally among 4 boys, what will be the share of each? SOLUTION.-TO give each boy 1 apple, will require 4 apples; hence, each boy will receive 1 apple as often as 4 apples are contained in 28 apples; 4 in 28, 7 times. Ans. 7 apples. 9. The product of two numbers is 612451, and one of' them is 203: what is the other? Ans. 3017. 10. The dividend is 395631145 and the quotient 4007: what is the divisor? Ails. 98735. 11. The product of three numbers is 195318005; two are 307 and 703:' what is the third? Ais. 905. 12. rWhat number, divided by 473, will give the quotient 8061 and remainder 365? _Ais. 3813218. 13. The dividend is 7781174, the quotient 8216, the remainder 622: what is the divisor? Ans. 947. ART. 87. AMISCELLANEOUS EXERCISES. 1. A grocer gave 153 barrels of flour, worth $6 a barrel, for 54 barrels of sugar: what did the sugar cost per bar rel?? AAs. $17. 2. When the divisor is 35, quotient 217, and remainder 25, what is the dividend? A ns. 76290. 3. What number besides 41 will divide 4879 without a reimainder? Ais. 119. 4. Of what number is 103 both divisor and quotient? Anes. 10609. 5. What is the nearest number to 53815, that can be divided by 375 without a remainder? Ans. 54000. 4 RAY'S 1IG IIER ARITHMETIC. 3. A fatrmer bought 25 acres of land for $2o675: what did 1S acres of' it, cost?' its. 4 203-. 7. toulght 15 horses at $75 a head: at how much per head must I sell thei to gain $;210? Aus. $89. 8. A locomotive has 391 miles to run in 11 hours: after running 139 miles in 4 hours, at what rate per hour must the remaining distance be run? Ans. 36 miles. 9. A merchant bought 235 yards of cloth at $5 per yard: after reserving 12 yards, what will he gain by selling the remainder at $7 per yard? Ans. p386. 10. A grocer bought 135 barrels of pork for $2295; he sold 83 barrels at the same rate at which lhe purchased, and the remainder at an advance of $2 per barrel: how much did he gain? Ans. $104. 11. A drover bcought 5 horses at $75 each, and 12 at $68 each; he sold them all at $73 each: what did he gain? AIis. 850. At iwhat price per head must he have sold them to have gained $118? Ais. $77. 1 2. A merchant bought 3 pieces of cloth of equal leng(ths at $14 a yard; he gained $24 on the whole, by selling 2 pieces for 1$240: how many yards were there in each piece? i_1s. 18. i3. If 18 men can do a piece of work in 15 days, in how many days will one man do it? S OL a JT o N.-It will require 1 marn 18 times as long as 18 men. Eighlteen times 15 days are 270 days. Ans. 1I. If 13 men can build a wall in 15 days, in how many days can it be done if 8 men leave? in's. 39. 15. If 1-4 men can perform a job of work in 24 days, in how many days can they perform it with the assistance of 7 more men? Ains. 16 days. 1;. 1A company of 45 men have provisions for 30 days: how many men lust depart, that the provisions may last the remainder 50 clays? Ans. I8 men. 17. A horse worth $8"5, and 3 cows at $18 each, were exchanged for 14 sheep and $41 in money: at how much each were the sheep valued?.As. V7. 1S. A drover bought an equal number of sheep and hogs for $1482; he gave $7 for a sheep, and $6 for a hog: what number of each did he buy? Ans. 114. SuGoasTIoN. —1 sheep and 1 hog cost $7- + $6 = $13. PROPERTIES OF N UIBERS. 55 19. A trader bounrht a lot of horses and oxen for p?1i-2(0; the horses cost $50, and the oxen $17. a head; thlere were twice as many oxen as horses: how many were there of each? As. 15 horses and 30 oxen. 20. In a lot of silver change worth 1050 cents, oneseventh of the value is in 25 cent pieces; the rest is made up of 10 cent, 5 cent, and 3 cent pieces, of each an eqpial number: how many of each coin are there? Ans. of 25 cent pieces, 6; of the others, 50 each. 21. A speculator had 140 acres of land. which he miight have sold at $210 an acre, and gained $G300, but after holding, he sold at a loss of $5600: how mnuch an acre did it cost him, and how much atn acre did he sell it, for? Ans. $165, cost; and $125, sold fbr. VII. PROPERTIES OF NUMBERS. ART. 86. DEFINITIONS. L. An integer is a wlhole number; as, 1, 2, 3, &e. 2. Wlhole numbers are divi-ded into two classes —prrme numbers and col?.cJtre inumb hers. 3.A pr/?e, number is one that can be exactly divided by no other whole number but itself and unity, (1); as, 1, 2, 3, 5, 7, 11, &c. 4. A comnposite number is. one that can be exactly divided by some other whole number besides itself and unity; as, 4, 6, 8, 9, 10, &e. R E 31 a R.-Every composite number is the product of two or more other numbers. 5. Two nunmbers are p)rhimte to each other, when unity is the only number that will exactly divide both; as, 4 and 5. RE ani A R a. —T-wo prime numbers ajre always prime to each othe:' sometimes, also, two composite numbers: as, 4 and 9. 6. A n even, number is one which can be divided by 2 without a remlainder; as, 2, 4, 6, 8, &c. R1vt e rvw.-s88. What is man integer? Into what classes are integers divided'?'What is a. ptlinie nulmber? a composite number? When are numbers prilme to each other? WhaVit kind' of nurbers mnust be prime to each other? What kind. may be? What is an even number? 56 RAY'S HIGHER ARITHMETIC. 7. An odd number is one which can not be divided by 2 without a remainder; as, 1, 3, 5, 7, &c. REr A RIA.-All even numbers except 2 are composite: the odd numbers are partly prime and partly composite. 8. A diviisor or measure of a number, is a number that will divide it without a remainder: 2 is a divisor of 4; 5 ^,f 10, &c. 9. One number is divisible by an-other when it contains that other without a remainder; 8 is divisible by 2. 10. A multiple of a number is the product obtained by taking it a certain number of times; 15 is a multiple of 5, being equal to 5 taken 3 times; hence, 1st. A mzulti)lle of a number caa alwayjs be divided by it without a r9emainder. 2d. Every gmultiple is a convmposite number. 11. Since every composite number is the product of factors, (Def. 4), each factor must divide it exactly; hence, every factor of a number is a divisor of it. 12. A prime factor of a numniber is a prime number that will exactly divide it: 5 is a pr'imne factor of 20; while 4 is a factor of 20, not a prime factor; hence, 1st. The prime factors of a number are all the prime numbers that will exactly divide it; 1, 2, 3, and 5, are the prime factors of 30. 2d. Every composite number is equal to the product of all its prime factors. All the prime factors of 15 are 1, 3, and 5; and lX3X 5=-15. 13. Any factor of a number is called an altquot part of it; 1, 2, 3, 4, and 6, are aliquot parts of 12. FACTORING. ART, 89. Factoring depends on the following PRINCIPLES and PROPOSITIONS. P12NCIPITE 1. A factor of a number is a factor of (ny m ultiTple of that number. D E at o s T A T I O N. —Since 6 - 2 X 3, therefore, any multiple of 6 =2 3.X some number; hence, every factor of 6 is also a factor of the multiple. FACTORING. / PRINCIPrr E 2. A factor of any two numbers is also a efactor' of t7hcir sui,. D E oN S T R T I N.-Since each of the numbers contails the factor a certain number of times, their sum must contain it as often as both the numbers; 2, which is a factor of 6 and 10, mnult be a factor of their sum, for 6 is 3 twos, and 10 is 5 tzos, and thei sum is 3 twos t- 5 twos - 8 twos. ART. 90. From these principles are derived six PROPOSITIONS. PROP. I.-Every numbler ending with O, 2, 4, 6, or 8, is divisible by 2. D E I,;S T A T TN.-Every number ending with a 0, is either 10 or sone number of tens; and since 10 is divisible by 2, therefore, by Principle lst,, (Art. 89), any number of tens is divisible by 2. Again, any number ending wvith 2, 4, G, or 8, may be considered as a certain number of tens plus the figure in the units' place; and since each of the two parts of the number is divisible by 2, therefore, by Principle 2d, (Art. 89), the number itself is divisible by 2; thus, 36 30 + 6 -= 3 tens + 6; each part is divisible by 2, lience, 36 is divisible by 2. Conversely, no nu~mber is divisible by 2, unless it ends with 0-, 2, 4, 6, or 8. PROP. II.-A n.ultmber is divisible by 4, when the number deznoted by its two right ha nd digits, is divisible by 4. DEAOsN STRATI0No.-Since 100 is divisible by 4, any number of hundreds will be divisible by 4, (Art. 89, Principle 1st); and any number consisting of more than two places may be regarded as a certain number of hundreds plus the number expressed by the digits in tens' and units' places, (thus, 384 is equal to 3 hundreds + 84); then, if the latter part (84) is divisible by 4, both parts, or the number itself, will be divisible by 4, (Art. 89, Prin. 2d). Conversely, no n ~umber is divisible by 4, ~unless the tumber denoted by its two right-hasnd digits is divisible by 4. R E v I w. —88. What is an odd number? Are the even numbers prime or composite? Are odd numbers prime or composite? What is a divisor of a number? When is one number divisible by another? WhaLt is a multiple of a number? 1lWhat two axioms concerning multiples? What is a factor of a number? A primne factor? What two axioms concerning prime factors? What is an aliquot part of a number? Give examples illustrating the definitions. $58 iRAY'S HiIGHER ARITHMETIC. PnroI. ITT. —A number ending in 0 or 5 is dvis'sble by 5, D Ic Ir o r\ s T R X T O'u. —Ten is divisible by 5, and every number of two or more figures, is a certain number of tens, plus the right h:and digit; if this is 5, both parts of the number are divisible by 5., and, hence, the number itself is divisible by 5, (Art. 89, Prin.'2d). Conversely, no number is divisible by 5, unless it elnds in 0 or. PRtoP. IV.-Every number ending iln 0, 00, &e., is diisib)le by 10, 1 00, &c. D E Pr o N S T It' T I O N.-If the number ends in 0, it is either 10 or a multiple of 10; if it ends in 007 it is either 100, or a multiple of 100, and so on; hence, by Prin. 1st, Art. 89, the proposition is true. PROP. V.-A composite number is (divisible by tlhe product of any two or more of its plrime fictors. D E 0 N S TR A T I 0 N.- Since 2 X 3 X 5 -- 30, it follows that 2 X 3 taken 5 times, makes 30; hence, 30 contains 2 X 3 (6) exactly 5 times. In like manner, 30 contains 3 X 5 (15) exactly 2 times, and 2 X 5 (10), exactly 3 times. Hence, if any even numnber is divisible by 3, it is also divisilble by 6. D E S o N S T R A T I O N.-An even number is divisible by 2; and if also by 3, it must be divisible by their product 2 X 3, or 6. PurOp. VI. —E, very prime number, except 2 and 5, ends with 1, 7, 7, or 9. DEM3 NSTRATIO N.-This is in consequence of Prop. I and III. AnT. 91. To find the prime factors of a composite number, RUL,E.-Ditvide the given number by any prime number that uill ex.actll divide it; divide the quotient in like mannle)J, a'nd so con'inve unt.l the qpotient is a prine t1mlber; the last q uotient and the several divisors are the prime factors. REVIsTw.-89. What is the first principle used in factoring? Prove it. What is the second principle used in factoring? Prove it, 90. When is a number divisible by 2, and when not? WThy? 11When by 4, anid w rhe n not? Wbly? When by 5, and when not? WIVlhy? When by 10, 1(0, &e., anud when not?'Why? What is every composite number divisible by? Why? If an evens number is divisible by 3, what else imust dividle it? Why? Howr do the pritme numbers end? Why? 91. \W\hat is the rule fox QPding the prime factors of a number? FACTORING. 59 RE: AT RKS.-I. Divide first by the smallest prime factor. 2. The leastZ divisor of any number is a prime number; for, if it were a composite number, its factors, which are less than itself', would also be divisors, (Art. 89), and then it would not be the least divisor. Therefore, the prime factors of any number may be found by dividing it first by the least number that will exactly divide it, then dividing this quotient in like manner, and so on. 3. Since 1 is a factor of every number either prime or composite. it is not usually specified as a factor. Find the prime factors of 42. DEMO NSTRAT TON. —y trial, 2 is found to be a factor 2)42 of 42. Also, 3 is found by trial to be a factor of 21, and consequenitly a factor of 42, which is a multiple of 21, 3) 2 1 (Art. 89). In like manner, 7 being a factor of 21, must f be a factor of 42; and since 2 X 3 X 7 42, there can be no other factors of 42 besides 2, 3, and 7. SEPARATE INTO PRIMTE FACTORS, 1. 45.. Ans. 3,3, 5 8. 72 Ans. 2,2,2,3, 3. 2. 48 Ans. 2,22,2,3. 9. 75. As. 3,5,5. 3. 50.. Ans. 2,5, 5. 10. 80 A s. 2 2, 5. 4. 54. Ans. 2,3,3,,3. 11. 84. As. 2, 2,3,7. 5. 56. Acns. 2,2,2,7. 12. 96 Anrs. 2,2,2 2,2)3. 6. 60. As. 2, 2,35. 1 3. 98. Al1s. 2,7,7. 7. 63.. Ans.. As. 3,3,7. i 14. 99. As. 3,3,11. THE PUPL who desires to be an expert Arithmetician, should be able to give the prime factors of all numbers under 100 by mere inspection, the operation being performed mentally. 15. Factor 210... Ans. 2, 3, 5, 7. 16. Factor 1155...... Ans. 3, 5, 7, 11. 17. Factor 10010.... Ans. 2, 5, 7, 11, 13. 18. Factor 36414... Ans. 2, 3 3, 7, 17, 17. 19. Factor 58425..... Ans. 3, 5, 5, 19, 41. The prime factors common to several numbers may le found by resolving each into its prime factors, then taking the prime factors alike in all. REVIEW.- 91. Which number is it most convenient to divide by first? What is said of the least divisor of a number? Of 1, as a factor? Explain the example, and show the reason of the rule. 60 RAY'S HIGHER ARITHMETIC. Find the prime factors common to 20. 42 and 98.. Ans. 2, 7. 21. 45 lLand 105.... e As. 3, 5. 2'2. 90 and 210... AIs. 2, 3, 5. 23. 210 and. 315...... As. 3, 5, 7. ART. 92. Since any composite number is divisible, not only by each of its prime factors, but also by the product of any two or more of them, (Art. 90, Prop. V.); hence,. To find all the divisors of any composite number, Rui,.-Resolre the number into its prime factors; and then form f;oom these factors all the ciffereJnt products of Vwhch they will adil; the prime factors and their _products will be (cll the divisors of the gicen nzumber. 42 = 2 X 3 X 7; and all its divisors are 2, 3, 7, and 2 X 3, 2 X 7, and 3 X 7; or, 2, 3, 7, 6, 14, 21. Find all the divisors 1. Of 70.. Ans. 2, 5, 7 and 10, 14 and 35. 2. Of 30.. Ans. 2, 3, 5 and 6, 10 and 15. 3. Of 196.. Ans. 2, 7, 4, 14, 28, 49 and 98. 4. Of 231,. Ans. 3, 7, 11 and 21, 33 and 77. GREATEST COMMON DIVISOR. ART. 93. A common divisor of two or Imore numbers, is a number that will divide each of them without a remainder; 3 is a common divisor of 12 and 18. The greatest common divisor of two or more numbers, is the greatest number that will divide each without a remainder; 6 is the greatest common divisor of 12 and 18. REMxARtK. —1. Two numbers may have several common divisors, ut they can hare only one greatest common divisor. 2. The greatest common divisor is often termed the greatest common. mneasure, or greatest commzon factor. Rn v rW.-92. What is the rule for finding all the divisors of a numiber? Why can there be no other divisor than these? 93. What is a common divisor of several numbers? the greatest common divisor/ Give exalmples. GREATEST COMMON DIVTSOR. 61 ART. 94. To find the greatest common divisor of two nuIImlbers, Ru.LE I. —esolve tlhe given ntu1mbers into their p'i7me factors; the proclwt o(' the factors common, to both nzmbersi will be the greatest contmon divisor soug7ht. Find the. greatest common divisor of 30 and 105. 30 = 2 X 3 x 5. ) 3 x 5 15, the greatest 105 3 X X 7. j common divisor. DE nfO S T I AT I 0 N.-The product 3 X 5 is a divisor of both the numbers, since each contains it; and it is their greatest common divisor, since it contains all the factors common to both. FIND T-IE GREATEST COMMON DIVISOR 1. Of 30 and 42.. As. 2 X 3= 6. 2. Of 42 and 70... Ans. 2 X 7 = 14. 3. Of 63 and 105 Ans. 3 X 21. 4. Of 66 and 165... Als. 3 X 11 = 33. 5. Of 90 and 150... Ans. 2 X 3 X 5 =30. The greatest common divisor contains, as factors, all the other common divisors; thus, 30, which is 2 X 3 X 5, contains 2, 3, 5, 2 X 3 -,G- 2 X 5 = 10, and 3 X 5 = 156, the only remaining common divisors of 90 and 150. 6. Of 60 and 84... Ans. 2X X 3 12. 7. Of 90 and 225... Ans. 3 X 3 X 5=45. 8. Of 112 and 140.. Ans. 2 X 2 X 7-= 28. The greatest common divisor of more than two numbers may be found, by resolving each into its prime factors, and taking the product of the factors common to all. 9. Of 30, 45, and 75.. Ans. 3 X 5 — 15. 10. Of 84, 126, and 210. Ans. 2 X 3 X 7- =42. ART. 95. Rule 1 is generally used when the numbers are small; but when they are large, apply RULE II.-Divide thie greater mtmber by the less, and thIe dcliisor by the remainder, and so onl; aloays dividing the last divisor bp tle last remainder, till nothing remains; the last divisor will be the greatest common divisor sought. 62, RAY'S HIG-GI IE~R ARI.ITIHM:ETIC. Find the greatest common divisor of 24 and 66 SoLuTioN. —-Divide 66 by 24; the 24)66(2 quotient is 2, and the remainder 18. 48 Next, divide 24 by 18; the quotient is 1, and the remainder 6. Lastly, divide 1) 24 ( ItS by 6; there is no remainder; hence, 18 6 is the greatest common divisor of 24 6)1 8 (.til 6. 1 8 IRule 2 depends on the following PRINCIPLES. 1. A divisor of a lnumber is a divisor of any nzciteple of that nlinober. As shown in Art. 89, Principle 1st. 2. A comimon divisor of two,neumbers is a divisor of their suM. As shown, Art. 89, Principle 2d. 3. A common divisor of vwo ~numbers is as divisor of their DIFFERENCE. DE.nONSTRATION.-Since each of the numbers contains the coinmon divisor a certain number of times, their difference mnust contain it as many times as the larger contains it more times than the smaller; 2 being a divisor of 16 and 10, must be a divisor of their difference; for, 16 is 8 tivos, and 10 is 5 twos, and their difference is 8 twos linus 5 twos - 3 twos. 4. The greatest common divisor of two nezmbers is ialso the grea.test conmmon divisor of the smaller, and thceir recnailder aIter division. OBSERVE, that in the followving demonstration, G. C. D. signifies greatest common divisor. DENIONSTRXT1ION. —The G. C. D. of 24 24) 6 6 (2 and 66 divides 24, and must, therefore, 48 divide 48, which is 2 times 24, (Principie lst); as it divides both 66 and 48, 1 ) 2 it must divide their difference 18, (Prin- _ _ ciple 3d); therefore, the G. C. D. of 6 24 and 66, is a common divisor of 24 1 8 and 18. It now remains to show that it is their greatest common divisor. If there could-be any greater common divisor of 24 and 18, as it would divide 24, it would also divide 48 (Principle 1st); and as it. REVIEw.-94. What is Rule I for finding the greatest collimon divisor of two numbers? Prove it. What are contained in tile greartest common divisor? How is the rule applied to more than two numbers? UREi'J'ATEST COTMMAON DIVISOR. 638 would divide botuh 48 anid 18, it would divide their sum 66, (Principle 2,1), and would. cunsequntly. be a conmmon divisor of 24 and 66. IV e shoutld then have a colmmon divisor of 24 and 66 greater than their gl(atest conimon divisor, which is absurd. Hence, the G. C. D. of 24 iand 66 is not only a com. divisor of 24 and 18, but is their G. Co D.; and the proposition is proved. DEMIONSTRATION OF RULE II. The reason of the rule follows immediately from Principle 4th; for, as the G. C. I). of the two given numbers is the same as the G. C. D. of the smaller and their remainder after division, and as this is the same as the G. C. D. of the smaller of those two and their remainder after division, and so on; it follows, thlat when we get the G. C. D. of any remainder and the previous divisor (which is,always the smaller of the two numbers used in the division), thlis will also be the G. C. D. of tile original two numrnbers; but, whenever a remainder is exactly contained in the previous divisor, it will necessarily be the G. C. D. of those two, since no number can have a greater divisor than itself; and, therefore, whenever this exact division takes place, the divisor will be the G. C. D. not only of tile two numbers last used, but also of the two lunubers first given. NOT'ES.-1. To find the G. C. D. of more than two numbers, first find tihe G. C. D. of any two; then of that G. C. D. and one of the remaining numbers, and so on -for all the numabers; the last C. D. will be the G. C. D. of all the numbers. 2. If two given numbers be divided by their G. C. D. the quotients will be prime to each other. FIND TIIE GREATEST COMMON DIVISOR OF ANS. ANS. 1. 85 and 120.. 5. 7. 597 and 897.. 3. 2. 91 and 133.. 7. 8. 825 and 1287.. 33. 3. 357 and 525.. 1. 9. 423 and 2313 9. 4. 425 and 493. 17. 10. 18607 and 24587. 23. 5. 324 and 1161.. 27. 11. 105, 231 and 1001. 7. 6. 589 and 899.. 31. 12. 165, 231 and 385. 11. 13. 816, 1360, 2040 and 4080.... Ans. 136. 1-4. 1274, 2002, 2366, 7007 and 13013. Ants. 91. TRE vIE Y.-95. When is Rule I generally used? What is Rule 2? Illustrate. What are the 1st and 2d principles on which the demonstration of the rule depends? The 3d? Prove it. The 4th? Provo it. Damonstrate the rule. 64 RAY'S HIGHER ARITHIMETIC. LEAST C01IMMON MULTIPLE. ARtT. 96. A common multiple of two or more numbers, is a number that can be divided by ea ch of them without a remlainder; 24 is a common multiple of 3 and 4. The least common multiple of two or more numbers, is the least number that is divisible by each of them. 12 is thie least common multiple of 3 and 4. It E 3ARE x s.-l. Since the common multiple of two or more numbers contains each of them as a factor, every common multiple is a composite number. 2. Since the continued product of two or more numbers is divisible by each of them, a common multiple of two or more numbers may always be found by taking their continued product; and since any multiple of this product will be divisible by each of the numbers, (Art. 89, Principle lst), an unlimited number of common multiples may be found for the same numbers. TO FIND TIlE LEAST C033OMMON MULTIPLE. ART. 97. RUaE I.-Separate the given numbers into t7eter prime factors; then nueltiply together sulch/, and only such, of those factors, as are necessary to fornm a product thLat will containb all the prinze factors of each Snumber. Find the least common multiple of 10, 12, 15. SOLUTION. -Factor the num- 1 0 -- ~ X bers; the prime factor 2 occurs 2 2 2 once in 10 and twice in 12; strike out the first 2 and reserve 2 X 2 as factors of the least common 2 X 2 X 3 X 5 - 60. Ans. multiple. The factor 5 occurs once in 10 and once in 15; strike out the first 5, and reserve the second as a factor of the least common multiple. The factor 3 occurs once in 12 and once in 15; strike out the' first 3, and reserve the second 3 as a factor of the least common multiple. Lastly, take the product of the reserved prime factors 2< 2 X 3 X 5 = 60, for the least common multiple. DExicONSTRATION.-The product 2 X 2 X 3 X 5 = 60, is a common multiple, because it contains all the factors of each of the RrVIlEW. —96. What is a common multiple of several numbers? Wh'at is the least common multiple? Give exarmples. IHow many common multiples may be found for several numbers? Why? 97. Give Rule 1. Prove it. How often must a factor be taken in the least common multiple?' Why? LEAST COIMMON MULTIPLE. 65 numbers; it is the least common multiple, because it does not contain afny primle factor not found in some one of the numbers. REIMARKs.-Each factor must be taken in the least common multiple the greatest number of times it occurs in either of the numbers. In the preceding solution, 2 must be taken twice, because it occurs twice in 12, the number containing it most. 2. To avoid mistakes, after resolving the numbers into their prime'factors, strike out the needless factors. FIND TIlE LEAST COMIMON MULTIPLE OF ANS. ANS. 2. 8, 10, 15. 120. 5. 8, 14,21, 28. 168.. 6, 9, 12). 38 6. 10,15, 20, 30. 0. 4. 12, 18, 24. 2. 7. 15, 30, 70, 105. 210. AIT. 98. r;LE II.-1. Place the numibert inl a line; strike out an1y tnlt wCill exacll y divide any of the others; divide by any prime ntumlJer lthat will divide two or?or0e of the wit,7hout a remeainder; set the quotien/s and ulzdivided nlumbers in ct line benea t7h. 2. Prloceed with this line as before, and continue the operation till no n7znber greater than 1 will exactly divide two or snore of the n11nbers. 3. 1ultiply/ together the divisors and the ntmbers in the last line; their jlrodlct will be the least comm07on mUnlliple required. Find the least common multiple of, 10, 20, 25 and 30. SoLUTION..-Strike 2 20, 25, 30 out the 10, because it is exactly contained in 20; 5) 10 2 5, 15 then divide by 2, because it is a factor of two of thle numbers, 20 2 5 2 X 5 X 3 = 300 0 Ans. and 30. Next divide by 6, because it is a factor of more than one of the remaining numbers. Lastly, multiply together the divisors 2 and 5, and the remaining numbers, 2, 5 and 3; their product, 800, is the least com. multiple. D E s o N s T A T I O N.-The 10 is marked out to shorten the operation. The least common multiple of 20, 25 and 30, contains 20, and therefore contains 10, which is a factor of 20, and will be the least REvIEI.-9S. Give Rule 2. Prove it. What kind of numbers must be used as divisors? 1Why? Whren the numbers are prima to each other, how is the least common multiple found? RAY'S HIGHER ARITHMIETIC. common multiple of all the numbers. The rest of the demonstration is similar to the previous one; for the division by the prime numbers serves to strike out the needless factors, and those divisors with the factors remaining in the last line, are evidently the reserved fac. tors necessary to constitute the least common multiple. For a full analysis, see Ray's Arithmetic, Third Book, Art. 120. NOTE.-In dividing to cancel the needless factors, divide by a primne number. Dividing by a composite number would not, in all cases, cancel all the needless factors; in the preceding example, if we divide first by 10, the 5 in the number 25 would not be canceled. ART. 99. RULE III. —1. Set the numbers in a line, striking out such as are contained in any of the others, and sepalate any convenient one, generally the largest, fromn the others, by a cuerved line. 2. Divide each of the remaining nuntbers by tihe greatest divisor common to it and the number cut of, and set the quotients in a line beneath. 3. Proceed with the second line exactly as swith the first, and continue so until all the quotients are observed to be prime to each other; the continued product of these qutotients and the nu?,bers cut oif, will be the least common multiple requtired. N o T E S.-I. If the number cut off is found to be prime to all therest in the same line, cut off another, and proceed with it, reserving the first as a factor of the least common multiple. 2. When the given numbers contain no common factor, it is evident their product will be the least common multiple required; the least, common multiple of 4, 9 and 25, is 4 X 9 X 25 = 900. 3. The least common multiple of several numbers is equal to the product of their greatest common divisor, by those factors of each number not found in the others. 4. If the least common multiple of several numbers be divided by either of them, the quotient will be the product of all the factors of the others not found in the divisor. Resume the example under Rule 2. S LU T IO N.-After striking out OPERATION. ttue 10, cut off 30 from the rest by Q, 2 0, 2 5 (3 0 a curve, and then divide 20 by 10, 2 5 which is the greatest divisor corn- 2t 5, 03 0 0 mon to it and 30; also divide 25 by 6, the greatest divisor common to it and 30. The quotients are sel CASTING OUT THE NINES. 67 beneath, and being prime to each other, further cutting off and diE viding are unnecessary, and the product of 2, 5 and 30, gives 300, the least common multiple required. D E MO N S T R A T I O N.-The least common multiple must contain the number cut off (30), and such factors of the rest as are not found in the -30; on this account, divide each of them by the greatest factor ~common to it and the number cut off, thus getting rid of the needless f':ctors; in like manner, the least common multiple must contain the number cut off in the second line, and such factors of the rest as are not found in the one cut off; therefore, we repeat the process for each line, until all the numbers in the same line are found to be prime to each other; when this is the case, there will be no more needless factors, and the least common multiple will be the continued product of the numbers cut off and those in the last line. FIND THE LEAST COMMON MULTIPLE OF ANS. ANS. i. 4,6,15.. 60. 5. 35,45,63,70. 630. 2. 6,9,20... 180. 6. 10,14,20,35. 140. 3. 15,20,30 60. 7. 8,15,20,25,30. 600. 4. 7,11,13,5. 5005. 8. 15,24,40,140. 840. 9. 30,45,48,80,120,135.... Ans.2160. 10. 77,91,105,143,165,195... Ans. 15015. 11. 174,485,4611,14065,15423. Ans. 4472670. 12. 498, 85988,235803,490546. Ans. 244291908. PROOF OF THE ELEMENTARY RULES BY CASTING OUT THE NINES. ART. 100. To cast the 9's out of any number, is to divide the sum of its digits by 9, and find the excess. To do this, begin at the left, add the digits together, and when the sum is nine or more, drop the 9, and carrythe excess to the next digit, and so on. For example, to cast the 9's out of 768945, say 7 and 6 are 13, which is 4 above 9; drop the 9, and carry the 4; REVIEW. —99. Give Rule 3. If a number cut off is prime to the rest in the same line, what must be done? Prove the rule. 1 00. What is meant by casting the 9's out of any number? 68 RIAY'S HIG HER ARITIIMETIC. 4 and 8 are 12, which is 3 above 9; then, 3 (the digit 9 bein, passed over) and 4 are 7 and 5 are 12, which is 3 above 9; consequently, the excess in this instance, is 3 All the methods of proof are founded on tliis PRINCIPLE.-Any nlumber dividedl by 9 will leave the same remnainller as the szum of its dligits divicded by 9. For example, take 3456. 3000=3(1000)=3 x (999+1)=3 x 999+3 3456 400= 4(100)= 4x(99+1)= 4x99+4 50= 5(10)= 5x(9+1)= 5 X 9+5 6=.6 DEO N S T RAT I O N.Observe that 3000 is a certain number of 9's, with a remainder 3; 400, a certain number of 9's, with a remainder 4; 50, a certain number of 9's, with a remainder 5; and 6, being less than 9, may be termed a remainder. The remainders, 3, 4, 5, 6, are the digits of the number 3456; hence, the excess of 9's in the number 3456 is the same as that in its digits, 3, 4, 5 and 6. PROOF OF ADDITION. Cast the 9's (or 1's)"' out of each of the 2numbers addecd, also out of their suin; the last excess mutst equal the sum7 of the others after droppinlg all 9's (or Il1's).' To cast the Il's out of any number, see page 70. The excesses in the numbers are 8, 2, 4 EXCESS. and 3, and the excess in the sum of these 4 8 excesses is 8. The excess in the sum of 5 84 2 the numbers is 8, the two excesses being ( 6 41 4 the same, as they ought to be when the 73 0 3 work is correct. 2 6 7 2 9 8 NOTE.-In proving Addition by this method, it is not necessary to stop at each number and wri.te the excess, but regard all the numbers as forming one horizontal line. DE OSTRATI ONx.-Divide each number by 9, and add up the remainders, omitting the 9's, if any; the result must be the same as the remainder obtained by dividing the sum of the numbers by 9; and as these relnainders are most easily obtained by casting the 9's out of each number, (see Principle), the reason of the proof is apparent. R E V IEn W. —100. On what principle do the proofs by casting out the 9's depelnd? Provo it. What is the proof of addition? CASTING OUT ELEVENS. 69 PROOF OF SUBTRACTION. Cast the 9's (or l's) out of the subtrahend, the remainder, and the minuend; the last excess will be equal to the sean of the other two, after dropping all 9's (or l's). E X AMIL E.-Proceeding as directed in the- note to last proof, we find the excess Minuend, 7 6 4 0 in the subtrahend and remainder together Subtrahend, 1 2 3 4 to be 8; the excess in the minuend is also Remainder 6 4 0 6 8, as it should be when the work is right. D EMONST RX AT IO N.-As the minuend is the sum of the subtrahend and remainder, the reason of this proof is seen from that of Addition. PROOF OF MULTIPLICATION. Cast the 9's (or Il's) out of the mutltiplicand and multiplier; m2nltipl/ the two excesses together; cast the 9's (or l's) out of the result, and also out of the product; when the work is correct, the last two excesses will agree. Multiply 835 by 76; the pro- 8 3 = 92 9 + 7 duct is 63460. The excess in 7 6 8 >< 9 4 the nimultiplicand is 7, in the multiplier 4, and in the pro- 368 X 9 + 2 8 duct 1; the two former multi- 7 3 6 x 81 + 5 6 x 9 plied give 28, and the excess 736 X 81 + 4 24X 9 +28 in 28 is also 1, as it should be. DE MONSTRATION.-Since each of the numbers to be multiplied contains a certain number of 9's and an excess, their product, as seen from the work, will consist of three parts, two of which are divisible by 9; therefore, the excess of 9's in the product must be the same as the excess of 9's in the 3d part, (28), which is the product of the excesses in the multiplicand and multiplier; and since these excesses are most easily obtained by casting the 9's out of the digits of each number, (see Principle), the reason of the proof is apparent. PROOF OF DIVISION. Cast the 9's (or l's) out of the divisor, dividend, quotient, and remainder. Nultipliy together the excesses in the divisor and quotient, and cast the 9's (or l1's) out of the result; then, to this excess, add the excess in the remainder, and cast the 9's (or I l's) out of the result; when the work is correct, this excess will be the same as the excess in the dividend. REsvIEw.-What is the proof of subtraction? Of multiplication? 70 RAY'S HIGHER ARITHMETIC. Divide 8915 by 25; the quotient is 356 and the re. mainder 15. Excess of 9's in the divisor.7. Excess of 9's in the quotient..... 5. 7 X 5 = 35. Excess of 9's in 35. 8. Excess of 9's in the remainder.... 6 6 + 8 =14. Excess of 9's in 14..... 5. Excess of 9's in the dividend is also... 5. DEaoNoSTRATrIN.-Since the dividend is the product of the divisor and quotient, with the remainder added, the reason of this proof is seen from those of Multiplication and Addition. PROOF BY CASTING OUT THE ELEVENS. To cast the 11's out of any number, add its alternate figures, commencing at the right, and dropping 11 when the sum exceeds that number; then do the same with the remaining figures, and subtract the last result from the former, increased by 11 if necessary. For example, to cast the II's out of 30752486, say 6 and 4 are 10 and 5 are 15; drop 11 and 4 remains, then 4 and 0 are 4; also, 8 and 2 are 10 and 7 are 17, drop 11 and 6 remains, and 6 and 3 are 9; 9 from 4 can not be taken, but 4 increased by 11 is 15, and 9 from 15 leaves 6, which is the excess required. The Proofs by casting out the Ii's depend on this PRINCIPLE.-Any number divided by 11 -leaves the same remainder as the excess obtained by casting out its 1I's. This principle, and the proofs to which it leads, can be demonstrated in a manner similar to those regarding the 9's, and, by putting 11 for 9 in the proofs by 9's, we have the proofs by l's. (See Proofs.) REMnARK.-These methods of proof are liable to this objection; two figures may be substituted for each other, or the correct figures nlay be replaced by wrong ones having the same sum, and the work appear to prove when it is wrong. These, however, occur so rarely as not to detract much from the merits of the methods. R E V I E w.-What is the proof of division? Illustrate and give the reason for these proofs. 100. Explain the proofs by casting out the 11's. When will these proofs fail to detect an error? CANCELLATION. 71 CANCELLATION. ART. 101. CANCELLATION is a short method of obtain. ing results by omitting equal factors from. a dividend and divisor. Its use is seen in the following examples. Exchanged 24 hats at $5 each, for coats at $24 each: how many coats did I get? So L UT IO N. —To solve this question, multiply 24 by 5, and divide the product by 24. Instead of performing this work, indicate it thus,; then, since dividing either factor of a product divides the product, (Art. 71); the result is 1 X 5 = 5; the same as would be got by canceling the 24 from both dividend and divisor. Multiply 105 by 18, and divide the product by 30. So L UT IO N.-Indicate the operations as in the margin. Divide both dividend and 2 1 3 divisor by 6; this gives 105 X 3 (Art. 71,) X above, and 5 below, and does not alter the quotient, (Art. 76).,The quotient is found, 21 3 63. as in last example, to be 21 X 3 (Art. 71) or 63. As each factor is used in canceling, it is crossed to indicate that there is no further use of it; and each quotient is placed beside the number from which it is obtained. The 21 and 3 being the factors left of the dividend after cancellation, are multiplied together; their product is 63, and as no factor of the divisor is left, the 63 is not to be divided, and is, therefore, the quotient required. Multiply 75, 153 and 28 together, and divide by the product of 63 and 36. S oL T IO N. Indicate the 2 5 17 operations as in the margin. P XA3X p Cancel 4 out of 28 and 36, leaving 7 above, and 9 below. Can- 3 eel this 7 out of the dividend and out of the 63 in the di- 25 X 17 425 visor, leaving 9 below. Cancel 3 3 a 9 out of the divisor, and out of 153 in the dividend, leaving 17 above. Cancel 3 out of 9 and 75, leaving 25 above and 3 below. No further canceling is possible; the factors remaining in the dividend are 25 and 17, whose product, 425, divided by the 3 in the divisor, gives 141:'. 7 2 RAY'S HIGIIER ARITHXMETIC. ART. 102. If the dividend or divisor, or both, is the product of several factors, the quotient can often be easily obtained by this RULE FOR CANCELLATION. 1. Indicate the e mldtiplications which produce the dividend, and those, if any, wchich prodluce the divisor. 2. Cancel equal factors from dividend and divisor; vnslttiply together the factors 1remaining in the dividend, and divide the product by the product of the factors left in the divisor. NOTES.-1. If no factor remains in the divisor, the product of the factors remaining in the dividend will be the quotient; if only one faictor is left in the dividend, it will be the answer. 2. Cancellation can only take place between a factor of the dividend and a factor of the divisor; not between two factors of the dividend, or two factors of the divisor. 3. Canceling equal factors is dividing both dividend and divisor by the same number, which does not affect the value of the quotient, (Art. 75.) EXAMPLES FOR PRACTICE. 1. Holw many cows, worth $24 each, can I get for 9 horses worth $80 each? A ns. o0. 2. I exchanged 8 barrels of molasscs, each containing 33 gallons, at 40 cents a gallon, for 10 chests of tea, each containing 24 pounds: how much a pound did the tea cost me? Ans. 44 cents. 3. HIow many bales of cotton, of 400 pounds each, at 12 cents a pound, are equivalent to 6 hogsheads of tugar, 900 pounds each, at 8 cents a pound? Ans. 9. 4. Divide 15 X 24 X 112 X 40 X 10 by 25 X 36 x 56 x 90. Ants. 3 VIII. COMMON FRACTIONS. ART. 103. A fraction is a part of a unit or whole thing, when it is divided into equal parts. If an apple is divided into 3 equal parts, each part is a fraction of the apple. NOTE. — raction is derived from the Latinfractls, broken. REVIES.-101. What is Cancellation? Illustrate it. 102. In what ease may cancellation be employed? Give the rule. COMM1DON FRACTION S. 73 ART. 104. The aname of a fraction, acn the size of its rac rts, de]:en(l on the numnber of poarts into whlich the un'it is dividled; thus, when the aunit is divided into 2, 3, or 4 equal parts, the fractions are named halves, thirds and fbirtrEhs (Art. 62): moreover, a ht!lf is evidently larger tuan a t/lhi'd, a,air d t han afou-rth, and so on. AirtT. 105. Fractions are divided into two classes, (,87nlnor, and Descinal. A common fraction is expressed by two numbers, one above the other, with a horizontal line between theim; onehalf is expressed by I, two-thirds by d. The number below the line is called the de7nominator, because' it denominates or gives name to the fraction: it shows into how many parts the ounit is divided. The number above the line is called the nutmz7erator, because it -ntbers the parts: it shows how many parts the frau.cttloa contains; in the fraction 4, the denominator 7, shows that the nnit contains seven equal parts; the numerator 4, shows that the fi'ac'tion, contains 4 of those parts. The term~,s of a fraction are the numerator and denominator; the terms of - are 5 and 8. ART. 106. There are two methods of considoring a friaction whose numerator is greater than 1; thus, to divide 2 apples of the same size, equally among 3 boys, divide each apple into three equlal parts, making 6 parts in all, and then give to each boy two of the parts, expressed by'. The 2 parts may be taken firom one apple, oir I part fromll each of the two apples; hence, 2 expresses either 2 thirds of one thing, or 1 third of two things. Also, - expresses either 4 fifths of one thing or 1 fifth of four things; therefore,'The nu?,meractor of a fraction macy be reg-arded as showing t:he nutllber' of units to be divided; the denominnator, the nmln5er o/f arts i~nto which the numerator is to be divided: the f/r'action, itself being the value of one of those parts. Ri STE w.-l102. If no fvitor remains in the divisor, what is the quoient? Betwseen whav, fietors only can cancellation take place? What Jtoes canceling equal factors amount to? 103. What is a Fraction? Why ao c:lled? 104. t-low are fractions named? What does the value of a?ractional part depend on? 105. HIow are fractions divided? ITow are eorlainon fractions expressed? What is the denominator? What is the numerator? Why are they called so? What are hoth together called? 7 74 RA Y'S HIIGHER A RITHIMETIC. Hence, a fraction may be considered as an expression of unexzecuted division, in which The DIVIDEND is called the NUMIERATOR; The DIvisoR is called the DENOM3INATOR; The QUOTIENT is called the FRACTION itself. - is the quotient of 1 (num'tor) divided by 4 (denol.) 4 _ is the quotient of 3 (num'tor) divided by 5 (denom.) 7 is the quotient of 7 (num'tor) divided by 6 (denom.) NUMERATION AND NOTATION. ART. 107. Since Fractions arise from Division, one of the sigsns of Division, (Art. 59), is used in expressing them; that is, the numerator is written above, and the denominator below a horizontal line; hence, TO READ COMMON FRACTIONS, RULE.-R- ead the number of parts taken as expressed by the numerator, and then the size of the parts as expressed by the denominator. I is read seven ninths. REMARxK.-S-een ninths, (-), signifies either 7 ninths of one, or. of 7, or 7 divided by 9, (Art. 106). Fractions to be read; that is, expressed in words: 5 3, 7 9 13 7 14 9 13, 7 1, 8 7 8 10 1 17 27 41 65 90 100 230 134.5 Of these friactions, which expresses parts of the largest size? Which, the smallest size? Which, the least nuinber of parts? Which, the greatest number of parts? Which, the same number of parts? Which, parts of the same size? TO WRITE COMMON FRACTIONS, ART. 108. RULE. — Write the number of parts; place a hor-z gontal lisze below it, under which write the number which inldi. cates the size of the parts. R VIE w. —106. Ilow many methods of considering a friaction? Illus. trate them. What operation is expressed by a fraction? What is the dividend? the divisor? the quotient? COMMON FRACTIONS. 76 Fractions to be expressed in figures: Seven eightihs. Four elevenths. Five thirteenths. One seventeenth. Three twenty-ninths. Eight twenty-oiethls. Nine fiortly-woths. Ninetee ty-thirds. Thirteen on? hndredCths. Twenty-four one huznclred (a,d fijteeaths. To express a whole number in the form of a fraetion, write 1 below it for a denominator; thus, 2 7, &7 =_, &c. PROPOSITIONS. ART. 109. Pimp. 1.- When t7he nznzerator of a fraction is less than the denzozinator, the fraction is less than 1. PROP. 2.- When the,1umzerator of a fraction is eqult to the denominaztor, the fraction is equal to 1. PRor. 3.- When the numerator of a fraction, is greater than the denomlinator, the fraction is greater th/an 1. DE O N S TR T I O N.-Prop. 1 is true, because, in that case, tlhe number of parts in the fraction is less than the number of parts in a unit. Prop. 2 is true, because, in that case, the number of parts in the fraction is the sane as tile number of parts in a unit. Prop..3 is true, because, in that case, the number of parts in the fraction is greater than the number of parts in a unit. ART. 110. DEINrITIONS. —1. A proper fraction is one whose numerator is less than the denominator; as, _; -. 2. An i ]7)proper fraction is one whose numerator is equal to, or greater than, the denominator; as, -3 and 6. REAsruaRtx.-The word fraction primarily signifies a part. A proper fraction is properly a fiaction expressing a value less than the whole. An inmproper fraction is not properly a fraction, the value expressed being equal to, or greater than, the whole. 3. A simple fraction is a single fraction, proper or improper; as,, 3 or. R v I Ue w. — 10. What is the rule for reading fractions? for writing them? 108.'What two other methods of reading a fraction are there? low may a lwhole number be expressed as a fraction? 109. When is a fraction less than 1? equal to 1? greater than 1? Give the reasons. 110. What is a proper fraction? an improper fraction?.Why so c"led? What is a simple fraction? ] 6 RAY'S HIGIHER ARITHMETIC. 4. A compoand fraction is a fraction of a fraction, or several fractions connected by qf; as, 3 of 3 of 4. 5. A mixed nunmber is a fraction joined with a whole number; as, 11 and 24. 6. A conmplex fraction is one having a fraction either in one or both of its terms; as,-, 2 1 and 1 4' 01 4 3' ART. 111. To show that 1 of =-4; that ~ of o -f; that 1 of 4 — 1 and so on. DEMONSTRATION.-Divide a A.-.-. line as A B into two equal parts; C one of the parts, as A C, is termed one half (1): divide a half into 3 equal parts, as in the figure; one of the parts is called one third of one half, which is expressed by figures thus, 1- of A. This is evidently one sixth of the whole line, that is, ~ of 1= G. In like manner, of = of -1 1 and so on. What is I of'? Why? What is I of I? 3 of -? What is 4 of? 4 5 PROPOSITIONS. ART. 112. PROP. I.OlMultply ng the numerator of a fraction likewise f multiplies the fraction. Multiply the numerator of the fraction' by 3; the result, 6, is three times as great as ~. ART. 113. P RO P. I I. - Dividing the numerator of a fraction likewtise divides the fraction. Divide the nunierator of the fraction s by 2; the result, 4, is only one half as great as 8. ART. 114. PROP. I II.-i-Multiplyibg the denominator oJ. a fraction, on the contrary, divides the fIraction. Multiply the denominator of the fraction 12 by 3; the result, 1~ is one thirhd as great as 1a.2 ART. 115. P R o P. IV.-Dividin,g the denominator of a fraction, on, the contrary, mtulytiplies the.fraction. Divide the denominator of the fraction, 1_2 by 2; the result, I is twice as great as 12. H N v I w. -.What is a compound friaction? a mixed number? a complex fraction? 112. What is the effect of multiplying the nnncerator of a fraction? 113. Of dividing the numerator? 114. Of multiplying the denominator? 115. Of dividing the denominator? COMIAMON FRACTIONS. 77 ART. 116. PR OP. V. —ultiplying both terms of a frac. tiol by tihe sacze 7umber, chanrges its form but does not talter its vt(l[ue. Multiply both terms of the fraction 5 by 3; the resuit, s has the same value as inART. 117. P R o P. V I. —Dividing both terms of a frac. tiot by the same VInumber, changes its form?", but does vot alter ifs v(/te. Divide both terms of the fraction 4 by 4; the result,', has the same value as. DE N ST R AT I ON.-Observe that the numerator of a fraction is a dividend of which the denominator is the divisor, and the value of the fraction the quotient. (Art. 106). Prop. 1 is true, because, (Art. 73), multiplying the dividend by any number, multiplies the quotient by the same number. Prop. 2 is true, because, (Art. 74), dividing the dividend by any number, divides the quotient by the same. Prop. 3 is true, because, (Art. 74), multiplying the divisor by any number, divides the quotient by the same. Prop. 4 is true, because, (Art. 73), dividing the divisor by any number, multiplies the quotient by the same. Prop. 5 and 6 are true, because, (Art. 75), multiplyingor dividing both dividend and divisor by the same number, the quotient remains the sarme. R E A A R IK.-TIence, there are two ways of multiplying a fraction, two ways of dividing a fraction, and two ways of changing its form without altering its value; that is, multiplying or dividing the numerator does the samne to the fraction; but multiplying or dividing the denonminator does the opposite to the fraction; whle multiplying or dividiEg both terms alike makes no change in its value. REDUCTION OF FRACTIONS. ART. 118. Reduction of Fractions consists in changing their forms without altering their values. CASE I. TO REDUCE A FRACTION TO ITS LOWEST TERMS. ART. 119. A fraction is in its lowest terms when the numerator and denominator are prime to each other, (Art. 88, D)ef. 5); as, 3, but not 4. RE vIE w.-116. What is the effect of multiplying both terms alike? 117. Of dividing both terms alike? Illustrate and prove these propositions. 7 B. h - AY'S G HIGHER ARITHMETIC. RuIeE.- Divide both terms by any common factor, do the same to h/e resullting fraction, and so on, till both terms are prime to each other. Or, divide both terms of the fraction by their greatest common divisor. Reduce 2o- to its lowest terms. S o L T IO N.-Dividing both terms by the common factor 2, the result is 2 A d-; clividing this by 5, the result is 2) 5 Ans. ~, which can not be reduced lower. Or, dividing at once by 10, the greatest common divisor of both 1 0) — 3 Ans. terms, the result is'z as before. D E 0 N S T RAT I N. -The value of the fraction is not changed, because both terms are divided by the same number (Art. 117). REDUCE TO THEIR LOWEST TERMIS, ANS. ANS. ANS..1. A.. 4. 4- 3 7 253 11 3-.. 54 1 1 7 g 7 2 2. 3" 4 5. _21 T - T 1. -S * 3 - 2~.4. o. I1. 4 3. TO. 6. 9 ~9 EXPRESS IN THEIR SIMIPLEST FORMIS~ 10. 923 ~1491 = 12. 2261 <-4123 -17 11. 890 1691 =- 13. 6160 -40480 = a CASE II. TO REDUCE AN IM1PROPER FRACTION TO A WHOLE OR MIIXED NUMIBER. ART. 120. RULE.-Divide the numerator by the denom? nator; the qpotient will be the whole or mixed number. NOTE.-If there be a fraction in the answer, reduce it to its lowest terms. To reduce 1 of a dollar to dollars, divide 13 by; meaking 2 dollars. D aE I O N s T R A TI O N.-Since 5 fifths make 1 dollar, there will be as many dollars in 13 fifths as 6 fifths are contained times in 13 fifths that is, 23 dollars;' and so in all such cases. REVIEw.-118. What is Reduction of Fractions? 119. When is a fraction in its lowest terms? COMMON FRACTIONS. 79 1. In 8 of a dollar, how many dollars? Ans. 4 g 2. In i'7 of a bushel, how many bushels? Ans. 34 4 3. In 7,o% of an hour, how many hours? Ans. 13REDUCE TO WHOLE OR 3MIXED NUi1BERS, 4. ~ o Ans. 1. 7 62 C ~. Ans. 1051-7f 5. 1 5 Ans. 85. 8. 4 6 o. Ans. 32 7 1"3 5d 0 Ans. 888 9. 1570 Ans. 509 ti CASE III. TrO REDUCE A WHOLE OR MIIXED NUMIBER TO AN IMPROPER FRACTION. ART. 121. RULE.- tZltij)lly together the whole number and the denominator of the fraction; to the product add the nume rator, and 1write the sum over the denominatoro Reduce 33 to an improper fraction; to fourths. DEMONSTRATION.-In 1 (unit), 3 there are 4 fourths; in 3 (units), 4 there are 3 times 4 fourths, - 12 fourths: and 12 fourths +-3 fourths 1 2 = fourths in 3 15 fourths. 3 = fourths in fraction. R E MA A R K. —I. This demonstra- 1 5 = fourths in 3 tion shoes that the whole number Ans. 1 5 is really the multiplier, and the denominater the multiplicand; but, multiplying by the denominator Es more convenient, and gives the same result, (Art. 47). 2. This, and the preceding case, are the reverse of, and mutually prove each other. 1. In $7M, how many 8ths of a dollar? O Ans. a9 2. In 19-3 gallons, how many fourths?. Ans. 79. In 13637 hours, how many sixtieths?. Ans. sa REDUCE TO IMPROPER FRACTIONS, 4s o..',4 A. 10919 5.. Ans. 2?~ 5 1 S As. An7s 5 2O7 12T2 6. 127 -' Ans. 2170 9. 13, Ans. 1000 REVIEsW,-119. What is the rule for reducing a fraction to its lowest terms? Prcve the rule. 120. Give the rule for reducing an improper fraction to a whole or mixed number. Prove it 59~) RAY'S HIGHER ARITHMETIC. ARtT. 122. To reduce a whole number to a fraction 1ihaving a given denominator, is merely an example of the prIcedinc case, the numerator of the fractional part being erro, (0). It is done by multiplying together the wholc inumber and the denominator, and writin,'lie ploductl ove thec denominator. To reduce 4 to a fraction whose denominator is 5, i tlec same as to reduce 4~- to an improper fraction. 1. Reduce 7 to 4ths. Ans. 4s 2. Reduce 9 to sevenths. 63.Ans. 6 3. Reduce 23 to twenty-thirds. Ans.'5239 4. Reduce 19 to a fraction whose denominator is 29. Ans. 5i 5l CASE IV. TO REDUCE COMPOUND TO SIMPLE FRACTIONS. ART. 123. RULE.-Niultiply all the numlerators together for a new numerator, and all the denwminators together Jfr a new denominator. Reduce 2 of 4 to a simple fraction. 2- of 4 2=3X4= _s4 A ns. DEMONSTRATION.-2 of 4 =2 times of 45-=2 times 1 of of 4, (since i of 4 is the same as 4, by Art. 106,)-2 times l-' of 4, (since 1 of 4 is the same as 1, by Art. 111,) =2 times -4A (since T-l of 4 is the same as 4 by Art. 106,) -I8 (since multiplying the numerator multiplies the fraction, by Art. 112.) NOTE. —Before applying the rule, reduce mixed or whole numbers to a fractional form. 1. Reduce 4 of 4 of 38 to a simple fraction. SOLUTION.-3q1- q 22 and 1 of 2 of 22 - 4 Ans. 2. - of 51 to a simple fraction. AAns. 45 3. 2 of - Of' 27 to a simple fraction. An. s. -. of of f 34 to a simple fraction.. Ans. l N o T E.-Equal factors may be canceled out of the numerator an!u denuaninator, as out of any other dividend and divisor, (Art. 101), before the multiplications are performed. RE~vIEW. —l12. Give the rule for reducing a whole or mixed number to an improper firaction. Prove it. Which number is really the multiplier? COMMON FRACTIONS. P1 5. IReduce 3 of Ao of i7, to a simple fraction. SoLuTIo N.r-The factors 2, 37 and 3 are common to both terms. X X Canceling them, and multiplying - - - A X ns. together the remaining factors,'; 4X 5 4 the result is 7 twentieths. REDUCE TO SIMPLE FRACTIONS, 6. - of 3 of 7.. Ans. 7.. 4 of _ of 23. Ans. 2. 7. of 4 of 2. Ans. 53 9. of 4 of 34. An.s. 1 10. 4 9 4 Ans. 3 11. 1 of 5 of -a Of - of 42 3 ns. 12. 8 o f A of 77 of 7.....e Ans. 1+ T T l 2 3 13. P of -19 of ~1 of 2- of 1-4... Ans. us N T E.-To reduce complex to simple fractions, see Art. 132. CASE V. ART. 124. To reduce fractions of different denominators to equivalent fractions of a common denominator: RuLE.-Afultiply bloth terms of each fr'action by the product of all th/e denomilators except its own. D E aI O N o s T R A T I ON.-Multiplying both terms of each fraction by the same number, does not alter its value; the new denominator of each fraction will be the same, since it will be the product of the same numbers; viz., of all the denominators. NoTE.-Reduce compound to simple fractions, and whole or mixed numbers to improper fiactions, before applying the rule. Reduce 1, 2 and 3 to a common denominator. 1 X 3 X 4 12 new num. Both terms of the first frac- 4 2 tion are multiplied by 3 X 4 = 12; of the second, by2 X 2 X 2 X 4 1 6 new num. 4 8; and of the third, by 2 3 X 2 X 4 2 4 new denom. X 3 —6. 3 X 2 X 3 18 new num. 4 X 2 X 3 2 4 new denom. Since the denominator of each new fraction is the product of the same numbers; viz., all the denominators of the given fractions, it RtFVieav.-122. HIow is a whole number reduced to a fraction having a given denominator? 123. HIow are compound fractions reduced to simple ones? Prove the rule. 82 b~tiRAY'S HIGHER ARITHMETIC. is unnecessary to find this product more than once. The operation is generally performed as in the following example. Reduce 1, 5 and 7 to a common denominator. 1 X 5 X =3 5 lst num. 3 x 2 x 7=42 2d num. 35 42 BO 6 X 2 X 5 = 60 3d num. 2 X 5 X 7 - 0 com. denom. REDUCE TO A COMMON DENOMINATOR, 2.1 2 3) "! eeo 41.,,....n.... 3., 4 4 3 40 2. 3 6 Ans. 3'1 7~!0 5. 3.... Ant, v8,.160. Tf, 4. n, -4 -, 0..... Ans. 4,8s~, 6..] of, _i of,, of oA of 2of. Ans. 2j0 280 189 ART. 125. When the terms of the fractions are small, and one denominator is a multiple of the others, reduce the fractions to a common denominator, by multiplying both terms of each by such a number as will render its denominator the same as the largest denominator. Tiis nlumber will be foultd by div idinlg the larC1gst denZominator by the denomiznator of the fraction to be edlzuced. Reduce 1 and 6 to a common denominator. SOLUTIOeN-The largest denominator6,, is a x 2 2 multiple of 3; therefore, if we multiply both 3 X 2 6 terms of 1 by 6 divided by 3, which is 2, it is 5 5 reduced to. 6 REDUCE TO A COMMON DENOMINATOR, 1. 1 4.nd. Ans. 8, -j, 2. 3, 5 and 1.... Ans. A, 12 3., o and...As. Ans 0, 0,, Rl EV EW.-123. If there are any mixed numbers, what must be dlone? What may be done before multiplying? 124. How are fractions of different denominators brought to a common denominator? Prove the rule. What must be done with compound fractions? With whole or mixed numbers? COMMON FRACTIONS. 8 CASE VI. ART. 126. To reduce fractions of different denomina. tors, to equivalent fractions of the least common denominator. RULE.-Find the least common multiple of the given denomninacors; multiply both terms of each fraction by the quotient obtained by dividing this least common mulliple by the denomi nator of the fraction. Reduce,, 3 and - to the least common denominator. 2)2 4 6 2)12 4)12 6)12 2 3 6 3 2 W? 9 2x2x3 1 Z:6X 6 3X3 9 5 X 10 least com. mul. Ans. T6A, 9 and 1 DEMroNsTRATIoN.-Since multiplying both terms of a fraction by the same number does not alter its value, (Art. 116), each of the given fractions may be reduced to an equivalent fraction, whlose denominator is any multiple of its own; and they may all be reduced to equivalent fractions of the least comnimon denominator, by taking for that denominator the least common msultiple of' the given denominators. After getting the least common denominator, both terms of each fraction are multiplied by the quotient of the least common denominator divided by its own denominator, as in Art. 125. N o TES.-1. Before commencing, each fraction must be in its lowest terms. 2. Reduce compound to simple fractions, and whole or mixed numbers to improper fractions. 3. When the pupil is acquainted with the principles of the operation, the multiplication of the denominators may be omitted, as the new denominator of each fraction will be equal to the least common multiple. 4. The object of reducing fractions to a common denominator, is to prepare them for Addition or Subtraction. RE VIEw. —125. When one denominator is a multiple of the others, how can the fractions be reduced to a common denominator? 126. How are fractions of different denominators reduced to a least common denominator? What must be done if a fraction is not in its lowest terms? If there are compound fractions? If there are whole or mixed numbers? What will the least common denominator be? 84 RAY'S HIGHER ARITHMETIC. REDUCE TO LEAST COMMON'DENOMINATOR, 3 4) Ars 127 1 T 1 1 3 9 3 10 12 18 15 3.4 o o o 1 An1,. 5 A, 44 4. 80, 20,, 4* 4. 4 6 9 12 74 4 As.,4 cA4 6. 2 I ~of {, 54. n. Ao, 1821 17 6 TO 4 o f 14 34 0 2' 0 SO2 30 7. 1-,) 3- and -l of 3 As. 147, 30o8 A90 * See Note 1, preceding page. ADDITION OF FRACTIONS. ART. 127. Addition of Fractions is the process of adding together two or more fractional numbers. RULE.-Reduce the fractions to a common denominator; add their nullerators together, and place the sum over the common denominator. z4a 4, 8, 17 24 / 8 (12, 4 6x3=18 ~2 8 3x5-=15 new 2 x12 24 12 2x =1 numerator. least com. mul. 1 = 4 Ans. DEMtONSTRATION.- When the denominators of two or more fractions are the same, they express parts of the same size; we can, there.fore, add fourth 1 cent. 2 fourths as we would add, 2 cents. 3 fourths) 3 cents. The sum being 6 fourths, (i), in one case, and 6 cents in the other. That is, to add fractions having a common denominator, find the sumb of the numerators and write the result over the denominators. But, if the denominators are different, fractions do not express thincgs of the same unit value, and can- not, therefore, be added together (Art. 40); in the preceding example we can not add fourtla, eighths and twzoefths, but, by reducing them to twenty-foztrlhs, they express things of the same denomination, and can then be added. RnvS E w-127. Why are fractions reduced to a common denominator I What is Addition of Fractions? Wh;at is the rule? Prove it. SUBTRACTION OF FRACTIONS. 85 NOTES. —1. Before commencing thie operation, each fraction should be in its lowest terms, and compound fractions must be reduced to simple ones. 2. Mixed numbers may be reduced to improper fractions, and then added; or, the fractions may be added, and thei t[he whole numbers, and the results united. 3. After adding, reduce the result to its lowest terms. 1. Add 4 8 129 and 1815 2 9 2., da. Ans. 1 4 5. c, * and t'... Ans. 34 6. 1' and 23...... Ans. 44 7. 2,, 3a and 4~... A_4,. 10o 8., - o and 14.Ans. 3I 4Oa( - 5 ns 91 -16 1-, and 2' o c Ans. 4 a 6 4 8 and 2 91. y, T, I, a,d 2. Ans. 3: - 10. 1, 23-, 831 and 4* o.. Ans. 11' o 11. of,and 3 of 5 of2.. Ans. ll' 12. WThat is 5 + -1 4- ~?. Ans. 1 ~-j8 j + I + 77 +- 29?s1 9' 13. 2 + +., 4 4As. 1 I 14. + 1- +4+1?.... Ans. 417 15. 2 +4 4 + 5 1 ~' Ans. 123" 16. *+3 of s of 6?.Aos. 2+]04**~ 17. ) + - + -4 +?.. Ans. 4T4 iS. - of 6f3 +- of ~o 7.? Ans. 4s. 19. 4 of 961 +8 of 5?. Ans. 59*+ 20. 3 9SU BTRACT ION80 42 + 728? Ans. 5385 SUBTRACTION OF FRACTIONS. ART. 128. Subtraction of Fractions is the process of finding the difference between two fractional numbers. Ru:s. —Reduce the fractions to a common denomzinator;.fiznd the difference of their numerators, azd place it over the common denomina/of RE vIEw.-127. Before commencing, what should be done? W-hat Nvith mixed numbtey,? What should be done with the answer, when obtained? 86 RAY' S IIGIHER ARITH-IMETIC. Find the difference between c, and -8. SOLUTION. —The fractions, when reduced to the least cornrmon denominator, become 3 aud -30; their difference is ) — jo* D E M O N S T R A TI o N.-VWhen the denominators of two fractions:e llte same, they express parts of the same size, and their diffelrence,nl be found as in the case of whole numbers. Thus 5 sevenths, 5 cents. 3 sevenths, 3 cen ts. Difference, 2 sevenths (2q) in one case, and 2 cents in the other. But, if the denominators are different, the fractions Clo not ex press things of the same kind; therefore, one can not be subtracted from the other, any more than 3 cents can be taken from 5 apples, (Art. 43); in the preceding example, fifteenths can not be taken frosn sixths; but by reducing them both to thirtieths, theil difference can be found. NOTES.-1. Before commencing, reduce compound to simple fractions, and see that each fraction is in its lowest terms. 2. After subtracting, reduce the result to its lowest terms. WItAT IS ANS. rI-IAT IS A NS. 1. T-G.. 2 6 14?.. 3 2. T8T —3 of 1?. 33 6. 9 16 4. -'3 Of 4?.. - — d?... 4. IfT T of 4?.T~ 8. ( - -~,?... yW ART. 129. Mixed numbers may be brought to improper fractions, and then subtractedl; or, the fractions may be subtracted, and then the whole numbers and the results united. In the latter method, if the lower fraction is the larger, increase the upper fraction by the number of parts in 1 unit, and carry I to the first figure of whole nulnbers in the lower line. Thus, 6 1 + 8= 9 -8 4 -12 Wa s S1 or, 4a s h 1 R EV I E V. —128. What is Subtraction of Fractions? Whlfat is the 1(ule?9 Prove it.'What should be done before conmmencing? Whalt should be done with the answer, when obtained? 129. How are rmixed numbers subtracted? MULTIPLICATION OF FRACTIONS. lWHAT IS ANNS. WHAT IS ANS. 10.. 15 9 3 l' 14 18?.. 11. 5.3. o - 15. 3 of 2 3I?3a 12... 3f? 3.. 16. 3 2- oof 14? 19 17. L6 of 41_ -3 of 3? 3. Azs. 13 18 81 919?.2ZS. 10 -I-I 19. A man owned -. of a ship, and sold -4 of his share: thow much had he left? A as. s 20. After selling 4 of 5 + 5 of 3 of oa farm; what part of it remains? A2ls. 3 21. 34 + 44-54 ++ 16'- 7- + 10- 14, is equal to what? Auls. 6 22. 5O 2- + 1 -3 — i + 38i-+ -1 81-16r=whalat t? A ns. 74 2z. 1-4 of 6-3 of 7 -how mluch? Ans. 1A MIULTIPLICATION OF FRACTIONS. ART. 130. 7[Multiplication of Fractions is the process of multiplication, when one or both of the factors are fractional numbers. It embraces three operations: 1. To multiply a fraction by a whole number. 2. To multiply a whole number by a fraction. 3. To multiply one fraction by another. Since any whole number may be expressed in the form of a fraction, (Art. 108), these 3 cases may all be performed by this GENERAL RULE FOR MULTIPLYING FRACTIONS. ifultiptly the nlamerators toyether for a.new numerator, and the denominators together for a new denomilator. Multiply 4 by.. 4 X S AS. DEMONSTRATION.-We can attach no other idea to the product of 4 fifths by 2 thirds, than that signified by 2 thirds of 4 fifths, RE vI E wV. —130. What is Multiplication of Fractions? What does it embrace? What is the general iule? Prove it. ;388 ItRAY'S HIGHER ARITHMETIC. But, 2 thirnts of 4 fifths is 8 fifteenths, (Art. 123); therefore, the product of 4 fifths by 2 thirds, is also 8 fifteenths. IHence, these COROLLARIES. I To MIULTIPLY A FRACTION BY A WHOLE NuMBEIR.-MUltiply the,ulmerator of the fraction by the whole number, and write the product over the denominator, (Art. 112). Or, Divide the denominatol of the fraction by the whole E1number, when it can be done without a remainder, anzd over the quotienrt write the numerator, (Art. 115). II. To MIULTIPLY A WHOLE NUM3BER BY A FRACTION.-fnltUl61pl7y the whole number by the numerator of the fiaction, and divide the product by the denominator. REARIts. —1. After indicating the operations, if the numerator and denominator contain common factors, cancel them before multiplying. The result will be in its lowest terms. 2. MIultiplying one fraction by another is the same as reducing a compound to a simple fraction, (Art. 123). EXAMPLES FOR PRACTICE. 1. X 12.. = 9 4. -9 X 28.. =153 c} 3 i 13 2. X18.. = 8 5. 5 13X 30 =26 3. 4 x 24.. =1 6.. 3 X5. =18k SUGaESTION.-In multiplying a mixed number by a whole number, multiply the fraction and the whole number separately, and add the products; or, reduce the mixed number to an improper fraction, and multiply it; as, 2X 5 -=3, and 3 X 5 = 15; and 15 3 -- 1830 Or, 3 = 1, and' x 5 5=5 18-. 7. 45 x. =35. 11. 28 X 3. =102 8 50 X 1. =39q 12. 32X. O 9. 25 x.. 18 13. 16X a. =T~ 10. 32 X 2 = 176 14. 4.., 15. What will 3~ yards of cloth cost at 0$41 per yard? REVIEw. —1i0. HIow is a fraction mnultiplied by a, whole number? Htow is. az whole number multiplied by a fraction? lIow may the work be shortened? Multiplying fractions is equivalent to what case of reductioni How may a mixed number be multiplied by a whole number? DIVYISION OF FRACTIONS. 8B SUoGEsTtoO. —In finding the product of two mixed numbers, it is genelrally best to reduce them to inmproper fractions; thus, 419; 39 0; I3 0 9 The operation may be performed without reducing 3; to improper fractions; thus, 3 yards will cost $13, 1 3 and ~ of a yard will cost I of $41 ].; hence, the whole will cost $15. 1 5 10. 6] x4-.. =30. 118. 1~2 X s. 40 17. 4 X. = 19. 7X X 32 -. - 20. -Multiply a of 8 by -- of 10O... Ans. 21. Mlultiply s of 5 by' of 31.. Ans. 7 22. MIultiply 3 of 2 of " by 7 of 3.. A s. 40 23. MIultiply 5 ,4 2 and - of 4.. As. 94:s 24. Multiply I 7, i, of y, 7- of An. ils. 3ii-3 25. Mlultiply 7, -,9, 3 and.. Als. 4. 26. MIultiply 83, 4, 5a 9 of -, G. Anls. 49. 27. At 8 of a dollar per yard, what will 95 yards of cloth cost? A is. 2.A 17 28. A qluanltity of provisions will last 25 men 129 days: howr lon l will thle same last one man? Ans. 318 days. 29. At 3, cents a yard, \what will 2-L yar ds of tape cost?' An7s. 9- cts. 30. What must be paid for 3 of 3 of a lot of grouncl that cost $18? Aibrs. $7 1 31. K owns of a ship, and sells 4 of his share to L: what part has hle left? Aurs. -, a) 3i. B bought 3 of a farm of 219, wacres, and soldl - of his part to Cs: ilat part of the whole, and how niany acres, did he sell? Ans. -, andc 294 acres. DIVISION OF FRACTIONS. ART. 131. Division of Fractions is the process of diviwson, (Art. 58), when the dividend or divisor, ol botl, are fractional nuImbers. Tt enabraceS three operations: 1. rl'o divide a firaction by a whole number. 2. To divide a whole nuiaaber by a fraction. 3. To divide one fraction by another. 2.TSiieawoenme yaf'cin 90 RAY'S HIGHER ARITHMETIC. Since any whole number may be expressed in tlhe form of a fraction, (Art. 108), these 3 cases may all be per. formed by this GENERAL RULE FOR DIVIDING FRACTIONS. hIvert the divisor; then multiply the numerators together for o new numerator, and the denominators for a new denominator. Divide 4 by 3. inverted 3 3 3 9 2inverted _, and X 2 = -1 Ans. 1st DEMONSTrItATION.-SUPppose the divisor were 2 instead of 2; the quotient would be X -, (Art. 114). But, the real divisor, (i), is only one third as large as the supposed divisor (2); therefore, the real quotient must be three tinzes as large as the supposed quotient, (Art. 73), and 3 times 9-= 8, (Art. 112); which agrees with the rule. 2d DE n.-It has been shown, (Art. 60), that the divisor and dividend must be of the same denomination; hence, to find how often 2 thirds is contained in 3 fourths, reduce them to twelfths. But, - t, and 3 -9-9 and 8 twelfths in 9 twelfths is the same as 8 in 9; that is, 9 = 1 times. Hence, the rule might have been expressed thus: To divide one fraction by another, reduce them both to a common denominator, and divide the numeralor of the dividend by the numerator of the divisor. COROLLARIES. I. To DIVIDE A FRACTION BY A WVIOLE NunMBE. —Zlltiply the denorminator of the fraction by the whole nminber, and over the product write the numerator. Or, Divide the numerator of the fraction by the whole lumber, when it can be done without a remainder, and under the qiuotient write the denominator. II. To DIVIDE A WHOLE NUMBER BY A FRACTION.-MU l1tiply the whole numzber by the denominator of the fraction, aned divide the 7produlct by the numerator. REIVIE w.-130. IHow may mixed numbers be multiplied together? 131. What is Division of Fractions? What does it embrace? What is the general rule? Illustrate it. What is the 1st demonstration? the 2(1? flow is a fraction divided by a whole number? How is a whole numlbe divided by a friaetion? DIVISION OF FRACTIONS. 9 The rule may be simply demonstrated, as follows: 3d D Ez.-Inverting the divisor, shows how often it is contained in a unit, and this multiplied by the dividend, shows how often ii contains the divisor; in the example given, since 1 is contained in a unit 3 times, 2 is contained in a unit 3 times, (Art. 74), and in 3, it is contained 3 X -a = times, (Art. 130). N o TE S.-1. Before commencing, reduce compound to simple fraoo tions, and mixed numbers to improper fractions; also, express whole numbers in the form of fractions. 2. In all cases, reduce the result to its lowest terms. EXAMPLES FOR PRACTICE. 1. -1 J.3... = ~ ~ 1 G 2|. 14 9 2. g 3, 9. 3 5 7 7 3. 10... 5.. 4. 6.. 9. 11. 81 * 12 5. 21. =23 12. 19 1 ='- ~. ~1 ~ -- ~ ~. -- — ~U 6. -.. 13. 3 94 6. 4....1 13. 7-. - - 7. 7. -— ~..-. 26 j 14. 5444 —25a. = 28 15. Divide l by Z of * of 7..... Ans. 16. Divide f of T3 of - by 77 of I1.. Ans. 45 NoTE.-After indicating the operations, if the numerator and denominator contain common factors, cancel them before multiplying. The result will be in its lowest terms. 17. Divide I of 23 by 8 of 1I SOLUTION. —2 = 7, and 13 = 2 Then, 1 X X X 2 Ans. ~4 3 3 S u o a E S T I O N.-In the division of fractions, or other 3 2 operations by cancellation, it is often convenient to place the divisors on the left of a vertical line, and the multipliers on the right. REVIE W.-131. What is the 3d demonstration? Before commeneing what must be done? HRow may the work be shortened? 92 EiL 1'S 1HIGTIER ARITHIMETIC. 18. Divide - of by 4 of... An7s. 2zL i9. Divide of i1 by 3ofr 7..... s. 4 3 f 9 20. Divide' ofX:.3 by, of3.. As 21. Divide of S by of,. of...1 Aiis. 1 22. Divide of of -4 by Dof- of4.. As. i 23. Divide 1- times 42 by lI7 times 3 Ans. 18 24. Divide 3} by 4 of 84 times If of O3o 0 Ans. 4 25. Divide o- 3 f f 27 - by -4of,3 o; An.' 34.1 13 Of:,1 5'3 2 9 1 26. What is 2 X -of 19-. 4 X - f 8? AnS. 3 ART. 132. To reduce complex to simple fractions. Rure.-Diide the nle nieraltor by the denominaltor as in D)ivzsion of fr'actions, (Art' 131). Or, fittllily both terms of the complex fraction1s by the least Common110 mtlliple of the denominators of their fJiactional parts. Reduce i to a simple fraction. 11 1 4 OP ERATION. 1; then, - 4 Ans......- 6 4 30 The least common multiple of 6 and 4 is 12; hence, 1E. X12 22 2 29 OPEATIONT. = - - lns. 2: x 12 33 3 DE MONSTRATION.-Since every fraction indicates tllat the nucrator is to be divided by the denominator, the 1st Iule needs no demonstration. Since the value of a complex fraction is not changed when both terms are multiplied by the same number, (Art. 116), and since 12 is the most convenient multiplier that will cause the small fractions to disappear, the reason of the 2d Rule is evident. REDUCE TO SIMPLE FRACTIONS, a3 4a 4-] ]_. 7 As..,. j3. I'' Ans. 13 5.:tS~ Ans. 2Ans. 2 A4is. lA G. Anls. 324 2(fll ItEVIEW.-132. IWhat are the rules for reduoin g a complex to a simple fraction? Illustrate and provea them. DIVISTON OF FR.ACTIONS. i: Comiplex fractions may be multiplied or divided, by reducing theml to siumple fl.ractions. The operation may oftell be shortened by cancellation. 41 -X-. o AS. 63-t 17,.? x. Ans. 9 1. A731 6. 7 A. 7 10 8 2, 12 31 8I aI 11 ___ - 6. A: C Ans. 1' 11. 10;- 40~ 7 t 3 11;e11 94fi8'0 a 21 i ll". 50 la l ~ x A1. 2 I 1..... nv.. 97} o,2 1 24 23 8 A TO FIND THE GREATEST COM. DIVISOR OF FRACTIONS. ART. 133. RULE. —7educe the t numberfs to simple fractions, and to their lowest terms;.find the grealest common divisor of thie,nuerators, and divide it by the least com, mon mnfr ltple oqf' /Ihe denomiiators: lhe quotienlt will be tle greatest common divisor of the f'tactions. NoTrEs.-. If the numerators are prime to each other, and the denominators are also prime to each other, the greatest common divisor of the numbers will be, 1 divided by tihe product of the denomn ilatos. 2. Thle greatest common divisor of more than two fractions can be obtained by first finding t hile greatest common divisor of two of them, thien of this and a third, and so on; the last divisor will be the greatest common divisor of all. Find the greatest corn. divisor of 26. and 10859So a, utr T0 N.-The numbers when reduced to the form of fractions, are 15 and 5 the g reatest common divisor of thile numerators 10;5 and 8685, is 15, (Art. 95), and tie least common multiple of thle denominators, 4 and 8, is 8, (Art. 99); hence, tile gretest common divisor of the given numbers is I- = 1 DEZMON'STRATTO N.-If the fractions, 0o and 86i85, are divided by 15 the greatest, common divisor of their numerators, the quotients ic fractious, viz: 2 and j If the divisor 15 be divided by 8, the quotients must be mudltiplied by 8, (Art. 73), and become whole numbers, viz: 14 and 579. Now, 8 is the smallest possible number, wIichI Iued as n multiplier, will convert the quotients, and 5a9n R s v r w.-133. IThat is the rulo for finding the greatest common divisor of fra:etions? Illustrate and prove it. 94 RAY'S HIGHIER ARITHMETIC. into whole numbers at the same time, since it is the least common multiple of the denominators 4 and 8; therefore, as ~1, when used as a divisor, gives the smallest possible whole numbers for quo tients, it must be the greatest common divisor of the given numbers. FIND THE GREATEST COMMION DIVISOR 1. Otf 83~, and 2684.... Ans. 2T1 2. Of 1417 and 95.3.. Ans. 24 3. Of 594 and 7354. As. 243 4. Of 23-7o and 213...... Ans. 2-4 5. Of 4183 and 1772. An1s. 3 6. Of 237] and 1751a.. Ans. 54 7. Of 2611-3 and 652... Ans. 4ai: 8. Of 444, 5462 and 3160.. Ans. 44 9. Of 1374, 4784 and 2093.... Ans. 34 10. Of 39774, 1022-7, 2954,7 168014.Ans. 16 3 TO FIND THE LEAST COMMION MULTIPLE OF FRACTIONS. ART. 134. RULE. —Reduce the numbers to simple fractions and to their lowest terms. Find the least common mulltiple of the numerators, and divide it by the greatest common divisor of the denominators; the quotient will be the least common multiple of the fractions. Find the least com. multiple of 34, 44, 148,.15 SOLUTION.-The numbers, when reduced to the form of simple fractions, become 15 2X57 and -15a; the least common multiple of the numerators is 225; the greatest common divisor of their denominators is 2; the former divided by the latter gives 1124, the least common multiple of the given numbers. DEMONSTRATION.-If the given numbers were the whole numbers 15, 25, 9, 5, their least common multiple would be 225, which would contain them' respectively 15, 9, 25 and 45 times; moreover, 225 contains 145, 25 89 a5 respectively 4 X 15, 6 X 9, 8 X 25, 4 a 9' T2 15 X 2 12 X 45 times, since dividing the divisor multiplies the quotient, (Art. 73); and 4 of 225 —-1121 will contain the same numbers half as many times respectively, viz: 2 X 15, 3 X 9, 4 X 25, 6 X 45 times; and since no other number will divide all these quotients exactly, 1121 is the least number which will contain the fractions exactly, and is, therefore, their least common multiple. DIVISION OF FRACTIONS. 95 R E A R K. —The least common multiple of fractional numbers can'also be found by Rule 3, Art. 99, taking care to obtain the greatest common divisors required, by the rule in the last article. FIND THE LEAST CO0IMMON MULTIPLE 1. Of 2, 4,3 a, nd I...... Ans. 60. 2. Of 41, 64, 5 and 10*... Ans. 472k 3. Of 31, 4,,4, 55 and 12A.. Ans. 350. 4. Of 14-, 9/T, 16' and 25... Ans. 100. 5. Of 184, 664, 213 and 15... As. 600. 6f. Of 98, 221-, 25 and 12g%.. Ans. 167006 7. Of 8, If 1 67. 2Of 8, -O 5 and 3.. Ans. 14373. Of -i,, 3 and 3.. Ans. 22458 9. Of 12 18, 13, 15- and 17.. Ans. 11400. ART. 135. PROMISCUOUS EXERCISES. 1. Add together 3, 4~,51, 4 of S, and' of 1 of o. Ans. 1392 2. The sum of 1-P and is equal to how many times their difference? 1 Ans. 5 times. 3. What is {2+ of 24 I? Ans. 5. 2 3 -~ + 2. 471 of 2 5 200 7~-la 4. Reduce 4 o; and X (100- + to 51 41 24 their simplest forms. Ans. 16 and 26T ->%s 5. What is 4 of 54 1 of 3? Asss. -4 4 4a i- 3a 6. What is 5 2 X — 12 X i X 236 equal to? Ans. -s 2 —i 7. Also -X - X-? Ans. 1 2 2 3 1 1 - 2 31- 4 - 8. Also X 2 X X 3 X? 3 2 3 5 4 (2 +)+ (3 + b7 147 s.Ans #4o 9. (2 ) (4- = what? Ans..4.4 REViE.w.-134. IWhat is the rule for finding the least common mul. tiple of fractions? Illus-taato and prove it. 06 RAY'g HIGHER ARITIIMETIC. 41 x 41 x 4: —1 10. 41 41 1 hat? Ans. 41 11. Add of o f, X of, and 1 Alns. +4 -3 30 12. 3 of 1o of what number, diminished by 9 ~ leaves? At., 13. Find the least common multiple of the numbers -from 10 to 20 inclusive. Ans. 232792560. 14. K leaves L for N, (109 miles apart) at the same time that B leaves N for L. K travels 7 les per hour, a.nd B1, 8I miles per hour: in how many hours will they meet, and how far will each have traveled? Ais. 688 hours. K, 51'- miles; B, 57., nmiles. 15: What number multiplied by 4 of 4 of 8 will produce 21? A,s., 2 13 16. What, divided by 1, gives 14? A1ns. 233' 17. Vhat, added to 14s-, gives 291?7 Ans. 15S 18. I spend 5 of my income in board, - in clothes, and save $(60 a year: what is my income? Ains. $216. 19. Find the G. C. D. (greatest common divisor), of 96, 120, 160 and 200, and the least com. inul. of 1-3, 19, 57 and 65. Ais. 8 a nd 3705. 20. The least comrn. mul. of 10, 24, 35 and an unknown number prime to these three, is 9240. What is the unknown number? (See erem. 1, Art. 97). Ains. 11. 21.'What two numbers between 35 and 840, have the former for their G. C. D., and the latter for their least corn. mul.? (See Note 3, Art. 99). Ans. 105 and 280. 22. Find a number between 697 and 731, which shall have with each, the same greatest common divisor that they have with each other. Ans. 714. 23. The G. C. D. of three numbers is 15, and their least corn. mul. is 450. What are the numbers? Ans. 30, 45 and 75. 24. -l is what part of? A7is. 4 7 3 13 25. Divide 4 of 3; bY 11 of 7; and of1 of 274. by g of 7 Of 51. Arns. a5 d and 34-1 26. nMultiply Y7 of 21 by A3 of 19-; and divide 4 of s of 144 by 3T of of 134. Ans. Ad74 and 6A I I I I 3_9 -s4 ITk~7ana iSZ~ DECIMAL FRACTIONS. 97 27. Add togethcr 23~, 3~4, 3 2 and 5-%,, and divide the sum by 28. Ans. 4,-y-Vo7 28. Express -5 + as, a simple fraction; and also multiply a of' 21 by 3 of -7a. Aats. 2" and 7> 29. A bequeathed {1 of his estate to his elder son; the rest to his younger, who received $525 less than his brother. What was the estate? Ans. $5250. 30. Find the sum, difference, and product, of 37 and 2-9s; also thle quotient of their sum by the difference. A.;1s. sum 571 d 13& s, sum 5q dff., prod. 8 1., quot. 3 5O27 31. A cargo is worth 7 times the ship: what part of the cargo is -i-a of the ship and cargo? Ans. ry4 32. If a railroad car runs 1123 miles in 5' hours, what is the rate per hour? Ais. 21.~, 33. Subtract - of y5 of 64 from 8 of 5.; and mul3 2. tiply 19$ by 3 of 5 Ans. 3 and 5' 34. 6.4 is what part of 10T1? and reduce to its simplest form - -- 1 AiLs. 18 and 0'4 4fr 92 5 1) 35. Multiply, 14,',' and 6. Ais. 13'g 36. 7 of - of what number equals 9 A-n? Ains. 20. 37. A 63 gallon cask is 8 full: 9 I. gallons being drawn off, how full will it be? Asis. 50 4 38. If a person going 3. miles per hour, performs a journey in 144 hours, how long would he be, if li he traveled 54i miles per hour? Ans. 10.- hours. 39. A man buys 324 pounds of coffee at 17e cts. a pound: if he had got it 4` ets. a pound cheaper, how many more pounds would he have received? Ans. 11347 IX. DECIMAL FRACTIONS. ART. 136. A Decimal fraction derives its name from the Latin word decemn, meaning ten, and is so called, because its denominator is always 1 with ciphers annexed: being 10, or the product of several 10's; thus, 37 84015 692 -— oo - and are decimal fractions. 100 1000 1000000 19 98 tAY'S HIGtHER ARITiiMETIC. All tlie rules and operations in the several cases of Common Fractions apply as well to Decimal Fractions when thus written. But, a more simple and convenient Notation has been devised for them similar to that of whole numbers. This notation consists in writing the numerator, and placing a, point, (.), so-that the number of fiures on tlhe rihlht of' it shall be equal to the number of ciphers in the denominator. The deeimal fractions before given, when expressed in this way are.37 and 84.015 and.000692 The use of the point instead of the denominator, saves time and space, while no mistake i' likely to occur on this account, since the denominiator can be obtained asfollows: The denominator of any decimal fraction is 1 with as many ciphers a/nnexeId as there are figzres onz th7e righ7 of the p2oi~nt. ART. 137. A decimal fraction, when written with a point, is simply called a decisnal. The places and figures on the right of the point are called d(lCintal places and decimal figures, to distin-guish them fronm the places and figures of whole numberls. The point, (.), is called the decimal point or sepl(tra.mtix; it separates the decimal places firom the places of whole numbers. A llure decimal has only decimal figures; as,.02.319 A nzixed decimnal has figures of whole numbers; as. 281.63 A comp2lex decimal has a common fraction in its righthand place; as,.8} and 2.62:z A whole number may be regarded as a decimal, by supposing a point to be on the right of its units' place; as, 154 = 154. RE VI a W.-136. What is a Decimal Fraction? Why so called? Give examples. What mode of expressing them has been adopted? Illustrato it. AWhat is the advantage of writing decimal fractions with a point? W hon thus written, how can the denominator be known? 137. What is a decimal fraction called, when written with a point? Whant are the figures on the right of the point called? Whaltt is the poinlt callled? What is a pure decimal? What is a mixed decimal? a complex decimal? Give examples. How may every whole number be regarded? NUMERATION OF DECIMALS. 99 NUMERATION OF DECIMHALS. ART. 138. Since.6 - %;.06 6- T —,; and.006 _ a 0 any figure expresses tenths, hl udredths, or t oS:U? tahs, accordingo as it is in the 1st, 2d, or 3d decimal place hence, these places are named respectively the ternths', the Zcodleths', the thousaindths' place; other places are named i[ the same way, as seen in the TABLE OF DECIMAL ORDERS. 1st place.2.... read 2 tenths. 2d...08.8 Hundreths. 3d...0 0 5... 5Thousan ths. 4th...0 0 0 7... 7 Ten- thousa l n dths. 5th...0 0 0 0 3.... 3 Hundred-thousandths. 6th...0 0 0 0 0 1... 1 ilionth. 7th.. 0000009. 9. 9Ten-Millionths. 8th...00 000004. 4 Hundrecd-lmillionths. 9th...000000006.. 6 Billionths. The names of the decimal orders are derived from the names of the orders of whole numbers. The table may therefore be extended to Trillionths, Quadrillionths, &c. ART. 139. By inspectinc the table, and recollecting that 1 tenth = 10 hundredths, and 1 hundredth 10 thousandths, it is clear that the decimal places decrease in value from left to right, like the places of whole numbers, and according to the same law, viz: 1 in aCnyplace equals 10 in the next right-hand place. This law holds good in every mixed decimal, like 715.2309; for, in all such, the last ffiure of whole rumbers, (5), is units, and the first decimal figure on the righlt is t-t7hs, and 1 unit= 10 tenths. REVIEWV.-13S..What is the name of the lst decimal place? The 2d? Third? Why? Repeat the Table of Decimal Orders. 100 RAY'S HIGHIER ARITHMETIC. ART. 140. Since decimals are subject to the same law of local value as whole numbers, like them, also, they can be read in either of two ways. RULE FOR READING DECIMALS. 1st. IleadC in succession thle value of li]e separate figures which cosmpose the decimal; or, which is much more convenient, 2d. Read the decimal as a whole nuzmber, and annex the name of the right-hand place. The decimal.004038, is read, 4038 millionths. DE I. —The reason of the rule depends on the law of local value, viz: "1 in anzy place equals 10 in the next right-hand place." Commencing with the first significant figure, 4 of the 3d place equal 40 of the 4th place; 40 of the 4th place equal 400 of the 5th place, which, with the 3 already there, make 403 of the 5th place; finally, 403 of the 5th place equal 4030 of the 6th place, which, with the 8 already there, make in all 4038 of the 6th place; or, 4038 millionths, since the 6th place expresses millionths. A mixed decimal may be read altogether as a decimal; or, the whole number may be read, then the decimal. Thus, 71.062 may be read 71062 thousandths; or, 71 units, and 62 thousandths. EXAMPLES TO BE READ. 1..9 9. 00.100 17. 41.14414 2..0 10. 180.010 18. 41 1.4414 3..305 11. 20300.0 19. 4.114414 4..7200 12. 40.68031 20. 15.0046$ 5..5060 13. 207.2007 21. 73002.1 d 6. 1.008 14..0900001 22..000000k 7. 9.001 15. 61.010101 23..200006 8 105.07 16. 9230010.0 24. 526.000 25. 12.3333333 28..437800629 26. 643000.643 1 29. 1000000'. 0303 27. 4009.62007 30. 200000.000006 E vI E w.-1)9. IIow do decimal places resemble those of whole numbers? What is the law thalt governs both? Does this law apply to mixed decimals? Why? 140. In how many ways may decimals be read?. What is the 1st? The 2d? Prove the 2d rule. How may mixed decimals be read? Give an example. NOTATION OF DECIMALS. 101 NOTATION OF DECIMALS. AiT. 141. Writc, eighty-three thousand and one bil. lionths. DaEo-ST srsTaTIox.-The Ist step 83001 = Numerator. numnber of parts must be 2d step.000083001 = Decimal. written as a whole number, (Art. 1386); the right-hand figure must express parts of the given size, (See last Rule); hence, the RULE FOR WRITING DECIAIALS. IWrite the numerator; fix the point so t7Lat the right-fhand figzure shall be of the same name as the decimal. R E M1 A rI K s.-1. In fixing the point, it may be necessarD to prefix ciphers to the numerator, as in the example just given; but in the case of an improper decimal fraction, the point will fall between two of the figures; thus, 346 tenths is written 34.6, which may enlso be read 34 units and 6 tenths. 2. The operations under this and the preceding rule se'7e to prove each other. EXAMIPLES TO BE WRITTEN. 1. Five tenths, 11. Four hundred and twenty2. Twenty-two hundredths. one tenths. 3. One hundred and four thou- 12. Six thouland hundredths. sandths. 13. Eight units and a half a 4. Two units and one hun- hundredth. dredth. 14. Forty-eight thousand three 5. One thousand six hundred hundred and five thouand five ten-thousandths. sandths. 6. Eighty-seven hundred-thou- 15. Tlirty-three million ten sandths. millionths. 7. T1'enty-nine and a half ten- 16. Four hundred thousandths millionths. 17. Four hundred-thousandths. 8. Nineteen million and one 18. One unit and a half a bilbillionths. lionth. 9. Seventy tllousand and forty- 19. Sixty-six thousand and three two units and sixteen hun- millionths. dredths. 20. Sixty-six million and three 10. T'wo thousand units and thousandths. fifty-six and a third rnil- 21. Thilty-four and a third liornths. tenths. RE v I E, w.-141. What is the rule for writing decimals? Explain it. 102 RAY'S HIGHER ARITHMETIC. 22. Forty-foiur million units and 31. Three hundred ncld fiftyfour mlillionthss cighlt thousandc and six ten23. Two hundred rand eighteen thousantlr ad six billionths. 32. Two niillion millionths. 24-. Ninety-six tiousands. 33. Four -million units anm( four 25. Ninety-six hIundreds. miillionths. 26. Ninety-sis tens. 34. Four nlillion andcl four nlil27. Ninety-six units. lionths. 35. Fifty-thousand and seven 28. Ninety-sis tenths. hundrce-thlousandths. 29. Ninety-six hundredths. 36. Three million and a half 30. Ninety-six thousandths. billionths. ART. 142. Decimal Fractions are distinguished from conmmnon fractions, by not having written denominators; and from lwhole numbers, by the decimal point. The (Ienonmnation,, or size of the p)arts in any decimal, depends on the position of the poinat; so that the set of fifgures which expresses but one value as a whole number, may, as a deCimatl, express different values, according to the situation of the point. Great care miist be exercised, theC,, iin placiLng the1 poi-nt correctly aLnt distinLctly. ART. 143. PROPOSITION I.-Decimal cipehers may be annzzexed to, or omittcd fo om, the rigfht of any numbe-e, and not allter its value. D E r ON STRATIO. —The ciphers EXAMPLEs.. themselves are of no value; the.2 5.25 0 other figures retain their places 1 6 - 1 6. =1 ].0 0 0 and, consequently, their values. 1 9.0 S 3 0 0 1 9.0 8 3 Ilence, the vable of the number is not 2 0 0.0 0 = 2 0 0 altered, but merely its denoiinaetion. R uE:AnRr.K-Be careful that the ciphers annexed or omitted are decimnal ciphers. N o T. —Annexing or omitting ciphers is equivalent to multiplying or di)idi2g both terms of the decimal fraction by 10, 100, 1000, &c. ART. 144. PRoPOSITION II. —I iin any decimal, the Ipoint be. moi,ed to tMle RIGHT, the lnulnber is MULTIPLIED continually by 10 as of/el, as a fgcure is passed over. R si v t F w. -142. IIow are decimals. distinguished from common fractions? Ilow from whole numbers? Itow is the size of the pnrts in any decimal known? Why should great care be exercisod in placing the point correctly? 143. What is Proposition 1? Pxovoe it. REDUCTION OF DECIMALS. 103 DEtO N STPATION.-For each place passed over EXAMPLES. by the point in moving to thie right, every figure.0 5 6 7 is sdroanced one step in the scale of notation, and 0.5 6 7 is worith 10 times as much as before; hundreths 5.6 7 arc chlllged into tenths, tenths into units, units 5 G. 7 into tens, tens into hundreds, and so on; hence, 5 6 7. th.e proposition is true. 5 6 7 0. AIT 145. PPRoPosITION III. —If in any decimal, the point be vmoved to the LEFT, the number is DIVIDED conlinually by 10 as often as a figure is _passed over. D E i o i S T It l T x.-For each place passed EXAMIPLES over by the point in moving to the left, every 2 3 4 0. figure is degraded one step in the scale of nota- 2 3 4. tion, and is worth only -1 - as much as before; 2 3.4 hundreds are changed into tens, tens into units, 2.3 4 units into tenths, tenths into hundredths, and so.2 3 4 on; hence, the proposition is true.. 0 2 3 4 NOTE.-If the point is to be moved in either direction over more places than the number can furnish, supply the deficiency by taking in ciphers, as in the last examples of these propositions. RE rI A.-These 3 propositions apply to any whole number, supposing a point to stand on the right of its units' place. REDUCTION OF DECIMALS. ART. 146. CASE I.-To convert a decimal into its simplest equivalent common fraction. rULnE. —Take thle decimal as it stands, for thle lnuimerator, commencitg wvit/l thle first siglijicalt figlure; for the denominator, werite 1 with as nanny ciphers annexed as there are decimal places; then redulce this fraction to its lowest terms. NO T E.-In retlucing the fraction to its lowest terms, use no -iivisors but, 2 and 5; since they are the only prime factors of 10, and, therefore, the only prime number that will divide the denomriR E r E Wv.-144. State Proposition 2. Prove it. 145. State Proposition 3. Prove it. Ilow are deficient places supplied? HIow can these propositions apply to whole numbers? 146. What is the rule for reduce lug a. decimal to its simplest equivalent common fraction? 104 RAY'S HIGHER ARITHMETIC. nator, which is either 10, or the product of 10's. When the nu. merator ends in any odd number except 5, neither 2 nor 5 will divide it, (Art. 90), and the fraction will be in its lowest terms. Reduce.039375 to its equivalent common fraction. S o L U T I 0 N.-After obtaining the common fraction, divide 39375 blotl terms by 5, four times in 1000000 tuccesnion; the numerator then 7875 1575 ends iu 3, and the fraction can =) _ 5) be reduced no further. (Note.) 200000 0-40000 A mixed decimal like 18.0067 315 63 may be written as an improper 5) - As. fraction!-jo6-o 6o7; or, as a mixed number thus: 181 660764O By this rule,. a complex decimal is converted into a complex common fraction, which may be simplified as in Art. 132. T 0 1 25 5 1 Thus,.08- 3 - _ Ans. 3 100 300 60 12 The rule also serves to change a part of a pure decimal into a common fraction, thereby rendering the decimal complex, as, 6.87=87 =-6.871; in this way a decimal may sometimes be easily reduced to a common fraction, if the student is familiar with the aliquot parts of 100. Thus, ___ 575 3 937 15.039375=.03983-0315-'63=- since —-and -=-. 4100 1600' 100 4 100 16 REDUCE TO COMMON FRACTIONS. 1..25625.. Ans. 1'0I 9. 11.0. Alls. 11-'s 2..15234375 Ans. o 10..390625. Ars. 4 3. 2.125. As. 2 11.19444.. Ans. -3 4. 19.01750 Ats. 19y 12..24... Ans. 4a 5. 16.00-. Ants. 16&o- 13..33.. rs. As. 6. 350.0284 Ans. 35030S 14. 6 6.. AsS. ] 7..6666662.. Ans. 15..25 -A rts. 4 8.003125.Ans. 16..75. A i E v is w. —146. What divisors only need be used in reducing to the lowest terms? Why? IIow can it be known, by inspaction, th~at the fraction is in its lowest terms? Why? How may a mixed decimal be changed to a common fraction? What does a complex fraction become by the application of the rule? How can a pure decimal be rendered complex? Give examples. REDUCTION OF DECIMALS. 1 0 17. 16. Ans. 26..91. Ans. y1 18.833... As. - 27..6 o. Ans. 1 19..12 1 or.125 Ans. - 28..183.. Ans. n; 20..3) or.375 Ans. a 29..314.. Ans. 5, 21..62' or.625 Ans. 380..43... Ans. - 22..87 o.875 Ans. 31..56.. Ans. 9 23..08... Ans. - 32..68 3. Ans. 1} 24..41. Ans. 1~. 33..81 ns. 1. 25..58. Ans..T 34..93.. Ans. -1 As much work is sometimes saved, by using a common fraction instead of its equivalent decimal, the pupil should be familiar with this transformation. R EM X a.-If the decimal contain a great many places, a simple approximate value can sometimes be obtained by using the first one or two figures only; for example,.3260873 is nearly -3 3 - nearly; and 1.7689 -- 1 - nearly -- 14 nearly. CASE II.-TO CIIANGE ANY FRACTION INTO A DECIMAL. ART. 147. If the fraction has 10, 100, 1000, &c., for its denominator, it can be written as a decimal, by writing the ntzumerator, and placing the point so that the number of figures on, the right of it, shall be eqgual to the number of ciphers in the denominator. EXPRESS IN DECIMAL FORIM, 3 8 109 120056 52' 1600 T0' 100' 1000' 10000' 100000' 1000000' Ans..3,.08,.109, 12.0056,.00052q,.001600, If the denominator of the fraction, is not 10, 100, 1000, use this RuLE.-Divide the numerator by the denominator, annexing decimal ciphers to the former as they are needed; for each cipher annexed, make a decimal place in the quotient. N o T E s.-1. Before annexing ciphers to the numerator, be careful to place a point on its right. R EvIEw.-147. How can a fraction whose denominator is 10, 100, 1000, &c., be written as a decimal? If the denominator is not 10, 100, 1000, &c.? What should be done before annexing the ciphers? 106 RAY'S HIGHER ARITHMETIC. 2. The point may be fixed in the quotient at any time, by making the figure last written iz the quotient occupy tlhe samle decinmal place as the fiyure of the. dividend last used; in doing so, prefix ciphers to the quotient, if necessary, to make the requisite number of places. 3. The operations under this and the preceding rule serve to prove each other. Ireduce 27 to its equivalent decimal. OPEI-; rION. DE31iNSTRATTION.-The numerator -with 8) 7 000 the decimal ciphers annexed, is of the same value as before, but of a lower denomination,. 8 7 5 (Art. 143), and, when it is divided by the -.87 Ans. denominator, the quotient must be of that denomination also, and must therefore contain the same number of decimal places. The analysis of the operation is as follows: 8 - of 7 units = of 7000 thousandths = 875 thousandths —.875 REDUCE TO DECIMALS, 2.....495 3..... =.05 8. Q....078125 4. T.- _ S.46875 9. _. =.05078125 5. TO9..=.005625!10. - ol =.000,9765625 The rule converts a mixed number into a mixed decimal, and a complex into a pure decimal; thus, 9- 9.375, since 8- -.375; and.26.-3, -.2(512, since 3 =.12 11. 16... = 16.5 1 13..015. - 0155 12. 4-2-.. 42.1875 14. 101.01} 3 1O 01. 0175 5s. 7511........ 75119.0375 16. 200...=.2.00003125 ART. 148. Sometimes annexing ciphers does not render the numerator exactly divisible by the denominator; in that case, after the quotient has been carried out as far- as desirable, the sign (+-) plues is annexed to show that tlicie is still a remnainder. Such, h7avinl.g o end, are called RevIiw.-147. When may the point be fixed in the quotient? Ilovw? Explain the example. What effect does the rule have on a rmixed number? On a complex decimnal? ADDITIONr OF DECIMALS. 1 0 internt7h..ate or infinfmie decimals; thus, 2 =.666666 -+, an inic'iatnCtic; while -.1 25 a te7mi. tate decimal. Sometimlles, when the remainder omitted in an'ilter. micnute decimal is large enough to give thle next quotient figure more than 5, the last quotient figure is written 1 larger than it really is, and the sign (-)'minz1s is annexed instead of (+) pllus, to show that the quotient is written a little too large; as, = 66 -, or.671 27....259259+ 2. 104.. =.. 10. 604173..065 =.0652857+ 4. 430.18)-t- i = 430.1809-41ADDITION OF DECIMALS. ART. 149. R149 LE. -TWrite the nnumbers to be added so tiaut Jf7lg.res o/' the sam7le dCenol0 iation. m?n! be in C0oh711.nns' C (I.s i whole ntlmbers, commencing at the?right, and point off the resull to agr'ee willh that one of the given nn1nlers. hatinlg the 9?ost decinmal llaces. P R o o F.-Same as in addition of whole numbers. NoTE. —Complex decimals, if tlhere are any, must be made pure, (Art. 147), as far, at least, as the. decimal places extend in the other numbers. If, after* this, there e common fractions in the righthand column, add them; or, neglect them, using the signs + or - as in Art. 1.48. Add 23.8 and 171 and.0256 and.41: SOLUTION. —After reducing, set OPERATION. figures of the same order in column; 2 3.8 then add -and carry as in whole num- 7 17.5 bers. As the right-hand figures, when 5 6 added, must make the rigllht-hland firgure of the ansrver of the same de-. nomination as those in the column, Anls. 41. 7 42 3 the point should be fixed as the rule directs. REVIEZr.-148. When the quotient can not be made exact, what is done? W'That are such decimals called? Why? When may the sign milzus be used? 108 RAY'S HIGHER ARITHMETIC. 1. Find the sum of 1 +.9475 Ans. 1.9475 2. Of 1.331 added to itself twice. Ans. 4. 3. Of 14.034, 25,.000062t,.0034 Ans. 39.0374625 4. Of 83 thousandths, 2101 hundredths, 25 tentlhs aid 94: units. Ans. 118.093 5. Of.1, 37, 5, 3.4a,.0008 Ans. 8.980j 6. Of 4 units, 4 tenths, 4 hundredths. Ans. 4.44 7. Of.11 +.66661- +.222222 Ans 1. 8. Of.14:,.0183, 920,.01392 Ans. 920.1754 5f 1~.08 1 9. Of 16.008 007~0074,.2,.00019042k Ans. 16.299768199' 10. Of.675, 2 millionths, 64~, and 3.49000107 Ans. 68.29000307 11. Of four times 4.0677 and.000. Anzs. 16.272 12. Of 216.86301, 48.1057,.029, 1.3, 1000. Ans. 1266.29771 13. Add 35 units, 35 tenths, 35 hundredths, 35 thousandths. Aims. 38.885 14. Add ten thousand and one millionths; four hundred-thousandths; 96 hundredths; forty-seven million sixty thousand and eight billionths. Ans. 1.017101008 SUBTRACTION OF DECIMALS. ART. 150. RULE-W7rite tze subtrahend uzder the minuend, placing.figures of the same denominlation in columns. Subtract as in whole numnbers, commencing at the r'ight, and point the result as in addition of decimals. Pnoo. —As in subtraction of whole numbers. NrTES.-If either or both of the given decimals be complex" proceed as directed in the note to last rule. REVIEvw.-149. What is the rule for adding decimals? The proof? What must be done with complex decimals? If there are common fractions in the right-hand column, what should be done? Explain the exsample. 150. What is the rule for subtraction of decimals? The proof? SUBTRACTION OF DECIMALS. 109 2. If the minuend has not as many decimal places as the subtra. hend, annex decimal ciphers to it, or suppose them to be annexed until the deficiency is -supplied. From 6.8 subtract 2.057 SO L UTON.-Write the numbers as the rule 6. directs; suppose ciphers to be annexed to the 8, 2.0 5 7 and subtract as in whole numbers; saying 7 from 10 leaves 3, carry 1; 6 from 10 leaves 4, and so on. A its. 4.7 4 3 From 13.256] subtract 6.773 In this example, the complex decimals 1 3.2 5 6 3 are rendered pure to,the same extent, and 6. 7 7 a _ 6.7 7 3 the common fractions subtracted. Ans. 6.4 8 3 EXAMPLES FOR PRACTICE. 1. Subtract 8.00717 from 19.54 Ans. 11.53283 2. 3 thousandths from 3000. Ats. 2999.997 3. 72.0001 from 72.01 Anzs..0099 4. Subtract.93' from 1.169a7 Anrs..238 5. How much is 19 less 8.9999? Ants. 10.000'9 6. How much less is.04.' than.4? Airs..35' 7. How much is.65007 --? A)s..15007 8. What is 2-3-1j4 in decimals? Ars..95 9. Take.007601 twice from.02 A.Ans..004798 10. Take 1.98~ three times from 6. Atns..05 11. Subtract 1 from 1.684 Ans..684 12.! of a millionth from.0004 Ans..000443UT 13. 1- hundredths from 49H tenths. Aii.s. 4.9225 14. 10000 thousandths from 10 units. A.7.ts. 0 15. 24V tenths from 3701 thousandths. Arts. 1.251 16. 1V units from 1875 thousandths. Als. 0 17, 5 of a hundreth from Yg of a tenth. A-ls. 0 18, 64L hundredths from 100 units. Ans. 99.35' RIt E VrTE.-150. What must be done with complex decimals? If the minuend do not contain as many decimal places as the subtrahend, whal must be done? Explain the examples. 110 RAY'S HIGIIER ARITHMETIC. )MULTIPLICATION OF DECIMALS. ART. 151. Ruir,. — lftltipl/ as in whole numbers, and point t/te product, so htCat it s/hall hiave as many decimal l(ices as lh/e nmtitltplicatld and zulltiplier togethelr. ITtoo F.-As in multiplication of wvhole numbers. l sMrARn. —If the product has not as many decimal places as reulnired, supply the deficiency by prefixing ciphers. Mlultiply 2.56 by.184 DE IoNST RATIO N. —Express 256 184 471 04 the decimals as common frac- X i00 1000 100000 tions; their product will be the product of their numerators di- Hence, vidled by the product of their 2.56 X.1A84 -.4 7104 denominators, (Art. 130). Write the fractions in decimal form; the denominator of the product has as many ciphers as both the other denominators: andl as each of these ciphers make a decimal place, (Art. 136), the product will have as many decimal places as both factors. EXAMPLES FOR PRACTICE. 1..X.1........... 2. 16 x.03}.533 3..01 X.1. =.0015 4..080 X 80.......=6.4 5. 37.5 X 8.;2. =3093.75 6. 64.01 x.32. o. = 0.4 832 7. 48000. X 73....... = 3504000. 8. 64.66 X 18. = 1164. 9..561 X.03j.. 017o'2 1 10. 738 X 120.4..... 88855.2 11..0001 X 1.006.....00010010 12. 34 units X.193, 5.. 6 562 13. 27 tenths X.41..... 1.134 14. 43."7004 X.008..3496032 15. 21.0375 X 4.44.... 93.5 tEV:Ew. —151. VWhat is the rule for Multiplication of Decimals? Ths proof? How are deficient places in the product to be supplied? Prove the rule. MULTIPLICATION OF DECIMALS. 1 16. 9300.701 X 251... 2334475.951 17. 430. 0126 x 4000. =17 200 50.4 18..059 x.059 x.059..000205379 19. 42 units X 42 tenths..... 176, 4 2()0. 221 hundredths X 600. 3...... 21. 7100 X 8 of a millionth....0008875 22. 26 millions X 26 millionths.... G i, 23. 2700 hundredths X 60 tenths.. 162. 24. What denomination will a figure in the tenths' place, multiplied by a figure in the'hundredths' place, give? A.s..1 X.01 =.001 or thousandths. 25. A figure in the units' place, by one in the hundred-thousandclths' place? Ans. Hundred-thousandths. 26. 1 thousandth by 1 thousandth? Ans. Millionths. 27. 1 hundred by 1 tenth? Ans. Tens. 28. 1 in thousands' by I in tenths'? Ans. Hundreds. CONTRACTED MULTIPLICATION OF DECIMALS. ART. 152. The methods of contracting multiplication in iwhole numbers, apply also to multiplication of decimals. It is only necessary to call attention to two cases. CASE I.-TO BIULTIPLY A DECIMIAL BY 10, 100, 1000, &c. RULE. — Ree ove the decinal point of the multiplicand to the righl, over as mc1ay places as there are ciphers in tlhe mzultplier; the result will be the produlct reqtired. NoTEs.-1. If there are not as many places to the right of the point as are required in the operation, supply the deficiency by taking in ciphers, unless the multiplicand be a complex decimal; in tlhat case, the proper figures should be ascertained and set down, (Art. 147). 2. The rule given in Art. 564, for multiplying a whole number by 10, 100, 1000, &c., is a particular case of this, since a whole number may be regarded as a decimal, the point being on the right of the units' place, (Art. 137). E vIe nw. —152. W'hat methods of Contracted Multiplieation are used for decimals? What is the 1st Case? Give the rule. Ilow are deficient places supplied? I-low does this rule apply to whole r umbers? 112 RAY'S HIGHER ARITIM!IETIC. DEMONSTRATION.-To multiply by 10, 100, 1000, h&c., is the same as to multiply continually by 10; this is done by moving the point to the right, over as many places as there are ciphers in the multiplier, (Art. 144). EXAMPLES FOR PRACTICE. 1. 56. x 100........ — 5600. 2..075 x 100... =7.5 3..014 X 1000....17.5 4. 16.083 X 10...... 160.83 5. 10.34' x 100000.. = 1034400. 6. 98.047? x 1000000.. =98041783331 ART. 153. Whenever the product of two decimals is not required to contain figures below a certain order, the work may be shortened. CASE II.-TO MULTIPLY, RESERVING A CERTAIN NUMBER OF DECIMAL PLACES IN THE PRODUCT. RULE.-Count off in the multiplicand, the number of decimal places to be reserved, draw a vertical line through the lowest, and write the multiplier so that its units' figure shall fiall ont this line. Begin at the left of the muzlliplier to form the partial produlcts, always starting at that figure of the multiplicalnd which is as far on one side of the line, as the figure of the multiplier then in use is on the other side, carrying the tens, however, obtained by multiplying the next lower figlure. Set the right-hand Jigures of' these partial products in a column, add and point off the numLber of decimal places requeired. NOTES.-1. The multiplicand should extend one figure further to the right of the line than the multiplier does to the left of it. To accomplish this, it may be necessary to annex decimal ciphers, or to convert a common fraction into a decimal. 2. If the multiplier contain a common fraction, and it becomes necessary to multiply by it, start at the same figure of the multiplicand as in the previous multiplication. REMA &R..-In carrying tens from the nearest rejected figure of the mlnltiplicand, carry 1 ten also for any number of units over 5; thus, f1r 55, carry 6; for 18, carry 2. RIEVIE W.-152. Demonstrate it. 153. What is Case 2? The ru.o How tar should the multiplicand extend to the right of the line? MULTIPLICATION OF DECIMALS. 113 Multiply 1.89361 by 3.5672, reserving 3 decimals in the product. So L U T I 0 N.Count off SHORT METHOD. ORDINARY METHOD. 3 decimals in the multi- 1.8 9 3 61 1.8 9 3 61 plicand, draw a vertical 3.5 6 7 2 3.5 6 7 2 line through the lowest 5 6 8 1 3 7 8 7 2 2 (3), and proceed thus: 9 4 7 13 2 5 5 2 Commencing with 3, 113 1 3 1 6 6 the left-hand figure of 1 3 9 4 6 8 0 5 the multiplier, and 3 in 65 6 8 08 38 the multiplicand, which 6 5 8 are both on the line, say, 6 5 4 8 8 5 5 9 2 3 timles 3 are 9; but 3 times 6 (the next lower figure) are 18, which is nearer 2 tens than 1 ten; carrying these 2 tens to 9 we have 11; set down 1 and carry 1. Keeping to the left, the first partial product is 5681. Next, take 5 in the multiplier, and start at 9 in the multiplicand, carrying 2 for the 15 obtained by multiplying the next lower figure, 3. The second partial product is 947, whose first figure is set under the first figure (1) of the previous product. Thus go on, using 6 and 7 in the multiplier successively, and starting at 8 and 1 in the multiplicand. As the figures of the multiplicand toward the left are then exhausted, the operation, after adding and pointing off 3 figures, is complete. DEAToNsTRATION.-The reason of the rule consists in the fact, ~that all the multiplications which would give rise to denominations lower than those required in the product, are omitted. The rule will not always give the la'n figure correct. To insure perfect accuracy, reserve one more figure than tz required, and omit the.last figure of the product. What is the product of.054363-7r by 6458.19 true to 3 decimal places? So LU'Io N.-Reserve 4 decimals, one more than is required; carry out the figures of the multiplicand, (Note 1): point off 4 places in the product, 351.0906, writing the next to the last, 1, instead of 0, because the figure omitted is over 5. This gives 351.091 fr the answer. R E v I E w.-153. How are deficient places supplied? What, if the multiplier contains a common fraction? In carrying tens from the rejected figures, what directions must be observed? Solve the example. What denomination in the product, is obtained by multiplying the marked figure of the multiplicalnd by the units' figure of the multiplier? Ana. The lowest. 10 114 RAY'S HIGH-ER ARITHMETIC. EXAM3PLES. DecimaZs reserved. ANSWFRS 1. 4.7501'X.002866 5.01361 2. 804.57] X 17.08132 4 13743.2315 3. 75.062 X 460.8917 2 34595.45 NOTE.-Take 75.062 for the multiplier. 4. 9.012 X 48.75 1 439.3 5, 4.804136 X.010759 6.051688 6. 8147 3 X 2263 3 21813.475 7. 702.61 x 1.258-7 3 884. 020 8. 849.93a X.0424444 3 36.075 9. 880.695 X 131.72 true to units. 116005. 10..025381 x.004907 5.00012 11. 64.01082 X.03537' 6 2.264063 12. 7.24651 X 81.4632 3 590.324 13..681472 x.01286 5.00876 14. 7.9444 x 3.69 4 29.3150 15..053497 X.047126 6.002521 16. 1380.371 X.2345 2 324.16 DIVISION OF DECI3MALS. ART. 154. RusLe.-Mlalke the decimal places of the dividend as many as those of the divisor, if they are less. Divide as in whole numbers, annexing other decimal Jfigures to the dividend as they are needed; point the quotient so that it shall have as many decimal places as the dividend has MORE than the divisor. P RooF.-Same as in Division of whole nunbcrs. NOTES.-1. To extend the dividend, decimal ciphers are used; but, if the dividend is a complex decimal, make it, pure. 2. If the quotient has not the number of figures required for decimal places, prefix ciphers. 3. If the dividend has the same number of decimal places as the i vi so, the quotient is units. i E V I E w. —153.'What, by multiplying each figure of the multiplier by the figure of the multiplicand as far on the other side of the line? Anls. The same. IWhat multiplications are omitted? 154. What is the rule for division of decimals? The proof? How are deficient places in the dividend supplied? How in the quotient? DIVISION OF DECIMALS. 115 4.'Make complex decimals pure; or, divide them like common mixed numbers; or, multiply both by the least common multiple of the denominators of the common fractions, and then divide. 5. If the division is not exact, it may be continued by annexing other decimal figures to the dividend; or the remainder may be written with the divisor under it, as a common fraction. Divide.50312 by.19 D E 0 T AT I 0 N.-Since the di-.19).5 0 312 (2. 648 vidend is the product of the divisor 1 2 3 AlAs. and quotient, it must have as many 9 1 decimal places as both of them, (Art. 1 5 2 151); hence, the quotient must have 0 decimal places enough to make with those of the divisor as many as are in the dividend, which is just as many as the dividend has more than the divisor. Divide 24 by 3.2 Annex a decimal cipher to the dividend, 3.2) 2 4.0 0 (7.5 to make it. have as many decimal places as 1.6 0 the divisor; afterward, annex another, to 0 contin e the division. Divide.07 by 21.6 In long division, annex the 2 1.6).0 00(. 0 0 3 2 4 0 7+ ciphers to the remainders as 5 2 they occur, reckoning them 880 still as decimal places of the 16 0 0 dividend; here the dividend 8 8 has eight decimal places, counting the ciphers annexed to the remainders, 1. Divide.0029 by.06 3 Ans..0375 S u c E S T I O.- Convert the numbers into the pure decimals.002475 and.066; or, multiply both by 40, the least common multiple of the denominators 5 and 40, making the dividend.099, and the divisor 2.64: use the latter method generally. EXAMPLES FOR PRACTICE. 2. 3.18........... 16. 3. 4.2 ~.31. =13.44 REE Tv w.-154. If the dividend and divisor have the same number of decimal places, what is the quotient? IHow may complex decimals be divided? If the division is not exact, what may be done? Demonstrate the rule. Explain the examples. 116 RAY'S HIGHER ARITHMETIC. 4. 63 - 4000..01575 5. 3.15 375.... =.0084 6. 1.008 18......056 7. 4096 ~.032.... 128000. 8. 9.7 - 97000.....0001 9..9 -.00075 1. -'12000 10. 13 —78.12..... =.1664 11. 12.9 8.256.. =1.5625 12. 81.2096 ~1.28..... 63.445 13. 12755 81632.... =.15625 14. 2401 21.4375...112. 15. 21.13212 -.916..... 23.07 16. 36.72672-.5025... - =73.088 17. 2483.25 5.15625..... 481.6 18. 142.0281-9.2376..... — 15.375 19. 1 - 100... 01 20. 10.117..... =.5941221..001~1 00...... o01 22..08 ~.12-...... 23..0001.01..... - 01 24. 95.3 -.264. - 360.984848-+ 25. 1000 ~.001..... 1000000. 26. Ten -- 1 tenth..... = 100. 27..000001 ~.01......0001 28..00001 -1000.... -.00000001 29. 16.275~.41664.....= 39.0625 30. 1 ten-millionth. 1 hundreth..00001 ART. 155. If the dividend is less than the divisor, the quotient may be expressed as a common fraction, by taking the dividend for the numerator, the divisor for the denominator, making both terms have the same number of decimal places, and then omitting the points, which is equivalent to multiplying them by the same number, (Art. 144); thus, 3.737 divided by 99.9, equals 37-'I7 - 3.737 - 3737 - 101 REVIEV. —155. If the dividend be less than the divisor, how may the quotient be writ-ten? DIVISION OF DECIMALS.' 117 1..75 2. — 125 1 3. 13' 4 4 2. 1 5.5 1 4..041 e.15 ART. 156. To find the denomination of the 1st quo. tient figure when obtained, count oq it the dividcn/,,, s many decimal places as are in the ditisor, placing a dA4l after the last; numerate from this dot as a deciqltl po:,/. either way, up to that figure of the dividend, under whiLc/ the right-hand figure of the 1st product falls. Th is wilt give the denomination of the 1st quotient figure. Thus, in dividing.07 by 21.6, (page 115), mark the dividend,.0'700; numerate from the dot as a decimal point to the right as far as the 0, under which the right-hand figure of the first product falls: this gives thousandths, which is the denomination of the 1st quotient figure, 3. CONTRACTED DIVISION OF DECIMALS. ART. 157. The methods of contracting division in whole numbers apply also to decimals. Only three cases need be noticed. CASE I.-TO DIVIDE A DECIMAL BY 10, 100, 1000, &c. RuLE.-Remove the decimal point of' the dividend to the left, over as many places as there are ciphers in the divisor; the result will be the required quotient. NOTE.-If there are not as many places to the left as are required, prefix ciphers. RErAR K. —The rule in Art. 67, for dividing a whole number by 10, 100, 1000, &c., is a particular case of this, since a whole number may be considered a decimal with a point on its right. (Art. 137.) Divide 68.075 by 10000. Ans..0068075 DE M N STR AT IO N.-To divide by 10, 100, 1000, &c., is the same as to divide continually by 10. This is done by moving the point to the left, over as many places as there are ciphers in the divisor. (Art, 145.) R E V IE W.-155. Before reducing the common fraction, what must be done? Why? 156. How is the order of any quotient figure determined as soon as it is set down? 157. What methods of contraction are used in division of decimals? What is the tst case? The rule? How are deficient plar.ces supplied? Why can the rule be applied to whole numbers? Prove thebo rule. 118 RAY'S HIIGIIER ARITHMETIC. 1. 65 ~ 1000. o =.065 2..072 10.. —.00723 3. 1 -- 1000000... =.000001 4. 2001.2 100.. 20.P.012 5. 93000 1000.. = 93.000 =- 93. 6; 4.472 —10000.. =.o.0004772 ART. 158. It often happens that the quotient is not rcquired to contain decimal figures below a certain denomination; if so, the work may be shortened. CASE II.-TO DIVIDE, RESERVING A CERTAIN NUMBER OF DECIMALS IN THE QUOTIENT. RuniE. —Mfark that figure of the dividend, whose denomin7ation would resultt froim mulltiplying a unit of the highest denomination in the divisor, by a unit of the -lowest denominalion required in the qzotient. Divide as usual, until this figqure is reached; then, stop bringing down from the dividend, and at each subsequent division, drop a,figure from, tlhe divi.sor, carrying for its tens. Continue thus, utntil the divisor is reduced to a single figure, and thenz point off the q uotient as required. NOTE.-If the marked figure of the dividend is reached at the first multiplication, mark the figure of' the divisor, whose product by the first quotient figure falls under the marked figure of the dividend; and at the next step, reject this figure with those on its right. REIMARKS. I. If the dividend has no figure of the denomination to be marked, annex ciphers to it, or continue it further, if it is an interminate decimal, until it does. 2. In carrying tens from the rejected figures, obs:rve the directions in Case 2 of contracted multiplication of decimals. (Art. 153, Rem.) Divide 1.078543 by 319.562 true to 5 decimal figures. SIIORT METIOD. ORDnNARY METIJOD. 65,p.A,)l.o0T7 @43(.00337 3190.562)1.078543(.00337 120 11918570 96 95 8686 24 23 98840 22 22 36934 2 1161906 I/IVISIfON iF It)ECIMAI1,S. 119 SOLvUTION.-The highest denomination in the divisor is hundreds, the lowest required in the quotient is hundred-thousandths;;and 100 X.00001 -.001; therefore, mark the thousandths' figure (8) of the dividend. As this marked figure is reached at the first multipliclation mark 9 in the multiplier, since it gives the figure that fialls under 8 in the dividend. Cross the rejected figures of dividend and divisor; cross one wmIre from the divisor at every new multiplication, carrying tens as directed: 337, with the necessary ciphers and point prefixed, is the quotient required. DE 3 ONST RATION.-Tho reason of the rule consists in the fact, that all operations, which would involve denominations lower than thlose required in the quotient, are omitted. EXAMPLES. Decizmals reserved. ANSWEERs. 1. 1 — 2.6783 3.373 2. 10.008371.056248 2 177.85 3..187564-.00043129 true to units. 435. 4..007516362. 652.18 8.00001152 5. 1000.86. 3.1415926 7 318.5836381 6. 61.0598314. 4278 6.014273 7. 421.331 - 9.1043 4 46.2778 8. 100 3.7320508 5 26.79492 9. 7912.5043~ 181.34 1 43.6 10..91 - 216.52 10.0042233717 11. 555 — 123456789 10.0000044955 12. 1 - 111111 6.000009 CASE III; —TO DIVIDE BY A DECIMAL LITTLE LESS THIAN I, RESERVING DECIMALS IN THE QUOTIENT. ART. 159. RULE.-LeUltiply the dividend by what the divisor wants of being a unit; multiply this product in like manner, and continue so u'ntil the product becomes too small to affect the result as required; thern add to obtain th7e qtotient. Divide 3815.64 by.994, reserving 2 decimals in the quotient. RrEvvirFw.-158. What is Case 2? The rule? Why should no de. nominations of the dividend be used below the one marked? Ans. Be. cause, when divided by the divisor, they give lower denominations than are required in the quotient. 120 RAY'S HIGHER ARITHMETIC. SOLUTION —Multiply 3815.64 by.006, the 3 81 5.6 4 difference between.994 and a unit; write the 2 2.8 9 4 product, 22.894, neglecting all denominations.1 3 7 below thousandths. Do the same to this pro- 1 duct, and to the next. The rest of the products are too small to affect the answer as re- 3 7 quired; therefore, add and obtain the quotient, 8338.67, true to 2 decimals. D E M0 N ST RAT I O N.-The operation and demonstration are similar to those in Art. 69, for the corresponding case of whole numbers, except that no remainders occur, the product being extended in decimals as far as is necessary to obtain the correct answer. EXAMPLES. Decimals reserved. ANSWERS. 1. 1000~.98 2 1020.41 2. 6215.75 --.99. 3 6246.985 3. 28012-.993 2 28209.47 4. 52546.35-.991 3 52678. 045 5. 4840 v.9875 2 4901.27 X. CIRCULATING DECIM~ALS. ART. 160. Many common fractions, when transformed, become interminate decimals. (Art. 148.) These have some curious and useful properties worth considering. PROPOSITION I. The only common fractions which can be changed into termilzate decimnals, are those which, reduced to their lowest terms, have no factors but 2 and 5 in their denominators. D) E M O N S T R A T I O N. -The numerator must EXAMPLES. contain all the prime factors of the denomi- 3 -.009375 nator to be divisible by it. Every cipher 320 annexed to the numerator multiplies it by - 3125 10, introducing 2 and 5, the factors of 10, 16 as factors of the numerator. If the denomi- 2 —.66666+ nator has no factors but 2's and 5's, enough ciphers may be annexed to the numerator to =.72727+ give it as many 2's and 5's for factors as the denominator, and then the quotient is exact. But if the denominator have any factor besides 2 and 5, this factor never can be introduced into the numerator by annexing ciphers, for 2 and 5 are the only factors that can be so introduced. In such cases, the exact division is not possible. CIRCULATINUG DECIMALS. 121 Tell whether the following common fractions can be changed into terminate or interminate decimals: 2 3 9 11 95 105 21 41 561 9 8' 32' 12' 152' 112' 30' 288' 87~ PIOPOSITION II..kRT. 161. Every interminate decimal arising from the rdnsfromzzation of a comm.zon fraction will be found, if the diitision 5be carrLed far enozigh, to contain the same figure, or. set of figures, repeated in the same order withoutl end. DEMONSTRATION. —In converting into a decimal, the remainders EXAIIPLES. are successively 3, 2, 6, 4, 5, 1, and 7' 42857142851+ as these are all the whole numbers less than 7, the next remainder must.5 555 _ be one of these repeated; on trial it 9 is found to be 3, to which if a 0 be 2 added, we have the same dividend 27' 74 and divisor as once before, and therefore must have the same quotient figure and remainder; and this remainder, with a cipher annexed, will yield another quotient figure and remainder, like a previous one, and so on continually. ART. 162. These interminate decimals on this account have received the name of circulhtt'i.l or recurring decimals; also, repeatilg or perio(lical decinmal(s. The figure, or set of figures, which is constantly repeated, is called the rpetel,, which signifies to be repeate. Such decimals are expressed by placing a dot over the first and last figures of the first repetend, and omitting those which follow; thus, =.6666 + =. 6 and I,.i 42857 =.142857142857+ ART. 163. A circllate or circulating decimal has one or more figures constantly repeated in the same order. A repeteed is the figure or set of figures repeated. REvIEw.-158. If the marked figure of the dividend is reached after tho first multiplication, what should be done? If the dividend has not the proper place to be marked, what should be done? What directions are to be observed in carrying tens from the rejected figures? Solve the e-:ample. Prove the rule. 159. What is Case 3? The rule? Illus. trate and pro ve it. 11 122 RAY'S H1 GHER ARITH5IETIC. A p2ll'e ciSCculate has no figures but the repetend; as,.5 and o124 A m/iXec cUir:cuate has other figures before the repetend' as,.20803 and.31247 A simple repetend has one figure; as,.4 *A <.olfncl.nd repecIend has two or more figures; as,.5~3 Siintiltr repetemrls begin at the same place; as,.356231 and.0178, which both begin at thousandths. Dissimilar repetends begin at different places; as,.205 and.312468 Similar and conterminotzs repetends begin a1nd end at the same places; as,.50;397 and.42618 ART. 164. Any terminate decimal may be conlsidered a circulate, its repetend being ciphers; as,.5 =.50.350000. Any simple repetend may be made compound, and any compound repetend still more compound, by takincg in one or more of the succeedincg repetends; as,'.3:333, and.0562 -.05626l2, and.25 7=.25725725 7 TWhen a repetend is thus enlarged. be careful to take in no plart of a repetend without takling thle whole of it; thus, if we take in 2 figures in the last examnple, the result,.25725, would be incorrect, for the next fiiiare understood being 7, shows that 25725 is not repeaeted. A repetend may be made to begin at any loweer pIlace by carrying its dots forward, each the same distance; thus,. =.555, and.2941=.29414, and 5.1836=5.1 836(35 Dissimilar repetends can be made similar, by carrying the clots forward till they all begin at the same place, as the one furthest from the decimal point,. Similar repetends may be made conterminous by enlarging the repetends until they all contain the same number of figures. This number will be the least common multiple of' th~ numbers of figures in the given repetends. For, suppoe one of the repetends to have 2, another, 8, another, 4, and the last, 6 figures; in enlarging the first, figures must be taken in, 2 at a time; and in the others, 3, 4, and 6 at a time. The number of figures which may be taken in 2, 3, 4, and 6 at a time, is a common mul REDUCTION OF CIRCULATES. 1'2 tiple of these numbers, and the least common multiple is preferred for convenience. REDUCTION OF CIRCULATES. CiASE I.-TO REDUCE A PURE CIRCULATE TO A COMMOuN FRACTION. ART. 165. RULE. — V'Tite the repetend for the numerator, and.fbr the denom.inator take as miany 9's as there are figures in the repetenLd. DEMO NSTRATION.-Take the pure circul'te.56 —.456456456, &c., and removeithe decimal point to the right over one repetend; the result 456.456456, &c. = 456.456 is 1000 times. 4,55, (Art. 144); hence, the part 456 = 999 times.456; and.456 wh 4 - s wich agrees with thle rule. NOTE.-If the repetend begins before the decimal point at some place of whole numbers, carry the dots forward until the repetend begins at the tenths' place, and then apply the rule; thus, 25.6 = 25.6is2 = 25. CASE II. —TO REDUCE A MIXED CIRCULATE TO A COIMMON FRACTION. ART. 166. RULE.-Sub/tract the figures which precede the repeteind fromn the whole circulate for the numerator; for t/he denolinlator take as many 9's as there are figlures in7 the repelend, with as manly ctphers ainnexed as there are decimal figyures before the repetend. Change.8214337 to its equivalent common fraction. OPERATION. DE nr. —The work, 3.81437. 1 by the rule in Case 1, results in 82143w7 821 x 999+437 821437-821 1 00 999000 99901000 999000 99,900t 821(1000 -1) + 437 for the answer, and __ — this is what would 999000 be got, at once, if the 821000- 821 + 437 821437- 821 rule last given were -- applied; and so with 999000 999000 all other mixed cir- 820616 102577 culates. S.- -. 999000 124875 i 124 RAY'S HIGHER ARItTHMETIC. REDUCE TO COMMON FRACTIONS, 1..3... =1 8. 1.001 1o.. 2..0~ 59. 05 o o — 1. 9 — 3. 3..123.. = 10. 2083. 4 2.63 2, -21 11. 85.7142 = 857 5..31i 4 12. 063492, 6..0216.. 13..4476190 = 7. 48.i. =48W 14..09027 143 ART. 16 7. Circulates may be added, subtracted, multi plied, or divided, by this GENERAL RULE FOR CIRCULATES. Reduce the circulates to common fi'actions, andc perform on them the opeiration required. R E I A R. —Circulates may be carried forward far enough to avoid any sensible error in the result, and then treated as other decimals. They can be added, subtracted, multiplied, and divided, without this preparation; as will now be explained. ADDITION OF CIRCULATES. ART. 168. RULE.-Make the repetends sinzilar anzd contermiznous, if' they be not so; add, and.point og as in ordinary decimals, increasilng the righlt-han(l coltumn by the amount if Ianl/,,which would be carried to it if the circulates were conltiluted;,then ma7ke a repetend in the sum, similar and conlterminzos wc1ith those above. Add.256, 5.3472, 24.815, and.9098 DE I o 4 s T' A T I O N.-AlMake the cir-.2566666666 culates similar and conterminous, as 5.3 4 7 2 7 directed in Art. 164. The first column of figures which would appear, if the 2 4.8 1 5 8 1 5 81 5 circulates were continued, are the same.9 0 9 8 O 0 0 0 0 0 as the firlst figures of the r epetends, 6, 7, 1, 0, whose sum, 14, gives 1 to be 1.3 95552097 carried to the right hand column. Since the last six figures in each number is a repetend, the last six figures of the sum is also a repetend. SUBITTR ACTION OF CIRCULATES. 12b RiEarARK. —TIn finding the amount to be carried to the rigbt hand column, it may be necessary, sometimes, to use the two sumceedling figures in each repetend. 1. Add.453,.068,.327,.946 Ans. 1.790 2. Add 3.04, 6.45(;, 23.38,.248 Ans. 33.133J 3. Add.25,.104,.61, and.5635 An s. 1.536 4. Add 1.0o,.257, 5.)4, 28.0445245 Ans. 34.37~ 5. Add.6,.138,.05,.0972,.0416 Ans. 1. 6. Add 9.21i07,.65, 5.004, 3.5622 Ans. 18.43 7. Add.2045,,., and.25 Ans..54 8. Add 5.07707,.24, and 7.i24943 Aiis. 12.4 9. Add 3.4884, 1.6l3, 130.81i,.066 As. 136.00 SUBTRACTION OF CIRCULATES. ART. 169. RuiE.-Make the repetends siimilar and conteTminous, if they be not so; subtract and point oft as it ordinary decimals, carryilg one, however, to the right-lhanld figure of the subtrahend, if on continuilg the circulates it be fmould necessary; then mnake a repetend in the remainder, similar and conlterminous with those above. Subtract 9.3156 firom 12.902i DEzION STRATION.-Prepare the num- 1 2.9 0 ~ 1 12 12 bers for subtraction. If the circulates were 9.3 1 5 6 1 5 6 i continued, the next figure in the subtrahend 3. 5 8 0 5 (5) would be larger than the one above it (2); therefore, carry 1 to the right-hand figure of the subtrahend. RE A R K.-It may be necessary to observe more than one of the succeeding figures in the cirQulates, to ascertain whether 1 is to be carried to the right-hand figure of the subtrahend or not. 1. Subtract.0074 from.26......259 2. Subtract 9.09 from 15.35465... 6.25 3. Subtract 4.51 from 18.230673... 13. 7 4. Subtract 37.0128 from 100.73. = 63.7i LRAY'S HIG HER ARITHMETIC. 5. Subtract 8.27 from 10.0563.. = 1.7936290 6. Subtract 190.476 from 199.6128571 = 9.16 MULTIPLICATION OF CIRCULATES. ART. 170. RULE.-If only one of the numbers be a circulate, malce it the multiplicand, and peiform the work as in ordinary decimals, carrying lo the right hand figure of each pirodtuct, the amoItut that would be necessary if the multiplicand were continued ]Jtrther; make the repetelnds in the partial products similar and contermzinous, and add according to the rule already given. If the mllltiplier have a repetend, reduce it to a commonz fraction, and add in the result obtained by using this fjaction. kMultiply.3754 by 17.43 SoLu T roN.-In forming the par-.3754 tial products, carry to the right- 17 4 =17.4, hand figures of each respectively, the numbers 1, 3, 0, arising from the.1 5 0 1 7 7 multiplicat;ion of the figures that do 2. 6 2 8 1 i not appear. The reTpetend of the 3 54, 4 4 multiplier being equal to ), I of the multiplicand is 125148, whose figures 125 148 are set down under those of the inul- 6.5 4 5 4 8 i Ans tiplicand from which they were obtained. Point the several products, carry them forward, until their repetends are similar and conterminous, and add for answer. 1. 4.75 x 7.349.... =34.800113 2..o 0T6 x.943........06666 3. 74.32 x 3.456.. =2469.173814 4. 16.204X 32.75. = e 530.810446 5. 19.072 X.2083. = 397348 6. 10.0512 X 4.2t3 - =42.854i88033, 7. 3. 54x4 X 4.7157... 17.704508 8. 1.256784x 6.42081.. =8.069588206 DIVISION OF CIRCULATES. 127 DIVISION OF CIRCULATES. ART. 17I. RULE. —iake thle repelends similar and contetr mzlnous. Subtract from each circulate the figures preceding itz repetend, ancd'use the remainders for the dividend and divisdr respectively, omiltting the dols. NOTE. —If the divisor is not a circulate, it will be shorter t divide as in ordinary decimals, bringing from the dividend the figures of the repetend instead of ciphers, to continue the division. Divide 2.65o by 1.8 FIRST OPERATION. SECOND OPERATION. 1.800 2.653 1.8)2.653(1.4L740 180 265 85 133 620 )2388( 1.4740 7 12]~00 -133 660 1200 DEos N sT R AT I oz.-The remainders 1620 and 2388, in the 1st operation, are the numerators of the common fractions to which the circulates are equivalent, (Art. 166); and as they have the same denominators (each being as many 9's as there are figures in the repetend, with as many ciphers annexe:l as there are decimal rgures before the repetend, Art. 166), dividing these numerators will give the same quotient as if the fractions themselves were used, and therefore the dots may be omitted. EXAMPLES FOR PRACTICE. i..75 -.I.......... =6.81 2. 51.49i- 17.... = 3.02$ 3. 681.559887 —9 94... = 7.250637i 4. 90.50374 - 9 6.754... =13.401 5 11.068763540 —.274..... =245.13 6. 9 53306639976.21.... _ — 1.53 7. 3.500691 358024 - 7.684.. =.45 M1ultiplication and division of circulates can be frequcntly performed wvith advantage by the general rule Cor circulates, (Art. 167). 1 28 RAY'S HIGHER ARIITHMETIC. ART. 172. There is a short method of converting a commoon fiaction into a circulating decimal, when the denominLator is a prime number, which is worth considering. By actual division, =.i142857, which has this property, viz: 1st Pr'operty. The number of figures in the repetend is either one less than the denominator, or a half, a third, or some other exact part of this one less; thus, T-T 63, the number of figures in the repetend (2) being 4 of 10. This is true of any fraction, whose denominator is a prime number other than 5. If the number of figures in the repetend is one less than the denominator, two other properties are observed. 2d Property. Each figure in the first half of the repetend added to the corresponding figure in the last half, makes 9; thus, in.J42857, 1+-8, 4+5, 2+-7, each equals 9. 3d Property. The same repetend serves for all fractions having the -same prime denominator, whatever be their numerators, by starting at different places; thus, -=.i42857, =.285714, =.42857i. Convert ~s into a circulating decimal. So u TI o N.-First, by actual division, -1 =.043478 56; instead of A, put 6 times the value of 2g just given, viz:.043478260869'3; having 12 figures without repeating, the repetend must be of 22 figures by property first: obtain the other figures according to the 2d property, by subtracting each of the first 11 from 9; thus, 2l g_.0)434782608695652173913; to get.( use this same repetend, and to ascertain the starting place, multiply the first 3 or 4 figures by 16; thus,.0434 multiplied by 16, gives.6944, showing that, we must commence at the eleventh figure; doing so, the value of = 6956521 739130434782G08. CONVERT INTO CIRCULATING DECIMALS,. 5 1 1 8 7 14 24 49 2 10 2O 55 AT5 i) Itt 2-9) 2' St, 4 1' 607'T1 g 19') A 9 XI. COMPOUND NUMAiBERS. ART. 173. A simple number is of one denomination; as, 95 dollars; 2000 bushels; 4 apples. COMPOUND NUMBERS. 129 A coempond number consists of several simple numbers of different denominations; as, 16 dollars 75 cents; 3 yards 2 feet 8 inches. Compound numbers are often called denominate numbers; they are used for measures, weights and money. LONG OR LINEAR MEASURE ART. 174. Is used for measuring distance, also the length, breadth, and hight of bodies, called their linem dimensions. TABLE. 12 inches, (in.)................... make 1 foot, (ft.) 3 ft....................................... 1 yard, (yd.) 5. yd. or 16, ft...................... 1 rod, (rd.) 40 rd.................... 1 furlong, (fur.) 8 fur........................ 1 mile, (mi.) RE M A R K.-The rod is sometimes called pole or perch. NOTES. —1. 4 in. = 1 hand, used in measuring the hight of horses; 9 in. 1 span; 3 feet = 1 pace; 18 in. = 1 cubit; 3 in. I palm. 2. The scale used by carpenters has the foot divided into 12 in.: each inch divided into 12 equal parts, called lines; each line into 12 equal parts, called seconds; and each second into 12 equal parts, called thirds. The inch in these scales is also divided into eighths and sixteenths, and tenths. MARINERS' 3MEASURE ART. 175. Is a kind of Long Measure used in estimating distances at sea. 6 feet....................................make 1 fathom. 120 fathoms............................... 1 cable-length. 880 fathoms, or 7-1 cable-lengths......... 1 mile. NoTE. —1 nautical league = 3 equatorial miles = 3.45771 statute miles. 60 equatorial miles - 69.1542 statute miles =-1 equatorial degree; 360 equatorial degrees = 1 great circle, or circumference of the earth. REVIEW.-173. What is a simple number? a compound number? Give examples. What are compound numbers often called? What are they used for? 174. What is long measure used for? Repeat the table. 175. What is mariners' me'asure? Repeat the table. 13(0 RAY'S HIGHER ARITHMETIC. SURVEYORS' AND ENGINEERS' BIEASURE AnRT. 176. Is a kind of long measure, used in laying out roads, and running the boundaries of land. 100 links (]k.).....mak..............ale 1 chain, (c1h.) 80 chains (ch.)......... mile, (nii.) NOTES.-i. The chain is called surveyors' or Gunter's chain, from its inventor, and is 4 rods, or 66 feet in length. As it consists of 100 links, each link must be 7.92 in. long; hence, to change these denominations to the ordinary linear measure, recollect, that 1 chain - 4 rd. or 66 feet. 1 link = 7.92 in. 2. Since each link is yTP1 of a chain, the number of links can be written as decimal hundredths with the whole chains. as 2.56 chains = 2 chains 56 links. REMARKC.-Inches and yards are not. used in the last two kinds of measurement. CLOTH MEASURE ArT. 177. Is a kind of long measure used for dry goods. 21. in...............................mke 1 nail (Tia), 4 na. or 9 in.................... 1 quarter, (qr.) 4 qr.................................. 1 yd. NOTES.-1. 1 ell Flemish = 3 qr. or 4 yd.; 1 ell English 5 qr. or 11 yd.; 1 ell French — 6 qr. or 1-. yd. 2. At the custorn-houseS, the yard only is used, being divided into tenths and hundredths; in mercantile transactions, the yard is the unit, and the fractional parts employed are quarters, eighths, sixteenths, and half-sixteenths. ART. 178. The standard of all our linear measure is the yard, being identical with the imperial yard of Great Britain, which was determined as follows: By accurate experiment at London, the length of a pendulum was ascertained, which, in a vacuum, at the level of the sea, vibrated 86400 times in a mean solar R E v w. -176. What is surveyors' measure? Rlepeat the table. What is the chhain called? Why? HIow long is it? Hlow long is a link? Why? owv are chains and links written together? 177. What is cloth measure? Repeat the table. What is an ell Flemish? An ell English? An ell French? COMPOUND NUMBERS. 131 day, or once every second. This pendulum was divided into 391393 equal parts, and 360000 of these were taken to be a yard, the pendulum itself then being 39.1393 inches long. 1E I; A Ri.-Linear measurement, as being especially necessary, was used, and to a certain degree fixed, at an earlier period than the measures of volume and weight, which have therefore been made to depend upon and be verified by the former. The standards of long measure were at first very imperfect, being derived from different parts of the human body, such as a fingerjoint, finger, hand, span, cubit or fore-arm, and yard or whole arm; but as commerce increased, and convenience demanded a change, they were rendered precise and uniform. The ancient yard of Great Britain is said to have been determined by the length of the arm of King Henry I. SQUARE OR SURFACE MEASURE ART. 179. Is used in estimating the contents of land, painters' and plasterers' work, and other surfaces. A square is an even surface, bounded by four straight lines or sides. Each side is perpendicular to two others. The size or name of any square depends upon that of its side; a square inch is a square, whose side is an inch long; a square foot, one whose side is a foot long, and so on. AnT. a80. The unit by which all surfaces are measured is a square, whose side is a linear inch, foot, yard, rod or mile; and the size of any surface will be the number of times it contains this unit. The simplest surface is a rectangle, which is an even surface, having four straight lines for sides, each opposite pair being equal, and perpendicular to the other pair. The ceiling and sides of a room, and sheets of paper, are examples of rectangles. If the length and breadth of a rectangle are the same, the sides are all equal, and it is a square. The size, or area of a rectangle, being the number of square measuring units it contains, can be ascertained as follows: R EVIE E w.-178. AWThat is the standard unit of length in the United States? Htow is it determined?'What measures of length were used at first? 179. What is square measure? Whatt is a square? A sq. inch? A sq. ft.? A sq. yd.? 180. What is the unit of measure for all surfaces? 132 RAY'S HIGHER ARITHMETIC. Take a rectangle 4 inches long by 3 inches wide. If upon each of the inches in the length, a square inch be conceived to stand, there will be a row of 4 square inches, extending the whole length of the rectangle, and reaching 1 inch of its width. As the rectangle contains as many such rows as there are inches' in its width, its area must be equal to the number of square inches in a row (4) multiplied by the number of rows (3), 12 square inches; hence, to find the area of a rectangle, RUIlE.-VUliltiply the number ofj linear units in' the length by the numzber of linear units in the breadth, after expvressing them in the same denomnilation. Tile product will be the area in square units of the same denominzation. It is easy to determine the number of sq. inches in a sq. foot, of sq. feet in a sq. yd., and so on; for, since 1 sq. foot is 12 inches long by 12 inches wide, it must contain 12 >X 12 144 sq. in.; and since 1 sq. yd. is 3 feet long by 3 feet wide, it must contain 3 X 3 = 9 sq. feet; and since 1 sq. rod is 5,yd. long by 5A yd. wide, it must contain 5~ X 5.} 301 sq. yd. 144 square inches (sq. in.) make 1 square foot, (sq. ft.) 9 sq. ft............................... 1 square yard, (sq. yd.) 30'4 sq. yd.............................. I square rod, (sq. rd.) LAND MEASURE ART. 181. Is a kind of surface measure, used to express the contents of land. 40 perches (P.) make 1 rood, (R.) 10 sq. chains, or 4 It.................. 1 acre, (A.) 640 A...................... 1 sq. mile, (sq. mi.) N OTE-. Since I chain = 4 rods, 1 sq. chain = 4 X 4 =16 sq. rods. Since links are written as decimal hundreths of a chain, chains and links can be considered as chains only; thus, 7 chains and 9 links -7.09 chains, and the area of a square whose side is 7.09 chains, will be expressed in square chains and a decimal, as follows: 7.09 X 7.09 = 50.2681 square chains; there is no necessity, then, of using the denominations link or square link in practice. REVIEw.-1SO. What is a rectangle? What is the rule for the area of a rectangle? Prove it. How many square inches in a square foot? Why? IIow many square feet in a square yard? Why? Ilow many square yards in a square rod? Why? Repeat the table. COMIPOUND NUMBERS. 1333 CUBIC OR SOLID MEASURE ART. 182. Is used to measure the bulk of stone, timber, masonry, and other solid work; to find the contents of cellars, and to verify measures of capacity. A cube is a solid, bounded by six equal squares o.r faces, each opposite pair of which is perpendicular to th1e other four. Its length, breath and hight, then, are all equal, and each is called the side of the cube. The size or name of any cube, like that of a square, depends upon its side, as cubic inch, cubic foot, cubic yard. ART. 183. The unit by which all solids are measured is a cube, whose side is a linear inch, foot, &c., and th:eir size or solidity will be the number of times they contain this unit. The simplest solid is the rectangular solid, which is bounded by six rectangles, called its faces, each opposite pair being equal, and perpendicular to the other four; a bar of soap, a candle-box, are rectangular solids. If the length, breadth, and hight are the same, the faces are squares, and the solid is a cube. The size, or solidity of any rectangular solid is found as we obtain the area of a square (Art. 180.) Suppose the rectangle 4 inches long by 3 inches wide, in Art. 180, to be the bottom or lower base of a rectangular solid, its upper face being of the same dimensions, and its hight 5 inches. If upon each of the 12 square inches in the lower base, a cubic inch be conceived to stand, there will be a section of 12 cubic i inches covering the whole bottom of the solid, and reaching I inch of its hight; as the solid contains as many such sections i as there are inches in hight, its solidity i must be equal to the number of cubic *illI'i\!hl!,!,4iH inches in a section (12) multiplied by the number of sections (5) or 60 cubic inches. But the number of cubic inches in a section is the same as the number of square REvIEwW. —81. What is land measure? Repeat the table. HI-ow Ilia'.y square chains in an acre? Why? Why do we not use square links? 182. What is cubic measure? What is a cube? A cubic inch? A culbic foot? A cubic yard? 183. What is tho unit for all solids? What is a rectangular solid? 134 RAY'S ILG HER A R TI METIC. inches in the base, (12,) and this again is equal to the number or linear inches in the length'(4), multiplied by the number in the width, (3); hence, TO FIND THE SOLIDITY OF A RECTANGULAR SOLID, RuIE. —llttltiply the legilyth, breadth, and.hight torlether, u/'c7t expressing them in the saute denoumilatioul; the product will be thAe solidity iL czubic rUenits of the saine dezonLzatiltn. It is easy to determine the number of cubic incihes in a cubic foot, and of cubic feet in a cubic yard. For, since 1 cubic foot is 12 dnches long, 12 inches wide, and 12 inches high, its solidity will be 12 X 12 X 12 1728 cubic inches; and, since 1 cubic yard is 3 feet long, 3 feet wide, and 3 feet high, its solidity will be 3 X 3 X 3 = 27 cubic feet; hence, 1728 cubic inches (cu. in.) make 1 cubic foot (cu. ft.) 27 cu. ft............................ 1 cubic yard (cu. yd.) NOTES.-1. 1 tun of round timber=40 cu. ft.; 1 tun of hewn timber - 50 cu. ft.; 1 tun of shipping = 42 cu. ft. 2. 1 cord of wood, 8 ft. long, 4 ft. wide, and 4 ft. high, contains 8 X 4 X 4 - 128 cu. ft.; 1 cord foot or foot of wood, 1 ft. long, 4 ft. wide, and 4 ft. high, contasins 1 X 4 X 4 = 16 cu. ft. 3. 1 reduced foot, plank111 meCasure, 1 ft. long, 1 ft. iwide, and 1 in, thick, contains 12 X 12 X 1 = 144 cu. in.; all planks zand scantling less than an inch are reckoned 1 inch thick; but, if more than 1 inch thick, allowance must be made by multiplying by that dimension. 4. 1 perch of masonry, 1 rod long, 1 ft. high and 11 ft. thick, contains 16i X 1 X 1-l, = 3 X 3 - 949-'24- cu. ft., which is usually taken 25 cubic feet in practice. TROY OR MINT WEIGIT ART. 184. Is used for weighing gold, silver, jewels, in testing the strength of spirituous liquors, in philosophical experiments, and in comparing different weights. 24 grains ( gr.).............make 1 pennyweiglht, ( pwt.) 20 pwt................................ 1 ounce, (oz.) 12 oz.................................. 1 pound, (lb.) R E V I E w.- 13. What is the rule for its solidity? Prove it. flow many cubic inches in a cubic foot? Why? T-Iow many cubic feet in a cubic yard? Why? Repeat the table. What is a cord of wood? A perch of masonry? How many cubic feet in each? CO1MPOUND N UMBERS. 135 R E r A r, K s.- 1. The Troy pound is tbhe standard of weight in the U. S. Mints; it is identical' with the inmperial Troy pound of Great Britain, wlhich contains 5760 grains, 252.458 of which are equal in weight to a cubic inch of distilled water when the barometer is aO inches, and the thermometer ( Fahrenheit's ) 62~. 2. The name Troy is supposed by some to be derived from Troyes, a city of France, where this weight iwas first introduced fromrn the'East, du:ing the Cruisades;' and by others from Troy Novant, the ancient name of London. The ounce Troy is the only denomination used at the mint, all amounts of gold and silver being expressed in it and its. decimal divisions. 3. The Troy pound is equal to the weight of 22.794422 cubic inches of distilled wtater, at the temperature of 390.83 Fahrenheit, the barometer being 30 ii.ches. 4. The grain was originally fixed by taking a grain of wheat from the middle of the ear, a.lnd thoroughly drying it; at first, 32 of these grains made a pennyweight, but afterward 24. 5. The pennyweight was the weight"df the silver penny in use at that time, and. is marked (pwt.) from (p.) for penny, and (wt.) for weight. DIAMOND WEIGHT ART. 185. Is used for weighing diamonds and other precious stones. 16 parts............make 1 carat grain =.8 Trony grain. 4 carat grains,........ 1 carat. = 3.2 do. do. REirMAicK.-Tlhe carat in this table is an absolute weigfht, and must be carefully distinguished from the same word used in speaking of the fineness of gold, for then it indicates the proportion of pure gold in a mass. APOTHECARIES' WEIGHT ART. 186. Is chiefly used in mixing medical prescriptions. 20 grains (gr.).................. ke 1 scruple, (.) 3....................................... I dram, (.) 8 3....................................... 1 ounce, (s.) 12.. 1 pound, (l.) The pound, ounce, and grain. of this weight, are the same as those of Troy weight; the pound in each contains 1!2 oz. = 5760 gr. RFvrIEw.-1S4. What is Troy Weight? Repeat the table. Wha.t is said of the Troy pound? The name Troy? What denomination only is used at the mint? What is said of the grain? 136 RAY'S HIGHER ARITITMETIC. AVOIRDUPOIS OR COMMERCIAL WEIGHT ART. 187. Is used in commercial transactions when goods are bought or sold by the quantity. Heavy and bulky articles, as groceries, the coarser metals, drugs, Pzc are weighed by it. 16 drais (dr.)......... make 1 ounce, (ou.) 16 oz............................ 1 pound, (lb.) 25 lb............................. quarter, (qr.) 4 qr............................. 1 hundred-weight, (cwGt.) 20 cwt........................... 1 tun, (T.) NOTES.-1. A stone= 14 lb.; but a stone of fish, or butcher's mnat= 8 lb., and a stone of glass = 5 lb.; a seam of glass = 24 stone = 120 lb.; 1 pack of wool =240 lb. 2. In Great Britain, the qr. = 28 lb., the cwt. = 112 lb., the tun 2240 lb. These values are used at the U. S. custom-houses in invoices of English goods; but generally in this country the qr. 25 lb., the cwt. = 100 lb., and the tun =-2000 lb. 3. The lb. avoirdupois is equal to the weight of 27.7274 cu. in. of distilled water at 62~ (Fah.); or 27.7015 cu. in. at 390.83 (Fah.), the barometer att 30 in. For ordinary purposes, 1 cubic foot of water can be taken b21 lb., or 1000 oz. avoirdupois. 4. The terms gross and net are used in this weight. Gross weight is the weight of the Foods, together with the box, cask, or whatever contains them. Net weight is the weight of the goods alone. R E I A RIr. —The word avoirdupois is from the French avoirs, du, pois, signifying goods of sweiht. The lb. avoirdupois differs from the lb. Troy or apothecaries,' the former being 7000 gr., the latter, each 5760 gr. The oz. avoirdupois also differs from the oz. Troy or apothecaries.' COMPARISON OF WEIGHTS. ART. 188. Since 1 lb. av. = 7000 gr. Troy, 1 oz. av. 1 of a lb. av., = - of 7000 gr. Troy = 437 gr. Troy; and 1 dr. av.' of an oz. av.. =' of 437 gr. Troy - 27-31 gr. Troy; and in a similar way 1 oz. Troy REVIEw.-185. Repeat the table of diamond weight. What is said of a carat? 1S6. 5What is apothecaries' weight? Repeat the table. What denominations are identical in Troy and Iapothecaries' weight? 187. TVWhat is avoirdupois weight? Repeat the table. What is said of the qr., cwt., and tun, in Great Britain and the U. S. custom-houses?'What elsewere? What is said of the lb. avoirdupois? What is gross weight? Net weight? COMPOUND NUMBERS. 137 and apothecaries' = 480 gr. Troy; and 1 dr. apothecaries' 60 gr. Troy. Hence, 1 lb. avoirdupois.............. = 7000 gr. Troy or apoth. 1 oz........ do.................... = 437 do...do........do. 1 dr............... = 27 do...do........do. 1 lb. Troy or apoth.......... - 5760 do...do........do. 1 oz. do..........do............ -- 480 do..do........do. 1 dr. apothecaries'.......... = 60 do...do........do. This table serves to convert denominations of one kind of weight into those of another. WINE OR LIQUID MEASURE ART. 189. Is used for, measuring all liquids, except ale, beer, and milk. 4 gills (gi.)..........make 1 pint, (pt.) 2 pt................................ 1 quart, (qt.) 4 qt........... 1 gallon, (gal.) — = 231 cu. in. NOTES.-I. 31-1 gal. = 1 barrel (bbl.); 63 gal. =1 Ihogshead (hhd.); 42 gal. 1 tierce; 84 gal. _ 1 puncheon; 126 gal. = 1 pipe or butt; 2 pipes = I tun. These are generally mentioned as measures, but are only vessels, and are gauged and sold by the gallon, not being of uniform capacity. When the contents of cisterns, wells, &c., are expressed in hhd. or bbl., they have the values given a-bove. 2. In England, 1 anker= 10 gal.; 1 runlet 18 gal. ART. 190. The standard unit of liquid measure in the U. S. is the wine gallon, formerly used in Great Britain, containing 231 cu. in., and which contains a weight of 58372.1754 gi., = nearly 8~ lb. av., of distilled water at 39~.83 Fahrenheit, the barometer at 30 inches. A pint of water is generally considered 1 pound. ALE AND BEER MEASURE ART. 191. Is used in measuring ale, beer, and milk, though milk is frequently sold by wine measure. 2 pints (pt.) make 1 quart, (qt.) 4 qt.................... 1 gallon, (gal.) = 282 cu. in. RE IE w.-1-87. What does avoirdupois mean? How does the lb. avoirdupois differ from the lb. Troy and apothecaries? 188. Repeat the table for comparison of weights. 189. What is wine measure? 12 138 RAY'S HIGHER ARITHMETIC. NOTE S.-1. 1 bbl.=3G gal.; hhd. -- 54 gal. These are mentioned as measures, but are only vessels, and are gauged and sold by the gallon, not being of uniform capacity. 2. In Englandl, 1 firkin = 9 gal.; 1 kilderkin = 2 firkins; 1 bbl. 2 kilderkins. REMAaRK.-The beer gallon contains 282 cu. in., or 10.179933 lb. av. o? distilled water at 390.83 Fahrenheit, the barometer at 30 inches. DRY MEASURE ART. 192. Is used for measuring grain, fruit, vcgetatiles, coal, salt, &c. 2 pints (pt.) make 1 quart, (qt.) 8 qt................... 1 peck, (pk.) 4 pk.................. 1 bushel, (bu.) = 2150.42 cu. in. N T E.-1. 4 qt. or a peck = 1 dry gal. = 268.8 cu. in. nearly. 2. 1 qr. = 8 bu. = 480 lb., used in England in measuring wheat. 36 bu., and in some places 32 bu. make 1 chaldron, used in some of the United States, and formerly in Great Britain, in measuring coal; but now coal is bought and sold by weight in England, and in many parts of this country. 3. Grain is often bought and sold by weight. In England, and in many of the United States, 60 lb. of wheat, 48 lb. of barley, 56 lb. of rye or corn, and 32 lb. of oats, are each declared to make a bushel. ART. 193. The unit of our dry measure is the Winchester bushel, formerly used in England, and so called firom the town where the standard was kept. It is 8 in. deep, and 182 in. diameter, and contains 2150.42 cu. in., or 77.627413 lb. av. of distilled water at 390.83 Fahrenheit, the barometer at 30 inches. The New York bushel is equal to the in'perial bushel of Greaf Britain, and therefore contains 2218.192 cu. in. COMPARISON OF MIEASURES. ART. 194. The wine gallon contains 231 cu. in.; the Leer gallon 282 cu. in.; and the dry gallon 268.8 cu. in. REVIEw.-1S9. What is said ot the bbl. and hhd.? 190. Whlat is our standard of liquid measure? What does it contain? 191. Whallt is beer measure? Repeat the table. What does the beer gallon contain? 192. \What is dry measure? Repeat the table. HIow much is a qr. of wheat? IIow is grain measured? 193. What is the unit of dry measure? What does it contain? What is the New York bushel? COMPOUND NUMBERS. 13'9 These were superseded in Great Britain, in 1826, by the imperial gallon, both of dry and liquid measure, which contains 2077.274 cu. in., or 10 lb. av. of distilled water at 62~ Fahrenheit, the barometer at 30 inches, At the same time the dry or Winchester bushel was replaced by the imperial bushel of 8 imperial gallons, con.taining 2218.192 cu. in. 1 wine gallon of United States - 231 cubic inches. 1 beer gallon......................... 282 do. 1 dry gallon........................... 268.8 do. 1 imperial gallon of G. Britain for dry and liquid measure = 277.274 do. I dry bushel of U. S................ = 2150.42 do. 1 imperial bushel of G. Britain - 2218.192 do. This table is useful in converting denominations of one measure to those of another. APOTHECARIES' FLUID MEASURE ART. 195. Is used for measuring all liquids that enter into the composition of medical prescriptions. 60 minims (nl)...............make 1 fluid drachm, (f3). 8 f3.................................. 1 fluid ounce, (f3l. 16 f3.......................1......... pint, (0). 8 0.................................... 1 gallon, (Cong.) NOTES.-I. Cong. is an abbreviation for congiariunz, the Latin for gallon; 0. is the initial of octans, the Latin for one-eighth, the pint being one-eighth of a gallon. 2. For ordinary purposes, 1 tea-cup = 2 wine-glasses = 8 table. spoons = 32 tea-spoons = 4 f 3. ART, 196. TIME. 60 seconds (sec.)..............make 1 minute, (min.) 60 rain.................................. 1 hour, (hr.) 24 hr..................................... 1 day, (da.) 7 da.................1................. 1 week, (wk.) 4 wk................................... 1 month, (mon.) 12 calendar mon...................... year, (yr.) 365 ida.................................... 1 common year. 366 da.................................... 1 leap year. 100 yr.................................... 1 century, (cen.) NOTE — 1 Solar year-= 365 da: 5 hr. 48 min. 48 sec.= 3651 d, nearly. 140 RAY'S HIG HTER ARITHMETIC. The denomination fiom which the preceding table is constructed, is the Cdcy, which is the interval of time be tween one mean noon and the next. The apparent noon is the moment when the sun comes to the meridian of any place, and appears exactly half-way between rising and setting. Owing to the unequal motion of the earth around the sun, and the oblique position of its axis to its orbit, the interval between any apparent noon and the next is not uniform. The average, however, is taken of all these intervals that occur in a year, and this average interval is called the mean solar day, which is divided as above. A year is the time during which the earth makes a complete circuit about the sun, and reaches again a given point in its orbit; it contains 365 da. 5 hr. 48 min. 48 sec., or nearly 365- days. The ancients were unable to find accurately the number of days in a year. They had 10, afterward 12 calendar months, corresponding to the revolutions of the moon around the earth. In the time of Julius Cesar the year contained 365k days; instead of taking account of the I of a day every year, the common or civil year was reckoned 365 days, and every 4th year a day was inserted, (called the intercalary day,) making the year then have 366 days. The extra day was introduced by repeating the 24th of February, which with the Romans was called the sixth day before the kale7ids of March. The years containing this day twice, were on this account called bissextile, which means having twzo sixths. By us they are generally called leapl. years. But 365~ days, 365 days and 6 hours, is a little longer than the true year, which is 365 days 5 hours 48 minutes 48 seconds. The difference, 11 minutes 12 seconds, though small, produced, in a long course of years, a sensible error, which was corrected by Gregory XIII., who, in 1582, suppressed the 10 days that had been gained, by decreeing that the 5th of October should be the 15th. To prevent difficulty in future, it has been decided to adopt the following rule. REvrEW. —194. Repeat the table for comparison of measures. 195. What is apothecaries' fluid measure? Repeat the table. 196. Repeat the table of time. What is the unit of this table? What is a mean solar day? A year? What did the months correspond to? What was the Julian calendar? How many days were taken for a year in it? What is the true length of the solar year? What error was committed in the Julian calendh.tr? When and by whom was it corrected? What is a leap year? COM:POUND NUMBERS. 141 RULE FOR LEAP YEARS. Every year that is divisible by 4 is a leap year, uznless ih ends with a double cipher; in which case it must be divisible b9 400 to be a leap year. Thus, 1832, 1648, 1600 and 2000 are leap years; but 1 857, 1700, 1800, 1918, are not. The Gregorian calendar was adopted in England in 1752. T'he error then being 11 days, Parliament declared the 3d of September to be the 14th, and at the same time made the year begin January Ist, instead of March 25th. Russia, and all other countries of the Greek Church, still use the Julian calendar; consequently their dates (Old Style) are now 12 days later than ours, (New Style). The error in the Gregorian calendar is small, amounting to a day in 3600 years. ART. 197. The names of the months in their order, are January, February, March, April, May, June, July, August, September, October, November, December. The year formerly began with March instead of January; consequently, September. October, November and December were the 7th, 8th, 9th, and 10th months, as their names indicate; being derived from the Latin numerals Septemn (7), Octo (8), Novem (9), Decem- (10). ART. 198. MISCELLANEOUS TABLE. 12 things...... make.................. dozen. 12 dozen or 144 things..................................1 gross. 12 gross or 144 dozen...........................1 great gross. 20 things..................................... score. 56 lb. 1 firkin of butter. 100 lb......................................l quintal of fish. 196 lb.................................................... bbl. of flor. 200 lb.................................................... bbl. of pork. 14 lb..........................................I. stone. 21~ stone.................................... pig of iron or lead. 8 pig......................................................... forne. ARTr. 199. The words folio, quarto, octavo, &c., used R e vnE W. —1i96. What is the rule for leap years in the Gregorian calendar? Where does the Julian calendar still prevail? What is the error in the Gregorian calendar? 197. Name the months in their order? What is the origin of the names, September, October, &c.? 142 RAY'S HIGHER ARITHMETIC. in speaking of books, show how many leaves makle a sheet of' paper. A sheet folded into 2 leaves forms a foio.................... size. Do...................4.. do...............quarto or 4to........ do. Do...................8 do.octavo or Svo......... do. Do.................12...do...............duodecimo or 12mo. do. Do.......18 do.. 18mo............... do. Do.................36..do...............36mo............... do. Also, 24 sheets of paper make................................ 1 quire. 20 quires..I......................................1 ream. 2 reams...................................................... bundle. 5 bundles...............................1 bale. CIRCULAR OR ANGULAR MEASURE ART. 200. Is used for latitude and longitude, and for expressing the distances between any two points on the surface of the globe, or in the heavens. 60 seconds (//) make 1 minute, (/) 60/.......................... 1 degree, (o) 300....................... 1 sign, (s.) 3600 or 128................ 1 circumference, (c.) N 0 T E. —90 = 1 quadrant or quarter of a circumference, 1800 — 1 semi-circumference. The degree being - ~ of a circumference is of different lengths in different circumferences; thus, the equator being larger than the polar circles, a degree of the former is larger than a degree of the latter. COMPARISON OF TIME AND LONGITUDE. ART. 201. The difference of longitulde of two places is the distance in degrees, minutes, and seconds, that one of them is further east or west of the established meridian than the other. The sun appears to go entirely round the earth, (360O), from east to west, in 24 hours, crossing in succession the meridians of all places on its surface. R E v I Wv. —-200. What is circular or angular measure? Repeat the table. 201. What is the difference of longitude of two places? At whlat rate per hour does the sun appear to travel around the earth? At what rate per minute? What per second? COMPOUND NUMBERS. 143 A place'further east than another will have tile sun on its mericdian sooner, and, therefore, its time will be later at the rate of 1 hour for every 150 of longitude; or, (by takirlng of each of these quantities), at the rate of 1 minute for every 15' of longitude; or (by taking,.o of these); at the rate of 1 second for every 15" of longitude -Ieence, 1 hour of time........................... 15~ of longitude. l minute do........................... 15' do. 1 second do.......................... = 15 do. NOTE.-Recollect that if one place has greater east or less west longitude than another, its time must be later; and conversely, if-one place has later time than another, it must have greater east or less west. longitude. MONEY TABLES. FEDERAL OR UNITED STATES MONEY ART. 202. Is the currency of the United States. 10 mills (n.)........................... make 1 cent, (ct.) 10 ct.............................................. 1 dime, (d.) 10 d.............................................. 1 dollar, ($). $10....................................... eagle. NOTE.-The cent and mill, which are T-1 and ~O of a dollar, derive their names from the Latin centumn and mille, meaning a hundred, and a thousand; the dime which is Ili of a dollar, is from the French word disme, meaning ten. ART. 203. The Federal currency was authorized by act of Congress, August 8th, 1786. It has great simplicity, being on the decimal basis, and subject to the law that on.e of atly de'nomintation is equal to 10 of the next lower; therefore, the same Notation, Numeration, and general order of operation, can be used for Federal money as for simple numbers. Any sum of Federal money of several denominations can be expressed as one denoiination, by writing those of that denomination in units' place, those of higher denominations in places of whole numbers, and those of lower denominations in decimal places; thus, REVIEw.-201. Why will a place that is east of another have later time? What is the table for comparing time and longitude? HIow do wo know from the longitudes of two places, which has later time? 202. What Is Federal or U. S. mruoney? Repeat the table. 144 RAY'S IIIGHER ARITHMETIC. 9 eagles $7 2.imes 4 ct. and 5 m. can be written 9.7245 eagles, or $97.245, or 972.45 dimes, or 9724.5 ct. or 97245 m. It is custonmary to consider the dollar as the unit, and express all sums of U S. money in that denomination and its decimal divisions. In reading U. S. money, name the dollars and all higher denomin-ations together as dollars, the dimes and cents as cents, and the next figfure, if there is one, as tmills; Or, name the whole numbers as dollars, and the rest as a decimal of a dollar. Thus, $9.124 is read 9 dollars 12 ct. 4 mills, or 9 dollars 124 thousandths of a dollar; $175.06238 is read 175 dollars 6 ct. 2 mills, and a remainder, or, 175 dollars, 6238 hundred-thousandths of a d(ollar. ART. 204. The national coins of the United States are of gold, silver, and copper. The gold coins are the doubleeagle, eagle, half-eagle, quarter-eagle, and one dollar piede. A 3 dollar piece has also been authorized. The silver coins are the dollar, half-dollar, quarter-dollar, dime, half-dime, and 3 cent piece. The copper coins are the cent and half-cent; the latter is now obsolete. The mill never has been a coin; it is only a convenient name for the tenth of a cent, or thousandth of a dollar. Pure gold and silver being too soft for coins. are mixed with baser metal, called alloy. By act of Congress, in 1837, our standard gold and silver is -9 pure and 1 alloy, (by weight). The alloy in the silver coins is pure copper; in the gold coins it is copper and silver, the latter not to exceed the Cormer in weight. The 3 cent piece is not standard silver, being one-fourth copper. It weighs 12- grains. The cent weighs 168 gr. RrVIE w.-202. Why is the cent so called? The mill? The dime? 203. How do the denominations of Federal money resemble those of simple numbers? How can sums of Federal money be read, written, added, subtracted, &c.? In writing any sum of Federal money, what single denomination is generally used? In reading any sum of Federal money, which denominations only are mentioned? What single denomination may be employed? 204. Which coins are of gold? Which of silver? Which of copper? What is the mill? What are standard gold and silver? What is the-alloy for gold? What for silver? COMIPOUND NUMBERS. jl4b The eagle weighs 258 gr. The half-dollar since April 1st, 1853, weighs 192 gr. The half dollar coined before April 1st, 1853, weighs 20G6- gr.; it contains more silver than the one now coiued, and is worth 53-9iga, instead of 50 cents. ART. 205. The fineness of manufactured gold is estimated in ceratts, an Arabic or Abyssinian word, signifying, a small weight. In determining the purity of gold by analysis, a portion of it is taken, and, without reference to its actual weight, is called the assay poLzd, which is thus divided: 4 quarters (qr.).................make 1 assay grain, (gr.) 4 gr....................................... carat, (car.) 24 car................................... 1 assay pound NoTES. —. Each carat is FL of the mass used; if it, is 18 carats gold, it is X8 pure gold, and a6e alloy; if it is 22 carats gold, it is 2 pure gold, and 24 alloy; if it is 24 carats gold, it is entirely free from alloy. 2. The quarters are written as fourths of a carat grain; thus, 19 car. 3 3 gr. ENGLISH OR STERLING MONEY ART. 206. Is the currency of Great Britain. 4 farthings (qr.) make 1 penny, (d.) 12 pence................... I shilling, (s.) 20 shillings........ 1 pound, (E) - $4.84* NOTEs.-l. The Guinea, (gold), = 21s.; crown, (silver), = 5s.; half-crown = 2s. 6d.; noble = 6s. 8d.; angel = 10s.; mark = 13s. 4d.; pistole= 16s. 10d.; moidore = 27s.; I sovereign = 20s., (gold) = $4.84, - by act of Congiress, 1842. 2. The farthing is not a coin, but stands for a quarter of a penny; thus, 5-d.-= 5 pence 3 farthings. When the penny was of silver, it was usual to mark it with a cross so deep that it could be easily broken into halves and quarters, called half-pennies and fourlhinzgs, finally, farthings. REVIEW.- 204. What is the weight of the eagle? The present halfdollar? What was the weight of the half dollar before 1853? How much is it now worth? 205. How is the fineness of manufactured gold estimated? Repeat the table. 206. What is sterling money? Repeat the table. How are fiarthings written? Which denominations are not coins? What is the origin of the name farthing? 13 a46 RAY'S HIGHER ARI'THMETIC. 3. The pound (~) is not a coin, but stands for 20s.; it is repre' sented by the sovereign, or the bank note of ~1. The pound is so called, because its equivalent, 240d. or 20s., formerly contained a pound weight of silver, the pound then being smaller than at present. A pound of standard silver is now coined into GGs. REMAjrIS.-1. The symbols, ~.,.,, d., q., axre the initials of the Latin words libra, solidarius, denlarius, qzadrans; signifying, respectively, pound, shilling, penny, and qu-arter. 2. The sovereign, the standard gold coin, weighs 123.274 gr., and is 22 carats fine. The shilling, the standard silver coin, is ipure silver, and 43 copper, and weighs 87.27 gr. Pence and hall'pence are made only of copper now, and each penny weighs 240 gr. = oz. Troy. OF STATE CURRENCIES. ART. 207. Before our present currency was established, our accounts were kept in pounds, shillings, and pence. In many States, the denominations shillings and pence are still retained, but not with the same values. In New IIampshire, IMassachusetts, Rhode Island, Connecticut, Virginia, Kentucky, and Tennessee, 12d.= 1 shilling=16 cents. 6 s. =,$1 = o00 cents. In New York, North Carolina, and Ohio, 12 d.= 1 shilling-= 12A cents. 8 s. = $1 = 100 cents. In New Jersey, Pennsylvania, Delaware, and Maryland, 12d. = 1 shilling- 13- cents. 7s. 6d. or 90d. = $l — 100 cents. In South Carolina ancd Georgia, 12 d. - 1 shilling 21 3 cents. 4 s. 8 d. or 56 d. $1 1 100 cents. In Canada and Nova Scotia, 12 d. = 1 shilling 20 cents. 5 s. =$1 - =100 cents. REViEw. —206. What is the origin of tile name polzd? Into how many shillings is a pound of silver now coined? 207. IWhat is the New England State currency? New York? Pennsylvania? Georgia? Canada? COMPOUND NUMlBERS. 147 FRENCH VWEIGHTS AND )MEASURES. AnT. 208. The present system of weights and measures in France, adopted in 1795, is on the decimal basis. After the unit of any measure has been determined and named, the higher denominations are made by prefixing the Greek numerals, (lcca (10), hecto (100), Adilo (1000), my2r'ict (10000) to the name of the unit; the lower denominations are formed by prefixing the Latin numerals deci (r'o), centi ( -rO), mllid ( Ion ). l FRENCH LONG MEASURE. ART. 209. The unit of lon, measure in France is the metre, which is the ten-millionth part of the quadrant, extending thlrough Paris from the equator to the pole. 1 metre........ -39.371 U. S. in. NOTE. —1 decanmtre =10 metres; 1 hectom~tre = 100 m6tres; 1 ikilomftre = 1000 matres; I myrianmtre = 10000 metres; 1 decimnttre-' meitre; 1 centim6tre - o0 mdtre; I millimatre -- ~0 metre. FRENCH SURFACE MEASURE. ART. 210. The unit of surface in France is the are, which is a square decametre, 1 are...... 119.6046 U. S. sq. yd. NOTE.-1 decare = 10 ares; 1 hectare = 100 ares; 1 centiare - UT are. FRENCH SOLID rMEASURE. ART. 211. The unit of solidity in France is the stere, which is a cubic meitre. 1 stere....... 35.31741 U. S. cu.'t. NOTE.-1 decastre = 10 stares. 1 decistere - y1O stere. FRENCH WEIGHTS. ART. 212. The unit of weight in France is the gramnme, whlich is the weight of a cubic centimetre of distilled water at the temperature of melting ice. 1 gramme.... 15.434 Troy gr. NOTE.-1 decagramme = 10 grammes; 1 hectogramme -- 100 grammes; 1 kilogramme 1000 grammes lb. av. nearly; = 00 grme g. ar. nearly; 14-' RAY'S HIGHER ARITHMETIC. 1 myriagramme - 10000 grammes. I decigramme = 1 gramme; 1 centigramme = To gramme; 1 milligramme - 01 gramme; I quintal = 220.55 lb. av.; ~1 millier or bar =- 2205.5 lb. av. FRENCH DRY AND LIQUID. MEASURE. Anu 213. The unit of capacity in Fran-ce is the litre, which is a cubic decimetre. 1 litre.... =2.1135 pt. wine measure, U. S. NOTE.-1 decalitre = 10 litres; 1 hectolitre = 100 litres; 1 kilolitre = 1000 litres; I myrialitre = 10000 litres. 1 decilitre-1 litre; 1 centilitre = litre; 1 millilitre - 1 litre. ART. 214. Some of the old weights and measures are used; as, 1 livre a kilogramme; 1 mare = 1 a livre; 1 once = Ilmarc; 1 gros - = once; 1 grain = Q gros: 1 toise = 2 maetres; 1 pied or foot - 1 matre; 1 inch = t pied or foot; 1 aune = 1 meltres; I boisseau or bushel = 12{ litres; 1 litronl = 1.074 Paris pints. When these are employed, the word usuel is annexed to them, signifying cuzstomary. FRENCIH MIONEY. ART. 215. The unit of money is the franc, which Is 9 pure silver, and -jo alloy, like our silver coins. 1 franc.. = $.1875 or 18 cents. NOTE.-1 decime =- l- franc; 1 centime = 1~0 franc. The livre tournois, the former unit of French money, -- l8 cents. FRENCH CIRCULAR OR ANGULAR MEASURE ART. 216. Is the same as that of the United States and other countries. FOREIGN WEIGHTS AND MEASURES. ART. 217. The pounds here mentioned are lb. avoirdupois, when not otherwise specified. Alexandria, _Egy/pt.-1 pik —: 26.8 in.; 1 rhebebe = 4.364 bu.; 1 quillot or kisloz = 4.729 bu.; 1 rottolo forforo.9347 lb. av.; I rottolo zaidino — 1.335 lb. av.; 1 rottolo zaro = 2.07 lb. av.; 1 rottolo mina 1.67 lb. av.; 1 quintal or cantaro _ 100 rottoli. FOREIGN WEIGHTS AND MEASURES. 14t9 Amsterdam?, folland. —French system adopted in 1820. Old measures as follows: 1 lb. =1.08923 lb.; 1 last =- 85.25 bu.i 1 alm -- 41 wine gal.; 1 stoop = 5' pints; 1 anker - 101 gal.; 1 foot = 11- in.; 1 ll = 27 7l in. Anxtwerp, Belgium.-French system adopted in 1816. Old mea. sures as follows: 1 lb. =-1.033 lb. av.;'1 schippound = 3 qullintals 310 lb. av.; 1 aam = 36A w. gal.; 1 viertel'2.125 bu.; 1 stoop = 5.84 pt. Bar'celona, Nlorlth of S)ain.-i lb. =.88215 lb. av.; 1 cana.58514 yd.; 1 quartera = 1.88288 bu.; 1 carga 32.7 w. gal. Bombay, East Indies.-i maund = 28 lb. av.; 1 candy 560 lb. av.; 1 tola = 179 Troy gr.;. 1 tank for pearls = 72 gr. 1 guz- 27 in.; 1 hath = 18 in.; 1 tursoo = 18 in. Bremein.-1 lb. — 1.098 lb. av.; 1 last — 80.7 bu.; 1 aam or 4 ankers = 373 w. gal.; 1 foot - 11.38 in.; 1 ell =.632 yd. Batavia, East Indies.-i pecul - 136 lb. av.; 1 catty = 1.36 lb. av. Bencoolen, East Indies. —i bahar = 560 lb. av. Ba/lia and.Rio de Janleiro, Brazil.-l alquiere of grain-= 1 bu., U. S.; 1 frasco = 4.5 pt. Cadiz, South of Spain. — lb. = 1.015 lb. av.; 1 ara =.9275 yd.; 1 arroba of wine -4, w. gal.; 1 arroba of oil = 33 w. gal.; 1 moyo = 68 w. gal.; 1 botta, = 127' w. gal. Calcutta and Bengal Factory, Eas! Inldies. —1 maund - 74' lb. av.; 1 bazaar maund = 82T —2 lb.; 1 tolan = 224.588 T. gr., 1 pallie — 9.08 lb. av.; 1 chittack = 45 sq. ft.; 1 biggabh = 14440 sq. ft.; 1 guz = 1 yd.; 1 coss = IT3.z miles. Canton, Chinla. — tael 1-3 oz.; 1 catty- 1V lb. av.; 1 pecul = 133k lb. av.; 1 covid or cobre = 14.62'5 in.; 1 li = Il971 ft. Coonstantinople, Turkey. — quintal or cantaro-= 124.457 lb. av.'; 1 quintal of cotton = 127.2 lb. av.; 1 pik of silk = 27.9 in.; 1 pik of cotton= 27 in.; 1 kisloz.741 bu.; 1 alma of oil 18 gal. Copenhagen, Denmark.-I lb. = 1.1025 lb. av.; 1 anker - 10 w. gal.; 1 pot 1.02 qt.; 1 last = 380 bu.; 1 Rhineland foot 12 — in.; 1 Danish ell = 2.06 ft. Dantzic, East Prtssia.-l lb. = 1.033 lb. av.; 1 last = 620.4 w. gal.; 1 ahm of wine = 392 w. gal.; 1 scheffel = 1.552 bu.; 1 Dantzic foot - 11.3 in.; 1 Rhineland or Prussian foot — 12.356 in.; I Prussian ell = 26.256 in.; 1 last of corn -- 91 bu.; I last of wheat, rye - 87 bu.; I Pru. mile =- 4.8 miles 15(0 RAY'S HIGHER ARITHIMETIC. Genoa, Sardiilia.- 1 lb. peso sottile, =.6989 lb. av.; I lb., peso go2sso,-.7G6875 lb. av.; 1 mina = 3- bu.; 1 mezzarola. 39,1 w. gal; 1 barilla of oil =17. gal.; 1 palmo = 9.725 in.; 1 canna- 9, 12 or 10 palmi, as it is used by manufac. turers, mcrchants, or custom-house officers. Iia in'bt7rgh.-1 lb. = 1.068 lb. av.; 1 ahm = 38- w. gal.; 1 fuder -- 299 gal.; 1 steckan of oil = 5 gal.; 1 last- 89.6 bu.: 1 foot - 11.289 in.; 1 Brabant ell - 27.585 in. Hctavana, Cuba. — arroba of wine or spirits = 4.1 w. gal.; 1 fanea -- 3 bu., nearly; 1 arroba of weigrht or 25 lb. - 25.43 75 lb. axv.; 1 vara -,2 ft. Konigsber.q.-LSame as Dantzic. La GCluyra, Venlezuela.-Same as Spain. Leghorn, Tscany. —I lb. = -.74864 lb. av., generally reckoned, -.77 lb. av.; I sacco of corn - 2.0739 bu. 1 barile =- 12 w. gal.; 1 braccio- = 22.98 or 23 in.; 1 canna = 92 in. Lima, Peri. —Same as Spain. Lisbone, Porttgal. — lb. or arratel = 1.10119 lb. av.; 1 moyo = 23.03 bu.; 1 almude 4.37 w. gal.; 1 tonelada.- 2271 w. gal.; 1 pipe of Lisbon -140 v. gal.; 1 pipe of port= 168 w. gal.; 1 pe or foot = 12.44 in.; 1 vara - 43.2 in.; 1 Almude of Oporto = Gw IsV. gal. Madras, East Indies. —1 atiund = 25 lb; 1 candy = 500 lb,; 1 garce = 1' b)u.; 1 Company maud -241 lb.; 1 varahun 52 gr.; 1 visay = 3 lb. 3 dr.; 1 baruay = 4892 lb.; 1 gursay 9= 9645 lb. Monllevideo, Bauetns Ayres.-Same as Spain. tlllscat, Arabia.-i maund or 24 cuchas = 83 lb. av. Naples, Xalples. — cantaro grosso =-1908 lb. av.; 1 cantaro piccolo = 100 lb.; 1 tomolo = 1.45 bu.; I carro = 264 w. ~gal.; 1 pipe wine or ludy = 132 w. gal.; 1 salma = 424 w. gal.; 1 canna- = 6 ft. 11 in.; 1 palmo = 10.375 in. Odessa, lutssia.-Same as St. Petersburg. Palermno, Sicily.-I1 oncie = - oz.; 1 salma grossa = 9.48 bu.; 1 salma generale = 7.62 bu.; 1 barile= 9 T w. gal.; 1 caffiso of oil - 4 w. gal.; 1 canna- 3-3 yd.; 1 palma = ft. Pourl-au-t1 ince, L/ayti. -Measures same as France-weights same as England, but 8 per cent. heavier. Porto-l:ico.-Same as HIavana.?anygoonl, East lldies.-l kyat or tical =.584 lb. av.; 1 Paiktha or vis = 3.65 lb. av.; 1 ten or basket = 58.4 lb. av., generally reckoned - cwt. REDUCTION OF COMPOUND NUMBERS. lbl Rigfa, Russia. — lb. =.9217 lb. av.; 1 loof = 1.9375 bu.; 1 anker = 10} w. gal.; 1 foot = 10.79 in., U. S. otllerllrm, Hlollaitd. — last = 10.6429 bu.; I ahlm = 40 w. gal., nearly; 1 stoop =.6775 w. gal.; 1 foot;- 1.02 ft., U. S. The rest like Amlsterdamnl. sinlga)ore, East ladies.-l maund of rice = 82.125 lb. av.; I bungkal of gold-dust = 832 gr. The rest sanme as Canton. Sniyrna, Turtley. —Same as Constantinople. 1 rottolo 1.27,18 lb. av.; l oke = 2 lb. 13 oz. 5 dr.; 1 tepper of silk = 4- lb., av.; I ehequce of opiumL = 1 lb. av.; 1 chequee of goats' wool = 5- lb.; i kellow 1.456 bu.; pie = 27 in., U. S. 5tsockeol1n, Siweden. — lb. or pund =.9345 lb., av.; 1 lb. of iron lb. av.; 1 tun= 4 bu.; 1 ahm = 41i'.:~ w. gal.; t pipe = 124- w. gal.; 1 foot 11.684 in., U. S.; I kannor -.692 w. gal.:S. Petersburg, Russia. — lb. =.9026 lb. ar.; I pood = 36. 1041 lb. generally reckoned 36 lb. av.; 1 wedro -. 3.14 w. gal.; I chetwert - 5.952 bu.; I sashern 7 ft.; I arsheen = 28 in.; 1 foot = 1.145 ft., U. S.; 1 verst or mile- 5.3 fur. Trebisond, Turkey. —Same as Constantinople. 2'Trie.ste, Austriac. —I lb. - 1.236 lb. atv.; I staro = 2.34 bu.; I Vienna metzen -= 1.723 bu.; 1 polonick -.861 bu.: i orna or eiter = 15 w. gal1.; barile = 173 w. gal..; I orna of oil = 17 w. gal.; 1 ell (for woolen goods) = 2.27 in.; I ell of silk - 25.2 in. Vlalparaiso, Chili. -Same as Spain. elnice, Lombardy. — I lb., peso sottile --.66428 lb. av.; 1 lb., peso grosso - 1.05186 lb. av.; I staja = 2.27 bu.; 1 anfora = 137 w. gal.; I miro = 4.028 w. gal.; I braccio of wool - 26.6 in. era, CGruz, 3exico. —Same as Spain. REDUCTION OF COMPOUND NUMIBERS. ART. 218. Reduction is changing the form of a number without altering its value. It has three cases. CASE I.-To reduce a simple number of any denomination to another denomination. GENERAL RULE. lltltiply the given number by its unit value in the denomsnation required; 152 RAY'S HIGHER ARITI-tIMETIC. Or, Divide it by the unit value of the required denomirnation its the o ne given. N o T E.-When there are one or more denominations between the one given and the one required, reduce the given number to each of theln in succession, until the required denomination is reached. Reduce 18 bushels to pints. SoLUTION. —Since I bu. = 4 pk., 18 bu. =18 1 8 bu. times 4 pk. = 72 pk., and since 1 pk. = 8 qt., 72 4 pk. = 72 times, 8 qt. = 576 qt., and since 1 qt.= -2 -p 2 pt., 576 qt. = 576 times 2 pt = 1152 pt. Or, find P the unit-value of bushels in pints, thus: I bu. 8 pk. = 32 qt. = 64 pt.; then multiply 18 by 64 pt. 5 7 6 qt. which gives 1152 pt. as before. This is called Re- 2 duction Descending, that is, going from a higher de- 1 5 pt. nomination (bu.) to a lower (pt.) and is performed by the 1st part of the rule, that is, by multiplication. Reduce 236 inches to yards. SoLuTION.-Since 12 inches =1 ft., 236 12)236 in. inches will be as many feet as 12 in. is con- 3 ft. tained times in 236 in., which is 190 ft., and ) 9 since 3 ft. = 1 yd., 19: ft. will be as many yd. 6~ yd. as 3 ft. is contained times in 192 ft. which is 65 yd. Or, find the unit-value of yards in inches, thus; I yd. = 38 ft. = 36 in.; then divide 236 in. by 36 in., which gives 6~ yd., as before. This is called Reduction Ascending, that is, going from a lower denomination (in.) to a higher (yd.), and is generally performned by the 2d part of the rule, that is, by division. ART. 219. Reduction Ascending is similar in principle to Reduction Descending, and can be performed by the 1st part of the rule. Thus, in last example, inst'ead of dividing 236 in. by 36 in. the unit value of yards, the 236 may be multiplied by <6 yd., the unit-value of inches; for, 236 X 1 yd. — 29 yd. - )1 y. The operation by division is generally-more convenient. R E A AtR. —Reduction Descending diminishes the size and, theref're, increases the nusmber of units given; while Reduction AscendR E V I E w. -218. What is reduction? 1Iow many cases? Whqlt is Case lst? The rule? When there are several denomilnations between the one given and the one required, what should be (lone? Explain.examples I and 2. What is Reduction Descending? How is it generally performed? What is Reduction Ascending? How is it generally performed? 219. How may it be performed? REDUCTION OF COMPOUND NUMBERS. 1, 3 ing increases the size, and, therefore, diminishes the enzmber of units given. This is further evident from the fact, that the multipliers in Reduction Descending are larger than 1; but in Reduction Ascending smaller than 1. Pteduc 3 *SOLUTION. Reduce 8 gal. to gills. REM A R KIr. -The rule also ap- 3 X X _ X 4 1 2 gills plies when the number to be re]uced is a common or decimal f'raction. Indicate the operations and then cancel. Reduce 5 - gr. to S. 2 SOLUTIO N.-Alt-ough 5 this is Reduction Ascend- 57 g r X -8 ing, use the first part of0 3 8 the rule, multiplying by the successive unit values, 0, - and I. Reduce 9.375 acres to perches.'9.3 7 5 A 4 37.500 R. 40 Alzs. 1 5 0 0.0 15 00 P. Reduce 2000 seconds to hours. Ans. 5 hr. 1 1 20 5 20PP X x - -9 hr. Also,.6428 dr. av. to lb. av. Ans..0025109375 lb. 16).642800 dr. 16).040175 oz..0025109375 lb. 1. Reduce 2{ years to seconds. Ans. 70956000 sec. 2. 49 hours is what part of a week? Ans. 5-2 wk. 3. Bring 1 circumference to seconds. Ans. 1296000". 4. 25" to the decimal of a degree. Ans..006940 5. 17.0625 rd. are how many inches? Ans. 33788 6. Bring 4A ft. to chains. Ans.'7 chain. 7. Reduce 192 sq. in. to sq. yd. Ans. -4 sq. yd, RE VIEEw.-219. Does the rule apply to fractions? How can th'e operation be shortened? In reducing common or decimal fractions, what rules must be borne in mind? An8. The rules for the multiplication and division of common and decimal fractions. 154 RAY'S IHIGHER ARITHMETIC. 8. 6 cu. yd. to cubic inches. Ans. 311._040 cu. in. 9.:'P 117.1:4 to mills.. 117140 mills. SuGGESTo N. -.In all reductions of U. S. money, or aty SySteml formed on the decimal basis, t.he multiplicatiolls or divisionis are readily performed by moving the point to the r1ighlt or left, siorce the multipliers and divisors are 10, 100, 1000, &c., (Arts. 1.44 and 14-)4' 10. Reduce 6.19 cents to dollars. Ans.,o0(;19 11. 1600 mills to dollars. Ans J 1.60 12. $5- to mills. Anzs. 5375 mills. S UGESTIO N.-Change - to a decihnal, then move the point. 13. Reduce 12 lb. av. to lb. Troy. Ais. 14-7 lb. SocTIrtoN. -The 12. lb. av. are first redtuced / X /T X > 1 17;5 14-7 to gr. by the table in 175,() 12 Art. 194, anrd the tr. to 4~s 12 Troy lb. by same table. 14. Reduce 33 beer gal. to wine gal. Ails. 40, w. gal. 15. 36 yd, to ells Flemish. Ants. 48 E. Fl. 16. 1 in. to qr. Ails. i? qr. 17..1G gr. to oz. Troy. Anzs..0004.,) oz. Tr. 18..0732, to pence. Ans. 17.568 d. 19. v lb. to tuns. Ants. a -o T. 20. 47.3084 sq. mi. to P. Ans. 4844380.16 P. 21. 4 D to lb. Ants. o lb. 22. 7 dr. av. to lb. As. 31 lb. 23. 50 U. S. bu. to imp. bu. Ans. 48.47236 nearly. 24-. 1200 inches to chains. A ns. 1 ch. 25. 99 yd. to furlongs. I,/O,o fur. 2G. 6.34:19 C. to cu. in. Ans. 1402726.8096 cu. in. 27. 1.8 fathoms to miles. Ants.:!i.-T 1 Iit. 22. HIow many acres in a rectangle 24 l'rd. lonl.y 16.02 rd. wide? Ans. 2.4;5306(5 ac Cs. 29. How many cubic yd. in a box (36 ft. ],tl- b/, 2t ft. wide and 3 ft. high? Ans. I cu'. t REs vIJE -219. how can reductions in U. S. money bo performied? REDUCTION OF COMPOUND NUMBERS. 155 30. -low many perches in a rectangular field 18.22 chai'ns long by 4.76 cli. wide? Alus. 1387.6352 P. 31. lReduce 256 3 to dr. av. Ans. 561}' dr. av. 32. 1 nautical league to feet. Aas. 1825 6.7088 ft. 33. 16.02 chains to iles. s..20095 mile. 34. 4.29 chains to feet. Ans. 283.14 ft. 35. 4 of a link to rods. AAns. l4, rd. 36. 4 of a nail to ell Engiish. A7Es.' E. E. 37. 1.644 inches to ell Flemish. Als..OG608 E. Fl. 38. 35.781 sq. yd. to sq. in. Alos. 46372.176 sq. in. 39. 256 roods to sq. chains. Ans. 640 sq. ch. SUCGESTIO. —First bring to acres, then to sq. chains. 40. 13' tuns of round timber to cords. Anzs. 41 C. 41. 6.1_5 tuns of hewn timber to reduced feet, plank measure, 1 inch thick. Alos. 369,0. 42. ITow many perches cf masonry in a rectangular solid iwall 40ft. long by 7~ ft. hilgh and 22 ft. average thickness? AIs. 82S 9'. 7 ab. Troy to -r. 4 Reduceb. Tsoy to gr. s. 1040 gr. 44. 8 pwt. to lb. Als. a'I lb. 45. HIow many oz. Troy in the Brazilian Emperor's diamond, which weighs 16-80 carats? Ans. 11; oz. 46. Reduce 75 pwt. to 5. Alts. 30 3. 47. - gr. to 3. As. RSi o0 3. 48. 19 cwt. to oz. Anls. 30400 oz. 49. 22 dr. to T. Ans..000044140625 T. 50. 184 3 to dr. av. Anls. 414 dr. 51. 96 oz. av. to oz. Troy. Ans. 87. oz. Tr. 52. HIow many wine gal. in a tank 3 ft. long by 2c ft. wide and 1. ft. deep? Ats. 75- w. gal 53. Ilow many U. S. bushels in a bin 9.3 ft. long )y 83 ft. wide and'2. ft. deep? Ais. 61 bu., nearly. 54. Reduce 21 bbl. of beer (36 gal.) to bbl. of wine, (31 g-al.) lAs. 294 55. 1 bu. to wine gal. Ans. 9.31 gal., nearly. 156 RAY'S HIGHER ARITHMETIC. 56. 1 bu. to beer gal. Ans. 7.625 +-gal. S U o G E sTIO N.-Reduce to cu. in., and then to gallons. 57. 45 f3 to 0. Ans. -4. 58. 2'f, to m. Ans. 1200 nt 59. If - of a piece of gold is pure, how many carata fine is it? Ans. 20} 60. In 184 carat gold what part is pure, and what part alloy? Ans. 2; and ~:x CASE II. ART. 220. To reduce a compound number to a simple number of any denomination. RuIE. —Commence with the highest denomination, if it be Reduction Descending; with the lowest, if it be Reduction Ascending. Reduce those of that denomignation to the denomination required, and during this reduction add in at the proper times those of the other denblminations. NoTES.-1. If the denomination required lie between the highest and lowest of those given, reduce part of the compound number by Reduction Descending, and the rest by Reduction Ascending, and add the two results. 2. The numbers added in must be of the same denomination as those to which they are added; mistakes can be avoided by marking the denomination of each number as it is obtained. Reduce 5 lb. 2 oz. 13 pwt. 10 gr. to grains. OPERATION. SoLnTION.-Since this is lb. oz. p vt. gr. Reduction Descending, com- 5 2 1 3 10 mence at the 5 b., and reduce Ii 2 b. oz. it to 60 oz.; the 2 oz. added in 62 oz.= 5 2 made 62oz.; this is reduced to 20 pwt., and 13 pwt. added in, mak- lb. OZ. pwt. ing 1253 pwt.; this finally is 12 5 3 pwt.= 5 2 13 brought to gr., and the 10 gr. 24 sadded, producing 30082 g~. for 5 0 2 2 the answer. 2 5 0 6 lb. oz. prwt. gr. 30082gr. —5 2 13 10 RE VTIE w.-220. What is Case 2d? The rule? If the required denomination lies between the highest and lowest given, what is necessary? What care must be exercised in the additions? REDUCTION 01 C0OiIMPOUND NUMBERS. 157 Reduce 2 cwt. 3 qr. 18 lb. to tuns. S O UT I O N. —This being Re- OPERATION. duction Ascending, commence 25 lb8.0 with 18 lb. and reduce it to.72 qr., making 3.72 qr. alto- qr. gether; this is reduced to.93 of a cwt., making 2.93 cwt. in cwt. cot. qr. lb all; and this finally is reduced 20 2.9 30 - 2 3 I8 to.1465 T.-the result required. T. cwt. qr. lb. The successive quotients might.1465=2 3 18 be put in the form of common fractions, if it were desirable. For convenience, the different simple numbers that compose the compound number, are set in a column, the lowest at the top, and the others under it in order, so that each quotient, as it is obtained, can be placed beside those of its own denomination. If any of' the denominations are vacant, a cipher must be placed in the column to correspond. Placing each quotient beside the number of its own denomination is the adding in required by the rule; just as the 2 oz. 13 pwt. and 10 gr., are acdded in in the first example. REMrAnRK.-The rule may be stated thus: Reduce each of the simple numbers that comtpo;se the comlpouzld number, to the requZired denomination, and add the results-but in practice it is not so convenient as the one given. EXAMPLES FOR PRACTICE. 1. 4 cu. yd. 5 cu. ft. 256 cu. in. to cu. ft. S o L U T I ON. —4 cu. yd. 5 cu. ft. = ]13 cu. ft., and 256 cu. mn.= 256 X 1-sg cu. ft. =-4 cu. ft. Ans. 1134 U. ft. 2. 1 mi. 3 fur. 38 rd. to rd. Ans. 5598 rd. 3. 8rd. l ft. to in. Ans. 1602 in. 4. 43ft. 5in. to:d. A -s. 21,9 5. 9mi. 22rd. 10.6175ft. to fur. Ans. 72.566+ SUGGESTION..-When the divisor is 5,, 30k, &c., multiply both dividend and divisor by the denominator of the fraction (2, 4, (&c.), and then divide. The quotient will not be altered. 6. 464 yd. 2 ft. 82 in. to miles. Ans..26415+ R E VIE.-220. Explain Example 1st. What convenient fornl is adopted for Reduction Ascending? If any denomination is vacant, what is necessary? How might the rule have been stated more simply? 158 RAY'S H1CIHER Allii)M'l'IC. 7. 1 mi. 3 fur. 7.2 in. to yd. A1Zs. 2420.2 yd. 8. 29 fathoms 47 ft. to rmiles. Ans.. 03378 mi. 9. 17 yd. 3 qtr. 2 na. to nla Az.o.. 286 h 8. 10 5 yd. 4 in. to ells English. AIs. 4o0S0) _1 TE oe 11 44 E. Flem. 2 na. 2 in. to qr. A)s. 182o7i 2 q. 12. 1 qr. 2 na. 1.785 in. to -na. A4s. 6 7) t 13. 29 E. Fle. l r. 2 na. to yd. is 2? 1 5b yd. 11. 75 yd. 3 qr. 22 na. to feet. 4ls. 2.7 7t. 15. 4 sq. rd. 13 sq. yd. 5 sq. ft. 98 sq. in. to sq. in. Anzs. 174482 sq. in. 16. 7sq. ft. 120.54sq. in. to sq. yd. Als..870817. 63 sq. rd. 10 4sq. in. to sq. ft. A ts. 17151, 3- 18. 1650 A. 3 R. 24.64 P. to sq. miles. Aos. 2.5795375 sq. mi. 19. 301A. 1R. 18 P. to sq. ch. Ans. 3013.137;7 20. 1 sq. mi. 424 A. to' perches. Ais. 170240 P. 21. 21A. 35 P. to sq. chains. A),s. 212.1875 sq. c1. 22. 1 cu. yd. 24 cu. ft. 8 7 6 cu. in. to cu. yd. Alos. 1 3s 23. What part of a cord of wood is a pile 711 ft. lonE by 5 ft. wide and 17 ft. high?,los 52IC),5 24. 83 cords 115 cu. ft. 1600 cu,. in. to tuns of round timber. As. 268 - T SUGGESTION.-First to cu. ft.; then to tuns of round timber. 25. 7 tuns of hewn timber 32 cu. ft. 480 cu. in to cords. Aos. 2374 C. 26. lb. 7 oz. 14.238:pwt. to gr. Ans. 20981.6 r. 27. 19 put. 6 gr. to lb. Alns'-.7o lb. 28. 4 lb. 221 gr. to oz. Ans. 48T -j 0 oz. 29. 77oz. 12pwt. 10.464gr. to lb. Ans. 6.468483. 30. 68 lb. 8 oz. to gr. Alns. 395808 gr. 31. 49 carats 2-7- gr. to oz. Troy. Ants. 7-27- oz. Tr. 32. A diamond weighing 3 oz. 16 pwt. is how roana1 @arats? lAs. 573J carats: 33. 3 b 7 12.24 gr. to 3. Ans. 86.9005 3 34. 9` 1 z D0 to gr. lAos. 4350 gr. 35. 23 29 141 gr. to. IAs..363,'~3 3l6. 1 lb-sgr. to 5. Ans.. 3 03 REDUCTION OF COMi1POUND NUMBERS. 159 37. 45 2D 15.5 gr. to lb. Ans..051302-,1b. 38 10 T. 9 cwt. 2 qr. 23 lb. to lb. Anls. 20973 lb. 39 8qr. 181 b. 15 oz. 10.08 dr. to cwt. A s..93 9711 40. IT. 6cwt. 1qr. 24 lb. 2oz. 4 3dr. to lb. Ans. 2649.-14,i 41. 1 qr. 12' oz. to tuns. Ans..012829 T. 42. 762 lb. 8 oz. 3 dr. to cwt. Ains. 7.62$ 5,ct. 43. 12 lb. av. 9oz. 10 dr. to Troy gr. Ans. 88210 1 gr. SuG.-First reduce to dr.; then to gr. by 27 1. (Art. 188.) 44. 6.33 5 1 D 15.232 gr. to lb. av. Ans..442176. 45. 5 5 2 D 7 gr. to pwt. Ans. 14,) pwt. 46. 15 lb. Troy, 11 oz. 4 pwt. 9.085 gr. to dr. av. Ans. 3356.71168 dr. av. 47. 6 oz. 10.48 pwt. to 5. Arns. 52.192 5. 48. 37 gal. 2 qt. 1 pt. 3 gills to qt. Anzs. 150 3 qt. 49. 4 gal.. pt. to gills. Ans. 130 gills. 50. 3 qt. 1 pt. 2.3 gills to gal. Ans..946' gal. 51. 52 gal. 1 qt. 1.052 gills to pt. Ans. 418.263 pt. 52. 1 gal. 3 qt. 1) gills to gal. Ans. 1 I gal. 53. 47 bu. 3 pk. 2 qt. to pt. Ans. 3060 pt. 54. 2_pk. 6qt. 1.8 pt. to bu. Ans..71-T, bu. 55. 3 bu.' 2 pt. to pk. Alls. 12 2 pk. 56. 3 pk. 1.093 pt. to bu. Ans..767-, bu. 57. 2 pk. 7-,) qt. to pt. Azns. 47 pt. 58. 8 hbu. 3 3 pk. to qt. Ans. )86 qt. 59. 286 bu. 3 pk. 1 qt. to imp. bu. Ans. 2 78.02Suo.-First to U. S. bu.; then to cu. in.; then to imp. bu. 60. 99w. gal. 1 qt. I pt. to imp. gal. Ans. 82.7904. 61. 67beer gal. 3qt., 1.94 pt. to dry qt. Alzs. 285.32567 — 62. 56 w. gal. l qt. 2) gillstobeerqt. Ans. 184.603 63. 13bu. 1 pk. 7qt. 1.35pt. to w. gal. Ans. 12.258-. 64. 27 gal. 3 qt. 1 pt. of beer to gills. Anls. 108 -S'~.., _.. _... _ _ _..O RE VsE w.-220. IS it as convenient in this form? When the divisor ia. 65, 301, &c., what should be done? B0 RAY'S HIGHER ARITHMETItC. 65. 14f, 5 f, 48 m to Cong. Ans..115039,!' Cong. 66. 1 0 3f3 6',f3 to m. Ans. 9510 rn. 67. 3f3 35rm to fz. Ans.' ft. 68. 20 7f- 4f3 58 in to f3. Ans. 39.62 f3. 69. 2 yr. 108 da. 1S hr. 40 min. to sec. Ans. 72470400 sec. 70. 6 yr. 44 da. 8 hr. 35 min. to wk. Ales. 319 a3'J 71. 21 hr. 4 min. 54.6 sec. to da. Ans..87840972 da. 72. 3 wk. 5 da. 12 hr. 26 min. 11s see. to hr. Ans. 6 8 6 i-~o hr. 73. 74 da. 16 hr. 45 min. 23.028 sec. to yr. Ans..2046525567, nearly. 74. ITow many sec. in a solar year? Ans. 31556928. 75. 90~1.6" to minutes. Ans. 540.226' 76. 42' 574" to degrees. Ans..71597L~ 77. 163~ 28' 7" to seconds. Ans. 588487". 78. 7s 120 6' 20" to degrees. Ans. 222.105~ 79. 4' 32.756" to circuim. Ans..00021046, nearly. 80. $76 5et. 23 m. to mills. Ans. 76052381. 9 ct. 9.1054 mlills to $. Ans. $.099105.4 82. $8391 7mills to cents. Ans. 39100.7 ct. 83. $84 32' ct. to mills. Ans. 843257 m. 84. 20 eagles 86 1 dime 21 ct. to $. Ans. $2061 85. 4 eagles 3.142 mills to dimes. Ans. 400.03142 86. 5 dimes 9 ct. 6 mills to eagles. Anzs..0596 87. 3804 19s. 2Ad. to ~. Ans. ~304 -' 88. 12s. 10?d. to s. Ans. 12:4s. 89. ~58 7s. lid. to d. Ans. 14015d. 90. ~25 44d. to s. Ans. 5001'ss. 91. Reduce 1 cwt to pwt. Ans. 291663 92. How many ounces of gold weigh as much as t pound of lead? Ans. 14 — CASE III.-TO REDUCE A SIMPLE NUMBER OF ANY DENOMINATION TO. A CO-MPOUND NUMBER. ART. 221. If the given number is a whole number, it may be reduced to a compound number by this REDUCTION OF COMPOUND NUMBERS. 161 RvulE. —Redluce the cqi'enl 111(nmer to the next inyher denzominlationi, 1'eerciCu thle 1 remainlder; reduce the quotietl to the ntext hi bher denomiatlion, and reserve the reemainder. Cotlinze thus unttlil the highest denlominiation has been reached, or utitil the quotielnt is so small as not to admit of fJirther redutclionl. The last quotieCnt with the several renmainders, form the compound nulember required. NoTE.-Each remainder is of the same denomination as tile dividend from which it is obtained. R E t A R I.-The operations under this and the preceding rule serve to prove each other. ART. 222. If the given simple number is a common or decimal fraction, it may be reduced to a compound number of lowocr denominations by this RIUlE. -Jiede lce giveni fraction to the?ext lower denzomination, reserting the wchole ntummber, if a.ny, of that delnom7iation. 1j' there is a Jiactioin i7n this resvullt, re(duce it to the ntext lower nce'io.ilotioll, re'serv'e'ig the whole?lnumber, if any, of that delromninationz. C'nitinue this process unt2lil no Jf'ctlionL occlr'S, or unttil the lozwest denominaion has been reached. PTe whole 1num7bers reserved, with the last fractionl, if anly, will be the coij)lound 7tnum ber irequ irecd. NorE.-If the given simple number be a mixed number, the w]iole number whiclh it contains nmay be reduced to a compound nunmber by the first of the rules given above, and the fraction by the last rule, aLnd the two results united. Reduce 1 706 inches to a compound number. So UTr oN. —The operation is 19)1 06 in. sin-ilar, to thlat in the 2d example, Art. 218, except thait each 3) 1 4 2 ft. 2 in. reinaindler is written as a whole 4 7 yd. 1 ft. 2 in. numnber of its own denomination, instead of a fraction of the next higher denomination; as 142 ft. 2 in. itnstead of 142, ft., and 47 yd. 1 ft. instead of 47 yd. 1n vt E wiv. —221. What is Case 3d? What is the rule for reducing a simple wlihle number to a compound number? IIow is the denomination of cieh remnaiinlder known? Ilowv are operations undler t}his trule eritifed7 222. W\hilit is the rule for reducing a simp,!le fraetiu:nal nunliber to i) eopoundl number? Hlow is a simple mixed number reduced to a compound nulmber? 14 162 RAY'S HIGHER ARITHMETIC. Reduce a of a wine,allon to a compound number. w. gal. cqt. pt. gill So L U T I. —Multiply gal. by 4; the result is 22 qt.; cut off whole number 2 2 13 as part of the compound number required. 4 Multiply a qt. by 2; the result is 11 pt.: qt. 2 cut. off the 1 pt. and reduce the I pt. to 1l 2 gills. This being the limit of the table, pt. 3 the operation must stop, and the compound 4 number required is 2 qt. 1 pt. 1 gills. gills 1 Reduce 4905.06185 lb. av. to a compound number. 25)4905 lb..06185 lb. 4)196 qr. 5 lb..9896 oz. 20)49 cwt. 16 2 T. 9cwt. 51b. 15.8336 dr. Ans. SoLU T IO N.-The fraction.06185, reduced to lower denominations, is 15.8336 dr.; thle whole number 4905 lb. reduced to higher denomi. nations, is 2 T. 9 cwt. 5 lb. These results united give 2 T. 9 cwt. 5 lb. 15.8336 dr. for the answer. Reduce 1657 yd. to a compound number. SOLUTION. —Divid te 5t le 1 6 5 7 yd. 1657 yd. by 51i, to reduce it to 2 2 rd. To do this conveniently, multiply both dividend and 11 ) 3 314 half yd. divisor by 2, making the di- 4 0) 3 0 1 rd. 3 half yd. visor 11, and the dividend7 fur. 21 rd 3314 htlf-vards, without al- 7fur. teiiung the quotient. The re- But, yd. = 1 ft. 6 in. mainder 3 is also half-yards Ans. 7 fur. 21 rd. 1 yd. 1 ft (Art. 221. Note), andt is there- 6 in. fore written yd. = 1-1-yd.; the, yd. is then reduced to a compound number by the last rule, and joined to the result already obtained. REDUCE. TO COMPOUND NU3IBERS, LXAmPLEs. ANxswERS. 1. 44753A ft... = 8 mi. 3fur. 32 rd. 5 ft. 6 in. 2. 99.75yd..... =18 rd. 2 ft. 3 in. 3. -fir..... 33 rd. l yd. 2 ft. 6 in. REDUCTION OF COMPOUND NUMBERS. 163 EXAMPLES. ANSWERS. 4. 8760531in.. =138 mi. 2 fur. 5 rd. 1 ft. 9 in 5. 904.178 fath. =904 fath. 1.068 ft. 6. 8873 na. = 55 yd. 1 lqr. 3 3 na. 7. 69.1525 yd. =69 yd. 2.44 na, 8. 60l1 jlj qr. 150 yd. I qr. -3 na. 9. 52.003 E. Fl. -52 E. Fl..036 na. 10. 1811.0625 sq. ft. = 201 sq. yd. 2 sq. ft. 9sq. in. 11. 300027sq. in. 231 sq. yd. 4 sq. ft. 75 sq. in. 12. 64.10826P..=1R. 24.10826P. 13. 8325 sq. yd. = 832 sq. yd. 2 sq. ft. 151 sq. in. 14. 1.10475 sq. mi... 1 sq. mi. 67 A. 6.4 P. 15. 322.7372A.... = 322A. 2R. 37.952P. 16. 706.2814 sq. ch.. =70A. 2 R. 20.5024P. 17. 1567804sq. in. = 1209 sq. yd. 6 sq.ft. 76 sq. in. 18. 87 sq. ft... 9s q. yd. 6 sq.ft. 90 sq. in. 19. 9311 A... =1 sq. i. 291A. 2R. 5P. 20. -3 sq. mi.. = 22 A. 2 R. 14 " P. 21. 9 sq. rd....=16 sq. yd. 7sq. ft. 36 sq. in. 22. T cu. yd. 17 cu. ft. 314-', cu. in. 23. 1013854 cu. in. = 21 cu. yd. 19 cu. ft. 1246 cu. in. 24..0038yr... =da. 9 hr. 17m min. 16.8 sec. 25. 65.387 cu. ft. = 2 cu. yd. 11 cu. ft. 668.736 cu. in. 26. 4.2045 cu. yd. =4 cu. yd. 5 cu. ft. 901.152 cu. in. 27. 18.9142mi. = 18mi. 7fur. 12rd. 2 yd. 2ft. 11.712in. 28. 40152387 min. (3651 da. to a yr.) Ans. 7 yr. 231 da. 14 hr. 38 min. 52 sec. 29. 13'- C. 13 C. 1=. 0 ou. ft. 30. 8.5646T. hewn timber... =T. 28ecu. ft. 31. 4000 gr. Troy.. = 8 oz. 6 pwt. 16 gr, 32. Troy. = 78 lb. 8 oz. 11 pwt. 10} gr. 33. 267.3 pwt. = 1 lb. 1 oz. 7pwt. 7.2 gr. 34.' -oz. Troy. 15 pwt. 9- gr. 35. 45.54 oz. Tr. 3 lb. 9 oz. 10 pwt. 19.2 gr. 36. 692 pwt.... 2 lb. 10 oz. 12 pwt 164 RAY'S HIGHER ARITHIIMETIC. ]EXAMPLES. ANSWERS. 37. 1i.7644lT... =13Tb. 913 13 2.944gr. 38. 805,5... 8 b.. 43 55 13 9 gr. 39. e.......5 1 1 3 gr. 40. 905629...... 14:). 5= 35 13. 41. Tlb...8 3 1 5.. 13 7 gr. 42. 170053.62gr. Apoth. 291lb. 63 25 13.62gr. 43. 4673.....1lb 75 35 2D 12gr. 44..37015...=1D 2.206 gr. 45. T.. =Ocwt. 2 qr. 16 lb. 10oz. 10] dr. 46..4815 lb. av. -- oz. 11.264 dr. 47. 809 " cwt. 40 T. 9 cwt. 3 qr. 16 lb. 1O oz. 10` dr. 48. 4.22031 lb.=.211 T. 1 qr. 6 lb. 49. 733 qr...... 9T. 3cwt. lqr. 181b. 12oz. 50. 7cwt. = 2qr. 20 lb. 13oz. 5, dr. 51. 8 gal -8 gal 3l gills. 52. 10072 gills... = 314 gal. 3 qt. 53. 952 qt...... 3 gal. 3qt. 17 gills. 54. 301.46 pt.. = 37 gal. 2 qt. 1 pt. 1.84 gills. 55. 808 qt. dry measure.. =25bu. 1 pk. pt. 56. 2191009.3dr.= 4T. 5 cwt. 2qr.81b.10 oz. 1.3dr. 57. 5600523 z. av.=17 T. 10 cwt. 3 lb. 4oz. 13 9 dr. 58. 365.2422414 days. Ans. 365 da. 5 hr. 48 min. 49.65696 sec. 5 9....3 6s. tad. 60. 33........... $833. 6255. 61.' wk....4 da. 16 hr. 62..... f.. 5 f 3 6 m. 63. c hr..... min. 45 sec. 64.: bu. I pk. l6 pt. 65..555~. 1....s. 1s d. 66. al.= 2qt. qt1 pt. 67. 67.76s. =~3. 7s. 9.12d. 68. 27 S.350. o 21'. 69. 32.4 0... 4 4Cong. 6 f5 3 f5 12 tit 70.' Cong. = 3 (). 3 f3 1 f5 36 in. REDUCTION OF COMPOUND NUMBERS. 165 EXAMPLES, ANSWERS. 71.0 7 5 qt.......... =.6 gill, 72. 3.07 pk....... D 3 pk. 1.12 pt. 73. 469f53. ~...=55f 6 f 57n. 74. 260234"...... 7 =20 17' 1 4". 75. 1246.05'... 2..0 46' 3". 76. 1856d...... 7 14s. 83d. 77. 28 98 S. =~140 19s. 1 d. 78..8054 bu..... = 3 pk. 1 qt. 1. 5456 pt. 79. 100000m.... =130 2 f3 40m. 80..1934 si(gn.. = 50 48' 7.2'" 81. 75 dims.,..... $7.57 et. 5 m. 82. 17.052pk. =4bu. lpk..832pt. 83. 2'0 circum. 2 c. 327~ 16' 21,-. 84. 19019.2 2m0. 2 0. 7 f5 4f5 59.2m. 85. 84312 mill s.. = $84.312. 86. 2 9 of a dollar..... 44 ct. 6-' m. 87.. of an eale a. c e4. = 361 { 88. 2093.57 cents..... 20.9357. 89. -iaof adegree...... =28' 25 5-". 90. 640 7-0 pt. dry meas.. =100bu. 3qt. 1 pt. 91. 10808107.87 sec. =125 da. 2 hr. 15 min. 7.87 sec. 92. 6.045964 yr. =6yr. 16da. 18hr. 38min. 40.704sec. 93. 17000.12 da. (3651 da. to a yr.) Ans. 46 yr. 198 da. 14 hr. 52 min. 48 sec. 94. 22303.5d. Ans. ~92 18s. V7d. PREARtK. -If our tables of Weights and Measures were on the decimal basis, like U. S. Money and the French tables, the same rules and methods would do for compound as for simple numbers. Common fractions would also occur less frequently, the comparatively complicated processes that arise from their use would be avoided, and the computations required in ordinary business transactions would be much shortened and simplified. As it is, Compound numbers must be treated somewhat differently from Simple numbers; though the rules and operations are not entirely new, but rather modifications of those already explained. 16(id6 RAY'S HIGHER ARITHMETIC. ADDITION OF COMPOUND NUMBERS. ART 223. RULE.- W'rite the numbers to be added, un2its of the same deno7miation in a columnz, reducing:any fractions to lower denominations, until none are found in any but the right. hand column. Add the right-hzand column, and77 reduce the result, jf' large. enough, to the ltext higher denomination; write the remaizlder, if any, under the column added, anzd carry the quolieznt to the next column. Add the next colztmn, reduce, set dtown, and carry as before, and continue so until all the col2umns lhave been added. PROOF. —Same as in addition of simple numbers. NOTE. —If the right-hand column contain common or decimal fractions, add them according to the usual rules; if any of the higher denominations in the answer has a fraction, reduce it to lower denominations, and add it in. Add 3bu. 2l-pk.; lpk. l1pt.; 5qt. Ipt.; 2bu. 13 qt; and.125 pt. SoLUTION.-Reduce the bu. pk. qt. pt. fraction in each number to 3 2 2 0 = 3 bu. 21 pIk. lower denominations, (Art. 1 0 1 = 1 pk. 1 pt. 222,) and place units of the 5 1 5 qt. 1 pt. same kind in columns. The 2 0 1 = 2 bu. 1A qt. right-hand column, when 1.125 pt. added, gives 3 7 pt. —1 qt. 6 0 1 17 = Ans. 124 pt.; write the 14 and add the 1 qt. with the next column, making 9 qt. 1 pk. 1 qt.; write the 1 qt. and carry the 1 pk. to the next column, making 4 pk.-= 1 bu.; as there are no pk. left, set down a cipher and carry 1 bu. to the next column, making 6 bu. Add 2rd. 9ft. 7i in.; 13 ft. 5.78in.; 4rd. 1lft. 6in.; 1 rd. 10] ft.; 6 rd. 14 ft. 6 in. rd. ft. in. SoLUTION.-The numbers are pre- 2 pared, written, and added, as in the last example; the answer is 16 rd. 3 5 8 9, ft. 9.655 in. The A1 foot is then re- 4 11 6 duced to 6 inches, (Note), and added to the 9.655 in., making 15.655 in.= 6 14 6 1 ft. 3.655 in. Write the 3.655 in., 16 91 9.655 and carry the 1 ft., which gives 16 rd. but' ft. = 6. 10 ft. 3:655 in. for the final answer. 6 10 3.65 16 10 3. 655 ADDITION OF COMPOUND NUMBERS. 167 1. Add mi.; 3 fur. 263 rd.; 10 mi. 14 rd. 7 ft. 6 in.; 5.24 fur.; 37 rd. 16ft. 2l in.; rlmi. 12ft. 8.726 in. Ans. 12mi. 4 fur. 20 rd. 9 ft. 4.633$ in. 2. 6.19yd.; 2yd. 2ft. 94 in.; lft. 4.54in.; 10yd. 2.376 ft.; - yd.; 1' ft.; in. As. 21 yd. 2 ft. 3.517in. 3. 3yd. 2qr. 3na. 1in.; lqr. 24na.; Gyd. 1na. 2.175 in.; 1.63 yd.; ~ qr.; na. Ans. 12 yd. 1 na. 0.755 in. 4. 4 E.Fr. 4qr. 2na.; 7 E. Fr. 5qr. 1 na.; 3qr. 3na. 1 in.; 7 E. Fr.; 1.6 na.; -"in. Ans. 14 E. Fr. 3 na. 7 in. 5. 2E. Fl. 1qr. 13a.; 5 E.F. 3na.; 2qr 2na.; E. P. 2 qr. 3.8 na.; E. F1. Ans. 11 E.Fl. 1 qr. l0 na. 6. 15sq.yd. 5sq.ft. 87 sq. in.; 16-1sq. yd.; 10sq. yd. 7.22 sq. ft.; 4 sq. ft. 121.6 sq. in.;'3 sq. yd. Ails. 43 sq. yd. 7 sq. ft. 37.78 sq. in. 7. 101A. 21R. 18.35P.; 66A. 1 R. 34.P.; 20A.; 12 A. 2.84R.; 5 A. 13.331 P. Ans. 205 A. 3 R. 19.78 P. 8. 23 cu. yd. 14 cu. ft. 1216 cu. in.; 41 cu. yd. 6 cu. ft. 642.132 cu. in.; 9 cu. yd. 25.065 cu. ft.; I -cu. yd. Ans. 75 cu. yd. 4 cu. ft. 1279.252 cu. in. 9. tC.; cu. ft.; 1000cu. in; Ans. 107 cu. ft. 1072 cu. in. 10. 2 lb. Tr. 61oz.; 1- lb. 12.68 pwt.; 11 oz. 13 pwt. 19' i,. 15 5 l9~gr.; lb., OZ.) pwt. Ans. 5 lb. Tr. 9 oz. 9 pwt. 2.853 gr. 11. 85 14.6gr.; 4.183; 7 2$,5; 23 2D 18gr.; 15 12gr.; -D. Ans. 1 M. 23 43 1D. 12. T T.; 9 cwt. lqr. 221b.; 3.06 qr.; 4 T. 8.764 cwt.; 3 qr. 6 lb.; 7 cwt. Ans. 5 T. 6 cwt. 2 qr. 14-3- lb. 13..3 lb. av.; oz.; 6 dr. Ass. 5 oz. 617 dr. 14. 6 gal. 38 qt.; 2 gal. 1 qt. 3.32gills; 1 gal. 2 qt. 1 pt.; gal; qt.; t. pt. Ans. 11 gal. 2 qt..46-} gill. 15. 4gal. 3 gills; 10gal. 3 qt. 1 pt.; 8gal. s pt.; 5.64 gal.; 2.3 qt.; 1.27 pt.;' gill. A1s. 29 gal. 2 qt..224 gill. RavIe w.- 222. What would be the ad;antages if our weights and mneasures were on the decimal basis? 223. What is the rule for addition of compound numbers? What is the proof? If the right-hand column contains common or decimal fractions, what must be done? If any of the higher denominations of the answer has a1 fraction, what must be dolnel 168 RAY'S HIGHER ARITHMETIC. 16. Add ] bu. +pk.; pc bu.; 3pk. 5(t. 11 pt.; 9bu. 3,28 pL.; 7 qt. 1.1 6 p t. Ais. 12 bu. 3 pk.. 463 pt. 17. Add 3 bu.; pk.; ~ qt.; 3 pt. Ans. 2 pk. 4 pt. 18. Add 6f 2f3 25m; 2+f5; 7f5 42m; 1f3 2 f3; 3f, 6'f3 51m. Ans. 14f3 7f3 38 tl. I)) 0I 1 19. Add +wk.; da.; d.; hr.; min.; sec. Ans. 4 da. 30 min. 302 sec. 20. Add 3.26yr. (365da. each); 118 da. 5hr. 42mmill. 3,7-)sec.; 63.4 da.; 74a hr.; lyr. 62 da. 19 hr. 24; lin.; 73, da. Ans. 4 yr. 340 da. I hr. 14 min. 551 sec. 21. Add 270 14' 55.24"; 9~ 18V"; 1~ 15'; 116~ 44' 23.8" Ans. 154~ 14' 57.29" 22. Add $84 1 et. 5.27m.; 67 ct. 8 m.; $25 9ct. 2g m.; ioaofadol.; 253ct. Ans. $110 60ct. 5.645 m. 23. $i;:- ct.; s m. Ans. 50 ct. 2-' m. 24. Add $3 7m.; $5 20ct.; $100 2ct. 6mi.; $19 c et. Ans. $127 23 ct. 4. m. 25. Add ~21 6s. 3Vd.; ~5 17'4s.; ~9.085; 16s. 71d.; 12-5. Ans. ~37 10s. 8.15d. SUBTRACTION OF COMPOUND NUMBERS. ART. 224. RuLE.-Prepare and write the nu1mbers as in addilion. of compound numbers, placing the subtrahend below: Comznmence at the riglht, and proceed to the left, subtbacli,ng each lower tlumber fr'om the one above, and settllng the remzainder below. If a lowerl numbter is larger than, fte- one above it, add to the upper as 1any tunits of its denomiZnation as make one of the next higher; subltract and carry 1 to the next figure. PRoo. -Same as in subtraction of simple numbers. NOT E.-If fractions are in the right-hand column, subtract them by the usual rules; if a fraction is in any of thle higlher denominations of the answer, reduce it to lower denominations, and add it in. Subtract 1 yd. 2.45 ft. from 9 yd. 1 ft. 6, in. SoLuTTON.-Change the in. to a decimal, yd. ft. in. making the minuend 9 yd. 1 ft. 6.5 in.; reduce 9 1 6.5 the.45ft. to inches, making the subtrahend 1 2 5.4 1 yd. 2 ft. 5.4 in. To subtract 2 ft. from the 7 1 1 number above, add 3 ft. ( = 1 yd.) to the 1 ft., making 4 ft.; set the remainder, 2 ft., below, and to compensate for Ul ITRtACTIOIN OF COMPOUND NUMBERS. 169 the, 3 ft. added above, add 1 yd. to the next lower figure, which gives 2 yd.; the rlemainder is then 7 yd. and the answer 7 yd. 2 ft. 1.1 in. 1. Subtract 16 rd. 8 ft. 1 in. from 23 rd. 1.2 ft. Ans. 6rd. 9ft. 7.15 in. 2. mi. from 3 fur. 24.86 rd. Arls. 16.86 rd. 3 1 35 yd. from 4 yd. 2 qr. I na. 14 in. Ats. 3 yd. iqr. in. 1. 2i E. Fl. from 2 E. E. Ans. 1 yd. 1 qr. 1in. 5. 2 sq. rd. 24 sq. yd. 91 sqin. in fom 5 sq. d. 16 sq. yd. 6] sq. ft. Ants. 2 sq. rd. 22 sq. yd. 8 sq. ft. 41 sq. in. 6..56 sq. yd. from 7 sq. ft. 18.27 sq. in. ALs. 2 sq. ft. 3.87 sq. in. 7. 2 R. 19' P. from 11 A. Ats. 10 A. 1 R. 20 P. 8. 384 A. I R. 3.92 P. from 1.305 sq. mi. Ans. 450 A. 3R. 28.08 P. 9.' cu. ft. from a cu. yd. Asis. 4 cu. ft. 1503 cu. in. 10. 13 cu. yd. 25 cu. ft. 1204.9 cu. in. from 20 cu. yd, 4 cu. ft 1000 cu. in. Ars. 6 cu. yd. 5 cu. ft. 1523.l cu. in. 11. 9.362 oz. Troy from 1 lb. 15 pwt. 4 gr. An?s. 3 oz. 7 pwt. 22.24 gr. 12. 5 oz. 15 pwt. from 3 lb. 22 gr. Ans. 2 lb. 6 oz. 5 pwt. 22 gr. 13. o 35 from At's. 3 3 1 5 9-5- gr. 14 231D6gr. from 43gr. Arns. 3 531 142gr. 15 56 T. 9 cwt. 1qr. 23 lb. from 75.004 T.' Ans. 18 T. 10 cwt. 2 qr. 10 lb. 16 1 wt. from 3 qr. 11 lb. 14oz. 10 -dr. A.s. 2 qr. 5 lb. 15 oz. 6a3dr. 17. 5 lb. Troy from -Alb. av. Ans. O. 18. 3 gal. 2 qt. 1 pt. from 8 gal. 1.1 qt. Ans. 4 gal. 2 qt., pt. 19. 12 gal. 1 qt. 3 gills from 31 gal. 1' pt. Arts. 18 gal. 3 qt. 3 gills. 2G.0625 bu. from 3 pk. 5 qt. 1 pt. Ans. 3 pk. 3 qt. 1 pt. Pt E v E w.-224. What is the rule for subtraction of compound numbers? M hat is the proof? If fractions are in the right-hand column, what must be done? What, if a fraction is in any of the higher denominations of the answer? 15 170 RtAY'S HIGHER ARITIHMETIC. 21. 2 pk..84 pt. from 3 bu. 41 qt. Ats. 2 bu. 2 pk. 3 qt. 1.GG6 pt 22. 15 wine gal. from 15 beer gal. Anls. 3~} w. 7 gal. = 3 w. gal. I qt. qt. 1 gills. SuoaESTION.-ReduCe the beer gal. to w. gal. (Art. 194.) 23. 10 U. S. bu. from 10 imperial bu. of G. Britain. Alas. lpk. 2 qt. 0.17+ pt. SvuUGEaTIN.-Redluce the imp. bu. to U. S. bu. (Art. 194). 24. 1 f: 4 f3 38 in from 4 f3 2 f5. A is. 2 fS 5 f53 22 tnm. 25..9 of a day from 4 wk. Aas. 20 hr. 24 min. 26. 3 da. 16 hr. 47 min. 33.3 sec. from 1 wk. Ans. 3 da. 7 hr. 12 min. 26.7 sec. 27. 275 da, 9 hr. 12c nin. 59 sec. firon 2.4 816 yr, (allowing 36541 days to the year.) Ans. 1 yr. 265 da. 18 hr. 29 min. 21.16 sec. 28. 1832 yr. 8 mon. 18 da. fiom 1840 yr. 5 mon. 26 da., considering 1 mon. to be 30 days, as is customary in business involvinr time. Ans. 7 yr. 19 mon. 8 da. 2'.). What is the difference of time between Aug. 5th, 1848, and Mar. 14th, 1851? Ans. 2 yr. 7 mon. 9 da. SU ESTION.-Proceed as in last example, writing Aug. 5th in the subtrahend, as 7 mon. 5da., and Ilarch 14th in the minuend, as 2 mon. 14 da., since these dates are respectively thalt long aftei the beginning of the year; allow 30 days for a month. 30. Find the difference of time between Sept. 22d, 1855, and July 1st, 1856. Ans. 9 mon. 9da. 31. Between De. 31st, 1814, and April 1st, 1822, A ns. 7 yr. 3 m o n. 32. Between IMay 20th, 1855, and Oct. 15th, 1857. Ans. 2 yr. 4 mon. 25 da. 33. Subtract 430 18' 57.18" from a quadrant. Ans. 46~ 41' 2.82" 34. 17~ 29' from 24~ 521". Ans. 60 31' 52". 35. 1610 34' 11.8" from 180~. Ant. 180 25' 48.2" 36. $12.857 fiom $19. Ans. $6.143 37. I mills from $40. Ans.. $39.996 R E V I E w.-224. How is the difference of time between two dates found? MULTIPLICATION OF COMPOUND NUMBERS. 171 38. 1 ct. from $1 and 1 m. Ans. 99 et. 1 m. 39. 86ct. and.6 mill from $2.62~. Ans. $1. 7644 40..08 dime from 1 ct. 8 mills. Ans. 1 ct. 41. {a ct. f om. Ans. 8 et. 4' m. 42. $5 43 ct. 2 i m. from $12 6 ct. 8] m. Ans. $6 63 ct. 56 m. 43. ~9 18s. 61d. from ~20. Ans. ~10 is. 5d. 44. As. from 4e s.. d. MULTIPLICATION OF COMPOUND NUMBERS. ART. 225. Since every compound number can be reduced to a simple nuimber of' either of its denominations (A rt. 220), the multiplication of a compound number will only differ from the multiplication of a simple number by the reduction before and after multiplying. GENERAL RULE. TO MULTIPLY A COMPOUND NUMIBER BY ANY SIMPLE NUMBER WHIOLE OR FRACTIONAL, Rledlce the compoualnd to a sinple number of either of its denolinzationts, and mtulliply as in simple nmbers. The product will be a siimple n27mber of the same denonmination as the multiplicand, and may be reduced to a compounld ntmbe'r. NOTE.-It is generally best to reduce the multiplicand to its lowest denomination. Multiply 2 bu. 3 pk. 7 qt. 1'pt. by 10.8724. Bu. pk. qt. pt. 1 0.8 724 2 3 7 14 =1914pt. 4 a'27181 1.1 pk. 108 724 8 97851(6 95 qt. 108724 2 2)2 0 7 9.34 6 5 pt. 191 Pt. 8)10 3 9 qt. 1.3465 pt. 4) 1 29 pk. 7 qt. 3 2 bu, 1 pk. Ans. 32 bu. I pk. 7 qt. 1.3465 pt. 172 RAY'S HIGHER ARITHMETIC. ART. 226. If the multiplier is a whole number, the redtuctions may take place during the multiplication, instead of bcjbre and aJter it. TO MULTIPLY A COMPOUND NUMBER BY A SIMPLE WHOLE NUMBER, lu.E.-Begin at the lowest denomnination; mzltiply each of tie simple numbers that comvpose the co.mpound number in succession: reduce each prodtuct to the next higher, denomi2nation, se/tinlg the remainder below the number mtltiplied, and carrying the quotient to the 6ext PRODUCT. PnooF.- Same as in multiplication of simple numbers. NoTES.-i. If the multiplier is a composite number, we may multiply by its factors in succession, as in Art. 53. 2. When the multiplications and reductions can not be readily performed in the mind, do the work on one side, and transfer the results. 3. If there be a fraction in the lowest denomination of the multiplicand, multiply it first; if one occurs in any of the higher denominations of the product, reduce it to lower denominations, and add it in. Multiply 9hr. 14min. 8.17 sec. by 10. SOLUTION.-Ten times 8.17sec. hr. min. sec. -81.7 sec.=m1 min. 21.7 sec. Write 9 14 8.17 21..7 sec. and carry 1 min. to the da. 10 140 min. obtained by the next mul- 3 20 21 21.7 tiplication. This gives 141 min.2 hr. 21 min. Write 21 min. and carry 2 hr. This gives 92 hr.3 da. 20 hr. Multiply 12A. 3 RP. 2 8 P. by 84. A. R. P. So L U T I O N.-Since 84 —-7 X 12, multiply by one af these factors, and this product 12 2 by the other; the last product is the one 7 required. The same result can be obtained 90 1 388 by multiplying by 84 at once; performing 12 l he work on one side and transferring the results. REVIEmw. —225. What is the rule for multiplying a compound number by a simple number, whole or fractional? To which of its denominations should the multiplicand generally be reduced? 226. Wlhait is the rule for multiplying a compound number by a simple whole number? The proof? MULT'PLICATiON OF COM1POUND NUMBERS. 1 1. 3Iultiply 7rd. 10 ft. 5in, by 6. Ans. 45 rd. 13 ft. 2 2mli. 3fuir. 27rd. by 8. Ans. 19mi. 5fur. 16rd. 3. 16 yd. 2{in. by 21. Ans. 837 yd. 1 ft. 38 in. 4. l mi. 14rd. 8$ ft. by 97. Anls. 101 mi. 3 fur. 6 rd, 5 ft. 3 in. 5. 4 yd. 2 ft. 9.14 in. by 47~-.:i Ans. 231yd. 2 ft. 9.73 1 i in. t. 12 E.Fl. 3' na. by 18. Anis. 221 E, Fl. 7. 6 E.E. 4qr. 3.44 na. by 28. Ans. 195 E.E. lqr..82na. 8. 5 sq. yd. 8 sq. ft. 106 sq. in. by 13. Ans. 77 sq. yd. 5 sq. ft. 82 sq. in. 9. 41A. 3R. 26.1087 P. by 9.046. Ans. 379 A. 28.4593+ P. 10. 10 cu. yd. 3 cu. ft. 428.15 cu. in. by 67. Ans. 678 cu.yd. 1 cu.ft. 1038.05 cu. in. 11. 7 oz. 16 pwt. 54 gr. by 1174. Anrs. 113 lb. 3 oz. 5 pwt. 161 gr. 12. 25 1 13 gr. by 20. Ans. 6 3 3 5. 13. 16 cwt. 1 qr. 7.88 lb. by 11. Ans. 8T. 19 cwt. 2 qr. 11.68 lb. 14. 7 lb. 6oz. 12 dr. by 283.44. Arts. 1 T. 1 cwt. 3 lb. 15 oz. 8.98 dr. 15. 1 qt. 38 gills by 7. Ans. 2 gal. 2 qt. 1 gill. 16. 5gal. 3qt. 1 pt. 2 gills by 35.108. A7is. 208 gal. 1 qt. 1 pt. 2.52 gills. 17. 26bu. 2 pk. 7 qt..37 pt. by 10. Ants. 267 bu. 7 qt. 1.7 pt. 18. 3f5 48 m by 12. Ans. 5 f 3 5f 36 m. 19. 18da. 9hr. 42min. 29.3 sec. by 167gl. Ants. 306 da. 4hr. 25 min. 2 sec., nearly. 20. $1072 9 ct. 2m. by 424. Ans. $454567 8m 21. ~215 16s. 2id. by 75. Ans. ~16185 14s. 3d. 22. 100 28' 422V" by 2.754. Ans. 28~ 51' 27.765" REVIEaw.-226. If the multiplier is a composite number, what may be done? When the muiltiplying and reducing can not readily be performned in the mind, what should be done? 174 RAY'S HIIGHER ARITHAMETIC. Awr. 227. The difference of time betwe-n two places is 4 hr. 18 mill. 26 sec.: what is their difference of longitude? Sor L uT o N.-Every hour of time cor- hr. min. sec. IPespOLnds to 15~ Of long1itudes; every 4 18 C2. minute of time to 15/ of longitude; 1 5 every second of time to 15/ of longi- 4 3 3 0 tiltle, (Alrt. 201). Hence, multiplying Dui of Lolg. the hours in the difference of time by 15 will give the degrees in the difference of longitude, multiplying the minutes of time by 15 will give minutes (/) of longitude, and multiplying the seconds of time by 15 wi,1l give seconds (/) of longiltlde; since reducing seconds to nIin. andl i nin. to hr. are tle satnie as reducing (") to (') and (') to (o), the divisor in both cases being always 60, hence, TO CIIANGE DIFF. OF TIME INTO I)TFF. OF LONGITUDE. RurE.- lutlliply the diferelnce of time by 15, according to the rmtle jbr Cottpoltttl A//ll/)ipicatioun, and.mark the prodtuct o / 1/ instead of hr. mlintl. and see. REIIARKI. —Thle work can be shortened by cancellation, for 15 times 26 divided by 60;~ X; - 61 — 6/ 30"/, since - 30". WVrite the 30// and catry the 6(. Then 15 times 18 divided by 60 - Xl 18 4 0=-4 30', since II= 30l; tadd in the G/ to be car00 4 ried Nwith this 30', laking 38i'. Carry the 40 to the next product, 60~, making 640; the answer is 64~ 36/ 30//, as before. 1. When it is 4 o'clock r. AI. at New York, it is 3 hr. 1l8 in. 28.4 sec. p. Ni. at Cincinnati: the loungitude of New York is 74~ 1' 6" W.: what is the longitude of Cincinnati? Ans. 84~0 24' WV. SuoGEsTIrOo.-Of two places, the one having ltter time is cast of the other; the one having earlier time is west of the other. 2. When it is 1 r. it. at St. Louis, it is 8 hr. 14 min, 55;' see. t. mI. at the Cape of Good Hope: the longitudf;3 of' the latter place is 18~ 28' 45" EI.: what is tlhe lonugi tude of St. Iouis? Ans. 900 15' 10" VW. R T, v IE w.-2217. What is the rule for converting difference of time into difference of longitude? Illustrate and prove it. Show how the operations under this rule can be shortened by cancellation. DIVISION OF COMPOUND NUMBERS. 175 3. A man travels from Halifax to Chicago; his watch hows 9 A. sI., — while the timc at Chicago is 7 hr. 24 lirn. 24` scc. A. Mr. The loingitude of Halifax being 63~ 86' 40" AV.: what must be the longitude of Chicago?.i4ns. 87~ 30' 30/" W.' 4. When it is 10 A. i1. at Stockholm, it is 3 hr. 24 minl. 58 sec. A. m1. at Wheelin(g: the longitude of Wheling is 80~ 42' W.: what is the longitude of Stockholm? Aits. 180 3' 30" E. 5. Noon comes 47 min. 17 see. sooner at Detroit than at Galveston, whose longlitude is 94~ 47' 15" W.: what is the longitude of Detroit? Ans. 82~ 5' \V. 6.'ime is 7 hr. 57 min. 26 see. later at St. Petersburgh than at New Orleans, and the lonoitude of the forller is 30~ 19' 46" E.: what is the longitude of' the latter? Ans. 89~ 1' 50" NV. 7. When it is 1 P, Ai. at Utiea, whose longitude is 750 1t3' 6W. it is hr. 52 in.- 4 sec. A. mw. at Little Rock: what is the longitude of the latter? -As. 92 12' W S. When it is 3 r. at Pregent's Park, London, it is 9 hr. 46 min. 31.2 sec. A. xI. at the University of Virginia, lwhose longitude is 78~ 31' 29" AW,: what is the longitude of the former? As. 9' 17" WV. 9. Whien it is midnilght at Iadras, in India, it is 1 hr. 2 min. 16.2 sec.'. La. at Buffalo; the longitude of the fibrmier place is s80 15' 57" E.: what is the longitude of the latter? 7Ans. 78~ 55' VW. 1i0. W!enil it is 1 A. xI. at Constantinople, it is 11 hr. 13 m)in. 25 7, sec.. I. mi. of the previous day at Paris, and tlle longitude of Pl'aris is 2~ 20' 22" E.: what is that of Conustantinlople? A ns. 28~ 59' E. 11. A ship's chronometer, set at G-reenwich, points to 4 hr. 43 min. 12 see. P. M.; the sun on the meridian: lmhat is the ship's longitude? Ans. 700 48' W. DIVISION OF COMPOUND NUMBERS. ARnT. 228.. Division of compound numbers like division of simple numbers, has two cases: 176 nRAY'S HIGHER ARITHMETIC. CASE I. ART. 229. To divide a compound number by a simple numuber. Since every compound number can be reduced to a simple number of either of its denominations, (Art. 220), the division of a compound number by a simple number will only differ fiom the division of simple numbers by the reduction before and after dividing. GENERAL RULE FOR DIVIDING A CO3MPOUND NUMBER BY ANY SIMIPLE NUMBER, WHOLE OR FRACTIONAL. Reduce the conmpound znumber to a simple number of either of its deonominations; divide as in simple znumbers: the quotieszt will be a simple number of the same denomiination as the dividenzd, ancd may be reduced to a compound ulmber. NOTE.-It is generally best to reduce the dividend to its lowest denomination. Divide 17 da. 5hr. 24 min. 19.208 sec. by 8.07 SOLUTION.-17 da. 5 hr. 24 min. 19.208 sec. - 1488259.208 sec. which divided by 8.07 gives a quotient 184418.737 + sec., and thif reduces to 2 da. 3 hr. 13tmin. 38.737 A- sec., the quotient required. ART. 230. If the divisor is a whole number, the re ductionsl may take place durbing the division, instead ol before anld after it. TO DIVIDE A COMPOUND NUMBER BY A SIMPLE WHOLE NUM IBER. RULE.-Divide that part of the dividend 7which is of the hihec.sl denom iaation first, and set the quotieil below: reduce the,remailnder, if there is one, to the next lower denominaltion, add in those of that denomination in the dividend, alnd divide again. Continue so until the lowest delomination has been used; whenl, if ther'e is a remaindcer, it should be expressed as a comnozo1 (a decimnal fraction of that denlomiation. R E VIE W. —228. How many cases in division of compound numbers? 229. What is the 1st case? What is the general rule for dividing a compound number by a. simple number, whole or fractional? To which of its denominations is the dividend generally reduced?- 230. If the divisor is a simple wiocule number, what is the rule? DIVISION OF COMPOUND NUMBERS. 177 PRoo F.-Same as in division of simple numbers. N T. —If the divisor is a composite number, we may divide by its factors inl succession, as in Art. 66. RE lAAI. t -This rule may be used when the divisor has a common or decimal fraction, by multiplying both numbers by the de. nominator of this fraction, which will convert the divisor into a whole number, and yet will not alter the quotient, (Art. 75). Divide 106 lb. 9 oz. 14. dr. of sugar equally among 8 nmeln. So L U T 0 N.-8 into 106 lb. gives lb. oz. dr. a quotient 13 lb., and 2 lb. = 32 oz. 8) 1 0 6 9 14 to be carried to the 9 oz. making As 13 5 34 41 oz.; 8 into 41 oz. gives a quotient A 5 oz., and 1 oz.- 16 dr. to be carried to the 144 dr., making 304 dr.; 8 into 8O0- dr. gives 34 dr. and the operation is complete. If $42 purchase 67 bu. 2 pk. 5 qt. 1' pt. of meal, how much will $1 purchase? SOLUTION.-Since 42 = 6 X 7, di- bu. pk. qt. pt. vide first by one of these factors, and 6) 6 7 2 5 14 the resulting quotient by the other; the 7)1 1 1 0 123 last quotient will be the one required. 1 3 A man travels 1472 mi. 6 fur. 32 rd. in 59 days; how much a day does he average? mi. fur. rd. mi. 59)1472 6 32(24 118 292 236 SOLUTION.-As the divisor is 5 6 mi. larger than 12, and can not be 8 separated into suitable factors, 4 5 4 fur. (7 fur. proceed as in Ist example, per- 4 3 forming the work by long, instead of short division. 4 1 fur. 40 1 6 72 rd.(284 rd. 118 492 472 Ans. 24 mi. 7 fur. 284 4 rd. 20 178 RAY'S T-IGHER ARITHMETIC. 1. Divide 16 mi. 2 fur. 29 rd. by 7. A)s. 2.2 fu'. 27 rd. 2. 37 rd. 14ft. 11.28 in. by 18. Aiis. 2 r'd. 1 ft. 8.96 in. 3. 43 E. Fl. l qr. 3 na. by 33. A;zs. 1 E. Fl. 347 na. 4. 675 C. 114. 66 cu. ft. by 83. AiLS. 8C. 18.3453+cu. ft. 5. 10sq. rd. 29sq. y. 5 s. ft. 94sq. in. by 17. AkIs. 19 sq. yd. 4 sq. ft. 119'i- sq. in. 6. 1000A. by 160. A:l.s. G Av. 1 R. 7. Gsq. mi. 35 P. by 221 Ans. 170 A. 2. 28' P. 8. 124-15 cu. yd. 24 cu. ft. 1627 cu. in. by 11.303 Ais. 110 cu. yd. 6 cu. ft. 338.4+cu. in. 9. 881b. 16pwt. 17.6 gr. by 54. Alrs. lb. 7 oz. lI pwt. 10.1 - gr. 10. 3 75 1 8 gr. by 12. iA.s. 25 10 6-' gr. 11. 600T. 7 cwt. 81 lb. by 29.06 Anrs. 20 T. 13 cwt. 20 lb. 14 oz. 12+ dr. 12. 62 lb. av. 8 oz. by 96. iAis. 10 oz. 6, dr. 13. 312 gal. 2 qt. I pt. 3.386 gills by 72a:4 Alls. 4 gal. I qt. 1.79 + gills. 14. 19302 bu. by 6.215 A);s. 3105 bu. 2 pk. 6 qt. 1. 5 pt. 15. 53 bu.. pt. by 63. Auis. 3 pk. 2 qt. 1.85- pt. 16. 76 yr. 108 da. 2 hIr. 38 min. 26.18 sec. by 45. Aiis. 1 yr. 254da. 27min. 31.25- sec. 17. 19 hr. 53- sec. by 7a Aiis. 2 hr. 26 min. 16.06 — sec. 18. 1520 46' 2" by, 9. Ans. 16~ 58' 206" ART. 231. Since the difference of time between two places, multiplied by 15, gives thelir difference of longitude, the product being marked o -, instead of hr. min. and,eee.: conversely, TO CIIANGE DIFF. OF LONGITUDE TO DIFF. OF TI3ME, l ULiE.-Dividle the diTerence of longithtdce by 15, according to hle ride fbi7 Coni.pound Divisionl, anid mark1 the quotield, hozrs, finntutes, anzd secon7cds, instead of ~ / DIVISION OF COMPOUND NUMBERS. 179 N OTE. —The division required by the rule can be shortened by cancelingo, in a manner similar to that explained in Art. 227. The difference of longitude between two places is 810 39' 22"; what is their difference of time? 15)810 39' 22" 5 hr. 26min. 3 7 sec. SOLUTION. —15 into 810 gives 5 (marked hr.), and 6~ to be carried. Instead of multiplying 6 by 60, adding the 39' and then dividing, proceed thus: 1].5 into 60 is the same as 15 into 6 X GO/ 6 X =4 24', and as 15 into 39/ gives 2'/ for a qllotient and 9/ remaLinder, the whole quotient is 2G/ (marked min.), and remainder 9/ =-9 X 60//, which divided by 15 gives - -8 303", which with 1i7T obtained by dividing 22// by 1.5 gives 37i 7,T", (marked sec.) The ordinary mode of dividing will give the same result, and may be used if preferred. 1. What time at Columbus (luona. 83~ 3' 1W1.). when it is 4 p. mr. at Baitimore, (long. 76O 37' W.)? iAs. 3 hr. 34 min. 16 sec. p. -m. 2. What time at Copenhaomen (long. 12~ 34' 57" EJ.), when it is 10 P. ia. at ilobile, (long. 88o 11' AA.)? Anrs. 4 hr. 43 nisiri. sc. A... the day after. 3. What time at P)ittiburg r (lo7 n. 790 58' W.), when it is 3 A. 1. at Dublin, (lon,. o~ 90' 30" r.)? Ais. 10 hr. 5 nmi. 30 sec. r. mi. the day before. 4. When it is noon at Louisville (long. 85~ 30' V.), what time at Bangor, (long. 680 47' NV.)? Anis. 1 hr. 6 min. 52 see. p. m. 5. When it is 6 p. mr. at Havana (long. 82~ 22' 21"'W.), what time is it at Paramnatta, (lon. 151~ 1' 35" E.)? Ais. 9 hr. 33 min. 35 — sec.... the day after. 6. What time at Cambridge, Eng., (long. 5' 21" E.), when it is 9 P. I. at Cambridge, Mass., (long. 71~ 7' 20 1 WV.)? Ans. 1 hr. 44 min. 5004 sec. A. 3r. the day after. 11 E vI E w. —230. What is the proof? If the divisor is a composite number, what may be done? IIow can this rule be used when the divisor has a common or decimal fraction? 231. What is the rule for converting difference of longitude into difference of tinme? Illustrate and prove it. How can the operations under the rule be shortened? fIStO) RAY'S HIGHER AMITHAIETIC. 7. When it is 7 A. Mr. at Washington (long. 77~ 1' 30 W.), what time at Miexico, (long. 990 5' WV.)? Ans. 5 hr. 31 min. 46 sec. A. M. CASE II. AnT. 232. To divide one compound number by another similar compound number, the quotient being an abstract number. RuLE. —Reduce both dividend and divisor to sim1ple mzzbers of the saume demnoeLinatios, and themi divide. P oo. —Same as in division of simple numbers. N T E.-It is generally best to reduce the numbers to their lowest denomination; if, after reduction, one or both contain a fraction, proceed as directed in division of decimal or common fractions. IIow often can a keg of 2 gal. 3 qt. 1 pt. be filled from a barrel of molasses containing 37 gal.? S O -37gal= 300 pt. and 2 gal. qt. 3 qt. 1 pt. = 23 pt.; 70 3 90 then 300 pt. ~23 1 pt.- 300 --- 300 X - - 12- times. Ans. 3 79i 7 1. How many lunar months of 29 da. 12 hr. 44 min. 2.84 sec., in a solar year, (Art. 1.96, Note)? Ans. 12.368+ 2. How many steps, 2 ft. 9 in. each, will a man take in going 311 miles? Ans. 6240. 3. The wheels of a locomotive are 10 ft. 5 in. in circumf'erence, and make 8 revolutions a second; how soon will it run 100 miles? Ans. 1 hr. 45 minm.:6 sec. 4. The new half-dollar of the U. S. contains 7 pwt. 41 gr. pure silver; how many dollars in 1.1 oz. 2 pwt. of pure silver, and if it be coined into 66s. what is one shilling worth in U. S. currency? Ans. $15.411, and s. = 23 ~-9e Ct. 5. How many half eagles, each weighing 5 pwt. 9gr., and 21 car. 2' gr. fine, can be made of 1000 sovereigns, each weighing 5 pwt. 3.274 gr., and 22 car. fine? A. s. 9 73)2 REavriw.-232. What is the rule for dividing one compound number by another similar one? The proof? To which denomination is it best to reduce the numbers? A LIQUOT PARTS. 181 6. A comet moves 8~ 17" 22'" in one day; in what time will it complete the circuit of the heavens, or 360~? Ans. 43 da. 10 hr. 16 min. 19 sec., nearly. 7. How many persons can receive each $2 18et. 7 1., out of a fund of $59 61 et?;As. 27. 8. How many half-crowns, each worth 2s. 6d., are in ~18 7s. 104d.? Ams. 147,1'2 9. The Julian calendar assumed the year 365 da. 6 hr., instead of 365 da. 5 hr. 48 min. 48 sec., its true length; in how many years was a day gained? Anls. 128} yr. 10. In how many years is a day gained by the Gregorian calendar, which allows for the fraction of a day by adding in 97 days in 400 years? Ans. 3600 yr. ALIQUOT PARTS. ART. 233. Aliquot parts is a useful method of finding a product, when one or both of the factors is a compound nuniber. The following is an example of the sort of problems to which it is generally applied. What cost 28 A. 8 R. 25 P. of land at $16 per acre? SoL u T I O N.-Multiply $16, the price $ of 1 A., by 28; the product $448 is 1 6 the price of 28 A. 3 R. is made up 2 8 of 2 R. and 1 R.; the former is I of an A., and the latter 4 the former; 32 obtain the price of 2 R. by taking the price of 1A., or ~ of $16 =$8; $448 the price of 1 R. is 4 of this=$4. 2R _ 1 8 25 P. is equal to 20 P. and 5 P.; the 1 R. = 4 4 former is 4- of 1 R., and worth 1 of 20 P. - 4 2 $4= $2: the latter being I of 20 P., 5 P. =.5 0 is worthof $2 — $ 50 ct. These $462.50 results added, give the value of 28 A. 38. 25 P., =-$462.50. The same result could be obtained by re ducing 28 A. 3 R. 25 P. to acres, viz: 28.90625 A., and multiplying it by 16. ART. 234. This method can be applied when the n-multiplicand is a compound number, as in the following example: I b2 RAY'S I{i (1ii IR tA IITItIM51 iTC. A travels 3 mni. 5 fur. 16 rd. in 1 hr.; how far will he go in O3da. 9 hr. 18 min. 48 see., (12 hr. to a day)? mi. fur. rd. SOLUTION. —This exampie is solved like the pre- ceding, except that the multiplieations and divisions da. hrl. 3 3 0 24 are performed on a com- 6 = 8 x 9 264 4 32 pound instead of a simple 1 5 mii. i 7 1 4 number. 3 1 18 One of the most valuable 3 0 sec. 9 applications of aliquot parts 1 5.. - 4 is when the product is to be 1 49 U. S. money; for instance, 9 -2-9$-(.....24,2 g If $47.52 is paid for the use of money 1 yr., how much oulght to be paid for using it 4 yr. 7 mon. 19 da.? SoLuTIoN..-When money is $ 4:7.52 paid for the use of money, 1 month- 4 is reckoned 30 days, and the aliquot O. o parts taken accordingly. The mills un 9.76 n 6 in o(1 mn. —- ~,2 3.7 6 in the result are usually neglected -' ] men.- 0.96 if they are under 5; but, if they are 8 da. 3 6 5 or over, they are counted 1 cent. 1 da. 1 2 I da. The above result would be called 132 $220.31. 2 2 0.3 0 8 REMA R.x..-When the number of which parts are taken, ends in 0, the simplest way is to take the tenths, instead of halves, fourths, &c. In the example above, since 1 mon. = 30 da., separate 19 dt. into 18 cda. and 1 da.; the former is % of 30 da., and the. value corresponding is found by multiplying the value for I mon. (3.96) by T-i; -6., lwhich is the same as to multiply by 6, and set the figures of the product 2.376 one place farther to the right. ART. 235. In all questions in aliquot parts, one of the numbers indicates a rate, and the other is a compound number whose value at this rate is to be found. RULE FOR ALIQUOT PARTS. Mthltply/ the naumber indicating the rate by the,number of hal denominalion for whose unit the rate is given, and separate z'he Raivisw.-233. What is Aliquot parts? Explain its use. 234. What kind of a number may the multiplicand be? ALIQUOT PARITS. 183 numbers of the other deomi2inatiouts ito patIt.s whose values caJn be ob/aied directly by a simple dicisiiouL osr ul/ljtlicalioll o' olne of the prccedilg values. Add these differeltt values; the result will be the ellire value required. N'OTE.-Sometimes one of the values may be obtained by adding or szubtracting two preceding values instead of by multiplying or dividing. EXAMPLES FOR PRACTICE. 1. If a person travel 4 mi. 5 fur. 10 rd. 12 ft. 4 in. in 1 hr., how far will he travel in 7 hr. 37 min. 28 see.? Ans. 35 mi. 4fur. 6rd. 2ft. d ~in. 2. AWhLat cost 86 yd. 3 qr. 2 na. of cloth at $29.434 per yard? (Turn -4 into a decimal.) Auns. $211.76 3. Find the cost of 231 A. 1 R. 34 P. of land at $17.28 per A. Ans. $3999.672 4. 1What is the cost of 127 yd. of carpet at $1.87-~ per yd.? Ans. $238.12SoLUTIo0N.- Iere the only 1 2 7 compound number is the rate ex- 1 8 7 pressed in Federal money. Take 1 it as the multiplier. $1 7 The cost of 127 yd. at $1 per 5 0 ct. = 63.5 0 yd. is $127; by taking suitable 2 5 at. = 31.7 5 parts of thlis, the cost of 127 yd. 12 2. ct. = 1 5.8 7 7 is found at 5O0 ct., 25 ct. 12 et., a $ 2 3 8.12 yd. respectively. The sum of all these is the cost of 127 yd. at $1.871 a yd. Therefore, aliquot parts can be used to find the value of any number of articles, when the value of one is known in U. S. money, by finding their cost at $1 a piece, and taking such aliquot parts of this as are necessary to make the cost at the given price. 5. Find the cost of 42 cu. yd. 24 cu. ft. of earth at $1.25 a cu. yd. Ans. $53.61 6. Of 71b. 8oz. 16pwt. 1 gr. of gold at $15.46 an oz. Ans. $1435.04 RETr,,Ew.-234. What is one of its most valuable applications? Give an example. 235. In questions in aliquot parts what relation exists between the two quantities? What is the rule for aliquot parts? IHow can a value sometimes be obtainedc? Hlow can the value of any number of articles be found when the price of one is given in U. S. money? 184 RAY'S HIGlEIll A;iITIMTI C. 7. Find the cost of 6 T. 13 cwt. 2 qr. 21 lb. of' sugar, at $4.68$? a cwt. Ans. $626.77 8. Of 3 lb. 7 oz. of cheese, at 15 et. a lb. Ans. 51,-6 ct. 9. 1 yd. of cloth is worth 3 qt. 1 pt. 1 2 gills of wine what is 47 yd. 2 qr. 1 na. worth? Ans. 44 gal. 2 qt. 2g gill. 10. In 1 wine gallon are 231 cu. in.: how Cmany c. ill in 24 gal. 3 qt. 1 pt. 24 gills? A. 5764l 4 11. In 1 beer gallon are 282 cu. in.: how mlany cu. in. in 38 gal. 1 qt. I pt. of beer? Ans. i08213 12. In 1 bushel are 2150.42 cu. in.: how many cu. in. in 15 bu. 1 pk. 6 qt. 1 pt? Ans. 33230.7 + 13. What is the value of 29 gal. 2 qt. I pt. of wine,_ at $2.25 a gal.? A.s. $66.65 14. What is the value of 46 gal. 1 qt. 1- pt. of beer, at 30ct. a gal.? Aiis. $13.94-+ 15. What is the value of 10 bu. 3 pk. 5 qt. of corn, at 621 ct. a bu.? Aas. $6.82i6. If ~3 6s. silver weigh 1 lb. Tr., how much will weigh 17 lb. 11 oz. 16 pwt. 9 gr.? Aats. ~59 7s.+ 17. Built 8rd. 14ft. 10in. of fence in 1 da.; how much can I build in 3-wk. 5 da. 9 hr. 46 min., if 1 da. 10 hr., and 1 wk. = 6 da.? Als. 213 rd. 6 ft. i in., nearly. 18. If a man is 2 hr. 25 min. 38 sec. in digging a cu. yd. of earth, how long will he be in digging 44 cu. yd. 22 cu. ft.? Ans. 108hr. 46 min. 31,a sec. 19. A comet moves 24~ 6' 49" in 1 hr.; how far will it go in 6hr. 14min. 52 see.? Ans. 150~ 39' 23.3"+ 20. A pendulum beating 54000 times a day, beats how often in 4 da. 3 hr. 20 min. 5 sec.? ASns. 2235038 times. ART. 236. Aliquot parts can be applied to making out bills in U. S. money, when the prices of the items are given in State currencies. What cost 338 gal. of wine, at 14s.. 7Ad. a gal., New England currency? SoLUTrION.-The tables of State $3 3 =- $33.50 currencies, (Art. 207), show that in 2 the New England States, 6s.= $1. 1 2s. 2 $ 6 7 The cost at Gs. or $1 a gal. is $33.50, 2 s. = 11. 6 7 from which, by multiplying and taking c d. = 1 2. 7 9 suitable parts, the cbst at 12s., 2s., gd., 1 d. 1.6 9 8 1, &d. is found; the sum of thebe is the 1.6 cost at 14s. 7 d., as required. ALIQUOT PARTS. 18 5 Wh11at cost, in New Enl c1and currency, 1. 35 yd. cloth, at 7s. Gd. a yd.? Ans. g43.75 2. 11 3Cyd. of muslin, at 2s. 4d. a yd.? Ais..68 3. 2{3 caps, at Ss. Gd. a piece? Aa s. $G3.8:3 4. 1 0, doz. copy books, at 4s. 3d. a doz.? A as. $7.44 5. 1;58 lb. starch, at 9d. a lb.? Ans. $19.7 6. 45 lb. 10 oz. butter, at 2s. 8d. per lb.? Als. -20.28 S Uc a Es Tr I o N. —435 b. 10 oz. = 453 lb. - 45.G 25 lb. 7. 34 da. work, at 5s. 4d. a day.? Ans. $30.22 8. lSdoz. andl 8c -ggs, at 1s. 10 d. a doz.? As,. $5.70'9.'72 bu. 3 pk. corn, at 3s. 3d. a b u.? Ains. $'39.41 10. 864s yd. carpet, at 10s. 5d. a yd.? Anzs. $150.82 AntT. 237. What cost, in New York currency, 1. 141 yd. of calico, at Is. 2d. a yd.? Ans. $.15 2. 1 obbl. potatoes, containing 2o bu. each, at 4s. Gd a bu.? inAs. $16.8S7 3. 2 bu. 3 pk. 6 cqt. of dried peaches, at 17 s. 8d. a bu.? Alns. 8(3.49 4. 4gall. I qct. 1 pt. oil, at 2s. 31. a gal.? iAns. 1.23 5.: d(lictionaries, at 5s. Gdl. a-piece c Ans. $ 2.84 6. 49 boxes latches, at 4!)d. a-piece? Ans. $2. 30 NAr. 233. What cost, in Pennsylvania currency, 1. 3 rqt. 1 pt. mnolasses, at 2s. aa cjt.? A is. 93 et. 2. 1 box candlles (40 lb.), at Is. Sd. a lb.? Ais. $8.89 3. 12 lb. of coffee, at 10d. a lb.? nAs. $1.3:3 4. 9 gal. 2 qIt. 1 pt. of milk, at Gd. a (1t.? Ains. $2.57 5. 4 lwk. 5 da. wanges, at 13s. 4(d. a week? ul9As. $8.38 6. 23 rd. 101 ft. of fencing, at 20s. a rd.? A)is. $63.03 ArnT. 239. What cost, in S. Carolina currency, 1. 13g,al. 3 lqt. of oil, at 3s. 6d. a,al.? Ains. $10.31,1 2. A ham of 17. lb., at lid. a lb.'? isl. 43-1 3. 43 lb. of butter, at Is. 4 1d. a lb.?,r. $12.67 4. 16 y d. of sill, at 8s. 3d. a yd.? Anls. $29.83 i16 186 RAY'S HIGHER ARITHAMETIC. ART. 240. What cost, in Canada currency, 1 7 gal. l qt. of honey, at 6s. Od. a gal.? Ans. $9.91 2. 5 bu. 1 pk. 7 qt. of dried apples, at 16s. 8d. a bu.? Ans. $18.23 ART. 241. M1ISCELLANEOUS EXAIPLES. 1. 1bow long is a rope winding 276 times round a tree, whose circum. is 4 yd. 2 ft. 6- in.? Ans. 1339 yd. 4 in. 2. VWhat is the area of a field, length 67 rd. 8 ft. 5 in., breadth 89 rd. 11 ft. 2 in.? A 1s. 16 A. 2 R. 38+ I'. 3. Hiow many bbl. (31'- w. gal.) in a room c2 ft. 8 in. lon,, 16 ft. 6 in. wide, 11 ft. 4 in. high? Ans. 988:4. 4. HIow many bu. in. a bin, 8ft. 10 in. lon;g, 4ft. 6 in. wide, 3 ft. 2 in. high? Ans. 101 bu. 5 qt., nearly. 5. IHow much land in a rectangular field, 83.44 ch. long, 56.27ch. wide? Ans. 469 A. 2R. 2.7008 P. 6. A bought a pipe of wine (137 gal.), lost 9 gal. 2 qt. l Pt. by leakage, and sold the rest at $2.371, per gal. how much did he receive? Ans. $3062.37-. 7. How many quart, pint, and half-pint bottles, of each an equal number, can be filled out of a cask containing 44 al. 2 (t. pt.? Anis. 102. 8. A man can mow in 1 day, 2 A. 3 R. 20 P. of grass: in what time will he mow 78 A. 1 R. 36 P., allowing 10 hr. to a day? Als. 27 da. 2 hr. 57-'3 min. 9. What is the value of 16 lb. 7 oz. 12 pwt. 3 gr. of gold, at $15.85 an oz.? Ans. $3163.76. 10. If 1 cu. ft. of water weigh 62- lb., what is the weight of the water in a room 20 ft. long, 15 ft. 5 in. wide, 9 f't. 10 in. highll? Ans. 94 T. 14 cwt. 3 qr. 21.'lb. 11. If a ship sail 10 mi. 6 fur. 185 rd. per hour, how long will it be in going; 3236 mi. 2 fur. 36.508 rd.? Ans. 12 da. 11 hlr. 28 min. 6+ sec. 12. In a rectangular field, are 160 A. 2 R1. 36( P.; one side is 74.18 ch.: what is the other? Ans. 21.67-ch. 13. HIow thin is a cu. in. of gold, beaten so as to cover a space 46 ft. 10 in. by 41 ft. 8 in.? As. sT' OO in. 14. When I arrived at Cincinnati, my watch, which kept time correctly, was 42 nmin. fast: from whichl direction had I come, and how far in that direction had I traveled? Ans. From the east; 564- mi. NOT E.-A degree of longitude at Cincinnati is about 533 mi RATIO. 1 87 XII. RATIO. ART. 242. Ratio is a Latin word signifying relation or connection; in Arithmetic, it means the relation of one number to another, expressed by their quotient. The ratio of 2 bu. to 6bu. is.; of 10 yd. to 3 yd. is 3o; showing that 5 bu. are 5 of 2 bu., and 3 yd. -3 of 10 yd. Ratio exists only between quantities of the same kind, since only such can be divided, one by the other. Since a ratio is a quotients it is an abstract number, (Art. 60), showing how many times one number contains the whole or part of another. ART. 243. The ratio of two numbers is indicated by writing them in the order in which they are mentioned, with a colon (:) between them. The ratio of 4 to 9 is written 4: 9; of 2.: to 4.65, is written 2: 4.65; of 2 ft. 8 in. to 1 yd. 1 ft., is written 2 ft. 8 in.: 1 yd. 1 ft. Each number is called a term of the ratio, and both together a cozluplet or ratio. The first term of a ratio is the antecedent, which means goinyg before; the 2d term is' the consequznt, which means following. A sinmple ratio is a sinre ratio of two terms; as, 3 4-4. A compound ratio is the product of two or more simple ratios; as, 4 X 5: 3 7 7 is the product of the sim4X5 ple ratios 4: 3 = and 5: 7=. The value of a ratio depends not on the absolute, but on the relative size of its terms. TO FIND THE VALUE OF ANY RATIO, RULE.-Express the terms in. the same denomination; take the consequent as the numerator, awid the antecede7nt as the denomi. nator of a com-mon fraction; this fraction reduced to its sinipelst form, will be the ratio required. R E v itE W.-242. What is the meaning of Ratio? What is ratio in Arithmetic? Give examples. When can two quantities have ratio? What kind of a number is every ratio? Why? 243. I-low is a ratio indicated? Give examples. 188 55~RAY'S IIUG IHERP ARITHIAMETIC. N OTE.- -Thle Freonch methodl of pbtaining the ratio fhas been adopllel here. Tle Elglisll rietthod makes t.lhe anttecelnt tle nutmert ator, and thle conselttent the denorminator of the ftraction; tile ratio of 3 in. to 7 in., by tlhe French method, is h; y the Ellglisil,. Since the value of a ratio is equ-al to the consequent divided by the antecedent, it follows that The actcccd-let is equtal to tile conscalcnt dicidedltl )7. thli vacle of the ratio; and that The con.cqr'tlt'is e2lc1 to the antecccdclt mnZltti)lcte bey the value of the rcttio. HIence, if the value of a ratio is known, and one of its terms, tihe other can be found. What is the rat-o of 9 to 15? SOIUJTIO- 15 = _ 5 = I 1 Ans. What is the ratio of 2 A. 3 R. 25 P. to 1 A.? SJruT1oNS. —2A. 31R. 25P.: 1A. or 4651'..: 1GOP., =- o I tls. What is the ratio of 4] to o2? Ans. 4. 21 SUGGESTroX. —lIere the rule gives a complex fraction X- which 8 is relduced by ALt. 132. What is the ratio of 7.108 to 9.220,? Sor,. 7.108: ~.2G8 (Art. 1.3), 3 G- 7.108 2 21.894 27 8no00 _ 9 0 (At. 1 ns. FIND THTE RATIO 1. O()f'to5; ofrO:to I o 2 to; o f l11to4G; of 2Gto oo50; of 1 2 to I; of 2 to 4; of 7 to 13)O; of 1 to j; of 20 to I; of' s 8,to4. Aa is,; -; 2; 4; 1 1T; -; 1T; 1 3; Uo.; u1;, ~ 7 2. Of' 2: 4, of 6.5:.01 3, of 9: 17.8, of 116 18.75, of' 4: ~9.8, of, o: 1 of 10.08 33, of 1 2.17~ 14.3, of (., 37: 34, of' 94- 44.4. IS.. 1o- 6 5 4 ) 1,. R E v i P. -1.3. \\'lhat is eacah of tlhe ntlTnlers called? lWhat. are 1both together called!? 9Which is thie anttecedelnt? Whlich, the consequent? Why so calloed? What is a simple ratio? A compound ratio? What does the value of a ratio depond on? RIATI,. l89 3. Of 2 ft. 6 in. to 3 yd. 1 ft. 10 in. Azs. 4 4. Of 4 Ili. (3 fur. 20rd. to 1 Imi. 2fur. IGrd. Ans. i.5. Of | A 3. 3 I. 25 P.: 6 A. 10 P. A is. 4 G. Of 2 C. 1 ft.: 5..Ans. 7''S 7. Of 3 1b. 10 oz. 6 pwt. 10-1 gr.: 2 lb. 14, pt. 8.Of~~ 0c~5 35"As 1 4 3.54 I~ 8. Of 3 1 5g 1 l4.32 gr. An. 2 i2-4209 O9. Of 13 Ib.: 9 lb. 15.2 dr. Ans:. 3:2 10. Of 14 T. 12 cwt. 1 qr. 1 8.44 lb.: 7 cwt. 4" l1b. A nst. i 6' 11. Of 3 qt. 1 gills: 8 al. 1 pt. Aus. 10 3 0 1. Of 10 gal. 1. 54 pt.: 7 gal. 2 qt.. 98 pt. zAns. A 4 0 1 13. Of 50 bu. 2 pk. 1 qt.: 35 bu. 3 pk. 6.055 qt. 14I-.,A 5 4. 4. Of 5 hr. 26 min. 444 see.: da. As. 1 4 g 15. 0f 2 yr. 2 da. 7 yr. 216~ do as. 3bs_ _ I16. Of 420 a~ 15', 9* ~. A 0s 2. 8! 85 17. If the antecedent is 7 and the ratio V1, whatt i thie consequent? Aiis. 1 0 1S. If thle consequent is $13.42c and the ratio w, what iS tthe anteceldent? An s. 35.800 I 9. If the antecedent is 2 3 and the ratio 6.048, what is tile consequent? As. 15.7248 20. If thec consequent is G yd. 2 ft. 83 in. and tlhe ratio is 3.i, wl1at is the antecedent? Ais. 2 yd.1 in. 21. If the antecedent is 5 bu. 1.68 pt. and the ratio is 5*;, whitat is the conscquent? Anis. 29 bu. I pk. 2qt. 7' Pt. 22. If the antecedent is 24.075 and the ratio is.1664, what is the conlselquent? Ans. 4.00608.23. If tble consequent is - and the ratio, wlhat is the antecedent? Asis. 1,4 24. If' tlhe consequent is 2 71b. 5oz. 14dr. and the ratio is S, whlat is the antecedent? Ans. 45 lb. 9oz. 12' dr. 25. If the consequcnt is $7.43. and the ratio 231, what Lb the antecedent? Ans. 83.184 REI Ptv w.-243. What is the rule for finding the value of a rntio? HIow manny methods of valuing a, ratio? Explain the difference. What is the antecedent equal to? The consequent? HIow is the ratio of two mixed tumbners fclrld? I ow is the rat.io of two decima.ls found? 190 RAY'S HIGHER ARITHMETIC. 26. Find the value of $2.561: ~ $10, of 37 ct.: 33 ct., of $13.66s: $4', of $22: 22ct. Ans. 43, 8'a7, 10 ART. 244. Two ratios may be formed with the same two numbers, by taking each of them in succession as the standard of comparison; thus, the ratio between 5 and 7 is 7 or 4; the former is the ratio of 5 to 7, and the latter. the ratio of 7 to 5. One of the ratios which can be formed with two nurnbers will be an improper fraction, and the other a propet fraction; for the sake of distinction, call the former the increasif.j ratio, and the latter the decrcasi',ng ratio. If the two quantities are equal, the ratio, which ever way it is formed, will be equal to 1, and therefore neither increasing nor decreasing, but a ratio of equality. TO MAKE AN INCREASING OR DECREASING RATIO, RULE. —Write the two nlumbers in the form of an improper fraction, to express al increasing ratio; blut in the form of a proper fraction, to express a decreasing ratio. 1. Make an increasing ratio with 7 and 18, with 51 and 4}, with 3 yd. 2 ft. and 3 yd. 1ft. 5 in., with 6,( and 6.38. A3s. -A s.,, 32., 2. Make a decreasing ratio withi 21 and 21, with 12.45 and 9s, with 3 gal. I pt. and 2 gal. 2 qt., with 131 and 1 5 Ans. 7 4 ~, 4 6 4 ART. 245. Since every ratio is a fraction whose numerator is the consequent, and denominator the antecedent, whatever is true of a fraction is true of a ratio; hence, 1st. JMultriplyilng tlhecosequent or dWcviding the antecedent, multiplies the ratio. (Arts. 112 and 115.) The ratio 10: 4 is -4=; if the consequent be multiplied by 3 the ratio 10: 1.2 is 3 —- = -, which is the former ratio multiplied by 3; if the antecedent be divided by 2, the ratio 5: 4 is 4-, which is the former ratio multiplied by 2. 2d. Jllltpliyinyg the antecedent or dviuiding the conseqteant divides the ratio. (Arts. 113 and 114.) REviEW.-244. IIow many ratios can be formed with two numbers? Give an example. IHow are they distinguished? Whlat is a ratio of equality? What is the rule for making an increasing or decreasing ratio I 245. How is a ratio multiplied? Why? How is a ratio divided? Why? PROPORTION. 191 The ratio 7: 6 is 6; if the consequent be divided by 3, the ratio 7: 2 is a, which is the former ratio divided by 3; if the antecedent be multiplied by 2, the ratio 14: 6 is t- = -, which is the former ratio divided by 2. 3d. Jlltiplyilng or dividing both terms of a ratio by a enuwtber, does not alter its value. (Arts. 116 and 117.) The ratio 9:6 is -- -; if both terms are multiplied by 2, the ratio 18: 12 is -- = 2 still; if both -terms be divided by 3, the ratio 6: 4 is 6 —, the same as at first. XIII. PROPORTION. ART, 246. Proportion is an, expression of equal ratios. The ratios 3: 5 and 6: 10 are equal, each being of the value -6. Placing a double colon (:: between them, forms the proportion 3: 5 6: 10, read 3 is to 5 as 6 is to 10, or the ratio of 3 to 5. is equal to the ratio of 6 to 10. RE AI A R K.-Either ratio may.be written first; thus, 6: 10:: 3: 5 is the same as 3: 5: 6: 10, since each expresses the equality of the same ratios. Instead of the double colon, the sign of equality is sometimes used; as, 3: 5 =6: 10, is the same as 3: 5:: 6.: 10. A proportion with more than two equal ratios is called a conltnttle(l proportion., as 3: 5:: 6: 10: 9: 15; but a proportion in Arithmetic generally contains only two equal ratios, and has 4 terms, since each ratio has 2 terms. Since each ratio has an antecedent and consequent, every proportion has two antecedents and two consequents, the Ist and 3d terms being the antecedents, and the 2d and 4th the consequents. The first and last terms of a proportion are called the exitrrnes; the middle terms, the mteans. All the terms are called p?roportionals, and the last term is said to be a fourth prop9ortional to the other 3 in their order. Ratio is the relation between two numbers shown by their quotient: proportion is the relation between two ratios shown by their equality. The former has two terms, the latter four. REVIEW\V.-245. IHow is a ratio altered in form and not in value? Why? 246. What is Proportion? Give an example. How is it written? What other way? What is a continued proportion? 92 RAY'S IIIGIER ARITI-IETIC. Three numbers are in proportion, when tile 1st has the same ratio to thie 2dl as the 2d has to thle 3d; tlius, 4, 8 and 163 are in proportion, for 4: 8: 8: 16, each ratio being;,2. The second tQrm is then called a mea p)roportionatl between the other two; and the last term a tlird propvortional to the first and second. ART. 247. T'Vcrcation is a general method of expressing proPortion often used, and is either dlirec t or ivecrse. Direct variation exists between two quantities when they increase together, or decrease together. Thus, the distance a ship goes at a uniform rate, varies directly as the time it sails; which means that the ratio of any two distlances ts equal to the ratio of the corresponding times taken in the same order. Inverse variation exists between two quantities when one increases as the other decreases. Thus, the time in which a piece of work will be done, varies inversely as the number of men employed; whiclh means that the ratio of any two times is equal to the ratio of the numbers of men employed for those times, taken in reverse order. ART. 248. Since only eqtal ratios form a proportion, to idetermnine the truth of a proportion, Fittcl the value of each r(zetio ia the proportion); if tthese ratios are equtal, the proportion.is trtte; j' tIot, it is fatlse. -Thus, 8: 10:: 12: 15 is a tlrue plropolrtion, thle raltios 1O alndl 1: being each equal to 5; bit 9: 5:: 3:'2 is not, because the ratios nd are not equal, the former beinD 9, the latter WIIICII ARE TRUE PROPORTIONS, AND IIICII NOT? 1. 7: 10':: 8 12, and 4: 3:: 24: 18. 2. 2 ft. 1 in. 1 yd. 4 in.: G2 ct.: $1. 3. 16._ ~ 21 2 ~ bu. 3 pk.: 3 yd. c q. 4. 3 hr. 18mmin. 7 hr. 20 min.:: 2gal. 1 qt. ~ 4gnl. 2qt. REVIivw.-246. Ilow many ratios in a proportion generally? Towv mnany antecedents? lIow many consequents? What are the extremes? The means? Vhat are the termns called? The last term? IIow are It lltio and Proportion distinguished? When are three numbers in Proportion? What is the second term then called? What is the third term called? 247. What is variation? What two kinds? What is direct variation? Give an examplo. XWhat is inverse variation? Give an example. PR'U'O'I'RLON. 193 5. $5.33-: $12 3 Cwt.: 150 lb. 6. 76yd. Ift. 10in.: 13rd. 7uft.: $17.1843: $168 7. 16.208:94:: 9A. L R. 61 P.: 5A. 3R. 37.64P. 8. 23T. 5cwt. lqr. 15 lb. 8T. 2cwt. 3qr. 141b.:: 2gal. 2qt. 3 qt. 1pt. 9. ~8 17s. 84d.: ~8 10s. 8nd.:: 164 bu.: 35bu. 3pk. 10. 2.. ~1'41i': 58~7' 6.42"::1.hr.:3hr.44min. 11. 25lb. Tr. 24 lb. av.: $6: 7. 12 yd.t.yd.ft 9in.: 9E,F1. 8na.: $7.75 9 PROPOSITION. ART. 249. In every true proPortion, the product of the num1 bers i, t]he mvumans is equacl to the pfrodlct ofj those in the extremes. DEMO.NSTRATiON-Iln every true proportion, as 5: 3: 10: 6, the ratios are equdal, viz: = -1-6. If both terms of the first ratio be multiplied by 10, and both terms of the second ratio by 5, their values are not altered and they are still equal, viz: 3 I ~ =-160x 5 The denominators of these fractions, having the same factors, are equal; to make the fractions equal, the numerators must also be equall that is, 3 X 10 =6 X 5; but 3 X 10 is the product of the numbers in the means, and 6 X 5, the product of those in the eoxtremes; hence, the proposition is proved. COROLLARY 1. —Either extreme is equal to the product of the meaus divided by the other extreme. CoorOLLARY 2. —Either mean is equal to the product of the extremes divided(l by the other mean. The truth or falsity of a proportion can be determined by this proposition; the corollaries serve to find any term of a proportion, when the other three are known. Thus, if the Ist, 2d, and 3d terms of a proportion are 6, 10, and 15, it may be written 6: 10:: 15: ( ); the4th term is represented by a parenthesis, and found by Cor. 1 to be 0 1 5 = 25. The unknown term, 25, must be of the same denomination as the other term, 15, of the same ratio. REMARR.-Before applying the proposition or its corollaries, the 9trma of each ratio must be of the same denomination. RiE vr v w.-248. How do we determine the truth or falsity of any pV-oportion? 249. What relation exists between the extremes and means in every true proportion? Prove it. 17 !3'194 RKioAY' S HIGHIER ARITI1HIETTIC. FIND THE UNKNOWN TERM OF t 1 2: 1 *: (6 ) 5 and 83-: ((I )1 I @ As. 3'. and 5. 2. lbu. 2 pk. 6 qt.: 6bu. 3 pk. $3.87i ~ ( ). Aos. 15. 50 3. ( ): 4hr. 30 min.:: 2. mi.: 3.375 3mi. iAts. 3 h>r. 20 niin...C16A.:( ): ~13 Gs. 8d.: ~15. Ans. 2.43k. 5. 12yd. 3qr.: 46 yd. 3qr.:: ( ):'T. I cwt..An.s. 1 T. 13 cwt. 6. $162.56M': $270.934:: 234 men. ( ). A n s. 390 imen. 7. 46~ 31' 9": ( ): hr. 83 min.: 2 hr. 412 mia. Ans. 434- 53' 219 " 8. $16: 45ct.::1 lb. 9oz. av. (). Aos. l dr.I 9. 5 mi. 3fur. 30rd.: 7mi. 10rd.: ( ) 54 horses Ans. 42 lhorses. 10. 33 bu. 1 pk. of potatoes ( ):: 4 bu. 1' ppk. of apples 2 bu. 2 pk. of apples. Ans. 19 bu. of potatoes. SITMPLE PROPORTION. ART. 250. Simple Prop>orlion is a method of solving practical questionts by a ratio or proportion; it is sometimes called the RuLle of Three, because the answer is obtained by finding one term of a proportion whose other three terms are known. If 5 qt. of strawberries cost 75 ct., what are 9 qt. worth at the same rate? SOLUTION BY ANALrSIS.-If 5 qt. cost 75 Ct., 1 qt. costs - of 75 ct = 15 ct., and 9 qt. cost 9 times 15 ct. = 135 ct. = $1.35. SOLUTION BY PROPORTION.-Since the cost of each quart is the same, the whole cost varies directly as the number of qt.; that is, 9 qt. being ~ of 5 qt. are worth 9 as much, or 9 of 75 ct. = 7.. ct. — =1.35. Or, the ratio of the quantities being the same as tile ralti of their costs, we have 5 qt.: 9 qt.:: 75 ct.: ( ), in whichI tlle re. quired term is found, by Cor. lst, Art. 249, to be 9X7 = 135 ct. REVIEW. —249. What is Cor. lst? Cor. 2d? Of wh.at use is the pro position? Of what use are tho Corollaries? Give an examtle. SIMPLE PRIOP-OI' OT1N. 195.I EMA RT.- Snple Proportion is sometimes applied to questimns wlhich do not admtit of' solution in thsit way. Inll cass of doubt, resort to a close and careful ana:lysis, oi to the following test: A question to be solved by Simple Proportion, nlmst c(,ntain two liands of quantities, and two of' each kind tlhr(ce of the quantities mnust be known and one required. Thec two given quantities which are of different kinds, u11:Ist be so relatcd, that when one is dcloulled the other is necessarily (dotblctd or ha/recd; in fact, they furnish tlhe rte which is anpplied to the remaininog given quantity to obtain the answer. Ilence, TO SOLVE QUESTIONS BY SIMPLE PROPORTION, RULE. - TWTith the /two qien quiantities schich are of the same kilid, fJ)t' a01L inc)Ceasilg or decreaCsiJg ratio, accorc'dinl as the aJtnsler s'ho11tld be,Crealer or less tilal tlhe thiud given qyllncllily; mullll)ly tie t/il-d lqltculltity by this'atio, aud tle resullt will be tle anster,'eqli)'ed. NorT..-Express the ratio in its simplest form, and cancel when possible. For the benefit of those who prefer to state the question in the form of a proportion before commencing the operation, we give the following RULE. — r'ite that qta72itly for th7e 3cld term chich is of tlhe czanoe ldenlomnilltiol) as tie an.s'wer; lext to it, inl thle 2d term, patg tzie lar1cer or s'mller ll' Ite Otler t1wo quanltilies, accorCdinlg as the, al1s11er shotltld be larer or smaller t/hal the thilrd t1lerm.. T7he remaililng yf/icell q7lta.tity is thell plt ill tle first termn, adcl the f:lth0 o7r req7 ired term, is fJbnd by iillitiplillng tile second and third terms tojelther, alnd diciding tlheir _prodlct by thle fi) st. 1. If 2. lb. of sugar are worth 7. lb. of rice, how much sug;ar is 9. 8 lb. of rice worth? 19.8 19.8 11 217.8 SOLUTION.- of 2 lb. - -- X 7.26 lb. Ans. 4 80 19.8 X 23 Or, 7: 19.8:: 2- (); a4th term = 7- = 7.26 lb. It EvIIE. —249. Of what denomination will the required term be? Before applying the proposition or corollaries, what should be lone? 250. What is Simple Proportion? lVhat is it sometimes callc(1? Why? Give an example. W'hat is the solution by analysis? By prop-rtion What kind of questions properly come under simple proportion? j19f5 PRAY'S IIIGHERl ARITHMETIC. 2. If 15 men do a piece of work in 9O da., how long will 36 men be in doing the same? SoLuTION.-Since 36 men will require less time than 15 men to do the same work, the answer should be less than 9; da.; make a decreasing ratio, 3U, and multiply the remaining quantity by it: ~ i X93 -= I X458 -4 da. Ans. 3. If I walk 10' mi. in 3 hr., how far will I go in 10 hr., at the same rate? Ans. 35 mi. 4. If the fore-wheel of a carriage is 8ft. 2 in. in circun.ference, and turns round 670 times, how often will the hind-wheel, which is 11 ft. 8 in. in circumference, turn round in going the same distance? Ans. 469 times. 5. If a horse trot 3 mi. in 8 min. 15 see., how fitr can he trot in an hour at the same rate? Anls. 21-r'9- nli. 6. What is a servant's wages for 3 wk. 5 da., at $1.75 per week? Ais. $6.50 7. What is a bbl. of powder containing 132 lb. worth, if 15 lb. are sold for $5.4343? All>. $47.85 8. A body of soldiers are 42 in rank when they are 24 in file: if they were 86 in rank, how many in file would there be? Aics. 28. 9. If a pulse beats 28 times in 16sec., how many times a minute is that? Aei.s. 105. 10. On 15 successive squares are 614 houses: how far from No. I is No. 277? Ans. 64I, or nearly 7 squares. 11. If a cane 3 ft. 4 in. long, held upright, casts a shadow 2 ft. 1 in. long, how high is a tree whose shadow at the same time is 25 ft. 9 in.? Aces. 41 ft. 2i in. 12. If a horse draw 25 bu. of coal, each 80 lb., how many bu. of coke, each 96 lb., can he draw? Ails. 206 13. If a farm of 160 A. rents for $450, how much should be charged for one of 840 A.? Ans. $2362.50 14. If 18d. sterling, equals 25 et. U. S. money, what is a half-crown (2s. Cd.) worth? Ans. 413 ct. 15. A grocer has a false gallon, contlaining 3 qt. 11 pt.: what is the worth of the liquor that he sells for p240, and his gain by the cheat? Avis. $225, and $15 gain. 16. If he uses 144- oz. for a pound, how much does he cheat by selling sugar for $27.52? Ans. 42.15 R E v IE W. —250. What is the rule for solving cquestions by Simple Proportion? How should the ratio be expressed? What is the ordinary rule for simlple proportion r !S:iM1' L i';!']'0 PROWOTlON. 197 17. An equatorial degree is 365000 ft.: how many ft in 80so 24 37" of the same? Anss. 29349751k,79 I8. If a pendulum beats 5000 times a day, how often does it beat in 9 hr. 20 min. 5 see.? An s. 4867 3- times. 19. If I do a piece of work in 108 days, of 8': hr., how many days of 6'- hr. would I be? Acs. 136. 20. A man borrows $1750, and keeps it 1 yr. 8 mon.: ow long should he lend $1200 to compensate for the favor? Ans. 2 yr. 5 nion. 5 da. 21. A garrison has food to last 9 mon., giving each man 1 lb. 2 oz. a day: what should be a man's daily allowance, to make the same food last 1 yr. 8mon.? Ans. 8 i'o oz. 22. A garrison of 560 men have provisions to last during a siege at the rate of 1 lb. 4 oz. a day per man; if the daily allowance is reduced to 14 oz. per man, how large a reinforcement could be received? Als. 240 men. 23. A shadow of a cloud moves 400 ft. in.184 see.: what was the wind's velocity per hour? Ans. 14ITX mi. 24. If 1 lb. Troy of English standard silver be worth ~3 6s., what is 1 lb. av. worth? Ants. ~4 2 id. 25. If I go a journey in 123 days, at 40 mi. a day, how long would I be at 29' mi. a day? Ans. 174 da. 26. If. of a ship are worth $6000, what is the whole of it worth? Ans. $10800. S U E S TION.-The two similar quantities in this question, are 6 ninthls of the ship, and 9 ninths of the ship. 27. If A, worth $5840, is taxed $78.14, what is B worth, who is taxed $256.01? Ans. $19133.59 28. What are 4 lb. 6 oz. of butter worth, at 28 ct. a lb.? Ans. $1.224 29. If I gain $160.29 in 2 yr. 3 mon., what would I gain in 5 yr. 6 mon., at that rate? Ans. $391.82 30. If I gain $92.54 on $1156.75 worth of sugar, how much must I sell to gain $67.32? Ans. $841.50 worth. 31. If coffee costing $255 is now worth $318.75, what did $1285.20 worth cost? Ats. $1028.16 32.: If I gain $7.75 by trading with $100, how much ought I gain on $847.56? Ans. $65.6859 33. A has cloth at ~$3.25 a yd., and B has flour at $5.50 a bbl. If, in trading, A puts his cloth at $3.,62, what should B charge for his flour? Ans. $6.13,6 34. TWhat is a pile of wood, 15 ft. long, 101 ft. high, and 12 ft. wide, worth, at $4.25 a cord? Ans. $62,75 9 RAY'S HIGHER ARITHMETI[C. 35. Find 7 mon. rent, at $4(5 a yr. ins. $2t7.)08 3). If'1 ia oat is rowed at tile rate of 6 miiles all }lo1or and is driven 44 feet in 9 strokes of' tlle o;ar, hovw nllalny strokes are made in a llinute? inlls. 108 strokes. 37. II f (of la yd. of' cloth cost ~3zW, what is thle worth of J of an E..? AIs. I1i3 In Fahrenheit's thermomceter, tlhe freezing point of wvate is marked 32~, and the boiling point 2120: in the Cenltigrade, the freezing point is 0~, and the boiling point 1000: in IReaumer's, the freczing point is 0~, and the boiling poillt S0~. 33. From these data, find the value of a degrce of each thlermometer in the deglrees of the other two. Alis. 1~ P'. = -~ F. 39. Convert 108~ F. to degrees of the other two tihermometers. (First snlL.razt 32~.) ins. 3 37 It. "arid 420 C. 40. Convert 25~ it. to degrers of tHie othler two tilermoileters. iA ns. 1 C. anl d 8 8 F 4-l. Convert 43~C. to degrees of the other two tihermoimeters. As I. 3tS,~ BR. andI I 1 I4 F. F In the working of macllinery, it is ascertained that tlhe avercrlltble )ontoer is to the towecliylht ovcercoLec,'ilve' sely as the dishtmaces theyt p/ss over i[, the stlame tiie. 42. If the whole power applied is 1 80 lb. and moves 4 ft., how far will it lift a weight of' 930 lb.? uAs. t3 in. NoTe.-Tlle zavailble power is taken'D of the lwhole power, - being allowed for friction and other ilpedinlents. 43. If 51 2 lb. be lifted 1 ft. 3 in. by a power moving (6 ft. 8 in., whlat is tHle power? It s. 144 lb. SU(IeFEST'ION.-Filld the available power; thenl add of itself. 44. A lifts a weight of 1440 lb. by a wheel andl axle; for every 3 ft. of rope that passes thlrough itis llaltils, the weight rises 4) in: Nlwhat power does he exerts? Ats. 270 lb. 4.5. A man weighing 198 lb. lct himself down 54 ft. with a unif'orm motion, by a wheel and axle: if the weigh.t at the hook rises 12 ft., how mluch is it? i/ts. 594 Ibt. 4(. Two bodies ft ree to move, attracet each other with forces that vary i[versely as their weigllts. If' thle weighilts are (9 lb. anid 4 lb. and the smaller is attracted 10 ft., how far will thle larger be attracted? ins. 4 ft. 51 in. 47. Suppose the earth and moon to approach each other in obedience to this law, their weights being 49147 and CO.'P)OUIN1D) PRl'OPOl'RTON. 199 12o3 respectively. flow many miles would the moon move while the earLh moved 250 miles? As. 99892+ mi. Can the following questions be solved by proportion? (See ReTn. Art. 230.) 48. If 3 men rmow 5 A. of grass in a day, how mann 3en will imow 13} A. in a day? 49. If 6 nmen build a wall in 7 da., how long would iin men be in doing the same? 50. If I gain 15 cents each, by selling books at $4.SO a. doz., what is my gain on each at $5.40 a doz.? COMIPOUND PRIOPORTION. AnT. 251. Comlpound Proportion is a method of solving questions by the use of a compound ratio, or by more than one proportion. The questions to which Compound Proportion is applied resemible those under simple proportion; but the value of the quantity required depends not on one pair, but on tVwo or nore pairs of' similar quantities: for instance, If 3 men imow 8A. of grass in 4da., how long would 1t0 imen be in nmowing 36 A.? S o L,;r I o N. —If the 10 men 1st step. %, of 4 da. were to mow 8 A. instead of 2d step. 3_6 of 3 of 4 da. 36 A., the question would be 5, d_. _ _s one of Simple Proportion, and a the answer would be 30 of 4 d., as shown in the 1st step of the operation; but this result being the time in which 10 men mow 8A, thlle tihue in which they will mow 36 A. will be 3' of it=- 3 6 of 3 of 4 dt., as shown in the 2d step, which reduces, by cancellation, to G — dxt. lience, a question in Compound Proportion, is only a succession of questions iln Simiiple Proportion, each of which gives a resualt to be used in the next, and the last result is the answer required. After sufficient practice, the successive steps may i.e dispensed with, and the answer written as a compottl t1 fraction at once. TO SOLVE QUESTIONS BY COIMPOUND PROPORTION, RuLE. —Form an increasing or decreasing ratio qf each pair of similar quatlities, as ~f the answer depended only on those 200 RAY'S HIIGHER ARItTHMETIC. two and the odd term; multiply these ratios and the odd term togelher; the product will be the answer. N o r E. The odd term is the one which is unlike all the rest. To TEACHERS. —Pupils should analyze each example thoroughly, and give the reasons for every step: such a- course will be a valuaIble training to the mental powers, and is the only way to clear up -n otherwise obscure and difficult subject. If the proportional form is preferred, use a succession of simple proportions, as already explained, or the rule in " Ray's Arithmetic," 8d Book, Art. 205. A simple mode of stating questions in proportion is this RULE OF CAUSE AND EFFECT. Separate all the quantities contained in the question into two causes and their effects. Write, for the first term of a proportion, all the quantities that constitute the first cause; for the second term, all that constitute the second cause; for the thlird, all that constitute the effect of the first cause; and for the fourth, all that constitute the effect of the second cause. The required quanltity ay w be indicated by a bracket, and fonzd by one of the rdles in Art. 249. N o TE.-The two causes must be exactly alike in the,nuq7ber and kind of their terms; and so must the two effects. 1. If 6 men, in 10 days of 9 hr. each, build 25 rd. of fence, how many hours a day must 8 men work to build 48 rd. in 12 days? SOLUTION.-6 men 10da. and 9hr. constitute the 1st cause, whose effect is 25 rd.; 8 men 12 da. and ( )hr. constitute the 251 cause, whose effect is 48 rd. HIence, 6 men. 8 men. 10 da.: 12 da.:: 25 rd.: 48 rd. 9 hr. ( ) hr. And the required term is 6 X I ~ X 9 X 4 8= 104 hrh. (Art. 249, Cor. 2.) 2. A boy makes 8 steps, each 1 ft. 10 in., while a lmana makes 5 steps, each 2 ft. 8 in.: how far will the boy ~g while the man is going 33 miles? Auns. 48 mi. RE vs E w.-251. - What is Compound Proportion? What does the value of the quantity required depend on? Solve the example. What is the rule? XVh:t is the odd term? What is the rule of cause and effect What is said of the two causes? The two effects? COMPOUND PROPORTION. 201 3. If 18 pipes, each delivering 6 gal. per minute, fill a cistern in 2 hr. 16 min., how many pipes, each delivering 20 g(al. per minute, will fill a cistern V7 times as large as tile first, in 3 hr. 24 min.? AsS. 27. 4. The use of' 100 for 1 year is worth $8: what is the use of $4500 for 2 yr. 8 mon. worth? Ans. $960. 5. If 12 men mow 125 A. of grass in 2 da. of 103z hr., litw manly lhours.1 day must 14 men work to mow an 80 A. field in 6 days? Anvs. 9 hr. 6. If 4 horses draw a railroad car 9 miles an hour, how many miles an hour can a steam engine of 150 horsepower drive a train of 12 such cars, the locomotive and tender being counted 3 cars? Ans. 22.1, mi. per hr. 7. low many half eagles, each weighing 5 pwt. 9 gr. and made of gold -T~o pure, are equivalent to 1000 English sovereigns each weighing 5 pwt. 3.274gr., and made of gold l)ure? Ans. 973~.''S. If the use of $3750 for 8 mon. is worth $68.75, what sum is that whose use for 2 yr. 4 mon. is worth $250? Ans. $3896.10+. 9. If the use of $1500 for 3 yr. 8 mon. 25 da. is worth $336.25, what is the use of $100 for 1 yr. worth? Arns. $6. 10. If 240 panes of glass 18 in. long, 10 in. wide, glaze a house, how many panes 16 in. long by 12in. wide will glaze a row of 6 such houses? Ans. 1350. 11. If it require 800 reams of paper to publish 5000 volumes of a duodecimo book containing 320 pages, how many reams will be needed to publish 24000 copies of a book, octavo size, of 550 pages? Ans. 9900 reams. 1. 2. A man has a bin 7 ft. long by ft. wide, and 2 ft. deep, which contains 28 bu. of corn: how deep must he make another, which is to be 18ft. long by 1V ft. wide, in order to contain 120 bu.? Ans. 44 ft. 13. If it require 4500 bricks, 8 in. long by 4 in. wide, to pave a court-yard 40 ft.:bing by 25 ft. wide, how many tiles 10 in. square, will be needed to pave a hall 75 ft. long by 16 ft. wide? Ans. 1728. 14. If 150000 bricks are used for a house whose walls average 1, ft. thick, 30 ft. high, and 216 ft. long, how many will build one with walls 2 ft. thick, 24 ft. high, and 324 ft. long? Ans. 240000. 15. A garrison of 1800 men has provisions to last 42 mon. at the rate of 1 lb. 4 oz. a day to each; how long 202 R AY'S HIiGHER ARITHM-IETIC. will 5 times as much last 3500 men, at the rate of 12. oz, per (.ly to each maln? A2ns. i yr. 7- mno n. 113. Whazlt sum1 of money is that whose use for 3 yr., at thie rate of';4.2 for every hundred, is worth as mluch as the use of $540 for 1 yr. 8 mon., at the rate of 87 for every hundred? Ancs. 6$416. 36 XIV. PERCENTAGE. ART. 252..IEnRCENTAGE is derived from the Latin phrase per centuin, meaning ol tfle htntldredf. It is understood to emnbrace all those operations in which reference is nlmade to 1 00 as a unit of conmparison. Its applications are, Gain and Loss, Commission, Interest, Banikin, Stocks, Insurance, )Duties and Taxes. If I have $750 in business, and gain $180, what is my gain on every 100? SO LUT IO.- This iS a question in Simple Proportion, and is solved as follows: $70: $100: $180: (); the 4th term is found by Cor. 1st, Art. 249, to be 00 --- $2:4. Ans. 750 IHence, I gain;$24 on every $100, or 24 per cent., since per cent. means on thle hzlndred. The t$24, in thle solution above, is called the rate per cent. of the $180 to the $,750. Thoulgh rlt/e per citt. strictly means t1Le nulmber on a 100, yet as',,$S _ on a $100" conveys thle same idea as "4 hlundedths,'' and tllhe latter is simIpler and more generlal, it: is the prev(a'iling' prIactice to consider thle'r(Ite 2per C/t. that one numbher is of anotlier, as the oambe~r of hittdredtlhs it is of that other, and any >per ccit. of a aunumber is so many Atldredlllts of it. Thus, 7 per cent. of any thing is 7 hundredths of it. 2. do.... is 2, do. do. 100 do..... is 100 do. do. or the whole. R E v E iw.-252. What is the meaning of Percentage? IWhat does it embrace? What are its princip:al applications? Solve the example. What does rate per cent. mean strictly? EI' CENTAGE. 203 2. If I hnave 160 slheep, and sell 35 per cent. of them how malny sheep do I sell? SoLuT IO. —l The whole flock (GO0) being 100 per cent., the proportion is, 160 sheep (.) sheep:: 100 per cent.: 35 per cent.; the required term is I O s 56 sheep. Ans. 3. In a battle, 78 men arc killed, whlich is 13} per cent. of the whole force; how many were engaged? SonrTION.-The whole force is 100 per cent.; the slain (78) are 13 I per cent.; hlence, ( ) me: 78 men:: 100 per cent.: 13g per cent.; tilhe required term is 78 X 100 3 X 8 X 1Ans00 =.... -= o58 men. Ans. I'dt.}40 Icence, there are three cases of percentage, accordin- as thie first of the two numbers complarcd, the secondd, or their rate per cent., is to be found. All these cases can be solvdc by the GENERAL RULE FOR PERCENTAGE. Ani1 iwo numbers are to each oltier, as lhe rates p)er cent. i/ey rep)resenlt of tle same qtieoutity. N o T E.-If one of the numbers is the stanclar~d of comparison for tlhe other, it will be 100 per cent.; if they are both referred to a third1 quantity, it will be 100 per cent., andl their rates per cent. -will depend upon tile nature of their relhtions to it. Altlhourgh this rule covers the wlhole subject, it is too genleral for practical purposes; therefore a special rule will be given for each case. CASE I. TO FIND ANY GIVEN PErt CENT. OF A GIVEN NUM3BER, ART. 253. RULEZ.-Alulti)y the given nttumber by tfhe given. rate 2)er cent., and divide'ei prodluct by 100. NOTES.-I. It often sazves work to indicate and cancel. 2. The dividing by 100 may generally be done by pointing qef in the product two more decinzlu places than are in the mulltiplicand aoln lizp tt}licer. SOLUTIONT.-9 per cent. of S182.50 i — of 9 $182..50-0= 9 times $182.50, divided by 100, as the rule directs. $' t 0 204 RAY'S HIGHER ARITHMETIC. REMARnK.-We may divide by 100 first, and then multiply by the rate per cent.; or, multiply by the rate per cent. written as decimal hundredths, 1. Find 6 per cent: of $82.37' Ans. $4.941 2. 14+ pr. ct. of 6yd. 2ft. 8in. Ans. 2 ft. 11.96 in, 3. 42 per cent. of $1250. Ans. $525. REArtAIRI.-Business men use instead of the words "per cent,' the character %: thus 42 per cent. is written 42 1%. 4, Find 62- % of 1664 men. Ans. 1040 men. 5. 35% of X A-ns. 6. 180 % of 4! Arts. 7 +a 7. 98 %, of 14 cwt. 2 qr. 20 lb. - 14 cwt. 1 qr. 155 lb. 8. 93 %o of 48 mi. 6 fur. 16 rd. 4 mi. 4 fur. 24 rd. 9. 331 % of 127 gal. 3 qt. 1 pt. 42 gal. 2 qt. 1 pt. So L UTT ON.-Since 313 =, take g of the given number. 100 10. Find 11' % of $3283.47. Ans. $364.83 11. 40 % of 6 hr. 28 min. 15 sec. A-is. 2 hr. 35 min. 18 sec. 12. 675 ci of 3 lb. 10 oz. 16 pwt. 22 gr. Ants. 26 lb. 4 oz. 4 pwt. 44 gr. 13. 104% of 75A. 1 R. 35P. =78A. R11. 382P. 14. 158 % of a book of 576 pages. Anls. 90 pages. 15. 23o %? of 45 ct. Alns.10 ct. 16. How much is 183 % of a cargo that weighs 416 T. 15 cwt. 20 lb.? Ants. 78 T. 2 cwt. 3 qr. 10 lb. 17. Find 337+ % of 11 Ans. 418. 56.' % of 144 cattle. Ans. 81 cattle. 19 16] % of 1932 hogs. Ans. 322 hogs. 20. 87 % of 1634C. 72 cu. ft. Ans. 1430C. 31 cu. ft. 21. 1000 % of $5.434a Ans. $54.37 22. 6- % of 10 Ants. > 23. 2+ % of 4 Ans..u REVIEs.-252. What is the rate per cent. that one number is of another considered? What is any per cent. of a number? Give examples. Solve example 2. Also, example 3. How many cases of percentage? What are they? 7 hat is the general rule for percentage? If one of the given quantities is the standard of comparison for the other, what rate per cent, will it represent? PERCENTAGE. 205 24. Wbhat part is 25 Eo of a farm? Ans. 2- -- 25. What part of a quantity is 64I % of it? also 83 a; 10; 12A %; 16 o; 20%; 331 %; 50%o; 66; %; and 75 %? Ans. 167 -, i-) 89 6, 5 3,:,, 2-, 4 Of it. 26. What part of a quantity is 18 % of it; 311 %; 371%; 43 4%; 5 6 4 o 5% 6 % 68 1; j & 4 %; 830 %; JlS, 1e, a ~, TO, TO I.,, 16 of it.- 4 27. How much is 100 % of a quantity; 125?% of it; 250 %; 675; 1000; 9437-1? Ails. 1 time, 1~ times; 21, C6, 10,- 941 times it. 28. A man owning 3 of a ship, sold 40 % of his shalre: what part of the ship did he sell, and what part did he still own? Ais. Ao sold; 4 left. 29. A owed B a sum of money; at one time he paid him 40 S% of it; afterward he paid him 25 % of what he owed; and finally he paid' him 20 SO of what lie then owed: how much does he still owe? Ans. - 9p of it. 30. WVhat is 78 %o of 12 T. 6 cwt. 8lb.? Acns. 9 T. 12 cwt. 1. qr. 31. Out of a cask containing 47 gal. 2 qt. ipt., leaked 61 %': how much was that? Als. 3 gal. 1I" pt. 32. A has an income of $1200 a year; lie pays 13 % of it for board; 10 O % for clothing; 6- c% for books; 17 %S for newspapers; 128 % for other expenses: how much does he pay for each item, and how much does he save at the end of the year? A11s. $156, board; $124.80, clothing; $81, books; $7, newspapers; $154.50, other expenses; $676.70, saved. 33. Find,- % of $1200. Ans. $10. 34. 1 %0 of $47.75. A-s. 238 ct. 35. 10 % of 20 % of $13.50. Ans. 27 et. 36. 40 % of 15 % of 75 % of $133.3388. Als. $6. 37. 8% of 62 1%of 150% of $462.50. Ans. $34.68' 38. A man contracts to supply dressed stone for a court-house for $119449, if the rough stone costs him REIE w.-253. What is Case 1? The rule? How is the division by 100 performed? How may the operation be performed? 206 RAY'S HIG tI1ER A.lllft'IIETIC. 16 t. a cu. ft.; but if h e can get it for 15 ct. a cu. ft., he will deduct 3 %/ flron his bill; how nmany eu. ft. nouild be nceded, and what does he char(re for dressi]ng a cu. ft.? Azs. 358347 cu. ft., and 17. ct. a cu. i't. 39. 48 % of brandy is alcohol; how much alcollhol dlcs a mnan swvallow in 40 years if' he drinks a gill of brandy b times a day? Aas. 657 gal. I qt. I pt. 2.4 gills. CASE II. ART. 254. Two numbers being given, to find the rate per cent. one is of the other. RtuIE.-AMntlflipy the ztz1ber wh!ich is to be the rcote per cent. by 100, and divide the plrodlct by the other nlu)nber. Or, whlen the numlbers lare small, Pt/ke saic/ a arlt of 100 as tle 1tt7uber which is to be the rate per cit. is of thle otler. Pn oo.-With the rate per cent. thus obtained, and the nuimber which is the standard of coinparison, proceed by the last rule; if the result is the same as the othler number, the work is right. 18 is how many percent. of 276? SOLUTION.-18 is how many per cent, 2 7) 1 8 1 of 276, means 18 is how rmany hu/nclredlth s 1 3 of 276. Now 1 hundredth of 276 is I, }, and as often as 18 contains this, so many 1 1 4 hundlredths will it be of 276; but 18 — 22 0: 17 1 X O (Art. 131), 1800 Abs. 6 per cent. 276, as the rule directs. 6 is how many per cent. of 9? SoLUTIO.-6 is 6 or O of 9, and to find how many hundredths of 9 it is, convert a into lundrledtlis. To do this, say o- uf 1 unit _- of 100 hunLdredcths - 6 hiundlredths- (tG G; or, nimultiply both terms of a by a numiber that will make the denominator 100; hicre, the multiplier is 33 1, which changes 3 to - 66 as before. 100 1. $14.40 is how many % of $54? Ais. 26s- 2. T E s T O N.-Reduce both to ct.; and generally reduce conmpound numbers to the same denomination before applying the rule. 9. 9 is how many % of C? Ans. 150 %. RE VVIEw.-254. What is Case 2d? The rule? 1W1hen the numbers are small what rule can be used? Explain example 1. Example 2. PEI- tLC TAG E.:07 3. 1ct. is how many % of $2? A nS. 7A 4. 2yd. 2ft. 3in. is how many % of 4rd.? Ans. 1~2 5. 3 a qt. 3 is w hat % of' 31.; gal.? Al.s. 11' J 6. -2 is h1ow lnny O of A Ans. 833 SoaLTTIONS. —1-' and 74=1; the first is; or' of tLt: last, but = of 100 % 3o. 7. - is how many %0 of? Ais. 250, 8. - of 3 of 7 is what % of' 1? Aits. 10119. 27 is how many % of 33? A ns. 663 10. 7 is how many % of I? Aiis. 96`a 11. 2 is how many % of;%? Azs. 2227 3 10 12. $5.12 is what % of $640? Ans. 7 13. 143 is how many % of 217? Aibs. 5 17 14. $3.20 is what % of $2-000? Ans. A. 15. $45 is what % of $12? Aibs. 375 16. 750 men is what % of 1I000? Aibs. 64 17. 8.- is how many % of.? 1Ans. 1050. 18. $7.29 is what % of $216? An s. 37 19. -3 is how many % of 7? Aits. 192a 20. 3 qt. 1. pt. is wvhat %A of 5 gal. 27 qt.? Ans. 16'-' 21. 1l6bu. 3pk is what % of 7.125 bu.? Ans. 236f97 22. A's money is 50 % more than B's; then B's money is how lmany %c less than A's? Ans. 33} SoLUTIoN.- Call B's money 100 /%; A's money is then 1;50 %,: their difference is 50 %, whiclh comnpared with A's money, 130, 9i 3 I a -3 23. If A's money is 10 %; 127, %; 25; 3 1; 43 %; 62 C; 75 %; 100 %5; 150 %; 200 C/; 225 375 c; 1000 % more than B's; then B's is how many % less than A's? is 9; 11 1 17 21, 0;0; 23 1,; 3077; 38P; 1 50 60; 66; 6 9a; 7 8 - 9 01 - 2-4. If A has 5 %; 15; 177%; - 2 %; 25%; 0; 45 ~;50; 63; t 75; 84 ~; 98%; anJ /9~~ ,9 Case III. PROOF-.-Find the given per cent. of the answer, and add it to, or subtract it from, the answer, as may be neeessary; the result must be the same as the given number. P EtL CENT A GE 1. 1 i I p:ay $377 rent for my house, which is 16 % morc than I paid last year: what was the rent then? SOhUTION.-Call the rent' 1 00 last year 100 per cent.; the 1 - rent this year being 16 O more, is 116 O; divide $337 by 116, to 116) 3 7. 00 (3.2 5 get 1 %, and multiply the quotient, $3.25, by 100, to get 100%, 0 3 2 50 0 or the rent last year, $325. I am 28 yr. old, which is 142 per cent. less than rly brother's age: how old is he? So LU TI O N.-Call my brother's age 100 %; my age being 14'7- / less, is 85, ~; 28 yr. divided by 857 =, i; n 100 tre this, or 1 6= 32, is 100 %o, or my brother's age. 1. 136 is 20 % less than what? An s. 170. 2. $4.80 is 338 % - more than what? AIs. $3.60 3. is 50 ~ more than what? Ants. 9 4. TN is 28 % less than what? Avs. 7 5. 96 da. is 100 %v more than what? AIs. 48 da. 6. 2576 bu. is 60 % less than what? Ans. 6440 bu. 7. 87 cet. is 872 % less than what? AIts. $7. 8. 1a is 500 % more than what? Ans. -r9. 3 bu. 2 pk. 7 qt. is 8 % more than what? Ai1s. 3 bu. 1 pk. 6~ qt. 10. 42 mi. 1 fur. 20 rd. is 55 % less than what? Anls. 93 mi. 6 fur. 11. 2 lb. 9 oz. 4c dr. is 50 % less than what? Ans. 5 lb. 2 oz. 9 J dr. 12. T'( is 99- % less than what? Ans. 155` 13. 9$920.93} is 337, % more than what? AIs. $210.50 14. $4358.061 is 233-. % more than what? Ans. $1307.41 15. In 641 gal. of alcohol, the water is 7? % of the spirit; how many gal. of each? Ans. 60 gal. sp., 4dgal. w. 16. A coat cost $32; the trimmings cost 70 o less, and the making 50 % less, than the cloth: what did each cost? Ans. Cloth $17-.77~, trim. $5.333, mak. $8.88' SUGGEST'ION.-Cloth = 100 %0; trimming = 30 %; making = -50 % their sum 180 o = $32. Find 1 %11 and then the rest. 12 RAY' S ilIGIE R AR1TITMETIC. 17. If a, bu. of wheat makes 39-lb. of flour, and the cost of grinding is 4 /:; h fo many bbl. of flour can a farmler get for 80 bu. of wheat? A us. 15 r' bbi. 1S. H1ow many eagles, each containing 9 pwt. 16.2 gr. of pure gold, can I get for 455.5G 38 oz. pure gold at the mint, allowing 1 % for expense of coinage? los. 9) 8. 19., 0047 is 10 % of 110 ~% less than what numblher? Ao.s. )300. 20. 4946G is G % of 50 % of 4GG6; % mole thin whalt number? _Als. 3 7 2. 21. A drew out of bank 40 % of 50 % of hO of'i0 c of hIis moneoy, and had left 115357.2 0; how miuch had Ile at first?.nts. 1 70t0. 22. I gave away 4'2 % of my money, and had left $2; what had I at first? Alts. $3.50 2:3. A man dying, left 3a %7 of his property to his wife, CO % of tHle remainder to his son, 75 %-0 of' the remaiiilder to his daugllter, alnd the bialance, $5 00, to a servant; wlaLt was tle rwlole property, and each silare? Il.,s. Property,,7500; wile had $2500; son $3000; daughter l) (00. 24. In a company of 87, the clhildren are 378-. % of the women, who are 44- % of the men; how many of' each? Ajis. 54 men, 24 womer-, 9 children. 25. In a school, 5 % of the pupils are always absent, and the attendance is 570; how many on the roll, and how many absent? Als. 600; and 30. XV. APPLICATIONS OF PERCENTAGE. ART. 257. The importance of thoroughly understanding Percentace, can not be over estimated. In marking goods, in calculatini gain or loss, the value of' illvestlnlCnts, interest, commission, &C., its applications are vnarious, imiport anyt, and of d aily occurrence; and no one can be' a ready and coumplete accountant until he is fniiliar with its principles and met.hods. Although the special rules are the simplest and most satisfactory, especially in the first two cases, the general ItR E v w.-256. What is (Case 4? The rule? The proof? Solve examnple. GAIN AND LOSS. 213 t'ule (Art. 2952), is casy to recollect, and serves very well for the filst three cases, and even for the fourth case, after slighlt preparation. P1articular care should be exercised in ascertaining t l, quanltity which is the standard of comlparison on wlhic the rate per cent. takes effect; and bear in mind that t/,i: ctal;tty alwc ys represents 100 petr cent. R EIIAPRKS.-I. The words per cent. have no reference t,,oulleS., All they mean is, on, or by the hundred, and a rate per cent. is a numtiber of hundredths of the quant.ity, whatever it may be. It is tlue, the phrase is used more in speaking of money than of any other article. but it is perfectly proper to say, "Ten per cent. of the labor;" "Twenty per cent. of the cloth " "Seven per cent. of tlhe time;" meaning so many hundredths of the labor, cloth, and time, respectively.,2. The advantage of using rales per cent., or ratios in kundredths, instead of ordinary rratios, is that rates per cent., like all other, fractions having a common denomninator, are more easily compared tllan ordinary ratios. Thus, it is not easy to see wlhich of the two ratios - anld q is the gieater, but as soon as they are expressed in hundrl.edlhs, viz: 42$ % and 444g %, the difficulty vanishes. XVI. GAIN AND LOSS. ART. 258. The increase or decrease which any variable qulnrttity undergoes, is called its gain. or loss. The ratre o'jgai at or loss is the rate per cent. the gain or loss is of the quantity on which the gain or loss accrues. The quantity on which gain or loss accrues, is the standard of comparison in questions of gain or loss, and is therefore 100 per cent. GENERAL RULE. ipresent the qlcuanttitj on wlzich gain, or loss acceres by 100 pe? cent., and proceed by sucih rule of Percenltage as the szature of the qutestion? requires. NOT E.-In gain or loss of money, the sum of money invested, or cost, is the standard of comparison, and is therefore 100 per cent. RaviEw.-257. What single rule comprehends all the cases? What care should be exercised in all questions of percentage? What must the standard of comparison always represent? 214 RAY'S HIGIHER ARITHMETIC. ART. 259. There are 4 eases of Gain or Loss, solved like the 4 corresponding cases of Percentage. CASE I.-Given, the quantity on which gain or loss accr?ues, andt the rate of gain or loss, to find the gain or loss. Having invested $4800, my rate of gain is 137 4: what is my gain? $4800 ANALYSIs.-The $4800 being 13 the quantity on which the gain 144 0 0 accrues, is 100 ~%; the gain being 4 8 0 0 137 = l]37 hundreths of it, is 4200 found as in Case I of Percentage, by multiplying it by 1387, and di- ~ 6 6 6.0 0 Gain. viding the product by 100. $ 4 0 0 $ 5 4 6 6 Sum after gain. REMARKs. -When the gain or loss is known, the amount afte7 gain or loss, is obtained by an addition or a subtraction. 1. If my rate of gain is 25 %, how should I mark goods for sale that cost me $8; $7.50; $6.25; $4.75; $3.87.1; $2.62.; $1.93:3; 62.ct.; 15ct. a yard? Ans. $10; $9.37'; $7.81-4; $5.9334; $4.84-l; $3.28'; $2.42-3,; 78- ct.; 18- ct. a yard. 2. If I must lose 20 %, on damaged goods, how should I mark those that cost me 12 et.; 25 et.; 43 t.; 15 ct.; $1.10; $2.40; $3.50; $4.37a; $5.814; $6.564; $7.68,; $8.10 a yard? Ans. 10ot.; 20ct.; 35ct.; 60ct.; 88ct.; $1.92; $2.80; $3.50; $4.65; $5.25; $6.15; $6.48, a yard. 3. The population of a town was 1760 last year, and has increased 264 %: what is it now? Ans. 2222. 4. A city containing 42540 inhabitants, lost.11] % of them by cholera: how many died, and how many are left? Ans. 4963 died; and 37577 left. 5. A lb. Tr. contains 5760 gr., and a lb. av. 2119 % more: how many gr. does it contain? Ars. 7000 or. R E V I Ew..-257. What is said of the words per cent.? What n dvanitag:c in using rates per cent. instead of ordinary ratios? Give an example. 258. What is gain or loss? What is the rate of gain or loss? In all questions of gain or loss, which quantity is 100 %? What is the general rule? In gain or loss of money which quantity is 100 %? Why? 259. How many cases of gain or loss? What is Case 1? Anaiyze the example. How can the quantity after gain or loss be found? GAIN AND LOSS. 21 f Gi. A (.g:es 42 mi. 3 fur. 18rd. a day; B, 15 % faster: how fatr does B go per day? Ans. 48mi. 6 fur. 14.7 rd. 7. U. S. standard gold is o% pure, and English standard gold is 12 purer: how pure is it? Ans. -.4 pure. i. U. S. standard silver is -o pure, and English standard silver is 2- % purer: how pure is it? Ans. 3- pure. 9. The cost of publishing a book is 50 ct. a copy; if the expense of sale be 10 % of this, and the profit 25 cS: what does it sell for by the copy? Ais. 67E ct. 10. A began business with $5000: the 1st year he gained 14i3 %, which he added to his capital; the 2d year he gained 8 0,,which he added to his capital; the 3d year he lost 12 %, and quit: how much better off was he than when he started? Agis. $452.92 11. A bought a farm of government land at $1.25 an acre; it cost him 160 % to fence it, 160 % to break it up, 80 % for seed, 100 %a to plant it, 100 0 to harvest it, 112 c for threshing, 100 % for transportation: each acre produced 35 bu. of wheat, which he sold at 70 et. a bushel: how much did he gain on every acre above all expenses the first year? Ans. $13.10 12. -What must I sell a horse for, that cost me $150, to gain 35 a,?' Ans. $202.50 13. A gave $4850 for his house, and offers it for 20 % less: what is his price? Ans. $3880. 14. Bought hams at 8 et. a lb.; the wastage is 10 %: how must I sell them to gain 30 a%? Ans. 11- ct. a lb. SUGGE SrT ION.-Since a lb. wastes 10 %, or -O, I get only -TO lb for 8 et., and a lb. at that rate costs 88 et.; to which 30 %o must be added, to get the selling price. 15. I bought a cask of brandy containing 46 gal. at $2.50 per gal.; if 6 gal. leak out, how must I sell the rest, so as to gain 25 %? Ans. $3.b59 per gai. 16. I started in business with $10000, and gained 20 % the first year, and added it to what I had; the 2d year I gained 20 %, and added it to my capital; the 3d year I gained 20 W. What had I then? Ans. $17280. ART. 260. CASE II.-Given, the quantity on which ga'in or loss accrues, and the gain or loss, to find( the rate of gayn, or loss. 21 f6 RAY'S HHIGHER ARITHMETIC. REM rtKa.-If t1he quantity on -which gain or loss CCItIIces, an1( thle quantity o(If/r gain or loss, are given, thleir difference is the galin or ross, and tile (qns'ion woUtlJ come uttnder t.his case. rThe population of Cincinnati in 1840 was 45000, and in 18)0 was 165000; how many per cent. lihad it in. creased in the interval? ANALYSIS.-llerc the gain is 120000, which, compared w-ithl 45000, the quantity on which the gain accrues, is -1 0 0 - 8 - 3 of 100 = 266ai S. An7s. 1. If I double my money, how many per cent. do I gain? Ans. 100. 2. If I lose half my goods, how many per cent. do I lose? A4s. 50. 3. If I buy at'1 and sell at $9, how many per cent. do I gain? AnIs. 800. 4. If I buy at $1 and sell at $4, how many pCe cent. do I gain? Ais. 300. 5. f' I buy at $4 and sell at $1, how many per cent. do I lose? nA is. 7;. G. If I sell' of an article for what the whole cost me, how many per cent. do I gaiI1? A Is. 80. 7. If I sell - of an article for what 3 of it cost le, how many per cent. do I lose? SAs. 31 8. If' a person sell 14 oz. of candles for a pound, how many per cent. does he gain? Ats. 14 9. -low many larger is the ea sthls equatouial diaill eter (15850 mi.) than its polar diameter (15798 ini.)? APus. 3- %o nearly. 10. If I sell an article for - of its cost, how many per cent. do I lose? A as. ( G` 11. The U. S. half dollars coined since 185)3, contain 8 pwt. of standard silver; those coined bef'ore, coltaiil 8 pwst. 14. gtr. of stlandard silver; how niany per celt. nttore valuable are the latter than the forntier? 1 s. i; 112. A lo', 1 ft. 6 in. thick, is sawn into 13 bolards eah il 11 in. thick: what % is wasted? A)Is. 9S) 13. The U. S. wine gallon contains 231 cu, in., and the beer gallon 282 cu. in.: how many c% larger is the latter R F, V I 1 w.V-260. What is Case 2? The rule? If the quantity on which gain or loss accrues, and the quantity after gain or loss, are given, how do weo proceed? Analyze the example. GAIN ANl) LOSS. 21 tllan the former? and how many % smaller is the former tilan the latter? Ais. 227 ~ larrer; 1847 % smaller. 1 t. The U. S. dry gallon or half' peck, contains 268.8 etu. in.: how many % larger is it than the wine gallon? and how mally (% smaller than the beer gallon? Ais.. lGT %6 larger; 4i % smaller. 13. The imperial gallon of" G- reat Brittain contain 277.24 4cu. in.: how nmany (% is it larger tlhan our wine and dry gallons? and how many %/ smaller than our beer gallo11? A nS. 20i {- % larg er; 3 -0- % lacgel, 1 41 -1 o lcss. 1G. U. S. goold (~q pure) is how many %, less pure than Eng-lish gold, (Vz pure)? Ains. b, 17. Gold 22 carats is how many pe cent. purer than gold 18 carats fine? and how many ~% pure is each? A ns. 22( 2 I puler; fi rst, 91t Ic pure; 2d, 75 c, pure. 18. If I pay for a lb. of sugar, and get a lb. Troy, what % do I lose, and what % does the grocer gain by the cheat? Aits. 17 %' loss; 21:'', ain. 19. A, having ftailed, pays B $1T50 instead of $2.50.( which he owed him': what % does B lose? A2s. 30. 20. Sugar bought at o-r ct. a lb., is sold at 7 ct. a lb. what is the rate of gain? Ain. 14 j. 21. An article has lost 20 %v by wastage, and is sold for 40 % above cost: what is the gain per cent.? Alns. 12. SOLUTIoN. —— 20 of the cost is wasted; 80 SO of the cost remains ftvaila,ble, on which a gain of 40 % is realizecl; 40 %o of 0 80 8 32 %, which, added to 80 %, makes 11'2 7;' deduct the cost, 100 %; the net gain is 12 %. 2'2. If my retail gain is 33t %, and I sell at wholcsale for 10 %c less than at retail, what is my gain %0 at wholesale? Ans. 20. 23. Bought a lot of glass; lost 15 "% by breakage: at -,lwhat (% above cost must I sell the remainder to clear 20 % on the whole? Ans. 41 -i7 24. If a bu. of corn is worth 35 et. and makes 2 gal. of whisky, which sells at 24 et. a gal., -what is the profit cf the distiller? Ains. 71 0o. ART. 261. CASE III.-Given, the gatiri or loss, an(l the rote, of gntbb or loss, to find the quantity on which gain or loss (xCCute,($s. 19 21 8 ltn Y'R S IiAS 1611I iER A TIITtlMETIC. By selling a lot for 34} % more than I gave, my gain is $423.50: what did it cost me? ANALYSIS.-Since 34 -- = $423.50, 1 % = $423.50 - 3,' & - $12.32; and 100 %, or the whole cost - 100 times $12.32 -- $1232; as in Case 3 of Percentage. R ETMAR. —After the quantity on which gain or loss accrues is known, the gain or loss may be added to it or subtryacted from it, to get the quantity after gain or loss accrules. 1. I-ow large sales must I malke in a year at a profit of 8 a% to clear $2000? Anls. $25000. 2. I lost $50 by selling sugar at 2,, below cost: what was the cost? AnS. $222_. 2 2 j 3. If I sell tea at 13} % gain, I make 10 ct. a l.. how muCh a pound did I give? A ns. 5 ct. 4. I lost a 2h dollar gold coin, which was 7} of all I had: how much had I? Aus. $35. 5. A and B each lost $5, which waS 2 %~ of A's and 3- o of B's money: which had the most money, and how much? Ans. A had $30 mnore than B. 6. I gained this year,$2400, which is 120 c% of my gain last year, and that is 4443 % of my gain the year before; what were my gains the two previous years? Ais. $9000 last year.; $4500 year before. 7. The dogs killed 40 of my sheep, iwhich was 4t( % of my flock: how many had I left? Aus. 920. AiT. 262. CASE Iv.-Given, the qanzity,after,qlin or loss has accruedI, annd the rate of yain or loss, to jifad the quantity on whtich gaiT or loss accrues. Sold goods for $25.80, by which I gained 71 %: what was the cost? SoLUTION.-The cost is 100 %O; the $25.80 being 7 1 msnore, is 107. %O; then 1 % =$25.80 -107-= 924ct., and 100 % or the cost = 100 times 24 et. = $24; as in Case 4 of Percentage. REMnA, -- After the quantity on which gain or loss accerues is found, the difference between it and tlhe quantity tJfter gain or loss aecrues, wTill be the gain or loss. RerVIEW.-261. What is Case 3d? Analyze the exsnlple. Howr can the quantity after loss or gain be found? 262. What is Case fth? Analyze the example. How is the gain or loss found? GAIN AND LOSS. 219 1. Sold cloth at $3.85 a yd.; my ~gain was 10 %: how much a yard did I pay? AIs. $3.50 2. Gold pens, sold at.$5 a piece, yield a profit of 33 %: vwhalt did tliey cost a piece? A,,s. 3. 75 3. Sold out for $9'3.82 and lost 12 %: what was the cost? and what would I have got if I had sold out at a gain of 12 %c? Al is. $1082.75 and $1212.68 4. Sold my horse at 40 C, gain; Awith the-proceeds Il bought another-l and sold him for p238, losing 20 ~o: what did each horse cost mne? Aiis. $ 212.50 for 1st, $297.50 for 9d. 5. Sold flour at an advance of 1:3 c%; invested the proceeids in flour aoain, and sold this lot at a profit of 24 e, realizing $;390t.0: how much did each lot cost me? Aois. 1st lo t, lot, 3187.50 6. An invoice of goods purchased in New York, cost me S 5 for' tainsportta tion, and I sold them at a gain of 16 % otn their total cost on delivery, realizing $12 60: whalt were they invoiccd at? A.;. 41000. 7. Tile population of a village increased 50 % each year on the previous one, for four successive yeairs, and at the end of thle 5th'was 405: what was it at the end of eachl previous year? Ans. 80, 120, 180, 270. 8. 7For 6 years my property increased each year on the previous, 100 %, and became worth $100000: what was it worth at fir'st? Ails. $,1562.50 9. A lost at play 50 %/ of his money thle 1st garme, 50 ~% of the remainder the 2d, and so on for 10 successive gamIues, when he was reduced to his last dollar: what had lie at first? Ans. }10:24. XVII. COMMISSION AND BROKERAGE. ART. 263. One who buys or sells property, makes investmnents, collects debts, or transacts other business for the benefit, and at the advice of another, is a comInzs1ssio0 msccrchlzt, clgefft, or factor. AWhlen the commission merchant lives in a different country or part of the country fiom his employer, he is fequently called corres pondcent! or consignee; goods sent R EVI, w.-263. What is a commission merchant, agent, -or fictor? 2241 0 RAY'S HIGHIER ARITHMETIC. to hiln to be sold are called a consignmzent, and the person sending them a consigncor. The charge made by a commission merchant for transacting another's business, is called his conmntision,, and is estimated at a certain rate per cent. of the sum invested or realized for the other's benefit. This rateper cent. is the ralte of con27issiom. BROKIERAGE is a charge of the saime nature as commlssion, but generally smaller. ART. 264. The am ouint of the sale, pur(rhase, or collectio'n, is the standard of comparison, and is, therefore, 100 %. The net _proceeds of a sale or collection is the sum left after deductin, the commission and other charges. GENERAL RULE FOR COMMISSION. Represent by 100 per cent. the quantity on which tlhe comm'lssion or brokerage is chargeyed, (whic/h is always the a2moutnt of the sale, purchase, or collection,) and thlen, proceed, by snlch rule of Percentage as the natlure of the quiestion requires. Commission has 4 cases, solved like the corresponding cases of Percentage. CASE I. ART. 265. Given, the amount of the sale, purchase, or collection, and the rate of commission, to find the commission. A commission merchant makes sales during a year amounting to $168475.37l., on which his charge was 2-' 0: how much did his commission come to'? ANALYsis. —The amount of sales, $168475.37, is 100 %; the commission, being 2'. 0, is found by multiplying by 21, and dividing by 100, as in Case I of Percentage; thlis gives $4211.88. PiE AR KI.-If the commission ($4211.88) be subtracted from the amount of sales (1684-175.374), the remainder ($164263.49.-) will be the amount paid to the consiguors. 1. I collect for A, $268.40, and have 5 % commision: how much do I pay over? Ans. 8254.98 REIrEVmE.-263. What is a consignol? A consigument? A consignee? How is a commission merchant paid? What is the rate of commission? What is brokerage? C(OMI.\IISSIUN AND UI2t0KERIAGE. 2'2. I sell for 1, 6(350 bbl. of flour, at $7.50 a bbl., 28 bbl. of whisky, 35 gal. each, at 2e2 ct. a gal. wlht is mly commissiol at 2.2,? As. $ 1 i4. 15 3. Received on conmmissiot 25 hhd. sugar (36547 lb.) of whichll I sold 10 hhd. (16875 lb.) at 6 ct. a lb., and 6 hhd. (8246 lb.) at 5 et. a lb., and the rest at 51- et. a lb.' wuhat is my commission at 3 %? Ais. $61.60 4. A lawyer charged 8 % for collecting a note of $648.75: what is his fee, and the net proceeds? Ais. $51.90, and $,596.85 5. A lawyer, having a debt of $1346.50 to collect, compromises by taking 80 O, and charges 5 % for his fee: what is his fee, and the net proceeds? Ans. $53.86, and $10.23.34 6. Bought for C, a carriage for $950, a pair of horses for $575, and harness for $120; paid charges for keeping, packing, shipping, &c., $18.25; fieight, $36.50: whllat was my commission, at 3al %0, and the whole bill? Ais. $54.83, and $1754.58 NOTE.-Commission is charged only on the amount of the purchase. 7. Sold 500000 lb. of pork, at 5a et. a lb.: what is my commission, at 124 C%? Als. $83437.50 8. An insurance agent 11as 10 0 of all sums received for his company: what does he make in a year, if he receives for the company, $28302.75? A ts. $2830.27 9. An insurance agent has 5 ~c of all sums received for his conipany, and 5 K/ of what remains at the end of thle year after payment of losses: what will he imake, if' lie recceives for his company, $473063.87i, and )pays losses, $31344.50? As..3169.1 ]6 10. An architect charges 31s % for designing and superintending a building, which cost $27814.60: what is his fee? Ans. $973.51 11. A factor has 237 % commission, and 3 c % for guaranteeing payment: if the sales are $6231.25, what does he get? Ans. $389.45 R J, r E Wv.-261. What quan tity is the standard of comparison? What does it represent then? Wha.tt are the net proceeds of a scale or collection? What is the general rule for colnriission and brokerage? IHow inliny eases of commission? 265. What is Case 1? Analyze the example. What are the net proceeds? 222 RAY'S tIG IIER ARITHMETIC. 12. An architect charges 1} % for plans and specifica tiOnlS, 11nd 2 for sueri illteldi'ng: what does lie Iiialc, if til buildilg costs $14902.50? -1s. $611 73 13. A brokcr sells for nie 10 lild. sugar (9)2 G lb.), at 5 ct. a lb.; lwhat is his brokcraIe, at CZ, ald my 111' )ceds? itls. Brolkerage, $3.47; proceeds, -54.to)3 14. A sells a. house and lot for inc at. 3830, and claros %c brokerage: what is his fee? AlIs. 024.06. 15. I have a lot of tobacco on commission, anrd sell it throulghl a broker for $4642. 8 my commi ssio n is 2o is, the brokerag'o 12 / what do I pay the broker, and what do I keep? Ans. I keep $63.84; and brok. $59.23 1f. What does a tax gatherer get for collecting a tax of $37850, at 3 a-, rid Iiow mIuch does -le pay over? A1s. $113o.50 rec'd; $83(714. 0 paid over. 17. A tax collector is paid 4,i co for collccting a tax of $218096. 75 What is his f'ce, and( the net proceeds? An-s. $9814.35 f'cc, $208282.40 paid over. CASE II. ART. 266. Given, the commission, and the amount of the sale, purchase, or collection, to find the rate of comnmission. 1. An auctioneer's commission for sellingn a lot was $50, and the sum paid the owner was $1200: what was the rate of' commlission? A)s. 4 %oSUGoGsTiON..-Find the amount of the sale, $12.50; th1en find lhat $7o50 is of $1250, as in Case 11 of Percentage. 2. A commission merchant sells 800 bbl. of flour, at $6.43' a bbl., and reCiits the net plroceds, $5021.25: what is his rate of' commission? Ails. 2,. 3. The cost of' a building was 819017.92, including, the architect's commission, which was $5$53.9.2 what rate did the architect charge? Alns. 3 %. 4. lBought flour for A; my whlole bill was $5802.57, including charges, $76.85, and commiission, $148.72: find the rate of commlission. A)s. 2)'. 5. Chark'ed $~52.50 for collcting a debt of $1050: what was lmy rate of comlmission? Aiis. 5 G. An agent gets $169.20 for selling p-roperty for $,8460: what was his rate of' brokerage? Ais. 2 %. COM5IILSL1S!;b,A.NDA BL1iKERAG-E.. 3 7. MLy commission for selling books was $6.92, and the net- proeeds, $62u. 2: what rate did I charge? 8. Paidl $38.40 for selling goods worth $6400: what was the r1ate of' brokerage? Als. -- C. 9. Paid a broker;;24.16. an d retained as my p:4'V of' the commilssion $4 }2.28, for selling a consignment a{!$o 113:it' whatt rwas the'rate of brokerage, and nlmy rat, a-t' commllistsion? Aiis. Brok. I /; Coln. 2: 5 10, A tax gatherer is paid $8711 for colleeting a Ltax of iT74220:' what rate is allowed him? Ars. 5 %. 11. A tax collector retains $6826.45, and pays over 08 9 28.9.53 whas1 t c/ is Iiis commlission? AIs. 1i c0 CASE i.L ART. 267. Given, the colmmlission, and rate of corn., to fand the sum on whicth commiassion is charged. N oT E.-After finding tle sum on which commissionis charged, subtract tilhe commissioin, to find the net proceeds, or add it, to find lthe w/to1 cost, as the case imay be. 1. MBy commissions in 1 year, at 2. %, are $3500: what w ere the sales, and the whole neIt proceeds? A2?s. p$140000 and $136500. SUOE C T, s o.l o -s,- % $300; find 1(, then 100I; a.s in C ase ill ot I'ercenlltne. 2. An insurance agent's inconme is $17833.45, being O 0 1 on the sunms reeived for the collpal)y: what were the company's net receipts?'A"as. $15601 05 3. A pork' merchMe-a.lt charged 15 % comrmission, and cleared $ 7o 1, I after pavying out $1O20G.5 for all exy pemises of' p:tcli: blow imany pounds of pork did he paclk if it cost 4: ct. a potandl? Am,s. 530800 lb. 4-. An aeoat purichased, according to order, 10400 hbu. of wheat; h1is commissiol, at 1 5-, was $ 56, and cihai L,(S for storage, shipping, and f'reih t, $527.10: what did Ihe:ay 2 l)bushel? anrid what was the whole cost? AUs. $8.20 a bh., and $131683.10, 1whole cos;. 5. Paitd $64.05 for selling coffee, which was s % brok. c2ra(,e w what are the net proceeds? A ls $7255.o9t5 E v i n w, -2,66. What is Case 2? 267. 1What is Caseo? HI-ow are the net proceeds ficund? low, the,whole cost? 24 RY'S IHIGHER ARITtIMETIC. 6. PReceived produce on commission at 2} c%; my surplus commission, after paying i i brokerage, is $107.03: lwhiat was the amount of the sale, the brokerage, and net pro ceeds? Ans. Sale, $6116; brok., $30.58; pro., $59731.39 7. A is paid $861.27 for collecting taxes at 2 1: what were the taxes, and the sum received by the State?' Ans. Tax, $3445 0.80, paid to State, 8$33589. 53 8. Paid A $1952.64 for collecting, at 11 c v: what were my net proceeds? Ans. $109626.79 CASE IV. ART. 268. Given, the rate of commission, and the net proceeds, or the whole cost, to find the sum on which commission is charged. NOTES. —1. After the sum on which commission is charged is known, find the commission by subtraction. 2. When the divisor in this case is little less than 100, use the contracted method of Division. (Art. 69.) 1. A lawyer collects a debt for a client, takes 4 %c for his fee, and remits the balance, $207.60: what was the debt and the fee? Ans. $216.25 and $8.65 SOLUTION.-The debt being the quantity on which commission is charged, is 100 %; if 4 % is taken, there is left 96 G- = $207.60: find 1 o and then 100 %; as in Case IV of Percentage. 2. Sent $1000 to buy a carriage, commission 29 %. what must the carriage cost? Ans. $975.61 SUGGESTION.-100 - 21 =- 1021 - $1000; find 1 a then 100 %. 3. A buys per order a lot of coffee; charges, $56.85; commission, 1: (/; the whole cost is 8539.61: what did the coffee cost? An.s $4876.80 SuGoGFsTiON.-TaJke out the charges; the rest as beforle. 4. Buy sugar at 2. % commission, and 21 for guar Antecing payment: if the whole cost is $1500, fwhlat w7Js the cost of the sugar? Alas. $1431.98 SGCoo ESTION.-100- 2o + 22-1 -104 = $1500; rest as before RL l:l tw.-268. Wha;t is Case 4? Hlow can the commnission be -found If the divisor is little less than 100, what trn.y be done? ''OCKS AND DlIVIDENDS. ~b 5. Sold 2000 hams (20672 lb.); commission, 2- %, guarantee, 2 %,/ net proceeds due consignor, $4-48..34 what did the halis sell for a lb.? A)is. 12 1 et. 6. Whlat tax must be assessed, to yield'$26782.45 net proceeds, and pay the collector 3 c%; and what is the collector paid? ALns. Tax, $27(646.40; col., $868.95 7. What tax must be raised, to yield $1044073.50 nd pay for collection, at 8 (? Ais. $1050640. S. A sells 1000 bbl. (304684 gal.) of whisky; brokerage, ~ %; proceeds, $7254: how much a gallon was it sold for? i Ais. 24 et. 9. Sold cotton on commission at 5 %; invested the net proceeds in sugar, commission, 2 %; my whole commission was $210: what was the value of the cotton and sugar? Ans. Cotton, $3060; sugar, $2850. S U G G E S T I o.-Cotton = 100. Corn. on Cotton = 5 %. Proceeds -=. Corn. on sugar=. o 05 % (Art. 256) - 144 a 441 44 Whole Corn. =5 6 14 4 = B 2 = $210; find 1,%, &c. 10. Sold flour at 3- % commission; invested of its value in coffee, at 1 % commrission; remitted the balance,,$432.50: what was the value of the flour, the coffee, and my commissions? Als. Flour, $1500; Coffee, $1000; 1st Coin., $52.50; 2d Com., $15. 11. Sold a consignment of pork, and invested the proceeds in brandy, after deducting my commissions, 4 f% for selling, and 1 %o for buying. The brandy cost $2304.00 what did the pork sell for, and what were my commissions? Als. Pork, $2430; 1st Com., $97.20; 2 Com., $98.80 12. Sold 1400 bbl. of flour, at $6.20 a bbl.; invested the proceeds in sugar, as per order, reserving my commissions, 4 %/ for selling, and 11 % for buying, and the expense of shipping, $34.16: how much did I invest in sugar? iAns. $8176. XVIII. STOCKS AND DIVIDENDS. ART. 269. A joint-stock company is an association of individuals, empowered to transact a specified business under certain restrictions. RAY'S HIGHER ARITHMETIC. Thle business transacted by them is generally such as to exceed tlhe means of one person; as, Ba.nking, Mining, Insurance, &c. alilroads, canals, steamboats, turnpikes, bridges, telegrapls, &co, arle owned and managed by joint-stock companies. The capital of a company is called its stock, and, for conrvenience is usually divided into slhares of $100, or $50, for each share a certijIccIte is issued. Persons who own shares are called stoclkh7ldcers, or sharleholf lers; stock can be transferred from one person to anothlie, the certificates being evidence of ownership. The d iciviulct( is the ogain to be divided amonl- the stoc holders, in proportion to their anmounts of stock. Hence, a joint-stock company is in the nature of a partnership. On, account of' the great number ol' shares in such a concern, it is convenient to declare the dividend, as a certain rate per cent. of the whole stock; this rate per cent. may be called the ratce of diidelcld. CASE I.-GiveLa, tlhe stoc]k, andl crate of divif(lend, to find1 the (cliicldtCl for t/hat stockl. (See Case I, Percentage.) 1. I own 1 8 shares of $50 each, in the City Insurance Co., which has declared a dividend of 7 w- >: llat do I receive? A.ns. $3j7.50 9. I own 147 shares of railroad stock ($50 encllh), on wihich I amn entitled to a dividend of 5 c, payable in stock: holw nilay additional shllares do I receive? Ais. 7 shlares, and $17.50 toward anotlier share. 3. A has 218 sllares bank stock ($100 each), and gets a dividen d of 12 %i: how much is that? Anl.s. $2616. 4. The- Cincinnati Gas Co. declare a dividend of 18 -%: what do I get on 50 shares ($100 each)? A.qS. $(0 0. 5. The AWecstern Stage Co. declare a dividend of 41- per cent: if their whole stockl is $150000, how much is distributed to the stockholders? A) s. $6750. (;. A railroad Co., whose stock account is $4-256000, declared a dividend of 3.' % what sum was distributedl monllg the stockholders? Aits. $148960. i, 1c v i w. —219. What is a joint-stock company? What kind of business i(h they tlinsact? W'hIat is the stock? Ilowl is it di i(dled? What are the stockhold(trs? What is the dividend? IHowv is it divided amongl the stockholders? What is the rate of dividend? What is Case I? PAR;, DISCOUNT, AND PREMIUM.. 22' 7. A Telegraph Co., with a capital of $75000, declares a dividelnd of 7 %j, and has $6500 surplus what has it earnlled l? Als. $11750. S. I own 24 shares of stock ($t25 each) in a Fuel Co. which declares a dividiend of' t %c; I take nmy dividend in coal, at S Ct. a bu.: how much do I get? LAs. 450 bu. ARtT. 270. CASE IT. —Gicve, the stockl, atnl dicvieUCl,, to find the,1cate o/ (liicdlzd. (See Case II, Percentage.) 1. Mly dividend on 72 shares bank stock, ($50 eaech), is $324: what was the rate of dividend?. As. 9 /. 2. A. Turnpike Co., whose stock is $225000, earns $163884.50': what rate of dividend can it declare? Aias. 7 C/, nd $6384.50 surplus. 3.'The recipts of a Canal Co., whose stock is $3650000, in onOi year are $2536484; the outlay is $793883: what rate of dividend can it declare? LAts. 4' %, and $12851 sur. ART. 271. CASE III. —The dividcend, and rate of (l;videcl(, YiCa, o ie filld cthe stock corresp'C oidinlg. (See Case III, Percentage.) 1. An Insurance Co. earns $18000, and declares a 15%. dividend': what is its stock account? Al2s. $120000. 2. A malln gets $94.50 as a 7 c% dividend: how Imlny shares of stock ($50 each) lhas lie? 4A s. 27. 3. Ieeeived 5 shiares ($50 each), and $2G6 of anothier shcare, as an 8 $c dividend on stockl how iany shares had 1? A i. 6 9. AItT. 272. CAsE IV. —(See Gise V, Percentacge.) 1. Received, 10 c stock dividend, and tlhen had 102 shares ($50 each') and $15 of nother share: how many shares hICad I before the dividend? A)s. 93. 2. IHaving received two dividends in stock, one of 5 O, another of 8, llmy stock has increased to 5t67 shares: how many had I at first? Als. 500. XIX. PAR, DISCOUNT, AND PREMIUMI AnT. 273. P-AR, DISCOUNT, and PREMIUM, are imercantile terms, applied to mooney, stocks, bonLs, and (r'qcts. Miloacy is the cirdculating medium of trade, in the form of gold and silver coins, and bank notes. s228 RAY'S HIGHER ARIlTH-METIC. S/oorks, are money invested in Banks, Insurance compa. nies, &c., in the form of shalres, $50 or $100 each. Drf(/ts, bills of exchan:ctcc e, or chec/ks, are written orders for money. Bonds are written obligations to pay money at a fitture time. AR T. 274. All money, stocks, bonds, and draftJs, lhave value on their face, called nlomzial or par value. Their real vatlue, is what they are intrinsically worth and sell for. When their real is the same as their lomfnnal value, they are said to be par, (the word par meaning equal, in Latin). When they sell for less than their nominal value, they are ieCloowt par, or at a discounlt. When they sell for more than their nominal value, they are above par, or at a p)'enianm. ARr. 275. Discount is how much less, money, stocks, drafts, &Te., are worth, than their f./ace. Premaniuz is how much m7ore, they are worth, than their face. Rate of prenilnt, or lrate oq (liscoultt, is the rate per cent. the pc)'elmium? or discouhnt is of the face. ART. 276. The face or par value of money, stocks, drafts, &e., is the standard of comparison: hence, this GENERAL RULE. IRepresent the face by 100 per cent; and then paroceed by such rtlle of Percentage as the nature qf the questionz requi'es. This subject has 4 cases, solved like the 4 corresponding cases of Percentage. CASE I. —G;ven, the par value, and the rate of ]r-ciassm or d(iso"ILt, to findtc the premtiut or discount. (See Case I, Perce intage.) No T E. —If the result is a premium, it must be added to the par'o get the real value; if it is a discount, it must h3 subtracted. RP E vi w.-270. What is Case 2? 271. Case 3? 272. 1What does (Case 4 correspond to? 273. What are Par, Diseoount. and Pres1iun1? What is money? What arestocks? What are drftS? Whatotlt r tlna\t( do they hnve? What are bonds? 274. Whallnt is the par val1e of umomIn, stocks, &(e.? WThat is their real value? When are they par? At', so called? When are they below par, or at a discount? When ore they above par, or at a premium? 275. What is discount? Prerntiurm Rate of discount? Rate of premium? PAR, D1ISCOUNT, AND PREMIUM. 229 1. Buy 18 shares stoclk ($100) at 8 % cliseount: find the discount and cost. Ans. $144, and $1656. 2. Sell the same at 4, %, premn.: find the prem., the price, and gain. Als. $81, $1881, and $225. 3. Bought 62 shares Railroad stock ($50) at 28 5/0 premium: what did they cost? Al)s. $8 968. 4. What is the cost of 47 shares Railroad stock ($50) at 30 %/v discount? A is. $1645. 5. Bought $150 in gold at %7 premiuml: what is the prenlium, and cost? A.s. $1.121 ncid $151.12' 6. Sold a draft on New York of $2563;8.45, at; Q,~ premlium: what do I get for it? Ans. $2581.29 7. Sold $425 uncurrent money at 3 % discouit: lwhlat did I get, and lose? Ans. $412.25 and $12.7t 8. What is a $5 bank note worth at 6 7/ discount? Ains,. p4. 70 9. Exchanged 32 shares Bank stock ($50), 5 5/o piemium, for 40 shares tRailroad stock ($50), 10 % discount, and paid the difference in. cash: what was it? An4s. $120. 10. Bought 98 shares stock ($50) at 15 % discount; gave in paynment a bill of exchange on New Orleans for $4000 at ~,- premium, and the balance in casll: how much cash did I pay? Ans. $140. 11. Bought 56 shares Turnpike stock ($50) at 69 %; sold them at 76!. %: what did I gain? A.s. $21 0. R Ean.-it.-The cost of stocks is generally given as so many % of tlhe face, instead of so many % discount or premium. 12. Bought telegraph stock at 106 5%; sold it at 91 %: what was my loss on 84 shares ($50)? Aos. $630. 13. What is the difference between a draft on Philadelphia of $8651.40, at 14-' premium, and one on N. Orleans for the same amount, at, a discount? Aiis. $151.40 14. Bought 18 shiares Railroad stock ($50) at 12 % discount, paying ( brokerage: what did it cost me? Ami7. $7 i 95,96 15. Find the cost of 9'Ohio State bonds ($500) at 8 6% premium, %o brokerage. Ans. $4896.45 RIr vvl,w.-276ii. What is the standard of coinmparison? What is tht general rule for buying or selling money, stocks, drafts, &ec.? Ihw maunny cases in this subject? What do they correspond to? What is Case I liow is the real value found after the premium or discount is known? 3t) I> [AYS!tII tHEll ARITt_'MIETIC. 16. I chane $380 bank notes fbor gold at 11, v' - miuml; $25 of thle lotes are I ~q discounlta; $40 arle'2 4 discount;. t 65 are 5l discount; and,$10 ar.3 discout: if' I receive $370 in gold, how much chanCe should be given the broker? A). 1 5 et. 17. Buy 364 shares stock ($50), at 16 %c discoult blrokerage. 3 c; sell them at 10 %o premium, ii brokralg %c: what is my gain? - As. $-424. 2 Ar. 2. 277. CAsE II. —Give, the face, (rtcd the (Eisconl,.,t 0o, Pvem11t1ne, to jfi(l the rate of cliscocmalt or j)'nm, itmtt. (See Case II, Percentage.) NOTES.-1. If the face ancl the real value arie known, take their difference for the discount or premium; then, apply the ruile. 2. If the rate of gain or losr is required, the seal value or cost is the standaLcrd of comparison, not the fcice. 1. Paid $2401.30 for a draft of $2360 on New York: what was the rate of premium? A.ts. 11 2. Paid $2508.03 f'or 26 shares stock ($100), and brokerage, $25.03: what the rate of discount? Ails. 4~ (. 3. Bought 119 shares Railroad stock ($50) for $3'64110: what was the rate of discount? A,,s. 3.5. 4. If the stock in last example yields 8 $% dividend, what is my rate of gctin? Ats. 12 -4 5. I sell the same stock for $59936: what rate of premium is that? what rate of yueil?,cts. 6.; 63 6. If I count my dividend as part of the gain, a what is my rate of gain? A;ts. 75 c, 7. Exchanged 12 Ohio bonds ($1000),7 7 prem nium, for 280 shares of Railroad stock ($50): what rate of (discount were the latter? Ac1s. 8'} (. S. Grave $266.66 of notes, - % discount, for, 20 of oold:'what rate of premium was tile goold?. A9 i. 9. Bought 58 shares Mlining stock ($50) at 40 % I)enmium, and gave in payment a draft on Boston fobr q$4;00: what rate of premliuml was the draft? Acts. 1-2 /0. 10. Received $4.60 for an uncurrent $5 note: iwhal was the rate of discount? Acts. 8 ~. ltrevEw. -277. What is Case 2? What case of Percentoge dioes it correspond to? What, if the Jface and the real value are known? What, if the rate of gain or loss is required? P>AR, DISCOUNT, AND PREMIUMl. 231 ART. 278. CASE IIT. —Gice,, the (lis(co17lt or )r'12i177m,, Ctd th/ r/e /' (lis/co/lt0 orj ]tic)iit, to jiud thle jitcc. (See Case Il, Perl'clntage.) N o vTE S. —1. After the fa'ce is obtained, add to it the premium, or sulbtract the discount, to get the real value or cost. 2. If the gain or loss is given, an t.lle rate per cent. of the facts orresponding to it, work by Case 11I, Percentage. 1. Paid 36 et. premium for gold a ( above par: how much gold was there? -- A/ls. $48. 2.'Took stock at par; sold it for 24 07 discount, and lost $117: how many shares ($50) had I? Alis. 104. T. I'e discounti, at 7. c, on stocls was $93.75: how many shares ($50) werelc sold? A, s. 25. 4. Buy stoek at 4-, cY. premiumn; sell at 8: l r ])emiuml; gain, $,343: hlow many shares ($100)? Als. 92. 5. Buy stocks at 14 7v% discount; sell at 381 %(;/o prem.; gain, $19 2.50; how many shares ($30)? Ams. 22. 6. The preiliunm on a draft, at c, was $10.386: what was the face? A)ls. $1184. 7. Buy stocks at 6.c% discoufit; sell at 42 %/o discoulnt, loss, $6(66: how may sllhares (.o)? A,,s. 37. 8. Stocks at 12 ~o discount, brokerage, $6.03, cost $239.97 less thlan the face: how man-y shaIres (c0$ )? its. 41. 9. Bonds, at 20 6C premium, brokerage - /, cost $300.S87 more than the fh:-e.: what is the foace? Ats. $1450. 10. Buy uncurrent bank notes at 10. discount, 24-. brokerage; sell them at par, and gain $48.75: what was the face of the notes? Anls. $4500. 11. Buy stocks at 40 % discount, brokerage 1 %;/ sell therm at 20 %o discount, brolerage 14 c(, and gain $5374.87': how many shares ($50)? At/s. 63. ARr. 279. CASE IV. —Give1, the real vatlte, acdl the 7rate of l)/r'?/, iii or (1dsco(0nt, to find the face of the drujis, stock, (C c. (See Case IV, Percentage.) No'rzs. —. A fter the face is known, take the difference betw-een i.a Ind the real value, to fildld the dliscoulnt or premium. RE\EVw.-278. What is Cnseo 3? WhaVt does it correspond to? ITow can the real vatlue ur cost be found? When the yain or loss is given, and the rate per cent. of the face, how proceed? ;232 RAY'S HIGHER AlltITUMiBT'l''. 2. Bear in mind that the rate of premium or discount, and the rate of gain or loss, are entirely different things; the former is refered(l to the par value or face, as a standard of comparison, the latter to the real value or cost. 1. What is the face of a draft on Baltimore costino S_861.45, at 1- %c premium?.Ass. $2819.16 2. Invested $1591 in stocks at 26 % discoummt: how.naiy slhares ($50) did I buy? 4 as. 4:. 8-. Boug'ht a draft on New Orleans at, % discountt, for G1i398.80: what was its face? Aus. $6430.4 5 4. Notes at 65 %, discount, 2 % brokerage, cost $881. 79: whlat is their face? Ans. $-2470. 5. Exchanged 17 Railroad bonds ($500), 25 %, below par, for bank stock at 64 % premium: how many shares ($100), did I get? Amr s. 60. ti. T-low much gold, at s G premium, will pay a check for 875 t7'? Als. $7520. 7. Iow tmuch silver, at 1'- % premium, can be bought for 83252.96 of currency? Aes. $3212.80 8. klow larlge a draft. at { %O premium, is worth 54 city bollCds ($.100), at 12 % discount? Ans. $4740.15 9.'Exchanged 72 Ohio State bonds ($1000), at 6C % premium, for Indiana bonds ($500), at 2 % preiniul: how many of the latter did I get? Aas. 150. XX. INSURANCE. ART. 280. INSURANCE is of two kinds-Insurance on property and Life Insurance; the latter will be explained under the head of Annuities. Tsmera)mtce on Priopcrty is of two kinds-Fire and MIarine; the former takes effect upon fixed property, as houses and their contents; the latter applies to property tra nsported by water, as vessels and their cargoes. R l vI m E. —279. What is case 4? What does it correspond to? ffow is thie discount or preimium found? What is the distinction between the rate of premniumn or discount, and the rate of gain or loss? VWhat is the stalldrlcd of comparison fior the former? What for the latter? 2S0. Itowv munty kinds of insurance? How is insurance on property divided? What is Fire insurance? Marine? INSURANCE. 233 Property conveyed by railroads is insured after the manner of marine insurance. ART. 281. Insurance on property, then, is security againlst loss by fire or the dangers of transportation. The companies or individuals that guarantee against such loss, are called ucdelrwriiters, or icsurers. If' an iridividual insures, it is called out-cdoor insurance. The written contract between the insurers and the inurecd is called the policzy. The sum charged for insuring is called the pj renisunm, and is a certain rate per cent. of the amount insuied. This rate per cent. is called the rate of' iRnsurance. Insuring property is called ta/ciJlg a risk. ART. 282. Insurance has 4 cases solved like the 4 corresponding cases of Percentage. The amount insured is the standard of comparison; hence, this GENERAL RULE FOR INSURANCE. Repr?eselt the anount insured by 100 per cent., and then procced by sac7t rzle of Percentage as the nature of the question requires. CASE T.-Givcn, the rate of insucraince, and the aneounit intsuied, to find thel)2?iumbn.?z (See Case I, Percentage.) 1. Insured a house for $2500, and furniture for $600, at fo:''hat is the preimium? Ans. $18.60 2. Insured' of a vessel worth $24000, and -: of its cargo worth $36000, the former at 2} %, the latter at V': what is the premium? Ans. $607.50 3. What is the prenmium on a cargo of railroad iron worth $28000, at 13 %? Ans. $490. 4. Insured goods invoiced at $32760, for 3 mon., at sa %: what is the premium? Anis. $262.08 5. M y house is permanently insured for $1800, by a deposit of 10 annual premiums, the rate per year being: ~: R t v IEw.-2S0. What is said of property conveyed by railroads, &c.? 281. What is insurance on property? Who are underwriters? What is out-cdoor insurance? What is the policy? The premium? The rate of insuralnce? IWVhat is insuring property called? 2S2. IHow many cases in Insurance? Whatt is the standard of comparison? Whaht is the general rule? What is Case I? What does it correspond to? 20 234 RAY'S HIGHER ARITHMETIC. how much did I deposit? and if, on terminating the insurance, I receive my deposit less 5 %, how much do I get'? iAns. $135 deposited; $128.25 received. 6. A shipment of pork costing $1275, is insured at ~ %, the policy costing 75 ct.: what does the insurance cost? Ails. $7.83 7. An Insurance company having a risk of $25000 at T9o%, re-insured $10000 at 4 % with another office, and $5000 at 1% with another: how much premium did it clear above what it paid? Airs. $95. ART. 283. CASE II.-Given, the amounut i7slre(d, cnud )relnlium, to fitd the rate of i7tsurance. (See Case II, Percentage.) 1. Paid $19.20 for insuring' of a house worth $4800: what was the rate? As.: %. 2. Paid $234, including cost of policy, $1.50, for insurin, a cargo worth $18600: what was the rate? Ais. 14 %. 3. Bought books in England for $2468; insured them for the voyage for $46.92, including the cost of the policy, $2.50: what was the rate? Ans. 1 %.4 4. A vessel is insured for $42000; $18000 at 22, 0,T $15000 at 3;- %, and the rest at 4` %: what is the rate on the whole $42000? Ains. s3~ %. 5. I took a risk of $45000; re-insured at the same rate, $10000 each, in 3 offices, and $5000 in another; my share of the premium was $262.50: what was the rate? Ans. 2 8. 6. I took a risk at 1.,; re-insured - of it at "2 e, and:- of it at 2 - %: what rate of insurance do I get on what is left? Ans.:! a6 SUGGESTION.-Find what per cent. of the risk is paid for reinsurance; take this from my rate, and find what per cent. the remainder is of the partial risk taken by me. ART. 284. CASE III. —Given, the J7renmtint, an1d( rate of,.slnrancce, to find the amount i7nsnrecd. (See Case III, 1' Plcentag'e.) 1. Paid $118 for insuring, at what was'l amount insured? Ans. $14750. REVIE w.-2S3. What is Caseo 2? What does Case 2 correspond Lt, 284. What is Case 3? What does it correspond to? INSURANCE. 235 2. Paid $411.3371 for insuring goods, at 1 %~ what was their value? Als. $2 4925. 3. Paid $42.30 for insuring I of my house, at o o: what is the house worth? A Is. 7 520. 4. Took a risk at 2} %; re-insured D of it at 2, j; my share of the premium was $197.13: how large was the risk? Ans. $26284. 5. Took a risk at 3a %; re-insured half of it at the same rate, and 3 of it at 1 %S; my share of' the preiliuIn Was $58.11: how large was the risk? Ans $19370. 6. Took a risk at. 2 %; re-insured $10000 of it at 2} %, and $8000 at 14 %; my share of the premlium was $(207.50: what sum was insured? Ans. $28000. AnT. 285. CASE IV.-Given, the 7rate of isCurcance, cadt the aco7u1nt of property to be inzsured, to find the amontCtl to be inlstured so as to cover both property anwd pyremu7im. In this case; the amount insured is made up of the prolprlty and pre2mintm; so that if a loss occurs, both the value of the property insured, and the premium, shall be recovered. The value of the property is a certain rate per cent. less than the amount insured, since it is less than that amount by the premium, which is always a certain rate per cent. of' the amount insured; hence, the amount insured is found as in Case IV of Percentage. What sum must be insured, to cover property worth $4800, and premium at the rate of %? ANALYSIS.-The amount insured being the standard of comparison, is 100.%; the premium is I- %; the property, ($4800), is the remaining 990 SO: then 1 % - $4800 -- 990 = $48.2412+, and 100 S is 100 times this, = 84824.12, the amount to be insured. It generally happens in this case, that $4 8 0 0. the divisor wants but little of being a 24. unit, and the division can be performed.12 with advantage by Case III of Contracted $4 Division of Decimals (Art. 159). Hence, $4 4.12 this mlode of operation, when applied to this case of Insurance, is generally stated in the following Rule. R E VIgE wv.-285. What is Case 4? What does it correspond to? Why 7 Analyze the example. What is said of the divisor? How may the division be performed? Apply it to the example. 23 6 _RAY'S IGHE-R ARITHIMETIC. PRACTICAL RULE, FOR INSURING BOTH PROPERTY AND PREM3IUM. Fixnd the premiium onl the property to be insured; t7zen, the premium on this premiuml: then, the preminum o7n the 2d premium7, if n2(ecessary, and so on, neglectilg all fiyuEres below mills; the sum7 of the successive preminmos will be the whole premizm; to fnd tlhe a70ount to be inzstred, add the property to the pretz ium. 1. What sum must be insured to cover property worth $2600, and prem. at?o %? Ans. $2618.33 2. Insured to cover a shipment of pork, valued at $12368.50, at 1: find the amt. insured, and prenm. Au2s. $12493.43, am't ins'd; $124.93, prem. 3. Insured to cover my library, $1856.20, at f6 %: what was the premium? Ans. $11.20 4. Insured to cover property to the value of $4840, at' %:'what was the premium? Anzs. $36.57 XXI. TAXES. &RT. 286. TAXES are money paid by the subjects of a governmnent, to defray its expenses; and are either direct or hindirhct. Direct taxes consist of a property-tax, and poll-tax or ceupitatton-tax, and are generally collected once a year;a proJ)ertly-tax is levied on all property, but that exempt by law; it is estimated as a certain rate per cent. of' the assessed value of the property: this rate per cent. is called the rate of taxationz. A ploll-tax or ccapitation-tax is a fixed sum, charged on every citizen, without regard to his property. This subject has four cases, solved like the four correspondin, cases of Percentage: the taxable property is the standard of comparison. Hence, this GENERAL RULE. Rep resent the taxable property by 100 %; then proceed by such rule of Percentage as the natsure of the question, requires. RIEvIEWv.-WIaiftt i3 the practical rule? 286. What are taxes? What are direct taxes? What is a property-tax? lIow is it estimated? What is a poll-tax? What is the general rule? TAXES. 237 CASE I. —Given, the tacxable property, and the rate rae of tr atio,, to find the _operty-tax. (See Case I, Percentage.) NOTE.-If there is a poll-tax, the sum produced by it should', added to the property-tax, to give the whole tax. 1. The taxable property of a county is $486250, oa l the rate of taxation is 78 ct. on $100; that is, - 1, 6) what is the tax to be raised? Ans. $37921.75 R1 E r. - The rate of taxation being usually small, is expressedt most conveniently as so many cents on $100, or as so many mills on $1. 2. A's property is assessed at $3800; the rate of taxation is 96 ct. on $100 (OxN% %) what is his whole tax, if he pays a poll-tax of $1? Ans. $37.48 3. The taxable property of Cincinnati, in 1855, was $85330880, on which was charged a tax of -joo % for the State, 4do-4o % for the county, and To, o 0O for the township and city: in all, 1 —io %; how much was raised for each and for all? A~ns. State tax, $273058.82; county, $353269.84;.township and city, $636668.36; total, $1262897.02 In mnaking out bills for taxes, a table is used, containing the units, tens, hundreds, thousands, &c., of property, with the corresponding tax opposite each. To find the tax on any sunm by the table, take out the tax on each figure of the suni, and add the results. In this table, the rate is 1- %, or 125 ect. on $100. TAX TABLE. prop.l Tax. Prop. TaxD. Tx. Taro. Tax. Prop. Tax. | i$1.012.5 $10.125 $100 i$1.2i $1000 $12.50 $10000 $125 i2.025 20.25 200 2.50 2000 25. 20000 250 3.1 o7 30.375 300 3~75 3000 37.50 3005-10 I 3 5 4.05 40.50 400 5. 4000 50. 40000 500 5.025 50.625 500 6.25 5000 62.50 50000 625 6.075 60.75 c00 7.50 6000" 75. 00000 1 750 7.0875 70.875 700 8.75 7000 87.50 70000 875 o.10 80 1o 800 10. 8000 100. 80000 1000 9,.112 9 1.125 o 900 1.25 000 112.50 90000 1125 4, Wlat will be the tax by the table, on property asa sessed at $25349? It, v I w.-What is Case 1? To what does it eorrespond? 23S3 RAY'S HIGHER ARITHMETIC. S o u r o N.- The tax for 20000 is 250; for 5000 is 62.50; for 300 is 3.75; for 40 is.50; for 9 is.1125, which, added, give the tax for $25349 -- $316.86. 5. Find the tax for $6815.30. Ans. $85.19 NoTE. —The tax for cents is not in the table, as being too insignificant; if desired, it may be found from the dollars, by moving the point to the left, 2 figures: the tax for 30 ct. = $.00375. 6. What is the tax on property assessed at $10424,50 and two polls, at $1.50 each? As. $138.31 ART. 287. CASE II.- Given, the taxable ])rolperty, and the tax, to find the rate of taxation, (See Case II, Percentage.) NOTE. —If there is a poll-tax, find what it produces, and subtract it from the whole tax; the remainder is the property-tax, and is used to find the rate of taxation. 1. Property, assessed at $2604, pays $19.53 tax: what is the rate of taxation? Ans. -4 % = 75 ct. on $100. 2. The taxable property in a town of 1742 polls, is $6814320. A tax of $66913.54 is proposed. If a polltax of $1.25 is levied, what should be the rate of taxation? (See Note.) Anls. -5 %i =SO 95 ct. on 100. 3. An estate of $350000 pays a tax of $5670: what is the rate of taxation? Ans. 1`-1 % = $1.62 on $100. 4. A's tax is $53.46; he pays for 3 polls, at $1.50 each, and owns $8704 taxable property: what is the rate of taxation? Ans. -9t % - 56}- ct. on $100. ART. 288. CASE III.-Given, the tax, and the rate oj taxation, to find the assessed value of the property. (See Case III, Percentage.) NOTE.-If any part of the tax arises from polls, it should be first deducted from the given tax. 1.'What is the assessed value of property taxed $66.96, at 14 %? Ans. $3720. 2. A corporation pays $564.42 tax, at the rate of 4<'o %o, or 46 et. on $100: find its capital. Ans. $122700. R E v I E w.-If there is a poll-tax, iwhat should be done? 287. What is Case 2? What does it correspond to? If any part of the tax is produced from polls, what should be done? DUTIES OR CUSTOMS. 23 9 3. A is taxed $71.61 more than B; the rate is 1. %. or 1,.32 on $100: how much is A assessed more than B? Ans. $5425. 4. A tax of $4000 is raised in a town containing 1024 polls, by a poll-tax of $1, and a property-tax of Tdd'S (24 ct. on $100): what is the value of' the taxable property in it? Anl.s. $1240000. 5. A's income is 16 % of his capital; he is taxed 2} % of his income, and pays $ 26.04: what is his capital? Als. $6510. ART. 289. CASE IV.-(See Case IV of Percentage.) 1. A pays a tax of 17 % ($1.35 on $100) on his capital, and has left $125127.66: what was his capital, and his tax? Ans. Cap., $126840; tax, $1712.34 2. Sold a lot for $7599, which was cost and 2 %/ beside paid for tax: what was the cost? Ans. $7450. XXII. DUTIES OR CUSTOMS. ART. 290. DUTIES or CUSTOMS, are taxes upon foreign merchandise, paid by the importer; they are of two kinds, ad valorenm and specific. Ad valorern duties are so much on the value of the goods as shown by the invoice. Ad valorem is Latin for on the value. An invoice is a bill of the goods, showing the kinds, quantities, and cost. Specific duties are a fixed sum for a fixed quantity, without regard to cost; as, $10 a cwt.,' 30 a hhd., &c. ART. 291. In specific duties, certain allowances are usual, called drcft, tare, leakacce, and breakage. Draft is an allowance made, that the goods may hold out when retailed. It is calculated as follows: REvIsEw.-2SS. What is Case 3? What does it correspond to? Tt' any part of the tax is produced from polls, what should be done? 289. What does Case 4 correspond to? 290. What are Duties or Customs? Ihlcw many kinds? What are ad valorem duties? Why so called? 240 RAY'S HIIGHER ARITHMETIC. On each parcel weighing, 112 lb., or less, it is 1 lb. from 336 lb. to 1120 lb., it is 4 lb. from l12lb. to 2241b... 2lb. from 11201b. to 20161b.,.-. b. from 2241b. to 336 lb. 31b. over 2016lb.,.. 91b. T5(re is an allowance for the weight of the box, bale, cask, or whatever contains the goods. It is generally estimnated at a certain per cent. on the quantity remaining after the draft has. been deducted. Tare is sometimes estimated at a certain quantity for each cask package, etc., or a certain weight for each cwt., tun, &c. The weight of the goods, before draft and tare have been allowed, is called gross weight; after they have been allowed, it is called net weight. Leakage is an allowance of 2 %, on all liquors in casks paying duty by the gallon. Breakage, usually 10 %, is allowed on ale, beer, and porter in bottles; on other liquors in bottles, it is only 5 %. The common sized bottles are estimated at 22 gal. per dozen. RULE FOR CALCULATING SPECIFIC DUTIES. ART. 292. Fincl the net weight of the goods in that deno wnination for which the rate of duty is given, and rntliply it by the given rate. N o T.-Observe that in custom-house business, 28 lb. make 1 qr.. 112 lb. make 1 cwt., and 2240 lb. make 1 tun. ART. 293. In ad valorem duties, the calculation is one of Percentage, and since the invoiced value of the goods is the quantity on which the rate of duty takes effect, it is 100 per cent.. RULE FOR ALL QUESTIONS IN DUTIES AD VALOREMI. Re present the invoiced value of the goods by 100 per cent., and proceed by such rule of Percentage as the case requires. RE VI E ~W.-290. What is an invoice? What are specific duties? 291. What allowances are made in specific duties? What is draft? tIHow is it estimated? What is tare? HIow is it generally estimatedl? What is it ealculated on? How is tare sometimes estimated? What is gross weight? Net weight? What is leakIgre? Breakage? IHow is the quantity of liquor in bottles estimated? 292. What is the rule for calculating specific duties? In custom-house business, hcTw many lb. in a qr.? DUTIES AND CUSTOMS. 24 REr ARA.-According to the last tariff-bill of the U. S., passed L84., none but ad valorem duties are allowed. CASE I.-See Case I of Percentage. WHAT IS THE DUTY 1. On 45 casks of wine, of 36 gal. each, invoiced a $1.25 per gal., at 40 % ad valorem? Ans. $793.80 Allow for leakage. 2. On 12 boxes fancy soaps, each 981 lb., tare 10 %, at 5ct. a lb.? Ans. $52.65 3. On 36 boxes sugar, each weighing 6 cwt. 2qr. 18 lb., tare 16 lb. per cwt., at, 2.- et. a lb.? Artns. $572.40 SUGGESTION.-Tare = 161b. per cwt._ —' f of the lb. 4. On 460 E. E. of broadcloth, at $3.20 per ell, at 30 % ad valorem? and if I pay $41.40 freight, how much per yard should I charge to gain 20 %? Ans. Duty, $441.60; price per yd., $4.08 5. On 50 drums figs, each weighing 57 lb., draft as usual, tare 20 lb. to the cwt., at $12 a cwt.? Aos. $246.43 6. On 8hhd. of sugar, each 12cwt. 3qr. 141b., tar-r 12 lb. to the cwt., at 1 ect. a lb.? Ans. $153.7r 7. On an invoice of carpets, worth $1859.60, at 30 % ad valorem? Ans. $557.88 8. On 680 boxes of raisins, invoiced at $1.25 a bo,, at 40 % ad valorem? Ans. $340. 9. On 26 chests of tea, each 118 lb., tare 10 lb. per chest, at 10 ct. a lb.? Arts. $275.60 10. On 42hhd. of sugar, each 13cwt. 3qr., and the tare 71b. per cwt., at $2 a cwt.? Ans. $1077.89 11. On 30 bags of rice, each 7cwt. 2qr. 101b., tare 12 %, at $1.20 a cwt.? Ans. $239.30 12. On oil-cloth, 40 yd. long, and 3 /d. 2 ft. 8 in. wide worth 75 ct. a sq. yd., at 30 % ad valor, m? Ans. $35. ART. 294. CASE II.-See Case II,.Percentage. 1. If goods invoiced at $3684.50 pay a duty ot $1473.80, what is the rate of duty? Ans. 40 %. PE vIE w.-29:3. What is the general rule for questions in ad valoret duties? What kind of duties only are allowed in this country? Wha' does Case 1 correspond to? 21 242 RAY'S HIGHER ARlTIIETI(X 2. If laces, invoiced at $7618.75, cost, when landed, $9142.50, what is the rate of duty? An s. 20 y. 3. If 40 hhd. (63 gal. each) of molasses, invoiced at 12- ct. a gal., pay $92.61 duty, after allowing 2 % for leakage, what is the rate of duty? A7is. 30 %. ART. 295. CASE III.-See Case IN/, ]erceJeagc. I. Paid $806.12 duty on watches, at 35 %) what were they invoiced at, and what did they cost in store'? Ants. Invoiced, $2303.20; cost, $3109.32 2. The duty on 1800 yd. of silk was $3317.50, at 25 % ad valorem: what was the invoice price per yd.? and whal must I charge per yd. to clear 20 %? Ans. Invoiced at 75 et. per yd.; selling price $1.12' 3. The duty on 15 gross London porter, allowing 10 C breakage, was $40.50, at 20 % ad valorem: how much a dozen were they invoiced at? Ans. $1.25 ART. 296. CASE IV.-See Case IV, Perceentage. 1. French cloths, after paying 30 %, duty, and other charges, $73.80, cost in store $7389.03: what were they invoiced at? Auls. $5627.10 2. Imported from IHavre 80 baskets of champagne, 12 bottles each, 5 % breakage, duty 40 %, freight and other charges $67.20, and the whole cost $729.60: what did it cost a bottle at EIavre, what in store, and how much a bottle should I charge to clear 35 %? Ans. 50 ct. at Havre; 80 ct. in store; selling price $1. 08 3. The cost in store of 20 puncheons Jaimaica rum, 84 gal. each, is $631.43; duty 15% %, leakalge 2 %, charges $53.34: what did it cost a gal. in Jamaica? Ans. 30 et. XXIII. INTEREST. ART. 297. INTEREST is money charged for the use of money. The profits accruing at regular periods on permanent investments, such as dividends or rents, are called interest, since they are the growth of capital, unaided by labor. R E vI E w.-297. What is interest? What are sometimes called interest? INTEREST. 243 T'he lprincifpal is the sum on which interest is charged. The principal is either a sum loaned; money invested to secure an income; or a debt, which not being paid when due, is allowed by agreement or law to draw interest. The amount is the principal, with its interest added. ART. 298. Interest is payable at regular intervals yearly, half-yearly, or quzarterly, as may be agreed: if there is no agreement, it is understood to be yearly. ART. 299. The rate of in~terest is the rate per cent. of the yearly interest to the principal: it is called rate per cent. per annucnm; per annZ?n meaning by the year, in Latin. If interest is payable half-yearly, or quarterly, the rate is still the rate per ann7im, or late per year. In short loans, the rate per month is generally given; but the rate per year, being 12 timres the rate per month, is easily found: thus, 2 % a month = 24 % a year. If the rate of interest is -not specified, the rate established by the law of the country where the contract is made, prevails, called the lcyal 7rate, which is generally, but not always, the highest rate allowed by the law. If a higher rate of interest is charged than the law allows, it is called usury, and the person offending is subjeot to a penalty. ART. 300..The following table shows the legal rate is each of the States of the Union. LEGAL RATES. STATES. 8 % in Georgia, Alabama, Mississippi and Florida. 7 % in New York, So. Carolina, Alichigan, Wiscon. and Iowa. 6 %0 in Louisiana. 10 /% in Texas. 6 % in all other places in the U. S., and the U. S. courts. ART. 301. Interest is either Simple or Compound. R E V I E. - 297. What is the principal? What may it sometimes be? What is the amount? 298. How is interest payable? If there is no agreement, how is it payable? 299. What is the rate of interest? For what time is it given? When is it given per month? How is the rate per year then found? If no rate of interest is mentioned, what one prevails? What is usury? 301. What two kinds of interest? 244 RAY'S HIGHER ARPITHMETIC. Simscple Interest is interest which, even if not paid when due, is not convertible into principal, and therefore can not draw interest itself and accumulate in the hands of the debtor, however long it may be retained. Cornpound Interest is interest which, not being paid when due, is convertible into principal, and from that time draws interest itself, and accumulates in the hands of the debtor, according to the time it is retained. Compound interest is more favorable to the creditor than simple interest, as shown in this example: If I lend $100 at the rate of 10 % per annum, what should I receive at the end of three years, presuming no interest paid in the mean time? SOLUTION BY SIMPLE INTEREST.-The interest for 1 yr. is 10 of $100 $10; for 3 yr. it is $30, and the whole sum due is $130. SOLUTION BY COM3POUND INTEREST.-The interest for the 1st year is $10; which, not being paid, makes $110 then due and drawing interest: 10 % of $110 = $11, the interest for the 2d year; which, not being paid, makes $121 then due and drawing interest: 10 %O of $121 = $12.10, the interest for the 3d year; which makes the whole sum then due, $133.10, being $3.10 more than by Simple Interest. XXIV. SIMPLE INTEREST. ART. 302. SIMIPLE INTEREST differs from the other applications of Percentage by taking the time into consideration, which they do not. All questions in Percentage involve three different quantities, and may be solved by a simple proportion. All questions in interest contain four different quantities, and may be solved by a comnpouzed proportion. ART. 303. The four quantities embraced in every question of interest are, 1st, the prfinczpal; 2d, the ineterest; tREVIEW. —301. What is simple interest? Compound interest? Which Is more favorable to the creditor? Show the difference by the example. 302. HIow does simple interest differ from the other applications of percentage? If the time were left out of view, howv could questions in simple interest be solved? How may they be solved as it is? 303. What four quantities are involved in every question of simple interest? SlAIPLE iNTERBEST. 245 3d, the rate; 4th, the timne: of these four, the principafl being the,tand(lrrd of com parison, is 100 per cent. Any three of these quantities being given, find the 4th by this GENERAL RULE. Represent the principal by 100 per cent., and proceed ae in questions of compound p0roportion. N OTE.-If the amount is given or required, it may be necessary to perform an addition or subtraction before applying the rule, or after the result has been obtained. Though the general rule is sufficient for all the ordinary problems in Simnple Interest, it is too general for practical purposes, and a special rule will be given for each case. CASE I. ART. 304. Given, the principal, rate of interest, and time, to find the interest. RULE.-Find the yearly interest by Case I of Percentage; then ascertain fiom it the interest for the given time by aliquot parts. NOTE.-In calculations of interest, every month is regarded as 30 days, and every year as 12 months, or 3GO days. What is the simple interest of $354.80 for 3 yr. 7 mon. 19 da. at 6 %? SoL. - The yearly interest, $ 3 54.8 0 being 6 % of the principal, is 6 found, by Case I of Percentage; yr. = 21.2 880 the interest for 3 yr. 7 mon. 19 da. is then obtained by ali- 3 yr. = 3 63.864 6 Mon. = I 10.644 quot parts; each item of interest mon 2. 6 1 is carried no lower than mills, 1.74 18 da. 1 1.064 the next figure being neglected d 11 1 da. --,0 5 9 if less than 5; but if 5 or over - it is counted 1 mill. $ 7 7.4 1 Ans. REVIEW.-303. How many must be known? What is the general rule for questions of simple interest? If the amnount is given or required, what may be necessary? 304. What is' Case 1? The rule? How many days are counted a month in interest? IIow many a year? Solve the example. What custom prevails in computing interest? Why? 246 RAY'S HIGHER ARITHMETIC. If the rate of interest were 5 %, 7, %, 10 %, &c. the pro. cess would be similar; but, as the prevailing rate is either 6 J, or a simple aliquot part greater or less than 6 %, it is custwonary, whatever be the rate, to compute the interest, first at 6 %c; theTn increase or diminish the resmult by such a piart of itself, as may be necessary to obtains the interest at the rate given. A SHORT METHOD. ART. 305. As 6 % a year is the same as 1 % for 2 months, take 1 % of the principal, by pointing off the two right hand figures of dollars; the result is the interest for 2 mon., or 60 da.; and the interest for the given time can be ibund by aliquot parts. Applying this method to the last example, the operation will appear thus: 38 54.80 SOaLUTION. —Write the principal, $354.80; cut 7 0 9 6 off the 2 right hand figures of dollars by a vertical 1 7 7 4 line; $3.548 is the interest for 2 mon.: $70.96 is I 0 6 4 the int. for 40mon., being 20 times the int. for 0 5 9 2 mon. just above; $1.774 is the int. for 1 mon., s 7 7 4 1 being c the top number; $1.0t4 is the int. for 18 da., bein T30 of the top number; (see Rein., Art. 234), and the $.059 is the inet. for 1 day, being w.qi of $1.774, the int. for 1 moe. The sum of these is the int. for 43 mon. 19 da. = 3 yr. 7 mon. 19 da. ART. 306. After finding the Int. at 6%, observe, that the Interest at 5 % interest at 6 %, - - of itself. And the Interest at4h = Int.at6%O -'of it. t % ltat at 6% at 4 o = Int. at 6 % - of it. at 7 % =Int. at6 0 of it. at 3 %X -- Int. at 6. at 7-1 % 60%+- of it. at 2 = Int. at6. at 8 o= " 6+ of it. at 1 -- Int. at 6 a. at 9 %= 6 +1 of it. at 12 %, 18 o, 24 %- = 2, 3, 4 times Interest at 6 %. at 5 a, 10, 15, 20 % -- =, I, 1 1 Interest at 6 %, after moving the point one figure to the right. ART. 307. ANOTHER METHOD. Take the example already solved. REV Eiw. —305. Explain the short method. 306. After the interest at 6 % has been found, how is the interest at 5 %0o obtained? At 4 e? 4 %, &c. 7? 7 %, &c. 307. Explain "another" method. SIMPLE INTEREST. 247 $ 354.80 A N ALY s.-The interest of any sum $1 77.40 ($354.80), at 6 % equals the interest of half 4 3 6 ) that sum ($177.40) at 1.2 %. But 12 % s a 096 0 year equals 1 % a mon., and for 3 yr. 7 mon. 6 8 4 0 -.3 mon., the rate is 43 I/6, and for 19 da., 9 3 which is.3 9 of a mon., the rate is 8 r hence, the rate for the whole time is 431) $ 7 74 5 5 3 and 4398 % of the principal will be the $7 7.4 1 Ans. interest. To get 431 Si), multiply by 4381 $9 774 ( hundredths =.4383 =.4368 (Art. 147). By using the contracted multiplication (explained in Art. 153), the work may be 7 0 9 6 shortened, and the answer obtained correct- 6 3 9 ly to cents. 6 In the multiplier.436-, the hundredths (43) are the number of months, and the thousandths (61) are -. of the days (19 da.), in the given time, 3 yr. 7 mon. 19 da. TWO PRACTICAL RULES. RULE 1. —Expre.ss tle years, if alny, in months, and write the whole number o'f onths as decimal hundred/ls; qfte)r which, place s of the days, itf any, as thousandths: mnlt}poly half the principal by tlhis number. RUE 2. — Or, point off O'Onm the prilncipal two more decimals than il already ha.s. This gives the ilterest for 2 months, or 60 days, friom which the interest for the given time can be found by aliquot par'ts. Eithler of these?rules gives tihe interest at 6 %e, from which the interest at any other rate cane beJfbund by aliquot parts. NOTE.-In the 1st rule, when the number of days is 1 or 2, place a ciphler to the right of the months, and write the I3 or 3: otherwie, they will not stand in the thousandths' place: thus, if the time is 1 yr. 4 mon. 1 da., the multiplier is.160); for 9 mon. 2 da.,.09003. Brevisw. —, 07. How can the work be shortened? What are the huniredtths of the multiplier equal to? The thousandths? What is the first rule for co0mputing interest? The 2d rule? In using the 1st rule, when the days are 1 or 2, whllat is necessary?'When the fraction in the multiplier is ], how do we proceed? Which rule is generally most converient for short times? Ili taking aliquot parts for the days in the 2d rule, what should we make use of? 248 RAY'S HIGHER ARITHMETIC. If the fraction in the multiplier is 2, it is better to take of the whole principal, than of half the principal. REMARti. —The 2d rule will generally be found most convenient for short times, and will not require more than two aliquot parts for the days, by using the tenths, as well as the halves, fourt//s, &c. {bee Art. 234, Rem.) FART. 308. In New York and some other parts of the IJ. S., and in Great Britain, the interest for days is calculated at 365 days to the year. In those places, 1 day's interest = g — of a year's interest = a of 360 da. interest obtained by the rule =3 or ~ 7 of a day's interest, as commonly obtained; but 7 is is less than the whole: Hence, the interest for any nzumnber of days, comatig 365 da. to a year, is -l,; less than the interest for the same nzumbe-t of days, coun2ting 360 da. to a year. FIND THE SI3IPLE INTEREST OF ANS. 1. $178.63 for 2 yr. 5 mon. 26 da., at 7. = $31.12 2. $6084.25 for 1 yr. 3 mon., at 4, %. - $342.24 3. $64.30 for 1 yr. 10 mon. 14da., at 9.=$10.83 4. $1052.80 for 28 da., at 10 %. = $.19 5. $419.10 for 8 mon. 16 da., at 6 %. = 17.88 6. $1461.85 for 6yr.7mon. 4 da.,at 10 lO.=$964.01 7. $2601.50 for 72 da., at 71 -.= 39.02 RE MARK.-72 da. - 2 mon. 12 da. 8. $8722.43 for 5' yr., at 6%. - $2878.40 9. $326.50 for I mon. 8 da., at 8%. = $2.76 10. $1106.70 for 4yr. 1mnon. ada., at 6%. $271833 11. $10000 for 1 da., at 6%. = $1.67 12. $94642.68 for 5mon. 17da., at 15 %. = $323.05 13. $13024 for 9mon. 13da., at 10. = $1023.83 14. $615.38 for 4yr. Ilmon. 6da., at 20%.= $607.17 15. $2066.19 for 3yr. 6mon. 2da., at 30 %._$2172.94 16 892.55 for 3 mon. 22da., at 5. = $1.44 R E v IE W. —307. Into how many aliquot parts can the days always be divided? 308. What is said of interest for days in New York. Great Britain, &c.? How is interest for any number of days obtained in those places? Why? SIMPLE INTEREST. 249 Find the simple interest of ANS. 17. $1532.45 for 9 yr. 2 mon. 7 da., at 12 %. = 1689.27 18. $78084.50 for 2 yi'. 4 mon. 29 da., at 18,. = $339927.72 19. $512.60 for 8mon. 18da., at 7 %. = $25.72 20. $3278.12 forl yr. 6mon. 3 da.,at4 %.= $197.78 21. $8408.46 for 11 mon. 5 da., at 3 %. = $234.74 22. $126.75 for 2 yr. 24 da., at 8 %. $20.96 23. $1363.20 for 39 da., at 14- % a month. = 22.15 24. $402.50 for 100 da., at 2 % a month. =$26.83 25. $6919.32 for 7yr. 6mon., at 6 5%. = $3113.69 26. $990.73 for 9 mon. 19 da., at 7 %. = $55.67 Find the amount of 27. $757.35 for 117 da., at 11 % a mon. = $801.65 28. $1883 for 1 yr. 4mon. 21da., at 6 %.= $2040.23 29. $5000 for 10 yr. 10 mon. 20 da., at 9. = 900. 30. $4212.45 for 5yr.5mon.25da.,at 5 %.=$5367.95 31. $262.70 for 53da. at 1 % a month. = $267.34 32. $584.48 for 133 da., at 7a %. =$'600.67 33. $8291.56 for 294 da., at 6 %. = $8697.85 34. $16372.05 for 3 yr. 9 mon., at 8 9o. -$ 21983.67 35. $92001.25 for 86 da., at 6 %O. = $029.93 36. $392.28 for 71 da., at 2 a mon. = $415.49 37. $3032.90 for 7mon. 7da., at 7 %O. = $3160.87 38. Find the interest of $7302.85 for 365 da., at 6 %, counting 360 da. to a year. Ans. $444.26 39. Of $10000 for 360 da., at 6 %, counting 365 days to a year. (Art. 308.) Ans. $591.78 40. Same as last, 360 da. to a yr. Ans. $600. 41. If I borrow $1000000 in New York, at 7 %, and lend it at 7 % in Ohio, what do I gain in 180 da.? Ans. $479.45 42. Find the interest of $5064.30 for 7mon. 12 da., at 7 %, in New York. (Art. 308). Ans. $218.45 S E s T I O N.-Find the interest for years and months in New York as elsewhere; but for days, find the interest as usual, and diminish it by V7 of itself, before adding it in. 2,50 RAY'S t-IIGHER ARITHMETIC. Find the simple interest of 43. (681.75 for 98da., 7 %, in N. Y. Ans. $12.81 44. $1353.10 for 2 yr. 8 mon. 29 da., at 7 % in New York. Ans. $260.10 45. $6786.24 for 1 yr. 10 mon. 16 da., at 10 % in New York. Ans. $1273. 89 4 6. If I borrow $12500 at 6 %', and lend it tt 1i0 %, what do I gain in 3 yr. 4lmon. 4 da.? Ans. $1672.2:2 47. If I borrow $93275 at 12 %, and invest it at 7 c/, what do I lose in 2 yr. I mon. 23 da.? Ans. $2498.83 48. What is the interest of $3416.20, at 6 c, friom Feb. 3, 1847, to Aug. 9, 1851? Als. $925.79 RE IAR K.-Pind the time by Art. 325, Rem. 3. 49. If $4603.15 is loaned July 17, 1853, at 7 % what is due MIarch 8, 1855? AnBs. $5130.34 50. Find the interest at 8 of $136t82..45, borrowed from a minor 13 yr. 2 mon. da. old, and retailled till he is of age (21 yr)..ns. $8543.93 51..What is the amount of $5772 from Oct. 26, 1850, to April 12, 1855, at 10%7/? A.s. $8348.56 52. In one year, a broker loans $876459. 50 for 63 da., at 1.' % a mon., and pays 6 % on $106525.20 deposits: what is his gain?.A s. $21216.96 53. What is a broker's gain in 1 yr., on $1i00 deposited at {6 %, and loaned 11 times for 33 da. at 2.j % a mon.? A)Is. $18.2O 54. Find the simple interest of Y~493 16s. 8d. f'or I yr. 8 mon. at 6 %. Ans. ~49 7s. 8d. SuGGEEsrosN.-Find the interest of sterling money by Rule in Art. 304, multiplying and dividing as in compound numbers; or, reduce the principal to one denomination, as pounde, and apply a Rule in Art. 3070 55. Find the simple interest of t;24 18s. 9d. for 1 l mon. at 6t %. Ans. ~1 4s. 11Jd. 56. Of ~25 for 1 yr. 9 mon. at 5 %. Ans. ~.2 3s. 9d. 57. Of ~651 for 7 mon. at 4, %. Ans. ~17 Is. 9Jd. 58. Of ~648 15s. 6d. from June 2 to November 25, at 5 %. - (Art. 308.) Ans. ~15 12s. 10d. 59. Of ~14 from March 23 to Nov. 2, at 6 ~,. Ans. 1Os. 3M-d. SIMPLE INTEREST. 251 60. Find the simple interest of Z~66 8s. from May 6 to Au,. 21, at 5~ %. Ans. ~1 is. sd. 61. Of~98 for 3 yr. 122 da. at 6 %. Ans. Y~19 12s. Id. 62. Of ~374 5s. from April 1 to Dee. 29, at 4 %. Ans. ~11 3s. Vd. CASE II. ART. 309. Given, the principal, interest, and time, to find the rate. RULE.-Assume I per cent. for the rate; determine the interest on this sutpposition, and divide the given interest by it. PRooF.-Calculate the interest at the rate thus found; if it agrees with the given interest, the work is rilght. NOTE.-If the amount is given instead of the interest, the principal may be subtracted from it, to obtain the interest. If I loan $6875 at simple interest, and in 2 yr. 3 mon. 18 da. receive $7823.75, what is the rate? SOL UTION.-Assume 1 % for the rate; the interest of $6875 for 2 yr. 3 mon. 18 da., at 1 %, is $158.125, (Case I); the given interest = $7823.75 - SG875 -$948.75, and since this contains $158.125 six times, it must have accumulated at a rate six times as great; that is, 6 %. 1. What is the rate of interest, when I pay $119.70 for the use of $3325 for 10 mnon. 24 da.? Ans. 4 %. 2. If $65.47 be paid for a loan of $844.75 for 93 days, what is the rate per month? Ans. 2- %. Su o. —Find the rate per year by the rule, and divide it by 12. 3. If I borrow $5000 for 7 yr. 6 mon. 28 da., and return $10000, what is the rate? Ans. 133'4- %. 4. At what rate per annum will any sum of money double itself by simple interest, in 5, 6, 7, 8, 9, 10, 12, 15, 20, 25 years, respectively? Ans. 20, 16', 1412, 121i, 10, 8-, 6 5, 4 %. SoaaEsTioN.-Take $100 for the principal, and $100 for the interest. If 100 be divided by the time in years, the quotient will be the rate at which any sum will double itself at simple interest in that time. REVIEW. —309. What is Case 2? The rule? The proof? If the amount is given instead of the interest, what is necessary? Solve the example. 252 RAY'S HIGHER ARITHMETIC. 5. At what rate per annum will any sum treble itself at simple interest, in 5, 10, 15, 20, 25, 30 years, respeotively? Ans. 40, 20, 13L, 10, 8, 6) %. 6. At what rate per annum will any sum quadruple itself at simple interest, in 6, 12, 18, 24, 30 years, respectively? Ans. 50, 25, 16 12, 1, 10 %. 7. What is the rate of interest, when $35000 yields an income of $175 a month? Ans. G %. 8. When $29200 produces $6.40 a day? Ans. 8 %. 9. When $12624.80 draws $315.62 interest quarterly? Ans. 10. 10. When stock bought at 40 % discount, yields a semiannual dividend of 5 %? Ans. 16" % per annum. 11. A house that cost $8250, rents for $750 a year; the insurance is -i %, and the repairs 2 %, every year: what rate of interest does it pay? Ans. 8 %-. CASE III. ART. 310. Given, the interest, rate, and time, to find the principal. RuLE.-Ass'ume $1 for the princpal; determine the ilterest onil this supposition, and divide the given interest by it. PRooF.-Calculate the interest on the principal found; if it agrees with the given interest, the work is right. NOTES.-1. After the principal has been found, the interest may be added to it, and the amount thus obtained. 2. The contracted method of division in Art. 1658 may be generally used. WThat sum will yield $228.80 interest in 4 mon. 23 da., at 15 % per annum? SOLUTION.-Assume $1 for the principal; the interest of it for 4 mon. 23 da., at 15 % is $.059-,7- (Case I): as the given interest, $228.80, contains this 3840 times, the principal producing it'must jo 3840 times as large, that is, $3840. WHAT PRINCIPAL WILL PRODUCE 1. $1500 a year, at 6 %? Ans. $25000. 2. $1830 in 2 yr. 6 mon., at 5 %? Ans. $14640. REVIEw. —310. Wha.t is Case 3? The rule? The proof? How can the amount be afterward found? SIMPLE INTEREST. 253 What principal will produce 3. $45 a mon., at 9 %? Ans. $6000. 4. $17 in 68 da., at 1 % a mon.? Ans. $750. 5. $86.15 in 9mon. llda., at 10%? Ans. $1103.70 6. $313.24 in 112 da. at 7 %? Ans. $14383.47 7. $146.05 in 7mon. 14da. at 6%? Ams. $3912.05 8. $79.12 in 5mon. 25 da. at 7 % in N. Y.? AnS. $2329.72 CASE IV. ART. 311. Given, the amount, rate, and time, to find the principal. RUE. —Assume $1 for the princi2pal; deternmile Ieic amount on? that sepposition, and divide the gicenl anioullt. by it. P o o F. —Calculate the amount on the principal found; if the same as the given amount, the work is right. NoTes. —1. After finding the principal, subtract it from the amount, to get the interest. 2. The contracted method of division (Art. 158) may be used. What sum, drawing simple interest at 5 %, will amount to $819.45 in 1 yr. 8mon. 5 da.? SoLUTION.-Assume $1 for the principal; the amount of $1 for lyr. 8 mon. 5 da., at 5 %, is $1.084i'g (Case I); as the given amount $819.45, contains this 755.93 times, it must have arisen from a principal 755.93 times as large; that is, $755.93. 1. What principal in 2 yr. 3 mon. 12 da., at 6 %, will amount to $1367.84? Ans. $1203.03 2. What principal in 10 mon. 26 da., will amount to $2718.96, at 10 % interest? Ans. $2493.19 3. What principal, at 4{%0, will amount to $4613.36 in 3 yr. 1 mon. 7 da.? Aghs. $4048.14 4. What principal, at 7%, will amount to $562.07 in 79 da. (365 da. to a year)? Ans. $553.68 PRESENT WORTH ART. 312. Is an important application of Case IV. RE I E TV. -- 310. What method of division may be generally used? Solve the example. 311. What is Case. 4? The rule? The proof? How is the interest afterward obtained? What method of division may be used? Solve the example. 54 KRAY'S HGiIGER ARITHMETIC. -Preset- IwVor'lth is a pihrase used in speaking of' a debt before it is due, and is the.sumz which at the l)reuailino'rulte of interest, will aneoclnt to that debt when it is (due, The difference between the present worth of a debt and the delt itself, is the interest of the present worth from tdhe present time until the time the debt is due; it is c!led the disco7net, since it is the sum which must be deducted f'rom the debt or nominal value, to give the ]present worth orr r eal value. REnMARK.-The discount in this case, like that in Art. 27h, is an abatement or deduction from the apparent value; in ftact, the difference between the real and the nomzinal value: but the latter is -a certafin per cent. of the nominal value, while this is a certain per cent., not of the nominal value (the debt), but of the real value (the Present Worth). 1. Find the present worth and discount of $5101.75 due in 1 yr. 9 mon. 19 da., rate of interest, 6 cj Ans. P-res. wor., $4603.775; Dis., $497).975 2. Also, of $1476.81, due in 4mlon. 11 da., rate, 6 %. Ans. Pres. wor., $1445.26; Dis., $31.55 3. Find the present worth of $2906.30, due in 103 days, rate, 8 %0. Ans. $2841.27 4. Find the discount of $6344.25, due in 23 days, rate of interest, 5 %. Ains. $20.20 5. Find the present worth of $12720.40, due in 9 da., at 7 % in New York. Ins. $;12698.48 6. I can sell property for $7500 cash, or for $4250 payable in 6 mon. and $4000 payable in 1 yr.: which should I prefer? and what do I gain by it if money is worth 12 % to me? Ans. The latter; $80.86 CASE V. ART. 313. Given, the principal, rate. and interest, to find the time. RULE. —As.sume 1 year jbf the time; determzine the inltre.rst on this supfposition, and divide the given interest by it. PRoor. —Calculate the interest for the time thus found; if it is the same as the given interest, the work is right. P E VIEW.-312. Why is this case important?:What is present worth? What is the difference between a debt not dclue and its present worth? What is it called? Why? 313. What is Case 5? The rule? The proof? SIMPLE lN'TEREST. 255 NoTES.-i. The quotient will be the time in years: if it contain a fraction, reduce it to months and days by Art. 222. 2. If the amount is given, subtract the principal from it, to get tile interest; then, apply the rule. In what time will 8830 amount to $1000, at 6 %0 simple interest? S O LU T O N. - Assume 1 yr. for the time; the interest of $830 for l yr. at 6 % is $49.80 (Case I): the given interest - $1000 - $8;30 $170, and as this contains $49.80, 8-k~ times, it must have taken 8. times as long to accumulate, that is, 3 0- yr. = 3 yr. 4 mon. 29 da., by reduction. (Art. 222.) IN WHAT TIME WILL 1. $1200 anmount to $1800, at 10 9? Ans. 5 yr. 2. $415.50 to $470.90, at 10 %? Anus. lyr. 4mon. 3. $3703.92 to $4122.15, at 8 %? Ans. 1 yr. 4 mon. 28 da. NOTE.-A part of a day is omitted in the answer, not being recognized in interest; it must be taken account of, however, in tle proof. 4. In what time wnill any sum, as $100, double itself by simple interest at 41, 5, 6, 7, 7l, 8, 9, 10, 12, 1'2, 15, 18, 20, 25, 30 %? Atns. 22,2, 1 6, i 142, 13,a 12, 1, 10, 8 8 6 5k;, 5, 4, 3: yr. 100 diviced by MJe rate of tt(erest, will give tlhe nzumber of years in whliclh any sum will double itself: 5. In what time will any sum treble itself by simple interest at 4, 4-, 5, 6, 7, 7V, 8, 9, 10, 12 9? Ans. 50, 44~, 40, 33k, 28, 26, 25, 22~, 20, 16g; yr. 6. In what time will any sum quadruple itself by simple Int., at, 5, 6, 7, 8, 10, 12, 15, 20, 30, 40, 50, 100%? Ans. 60, 50, 42, 3748, 30, 25, 20, 15, 10, 7~, 6, 3 yr. 7. -ow long must I keep on deposit $1374.50, at 10 %, to pay a debt of $1480.78? Ans. 9 mon. 8 da. 8. How long will it take $3642.08 to amount to $4007.54, at 12 %? ARs. 10 mon. 1 da. 9. How long would it take $175.12 to produce $6.43 interest at 6 %? Ans. 7 mon. 10 da. R evi w. —313. What will the quotient be? If it contains a fraction, what is necessary? What, if the amount is given? Solve the example. What is said of a part of a day? 25a6 RAY'S HIGER ARITHMETIErC. 10. Htow long would it take $415.38 to produce $10.69 interest in New York at 7 %? Ans. 1334 da. R E I A R K. —In this example, multiply the fraction of the year by 365, instead of 12 and 30. (Art. 308.) ART. 314. All the Cases of Simple Interest, except the 4th, which treats of Present Worth, can be solved by Compound Proportion, after the following form: $100. Principal.::: Rate.: Interest. 1 yr. Time in yr. ART. 315. The last 4 cases show a striking similarity in their rules and modes of operation, and can be put into a GENERAL RULE FOR TIIE LAST FOUR CASES. Assume 1 for the qutantity reequtired; determine the interest (or amount if it be necessary), on this supposition, and divide the given interest (or amount) by it. XXV. BANKING. ART. 316. The most important application of Simple Interest is Banking, including the Discounting of Notes, Exchange, and the settlement of depositors' accounts. BANKS are corporations which deal in money. A bank of issue is one which issues its own notes as money. A bank of discouent is one which makes loans. A bank of deposit is one which takes charge of money belonging to others. REeArtii.-A bank is generally controlled by a board of directors, elected by the stockholders; the principal officers are the president and cashier. REVIFEW.-314. By what compound proportion may all the cases of interest, except the 4th, be solved? Why can not the 4th be solved thus? 315..W1hat general rule serves for the last 4 cases? In which case only will it be necessary to use the amount? Why? 316. lWThat is the most important application of simple interest?'WThat does it include? 17What are banks? What kinds? Define each. How is.a bank generally controll 1? What are the principal officers? BANKING. 257 PROMISSORY NOTES. ART. 317. A PROMISSORY NOTE is a written promise by one party to pay a named sum to another. The person by whom the note is signed is the maker of the note. The person to whom the money is promised is the payee. The owner of a note is the holder. A promissory note is negotiable, when it is payable to bearer, or to the order of the payee; otherwise, it is not negotiable. A negotiable note may pass from hand to hand; the payee or holder endorsing it by writing his name on its back, thus becoming liable for its payment, if the note is payable "' to order." If the note is payable "'to bearer," no indorsement is required on transferring it, and only the maker is responsible. It is essential to a valid promissory note, that it contain the words "value received," and that the sum of money to be paid should be written in words. The face of a note is the sum promised to be paid. If a note contain the words "with interest," it draws interest from date, and if no rate is mentioned, the legal rate prevails. The face of such notes is the sum mentioned with its ~nterest from date to the day of payment. If a note does not contain the words "with interest," and is not paid when due, it draws interest from the day of maturity, at the legal rate, till paid. REMARI-K.-The day of maturity is when the note is legally due. If a note is not paid at maturity, the indorsers must be notified of the fact, in writing, when it falls due, or they will not be liable. This notification is generally made by a Notary, and is called a Protest. R E VIE w.-317. What is a promissory note? Who is the maker of a note? The payee? When is a promissory note negotiable? What may be done with a negotiable note? If it is payable to order, what is necessary? What is indorsing? What effect does it have? What kind of a note needs no indorsement when transferred? What are essential to a valid promissory note? What is the face of a note? What is the face, when the note contains the words "with interest?" 22 258 RAY'S HIGHIER ARITHIIETIC. AnT. 318. If a note is payable "on demand," it is legally due when presented. Bank notes are of this sort, being payable "to bearer on demand." TIf a day of payment is specified in the note, it is due the 3d day aftcrward; in some places, it is due on the day specified. If a note is payable a certain time 1" after date," proceed thus, TO FIND TIlE DAY A NOTE IS LEGALLY DUE. RULE. —Add to the dale of t7ke notle, the numtber of years and montlths to elapse before p2aymelnt; if this gies the day of a monthl higlher tlhan that a07,month contains, takle the last dacly i that monzth; thet., cotiunt the nmlber of days nzeltionzedl in t1he nole a7nd 3 7morte: this will gite the day the note is legally cdile; but if'it i i a Slllday or a national holidnay, it mnust be paid thle day previous. NOxTES.-1. When counting in the days, do not reckon the one front which the counting begins. The three additional dclays are called "days of grace"; in some countries they are 4, 5, or more. The day before 1"grace" begins, the note is nzominallJ due; it is legally due on the last day of grace. 2. The months mentioned in a note are calendar months. Hence, a 3 mon. note would run longer at one time than at another; one dated Jan. 1st, will run 93 (lays, one dated Oct. 1st, will run 95 days: to avoid this irregularity, the time of short notes is generally given in days instead of. months; as, 30, GO, 90 days, instead of 1, 2, 3 months. DISCOUNTINGO NOTES. ART. 319. DISCOUNTING NOTES is buying them at a discount, and is chiefly done by banks and brokers. Notes to be discounted must be payable to order, and indlorsed by the payee. The proceeds or cost of a note is the sum paid for it. The difference between the cost and the face of the note is the dliscouttt; it is more or less, accordinog to the title the note has to run. REVIE w.-.317. If a note does not contain these words, and is not pa.id when due, what is the consequence? What is the day of maturity? 518. If a note is payable "on demand," when is it legally due? DISCOUNTING NOTES. 259 The time to'tun is the number of days from the day the note is discounted, to the day the note is legally due, counting the latter, but not the former. It EM ARRI. —The time to run is taken in days by banks, because it is to their advantage; if it were taken in months and days, each month would have to be considerod 30 days in calculating the discount, whereby a day would be lost in each month of 31 days. ART. 320. In determining the day of maturity and the time to run, it is convenient to use this TABLE SlIOWING TIlE NUMBER OF DAY.S FRO3I ANY DAY OF. FoY ] g r w G - 6J O Z To the'p CD i n sanme day td rJ of next 3 (653rt 069G 275 24055245l 184 15353 122 -92 61 31 Jan. 31 36o337 306 276 245 215 184 153 -)- 92 6 " Feb. 3-`45 304 127 3 243 212 18 3,051 4[30 1 O t. 59 _28 365 343042732432l (l81b1 Mar. _90 595 31 3G5 335 304 4274 243 212' 1851 1 1 A pril. 1"'0 -8-9 61 39 3t5 33341304 27 3 _42 2)1 181 151 1May. 151 1'4092 61 313 65 335 341273;)012 1823 June.! 334 303 j5.244121411831153 122 91 1 30.365 Dec. 1R a150.In leap years, if the last day of Feb2uary is included in the time, 1 day must be added to the number obtained from the table. 3 note maturing Sept. 13, is discounted June 243 pre vious: what is the time to run? S 4o nUT.-From June 24 to Sept. 24, by the table, is 92 d.; the 13th being 11 days before the 2th, will give 92 - 11, 851 dae, Anc. for the time to run. qvsuwu. —318. If it is payable a certain time after date, what is the rule. for finding when it is legally due? Which day is not counted? What are days of grace? Why so called? When is a note nominlluded indue? When legally ue?' Whatbe are the months? obWhat is said of notes drawn for months? How is this irregularity avoided in short notes drawn for months? How is this irregularit~y avoided in short notes? 260 RAY'S 3HIG-HER AR1ITII'ETJIC. ART. 321. CASE I. To find the discount and proceeds of a Note of short date, custom has sanctioned this RULE. Takce the face of the note as a principal, the rate of discount as the rate of interest, and calculate the interest /rbs the tile the note has to run; this will be t/he discount of the note, and if it is subtracted fioom the Jhce of the note, the remainder will be the proceeds or cost of the note. NOTES. —1. The discount on short notes, resembles the discount on money, stocks, bonds, &c., being calculated on the nonminal value or face, but differs from the discounts on debts and long notes, (Art. 312): the former is called Bank discount; latter, true discount. 2. Since the face of every note is a debt due at a future time, its cost ought to be the present worth of that debt, and the bank discount to be the same as the true discount. As it is, the former is greater than the latter; for, bank discount is the interest on the face of the note, while true discount is the interest on (he present worth, which is always less than the face. Hence, their difference is the interest of the difference between the Present Worth and face; that is, the interest of the true discount. 3. In calculating the discount on notes, use generally the 2d rule (Art. 307). The operation may often be shortened, by recollecting that when the two right hand figures of dollars are pointed off, the result is the interest of the principal, for 60 da. at 6 %, for 72 da. at 5 %, for 45 da. at 8 S%, for 36 da. at 10 So, or for. any other number of days and rate, whose product is 360; so that the interest at any rate, can be frequently got in this way by aliquot parts, without first getting it at 6 %. Find the day of maturity, the time to run, and the proceeds of the following notes. 1. $792 5-0~ CINCINNATI, Jan. 3d, 1854. Six months after date, I promise to pay to the order of Willis & Markham, seven hundred ninety-two 5 ~o0o dollars, al the Commercial Bank, value received. H. WHITTAIER. Discounted, Feb. 18, at 6 S. Ans. Due, July 316; 138 da. to run; Pro. $774.27 R E V IE. —319. What is discounting notes? By whom is it done? What kind of notes are discounted? What is the proceeds or cost of a note? The discount? What is the time to run? 320. Show the use of the table. What is said of leap years? 321. What is the rule for dis counting short notes? What does bank discount resemble? DISCOUNTING NOTES. 261 R E MA a K. —Th e date before the line, in July 3169 is when the note is nominally due, the other when it is legally due. 2. $10666 7 o LOUISVILLE, May 19, 1855. Value received, ninety days after date, I pronmise to paq, Thomas Beatty, or order, one thousand sixty-six -7o dollarTt, at the City Bank. G. W. ALEXANDER. Discounted June 8, at 6 %. lAns. Due, Aug. l71o0; 73da. to run; Pro. $1053.77 3. $1962 405 NEW YORK, July 26, 1850. Value received, four months after date, I promise to pay B. Thoms, or order, one thousand nine hundred sixty-two v4-' dollars, at the Chemical Bank. E. WILLIAMS. Discounted Aug. 26, at 7 %. Ans. Due, Nov. 2 1621; 95 da. to run; Pro. $1926.70 4. $543 -6 CINCINNATI, Oct. 80, 1855. Thirty days after date, I promise to pay to the order of Baker & Goodal, five hundred forty-three i6%5p dollars, value received. T. H. SHORT. Discounted Oct. 81, at I o a month. Ans. Due, Nov. 29 1 Dec. 2; 32 da. to run. Pro. $537.88 5. $2672-2o PHILADELPHIA, March 10, 1852. Nine months after date, for value received, I promise to pay Edward H. King, or order, two thousand six hundred seventy-two v'e5u dollars. JEREhIAi BARTON. Discounted July 19, at 6 %. Ans. Due, Dec. 1' 1; 147 da. to run. Pro. $2606.71 6. $804-,-o COLUMBUs, Aug. 12, 1854. Three months after date, I promise to pay at the City Bank of Columbus, eight hundred four 39y dollars to the order of Irwin & Lee, value received. JOSIAH NEEDHAMI. Discounted September 3, at 6 %. A's. Due, Nov. I'j l; 73 da. to run. Pro. $794.60 REvIEw.-321. What kind of notes are discounted by this rule? How are long notes running for one or more years discounted? What ought the cost of a note te be? What would the bank discount then be? Which is greater, bank discount or true discount? Why? How much? What rule of interest is generally used in notes? How may the operation often be shortened? 262 RAYS HIGHER ARITHMETIC. 7. $3886. ST. Louis, Jan. 3]1, 1853. One month after date, we jointly and severally promise to pay C. MIcKnight, or order, three thousand eighlt hundred eighty-six dollars, value received. T. MIONROE, Discounted Jan. 31, at 1. o a month. I FOSTER. Ans. Due, Feb. 28 1 MIar. 3; 31da. to run. Pro. $3825.77 REMARK. —A note, drawn by two or more persons, " jointly and ~verally," may be collected of either of the makers, but if the words " jointly and severally " are not used, it can only be collected of the makers as a firm or company. 8. $1425. NASHVILLE, April 11, 1853. For value received, eight months after date, we promise to pay Henry Hopper, or order, one thousand four hundred twenty-five dollars, with interest from date at 6 per cent. per annum. Nixon & MiARSHI Discounted June 15, at 6 %. Ans. Due, Dec. I 1,I4; 182 da. to run. Pro. $1437.73 N. B.-Observe that the face of this note is the amount of $1425 for 8 mon. 3 da. at 6 %. 9. $3703i4 0 BALTIMORE, June 6, 1850. For value received, four months after date, we promise to pay to the order of Jones & Newcome, three thousand seven hundred and three dollars and eighty-four cents, at the Savings Bank. THOMAS SHIRPE & Co. Discounted June 18, 1850, at 1 % a month. Ans. Due, Oct. 619; 113da. to run. Pro. $3564.33 10. $8131oG DAYTON, May 31, 1856. For value received, sixty days after date, I promise to pay to the order of Hiram Wells & Co., eight hundred thirteen dollars sixty cents. JAMIES T. FISHEIR. Discounted May 31, 1856, at 2 O a month. Ans. Due, July 30 1Aug. 2; 63 da. to run. Pro. $779.43 11. $737 -1- BOSTON, Feb. 14, 1856. Value received, two months after date, I promise to pa, to J. K. Eaton, or order, seven hundred thirty-seven,Tdollars, at the Suffolk Bank. WVILLIAM ALLEN. Discounted Feb. 23, at 10. Ans. Due, April 14,,7; 54da. to run. Pro. $726.34 DISCOUNTING NOTES. 263 12. $40851. 20 NEW ORLEANS, NOV. 20, 1855. Value received, six months after date, I promise to pay John A. Westcott, or order, four thousand eighty-five 02,% dollars, at the Planters' Bank. EE. WATERMAN. Discounted Dec. 31, 1855, at 5 %. Ans. Due i)ay 2, 1856; 144da. to run. Pro. $4003.50 13. $2623f%% CINCINNATI, Aug. 7, 1854. For value received, eihteen months after date, I promise to pay to the order of Jonathan Evans, two thousand six hundred twenty-three oa dollars, with interest at ten per cent. per annum. MORRIS TALBOT. Discounted June 24, 1855, at 1A O a month. Ans. Due, Feb. 71, o, 1856; 231 da. to run. Pro. $2728.61 ART. 322. It may be inquired, what rate of interest is paid, lwhen a note is discounted. The proceeds of the note is the sum received or borrowed, and is, therefore, the princi)tal, while the discount is its interest for the time the note runs; so that having the principal, time, and interest, the rate of interest can be found as in Art. 809. It is simpler, however, to leave the face of the note out of view, and proceed thus: TO FIND THE RATE OF INTEREST WIIEN A NOTE IS DISCOUNTED. RULE.-Assume,100 for the face of the note, and on this sulpposition determine the discount and proceeds for the time it has to run; the jf7mer will be the interest; the latter, the principal; and the time to run, the time: from which the rate of interest can be found by Case II of Simple Interest. What is the rate cr interest, when a sixty day note is discounted at 2 % a month? SOLUTION.-For every $100 in the face of the note, the discount for 63 lda. at 2 7% a month, is $4.20, and the proceeds $95.80; then $95.80 being the principal, and $4.20 its interest for 63 da., the rate of interest is found by Art. 309 to be 254- 2% per annum. REvIEw.-322. When a note is discounted, what is the principal or sum borrowed? What its interest? IIow may the rate of interest then be found? What may be left out of view in this calculation? 264 RAY'S HIGHER ARITHMETIC. WHAT IS THE RATE OF INTEREST, 1. When a 30da. note is discounted at 1,1 1, 1, 2 % a mon.? Ans. 12t-, - 15, l, 1%, 24-8s % per annunm. 2 When a 60da. note is discounted at 6, 8, 10 % per annum? Ans. 6,-1f9S,9 8 s63, 10 37-0o % per annum. 8. When a 90 da. note is discounted at 2, 2A, 3 % a imon.? Ans. 25467- 32- 94-, )6 % per annum. 4. When a note running 1 yr. is discounted at 5, 6, 7, 8, 9, 10, 12%? Ans. 5, 6, 7 4, 8 6, 9,7 11, 13 %. R E3iA R r. —It may seem unnecessary to regard the time the note has to run, in determining the rate of interest; but, a comparison of examples 1 and 3, shows that a 90da. note, discounted at 2 S a month, yields a higher rate of interest then a 30 da. note of the same face, discounted at 2 % a month. The discount, at the same rate, on all notes of the same. face, varies as the time to run, and if in each case, it was referred to the same principal, the rate of interest would be the same; but when the discount becomes larger, the proceeds or principal to which it is referred, becomes smaller, and ther-efore the rate of interest corresponding to any rate of discount increases with the time the note has to run. Hence, the profit of the discounter is greater proportionally on long notes than short ones, at the same rate. ART. 323. Discounting Notes is an application of Simple Interest: the face of the note is the princjpal; the dlscoiunt is the interest; the rate of discount is the rate of interest; and the tinme to ruIa is the time. Hence, it may be divided into cases corresponding to, and solved like those of simple interest. The only case of sufficient importance to require distinct notice, is CASE II. Given, the proceeds, time, and rate of discount, to find the face of the note. RULE.-Assume l$1 for the face of the note; determine the proceeds on this supposition, and divide the given proceeds by it. RvsiE I w. —322. Give the rule. Analyze the example. Can the time to run be left out of view in determining the rate of interest corresponding to any rate of discount? Why not? What does the rate of interest increase with? What sort of notes are most profitable to the discounter, then, provided the rates of discount are the same? 323. What is Case 2? The Rule? The proof? Solve the example. DL) SC'OU NT[ING NO'TES. 265 Pnoov. —Discount the note thus found; if it yields the given proceeds, the work is right. For what sum must a 60 da. note be drawn, to yield $1000, when discounted, at 6 % per annum? SOLUTION. —For every $1 in the face of the note, the proceeds, by Case I, (Art. 321), is $.9895; hence, there must be as many dol]axrs in the face as this sum, $.9895, is contained times in the givec proceeds, $1000: this gives $1010.61 for the face of the note. 1. Find the face of a 30 da. note, which yields,$1650 wlien discounted at 1- % a nion. As. $1677.68 2. The face of a 60 da. note, which, discounted at 6 % per annum, will yield $800. An.l:. $808.49 3. The fice of a 90 da. note, which, discounted at 7 % per anlnuam, yields $1235.40. A,s. $1258.15 4. The f lce of' a 4 mon. note, which, discounted at 1 % a month, yields $3375. Ails. $3519.29 5. The face of a 6 mon. note, which, discountec at 10 % per annum, yields $4850. Ans. $5109.075 6. The face of a 60da. note, discounted at 2~4 a month, to pay $768.25. Ai.s. 801.93 7. The face of a 40da. note, which, discounted at 8 %, yields $2072.60. Alls. $2092.60 8. The face of a 30 da. and 90 da. note, to net $1000 when discounted at 6 c. Ans. $1005.53 at 80 da.; $1015.74 at 90 da. ART. 324. To find the rate of discount corresponding to a given rate of interest. RULE.-Assume $100 for the proceeds;,fud its interest and amoutll at the gyiven rate, for the time the?ot/e r?inns; lhe latte2r is the face of the note, ancd the former is the discount, or interest on that face br tie te tie the note rrunts; the srate of illteest will then be found by Case II of Simple interest, adl will be the rate of discount requiired. tf I wish to get interest at the rate of 20 % per annum, a -}lhat rate should I discount 60 da. notes? SOLUTION.-For every $100 in the proceeds, I wish to get interest at 20 O per annum, which for 63 daysis $3.50, interest, and It E v I E w. —324. What is the rule for finding the rate of discount cor. ra:spond:1..in to a given rate of interest? Analyze tho example. 'tfi2($ RAY'S HIGHER ARl'ltJiETIC',. Si 03.50.1 amount'1 take $103.50 as the face of the note, or prinacipail, and'f3.50 as the discount, or interest of that principal for 68 days, and find the rate by Case 1 of Simple Interest, 19 207-%. WHAT RATES OF DISCOUNT 1. On 30 da. notes, yield 10, 15, 20, 30, 40, 50 (0 inter. e-t? An1S. 9 1" -1, 1 1, 29 -,, 3 %, - 2. On 60 da. notes, yield 6, 8, 10, 12, 18, 24 % iaterest? i-nszs 5f" $9 141 s. 5 7 1, 9 a 5 47, 94 7, 1 1' -,-1 7 -', 2'3 3. On 90 da. notes, yield 1, 11, 2, 2, 3, a mon int.? Ans. 11y~o ~'03 17, 22_6, 2 9"Y 732: 2 % ess. l l o a3 I, 172 0 s 3 7 2 2 31 7 7 7 2 7.- 3 -g9,8 4. On notes running 1 yr. without grace, yield 5, C, 7, 8, 9, 10, 12, 15, 18; 20, 25 %0 interest? A 4s -i, 5 6 5, 7 1, 8' 8s 97, 1o5 13=- 5`" 16s, 5o 20 PARTIAL PAYMENTS. ART. 325. A note, not paid at maturity, draws interest at the legal rate, from the day it is due till it is paid but if it contains the words " with interest, it draws interest from date. Any sums paid on ctccote of' the note, are indorsed with their respective dates, and are called Partial Payntents. In such cases, the balance due on settlement is for.,11 by the U. S. RULE FOR PARTIAL PAYI MENTS. "Apply the paymalt, in the first place, to the discharrge of the interest t te!, dlt,; if the payment EXCEEDS the ia/ei est, the surplus oes towvarc discharging the principeal, and the sulbsequeent inter,st is to be computed on, the balance of principal remaining due. "If the paygme t be LESS thian the interest, the surpu1s of interest minst not le taken to augment the principal; buit interest coottian'ts on the f rmner prilc-ipal, until the period whent the pay~. ments taken toget/;er exceed the interest dle, and then thle suriplts R,VIraVw.-325. ff a note is not paid at maturity, what is the conseqafncee? If it co: tains the words "with interest,"' when does interest commence?'WThat are partial payments? What is the United States rule for casting, interestl on notes and bonds when partial payments have been made? On what principle is this rule founded? PARTIAL XPAY AMEN'TS. 267 is to be applied toward diichlartyig the principal, and ilnterest ia to be colmpufed oul the balance as ajbresaid." R F3 It K.-1. This "'rule, is adopted by the Courts of the U. S. and of most of the States; it is on the principle that neither interest arl' pa':lment shall draw interest. 2. It'is worthy of remark, that the whole aim and tenor of legis* itive enactments and judicial decisions on questions of interest, i:yve been to favor the debtor, by disallowing compound interest, and y }t this vety rule fails to secure the end in view, and really maintains anud enforces the principle of compound interest in a most objectionable shape;.fo'r it,makes inzterest due, (not every year as compound interest ordinarily does) but as often as a 1paymenzt is nzade; by whlich it, hxappens tlhat the closer the payments lere toyether, the gyeater the loss of the debtor, who thus suffers a penalty for his very promnptness. To illustrate, suppose the note to be for.$2300, drcawing interest at g e, and the debtor pays every month $10, which just meets the interest then due; at the end of the year lie would still owe $2000. But if he had invested the $10 each month, at 6 /O lie would have hlad, at the end of the year, $123.30 available for payment, while the debt would have increased only $120, being a dlifflreiice of $3.30 in his favor, and leaving his debt $1996.70, instead of $'2000. 3. To find the difference of time between two cldates on the note, reckon by years and months as far as possible, and then count the days; as in the rule Art. 318. 850o. CINCINNATI, April 29, 1850. Ninety days after date, I promise to pav Stephen Ludlow, or order, eight hundred fifty dollars, with interest; value received. CHARLES K. TAYLOR. Inzdorsemeezts.-Oct. 13, 1850, $40; June 9, 18351, $32; Aug. 21, 1851, $125; Dec. 1, 1851, $10; March 16, 1852, $80. What was due Nov. 11, 1852? SoLubToN.-Interest on face ($850) from April 29 to Oct. 13, 1850, being 5 mon. 14 da., at 6 % per annum, $23.233 850. Whole suml due Oct. 13, 1850,...... 873.233 Payment to be deducted,....... 40. Balance due Oct. 13, 1850,......... 833.233 R i vIE v. -225. How does the rule allow compound interest? Illus. trate by an examjle. How find the difference of time between two dates. on the note? RAY'S I1G0 HEl: AItrUITMETIC. Interest on balance ($833.233) from Oct. 13, 1850, to June 9, 1851, being 7 mon. 27 da.,....... 32.913 Payment not enough to meet the interest, 32. Surplus interest not paid June 9, 1851;.....913 Interest on former principal ($833.233) from June 9, 1851, to Aug. 21, 1851, being 2mon. 12da,... 9.999 Whole interest due Aug. 21, 1851,.. 10.912 833.233 Whole sum due Aug. 21, 1851,........ 844.145 Payment to be deducted,.......... 125. Balance due Aug. 21, 1851,.......... 719.145 Interest on the above balance ($719.145) from Aug. 21, 1851, to Dec. 1, 1851, being 3 mon. 10da.,.... 11.986 Payment not enough to meet the interest,.... 10. Surplus interest not paid Dec. 1, 1851,..... 1.986 Interest on former principal ($719.145) from Dec. 1, 1851, to March 16, 1852, being 3mon. 15da.,... 12.585 Whole interest due March 16, 1852,....... 14.571 719.145 Whole sum due March 16, 1852,........ 733.716 Payment to be deducted,.......... 80. Balance due March 16, 1852,....... 653.716 Interest on Balance from March 16, 1852, to Nov. 11, 1852, being 7 mon. 26 da.,...... 25.713 Balance due on settlement, Nov. 11, 1852,..... $679.43 1. $3o04 76-a CHIICAGO, March 10, 1852. For value received, six months after date, I promise to pay G. Riley, or order, three hundred and four T7-60` dollars. No paymtents. What was due Nov. 3, 1853? Ans. $325.63 2. $2250. LouIsvrLLE, Dec. 6, 1850. For value received, one year after date, I promise to pay Albert Rogers, or order, two thousand two hundred and fifty dollars, with interest. NATHAN PHILIPS. hnlo,se(l: April 18. 1852, l1000. What. was due Aug. 19, 1854? Ans. $1634.63 2. $429 eo% INDIANAPOT,IS, -April 13, 1853. On demand, I promise to pay WX. il~organ, or order, fouT hundred and twenty-nine 3-i-0 dollars, value received. P T. VT VIISON. TIn A Ii -P)AYSlENTS. 69 Indorsed: Oct. 2, 1853, $10; Dec. 8, 1853, $60; July 17, 1854, $20i0. What was due Jan. 1, 1855? Ans. $195.06 4. $1750. NEW YORK, Nov. 22, 1852. For value received, two years after date, I promise to pay to the order of Spencer & Ward, seventeen hundred and fifty dollars, with interest at 7 per cent. JAcuB WINSTON. Indorsed: Nov. 25, 1854, $500; July 18, 1855, $50; Sept. 1, 1855, $600; Dec. 28, 1855, $75. What was due Feb. 10, 1856? Ans. $879.71 5. $4643% 00 DAYTON, Ohio, March 7, 1851. For value received, three years after date, I promise to pay R. Banks, or order, forty-six hundred and forty three 5 ( dollars, with interest at 10 per cent. W.N G. BROOKS. Indorsed: June 25, 1854, $1000; Nov. 1, 1854, $500; Jan. 12, 1855, $2500; Sept. 4, 1855, $1350; May 10, 1856, $150. What was due July 1, 1856? Ans. $1189.86 6. $540. BALTIMTIORE, Jan. 11, 1849. Eighteen months after date, we promise to pay Silas Greene, or order, five hundred and forty dollars, with interest, value received. EVANS & HART. Indorsed: Nov. 23, 1850, $125; March 5, 1851, $35; Feb. 27, 1852, $25; May 31, 1852, $80; Oct. 16, 1852, $100. What was due March 1, 1853? Ans. $291.60 7. $2500. PHILADELPHIA, Aug. 8, 1850. For value received, nine months after date, I promise to pay Abijah Warren, or order, two thousand five hundred dollars, with interest at 6' per cent. HENRY CROSS. Indorsed: Nov. 1, 1851, $400; Dec. 14, 1852, $100; July 6, 1853, $50; Oct. 21, 1854, $750; April 18, 1855, $500. What was due July 1, 1855? Ans. $1362.04 8. $6875. BOSTON, Oct. 22, 185'2. For value received, four months after (late, we promise to pay Augustus King, or order, six thousand eight hundile(i and seventy-five dollars. DAvIs & UNDERWVoo O. Indorsed: Feb. 25, 1853, $2000; Jan. 1, 1854,. $245; April 1, 18504 $75; July 10, 1854, $1500; Dec. 26, 1854, $95; May 12, 1855, $1200 Sept. 29, 1855, $50. What was due Jan. 1, 1856? Ans. $2376.64 270 RAY:S JilGitEit AII'iTi'icETIC. 9. $550 Sr. 50 LoIus, June 17, 1848. ];o1 valte received, one year after date, I promnise to pay to t|he order of IRoss & Wade, five hundred and'ty dollars, with interest..Tio-}ifv GomnON. Indorsed: Ang. 10, 1848, $100; Dec. 1, 1845, $40; Feb. 8, 1850, $30; Sept. 27, 1850, $200; Jan. 4, 1852, $10; Mar. 1, 1852, $60. What was due aug. 28, 1852? Ans. $195.87 CONNECTICUT RULE. ART. 832. "Uompute the interest to the time of the first payment, i/' chatt be one year or more frone the time that interest coitmenccd; add it to the principal, and deduct the payment from the saimu total. " If there be after' palments made, compute the interest on the balance due to the next payment, and then deduct the payment as above; and in like tman'ner from one payment to another, till all the payments are absorbed: Provided, the time between one pcayment and another be one year or more. "But if cany payment be made be/bre one year's interest hath accrued, then covmpute the interest on the principal sum dele on the obligation, for one year, add it to tlhe principal, and compute the intrerest on the sum paid, f.rom the time it was paid, up to the end of the year; add it to the sum paid, and decdact that sum from the principal and interest! added as above. (See Note.) "f any payment be mnade of a less sum than the interest arisen at the time of such payment,,no interest' is to be conmputedl, but only on the pri'ncipal sum, for any period." N o T E.-" if ai year doeq not extend beyond the time of payment; but if it does, then find the amount of the principal remaining uenpaid up to the time of settlement, likewise the amount of the payment or payments froom the tnime they were paid to the limne of settlement, and deduct the sum of these several anmunts from the amount of the princij)al. 1. What is due on the 2d, 3d, and 4th of the preceding notes, by the Connecticut rule? Ans. 2d, $1634.63; 3d, $194.54; 4th, $877. 95 VERMONT RULE. ART. 327. Ffind the simple interest qf the prinxcpal from the ti'me it begins to draw interest to the time of settle.,elet, canld add it to the principal. DO the samne for each payment. iddl logqether lhe amounts of the severctl pcymients, and dednct the seum froan the amount ofj the principal. The remainder will be the balance duce oce settlement. EXCIIA:' G h. 271 To'rE. —This.rule i; of frequent use, when settlement takes place a year o le.ss from the time interest begins. 1. What is due on the 2d, 3d, and 4th of the preceding notes, by thle Vermolont rule? Ais. 2d, $1608. 88; 3d, $193.50; 4th, $855.79 REP. a AR. —'ftle following has been recommended as a more equitable RULE for Plartial Payments, than any in use. l' Starting at the time interest begins, find the Present Morth by Simple fiterest of each payment; deduct the sum of these Present Tforths from the Pr incijpal: the amount oj the balance, by Simple Interest, to the day Crf settlemrent, will be the sum then due." The principle of this rule is, each payment discharges a part of the pr'inlipal with its simple interest to the day the payment is made; making interest and principal due at the same time, instead of the interest a,11 due firstS and the principal afterward. XXVI. EXCHANGE. ART. 328. EXCHANG(;E is the method of transmitting money from one place to another by mLeans of Bills of irechtange. If the places are in lifferent countries it is called Foreign ELr:chn7 7ye; if not, liontee, Domeestic or Inzlancd Exchange. Bills of ESchaltye, also called drafts or checks, are writen orders for the payment of money. A Sighte Bill, is one payable "at sight." A. Tine Bill is payable a specified time after sight, or after date. The signer of the bill, is the maker or drawer. The one to whom the draft is addressed, and who is requested to pay it, is the drawee. The one to whom the money is ordered to be paid, is the p.acyee. The one who has possession of the draft, is called the own?er or holder; when he sells it, and becomes an indolrsir, he is liable for payment. R E V Is Ew.-328. What is exchange? Foreign exchange? Home exchanrge? What are Bills of Exchange? What is a sight bill? A timrn bill? Who is the drawer? The drawee? The payee? Who is the holder? What, if he becomes an indorser? 272 tAY'S HI (G11ERl ARIIT!M1E'TI'C. A sjpectal indorsernent is an order to pay the draft to a particular person named, who is called the indorsee, as " Pay to F. II. Lee. —W. Harris.' and no one but the indorsee can collect the bill. When the indorsement is in blank, the payee merely writes his name on the back, and any one who has lawful possession of the draft can collect it. If the drawee promises to pay a draft at maturity, he writes across the face the word "Accepted," with the date, and signs his name, thus: "' Accepted, July 11, 1851.-H. Morton." The acceptor is first responsible for payment, and the draft is called an acceptance. ART. 329. A bill of exchange, like a promissory note, may be payable "to order," or "bearer," and is subject to protest in case the payment or acceptance is refused: it is also entitled to the 3 days grace, whether a sight or ticme bill. In some States, drafts are not entitled to days of grace. ART. 330. The following are forms of drafts or bills of exchange. INLAND DRAFT. $1 500. CINCINNATI, Feb. 8, 1855. Please pay, at sight, to Williams & Baker, or order, fifteen hundred dollars, value:eceived, and charge To KINO & CLAaK, Brokers, New York. THOMAS ATKINS. FOREIGN DRAFT. Exchange ~2000. New York, May 21, 1853. At sixty days sight of this first exchange (second and third of the same date and tenor unpaid), pay to the order of H. Watts, two thousand pounds, without further advice. To GEORGE SPENCE, Merchant, Liverpool. ELMER & BATES. No TE. —1. The words "value received" are not essential to a bill of exchange. 2. In foreign exchange, three separate bills are generally drawn, so that if one or two are lost, the other may reach its destination. When one bill of a set has been paid, the rest are void. R E vIE w. —328. What is special indorsement? Give an example. What is the consequence of a special indorsement? What is an indorselment in blhik? Its consequence? What is an acceptance? Give an example. What is the effect of it? 329. How does a bill of exchange resemble a promissory note? What is said of protest? Of grace? Of sight-bills? 330. Give an example of an inland draft. Of a foreign draft. What is said of the words " value received"?' What is a set of foreign exchange? When one of a set is paid, what is the consequence? EXCi 1 LANG;. 273 ART. 331. The rate of exchanye is a rate per cent. of the face of the draft. The rate of exchange between two places depends on their trade with each other; thus, at New Orleans, drafts on New York will be at a premiumn, more or less, according as the demand at New Orleans, for drafts to make payments in New York, exceeds, more or less, the supply furnished by parties drawing against sums due them in New York. If the demand is less than the supply, exchange on New York is at a discount in New Orleans, more or less, according to the excess. If the demand just epuals the supply, exchange is at par. ART. 332. The calculations connected with inland exchange have been explained in Art. 276-279. The computation of foreign exchange is similar, except that it is necessary to express the money of one country in that of the other. The sum mentioned in a draft on a foreign country, is expressed in the money of that country. In calculating the premium, discount, or cost of such a draft, in the home currency, it is necessary not only to have tables of foreign money, but also to know the comparative values of the home and foreign currencies. The par of exchange is the comparative value of the money of two countries, depending upon the amounts of pure gold or silver they contain: it must, therefore, remain fixed, as long as the coins retain the same weight and fineness; as, $4.86 (intrinsic value,) = ~1, and $1 = 54 francs. The course of excha.nge is the par of exchange after allowing for the rate of exchange; since the rate of exchange' is variable according to the balance of trade, the course of exchange will also fluctuate within certain limits, being sometimes above and sometimes below the par, as, 1 5.25 francs, $4.87 (exchangeable value,) = ~1. RlxVIEW. —331. What is the rate of exchange? What does it depend cmn 7 Example. 332. To what subject do computations in inland Exchango belong? How does the computation of foreign exchange differ fiom that of inland exchange? How is the face of a foreign draft expressed? What must be known to find its value in our currency? What is the par of exchange between two countries? What does it depend on? Can!d ever change? 274 ItAYT'S HIG HER ARITHMETIC. Since the cori'Se of exchc.an ge allows for the rate of ex. clh.<,ta'w.', when the formier is known, the3 cost of the bill is found by reduction, without using Percentage. If the s(ate of