DAY AND THOMSON'S SERIES. HI G IER ARITH METIC; OR TIE SCIENCE AND APPLICATION OF NUMBERS; COMBIINNG TtEX ANALYTIC AND SYNTHETIC MODES OF INSTRUCTION. DESIGNED FOE ADVANCED CLASSES IN SCHOOLS AND ACADEMIES. BY J~aES B. THOMSON, LL.D., AUONB OF MENTAL A MJITIICEfO; EEECISES IN ABITHMETIOAL ANALYogIe EAOUTICAL AEITrMETIC; EDITOB OF DAY'S SOHOOL ALGEEBA; LEGEiNDEE'S GEOMETRY, ETC. ONE HUNDRED AND TWENTIETH EDITION. NEW YORK: IVISON & PHINNEY, 48 & 50 WVALKER ST. (SUCCES SORS OF MAEXB H. NEWMAN a& C.) CHICAGO: S. C. GRIGGS & CO., 39 & 41 LAKE ST. BUFFALO: PHINNEY & CO. 1* DAY AND THOMSON'S MATHEMATICAL SERIES. FOR SCHOOLS AND ACADEMIES. I. MENTAL ARITHMETIC; or, First Lessons in Nsenmbers;-for Beginners. This work commences with the simplest combinations of numbers, and graduqally advances to more difficult combinations, as the mind of the learner expands and is prepared to comprehend them. II. PRACTICAL ARITHMETIC; -Uniting the Induective with the Synthetic mode of Instruction; also illustrating the principles of CANCELATION'. The design of this Work is to make the pupil thoroughly acquainted with the reason of every operation which he is required to perform. It abounds in examples, and is eminently practical. III. KEY TO PRACTICAL ARITHMETIC;-Containing the answers, with numerous suggestions, &c. IV. HIGHER ARITHMETIC; or, the Science and Application of Nnumbers; —For advanced Classes. This work is complete in itself, commencing with the fundamental rules, and extending to the highest department of the science. V. KEY TO HIGHER ARITHMETIC;-Containing all the answers, with many suggestions, and the solution of the more difficult questions. VI. THOMSON'S DAY'S ALGEBRA;-This work is designed to be a Lucid and easy transition from the study of Arithmetic to the higher branches of Mathematics. The number of examples is much increased; and the work is every way adapted to the improved methods of instruction in Schools and Academies. VII. KEY TO THOMSON'S DAY'S ALGEBRA;-Containing the answere, the solution of the more difficult problems, &c. VIII. THOMSON'S LEGENDRE'S GEOMETRY;-With practical notes and illustrations. This work has received the approbation of many of the most eminent Teachers and Practical Educators. IX. PLANE TRIGONOMETRY, AND THE MENSURATION OF' HEIGHTS AND DISTANCES; wit/; a smmzary view of Ote ANatu're and Use of Logarithms;-Adapted to the method of instruction in Schools and Academies X. ELEMENTS OF SURVEYING;-Adapted both to the wants of the learner and the practical Surveyor. (Published soon.) Entered according to Act of Congress, in the year 1847, BY JAMES B. THOMRSON, in the Clerk's Office of the Northern District of New York. PR E F A C E. Tine HIigher Arithmetic which is now presented to the public, is tho third and last of a series of Arithmetics adapted to the wants of different classes of pupils in Schools and Academies. The title of each explains the character of the work. The series is constructed upon the principle, that "there is a place for everything, and everything should be in its proper place." Each work forms an entire treatise in itself; the examples in each are all different from those in the others, so that pupils who study the series, will not be obliged to purchase the same matter twice, nor to solve the same problems over again. The Mental Arithmetic, is designed for children from six to eight years of age. It is divided into progressive lessons of convenient length, beginning with the simplest combinations of numbers, and advancing by gradc ual steps, to more difficult operations, as the mind of the learner expands and is prepared to comprehend them. The Practical Arithmetic embraces all the subjects requisite for a thorough business education. The principles and rules are carefully analyzed and demonstrated; the examples for practice are numerous, and the observations and notes contain much information pertaining to business matters, not found in other works of the kind. This is the FIRST SCHOOL BOOKs in which the Standard Units of Weights and Measures adopted by the Government in 1834, were published. The ITigher Arithmetic is designed to give a full development of the philosophy of Arithmetic, and its various applications to commercial purposes. Its plan is the following: 1. The work is complete in itself. It commences with notation, and illustrating the different properties of numbers, the principles of Cancelation, and various other methods of contraction, extends to the higher operations in mercantile affairs, and the more abstruse departments of the science. 2. Great pains have been taken to render the definitions and rules clear, concise, exact, comprehensive. 3. It has been a cardinal point never to anticipate a principle; and never to use one principle in the explanation of another, until it has itself been explained or demonstrated. 4. Nothing is taken for granted whichl requires proof. Every principle therefore has been reinvestigated, and carej.:'iljy iai~nyzed. Vi PREFACE. 5. The principles are airanged consecutively, and the dependence of each on those that precede it, is pointed out by references. Treated in this manner, the science of Arithmetic presents a series of principles and propositions alike harmonious and logical; and the study of it cannot fail to exert the happiest influence in developing and strengthening the reasoning powers of the learner. 6. The rules are demonstrated with care, and the reasons of every oper ation fully illustrated. 7. The examples are copious and diversified; calling every principle into exercise, and making its application thoroughly understood. 8. In the arrangement of subjects, the natural order of the science has been carefully followed. Common Fractions have therefore been placed immediately after Division, for two reasons. First, they arise from division, and a connexion so intimate should not be severed without cause. Second, in Reduction and the Compound Rules, it is often necessary to multiply and divide by fractions, to add and subtract them, also to carry for them, unless perchance the examples are constructed for the occasion and with special reference to avoiding these difficulties. For the same reason Federal Money, which is based upon the decimal notation, is placed after Decimal Fractions; Interest, Commission, &c., after Percentage. To require a pupil to understand a rule before he is acquainted with the principles upon which it is based, is compelling him to raise a superstruclure, before he is permitted to lay a foundation. 9. In preparing the Tables of Weights and Measures, no effort has been spared to ascertain those in present use in our country; and rejecting such as are obsolete, we havo introduced the Standard Units adopted by the Government, together with the methods of determining and applying those standards. 10. Great labor has also been expended in preparing full and accurate Tables of Foreign Weights and Measures, and Moneys of Account, and in comparing them with those of the United States. Such is a brief outline of the present work. In a word, it is designed to be an auxiliary to the teacher, a lucid and comprehensive text-book for the pupil, and an acceptable acquisition to the counting-room. It contains many illustrations and principles not found in other works before the public, and much is believed to be gained in the method of reasoning and analysis. No labor has been spared to render it worthy of the marked favor with which the former productions of the author have been received. J. B. THOMISON. New Yh'rk, August, 1847. SUG GESTIONS ON THE IODE OF TEACHING ARITHMIETIC. I. QUALIFICATIONs.-The chief qualifications requisite in teaching Arithmetic, as well as other branches, are the following: 1. A tl'orougsh knowledge of the subject. 2. A love for the employment. 3. An aptitude to teach. These are indispensable to success. II. CLAssIFICATION.-Arbitlmetic, like reading, grammar, &c., should be taught in classes. 1. This method saves much time, and thus enables the teacher to devote more attention to oral illustrations. 2. The action of mind upon mind, is a poweiful stimnulant to exertion, and cannot fail to create a zest for the study. 3. The mode of analyzing and reasoning of one scholar, will often sugg'est new ideas to others in the class. 4. In the classification, those should be put together who possess as nearly equal capacities and attainments as possible. If any of the class learn quicker than others, they should be allowed to take up an extra study, or be furnished with additional examples to solve, so that the whole class may advance together. 5. The number in a class, if practicable, should not be less than six, nor over twelve or fifteen. If the number is less, the recitation is apt to be deficient in animation; if greater, the turn to recite does not come round sufficiently often to keep up the interest. III. APPARATUs.-The Black-board and Nuimerical Framue are as indispensable to the teacher, as tables and cutlery are to the house-keeper. Not a recitation passes without use for the black-board. If a principle is to be demonstrated or an operation explained, it should be done upon the black-board, so that all may see and understand it at once. To illustrate the increase of numbers, the process of adding, subtracting, multiplying, dividing, &c., to young scholars, the NuTmerical Frame furnishes one of the most simple and convenient methods ever invented. Every one who ciphers will of course have a slate. Indeed, it is desirable that every scholar in school, even to the very youngest, should be furnished with a slate, so that when their lessons are learned each one may busy himself in writing and drawing various familiar objects. 1(dc7/less in school is the parent of 7lischi; f, and esmploymlenst is the best antidote against clisbcdievzce. Geometr'ical diacgra'rms and solids are also highly useful in illustrating many points in arithmetic, and no school should be without them. IV. RECITATIONs.-The first object in a recitation, is to secure the attenstion of the class. This is done chiefly by throwing liJc and var7iety into the exercise. Children loathe dullness, while animation an( variety are i7seii delight. vriii SUGGESTIONS. 2. Every example should be analyzed; tlie "why and the wherefore" of every step in the solution should be required, till each member of the class becomes perfectly familiar with the process of reasoning and analysis. 3. To ascertain whether each pupil has the right answer, it is an excellent method to name a question, then call upon some one to give the answer, and before deciding whether it is right or wrong, ask how many in the class agree with it. The answer they give by raising their hand, will show at once how many are right. The explanation of the process may now be made. V. OBJECTS OF THE STUDY.-When properly studied, two important ends are attained. 1st. Discipline of mind, and the development of the reasoning powers. 2d. Facility and accmenracy in the application of numbers to business calculations. VI. THOROUGIINESs.-The motto of every teacher should be thoroqg/leness. Without it, the great ends of the study of Arithmetic are defeated. 1. In securing this object, much advantage is derived from freqnuent r'eviews. 2. Every operation should be proved. The intellectual discipline and habits of accuracy thus secured, will richly reward the student for his time and toil. 3. Not a recitation should pass without practical exer'cises upon the blackboard or slates, besides the lesson assigned. 4. After the class have solved the examples under a rule, each one should be required to give an acc?'ate account of its principles with the r'easaon for each step, either in his own language or that of the author. 5. 1lelIcal EKxercises in arithmetic are ex;ceedinlg/ly useful in making ready and accurate arithmeticians; hence, the practice of connecting meental with writtevn exerCcises, throughout the whole course, is strongly recommended. VIT. SELF-nELIANCE.-The hrabit of selif-'eliatmce in study, is confessedly invaluable. Its power is proverbial; I had almost said, olniypotc'at,. "Where there is a will, there is a wrnay." 1. To acquire this habit, the pupil, like a child learning to walk, must be taught to depend upos? himnself. Hence, 2. When assistance is required, it should be given indirqectly; not by taking the slate and solving the example for him, but by explaining the mneanctisg of the question, or illustrating the principle on which the operation depends, by supposing a more familiar case. Thus the pupil will be able to solve the question himself, and his eye will sparkle with the consciousness of victory. 3. The pupil should be fencouraged to study out different solutions, and to adopt the most conscise and elegant. 4. Finally, he should learn to perform examples intdependent of the answer. Without this attainment the pupil receives but little or no discipline from the study, and acquires no confideszce in his own abilities. What though he comes to the recitation with an occasional wrong answer; it were better to solve one question uqn;derstlcandisgly and alone, than to copy a score of answers from the book. What would the study of mental arithmetic be worth, if the pupil had the answers before him? What is a young man good for in the co;un~tin'7g-room, who cannot perform arithmetical operations without looking to the ansi, ei? Every one pronounces him stqsfil to be trusted with bil, lLes:' calculations. C 0 N TE N T S. INTII *D JCTr ON, — brief survey of the science of Mlathematics, m 13 SECTION I. NOTJ rION,. 20 Nunmlration, - 24 Formation of different systems of Notation, - 29 SECTION II. GENERAL RULE for Addition, - - 33 Different methods of proving Addition,. 34 Counting-room Exercises, - 31 Adding numbers expressed by the Roman Notation, - - 40 SECTION III. GENERAL RULE for Subtraction, L 44 Different methods of proving Subtraction,. 44 Subtracting numbers expressed by the Roman Notation, 47 SECTION IV. GENERAL RULE for MIultiplication, - 53 Different methods of proving Multiplication, - 54 Contractions in Miultiplication, six methods,. 56 SECTION V. GENERaAL RULE for Division, - - 71 Different methods of Proving Division,.. 72 Contractions in Division, nine methods, - 74 General principles in Division, -. 81 CANCELATION, - - - 83 Applications of the fundamental rules, - 85 SECTION VI. PROPERTIES OF NUMBERS,.-.. 89 Method of finding the prime numbers in any series, - 93 Table of prime numbers froml 1 to 3413. 94 Numbers changed from the decimal to other seales of Notation, - 95 Analysis of Composite Numbers, - -. 97 Greatest Common Divisor, two methods, - 100 Least Common Multiple, three methods, 102 SECTION VII. FRACTIONS, - 107 General principles pertaining to Fractions, - - 109 Reduction of Fractions, - - - 111 Addition of Fractions, - H 119 Subtraction of Fractions, 122 Mlultiplication of Fractions, common method, - - - 124 Ml[ultiplication of Fractions by Cancelation, - 130 onltractions in Multiplication of Fractions, four methods, 131-132 I* X CONTENTS. PAUG Division of Fractions, common method,. 133 Division\of Fractions by Cancelation, 136 Contractions in Division of Fractions, three methods, - 138 Complex Fractions reduced to Simple ones, - - - 139.SECTION VIII. COMPOUND NUMBERS, -.. - 144 Tables of Compound Numbers, 144 The Standard for Gold and Silver Coin of the United States,. 145 The Standard UNIT of Weight of the United States, - 147 Origin of Weights and Measures,. 147 The Avoirdupois Pound of the United States and Great Britain, 148 The Standard UNIT of Length of the United States, - - - 150 99" " " of New York and Great Britain, - 151 The Standard UNIT of Liquid Measure of the United States, - 154 " " of Dry Measure of the United States, - 155 Method of determining whether a given year is Leap Year, - - 157 French Money, Weights and Measures, - - - 161 Foreign Weights and Measures, compared with United States, - 163 GENERAL RULE for Reduction, 166 Applications of Reduction, - 169 Cubic Measure reduced to Dry, or Liquid Measure, &c., - 172 Longitude reduced to Time, &c., - 175 Compound Numbers reduced to Fractions, - - 176 Fractional Compound Numbers reduced to whole numbers, - 179 Compound Addition, 181 Compound Subtraction, - - - -.183 Compound Multiplication, -. 186 Compound Division, - - - - - 189 SECTION IX. DECIMAL FRACTIONS, their origin, &c., - - 191 Method of reading Decimals, - - - - - 193 Addition of Decimals, - - - - - 195 Subtraction of Decimals, - - - - - 197 Multiplication of Decimals, 199 Contractions in Multiplication of Decimals, two cases, - - 201 Division of Decimals, - - v. 205 Contractions in Division of Decimals, - - 208 Decimals reduced to Common Fractions, - 210 Gommon Fractions reduced to Decimals, - -. 11 Compound Numbers reduced to Decimals, - - - 15 SECTION X. ADDITION of Circulating Decimals,. 222 Subtraction of Circulating Decimals, -.. 223 Multiolication of Circulating Decimals,. 224 Division of Ci.rculating Decimals,.. 25 CONTENTS. Xi SECTION XI. FEDERAL MONEY, - - - - 226 Addition of Federal Money, - 229 Subtraction of Fedoral Money, - ~ - - 231 Multiplication of Federal Money,.. 232 Division of Federal Money, - - - 234 Counting-room Exercises, Contractions, &c. - - - - 237 To find the cost of articles bought and sold by the 100 or 1000, - 238 Bills, Accounts, &c., - - - - - 239 SECTION XII. PERCENTAGE, Percentage Table, &c., - - - 241 Applications of Percentage, - - 244 Commission, Brokerage, and Stocks, - - - - 244 Commission deducted in advance, and the balance invested, - 247 INTEREST, - 2. - - 249 General method for computing Interest, - - - - 252 Second method, multiplying the Prin. by the Int. of $1 for the time, 255.To compute Interest by half the number of months, - - 257 To compute Interest by the number of days, - - - 258 Applications of Interest, remarks on Promissory Notes, &c., - - 258 Forms of Negotiable Notes, &c., 260 Partial Payments; United States Rule, - - - 261 Connecticut Rule, 263 Vermont Rule, - - 264 Problems in Interest, - - - 265 Compound Interest, - - - - - - 270 Discount, - - 272 Bank Discount, - - - 274 To find what sum must be discounted to produce a given amount, 277 Insurance, four Cases, - - - - - 278 The sum to be insured to recover the premium and the property, 282 Life Insurance, - - - - - - - 282 Profit and Loss, four Cases, - 283 Duties, Specific and Advalorem, -. 289 Assessment of Taxes, - - - - - 292 To find what sum must be assessed to raise a given net amount, - 295 Formation of Tax Bills, - - - - 295 Rate Bills for Schools, - - - - - 297 SECTION XIII. ANALYSIS, - 298 Analytic so.utions of questions in Simple Proportion, -- 299 " " " Barter, - - - 301 ct " " " Partnership, - - - 30'2 "s " " General Average, - - - 303 S, 6 "99 Alligation, - - - 304 X1 CO N'0NT,NTS. PAGR Analytic solutions af questions in Compound Proportion, - 307 " i Ad "d Position, - - 308'" " "e Practice, - - 309 SECTION XIV. RATIO, general principles pertaining to it, - - - 313 Simple Proportion, - - 321 Simple Proportion and its Proof by Cancelation, - - - 325 Compound Proportion, - - - - 328 Compound Proportion and its Proof by Cancelation, - - 330 Conjoined Proportion, - - - 332 SECTION XV. DUODECIMALS, its principles, &c., - 334 Multiplication of Duodecimals, - - - 334 SECTION XVI. EQUATION OF PAYMENTS, - - - 338 Partnership, - -.. 340 General Average, - -. 343 Exchange of Currencies, - - 345 Foreign Coins and Moneys of Account, -. - 348 Exchange, Form of Bills of Exchange, &c., -. 351 Arbitration of' Exchange, - - - - - 355 Alligation, Medial, and Alternate, - - - 356 SECTION XVII. INVOLUTION, and Evolution, - 360 Properties of Squares and Cubes, - - - 3(;5 Extraction of the Square Root, - - - 36;7 Demonstration of the Square Root, -. 368 Applications of the Square Root, - - - - 370 Extraction of the Cube Root, Horner's Method, - - 374 Demonstration of the Cube Root, - - - 376 Extraction of Higher Roots, - 378 SECTION XVIII. Arithmetical Progression,. 381 Geometrical Progression, -. 384 Annuities, - - 386 Permutations and Combinations, - -. 388 SECTION XIX, Application of Arithmetic to Geometry, -. / 389 Mensuration of Surfaces and Solids,. 389 Measurement of timber, - - 392 Gauging of Casks, -. 393 Tonnage of Vessels, -... 393 Mechanical Powers, -.. 394 Miscellaneous Examples, - - - 395 INTRODUCTION. ART, 1. Anything which can be multiplied, divided, or measuled, is called QUANTITY. Thus, lines, weight, time, number, &c., are quantities. Oss. 1. A line is a quantity, because it can be measured in feet and inches; weight can be measured in pounds and ounces; time, in hours and minutes; numbers can be multiplied, divided, &c. 2. Color, and the operations of the mind, as love, hatred, desire, choice, &c., cannot be multiplied, divided, or measured, and therefore cannot properly be called quantities. 2. MATHEMATICS is the science of Quantity. 3. The fundamental branches of Mathematics are, Arithmetic, Algebra, and Geometry. 4. Arithmetic is the science of Numbers. 5. Algebra is a general method of solving problems, and of investigating the relations of quantities by meal s cf letters and signs. OBs. FlPx.ions, or the Differential and Inltegal Ccicd'ls, may be considered as belonging to the higher branches of Algebra. 6. Geometry is that branch of Mathematics which treats of Magnitude. 7. The term magnitude signifies that which is extended, or which has one or more of the three dimensions, length, breadth, and Ihickness, Thus, lines, surfaces, and solids are magnitudes. QUEST.-1. What is Quantity? Give some examples of quantity. 01,. Why is a line a quantity? Weight? Time? Numbers? Are color and the operations of the mind qluantities? Why not?'2. What is Mathematics? 3. What are the fundamlental branches of mathematics? 4. What is Arithmetic? 5. Algebra? 6. Geometry? 7. What is meant by magnitudle? 14 INTRODUCTION. 03S. 1. A line is a magnitude, because it can le extended in length; a su/face, because it has length and breadth; a solid, because it has length, breadth, and thickness. 2. 3lotion, though a quantity, is not, strictly speaking, a magnitude; for it has neither length, breadth, nor thickness. 3. The term mag-nitiude is sometimes, though inaccurately, used as syncanymous with quantity. S. Triyonometry and Conic Sections are branches of Mathemates, in which the principles of Geometry are applied to triangles, lnd the sections of a cone. 9. Mathematics are eitherpure or mixed. In pure mathematics, quantities are considered, independently of any substances actually existing. In mixed mathematics, the relations of quantities are investigated in connection with some of the properties of matter, or with reference to the common transactions of business. Thus, in Surveying, mathematical principles are applied to the measuring of land; in Optics, to the properties of light; and in Astronomy, to the heavenly bodies. OBs. The science of pure matemmatics has long been distinguished for the clearness and distinctness of its principles, and the irresistible conviction which they carry to the mind of every one who is once made acquainted with them. This is to be ascribed partly to the nature of the subjects, and partly to the exactness of the definitions, the ax.ioms, and the demonstrations. 1 0. A definition is an explanation of what is meant by a word, or phrase. OBS. It is essential to a complete definition, that it perfectly disting-uishcs the thing defined, from everything else. I 1. A proposition is something proposed to be proved, or required to be done, and is either a Theorem, or a Problem. 1 2. A theorem is something to be provedl. 1 3. A problem is something to be done, as a question to be solved. QUEST.-Obs. Why is a line a magnitude? A surface? A solid? Is motion a magnitude? Why not? t. Of how many kinds are mathematics? In pure mathematics how are quantities considered? How in mixed mathematics? Obs. For what is the science of pure tmathematics distinguished? 10. WVhat is a definition? Obs. What is essential to a complete (efinitiln? 1i. WVhat is a proposition? 12. k theorem? 13. A problem INTR ODUCTIO 10 N. 15 OBs. 1. In t.ie statement of every proposition, whether theorem or problem, certain things must be given, or assumed to be true. These things are called the data of the proposition. 2. The operation by which the answer of a problem is found, is called a solutioln. 3. When ths given problem is so easy, as to be obvious to every one without explanation, a, is called a postllate. 1 4. One proposition is contrary, or contradictory to another, when what is aylrmed in the one, is denied in the other. OBs. A proposition and its contrary, can never bothl be true. It cannot be true, that two given lines are equal, and that they are not equal, at the same tine. 1 5. One proposition is the converse of another, when the ordel is inverted; so that, what is given or supposed in the first, becomes the conclusion in the last; and what is given in the last, is the conclusion, in the first. Thus, it call be proved, first, that if the sides of a triangle are equal, the angles are equal; and secondly, that if the angles are equal, the sides are equal. Here, in the first proposition, the equality of the sides is given, and the equality of the angles inferred; in the second, the equality of the angles is given, and the equality of the sides inferred. Oss. In many instances, a proposition and its converse are both true, as In the preceding example. But this is not always the case. A circle is a figure bounded by a curve; but a figure bounded by a curve is not necessarily a circle. 1 6. The process of reasoning by which a proposition is shown to be true, is called a demonstration. OBs. A demonstration is either dirdct or tndirect. A direct demonstration commences with certain principles or data which are admitted, or have been proved to be true; and from these, a series of other truths are deduced, each depending on the preceding, till we arrive at the truth which was required to be established. An indir'ect demonstration is the mode of establishing the truth of a proposition by proving that the supposition of its cont'ary, involves an absurdity. QUEST.-Obs. What is nleant by the data of a proposition. By the solntion of problem? What is a postulate? 14. When is one proposition contrary to another Obs. Can a proposition and its contrary both be true? 15. When is one proposition the converse of another Obs. Can a proposition and its converse both be true? 16. What is a demonstration? Obs. Of how many kinds are demonstrations? What is a direct demonstration? An indirect demonstration? 1.6 INTRODUCTION. This is commonly called r'eductio ad abhsur'duv. The former is the more common method of conducting a demonstrative argulment, and is the m( st satisfac tory to the mind. 1 7. A Lemma is a subsidiary truth or proposition, demonstrated for the purpose of using it in the demonstration of a theorem, or the solution of a problem. 18. A Corollary is an inference or principle deduced from a preceding proposition. 19. A Scholium is a remark made upon a preceding pIroposition, pointing out its connection, use, restriction, or extensions. 20. An Hypothesis is a supp2osition, made either in the statement of a proposition, or in the course of a demonstration. AXIOMS. 2 1. An;xiom? is a self-evident proposition; that is, a proposition whose truth is so evident at sight, that no process of reasoning can make it plainer. The following axioms are among the most common: 1. Quantities whllich are equal to the same quantity, are equal to each other. 2. If the same or equal quantities are added to equals, the sums will be equal. 3. If the same or equal quantities are subtracted from equals. the remainders will be equal. 4. If the same or equal quantities are addled to unequals, the sums will be unequal. 5. If the same or equal quantities are subtracted from unequals, the remainders will be unequal. 6. If equal quantities are multi2plied by the same or equal quantities, the products will be equal. 7. If equal quantities are divided by the same or equal quantities, the quotients will be equal. 8. If the same quantity is both added to and subtracted front another, the value of the latter will not be altered. QUEST -17. What is a leIlma? 18. What is a corollary? A9. Whaw is a schollum 20. What is an hypothesis? 21. What is an axiom? Name some of the most common o0ms, INTRODUCTION. 17 9. If a quantity is both vmultiplied and divided by the same or an equal quantity, its value will not be altered. 10. The whole of a quantity is greater than a part. 11. The whole or s quantity is equal to the sum of all its parts. SIGNS. 22. Addition is represented by {he sign (+), which is called plus. It consists of two lines, one horizontal, the other perpen. licular, forming a cross, and shows that the numbers between which it is placed, are to be added together. Thus, the expression 6+8, signifies that 6 is to be added to 8. It is read, " 6 plus 8," or " 6 added to 8." OBs.-The term plus is a Latin' word, originally signifying " more," hence " added to." 23. Subtraction is represented by a short horizontal line (-), which is called minus. When placed between two numbers, it shows that the number after it is to be subtracted from the one before it. Thus, the expression 9-4, signifies that 4 is to be subtracted from 9; and is read, " 9 minus 4," or " 9 less 4." OBs.-The term mi~nus is a Latin word, signifying less. 24. lJultiplication is usually denoted by two oblique lines crossing each other (X), called the sign of multiplication. It shows that the numbers between which it is placed, are to be multiplied together. Thus, the expression (9X6), signifies that 9 and 6 are to be multiplied together, and is read, " 9 multiplied by 6," or simply, "9 into 6." Sometimes multiplication is denoted by a point (.) placed between the two numbers or quantities. Thus, 9.6 denotes the same as 9X6. OBS. It is better to denote the multiplication of figures by a cross than by a point; for the latter is liable to be confounded with the decimal point. 24. a. When two or more numbers are to be subjected to the same operation, they must be connected by a line ( ) placed QUEST.-22. What is the sign of addition called 1 Of what does it consist? What does it show? Obs. What is the meaning of the termn plus?. 23. How is subtraction represented? What is the sign of subtraction called? What does it show. Obs. What does the tern minus signify? 24. How is multiplication usually denoted? What does the sign of mul. tiplication show? In what other way is multiplication somretimles denoted? 18 INTRODUCTION. over them, called a vinculum, or by a parenthesis ( ). Thus the expression (12-3)x2, shows that the sum of 12 and 3, is to be multiplied by 2, and is equal to 30. But 12+3X2, signifies that 3 only is to be multiplied by 2, and that the product is to be added to 12, which will make 18. 25. Division is expressed in two ways: First, by a horizontal line between two dots (-), called the sign of division, which shows that the number before it, is to be divided by the number after it. Thus, the expression 24 6 signifies that 24 is to be divided by 6. Second, division is often expressed by placing the divisor under the dividend, in the form of a fraction. Thus, the expression I, shows that 35 is- to be divided by 7, and is equivalent to 35- 7. 26. The equality between two numbers or quantities, is represented by two parallel lines (=-), called the sign of equality. Thus, the expression 5+3=8, denotes that 5 added to 3 are equal to 8. It is read, "5 plus 3 equal 8," or "the sum of 5 plus 3 is equal to 8." So 7+-5=16-4-12. QUEST. —24. a. When two or more numbers are to be subjected to the same operatio:,g what must be done. 25. In how many ways is division expressed t What is the first What does this sign show. What is the second X 26. How is the equality between two numbers or quantities represented X AR IT H ME E T I C. SECTION I. NOTATION AND NUMERATION. ART. 27. Any single thing, as a peach, a rose, a -book, is called a unit, or one; if another single thing is put with it, the collection is called two; if another still, it is called three; if another, four; if another, five, &c. The terms, one, two, three, &c., by which we express how many single things or units are under consideration, are the names of numbers. Hence, 28. NUMBER signifies a unit, or a collection of units. Ons. 1. Numbers are divided into two classes, abstract and concrete. When they are applied to particular objects, as two pears, five pounds, ten dollars, &c., they are called concrete numbers. When they do not refer to any particular object, as when we say four and five are nine, they are called abstract numbers. 2. Whole numbeis are often called integers. 3. Numbers have various properties and relations, and are applied to vanous computations in the practical concerns of life. These properties and applications are formed into a system, called Arithmetic. 29. ARITHIMETIC is the science of numbers. OBs. 1. The term Arithmetic is derived from the Greek word arithmge,~tr, which signifies the art of reckoning by numbers. 2. The aid of Arithmetic is required to make and apply calculations not only in businzess trannsactions, but in almost every department of mathematics. Q.UEST.-27. What is a single thing called? If another is put with it, what is the collection called? If another, what What are the terms one, two, three, &c.7 28. What does number signify? Obs. into how many classes are numbers divided? When are they called concrete? When abstract? To what are numters applied? 29. What Is Arithmetic 3 Obs. In what is the aid of arithmetic requirel i 2 20 NOTATION. [SECT. I Numbers are expressed by words, by letters, and by.fiqures. NOTATION. 30. The art of expressing numbers by letters or figyures, is called NOTATION. There are two methods of notation in use, the.Roman and the Arabic. 3 1. The Roman method employs seven capital letters, viz: I, V, X, L, C, D, M. When standing alone, the letter I, denotes one; V, five; X, ten; L, fifty; C, one hundred; D, five hundred; M, one thousand. To express the intervening numbers from one to a thousand, or any number larger than a thousand, we resort to repetitions and various combinations of these letters. The method of doing this will be easily learned from the following TABLE. - denotes one. XXX denotethirty. II " two. XL " forty. III " three.' L " fifty. IV " four. LX " sixty. V " five. LXX " seventy. VI " six. LXXX " eighty. VII " seven. XC " ninety. VIII " eight. C " one hundred. IX' nine. CI " one hundred and one. X " ten. CX " one hundred and ten. XI " eleven. CC " two hundred. XII " twelve. CCC " three hundred XIII " thirteen. CCCC " four hundred. XIV " fourteen. D " five hundred. XV "' fifteen. DC " six hundred. XVI " sixteen. DCC " seven hundred. XVII' seventeen. DCCC " eight hundred. XVIII " eighteen. DCCCC " nine hundred. XIX " nineteen. M " one thousand. XX " twenty. MM " two thousand. XXI " twenty-one. MDCCCLV, one thousand eight XXII " twenty-two,' &c. hundred and, fifty-five. QuRsT. —How are numbers usually expressed? 30. What is notation? hlow masny methods are there in use 7 31. What is employed by the Ronian method? AR'TS. 30-33.] NOT'rATION. 21 Oss. 1. This method of explessing numbers was invented by the Romlrans, and is therefbore called the Romn Notlation. It is now seldoln used; except to denote chapters, sections, and other divisions of books and tiiscourses. 2. The letters C and M, are the initials of the Latin words cenltuar, and mille, the former of which signifies a hamdrled, and the latter a thovsaszLnd: fbr this reason it is supposed they were adopted to represent these numbers. 31. a. It will be perceived from the Table above, that every time a letter is repeated, its value is repeated. Thus I, standing alone, denotes one; II, two ones, or two, &c. So X denotes ten; XX, twenty, &c. When a letter of a less value is placed before a letter of a greater value, the less takees away its own value from the greater; but when placed after, it adds its own value to the greater. 320 A line or bar (-) placed over a letter, increases its value a thousand times. Thus, V denotes five, V denotes five thousand; X, tenl; X, ten thousand, &c. OBs. 1. In the early periods of this notation, four was written III, instead of IV; nine was written VIIII, instead of IX; forty was written XXXX, instead of XL, &c. The former method is more convenient in perfobrming arithmetical operations In addition and subtraction; while the latter is shorter and better adaptedl to ordinary purposes. 2. A lthouscand was originally written CID, which, in later times, was changed into M; five hlLnd.red was written ID instead of D. Annexing 0 to ID increased its value ten times. Thus, ID0 denoted five tIoussand; 1D0), fi/l/' thousand, &c. 3. Prefixing C and annexing 0 to the expression CIO, makes its value ten times greater: thus, CCIO denotes ten thoIstsa.d; CCCIOD0, a hluntl're thomusancd. According to Pliny, the Romans carried this mode of notation no further. When they had occasion to express a larger number, they did it by repetition. Thus, CCCIODD, CCCID0D, expressed two h/,iqdr'ed thlousand, &c. 33. The common method of expressing numbers is by the A2rabic.Notation. The Arabic method employs the following ten claracters or figures, viz: 1 2 3 4 5 6 7 8 9 0 one, two, three, four, five, six, seven, eight, nine, zero. QuRsT.-Obs. tWThy is this method called Roman? 31. a. What is the effect of repeatitng a letter? If a letter of less value is placed biefore another of greater value, what is the effect? If placed after, what? 32. When a line or bar is placed over a letter, how does it affect its value? 33. Wha.t is the commnon way of expressing numbers? IIow maay characters does this method elnploy? 22 NOTATION. [SECT. I The first nine are called significant figures, because each one always has a value, or denotes some number. They are also called, digits, from the Latin word digittus, which signifies a finger. The last one is called a cipher, or naught, because when standing alone it has no valiue, or signifies nothing. OBS. 1. It must not be inferred, however, that the cipher is useless; for when placed oil the right of any of the significant figures, it increases their value. It may therefore be regarded as an acxcilicary digit, whose office, it will be seen hereafter, is as important as that of any other figure in the system. 2. Formerly all the Arabic characters were indiscriminately called ciphers; hence the process of calculating by them was called ciphering; on the samle principle that calculating by figures is called fig-uring. 34. It will be seen that nine is the greatest number that can be expressed by any single figure in the Arabic system of Notation. All numbers larger than nine are expressed by' combining together two or more of these ten figures, and assigning different values to them, according as they occupy different places. For example, ten is expressed by combining the 1 and 0, thus 10; eleven by two Is, thus 11; twelve by 1 and 2, thus 12; twenty, thus 20; thirty, thus 30; &c. A hundred is expressed by combining the I and two Os, thus 100; two hundred, thus 200; a thousand by combining the 1 and three Os, thus 1000; two thousand, thus 2000; ten thousand, thus 10,000; a hundred thousand, thus 100,000; a million, thus 1,000,000; ten millions, thus 10,000,000; &c. Hence, 35. The digits 1, 2, 3, &c., standing alone, or in the right hand place, respectively denote units or ones, and are called units of the first order. When they stand in the second place, they express tens, or ten ones; that is, their value is ten times as much as when standing QuEST.-What are the first nine called? ~Why? What else are they called? What Is the last one called? Why? Obs. Is the cipher useless? What may it be regardedl 1 What is the origin of the term ciphering? 34. Wltat is the greatest number that can be expressed by one figurle? How are larger numbers expressed?. 35. What do the digits, 1, 2, 3, &c., denote, when standi tg alone, or in the right hand place? What are they then called?. What do they delate when standing in the second pla(e I ARTS. 34-37.] NOTATION 23 in the first or right hand place, and they are called units of the second order. When occupying the third place, they express hundreds; that is, their value is ten times as much as when standing in the second place, and they are called units of the third order. When occupying the fourth place, they express thousands; that is, their value is ten times as much as when standing in the third place, and they are called units of the fourth order, &c. Thus, it will be seen that, Ten units make one ten, ten tens make one hundred, and ten hun. dreds make one thousand; that is, ten in an inferior order are equal to one in the next suzperior order. Hence, universally, 36..VzNumbers increase from right to left in a tenfold ratio; consequently each removal of a figure one 2lace towards the left, increases its value ten times. Nole.-l. The number which forms the basis, or which expresses the ratio of increase in a system of Notation, is called the RADIX of that system. Thus, the radix of the Arabic notation is ten. 2. The reason that numbers increase from righ/t to lcft, instead of left to right, is probably owing to the ancient practice of writing from the right hand;o the left. 37. The different values which the same figures have, are called simple and local values. The simple value of a figure is the value which it expresses when it stands alone, or in the right hand place. Hence the simple value of a figure is the number which its name denotes. The local value of a figure is the increased value which it expresses by having other figures 1laced on its right. Hence the local value of a figure depends on its locality, or the place which QuesT. —WVhat is their.value then What are they called? What is a figure called when it occupies the third place? What is its value then? What is it called when in the fourth place? What is its value? Iow many units are required to nlake one ten I How miany tens makle a hundred? How many hundreds. malie a thousand? Hlow many of an inferior order are required to make one of the next superior order? 36. What is the general law by which numbers increase? What is the effect upon the value of a figure to remove it one place towards the left?.sote. WVhat is the nulmber called wthich formns the basis xr the ratio of increase in a system of notation? What is the radix of the Arabic notation? Why do numbers increase from right to left? 37. What are the different values of She same figure called? What is the simple value of a figure? W hat tb e local I 24 NUMERATION. [SECT. I. it occupies in relation to other numbers with - hich it is connected. (Art. 35.) OBs. 1. This system of notation is called Arabic, because it is supposed to Lave been invented by the Arabs. 2. It is also called the decimal system, because nulmbers increase ill a ten.r fold ratio. The term decimal is derived from the Latin word decenm, which signifies ten. 3. The early history of the Arabic notation is veiled in obscurity. It is the opinion of some whose judgmnent is entitled to respect, that it was invented by the philosophlers of India. It was introduced into Europe from Arabia abou the eit]lth century, and about the eleventh century it came into general use,'oth in Fngland andc on the continent. The application of the term di,'it to the sin ificant figures, afilrds strong presumptive evidence that the system had its orliin in the ancient mode of counting and recltoning by means of the fir'e rs;. end that the ideza of employing ten characters, instead of twell;e or apy Dotlle nuinber, was suggested by the number of fingers and thumbs on both hands. (Art. 33.) NUMERATION. 38. T/,e art of recading numbers when expressed by figures, i8 called NUMERATION. The pupil will easily learn to read the largest numbers from the following scheme, called the NUMERATION TABLE. a,, O X 0, of tre fures eac, beginnn at the right and. e fi-st 685, 876, 389, 764, 391, 827, 218, 649, 853, 123, 234, 579, 793, 465, 623. XV. XIV. XIII. XII. XI. X. IX. VII1. VIi. VI. V. IV. IiI. I. [. 3,). The different orders of numbers are divided into periods of tasree figures each, beginning at the right hand. The frst period, which is occupied by units, tens, hundreds, is called unitd QursT.-Upon what does the local value of a figure depend? Obs. Why is this systeim of notation called Arabic? What else is it somietines called?. Why? NWVhat do yoe say of its early history? When was it introduced into Europe? What is the probable origin of the systernm? Why were ten chara;cters. rather thani any otlier nuiibler, adopted' 38. What is Numneration? 39. i1ow are the orders of numbers divided? WhLit is the first period called? By what is it occupied? ARTS. 38, 39.] NUMERATION. 25 period; the second is occupied by thousands, tens of thousands, hundreds of thousands, and is called thousands' period; the third is occupied by millions, tens of millions, hundreds of millions, and is called millions' period; the fourth is occupied by billions, tens of billions, hundreds of billions, and is called billions' period; and so on, the orders of each successive period being units, tens, and hundreds. The figures in the table are read thus: 685 tredecillions, 876 duLodecillions, 389 undecillions, 764 decillions, 391 nonillions, 827 octillions, 218 septillions, 649 sextillions, 853 quintillians, 123 qwadrillions, 234 trillions, 579 billions, 793 mIillions, 465 thousand, 6 hundred and twenty-three. ANote.-i. The terms thirteen, fowr'teegn, fifteen, &c., are obviously derived from three and ten, four and ten, five and ten, which by contraction become thirteen, fourteen, fifteen, and are therefore significant of the numbers which they denote. The terms elevent and twelve, are generally regarded as primitive words; at all events, there is no perceptible analogy between them and the numbers which they represent. Had the terms onteteen and twoteen been adopted in their stead, the names would then have been significant of the numbers one and ten, two and ten; and their etymology would have been similar to that of the succeeding terms. The terms twenty, thirty, forty, &c., were formed from two tens, three tens, four tens, which were contracted into twenty, thirty, forty, &c. The terms ttwenty-one, twenty-two, twenty-three, &c., are compounded of twenty and one, twenty and two, &c. All the other numbers as far as ninetynine, are formed in a similar manner. 2. The terms handred, thousand and iillhon are primitive words, and bear no analogy to the numbers which they denote. The numbers between a hundred and a thousand are expressed by a repetition of the numbers below a hundred. Thus we say one hundred and one, one hundred and two, one hundred and three, &c. 3. The terms billions, trillions, quadrilliotn, &c., are formed from million and the Latin numerals bis, tr'es, quLatiuor, &c. Thus, prefixing bis to million, by a slight contraction for the sake of euphony, it becomes billionm; prefixing tres to millions, it is easily contracted into trillion, &c. The Latin word bis signifies two; tres, three; qcatLuor, four; quiTque, five; sex, six; septen, seven; octo, eight; tn.ovem, nine; decent, ten; 6ndedecil, eleven; dtLodecmz, twelve; tredlecem, thirteen. QUtsT.-W-ihat is the second period called? By what occl yiod? What is the third called? By what occupied? VWhat is the fourt'a called? By w'hat occupied? What:s the fifth called? By what occuptied? Repeat the Nunicrnaticn Table, beginning at the right hand,'4 26 NUMERATION. LSEC[. I. Higher periods than those in the Table, may be easily forraed by following the above analogy. 4. The foregoing law, which assigns superior values to these ten characters, according to the order or place which they occupy and the use of so many derivative and compound words in forming the names of numbers, saves an inconceivable amount of time, and labor in learning Notation and Numeration, as well as in their application. 40. To read numbers which are expressed by figures. Point thert off into periods of three figures each; then, beginning at the left hand, read the figures of each period in the same manner as those of the right hand period are read, and at the end of each period, pronounce its name. OBS. 1. The learner must be careful, in pointing coff figures, always to begin at the right hand; and in readinwg them, to begin at the left hand. 2. Since the figures in the first or right hand period always denote units, its name is not pronounced. Hence, in reading figures, when no period is mentioned, it is always understood to be the right hand, or units' period.] EXERCISES IN NUMERATION. Note.-In numerating large numbers, it is advisable for the pupil first to apply to each figure the name of the order which it occupies. Thus, beginning at the right hand, he should say, "Units, tens, hundreds," &c., and point at the same time to the figures standing in the order which he mentions. Read the following numbers: Ex. 1. 3506 11. 706305 21. 967058713 2. 6034 12. 1640030 22. 32100040 3. 5060 13. 830006 23. 106320000 4. 90621 14. 70900038 24. 780507031 5. 73040 15. 3067300 25. 4063107 6. 450302 16. 12604321 26. 29038450 7. 603260 17. 70003000 27. 1046347025 8. 130070 18. 161010602 28. 20380720000 9. 2021305 19. 80367830 29. 8503467039 10. 4506580 20. 400031256 30. 450670412463 QUICST.-40. How do you read numbers expressed by figures? Cbs. Where begin te, point them off? Where to read them? Do you pronounce the name of the right hand period? When no Deriod is named, what is understood? ARTS. 40, 41.] NUMERATION. 27 31. 430812000641 36. 120340078910356 32. 5200240301000 37. 43601000345000 33. 98760000216 38. 506302870045380 34.. 82600381000000 39. 42008120537062035 35. 403070003462000.40. 653107843604893048 41. 210 256 031 402 385 290 845 381 467. 42. 361 438 201 219 763 281 572 829 318 278. 41t The method of dividing numbers into periods of three figtres, was invented by the French, and is therefore called the French Nfumeration. The English divide numbers into periods of six figures, in the following manner: o o 0.-n c4i 2 3 s5 6 3 4 o 2 c3 " c a 0 c-c-S O c - s - 0 n-l S:c,5Md o F *r;~ 4 2 3 5 6 1, 2 3 4 8 2 6, 4 7 9 3 6 5 Period III. Period II. Period I. According to this method, the preceding figures are read thus. 423561 billions, 234826 millions, and 479365. OBS. 1. It will be perceived that the two methods agree as far as hundreds of millions; the former then begins a new period, while the latter continues on through thousands of millions, &c. 2. The French method is generally used throughout the continent of Europe, as well as in America, and has been recently adopted by some English authors. It is very genet illy admitted to be more simple and convenient than thle Eng. fish method. QVEST.-41. What is the French method of numeration? Wi at the English inethod I Obs. Which is the more simple and convenient? 2* 28 NOTATION. [SECT. I. EXERCISES IN NOTATION'. 42. To express numbers by figures. Begin at the left hand, and write in each- order the figure which denotes t/he giveen number in that order. If any intervening orders are omitted in the plroposed number, wtrite ciphers in their places. (Art. 38.) Write the following numbers in figures: 1. Two thousand, one hundred and nine. 2. Twenty thousand and fifty-seven. 3. Fifty-five thousand and three. 4. One hundred and five thousand, and ten. 5. Seven hundred and ten thousand, three hundred and one. 6. Two millions, sixty-three thousand, and eight. 7. Fourteen millions, and fifty-six. 8. Four hundred and forty millions, and seventy-two. 9. Six billions, six millions, six thousand, and six. 10. Forty-five billions, three hundred and forty thousand, and seventy-six. 11. Five hundred and. fifty-six millions, three thousand, two hundred and sixty-four. 12. Eight hundred and ten billions, ten millions, and seventyfive thousand. 13. Nirnety-six trillions, seven hundred billions, and fifty-four. 1. Three hundred and forty-nine quadrillions, five tiillions, seven billions, four millions, and twenty. 15. Nineteen quintillions. 16. Six hundred and thirty sextillions. 17. Two hundred and ninety-eight septillions. 18. Seventy-four octillions. 19. Four hundred and ten decillions. 20. Eight hundred and sixty-three duodecillions. 21. Nine hundred and thirty-five tredecillions. 22. Six hundred and seventy-three quintillions, seventeen quad%difions, and forty-five. 23. Twenty triilions, six hundred and forty-eight billions, and t-centy-five thotlsand. ARTS. 42-44.] NOTATION. 29 Bas. The great facility with which large numbers may be expressed both in language and by figures, is calculated to give an imperfect idea of their real mag-gitnde. It may assist the learner in forming a just conception of a million, a billio?: a tillion, &c., to reflect, that to count a million, at the rate of a hundred a minute, would require nearly seventcen days of ten hours each; to colnt a billion, at the same rate, would require more than forty-five years; and to couXnt a trillion, more than 45,662 years. 43. From the preceding illustrations, the learner will perceive that a variety of other systems of notation may be formed upon the same principle, having different numbers for their radices. Thus, if we wished to form a quinary system; that is, a system in which the numbers should increase in a five-fold ratio, or has five for its radix, it would require four significant figures and a cipher. Let the figures 1, 2, 3, 4, and 0, be the characters employed; then five would be expressed by 1 and 0, and would be written thus 10; six by 1 and 1, thus 11; seven by I and 2, thus 12; eight by 1 and 3, thus 13; nine by I and 4, thus 14; ten by 2 and 0, thus 20; eleven by 2 and 1, thus 21, &c. 44. In the binary or diadic system of notation developed by Leibnitz, there are two characters employed, 1 and 0. The cipher when placed at the right hand of a number, in this system, multiplies it by two. Thus the number one is expressed by 1; two by 10; three by 11; four by 100; five by 101; six by 110; seven by 111; eight by 1000; nine by 1001; ten bjy 1010; eleven by 1011, &c. OGs, 1. In like manner other systems of notation may be formed, having tilree, fiour, six, eight, twelve, or aney given number' for their radix. When the radix is two, the system is called binary or diadic; when three it is called ternary; when four, qulaterznary; when five, qinarwy; when six, svnoary; when seven, septenla7?; when eight, octary; when nine, qwnoary, &c 2. It should be observed that every system of notation, formed upon the fioregoing principles, will require as many distinct characters, as there are uants in the irda,/ix, and th at one of them must be a cipher, and another a unit. For the method'of changing numbers from the decimal to other scales Qf notation, and the converse, see Arts. 162, 163. QvIcsT.-43. Is the decimal notation the only system that can be formed on the same airnciples? flow would you form a quinary system of notation? Wlite six in the qtiaary scale on the black-1hcard. 1Write seven, nine, teh, eleven, twelve Obs. How many iharacters will any s-ystem formled upon this principle require' 3cts 30 NOTATION. LSECT. 1. 45. About the commencement of the second century, Ptolemy introduced the sexagesimal notation, which has sixty for its radix. OBs. 1. It is said that the Chinese and some other eastern nations now employ this system in measuring time, using periods of sixties, instead of ceinturies. Relics of the sexagesimal notation may also. be seen in our division of the circle, and of time, where the degree and hour are each divided.into 60 minutes, the minute into 60 seconds, &c. 2. The Roman notation seems to have been commenced with V or five for its radix, which was afterwards changed to X or.ten. It may therefore be regarded as a kind of combination of the qusinary and decimal systems. 46. Since the number eight may be divided and sub-divided so many times without a remainder, some contend that a system of notation having eight for its radix, would be preferable to the decimal system. Others claim that the duodecimal notation; that is, a system with twelve for its radix, would be more convenient than either.* However this may be, the decimal system is so firmly rooted, it were hopeless to attempt a change. Ons. It may be doubted whether any other ratio of increase would, on the whole, be more convenient, than that of the present system. If the ratio were less, it would require more places of figures to express large numbers; if the ratio were larger, it would not indeed require so many figures, but the operations would manifestly be more difficult than at present, on account of the numbers in each order being larger. Besides, the decimal system is sufficiently comprehensive to express with all desirable facility, every conceivable number, the largest as well as the smallest; and yet it is so simple, that a child may understand and apply it. In a word, it is every way adapted to the practical operations of business, as well as the most abstruse mathematical investigations. In whatever light, therefore, it is viewed, the decimal notation must be regarded as one of the most striking monuments of human ingenuity, and its beneficial influence on the progress of science and the arts, on commerce and civilization, must win for its unknown author the everlasting admiration and gratitude of mankind. * Barlow's Theory of Numbers, Leslie's Plilosophy of Arithmetic, Edi: ca rg Ea.cy.!oroedia. ARTS. 45-50.] ADDITION. 31 SECTION II. ADDITION. ART. 49. Ex. 1. A man bought three lots of land; the first contained 23 acres, the second 9 acres, and the third 15 acres: how many acres did he buy? Solution.-23 acres and 9 acres are 32 acres, and 15 are 47 acres. Ans. 47 acres. OBs. It will be seen, that the solution of this example consists in finding a single number', which will exactly express the value of the several given nuombers unqited together. 50. The process of uniting two or more numbers together, so as to form a single number, is called ADDITION. The answer, or the number thus found, is called the Sum or Amount. OBs. When the numbers to be added are all of the sanme denwmination, as all dollars, all pounds, &c., the operation is called Simnple Addition. Ex. 2. A miller bought 7864 bushels of wheat of one man, 4952 bushels of another, and 3273 bushels of another: how many bushels did he buy of all? Write the numbers under each other, so that Operation. units may stand under units, tens under tens, 7864 &c., and draw a line beneath them. Then be- 4952 ginning at the right hand or units, add each 3273 column separately. Thus, 3 units and 2 units Ans. 16089 bu. are 5 units, and 4 are 9 units. Write the 9 in units' place under the column added. Next 7 and 5 are 12, and 6 are 18 tens. But 18 requires two figures to express it; (Art. 34;) consequently it cannot all be written under its own column. We therefore write the 8 or right hand figure in tens' place under the column added, and reserving the 1 or left hand figure, add it with the hundreds. Thus, 1 which was reserved, and 2 are 3, and 9 are 12, and 8 are 20 hundreds. Set the 0 or right hand figure under the column QUEST.-50. What is Addition? What is the answer called? Obs. When the numbets to be added are all of the same denomination, what is the opera lion called? 51. What,rderi of figures d o you add together? 32 ADDITION. [SECT. II. add, d, and reserving the 2 or left hand figure, add it to the next colunn as before. Thus, 2 which were reserved and 3 are 5, and. 4 are 9, and 7 are 16 thousands. Set the 6 under the column added; and since there is no other column to be added, write the 1 in the next place on the left. 5 1. It will be perceived in this example, that units are added to units, tens to tens, &c.; that is, figures of the same order are added to each other. All numbers must be added in the same manner. For, figures standing in different orders or columns express diferent values; (Art. 35;) consequently, they cannot be united together directly in a single sum. Thus, 3 units and 5 tens will neither make eight units, nor eight tens, any more than 3 oranges and 5 apples will make 8 apples, or 8 oranges. In like manner it is plain that 7 tens and 2 hundreds will neither make 9 tens, nor 9 hundreds. Ons. The object of writing units under units, tens under tens, &c., is to prevent mistakes which might occur from adding di4fferent orders to each other. 52. When the sum of a column does not exceed 9, it will be noticed, we set it under the column added; but if it exceeds 9, we set the units or right hand figure under the column added, and reserving the tens or left hand figure, add it to the next column. In adding the last column on the left, we set down the whole sum. OBS. The process of reservling tlA tents, or left hand figure, and adding, it to the next column, is called carrying tens. 5 3. The principle of carrying may be illustrated in the following manner. Take, for instance, the last example, 7864 and adding as before, write the sum of 4952 each column in a separate line. Thus, 3273 the sum of the units' column is 9 units; 9 sum of units. the sum of the tens' column is 18 tens, or 18' " " tens. 1 hundred and 8 tens; the sum of the 19** " "hund. hundreds' column is 19 hundred, or 1 14#** " thou thoustnd 9 hundred; the sum of the 16089 Amount. QUEsT. —Why not add figures of different orders together? ARTS. 51-55.] ADDITION. 33 thousands' column is 14 thousand. Now, adding these results together as they stand, units to units, tens to tens, &c., the amount is 16089 bushels, which is the same as in the solution above. Thus, it is evident, when the sum of a column exceeds 9, the right hand figure denotes units of the same order as the column added, and the tens or left hand figure denotes units of the next higher order. Hence, Thle reason we carry the tens or left hand figure to the next column, is because it is of the sanme order as the next column, and figures of the same order must always be added together. (Art. 51.) OBS. 1. The reason for setting down the whole s'ne of the last or left hand column, is because there are no figures in the next order to which the left hand figure can be added. It is, in fact, carrying it to the next column. 2. From the preceding illustration it will also be seen, that the object of beginning to add at the right hand is, that we may car'qy the tens, as we proceed in the operation. 5.4-. From the preceding illustrations and principles we derive the following GENERAL RULE FOR ADDITION. 1. 7r'ite the numbers to be added, under each other; so that units may stand under units, tens under tens, dc. (Art. 51. Obs.) II. Begin at the right hand, and add each column separately. tWhenz the sum of a column does not exceed 9, write it under the column; but if the sum of a column exceeds 9, write the units' figure under the column added, and carry the tens to the next column. (Arts. 52, 53.) III. Proceed in this manner through all the orders, and set down the whole sum of tw zlast or left hand column. (Art. 53. Obs.) 55. PRooF.-Beginning at the tokp, add each column downwards, and if the second 4result is the samze as the first, the work is sugpposed to be rigqht. QUEST. 54.-How do you write numbers to be added? Why place units under units, &c.. Where do you begin to add? When the sum of a columln does no'exceed 9, what do yola do with it? When it exceeds 9; how proceed? What is mea: t by carrying the tensI Why carry the tens to the next column I Why begin to add a the right hand? What do you do with the sumn of the last column? 55. How is additio e proved? 34 ADDITION. [SECT. II. Note. —The object of beginning at the top and adding downwards, is that the figures may be taken in a different order from that in which they were added before. The order being reversed, the presumption is, that any mistake which may have been made will thus be detected; for it can hardly be supposed that two mistakes exactly equal will occur. 56, Second Method —— Cut off the bottom line, and find the sumn of the rest of the numbers; then add this sum and the bottom line together, and if the second result is the same as the first, the work is supposed to be right. Note. —1. This method of proof depends on the axiom, that the wuhole of a quantity is equal to the sum of all its parts. (Ax. 11.) 2. The method of cutting off the top line, and afterwards adding it to the sum of the others, is objectionable on account of adding the numbers in the same order as they were added in the solution. (Art. 55. Note.) 57. Third Method.-From the amount, subtract all the given numbers but one, and if the remainder is equal to the number not subtracted, the work may be supposed to be right. ATote.-This method supposes the pupil to be acquainted with subtraction before he commences this work. It is placed here on account of the convenience of having all the methods of proving the rule together. 5~. Fourth Method.'-Cast the 9s out of each of the given numbers separately, and place each excess at the right of the number. Then cast the 9s out of the sum of these excesses; also cast the 9s out of the amount; and if these two excesses are equal, the work may be supposed to be right. Note.-I. This mode of proof is based.on a peculiar property of the number 9. For its illustration and demonstration, see Art. 161. Prop. 14. 2. To cast the 9s out of a number, begin at the left hand, add the digits, together, and, as soon as the sum is 9 or over, drop the 9, and add the remainder to the next digit, and so on. For example, to cast the 9s out of 4626351, we proceed thus: 4 and 6 are 10; drop the 9 and add the 1 to the next figure. 1 and 2 are 3, and 6 are 9; drop the 9 as above. 3 and 5 are 8, and 7 are 15; dropping the 9, we have 6 remainder. EXAMPLES FOR PRACTICE. 59, Ex. 1. A man paid 2468 dollars for his farm, 1645 dollars for a house, 865 dollars for stock, and 467 dollars for tools; how much did he pay for the whole? 2. A produce merchant bought 5 cargoes of corn; the first conQUEST. —Note. Why add the columns downwards, instead of upwards? Can addition be proved by any other methods? * Wallis' Arithmetic, Oxford, 1657. ARTS. 56-59.] ADDITION. 35 tained 6725 bushels, the second 7208, the third 5347, the fourth 12386, and the fifth 10391 bushels: how many bushels did he buy?' 3. A tavern-keeper bought six loads of hay which weighed as follows: 1725 pounds, 2163 pounds, 1581 pounds, 1908 pounds, 2340 pounds, and 1879 pounds: what was the weight of the whole? 4. A man gave 5460 dollars to his oldest son, to the next 4065, o the next 6750, to the next 8000, and to the youngest 7276 dollars: how much did he give to all? 5. A merchant, on settling up his business, found he owed one creditor 176 dollars, another 841 dollars, another 1356 dollars, another 2370 dollars, another 840 dollars: what was the amount of his debts? 6. The state of Maine contains 32400 square miles; New Hampshire, 9500; Vermont, 9700; Massachusetts, 7800; Rhode Island, 1251; and Connecticut, 4789: how many square miles are there in the New England States? 7. The state of New York contains 46220 square miles; New Jersey, 7948; Pennsylvania, 46215; and Delaware, 2068: how many square miles are there in the Middle States? 8; The state of Maryland contains 10755 square miles; Virginia, 65700; North Carolina, 51632; South Carolina, 31565; Georgia, 61683; Florida, 56336; Alabama, 54084; Mississippi, 49356; Louisiana, 47413; and Texas, 100000: how many square miles are there in the Southern States? 9. The state of Tennessee contains 41752; Kentucky, 40023; Ohio, 40500; Michigan, 60537; Indiana, 35626; Illinois, 56506; Missouri, 70050: Arkansas, 54617; Iowa, 173786; and Wisconsin, 92930; how many square miles are there in the Western Staites? 10. What is the whole number of square miles in the United States? 11. What is the sum of 75234+41015+19075+176+88350 + 10040? 12. What is the sum of 250120+30402+7850+465000+ 10046+65045? 36 ADDITION. [SECT. 11. 13. What is the sum of 85046+90045q-412260+125781+ 4060-+273048? 14. What is the sum of 1500267+45085+4652+4780400+ 90276+89760841? 15. What is the sum of 45702125+67070420+670856-F 4230825d —750642+8790845? 16. What is the sumn of 825760842+35620476+7800490+ 467243+98371+6425+740? 17. What is the sum of 2503+37621+475290+1223729+ 10671840+275600312? 18. What is the sum of 463270+2500+7200342+10271426345+ 6200705? 19. What is the sum of 80429+7562345+700100+85261798 +4000101+3007002? 20. What is the sum of 756+849+934+680+720+843-t657689 +989876498+8045685 +807266780. 21. What is the sum of 6457+29301+82406+7589+63489 -1-101364+46745? 22. Add together 786, 840, 910, 403, 783, 650, 809, 670, 408, 310, and 652. 23. Add together 16075, 250763, 7561, 830654, 293106, 2537104, and 3167925. 24. Add together 256, 40, 751, 302, 75, 831, 26, 43, 621, 340, and 510. 25. Add together 493742, 56710607, 23461, 400072, 6811004, 8999003, and 26501. 26. Add together 629405, 7629, 31000401, 263012, 1300512, 390217, and 13268. 27. Add together 286013, 4016702, 1971342, 6894680, 28945, and 26243Q2. 28. Add together 460167, 296345, 84634123, 64205, 96 73108, and 1931456. 29. Add together 432678902, 310046734, 2167005; 327861 and 293000428. ART. 60.] ADDITION. 37 COUNTING-ROOM EXERCISES, 60o To the accountant as well as the mathematician, accuracy and expertness in adding, are indispensable. These attainments can be acquired only by frequent exercises in footing up long columns of figures. No/le. —. Instead of saying 4 and 8 are 12, and 2 are 14, 30. and 7 are 21, and 4 are 25, &c., a skilful accountant, per- 86015 formilg the addition at a glance, simply pronounces the 25163 results. Thus, four, twelve, twenty-one, thirty-one, (4+6 85057 =10,) thirty-seven, forty-seven, (7+3=10,) fifty-two. 12236 2: When two or three figures taken together make 10, as 43026 6 and 4, or 2, 3, and 5, &c., it accelerates the process to 67084 add their sum at once. A little practice will enable the 21167 studlent to run'up a long column of figures with as much 54042 facility almost as he can count. 42158 3. When the columns are long, accountants sometimes 24034 set the figure to be carried below the other figure under the 459982 Amls. column added. Thus, the sum of the first column in the 3045 Car. example above being 52, set the 5 (the figure carried) below the 2. The sum of the second column being 48, set the 4 below the 8, &c. This method saves much time in reviewing an operation, and also enables us, when interrupted, to resume the process where we left off. Required the amount of each of the following examples: 31. 32. 33. 34. Dollars. Dollars. Yards. Pounds. 2425 46,519 607,253 421,536 3282 32,271 232,012 310,101 2793 17,436 211,849 797,019 2354 81,587 380,436 233,680 4262 28,333 578,551 124,402 9158 52,745 231,349 255,353 2653 23,052 145,763 852,057 3424 20,158 605,037 618,041 1266 71,232 760,155 100,266 8742 39,464 357,676 971,134 2126 18,643 544,844 536,920 5387 42,027 276,232 703,352 Ans. 47872 73,235 803;383 420,503 Car. 465 24,103 725,918 312.675 4 38 ADDITION. [SECT. II. 35. 36. 37. 38. 348,037 460,375 963,172 849,652 272,465 841,681 300,725 361,728 530,634 -239,724 463,248 412.,381 109,871 763,256 721,003 635,403 693,036 437,891 387,356 872,545 764,543 825,432 241,653 406,223 323,638 285,678 603,280 294,867 428,432 310,720 532,176 811,236 389,763 403,521 278,321 576,037 210,045 687,489 829,248 213,744 760,806 324,061 171,320 764,368 636,215 530,724 206,782 305,216 253,734 623,452 461,027 436,720 251,600 487,638 589,203 823,284 575,453 290,731 248,639 217,436 807,720 803,256 730,461 592,301 930,046 731,463 672,398 243,762 174,173 379,574 246,175 731,445 626,245 823,156 928,340 429,374. 342,734 928,348 731,629 684,569 61. Accountants often acquire the habit of adding two columns of figures at a time. The power of rapid addition is easily acquired, and is well worthy the attention of the student. The following examples will illustrate the principle. 39. What is the sum of 312817+527236+141625+462415 - 25181,8+234112? Operation. Taking the two right hand columns, we 312817 say, 12 and 18 are 30, and 15 are 45 and 527236 25 are 70, and 36 are 106, and 17 are 123 141625 Set down the 23 under the columns added, 462415 and carry the 1 or left hand figure to the 251818 column of hundreds. Proceed in the same 234112 manner with the other columns. -Ans. 19300.23 ART. 61.] ADDITION. 39 (41.) (42.) (43.) (44.) (45.) (46.) 21 22 44 1325 2610 344235 30 13 20 1510 1511 402321 11 40 25 1314 1021 141511 13 25 17 3141 1115 201250 20 14 50 1016 1513 154036 15 11 14 2233 4020 132212 34 33 16 1224 1316 181714 18 45 28 2415 1233 2130-25 12 12 11 1830 2515 111817 17 20 14 1814 1718 161518 23 18 37 1621 2142' 432733 What was the amount of exports and imports of the United States in 1840, and of shipping in 1842? (47.) (48.) (49.) States. Exports. Imports. Shipping. Maine,. Dolls. 1,018,269 Dolls. 628,762 T. 281,930 N. Hampshire, 20,979 114,647 23,921 Vermont,. 305,150 404,617 4,343 Massachusetts, 10,186,261 16,513,858 494,895 Rhode Island, 206,989 274,534 47,243 Connecticut,. 518,210 277,072 67,749 New York,. 34,264,080 60,440,750 518,133 New Jersey,. 16,076 19,209 60,742 Pennsylvania, 6,820,145 8,464,882 113,569 Delaware,. 37,001 802 10,396 Maryland,. 5,768,768 4,910,746 106,856 Dist. of Columbia, 753,923 119,852 17,711 Virginia, 4,y778,220 545,085 47,536 North Carolina, 387,484 252,532 31,682 South Carolina, 10,036,769 2,058,870 23,469 Georgia, 6,862,959 491,428 16,536 Alabama,. 12,854,694 574,651 14,577 Louisiana, e 34,236,936 10,673,190 144,128 Ohio,.. 991,954 4,915 24,830 Michigan,. 162,229 148,610 12,323 Florida,. 1,858,850 190,728 7,28B 40 ADDITION. [SECT. II. 50. The appropriations of the Government of the United States, for 1847, were as follows: for the Civil and Diplomatic expenses 4,442,790 dolls.; for the Army and Volunteers 32,178,461 dolls.; for the Navy 9,307,958 dolls.; for the Post Office Department 4,145,400 dolls.; for the Indian Department 1,364,204 dolls.; for the Military Academy 124,906 dolls.; for building Steam Ships 1,000,000 dolls.; for Revolutionary and other Pensions 1,358,700 dolls.; for concluding Peace with Mexico 3,000, 100 dolls.; for Light Houses 518,830 dolls.; Miscellaneous 540,243 dolls. What was the amount of all the appropriations? 62. It may sometimes be convenient for the learner, as well as gratifying to his curiosity, to be able to add numbers expressed by the Roman Ngcation. 51. A man paid MDCCCLXXXIII dollars for a farm, DCCXXIIII dollars for stock, and CCCLXVIIII dollars for tools: how much did he pay for all? Beginning at the right hand, we proceed thus: Operation. four Is and four Is are eight, and three Is make MDCCCLXXXIII dolls, eleven, whichis equal to two Vs and I. We set DCCXXIIII dolls. down the I, and adding the two Vs to one V CCCLXVIIII dolls. makes fifteen, which is-equal to X and V. Set- IMMDCCCCLXXVI dolls. ting down the V, we count in the X with the other Xs, and find they make seven Xs or seventy, which is expressed by L, and' XX. We set down the two Xs, and adding the L to the other Ls, it makes three Ls, or one hundred and fifty, which is expressed by C and L. Setting down the L, and counting the C with the other Cs, we have nine Cs or nine hundred, which is expressed by D and CCCC. We set down the four Cs, and counting the D with the other Ds, it makes three Ds or fifteen hundred, which is expressed by M and D. We set down the D, and adding the M to the other M, we have two Ms, which we set down on the left of the other letters. Hence, 63. To add numbers expressed by the Roman Notation. Beginning at the right hand, count all the letters of each kind together; set down the result, and carry on the 1rincaiple that five Ts mtake one V; two Vs, one X; five Xs, one L, &.c. Ous. The teacher can extend the exercises in the Roman Notation as fa. as he may deem it expedient. A single examniple is sufficient to illustrate the principle, and to show that the Roman is greatdly inferior to the Arabic method in its adaptation to business calculations. ARTs. 62-66.] SUBTRACTION. 41 SECTION III. SUBTRACTION. ART. 65. Ex. 1. A merchant bought 37 barrels of flour, and afterwards sold 12 of them: how many barrels had he left? Soluztion.-12 barrels from 37 barrels leave 25 barrels. Ans. 25 barrels. OBS. It will be perceived, that the object in this example, is to find thLe difference between two numbers. 66. The jprocess of finding the clderence between two numbers is called SUBTRACTION. The difference, or the answer to the question, is called the Remainder. Ons. 1. The number to be subtracted is sometimes called the subtrahend, and the number from which it is subtracted, the mninzenzd. 2. Subtraction, it will be perceived, is the rever'se of addition. Addition unites two or more numbers into one single number; subtraction, on the other hand, separates a number into t'wo parts. 3. When the given numbers are of the same denomivnation, the operation is called Simpyle Subtraction. (Art. 50. Obs.) Ex. 2. What is the difference between 5364 and 9387? Write the less number under the greater, 2Operation. units under units, tens under tens, &c, Then, 9387 beginning at the right hand, proceed thus: 5364 4 units fifm 7 units leave 3 units. Write 4023 -Rem. the 3 in the units' place, under the figure subtracted. 6 tens from 8 tens leave 2 tens; set the 2 in tens: place. 3 hundred from 3 hundred leave 0 hundred; we therefore write a cipher in hundreds' place. 5 thousand from 9 thousand leave 4 thousand; set the 4 in the thousands' place. The answer is 4023. Q(UEST.-66. -What is subtraction? What is the difference or answer called' Obs What is the number to be subtracted sometimes called? The number from which it is subtracted? Of what is subtraction the reverse? When the given numbers are of the amne denomination, what is the operation called? 42 SUBTRIACTION. [SECT. III. 67. It will be observed, that we subtract units from units, tens from tens, &c.; that is, we subtract figures of the same order from each other. This is done for the same reason that we add figures of the same order to each other. (Art. 51.) OBs. The less number is written under the greater, simply for convenience in subtracting; and nmits are placed under units, tens under tens, &c., to avoid mistakes which might occur from taking different orders from each other. 68. It often happens that a figure in the lower number i larger than that above it, and consequently cannot be taken from it. Ex. 3. What is the difference between 94 and 56? Analytic solution. It is manifest that we cannot take 6 94=80+14 units from 4 units, for 6 is larger than 4. 56=504+ 6 To obviate this difficulty, we may take Rem. 38=30+ 8 1 ten from the 9 tens, and uniting it with the 4 units, the upper number will become 8 tens and 14 units, or 80 +-14. Separating the lower number into the parts of which it is composed, it becomes 5 tens and 6 units, or 50+6. Now, subtracting as in the last example, 6 from 14 leaves 8, 50 from 80 leaves 30. The answer is 30+8, or 38. Or, we may simply take I ten from the 9 tens, and adding it, mentally, to the 4 units, say 6 from 14 leaves 8; set the 8 under the figure subtracted. Then, having taken 1 from the 9 tens, we have but 8 left, and 5 from 8 leaves 3. The answer is 38. PRooF.-38+56=94; that is, the sum of the remainder and smaller number being equal to the larger, the answer is right. He-nce, 69. When a figure in the lower number is larger than that above it; take 1 from the next higher order in the upper number, and add it to the upper figure; from the sum subtract the lower figure, and diminishing the next upper figure by 1, proceed as before. Oss. 1. The process of taking one from the next higher order and adding it to the figure from which the subtraction is to be made, is called borrooing ten. It is the reverse of carrying. QUEST. —67. What orders of figures do you subtract from each other? Wh no zalbtract different orders from each other? ARTS. 67-71.] SUBTRACTION. 43 2. This method of borrowing, it will be seen, does not acffee the diference between the two given numbers; for, it is simply transposing a part of one order to another order in the same number, which, it is obvious, will neither increase nor diminish its value. 3. It may be asked, how can we take one from the figure in the next higher order, when that figure is a cipher? How can nothing lend anything, and how can notAlinvg be diminished by onbe? The explanation of this apparent contradiction is this: when the next figure is a cipher, we go to the next higher column still, and take one, which, added to the figure in the next lower order, makes ten; we then take one from the ten and add it to the upper figure, and proceed as before. 70. There is another method of borrowing, or rather of paying, which, though perhaps less philosophical than the preceding, is more convenient in practice, especially when the figures in the next higher orders are ciphers. Thus, in the last example, adding.10 to the upper figure, it becomes 14, and 6 from 14 leaves 8. Set down the 8 as before. Now, instead of diminishing the next upper figure by 1, if we add 1 to the next figure in the lower number it becomes 6 tens; and 6 from 9 leaves 3, which is the same as 5 from 8. The answer is 38, the same as before. Hence, 7 1 When a figure in the lower number is larger than that above it, add 10 to the upper figure, and to compensate this, add 1 to the next left hand figure in the lower number. OBS. 1. This method of borr0owing' depends on the self-evident principle, that if any two numbers are eqznally ilncrea(sed, their difference will not be altered. That the two given numbers are equally increased by this process, is evident from the fact that the i added to the lower number is of the next ~superior order to the 10 added to the upper number, and is therefore equal to it. (Art. 35.) 2. The reason that we borrow 10, instead of 8, or 12, or any other number, is because the radix, or ratio of increase, in the Arabic notation, is 10. (Art. 36.) If the radix of the\system were 8, it would be necessary to borrow 8; if 12, it would be necessary to borrow.12, &c. 3. On account of borrowing, the learner will perceive it is always necessary to begin to subtract at the right hand. Ex. 4. A man bought a house for 23006 dollars, and sold it for 21128 dollars: how much did he lose by his bargain? Operation. PI' roof. Cost 23006 dolls. 21128 Less number. Rec'd. 21128 dolls. 1878 Remainder. Ans. 1878 dolls. 23006 Larger number. 44 SUBnrRACTION. LS ECT. lII. 7 2. Fruol the preceding illustrations and- principles we derive the following GENERAL RULE FOR SUBTRACTION. I. Write the less number under the greacer, so that units may stand under units, tens under tens, S&c. (Art. 67. Obs.) II. Beginning at the right hand; subtract each figure in the lower number from the figure above it, and set the remainder directly under the figure subtracted. (Art. 71. Obs. 3.) III. When a figure in the lower number is larger than that above it, add 10 to thie upper figure; then subtract as before, and add 1 to the next figure in the lower number, or consider the next upyper figure 1 less than it is. (Arts. 69, 70. Obs. 1, 2.) 7 3. PROOF.-Add the remainder to the smaller number; and if the sum is equal to the larger number, the works is right. OBS. This method of proof depends upon the principle, that the diff'reqnce between two numbers being added to the less, the surm must be equal to the greater. For, the difference and the less number are the two paits into which the greater is separated, and the whvole of a quantity is equal to the sum of all its parts. (Ax. 11.) 7 4. Second M2fethod.-Subtract the remainder from the greater of the two given numbers; and if the difference is equal to the less number, the work is right. 75. Thiird.Metthod.-Cast the 9s out of the larger number, and place the excess at the right. Next, cast the 9s out of the smaller number, and also out of the remainder; then cast the 9s out of the sum of these two excesses; and if this last excess is the same as the excess of the larger number, the work may be supposed to be right. Thus, Ex. 5. From 7843 Excess of 9s in the greater number is 4 Take 5675 " CC less " is 5 Rcm. 2168 " " " remainder is 8 Now, 8+5=13, and the excess of 9s in 13 is 4, the same as that of the greater number. QUEST.-72. How ldo you write numbers for subtraction? Why write the less number under the greater? WVhy place units under lunits, &c.? Where do you begin to subtract? When a figure in the lower line is larger than that above it, how do you proceed? What is meant by borrow-ing ten? flow many methods of hborowin, are mentioned? Iilustnate the first method upon the black-board. How does it appear that this tmethod of borr~cwing does not affect the difference between the two given numbers? Explain the seconl method. Upon what principle does this mnethod depend? Why do yol borrow 10, instead of 8, or 12, or any other number? Why do you begin to subtract at the right hand 73. Ifow is subtraction proved? Obs. Upon what principle does this method of proof depen?. Can subtraction be proved by any other methods? ARTS. 72-76.] SUBTRACTION. 45 Note.-This method,)f proof depends on the same property of the number 9, as that in addition. (Art. 58. Note.) For, since the sum of the smaller number and remainder is equal to the larger number, it follows that the excess of 9s in the larger number must be equal to the excess of 9s in the remainder and smaller number together. EXAMPLES FOR PRACTICE. 76. Ex. 1. A merchant bought a ship for 35270 dollars, and r,ld it for 42365 dollars: how much did he make by his bargain? 2. A miller bought 46235 bushels of wheat, and ground 17251 blushels of it: how many bushels had he left? 3. A speculator laid out 50000 dollars in wild land, and afterwards sold it at a loss of 19046 dollars: how much did he get for his land? 4. A man owning a block of buildings worth 155265 dollars, keeps it insured for 109240 dollars: how much would he lose in case the buildings should be destroyed by fire? 5. The distance from the Earth to the Sun is 95000000 of miles; the distance of Mercury is only 37000000: how far is Mercury from the Earth? 6. The imports of Massachusetts in 1840, were 16,513,858 dollars, the exports were 10,186,261 dollars: what was the excess of her imports over her exports? 7. The imports of New York in 1840, were 60,440,75,0 dollars, the exports were 34,264,080 dollars: what was the exaess of her imports over her exports? 8. The imports of Pennsylvania in 1840, were 8,464,882 dollars, the exports were 6,820,145 dollars: what was the excess of her imports over her exports? 9. The imports of South Carolina in 1840, were 2,058,870 dollars, the exports were 10,036,769 dollars: what was the exs cess of her exports over her imports? 10. The imports of Alabama in 1840, were 574,651 dollalrs, the exports were 12,854,694 dollars: what was the excess of her exports over her imports? 11. The imports of Louisiana in 1840, were 10,673,190 dcllars, the exports were 34,236,936 dollars: what was the excess of her exports over her imports? 46 SUBTRACTION. LSECT. III. 12. The tonnage of the United States in l 1042, was 2069857, in 1846 it was 2500000: what was the increase in 4 years? 13. 14. 15. From 253760 3856031 54903670 Take 104523 462702 504089 16. 9876102-1050671. 28. 10000000-999999. 17. 4006723-5001. 29. 99999999 —100000. 18. 3601900-1000000. 30. 83567000 —438567. 19. 5317004-3565. 31. 40600056-7632. 20. 1000000-456321. 32. 56409250 —1057245. 21. 2035024- 27040. 33. 20030000-72534. 22. 45563075-460001. 34. 83175621-5256360. 23. 67030001 -300452. 35. 70301604-250041. 24. 73256300-436020. 36. 60050376-6849005. 25. 56037431-735671. 37. 34200591-8888888. 26. 80200430-250. 38. 87035762 —753017. 27. 96531768-873625. 39. 95246300-9438675. 40. From 6764+3764 take 6500+2430. 41. From 2890+8407 take 4251+3042. 42. From 7395+4036 take 8297+1750. 43. From 8404+7296 take 3201-1562. 44. From 6008+9270 take 5136 —2352. 45. From 9234+6850 take 9320 —4783..46. From 8564-2573 take 4431-1735. 47. From 7284-5362 take 6045-5729. 48. From 9561-4680 take 7352-6178. 49. From 8630 —1763 take 2460+1743. 50. From 7561 —.2846 take 1734+2056. 51. From 9687-3401 take 3021+1754. 52. A man having 55000 dollars, paid 7520 dollars fol a house, 3260 dollars for furniture, 2375 dollars for a library, and invested the balance in bank stock: bow much stock did he buy? 53. A gentleman worth 163250 dollars, bequeathed 15200 dollars apiece to his two sons, 16500 dollars to his daughter, ani to his wife as much as to his three children, and the remainder to a hospital: how much did his wife receive, and how much the hos. pital? ARTS. 76, 77.] SUBTRACTION. I4 54. A man bought three farms: for the first he paid 5260 dollars, for the second 3585, and for the third as much as for the first two. He afterwards sold them all for 15280 dollars: did he make or lose by the operation; and how much? 55. What number is that, to which 3425 being added, the sum will be 175250? 506. A man being asked how much he was worth, replied, if you will give me 325263 dollars, I shall have two millions of dol~Jars: how much was he worth? 57. A jockey gave 150 dollars for a horse, and meeting an acquaintance swapped with him, giving 37 dollars to boot; meeting anlother, he swapped and received 28 dollars to boot; he finally swapped again and gave 78 dollars to boot, and then sold his last horse for 140 dollars: how much did he lose by all his bargains? 58. A speculator gained 3560 dollars, and afterwards lost 2500 dollars; at another time he gained 6283 dollars, and then lost 3450 dollars: how nluch more did he gain than lose? 59. A man bought a house for MDCCCCXXXVII dollars, and sold it for DCXVIIII dollars less than he gave: how much did he sell it for? We perceive that the IIII in the lower number Operatiozn. cannot be taken from II in the upper number; MDCCCCXXXVII dolls. we therefore borrow a V, which added to the II, DCXVIIII dolls. makes IIIIIII; then IIII from IIIIIII, leaves Ans. MCCCXVIII dolls. III, which we set down. Now since we have borrowed the V in the upper number, there are no Vs left from which we can take the V in the lower number. We must therefore boriow an X; but X is equal to VV; and V from VV leaves V, which we set down. Having borrowed an X from the upper number, there are but XX left, and X from XX leaves X. C from CCCC leaves CCC. D from D leaves nothing. And nothing from M leaves M. Hence, 7. To subtract numbers expressed by the Roman lNolation. H'rrite the less number sender the great'er; then, beginnling at the rioht and, take the n16mbeqr in the lower line fr'om that expressed by the same letters in the upper line, and set the ~remainder below. If the,number in' the lvwer line is larger than thMat exrressed biy the same letters in the iu)pper livne, borrolw a letter nest highAer and arid it to the vnumber in the uplper line; then, subtract as b&f (re, obseurving to pay whevn you borrow as in subtraclton of figures. (Art. 72.) Oss. Other examples expressed by the Roman Notation, can be added by the teacher, if deemed expedient. 48 MULTIPLICATION [SECT. [V SECT1ON IV. MULTIPLICATION. Ari,. 7 9. Ex. 1. WThat will 3 melons cost, at 15 cents apiece Analysis. —If 1 melon costs 15 cents, 3 melons will cost 3 times 15 cents; and 3 times 15 cents are 45 cents. -Ans. 45 cents. 2. What will 4 sleighls cost, at 21 dollars apiece? Analysis.-Recasoning as before, if 1 sleigh costs 21 dollars, 4 sleigchs will cost 4 times as much; and 4 times 21 dollars ala 84 dollars. Ans. 84 dollars. OBs. It is obvious that 3 times 15 cents is the same as 15 cents+-15 cents — 15 cents, or 15 cents added to itself 3 times; and 4 times 21 dollars is the same as'21 dolls.+21 dolls.+21 dolls.+-j21 dolls., or 21 dollars added to itself 4 times. 80. This repeated addition of a number or quantity to itself, is called MULTIPLICATION. The number to be repeated, or multiplied, is called the KMultiplican'd. The number by which we multiply, is called the multtplier; and shows how many times the multiplicand is to be repeated. The number produced, or the answer to the question, is called the product. Thus, when we say, 8 times 12 are 96, 8 is the multiplier, 12 the multiplicand, and 96 the product. 8 1. The multiplier and multiplicand together are often called factors, because they make or produce the product. OMs. 1. The term factor is derived from a Latin word which signifies a agenlt, a doer, or produlcer. 2. When the multiplicand denotes things of onle de-nominat/ion only, the operation iu called Simple Multiplication. QUEsT.-80. What is multiplication What is the number to be repeated called What the number by which we multiply? What does the mnultillier slhow? What is the number produced called? 81. What are the multiplicand and multiplier together called? Why? Obs. What does the term factor signify? ARTS. 79-82.] MULTIPLICATION. 49 MULTIPLICATION TABLE. 1 1 2c 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20l 2 -4 6 -810 _1214 _16 18 20 22 24 24 8 0 3234 3 3- T 6 3 40 3 6 91 15 18 21 24 27 309 33 36 39 42 45 48 51 54 57 60 4 8 12 16 2024283236 40 4448 52 56 606468727680 5 10 15 20 25 30 35 40 45 50 55 60 65 [70 75 80 85 90 95 100 612 18 24 30 36 42 48 54 60 66 72 8 84 90 96 102 108 114 120 7 14 21 28 35 42 49 56 63 70 77 84 91 98 103 112 119 126 133 140 8 1 24 32 40 48 56 64 72- 89/ 88 90 104 112 1~20 128 136 144 152 160 18 27 5 54 63 72 81 93 99 108 117 126 135 144 153 162 171 180 10 20 30 40 5o 60 70 80 90 100 110 120 130 140 1]50 160 170 180 1 90 200 1. 1 22 33 44 55 66 77 88 99 110 121 132 144 134 165 178 187 198 209 220 12 24 36 48 60 72 84 9 108 12) 132 144 1lOi6 8S 180 192204 2116 228 240 13 206 39 50 65 78 91 104 117 7 131 143 15l 1(6911 1951208 2t 221 234 247 260 14 28 42 56 70 84 98 112 126 14 0 154 108 182 198 210 224 238 2a2 2(56 280 15 301 4o 6(1) 75 90 10 120,135 150 |1 65 180 |1895 0-10 _29 25 2)40( 0 r5 55270 285 300 16 32, 48 64 80 995112 128 144 160 17 M21 08 2`4 2401256)/212288 304 320 171 34 51 65 1 85 102 119 136 153 10 78ii 7204 221 238 255 272 289 3-060 323 340 18 36 54 721 90 108 126 144 1621 180 198 216 2-34 252 270 288 306 324 342 360 19 38 57 761 9 14 133 152 171 1.40 209 228 247 276 6285 304 14323 342 361 380 20 40 60 801 101) 120 140 lS0i18012)00 20)0240 260280 300 320 340 360 380 400 Note.-This Table was invented by Pythagor'as, and is therefore sometimes called the Pythagorean Table. The pupil will find assistance in learning the Multiplication Table by observing the following particulars. 1. The several results of multiplying by 10 are formed by simply adding a cipher to the figure that is to be multiplied. Thus, 10 times 2 are 20, 10 times 3 are 30, &c. 2. The results of nlultiplying by 5 terminate in 5 and 0, alternately. Thus, 5 times 1 are 5, 5 times 2 are 10, 5 times 3 are 15, &c. 3. The first nine results of multiplying by 11 are formed by repeating thle figure to be multiplied. Thus, 11 times 2 are 22; 11 times 3 are 33, &c. 4. In the successive results of multiplying by 9, the right hand figure regularly decreases by 1, and the left hand figure regularly increases by 1. Thus, 9 times 2 are 18; 9 times 3 are 27; 9 times 4 are 36, &c. $2. Multiplying by 1, is taking the multiplicand once: thus, 4- multiplied by 1=4. Multiplying by 2, is taking the multiplicand twice: thus, 2 times 4, or 4+4=8. Multiplying by 3, is taking the multiplicand three times: thus, 3 times 4, or 44 -4+4=12, &c. Hence, QUES'. —82. What is it to nliltiiply by 1? By 2? By 3? 50 MULTIPLICATION. LSECT. IV Multiplying by any whole number, is taking the multiplicand as many times, as there are units in the multiplier. The application of this principle to fractional multipliers will be illustrated under fractions. OBs. 1. From the definition of multiplication, it is manifest that the prodquct Is of the same kivnd or deznomiination as the multiplicand: for, repeating a number or quantity does not alter its nature. Thus, if we repeat dollars, they are still dollars; if we repeat yanrds, they are still yards, &c. Consequently; if the multiplicand is an abstract qnmber, the product will be an abstract number; if noney, the product will be money; if barrels, barrels, &c. 2. Every mueltiplier is to be considered an abstract number. In familiar language it is sometimes said, that the price multiplied by the weight will give the value of an article; and it is often asked how much 25 cents multiplied by 25 cents, &c., will produce. But these are ablbreviated expressions, and are liable to convey an erroneous idea, or rather no idea at all. If taken literally, they are absurd; for multiplication is repcating a number or quantity a certain nuzmber of times. Now to say that the price is repeated as many times as the given quantity is 7wavy, or that 25 cents are repeated 25 cents times, is nonsense. But we can multiply the price of 1 pound'by a number equal to the number of pounds in the weight of the given article, and the product will be the value of the article. We can also multiply 25 cents by the number 25; that is, repeat 25 cents 25 times, and the product is 625 cents. Construed in this manner, the multiplier becomes an abstract number, and the expressions have a consistent meaning. Ex. 3. What will 6 houses cost, at 2341 dollars apiece? Write the numbers on the slate as Operation. in the margin, and beginning at the 2341 Multiplicand. right hand, proceed thus: 6 times 1 6 Multiplier. unit are 6 units; write the 6 under the Ans. 14046 Dollars. figure multiplied. 6 times 4 tens are 24 tens; set the 4 or right hand figure under the figure multiplied, and carry the 2 or left hand figure'.o the next product figure, as in addition. (Art. 52.) 6 times 3 hundreds, are 18 hundreds, and 2 to carry make 20 hundreds; set the 0 under the figure multiplied, and carry the 2 to the next product as before. 6 times 2 thousands are 12 thousands, and 2 to carry make 14 thousands. Since there are no QUEST.-What is it to multiply by any whole number? Obs. Of what denomination is the product? How does this appear? What must every multiplier be considered? Can you multiply by a given weight, a measure, or a sum of money? ART. 83.] MULTIPLICATION. 51 more figures to be multiplied, set down the 14 in full as in addition. (Art. 53. Obs. 1.) The product is 14046 dollars. 83. The product of any two numlbers will be the sacne, whichever factor is taken for the multiplier. Thus, If an orchard contains 5 rows of trees, and each row has 7 trees, as represented by the * * -* * * stars in the margin, it is evident the whole * * * * umber of trees is equal either to the number -* e * of stars in a horizontal row repeated five times, e * * * * * or to the number of stars in a perpendicular row repeated seven times, viz. 35. For, 7X 5=-35, also 5X 7 —35. Ons. 1. It is more convenient and therefore customary to place the larg er num ber for the multiplicand, and the smaller for the multiplier. Thus, it is easier to multiply 8468946 by 3, than it is to multiply 3 by 8468946, but the product would be the same. Ex. 4. What will 237 coaches cost, at 675 dollars apiece? Since it is not convenient to multi- Opera tion. ply by 237 at once, we multiply first 675 Multiplicand. by the 7 units, next by the 3 tens, 237 Multiplier. then by the 2 hundreds, and place 4725 cost 7 coaches. each result in a separate line, with 2025* cost 30 the first figure of each line directly 1350** cost 200 under that by which we multiply. 159975.cost 237 Finally, adding these results together, units to units, &c., we have 159975 dollars, which is the whole product required. (Ax. 11.),Note.-When the multiplier contains more than one figure, the several products of the multiplicand into tne separate figures of the multiplier, are called pa'rtial products. OBs. 2. The reason for placing the first figure of the several partial producis undler the figure by which we multiply, is to bring the same orders under each other, and thus prevent mistakes in adding them together. (Art. 51.) 3. The several partial products are added together for the obvious purpose:f finding the,whole product or answer required. (Ax. 11.) Qir.S'r.-83. Does it make any difference with the result. which of the given numnbeTr Is taken for the mlultiplier? Obs. Which is usually taken? Why? 3* 52 MULTIPLICATION. [SECT. IV. 8 4. The principle of carrying the tens in multiplication is the same as in addition, and may be illustrated in a similar manner. (Art. 53.) Thus, Ex. 5. 9382 Mult'd. Or, separating the multiplicand into 7 Mult'r. the orders of which it is composed, 14=units, 9382=9000+300+80+2, 56*=tens, and 9000 X 7=63000 21**=hunds. 300X7= 2100 63***=-thou. 80X7= 560 65674 Product. 2X7= 14 Adding these results together, we have 65674 Ans. Os. The reason for always beginning to multiply at the right hand of the multiplicand, is that we may carry the tens as we proceed in the operation. 85. From this illustration it will be observed that units multiplied into units produce units; tens into units, or units into tens, produce tens; (Art. 83;) hundreds into units, or units into hundreds, produce hundreds, &c. Hence, 86. Whlelz units are multiplied into any order whatever, the product will always be qf the same order as the other figure. And universally, the product of any two integers is of the order.next less than that denoted by the sum of the orders of the two given.figates. Thus, hundreds into tens produce thousands, or the 4th order, which is one less than the sum of the two given orders. OBs. When the multiplier contains more than one figure, it is customary to begin to multiply with its units' figure. The result however will be the same, if we begin with its hundreds or any other order of the multiplier, and place the first figure of the partial products, so that the same orders shall stand under each other. tFirst Operation. Second Operation. 1357 1357 3574 3574 4071 4071 6785 6785 9499 9499 5428 5428 4849918. Prod. 4849918. Prod. QUEST. —85. WVhat do units into unrits produce? Units into tens, or tens into unlts I ARt s. 84-89. ] MIULTIPLICATION. 53 Ex. 6. What is the product of 5690 into 3008? After multiplying by the 8 units, we next Operathon. multiply by the 3 thousands, since there are no 5690 tens nor hundreds in the multiplier, and place 3008 the first figure of this partial product under the 45520 figure 3 by which we are multiplying. 17070 17115520 Ans. 87. From the preceding illustrations and principles we derive the following GENERAL RULE FOR MULTIPLICATION. I. When the multiplier contains but one figure. Write the multiplier under the multip2licand, units under units, tens under tens, &c. (Art. 83. Ohs. 1.) Begin? at the right hand and multiply each figure of the multiplicand by the multiplier, setting down the result and carrying as in addition. (Art. 84. Ohs.) 11. When the multiplier contains more than one figure. Multiply each figure of the multipli-cand by each figure of the multiplier separately, beginning with' the units, and write the partial products in separate lines, placing the first figure of each line directly under the fjgure by which you multiply. (Art. 86. Obs. 2.)?Finally, add tie several partial products together, and the sum tvill be the uwhole product. (Art. 83. Obs. 3.) S. PROOF.-ji~ultiply the multiplier by the multiplicanct, cUtd if the plodect thus obtained is the sam/e as the other product, the worle is su2pposed to be right. OBs. This method of proof depends upon the principle, that the product of any two numbers is the same, whichever is taken for the multiplier. (Art. 83.) 89. Second Method.-Add the nmultiplicand to itself as many QuEs'. —86. When units are multiplied into any order, what order is the product? When any two integers are multiplied together, of what order is the product? 87. How do you write the numbers for multiplication? When the multiplier contains but one fig ure, how proceed? Why begin at the right hand of the multiplicand? When the multiplier contains more than one figure, how proceed? What is meant by partial products I Why pllace the first figure of each partial product untler the figure by which you multiply? ViWhat is to bte dolne withl the partial products? Why add the several partial pro. dicts together? WVhy should this give the whole product? 88. low is multiplicaitoa proved? Obs. On what pnricille does this proof depend'! _' S 54 MULTIPLICATION. [SECT. IV. times as there are units in the multiplier, and if the product obtained is equal to the amount, the work is right. Note.-When the multiplier is small, this is a very convenient mode of proof. 90. Third Method.-Cast the 9s out of the multiplicand and multiplier; multiply their remainders together, and casting the 9s out of their product, set down the excess; then cast the 9s out of the answer obtained, and if this excess be the same as that obtained from the multiplier and multiplicand, the work may. be considered right. Ex. 7. Multiply 565 by 350. Operation. Proof. 565 The excess of 9s in the multiplicand is 7. 356 " " 9s " multiplier is 5. 3390 7X5=35; and the excess of 9s is 8. 2825 1695 Prod. 201140. The excess of 9s in the Ans. is also 8. 91. Fourth Method.-Divide the product by one of the factors, and if the quotient thus arising is equal to the other factor, the work is right. Note.-This method of proof s'upposes the learner to be acquainted with division before he commences this work. (Art. 57. Note.) It is simply reversing the operation, and must obviously lead us back to the number with which we started: for, if a number is both multiplied and divided by the same numnber, its value will not be altered. (Ax. 9.) 92. Fifth Method.* —First, cast the 11s out of the multiplicand and multi)plier; multiply their remainders together, cast the 1Is out of the product, and set down the excess; then cast the 11s out of the answer obtained, and if the excess is the same as that obtained from the multiplier and multiplicand, the work is right. Note.-1. This method depends on a peculiar property of the number 11. For its further development and illustration, see Art. 161. Prop. 18. 2. To cast the 1 s out of a number, begin at the right hand, mark the alternale figures; then from the sum of the figures marked, increased by 11 if necessary, take the sum of those not marked, and the remainder will be the excess required. Thus to cast the Ils out of 39475025, mark the alternate figures, beginning at the right hand, 39475025, then the sum of QUEST.-Can multiplication be proved by any other methods? * Leslie s Philosophy of Arithlmetic. ARlrs. 90-93.] MULTIPLICATION. 55 5-t-0+-7+9=21. Again, the sum of the others, viz: 2-+5+4-3- 14. Now, ~21-14=7, the excess of Ils. Or, as soon as the sum is 11 or over, we may drop the 11, and add the remainder to the next digit. Thus, 5 and 7 are 12; dropping the 11, 1 and 9 are 10. Again, 2 and 5 are 7, and 4 are 11; drop the 11, and there are 3 left. Now, 10-3=7, the same excess as before. Ex. 8. Multiply 237956 by 3728. Oper'ation. Proof. 237956 Excess of 11s is 4. I Now, 4XI0=40; the excess of 11s 3728 " " 10. in 40 is 7. Ans. 887099968 Excess of 11s in the answer is also 7. EXAMPLES FOR PRACTICE. 93. Ex. 1. What will 435 acres of land cost, at 57 dollars per acre? 2. What cost 573 oxen, at 63 dollars per head? 3. What cost 1260 tons of iron, at 45 dollars per ton? 4. If a man can travel 248 miles in a day, how far can he travel in 365 days? 5. If an army consume 645 pounds of meat in a day, how much will they consume in 115 days? 6. If 1250 men can build a fort in 298 days, how long would it take 1 man to do it? 7. How many rods is it across the Atlantic Ocean, allowing 320 rods to a mile, and the distance to be 3000 miles? 8. What is the product of 463 X 45? 9. What is the product of 348X 62? 10. What is the product of 793X86? 11. What is the product of 75 X 42X 56? 12. What is the product of 7198 X216? 13. 31416X175. 22. 8320900X1328. 14. 8862X189. 23. 17500 X 732. 15. 7071X556. 24. 15607 X 3094. 16. 93186X4455. 25. 7422153 X 468. 17. 40930X779. 26. 9264397 X 9584. 18. 12345X686. 27. 4687319X1987. 19. 46481X936. 28. 9507340 X 7071. 20. 16734X'708. 29. 39948123X6007. 21. 7575X'7575. 30. 73885246X6079. 56 MULTIPLICATION. [SECT. IV. 31. 57902468X5008. 37. 58763718X6754. 32. 57902468X5080. 38. 73084163X7584. 33. 57902468X 5800. 39. 144X144X144. 34. 12481632X 1509. 40. 3851X3851X3851. 35. 79068025 X 1386. 41. 79094451 X 764094, 36. 92948789 X 7043. 42. 89548050 X 972800. CONTRACTIONS IN MULTIPLICATION. 94. The general rule is adequate to the solution of all examples that occur in multiplication. In many instances, however, by the exercise of judgment in applying the preceding principles, the operation may be very much abridged. 95. Any number which may be produced by multiplying two or more numbers together, is called a Composite Nlumber. Thus, 4, 15, 21, are composite numbers; for 4=2X 2; 1.5= 5X3; 21=7X3. OBS. 1. The factors which, being multiplied together, produce a composite number, are sometimes called the component parts of the number. 2. The process of finding the factors of which a given number is composed, is called resolving the number intofactors. Ex. 1. Resolve 9, 10, 14, 22, into their factors. 2. What are the factors of 35, 54, 56, 63? 3. What are the factors of 45, 72, 64, 81, 96-? 96. Some numbers may be resolved into more than two factors; and also into different sets of factors. Thus, 12 2 X 2 X 3; also 12-4X3-6X2. 4. What are the different factors and sets of factors of 8, 16, 18, 20, 24? 5. What are the different factors and sets of factors of 27, 32, 86, 40, 48? 96, a. We have seen that the product of any two nunmbers is the same, whichever factor is taken for the multiplier. (Art. 83.) In like manner, it may be shown that the product of any. three or Q TrEST.-95. What is a composite number? Obs. What are the factors which pro(dutce it sometimes called? What is me'-nt by resolving a number into factors? 96. Are numbers ever composed of more than two factors? 96. a. When three or lmore factors are to be multipliedt Tgether, does it make any difference in what order they are taken? ARTS. 94-97.] MULTIPLICATION. 57 more factors will be the same, in whatever order they are multiplied. For, the product of two factors may be considered as one number, and this may be taken either for the multiplicand, or the multiplier. Again, the product of three factors may be considered as one number, and be taken for the multiplicand, or the multiplier, &c. Thus, 24=3X2X2X2=6X2X2=x12X2=6X4= 4X2X3-8X3. CASE 1.-When the multiplier is a composite number. 6. What will 27 bureaus cost, at 31 dollars apiece? Analysis.-Since 27 is three times as much as 9; that is, 27 — 9 X 3, it is manifest that 27 bureaus will cost three times as much as 9 bureaus. Operation. T)olls. 31 cost of 1 B. Having resolved 27 into the factors 9 9 and 3, we find the cost of 9 bureaus, Dolls. 279 cost of 9 B. then multiplying that by 3, we have 3 the cost of 27 bureaus. Dolls. 837 cost of 27 B. 7. What will 36 oxen cost, at 43 dollars per head? Solution.-36-99X4; and 43X9X 4=1548 dolls. Ans. Or, 36=-3X3X 4; and 43 X 3 X 3 X 4=1548dolls. Ans. Hence, 97. To multiply by a composite number. Resolve the multiplier into two or more factors; multiply the multiplicand by one of these factors, and this product by another factor, and so on till you have multiplied by all the factors. The last product will be the answer required. OBs. The factors into which a number may be resolved, must not be confounded with the parts into which it may be separated. (Art. 53.) The former have. reference to multiplication, the latter to addition; that is, facto'rs must be mzultiplied together, but parts must be added together to produce the given number. Thus, 56 may be resolved into two factors, 8 and 7; it may be sep. arated into two parts, 5 tens or 50, and 6. Now, 8>X7_56, and 501-6-56. 8. What will 24 horses cost, at 74 dollars a head? QvUEST.-97. When the multiplier is a composite number, how do you proceed? Ohs. What is the difference between the factors into which a number may be resolved and the parts into which it may be separated? t55 MULTIPLICATION. I SECT. IV. 9. What cost 45 hogsheads of tobacco, at 128 dollars a hogshead? 10. What cost 54 acres of land, at 150 dollars per acre? 11. At 118 shillings per week, how much will it cost-a family to board 49 weeks? 12. If a man travels at the rate of 372 miles a day, how far will he travel in 64 days? 13. At 163 dollars per ton, how much will -72 tons of lead cost 9 14. What cost 81 pieces of broadcloth, at 245 shillings apiece? 15. What cost 84 carriages, at 384 dollars apiece? CASE II. —When the multiplier is 1 with ciphers annexed to it. 98. It is a fundamental principle of notation, that each re moval of a figure one place towards the left, increases its value ten times; (Art. 36;) consequently, annexing a cipher to a number will increase its value ten times, or multiply it by 10; annexing two ciphers will increase its value a hundred times, or multiply it by 100; annexing three ciphers will increase it a thousand times, or multiply it by 1000, &c. Thus, 15 with a cipher annexed, becomes 150, and is the same as 15 X 10; 15 with two ciphers annexed, becomes 1500, and is the same as 15 X 100; 15 with thr~e ciphers annexed, becomes 15000, and is the same as 15X 1000, &c. Hence, 99. To multiply by 10, 100, 1000, &c. Annex as many ciphers to the multiplicand as there are ciphers in the multiplier, and the number thus formed will b( the product rsequired. Note.-To annex means to place after, or at the right hand. 16. What will ten boxes of lemons cost, at 63 shillings per box? Ans. 630 shillings. 17. How many bushels of corn will 465 acres of land produce, at 100 bushels per acre? QuEST.-98. What is the effect of annexing a cipher to a number? Two ciphers? Three? FourS? )). Ihow do you proceed when the multiplier is 10, 100, 1000), &c.? blote, What is the lmeaning of the term annex! ARTS. 98-100.] MULTIPLICATION. 59 18 Allowing 365 days for a year, how many days are there in 1000 yeais? 19. Multiply 153486 by 10000. 20. Multiply 3120467 by 100000. 21. Multiply 52690078 by 1000000. 22. Multiply 689063457 by 10000000. 23. Multiply 4946030506 by 100000000. 24. Multiply 87831206507 by 1000000000. 25. Multiply 67856006109 by 10000000000. CASE III.- When the multiplier has ciphers on the'light hans. 26. What will 30 wagons cost, at 45 dollars apiece? Note.-Any number with ciphers on its right hand, is obviously a composite number; the significant figure or figures being one factor, and 1, with the given ciphers annexed to it, the other factor. Thus, 30 may be resolved into the factors 3 and 10. We may therefore first multiply by 3 and then by 10, by annexing a cipher as above. Solutions.-45X3=135, and 135 X 10=1350 dolls. Ans. 27. How many acres of land are there in 3000 farms, if each farm contains 475 acres? Analysis.-3000=3 X 1000. Now 475X Operation. 3=1425; and adding three ciphers to this 475 product, multiplies it by 1000. (Art. 99.) 3 Htence, Ans. 1425000 acres. 100. When there are ciphers on the right of the multiplier. Multiply the multiplicand by the significant figures of the multiplier, and to this product annex as many ciphers, as are found on the right of the multiplier. OBs. It will be perceived that this case combines the principles of the two preceding cases; for, the multiplier is a composite number, and one of its factors is 1 with ciphers annexed to it. 28 How much will 50 hogs weigh, at 375 pounds apiece? 29 If 1 barrel of flour weighs 192 pounds, how much 1500 balrrels weigh? 30. Multiply 14376 by 25000. Qutssr.-100. When there are ciphers on the right of the multiplier, how do vo ceed' Obs. What principles does this case combine? 60 MULTIPLICATION. LSECT. IV 31. Multiply 350634 by 410000. 32. Multiply 4630425 by 6200000. CASE IV.- When the multiplicand has ciphers on the right hand. 33. What will 37 ships cost, at 29000 dollars apiece? Analysis.-29000=29X1000. But the Operation. product of two or more factors is the same 29000 in whatever order they are multiplied. 37 (A rt 96. a.) We therefore multiply 29 203 by 37, and this product by 1000 by adding 87 thcee ciphers to it. Ans. 1073000 dolls. PRoor. —29000 X 37-1073000, the samne as before. Hence, 10 1. When there are ciphers on the right of the multiplicand. lIMultiply the significant figures of the multiplicand by the multiplier, and to the product annex as many ciphers, as are found on the right of the multiplicand. OmS. When both the multiplier and multiplicand have ciphers on the right, multiply the significant figures together as if there were no ciphers, and to their product annex as many ciphers, as are found on the right of both factors 34. Multiply 2370000 by 52. 35. Multiply 48120000 by 48. 36. Multiply 356300000 by 74. 37. Multiply 1623000000 by 89. 38. Multiply 540000 by 700. Alnalysis.-540000= 54X10000, and Operation. 700=7 X 100; we therefore multiply the 540000 significant figures, or the factors 54 and 700 7 together, (Art. 96. a,) and to this pro- Ans. 378000000 duct annex six ciphers. (Art. 99.) 39. Multiply 1563800 by 20000. 40. Multiply 31230000 by 120000. 41. Multiply 5310200 by 3400000. 42. Multiply 82065000 by 8100000. 43. Multiply 210909000 by 5100000. QUEST.-101. When there are ciphers on the right of the tmultiplicand, how proceed I Obs. IIow, when there are ciphers on the right both of the multiplier and multiplicand' ARTS. 101-104.]1 MULTIPLICATION. 61 1 02. There are other methods of contracting the op2erations in multiplication, which, in certain cases, may be resorted to with advantage. Some of the most useful are the following. 44. How many gallons of water will a hydrant discharge in 13 hourns, if it discharges 2325 gallons per hour? Operation. Multiplying by the 3 units, we set the 2325 X 13 first figure of the product one place to the 6975 right of the multiplicand. Now, since Anec 30225 gallons. multiplying by 1 is taking the multiplicanld once, (Art. 82,) we add together the multiplicand and the partial product already obtained, and the result is the answer. PROOF.-2325X13=30225 gallons, the same as above. Hence, 103. To multiply by 13, 14, 15, 8&c., or 1, with either of the other digits annexed to it. flultiply by the units',figure of the multiplier, and write each fi gure of the partial product one place to the right of that from which it arises; finally, add the partial product to the multi licand, and the result will be the answer required. Note. —This method is the same, in effect, as if we actually multiplied by the 1 ten, and placed the first figure of the partial product under the figure by which we multiply. (Art. 87. II.) 45. Multiply 3251 by 14. 46. Multiply 4028 by 17. 47. Multiply 25039 by 16. 48. Multiply 50389 by 18. 49. If 21 men can do a job of work in 365 days, how long will it take 1 man to do it? Operation. We first multiply by the 2 tens, and set 365 X 21 the first product figure in tens' place, then 730 adding this partial product to the multipliAns. 7665 days. cand, we have 7665, for the answer. PnooF.-365 X 21=-7665 days, the same as above. H(nce, 1 04. To multiply by 21,.31, 41, &c., or 1 with either (Df the other significant figures prefixed to it.,Multiply by the tens' figure of the multiplier, and write the first 6 62 MULTIPLICATION. LSECT. IV. figure of the partial product in tens' place; finally, add this p2artial product to the multiplicand, and the result will be t/he answer required. lote.-The reason of this method of contraction is substantially the same as that of the preceding. 50 Multiply 4275 by 31. 51. Multiply 7504 by 41. 52. Multiply 38256 by 61. 53. Multiply 70267 by 81, 54. How much will 99 carriages cost, at 235 dollars apiece? Analysis.-Since 1 carriage costs 235 0peration. dollars, 1.00 carriages will cost 100 times 23500 price of 100 C. as much, which is 23500 dollars. (Art. 235 " of 1 C. 99.) But we wished to find the cost 23265 " of 99 C. of 99 carriages only. Now 99 is 1 less than 100; therefore, if we subtract the price of 1 carriage from the price of 100, it will give the price of 99 carriages. Hence, I 05. To multiply by 9, 99, 999., or any number of 9s. Annex as many ciphers to thle multiplicand as there are 9s in the multiplier; from the result subtract the given multiplicand, and the remainder will be the answer required. No/te.-The reason of this method is obvious from the fact that annexing as many ciphers to the multiplicand as there are 9s in the multiplier, multiplies it by 100, or repeats it once mor'e than is required; (Art. 99;) consequently, subtracting the multiplicand from the number thus produced, must give the true answer. 55. Multiply 4791 by 99. 56. Multiply 6034 by 999. 57. Multiply 7301 by 999. 58. Multiply 463 by 9999. 59. What is the product of 867 multiplied by 84? Analysis.-We first multiply by 4 in the usual Operation. way. Now, since 8-4 X 2, it is plain, if the par- 807 tial product of 4 is multiplied by 2, it will give 84 the partial product of 8. But as 8 denotes tens, 3468 X 2 the first figure of its product will also be tens. 6936 (Art. 80.) The sum of the two partial products 7 8328 Anis. will be the answer required. ATote.-For the sake of convenience in multiplying, the factor 2 is placed at the right of the partial product of 4, with the sign X, between them. ARTs. 105, 106.] MULTIPLICATION. 63 60. What is the. product of 987 by 486? Operation. 987 Since 48 —6 X 8, we multiply the partial prod486 uct of 6 by 8, and set the first product figure 5922 X 8 in tens' place as before. (Art. 86.) 47376 479682 -Ans. PRooF. —987X486=479682, the same as above. Hence, 1 06. When part of the multiplier is a comlposite number of, which the other figure is a factor. Ffirst multiply by the figure that is a factor; then multiply this plartial product by the other factor, or factors, tahcing care to write the first fiyure of eackl.;i, tial product in its proper order, and their sum.will be tht:.nsZe',' required. (Art. 86.) Oi3s. When the figure:n thou-lands, ten thousands, or any other column, is a factor of the other par.. or;arts of the multiplier, care must be taken to place the first figure of its product under the factor itself, and the first figure ofeach of the other partial products in its own order. (Art. 86.) (61.) (62.) 2378 256841 936 85632 21402 X4 2054728 7 X 4 85608 14383096 2225808 Ans. 8218912 21993808512 Ans. 63. Multiply 665 by 82. 64. Multiply 783 by 93. 65. Multiply 876 by 396. 66. Multiply 69412 by 95436, 67. 324.325X54426. 68. 256721X85632. 69. What is the product of 63 multiplied by 45? Note.-By multiplying the figures which produce the same order, and adding the results mentally, we may obtain thb Naswer without setting down the partial prodlutcts. First, multiplying the units into units, we set Operation. down the result and carry as usual. Now, since 63 the 6 tens into 5,mnits, and 3 units into 4 tens will 45 both produce the same order, viz: tens, (Art. 86,) 2835 Ans. we multiply them and add their products men' 64 MULTIPLICATION. [SECT. IV. tally. Thus, 6 X 5=30, and 3 X 4= —12; now, 30+ 12=42, and 1 (to carry) makes 43. Finally, 6 X 4=24, and 4 (to carry) make 28. PRooF.-63X45-2835, the same as before. Hence, 1 07. To multiply any two numbers together without setting down the partial products. First multiply the units together; then multiply the figures w'hich produce tens, and adding the products mentally, set down the esult and carry as usual. Next multigply the figures which produce hundreds, and add the products, &c., as before. In like manner, perform, the multigplications which _produce thousands, ten thousands, c&c., adding the products of each order as you proceed, and thus continue the (peration till all the figures are multiplied. 70. What is the product of 12346789 into 54321? Analytic Operation. 2 3 4 5 6 7 8 9 5 4 3 2 1 2X1 3X1 4X1 X ) 1 6X6 7Xl 8 X1 9 X1 2X2 3X2 4X2 5X2 6X2 7X2 8X2 9X2 2X3 3X3 4X35X36X3 7X3X3 9X3 2X4 3X44X4 5X4 6X4 7X48X4 9X4 2X5 3X5 4X5l5X16XS 7X5 8X5 9X5 7 0 6 2 9 6 2 3 5 2 6 9 Explanation.-Having multiplied by the first two figures of the multiplier, as in the last example, we perceive that there are three multiplications which will produce hundreds, viz: 7 X 1, 8 X 2. andl 9X 3; (Art. 86;) we therefore perform these multiplications, add their products mentally, and proceed to the next order. Again, there are four multiplications which will produce thousands, viz: 6X 1, 7 X 2, 8 X 3, and 9 X 4. (Art. 86.) We perform these multiplications as before, and proceed in a similar manner througoh all the remaining orders. Ans. 706296235269. Note.-l. In the solution above, the multiplications of the different fiCgO'es are, arranged in separate columns, that the various combinations which produce the same order, may be seen at a glance. In practice it is unnecessary to denote these multiplications. The principle being understood, the process of A RTS. 107, 108. | MULTIPLICATION. 65 multiplying and adding may easily be carried on in the mind, while the final product only is set down. 2. When the factors contain but two or three figures each, this method is very simple and expeditious. A little practice will enable the student to apply it with facility when the factors contain six or eight figures each, and its application will afford an excellent discipline to the mind. It has sometimes been used when the factors contain twenty-four figures each; but it is doubltful whether the attempt to extend it so far, is profitable.'71, Multiply 25 X 25.'72. Multiply 54X 54. 73. Multiply 81 X 64. 74. Multiply 45 X 92. I''5. Multiply 194 X 144. 76. Multiply 1234X 125.'7'. Multiply 4825 X 2352. 78. Multiply 6521 X 5312. 1OS. By suitable attention, the critical student will discovet Various other methods of abbreviating the processes of multiplication. Solve the following examples, contracting the operations when practicable. 79. 42634X63. 99. 12900X14000. 80. 50035X56. 100. 64172X42432. 81. 72156X1000. 101. 26815678 X 81. 82. 42000 X 40000. 102. 85X85. 83. 80000X 25000. 103. 256X 256. 84. 2567345X1'7. 104. 322X325. 85. 4300450X 19. 105. 5234X2435. 86. 9803404X41, 106. 48743000X 637. 87. 6710045X71 107. 31890420X85672. 88. 3456710X18 108. 80460000 X 2763. 89. 7000541X91. 109. 2364793X 8485672. 90. 4102034X 99. 110. 1256702X 999999. 91. 42304X999. 111. 6840005 X 91X 61. 92. 50421 X 9999. 112. 45067034 X 17X51 93. 67243X 99999. 113. 788031245X 81X16. 94. 78563X93. 114. 61800000X23000. 95. 34054X639. 115. 12563000X4800000. 96. 52156 X 75. 116. 91300203X1000000. 97. 41907 X 54486 117. 680040000 X 1000000. 98. 263977X 24648. 118. 4000000000X 100000a, 6* (16 DIVISION. [SEcT. V. SECTION V. DIVISION. ART. 1 10. Ex. 1. How many barrels of flour, at 8 dollar per barrel, can you buy for 56 dollars? Analysis. —Since flour is 8 dollars a barrel, it is obvious you can buy 1 barrel as often as 8 dollars are contained in 56 dollars; and 8 dolls. are contained in 56 dolls. 7 times. Ans. 7 barrels. Ex. 2. A man wished to divide 72 dollars equally among 9 beggars: how many dollars would each receive? Solution.-Reasoning as before, each beggar would receive as many dollars as 9 is contained times in 72; and 9 is contained in 72, 8 times. Ans. 8 dollars. OBs. The learner will at once perceive that the object in the first example, is to find how manny times one number is contained in another; and that the object of the second, is to divide a given number into equal parts, but its solution consists in finding how many times one number is contained in another, and is the same in principle as that of the first. 1 1 1. The Process of Jinding how many times one number is contained in another, is called DIVISIOT\. The number to be divided, is called the dividend. The number by which we divide, is called the divisor. The number obtained by division, or the answer to the question, is called the quotient. It shows how many times the divisor is contained in the dividend. Hence, it may be said, 1 1 2. D)ivisionz is finding a quotient, which multiplied into the divisor, will produce the dividend. Note.-The term quotient is derived from the Latin word guoties, which sig. nifies how often, or how mcany times. QUEST.-111. What is division? What is the number to be divided called? The nuin. ber by which we divide? What is the nunlber obtained called? What does the quotient show 3 112. What then may division be said to be? ARTS. 110-114.] DIVISION. 67 1 13. The number which is sometimes left after divison, is called the remainder. Thus, when we say 5 is contained?n 38, 7 times, and 3 over, 5 is the divisor, 38 the dividend, 7 the quotient, and 3 the remainder. Ohs. 1. The remainder is of the same denomination as the dividend; for, it is a part of it. 2. The remainder is always less than the divisor; for, if it were equal to, or greater than the divisor, the divisor could be contained once more in the dividend. 1 1 4L. It will be perceived that division is similar in principle to subtraction, and may be performed by it. For instance, to find bow many times 7 is contained in 21, subtract 7 (the divisor) continually from 21 (the dividend), until the latter is exhausted; then counting these repeated subtractions, we shall have the true quotient. Thus, 7 from 21 leaves 14; 7 from 14 leaves 7; and 7 from 7 leaves 0. Now by counting, we find that 7 has been taken from 21, 3 times; consequently, 7 is contained in 21, 3 times. Hence, Division is sometimes defined to be a short way of performing repeated subtractions of the same number. OBs. 1. It will be observed that division is the reverse of multiplication. Multiplication is the repeated addition of the same number; division is the repeated subtraction of the same number. The product of the one answers to the dividend df the other; but the latter is always given, while the former is required. 2. When the dividend denotes things of one denomination only, the opera tion is called Simple Diviswio. SHORT DIVISION. Ex. 3. How many hats, at 2 dollars apiece, can be bought for 4862 dollars? Opercatio~n. V~We write the divisor on the left of the diviDivisor. Divid. dend with a curve line between them; then, 2) 4862 beginning at the right hand, proceed thus: 2 is Quot. 2431 contained in 4, 2 times. Now, since the 4 de QUEST.-113. What is the number called which is sometimes left after division. C2bs. Of what denomination is the remainder? Why? Is the remainder greater or less than the divisor? Why? 114. To what rule is division similar in principle? Obs. Of what is division the reverse. When the dividend denotes things of one denomination only, what is the operation called 1 4 68 DIVISION. [SECT. V notes thousands, the 2 nust be thousands; we therefore write it in thousands' place, under the figure divided. 2 is contained in 8, 4 times; and as the 8 is hundreds, the 4 must also be hundreds; hence we write it in hundreds' place, under the figure divided. 2 in 6, 3 times; the 6 being tens, the 3 must also be tens, and should be set in tens' place. 2 in 2, once; and since the 2 is units, the 1 is a unit, and must therefore be written in units' place. The answer is 2431 hats. 1 15. When the process of dividing is carried on in the mind, and the quotient only is set down, as in the last example, the operation is called SH0aOT DIVISION. 1 16. The reason that each quotient figure is of the same order as the Jigure divided, may be shown in the following manner: Having separated the dividend Analytic Solution. of the last example into the orders 4862=4000+800+60+2 of which it is composed, we per2)4000+800 +602 ceive that 2 is contained in 4000, 2000+400+30+1 2000 times; for 2X2000-4000, Again, 2 is contained in 800, 400 times; for 2X400=800, &c. Ans. 2431. Ex. 4. A man left an estate of 209635 dollars, to be divided equally among 4 children: how much did each receive? Since the divisor 4, is not contained in Operation. 2, the first figure of the dividend, we find 4)209635 how many times it is contained in the first Ans. 52408M dolls. two figures. Thus, 4 is contained in 20, 5 times; write the 5 under the 0. Again, 4 is contained in 9, 2 times and 1 over; set the 2 under the 9. Now, as we have 1 thousand over, we prefix it mentally to the 6 hundreds, making 16 hundreds; and 4 in 16, 4 times. Write the 4 under the 6. But 4 is not contained in 3, the next figure, we therefore put a cipher in the quotient, and prefix the 3 to the next figure of the dividend, as if it were a remainderi Then 4 in 35, 8 times and 3 over; place the 8 under the 5, and setting the remainder over the divisor thus a, place it on the right of the quotient. Note.-To prefix means to place before, or at the left hand. ARTS. 115-118.] DIVISION. 69 1 17. When the divisor is not contained in any figure of the dividend, a cipher must always be placed in the quotient. Os. The reason for placing a cipher in the quotient, is to pre-erve the trate local value of each figure of the quotient. (Art. 116.) 1 18. In order to render the division complete, it is obvious that the whlole of the dividend must be divided. But when there is a remainder after dividing the last figure of the dividend, it must of necessity be smaller than the divisor, and cannot be divided by it. (Art. 113. Obs. 2.) We therefore represent the division by placing the remainder over the divisor, and annex it to the quotient. (Art. 28.) OBs. 1. The learner will observe that in dividling, we begin at the left hanit, instead of the righct, as in Addition, Subtraction, and Multiplication. The reason is, because there is frequently a remainder in dividing a higher ordler, which must necessarily be united with the next lower order, before the division can be performed. 2. The divisor is placed on the left of the dividend, and the quotient under it, merely for the sake of convenience. When division is represented by the sign,, the divisor is placed on the right of the dividend; and when represented in the form of a fraction, the divisor is placed qtunder the dividend. LONG DIVISION. Ex. 5. At 15 dollars apiece, how many cows can be bou(llt for 3525 dollars? Opveration. Having written the divisor on the left of Divisor. Divid. Quot. the dividend as before, We find that 15 is 15) 3525 (235 contained in 35, 2 times, and place the 2 on 30 the right of the dividend, with a curve line 52 between them. We next multiply the di- 45 visor by this quotient figure, place the prod- 75 uct under the figures divided, and subtract 75 it therefrom.'We now bring down the next figure of the dividend, and placing it on the right of the remainder 5, we perceive that 15 is contained in 52, 3 times. Set the 3 on the right of the last quotient figure, multiply the divisor by it, and subtract the product from the figures divided as before. We then 70 1IV1SION. [SECT. V brijng down the next, which is the last figure of the dividend, to the right of this remainder, and finding 15 is contained in 75, 5 times, we place the 5 in the quotient, multiply and subtract as before. The answer is 235 cows. 1 1 9. WI7en the result of each step in the operation is written down, as in the last example, the process is called LONG DIVISION. Long Division is the same in principle as Short Division. The only difference between them is, that in the former, the result of each step in the operation is written down, while in the latter, we carry on the process in the mind, and simply write the quotient. OBs. 1. When the divisor contains but one figure, the operation by S&hort DivisioTn is the most expeditious, and therefore should always be practiced; but when the divisor contains two or more figures, it will generally be the most convenient to use Long Division. 2. To prevent mistakes, it is advisable to put a dot under each figure of the dividend, when it is brought down. 3. The PSrenc/h place the divisor on the rig/t of the dividend, and the quotif,nt below the divisor, * as seen in the following example. Ex. 6. How many times is 72 contained in 5904? Operation. 5904 (72 divisor. The divisor is contained in 590, the 576 82 quotient. first three figures of the dividend, 8 144 times. Set the 8 under the divisor, 144 multiply, &c., as before. Ex. 7. Hlow many times is 435 contained in 262534? Operation. Since the divisor is not contained 435)262534(603{ } Ans. in the first three figures of the divi2610"' dend, we find how umany times it is 1534 contained in the first four, the few1305 est that will contain it, and write 229 rem. the 6 in the quotient; then multiQuEsTr.-115. What is short division? 119. What is long division? What is the difference between them? * Ellnents D'Arithmntique, par M. Bourdon. Also, Lacroix's Arithinetic, translated by Professor Farrar. ARTS. 119, 120.] DlVISION. 71 plying and subtracting as before, the remainder is 15. Bringing down the next figure, we have 153 to be divided by 435. But 435 is not contained in 153; we therefore place a cipher in tho quotient, and bring down the next figure. Then 435 in 1534, X times. Place the 3 in the quotient, and proceed as before. Note. —After the first quotient figure is obtained, for each Jfigurqe of the divi dend owhich is br'ought down, either a sigificant fig4ure, or a cipher, must be pu, in the quotient. (Art. 116.) 1 20. Fionm the preceding illustrations and principles we derive the following GENERAL RULE FOR DIVISION. I. When the divisor contains but one figure. WTi-ite the divisor on the left of the dividend, with a curve line be otueen them. Begin at the left hand, divide successively each, Jigurle of the dividend by the divisor, and pilace each quotient filure directly under the figure divided. (Arts. 116, 118. Obs. 1, 2.) If there is cc remcainder after dividing any figure, 2reitx it to tho next figure of the dividend and divide this number as beJfre; and if the divisor is not contained in any figcure of the dividend, place a cip2her in the quotient and prefix this figure to the next one of the dividend, as if it were a remainder. (Arts. 117, 118.) II. When the divisor contains more than one figure. Beginning on the left of the dividend, find how many times the divisor is contained in the fewest figures that will contain it, and _place the quotient Jfigure on the right of the dividend with, a curve line between them. Then mnultijly the divisor by this figure and subtract the product from the Jfigures divided; to tihe right of the recmainder bring doown the next figure of the dividend and divide this nunmber as before. Proceed in this manner till all the figuret of the dividend are divided. QUEST.-12C. How do you write the numbers for division? When the divisor conlains but one figure, how proceed? CWahy place the divisor on the left of the dividend and the quotient under' the figure divided? When there is a remainder after dividing a figure, what is to be done with it? When the divisor is not contained In any figure of the divi dend. how proceed? Why? Why begin to divide at the left hand? When the divisor contains more than one figure, how proceed I 72 DIVISION [SECT. V. Whenever there is a remainder after dividing the last figure, write it over the divisor and annex it to the quotient. (Art. 118.) Dczems/tLra!ioe. —-The principle on which the operations ill Division depend, Is that ca part of the quotient is fobund, and the product of this part into thle divisor is taken from the: dividend, showing how much of the latter remains to be divided; then cezo'teir part of the quotient is found, and its product into the divisor is taken firom what remained before. Thus the operation proceeds till the whovle of the dividend is dividedl, or till the resmain,7der is less than th/e divis(. (Art. 113. Obs. 2.) Ohs When the divisor is large, the pupil will find assistance in determining the quotient fignure, by filling how many times the first figure of the divisor is contained in the first figure, or if necessary, tlle first two figures of the dividendl. Ttnis will give p)retty nearly the right figure. Some allowance must, however, be made for carrying from the product of tlhe other figures of the divisor, to tile product of the first into the quotient figure. 1 2. PRooF.-i~Ualtiply the divisor by the quotient, to the product add the remainder, and if the sum is equal to the dividend, the workC is right. Ors. Since the quotient shows how many times the divisor is contained in the dividend, (Art. 111,) it follows, that if the divisor is repeated as many times as there are units in the quotient, it must produce the dividend. Ex. 8. Divide 256329 by 723. OQieration. Proof. 723)256329(3543a4j Ans. 723 divisor. 2169 354 quotient. 3942 2892' 3615 3615 3279 2169 2892 387 rem. 387 rem. 256329 dividend. 122. Second.fethod.-Subtract the remainder, if any, fronm the dividend, divide the dividend thus diminished, by the quotient and if the result is equal to the given divisor, the work is right. Q,:EsT. —Vhen there is a remainder after dividing the last figure of the dividend, what murst be (ldone with it? 121. How is division proved? Obs. Iow does it appear that the product of the divisor and quotient will beequal to the dividend, if the work is right I Can division be provetl by any other methods? ARTS. 121-127.1 DIVISION. 73 123. Third M~ethod.-First cast the 9s out of the divisor and quotient, and multiply the remainders together; to the product add the remainder, if any, after division; cast the 9s out of this sum, and set down the excess; finally cast the 9s out of the dividend, and if the excess is the same as that obtained from the divisor and quotient, the work may be considered right. lNote.-Since the divisor and quotient answer to the multiplier and multipliulanld, and the dividend to the product, it is evident that the principle of casting out the 9s will apply to the proof of division, as well as that of multiplication. (Art. 90.) 124. Fourt)'lth el/oeltvd.-Add the remainder and( the respective products of she divisor into each quotient figure together, and it the S,'riL is equal to the dividend, the work is'ig/ct. Note.-This mode of proof depends upon the principle that the shtole of a /lieatitly is eqeal to tale sesc of all its pefs. (Ax. I1.) 1.2 i Fif/ Method.-First cast the 11 s out of the divisor and quotient, and multiply the remainders together; to the product add the remainder, if any, after division, and casting the 11s out of this sum, set down the excess; finally, cast the lUs out of the dividend, and if the excess is the same as that obtained from the divisor and quotient, the work is ringht. (Art. 92. Note 2.) EXAMPLES FOR PRACTICE. 127. Ex. 1. A farmer raised 29T5 bushels of wheat on 45 acres of land: how many bushels did he raise per acre? 2. A garrison consumed 8925 barrels of flour in 105 days: how much was that per day? 3. The President of the United States receives a salary of 25000 dollars a year: how much is that per day? 4. A drover paid 2685 dollars for 895 head of cattle: how muchl did he pay per head? 5. If a man's expenses are 3560 dollars a year, how much are they per week? 6. If the annual expenses of the government are 27 millions of dollars, how much -will they be per day? 7. How long will it take a ship to sail from New York to Liverpool, allowing the distance to be 3000 miles, and the ship to sail 144 miles per day? 8. Sailing at the same rate, how long would it take the same ship to sail round the globe, a distance of 25000 miles? 7 74 DIVISION. [SECT.' 10. 47839. +42. 25. 1203033. 327. 11. 75043 — 52. 26. 1912500 425. 12, 93840. 63. 27. 5184673 -102. 13. 421645 —7 4. 28. 301140 —478. 14, 325000. 85. 29. 8893810.37846. 15. 400000- 96. 80. 9302688- 14356. 16. 999999+-47. 31. 9749320- 365. 17. 352417 —29. 32..3228242 5734. 18. 47981 + 251. 33. 75843639426-+ 8593. 19. 423405 485. 34. 65358547823. 2789. 20. 16512-.344. 35. 102030405060- 123456. 21. 304916- 6274. 36. 908070605040+ 654321. 22. 12689-. 145. 37. 1000000000000000 111. 23. 145260- 1345. 38. 1000000000000000 — 1111. 24. 147735 + 3283. 39. 1000000000000000+ 11111 CONTRACTIONS IN DIVISION. 1 280 The operations in division, as well as those in multipli. cation, may often be shortened by a careful attention to the application of the preceding principles. CASE 1.- When the divisor is a conmposite number. Ex. 1. A man divided 837 dollars equally among 27 persons, who belonged to 3 families, each family containing 9 persons: how many dollars did each person receive? Analysis.-Since 27 persons received 837 dollars, each one must have received as many dollars, as 27 is contained times in 837. But as 27 (the number of persons), is a composite number whose factors are 3 (the number of families), and 9 (the number of persons in each family), it is obvious we may first find how many dollars each family received, and then how many each person received, Operation. If 3 families received 837 3)837 whole sum divided. dollars, 1 family must have 9)279 portion of each Fam. received as many dollars, as Ans. 31 " "' person. 3 is contained times ir 837 and 3 in 837, 279 times. That is, each family received 279 dllars ARTS. 128, 129.] DIVISION. 75 Again, if 9 persons, (the number in each family,) received 279 dollars, 1 person must have received as many dollars, as 9 is contained times in 279; and 9 in 279, 31 times. Ans. 31 dollars. PROOF.-31X 27=837, the same as the dividend. HIence, 1 29. To divide by a composite number. I. -Divide the dividend by one of the factors of the divisor, then tlivide the quotient thus obtained by another factor; and so on till ll the factors are employed. The last quotient will be the answer. II. To find the true remainder. if' the divisor is resolved into but two factors, multiply the last remainder by the first divisor, to the product add the first renmainder, if any, and the result will be the true remainder. TWhen more than two factors are employed, multiply each remainder by all the preceding divisors, to the sum of their prodzucts, add the first remainder, and the result will be the true remzainlder. OBs. 1. 1. The true remainder may also be found by multiplying the quotient by the divisor, and subtracting the product from the dividend. 2. This contraction is exactly the reverse of that in multiplication. (Art. 97.) The result will evidently be the same, in whatever order the factors are taken. 2. A man bought a quantity of clover seed amounting to 507 pints, which he w;shed to divide into parcels containing 64 pints each': 0how Ineay parcels can he make? Nole.-Since 64=/X2X8X4, we divide by the factors respectively. Operaltion. 2)507 8)253-1 rem. = 1 pt. 4)31 —5 rem. No;v 5X 2 = 10 pts. 7-3 rem. and 3X8X2 = 48 pts. Adns. 7 parcels, and 59 pts. over. 59 pts. True Rem. Demonsl-rat'iol,.-1. Dividing 507 the number of pints, by 2, gives 253 for the quotient, or distributes the seed into 253, equal parcels, leaving 1 pint over. Nowv the units of this quotient are evidently of a different value from those of the given dividend; for since there are but Italf as many parcels as at first, it CQuEsT. —129. tlow proceed when the divisor is a colmposite number? Itow find the true remlainder X 4* 76 DIVISION. SECT. V. is plain that each parcel must contain 2 pints, or I quart; that is, every unit of the fi:st quotient contains 2 of the units of the given dividend; consequently, every unit of it that remains will contain the same; (Art. 113. Ohs. 2;) therefore this remainder must be multiplied by 2, in order to find the units of the given dividend which it contains. 2. Dividing the quotient'253 parcels, by 8, will distribute them into 31 other equal parcels, each of which will evidently contain 8 times the quantity of the preceding, viz: 8 times I quart =8 quarts, or I peck; that is, every unit of th second quotient contains 8 of the units in the first quotient, or 8 times 2 of th units in the given dividend; therefore what remains of it, must be multiplied by 8X2, or 16, to find the units of the given dividend which it contains. 3. In like manner, it may be shown, that dividing by each successive factor reduces each quotient to a class of units of a higher value than the preceding; that every unit which remains of any quotient, is of the same value as that quotient, and must therefore be multiplied by all the preceding divisors, in order to find the units of the given dividend whi'ch it contains. 4. Finally, the several remainders being reduced to the same units as those of the given dividend according to the rule, their sum must evidently be the tree remainder. (Ax. 11.) 3. How many acres of land, at 35 dollars an acre, can you buy for 4650 dollars? 4. Divide 16128 by 24. 5. Divide 25760 by 56. 6. Divide 17220 by 84. 7. Divide 91080 by 72. CAxsi: II. —rlhen thle divisor is 1 with ciphers annexed to it. 1 30. It has been shown that annexing a cipher to a number increases its value ten tisnes, or multiplies it by 10. (Art. 98.) Reversing this process; that is, removing a cipher from the right hand of a number, will evidently diminish its value ten tines, or divide it by 10; for, each figure, in the number is thus restored to its original place, and consequently to its original value. Thus, annexing a cipher to 15, it becomes 150, which is the same as 15X 10. On the other hand, removing the cipher from 150, it becomes 15, which is the same as 150 10. In the same manner it may be shown, that removing tzwo ciphers from the right of a number, divides it by 100; removing tl)hee, divides it by 1000; removingfour, divides it by 100OO0, &c. Hence, QUFST -130. What is the effect of annexing a cipher to a number? ~What is the effect of removing a cipher from the right of a numlber? How does this appear X ARTS. 130-132.] DIVISION. 77 1. 31. To divide by 10, 100, 1000, &c. Cut of' as many figures from the right hand of the dividend as there are ciphers in the divisor. The remaining figures of the dividezd will be the quotient, and those cut of the?remainder. 8. In one dime there are 10 cents: how many dimes are there in 200 cents? In 340 cents? In 560 cents? 9. In one dollar there are 100 cents: how many dollars are there in 65000 cents? In 7605000 cents? In 4320000 cents? 10. Divide 26750000 by 100000. 11. Divide 144360791 by 1000000. 12. Divide 582367180309 by 100000000. CASE III. —Then the divisor has cithAers on the right hand. 13. How many hogsheads of molasses, at 30 dollars apiece, can you buy for 9643 dollars? Ors. The divisor 30, is a composite number, the factors of which are 3 and 10. (Arts. 95,96.) We may, therefore, divide first by one factor and the quotient thence arising by the other. (Art. 129.) Now cutting off the right Iand figure of the dividend, divides it by ten; (Art. 131;) consequently dividiiow the remaining figures of the dividend by 3, the other factor of the divisor,,1l give the quotient. Op1eration. We first cut off the cipher on the right 310)96413 of the divisor, and also cut off the right 321 - iAns. hand figure of the dividend; then dividing 964 by 3, we have 1 remainder. i~)ow as the 3 cut off, is part of the remainder, we therefore - nex it to the 1. Ans. 321J — hogsheads. Hence, 1 32. When there are ciphers on the right hand of the divisor. Cat ol' the ciphers, also cut of' as many figures from the right of the dividend. Then divide the other figures of the dividend by the remaining figures of the divisor, and annex the figures cut off from the dividend to the recmainder. 14. H-ow many buggies, at 70 dollars apiece, can you buy foa 7350 dollars? QU-EsT. —131. How proceedl when the divisor is 10, 100, 1000, &c. 3 1`2. When there are ciphers oil the right bhand of the divisor, how proceed. W hat is to bo done with figures cut off frolm the dividend? A.~b 78 DIVISION. [SECT. V. 15. HIow many barrels will it take tc pack 36800 pounds of pork, allowing 200 pounds to a barrel? 16. Divide 3360000 by 17000. 133. Operations in Long Division may be shortened by subtracting the product of the respective figures in the divisor into each quotient figure as we proceed in the operation, setting down the remainders only. This is called the Italian J~ethod. 17. How many times is 21 contained in 4998? Operation. 21)4998(238 This method, it will be seen, requires a much 79 smaller number of figures than the ordinary 168 process. 18. Divide 1188 by 33. 19.. Divide 2516 by 37. 20. Divide 3128 by 86. 21. Divide 7125 by 95. 22. A merchant laid out 873 dollars in flour, at 5 dollars a barrel: how many barrels did he get? Operation. We first double the dividend, and then di873 vide the product by 10, which is done by 2 cutting off the right hand figure. (Art. 131.) 110)17416 But since we multiplied the dividend by 2, it 174 - Ans. is plain that the 6 cut off, is 2 times too large for the remainder; we therefore divide it by 2, and we have 8 for the true remainder. Hence, 1 34:. When the divisor is 5. Multiply the dividend by 2, and divide the product by 10. (Art. 131.) Note.-I. When the figure cut off is a significant figure, it must be divided by 2 for the terue remainder. 2. This contraction depends upon the princ;ple that any given divisor is contained in any given dividend, just as many times as twice that divisor is contained in twice that dividend, three times that divisor in three times that dividend, &c. For a further illustration of this principle see General Pr'pciplea in Division. 23. Divide 6035 by S5. 24. Divide 8450 by 5. 25. Divide 32561 by 5. 26. Divide 43270 ty 5. A.RTS. 133-139.] DIVISION. 79 1 35. When the divisor is 15, 35, 45, or 55. D.ouble the dividn d, and divide the product by 30, 70, 90, or 110, as the case may be. (Art. 132.) Note.-This method is simply doubling both the divisor and dividend. We must therefore divide the remainder, if any, by 2, for the true remainder. 27. Divide 1256 by 15. 28. Divide 2673 by 35. 29. Divide 3507 by 45. 30. Divide 7853 by 55. 1 36. When the divisor is 25. M'ultiply the dividend by 4, and divide the product by 100. (Art. 131.) Note.-This is obviously the same as multiplying both the dividend and divisor by 4. (Art. 134. Note 2.) Hence, we must divide the remainder, if any thus found, by 4, for the true remainder. 31. Divide 2350 by 25. 32. Divide 4860 by 25. 33. Divide 42340 by 25. 34. Divide 94880 by 25. 137, To divide by 125. Multiply the dividend by 8, and divide the product by 1000. (Art. 131.) Note.-This contraction is multiplying both the dividend and divisor by 8. For the true remainder, therefore, we must divide the remainder, if any, by 8. 35. Divide 8375 by 125. 36. Divide 25426 by 125. 138. To divide by 75, 175, 225, or 275. M'ultiply the dividend by 4, and divide the product by 300, 700 900, or 1100, as the case may be. (Art. 132.) Note.-For the t'rue remainder, divide the remainder, if any thus found, by 4 37. Divide 1125 by 75. 38. Divide 2876 by 175. 39. Divide 3825 by 225. 40. Divide 8250 by 275. 1 39. The preceding are among the most frequent and useful modes of contracting operations in division. Various other methods might be added, but they will naturally suggest themselves to the inventive student, as opportunities occur for their application. 41. HIow long wvould it take a vessel sailing 100 miles per day to circumnavigate the earth, whose circumnference is 25000 miles? S0 DIVISION. [SECT. V. 42. The distance of the Earth from the Sun is 95,000,000 of miles: how long would it take a balloon going at the rate of 100,000 miles a year, to reach the sun? 43. The debts of the several States of the Union, in 1840, amounted to 171,000,000 of dollars, and the number of inhabitants was 17,000,000: how much must each individual have been taxed to pay the debt? 44. The national debt of Holland is 800,000,000 of dollars, and the number of inhabitants 2,800,000: what is the amount of indebtedness of each individual? 45. The national debt of Spain is 467,000,000 of dollars, and the number of inhabitants 11,900,000: what is the amount of indebtedness of each individual? 46. The national debt of Russia is 150,000,000 of dollars, and the number of inhabitants 51,100,000: what is- the amount of indebtedness of each individual? 47. The national debt of Austria is 380,000,000 of dollars, and the number of inhabitants 34,100,000: what is the amount of indebtedness of each individual? 48. The national debt of France is 1,800,000,000 of dollars, and the number of inhabitants 33,300,000: what is the amount of indebtedness of each individual? 49. The national debt of Great Britain is 5,556,000,000 of dollars, and the number of inhabitants 25,300,000: what is the amount of indebtedness of each individual? 50. Divide 467000000000 by 25000000000. 51. 568240-.42. 62. 462156- 75. 52. 785372-+-63. 63. 3562189+ 225. 53. 896736 72. 64. 685726- 32000 54. 67234568. 5. 65. 723564- 175. 55. 34256726- 15. 66. 892565 225. 56. 42367581 — 45. 67. 456212- 275. 57. 16753672 35. 68. 925673-. 125. 58. 3256385-+ 55. 69. 763421.175. 59. 45672400 25. 70. 876240 275. 60. 6245634-:-45. 71. 7825600- 80000. 61. 8245623- 125. 72. 92004578 — 100000. ARTS. 139-143.] DIVISION. 81 GENERAL PRINCIPLES IN DIVISION. 1 40. From the nature of division, it is evident, that the value of the quotient depends both on the divisor and the dividend. 1410. If a given divisor is contained in a given dividend a certain number of times, the same divisor will obviously be contained, In double that dividend, twice as many times. In three times that dividend, thrice as many times, &c. Hence, If the divisor remains the same, multiplying the dividend by any number, is in effect multiplying the quotient by that number. Thus, 6 is contained in 12, 2 times; in 2 times 12 or 24, 6 is contained 4 times; (i. e. twice 2 times;) in 3 tunes 12 or 36, 6 is contained 6 times; (i. e. thrice 2 times;) &c. 1 42. Again, if a given divisor is contained in a given dividend a certain number of times, the same divisor is contained, In hallf that dividend, half as many times; In a third of that dividend, a third as many times, &c. Hence, If the divisor remains the same, dividing the dividend by any number, is in effect dividing the quotient by that number. Thus, 8 is contained in 48, 6 times; in 48. 2 or 24, (half of 48,) 8 is contained 3 times; (i. e. half of 6 times;) in 48 3 or 16, (a third of 48,) 8 is contained 2 times; (i. e. a third of 6 times;) &c. 143. If a given divisor is contained in a given dividend a certain number of times, then, in the same dividend, Twice that divisor is contained only half as many timaes; Three times that, divisor, a third as many times, &c. Hen-ce,.If the dividend remains the same, multiplying the divisor lby any number, is in effect dividing the quotient by that number. Thus, 4 isa contained in 24, 6 times; 2 times 4 or 8 is ctn QUEST.-140. Upon what does the value of the quotient depend? 141. If the divisor remains the same, what is the effect on the quotient to multiply the dividend? 142. What is the effect of dividing the dividend by any given number? 143. If the divide ld remains the same, what is the effect of multiplying the divisor by any given number? 82 DIVISION. [SECT. V. tained in 24, 3 times; (i. e. half of 6 times;) 3 times 4 or 12 is contained in 24, 2 times; (i. e. a third of 6 times;) &c. 144. If a given divisor is contained in a given dividend a certain nuimber of times, then, in the same dividend, Half that divisor is contained twice as many times; A third of that divisor, three times as many times, &c. Hence, If the dividend remains the same, dividing the divisor by any number, is in efect multilplying the quotient by that number. Thus, 6 is contained. in 36, 6 times; 6- 2 or 3, (half of 6,) is contained in 36, 12 times; (i. e. twice 6. times;) 6.3 or 2, (a third of 6,) is contained in 36, 18 times; (i. e. thrice 6 times;) &c. 1 45. From the preceding articles, it is evident that any given divisor is contained in any given dividend, just as many times as twice that divisor is contained in twice that dividend; three times that divisor in three times that dividend, &c. Conversely, any given divisor is contained in any given dividend just as many times, as half that divisor is contained in half that dividend; a third of that divisor, in a thi'rd of that dividend, &c. Hence, 1 46. If the divisor and dividend care both multiplied, or both divided by the same number, the quotient will not be altered. Thus, 6 is contained in 12, 2 times; 2 times 6 is contained in 2 times 12, 2 times; 3 times 6 is contained in 3 times 12, 2 times, &c. Again, 12 is contained in 48, 4 times; 12 —2 is contained in 48- 2, 4 times; 12 -3 is contained in 48 -3, 4 times; &c. 1 47. If the sum of two or more numbers is divided by any number, the quotient will be equal to the sum of the quotients which will arise from dividing the given numbers separately. Thus, the sum of 12-+-18=30; and 30- 6=-5. Now, 12 — 6=2; and 18 —6-3; but the sum of 2 +3=5. Again, the sum of 32+24+40 —96; and 96- 8=-12. Now, 32 —8-4; 24 8-3; and 40+8=5; but 4,-34-5= J2. Qr!EST.-144. W~hat of dividing the divisor? 146. What is the efiTect nl.'J the quotient If the divisor and dlividend are both multiplied, or both divided by the srame number 7 ,RTS. 144-150.] CANCELATION. 83 CANCELATION.* 148. We have seen that division is finding a quotient, which, multiplied into the divisor, will produce the dividend. (Art. 112.) If, therefore, the dividend is resolved into two such factors that one of them is the divisor, the other factor will, of course, be the quotient. Suppose, for example, 42 is to be divided by 6. Now the fiactors of 42 are 6 and 7, the first of which being the divisor, he other must be the quotient. Therefore, Canceling a factor of any number, divides the number by that factor. Hence, t 49. When the dividend is the product of two factors, one of which is the same as the divisor. Cancel the factor common to the dividend and divisor; the other factor of the dividend will be the answer. (Ax. 9.) Note.-The term cancel, signifies to erase or reject. 1. Divide the product of 34 into 28 by 34. Common.Method. By Cancelation. 34 $X)24x28 28 28 Ans. 68 Canceling the factor 34, which is com34)952(28 Ans. mon both to the divisor and dividend, we 68 hlave 28 for the quotient, the same as be272 fore. 272 1 50. The method of contracting arithnetical operations, by rejecting equal factors, is called CANCELATION. Ons. It applies with great advantage to that class of examples and problems, which involve both multiplication and division; that is, which require the product of two or more numbers to be divided by another number, or by the product of two or more numbers. 2 Divide 76X45 by 76. 3. Divide 63X81 by 81. 4. Divide 65 X 82 by 82. 5. Divide 95 X 73 by 95. 6. Divide the product of 45 times 84 by 9. * Birk's Arithmetical Collections: London, 1764. 84 APPLICATIONS OF [SECT. V. AnaTysis.-The factor 45=5X9; hence the dividend is composed of the factors 84 X 5 X 9. We may therefore cancel 9, which is common both to the divisor and dividend, and S4 X 5, the )ther factors of the dividend, will be the answer required. Op eration. Proof. )8 x 5 x 84X5X9=-3780 420 Ans. And 3780 —9=420. 7. Divide the product of 45X6 X3 by 18 X 5. Operation. Proof.,lX545XOX$ 45X6X3=810; and 18X5= 90 9 Ans. Now, 810 — 90-9 Note.-We cancel the factors 6 and 3 in the dividend and 18 in the divisor; for 6X3=18. Canceling the same or equal factors in the divisor and dividend, is dividing them both by the same number, and therefore does not affect the quotient. (Arts. 146, 148.) Hence, 1 5 1. When the divisor and dividend have common factors. Cancel the factors common to both; then divide the product of those remaining in the dividend by the product of those remaining in the divisor. 8. Divide 15X7X12 by 5X3x7X2. 9. Divide 27X3X4X7 by 9X12X6. 10. Divide 75X15X24 by 25X3X6x4X5. Note.-The further development and application of the principles of Cancelation, may be seen in reduction of compound fractions to simple ones; in multiplication and division of fractions; in simple and compound proportion, &c. 1 5 1. a. The four preceding rules, viz: Addition, Subtraction, Multiplication, and Division, are usually called the FUNDAMENTAI RULES of Arithmetic, because they are the foundation or cadsis ot all arithmetical calculations. OBS. Every change that can be made upon the value of a number, must necessarily either increase or diminish it. Hence, the fundamental operations in arithmetic are, strictly speaking, but two, addition and subtractiorn; that is increase and decrease. Multiplication, we have seen, is an abbreviated form of addition; division of subtraction. (Arts. 80, 114.) QUEssT.-151. a. Name the fundamental rules of Arithmetic. IWhy are these rules called hfndimental? ARTS. 151-154.] FUNDAMENTAL RULES. 85 APPLICATIONS OF THE FUNDAMENTAL RULES. 1 5. Wihen the suml of two numbers and one of the numbers are given, to find the other number. From} the given sum, subtract the given number, and the remainder will be the other number. Ex. 1. The sum of two numbers is 87, one of which is 25: what is the other number? Solution.-87 —25=-62, the other number. (Art. 72.) PRooF. —62t+25=87, the given sum. (Ax. 11.) 2. A and B together own 350 acres of land, 95 of which belong- to A: how many does B own? 3. Two merchants bought 1785 bushels of barley together, one of them took 860 bushels: how many bushels did the other have? 1 5 3. When the difference and the greater of two numbers are given, to find the less. Subtract the difference from the greater, and the remainder will b,? the less number. 4. The greater of two numbers is 72, and the difference between them is 28: what is the less number? Solution. —72-28=44, the less number. (Art. 72.) PRooF. —44+28 —72, the greater number. (Art. 73. Obs.) 5. A man bought a horse and chaise; for the chaise he gave 2G65 dollars, which was 75 dollars more than he paid for the horse: how much did he give for the horse? 6. A traveler met two droves of sheep; the first contained 1250, which was 125 more than the second had: how many sheep were there in the second drove? 1 5 4. When the difference and the less of two numbers ar given, to find the greatern QUEST. —152. When the sum of two numbers and one of them are given, how is the other follund 2 153. When the difference and the greater of two numbers are giver, how is the less folund? 154. When the difference and the less of two numbers are given, how is the greater found? 8 86 APPLICATIONS OF [SECT. V Add the difference and the less number together, and the sum will be the greater number. (Art. 73. Obs.) 7. The difference between two numbers is 12, and the less number is 45: what is the greater number? Solution.-45-+12 =57, the greater number. PROOF.-57 —45= 12, the given difference. (Art. 72.) S. A is worth 1890 dollars, and B is worth 350 dollars ni(rc than A; how much is B worth? 9. A man's expenses are 2561 dollars a year, and his income exceeds his expenses 875 dollars: how much is his income? 1 5 5. When the sum and difference of two numbers are given, to find the two numbers. From the sum subtract the difference, divide the remainder by 2, and the quotient will be the smaller number. To the smaller number thus found, add the given difference, and the sum will be the larger number. 10. The sum of two numbers is 48, and their difference is 18: what are the numbers? Solution.-48-18=30, and 30.-2=15, the smaller number. And 15+18=33, the greater number. PRooF.-33+15=48, the given sum. (Ax. 11.) 11. The sum of the ages of two men is 173 years, and the difference between them is 15 years: what are their ages? 12. A man bought a span of horses and a carriage for 856 dollars; the carriage was worth 165 dollars more than the horses: what was the price of each? 156. When the product of two numbers and one of the mlmbers are given, to find the other number. Divide the given product by the given number, and the luotient will be the number required. (Art. 91.) QUEST. —155. When the sum and difference of two numbers are given, how are the numbers found? 156. When the product of two numbers and one of them are given, now is the other found? ARTS. 155-158.] FUNDAMENTAL RULES. 87 13. The product of two numbers is 144, and one of the n am. bers is 8: what is the other number? Solution.-144. 8=18, the required number. (Art. 120.) PRooF. —18X 8=144, the given product. (Art. 88.) 14. The product of A and B's ages is 3250 years, and B's age is 50 years: what is the age of A? 15. The product of the length of a field multiplied by its breadth is 15925 rods, and its breadth is 91 rods: what is its length? 1 57. When the divisor and quotient are given, to find the dividend. Multiply the given divisor and quotient together, and the product will be the dividend. (Art. 121.) 16. If a certain divisor is 12, and the quotient is 30, what is the dividend? Solution.-30 X 12=360, the dividend requiired. PnooF. —360 12=30, the given quotient. (Art. 120.) 17. If the quotient is 27.5 and the divisor 683, what must be the dividend? 18. If the divisor is 1031 and the quotient 1002, what must be the dividend? 1 5 8. When the dividend and quotient are given, to find the divisor. Divide the given dividend by the given quotient, and the quotient thtus obtained will be the number required. (Art. 122.) 19. A certain dividend is 864, and the quotient is 12: what is the divisor? Solution.-864-. 12=72, the divisor required: (Art. 120.) PRooF. —72X 12=864, the given dividend. (Art. 121.) 20. A gentleman handed a purse containing 1152 shillings, to QuEsve.-157. When the divisor and quotient are given, how is the dividend found 158. When the dividend and quotient are given, how is the divisor found? 8t PROPERTIES [SECT. V. a company of beggars, which was sufficient to give them 24 shil lings apiece: how many beggars were there? 2]. A farmer having 2500 sheep, divided them into flocks of 125 each: how many flocks did they make? 159. When the product of three numbers and two of the numbers are given, to find the other fiumber. Divide the given product by the product of the two given ntumers, and the quotient will be the other number. 22. There are three numbers whose product is 288; one of them is 8, and another 9: it is required to find the other number. Solution.-9 X 8=72; and 288-. 72=4, the number required. PROOF. —9 X 8 X 4=288, the given product. 23. The product of three persons' ages is 14880 years; the age of the oldest is 31 years, and that of the second is 24 years: what is the age of the youngest? 24. If a garrison of 75 men have 18750 pounds of meat, how long will it last them, allowing 25 pounds to each man per month? 25. The sum of two numbers is 3471, and the less is 1629: what is the greater? 26. The sum of two numbers is 4136, and the greater is 3074: what is the less? 27. The difference between two numbers is 128, and the greater is 760: what is the less? 28. The difference between two numbers is 340, and the less is 634: what is the greater? 29. The sum of two numbers is 12640, and their difference is 1608: what are the numbers? 30. The sum of two numbers is 25264, and their difference is 736: what are the numbers? 31. The sum of two numbers is 42126, and their difference is 176: what are the numbers? 32. The product of two numbers is 246018, and one of them is 313: what is the other number? ARTS. 159, 160.1 PROPERT1ES OF NIJMBERS. 89 SECTION VI. PROPERTIES OF NUMBERS.* ART. 160. The progress as well as the pleasure of the student iL Arithmetic, depends very much upon the accuracy of his knowledge of the terms, which are employed in mathematical reasoning Particular pains should therefore be taken to understand their true im-port. DEF. 1. Alln integer signifies a whole number. (Art. 28. Obhs. 2.) 2. Whole numbels or integerls are divided into prime and composite numbers. 3. A composite number, we have seen, is one which may be produced by multiplying two or more numbers together; as, 4, 10, 15. (Art. 95.) 4. A prime number is one which cannot be produced by multiplying any two or more numbers together; or which cannot be exactly divided by any w/hole number, except a unit and itself. Th1us, 1, 2, 3, 5, 7, 11, 13, &c., are prime numbers. OIs. 1. One number is said to be prime to another, when a unit is the only number by which both can be divided without a remainder. 2. The learnel must be careful not to confound numbers which are prime to each other with prime numbers; for numbers that are prime to each other, may themselves be composite numbers. Thus 4 and 9 are prime to each other, while they are composite numbers. 3. The number of' prime numbers is unlimited. For those under 3, see Table, page 9-4. 5. An even number is one which can be divided by 2 without a remainder; as, 4, 6, 8, 10. QAJEST, —?C0. Upon what does the progress and pleasure of the student in Arithmetic very much depend? CWhat is an integer? What is a composite numnlber? What is a prime numbuler? Are prime numbers divisible by other nulmbers? Obs. When is one number said to be prime to another? How many prime numbers are there? What is an even number? An odd number? Obs. Are even numbers prine or composite? What is true of oddl numbers in this respect? * Ba:low on the Theory of Nunlbers; also, Bonnycastle's Arithmetic. 90 PROPERTIES OF NUMBERS. LSECT. VI. 6. An odd number is one which cannot be divided by 2 without a remainder; as, 1, 3, 5, 7, 9, 15. Oss. All even numbers except 2, are somposite numbers; an odd number is sometimes a composite, and sometimes a prime number. 7. One number is a measure of another, when the former is contained in the latter, any number of times without a remainder.;Tus, 3 is a measure of 15; 7 is a measure of 28, &c. 8. One number is a multiple of another, when the former can be divided by the latter without a remainder. Thus, 6 is a multiple of 3; 20 is a multiple of 5, &c. OBs. A multiple is therefore a composite number, and the number thus contained in it, is always one of its factors. 9. The aliquot parts of a number, are the parts by which it can be measured or divided without-a remainder. Thus, 5 and 7 are the aliquot parts of 35. 10. The reciprocal of a number is the quotient arising from dividing a unit by that number. Thus, the reciprocal of 2 is j; the reciprocal of 3 is L-; &c. 11. The difference between a given number and 10, 100, 1000, &c., that is, between the given number and the next higher order, is called the ARITHMETICAL COMPLEMENT of that number. Thus, 3 is the complement of 7; 15 is the complement of 85. OBs. The arithmetical complement of a number consisting of one integral figure, either with or without decimals, is found by subtracting the number from 10. If there are two integral figures, they are subtracted from 100; if th/ree, from 1000, &c. 12. A perfect number is one which is equal to the sum of all its aliquot parts. Thus, 6 = 1 + 2 + 3, the sum of its aliquot parts, and is a perfect number. Oms. 1. All the numbers known; to which this property really belongs, are the following: 6; 28; 496; 8128; 33,550,336; 8,589,869,056; 137,438,691,328; and 2,305,843,008,139,952,128.* 2. All perfect numbers terminate with 6, or 28. QUEST.-When is one number a measure of another? What is a multiple? What are aliquot parts? What is the reciprocal of a number? * Hutton's Mathematical Recreations. ARTS. 16), 161.] PROPERTIES OF NUMBERS. 91 1 61. By the term prolperties of numbers, is meant those qualities or elements which are inherent and inseparable frem them. Some of the more prominent are the following: 1. The sum of any two or more even numbers, is an even number. 2. The difference of any two even numbers, is an even number. 3. The sum or difference of two odd numbers, is even; but the sum of three odd numbers, is odd. 4. The sum of any even number of odd numbers, is even; but the sum of any odd number of odd numbers, is odd. 6. The sum, or difference, of an even and an odd number, is an odd number. 6. The product of an even and an odd number, or of two even numbers, is even.'7. If an even number be divisible by an odd number, the quotient is an even number. 8. The product of any number of factors, is even, if any one of them be even. 9. An odd number cannot be divided by an even number without a remainder. 10. The product of any two or more odd numbers, is an odd number. 11. If an odd number divides an even number, it will also divide the half of it. 12. If an even number be divisible by an odd number, it will also be divisible by double that number. 13. Any number that measures -two others, must likewise measure their sum, their difference, and their product. 14. A number that measures another, must also measure its multiple, or its product by any whlole number. 15. Any number expressed by the decimal notation, divided by 9, will leave the same remainder, as the sum of its figures or digits divided by 9. Demonstration.-Take any number, as 6357; now separating it into its several parts, it becomes 6000+300-50+7. But 6000=6X1000=6X(999+1) =6X999+6. In like manner 300-3X99+3, and 50=5X9 —5. Hence 6357=6X999+3X99+-5X9-+6+3+5+t-7; and 6357- -9(6X9C9O-3X99sQUEsT. —16. What is meant by properties of numbers? 5 92 PROPERTIES OF NUMBERS, [SECT. Vii. 5X9+6+-3+5-l-7)-i-9. But 6X(99+{-3X99- -5X9 is evidently diisible by 9; therefore if 6357 be divided by 9, it will leave the same remainder as 6 —35-L7. —9. The same will be found true of any other number whatever. OBs. 1. This property of the number 9 affords an ingenious method of proving each of the fundamlental rules. (Arts. 90, 123.) The same property belongs to the number 3; for, 3 is a measure of 9, and will therefore be contained an exact number of times in any number of 9s. But it belongs to no other digit. 2. The preceding is not a necessary but an inzcidental property of the number 9. It arises from the law of inc'rease in the decimal notation. If the cladi. of the system were 8, it would belong to 7; if the radix were 12, it would belong to 1I; and universally, it belongs to the number that is one less than the radix of the system of notation. 16. If the number 9 is multiplied by any single figure or digit the sum of the figures composing the product, will make'. Thus, 9X4=36, and 3+6 —=9. 17. If we take any two numbers whatever; then one of then, or their sutm, or their dicfereazce, is divisible by 3. Thus, take 11 and 17; though neither of the numbers themselves, nor their sum is divisible by 3, yet their difference is, for it is 6. 18. Any number divided by 11, will leave the same remainder, as the sum of its alternate digits in the even places reckoning from the right, taken from the sum of its alternate digits in the odd places, increased by 11 if necessary. Take any number, as 38405603, and mark the alternate fig'ures. Now the sum of those marked, viz: 8+0+6+3 —-17 The sum of the others, viz: 3+4-+5+-0=-12. And 17-12=5, the remainder sought. That is, 38405603 divided by 11, will leave 5 remainder. Again, take 5847362, the sum of the marked figures is 14; the sum of those not marked is 21. Now 21 taken from 25, (=14+11,) leaves 4, the remainder sought. 19. Every composite number may be resolved into prime factors. For, since a composite number is produced by multiplying two or more factors together, (Art. 160. Def. 3,) it may evidently be resolved into those factors; and if these factors themselves are composite, they also may be resolved into otlher factors, and thus the analysis may be continued, until all the factors are prime numbers. 20. The least divisor of every number is a prime number. For, every whole number is either prime, or comnpcsite; (Art. 100, ART. 161. a.] PROPERTIES OF NUMBERS. 93 Def. 2;) but a composite number, we have just seen, can be re, solved into prime factors; consequently, the least divisor.of every number must be a prime number. 21. Every prime number except 2, if increased or diminished by 1, is divisible by 4. See table of prime numbers, next page. 22. Every prime number except 2 and 3, if increased or diminished by 1, is divisible by 6. 23. Every prine numiber, except 2 and 5, is contained without a remainder, in the number expressed in the common notation by as many 9s as there are units, less one, in the prime number itself.* Thus, 3 is a measure of 99; 7 of 999,999; and 13 of 999,999, 999,999. 24. Every prime number, except 2, 3, and 5, is a measure of the number expressed in common notation, by as many is as there are units, less one, in the prime number. Thus, 7 is a measure of 111,111; and 13 of 111,111,111,111. 25. All prime numbers except 2, are odd; and consequently terminate with an odd digit. (Art. 160. Def. 5.) Vote. —. It must not be inferred from this that all odd numbers are prime. (Art. 160. Def. 6. Obs.) 2. It is plain that any number terminating with 5, can be divided by 5 without a remainder. Hence, 26. All prime numbers, except 2 and 5, must terminate with 1, 3, 7, or 9; all other numbers are composite. 16 1 * a. To find the prime numbers in any series of numbers. Write in their proper order all the odd numbers contained in the series. Then reckoning from 3, place a _point over every third number in the series; reckoning from 5, place a point over every fifth number; reckoning from 7 place a point over every seventh num. ber, and so on. The numbers remaining without points, together with the number 2, are the primes required. Take the series of numbers up to 40, thus, 1, 3, 5, 57, 9, 11, 13, i5, 17, 19, 21i, 23, 25, 27, 29, 31, 33, 35, 37, 39; then adding the number 2, the primes are 1, 2, 3, 5, 7, 11, 13, &c. Note.-This method of excluding the numbers which are not prime from a series, was invented by Eratosthenes, and is therefore called Er'atosthenes' Sieve. * Thl.orie des Nombres, par M. Legendre 94 PROPERTIES OF NUMBERS. [SECT. VI TABLE OF PRIME NUMBERS FROM 1 TO 3413. 1 173 409 659 941 1223 1511 1811 2129j24232741 83079 2 179 419 661 947 1229 1523 1823 2131 2437 2749 3083 3 181 421 673 953 1231 1531 1831 2137 2441 2753 3089 5 191 431 677 967 1237 1543 1847 2141 2447 2767 3109 7 193 433 683 971]124911549 1861 2143 2459 2777 3119 11 197 439 691 977 1259 1553 1867 2153 246712789 3121 1311991443 701 983 1277 1559 1871 2161 2473 2791 3137 17 211 449 709 991 1279 1567 1873 2179 2477 2797 3163 19 223 457 719 997 1283 15711877 2203 250203 2801 3167 23 227 461 727 1009 1289 1579 1879 2207 2521 2803 3169 29 229 463 733 1013 1291 1583 1889 2213 2531 2819 3181 31 233 467 739 1019 1297 1597 1901 2221 2539 2833 3187 37 239 479 743 1021 1301 1601 1907 2237 2543 2837 3191 41 241 487 751 1031 1303 1607 1913 2239 2549 2843 3203 43 251 491 757 1033 1307 1609 1931 2243 2551 2851 3209 47 257499 761 1039 1319 1613 1933 2251 2557 2857 3217 53 263 503 769 1049 1321 1619 1949 2267 2579 2861 3221 59 269 509 773 1051 1327 1621 1951 2269 2591 287913229 61 271 521 787 1061 1361 1627 1973 2273 2593 2887 3251 67 2771523 797 1063 1367 1637 1979 2281 2609 2897 3253 71 281 541 809 1069 1373 1657 1987 2287 2617 2903 3257 73 283 547 811 1087 1381 1663 1993 2293 2621 2909 3259 79 293 557 821 1091 1399 1667 1997 2297 2633 2917 3271 83 307 563 823 1093 1409 1669 1999 2309 2647 2927 3299 89 311 569 827 1097 1423 1693 2003 2311 2657 2939 3301 97 313 571 829 1103 1427 1697 2011 2333 2659 2953 3307 101 317 -577 839 1109 1429 1699 2017 2339 2663 2957 3313 103 331 587 853 1117 1433 1709 2027 2341 2671 2'3633319 107 337 593 857 1123 1439 1721 2029 2347 2677 2969 3323 109 347 599 859 1129 1447 1723 2039 2351 2683 2971 3329 113 349 601 863 1151 1451 1733 2053 2357 2687 2999 3331 127 353 607 877 1153 1453 1741 2063 237112689 3001i3343 131 359 613 881 1163 1459 1747 2069 2377 2693 301113347 13367 617 883 1171 1471 17513 2081 2381 12699 3019 3359 139 3738 619 887 1181 14811759 2083 2383 270713023 3361 1491379 631 907 1187 1483 1777 2087 238912711 303713371 15113831641 911 119311487 1783 2089 2393 2713 3041 3373 15713891643[919 1201148911787[2099 239912719 3049 3389 1 163 397 6470929 121311493 1789 211112411 27291306113391 167 401 653i937 1217 14-99 1801 2113j24171273113067 3407 ART. 162.] PROPERTIES OF NUMIBERS. 95 DIFFERENT SCALES OF NOTATION. 162. A number expressed in the decimal notation, may be changed to any re quired scale of notation in the following manner. Divide the given number by the radix of the required scale continually, till the quotient is less than the radix; then annex to the last quotient the several remainders in a retrograde order, placing iphers where there is no remainder, and the result will be the numnin the scale required. (Arts. 43, 44.) Ex. 1. Express 429 in the quinary scale of notation. Ex.lanation.-By Dividing the given number 5)429 by 5, it is evidently distributed into 85 parts, 5) 85-4 each of which is equal to 5, with 4 remainder. 5) 1 7-0 Dividing again by 5, these parts are distributed 3 —2 into 17 other parts, each of which is equal to 5 Ans. 3204 times 5, and the remainder is nothing. Dividing by 5 the third time, the parts last found are again distributed into 3 other parts, each of which is equal to 5 times 5 into 5, with 2 remainder. Thus, the given number is resolved into 3 X 5 X 5 X 5 +2 X X 5 X 5 0X5 5+4, or 3204, which is the answer required. 2. Change 7854 from the decimal to the binary scale. Ans. 1111010101110. 3. Change 7854 from the decimal to the ternary scale. Ans. 101202220. 4. Change 7854 from the decimal to the'quaternary scale. Ans. 1322232. 5. Change 7854 from the decimal to the quinary scale. Ans. 222404. 6. Change 7854 from the decimal to the senary scale. Ans. 100210. 7. Change 7854 from the decimal to the octary scale. Ans. 17256. 8. Change 7854 from the decimal to the nonary scale. Ans. 11686. 9. Change 7854 from the decimal to the duodecimal scale. Ans. 4666. 96 PROPERTIES OF NUMBERS. [SECT. VI. 10 Change 35261 from the decimal to the quaternary scale. 11. Change 643175 from the decimal to the octary scale. 12. Change 175683 from the decimal to the septenary scale. 13. Change 534610 from the decimal to the octary scale. 14. Change 841568 from the decimal to the nonary scale. 15. Change 592835 from the decimal to the duodecimal scale. Note.-Since every scale requires as many characters as there are units in 1 e radix, we will denote 10 by t, and 11 by e. Ans. 2470 t e. 163. To change a number expressed in any given scale of notation, to the decimal scale. Multiply the left hand figure by the given radix, and to the 2roduct add the next figure; then multiply this sum by the radix again, and to this product add the next figure; thus continue the operation till all the figures in the given number halve been employed, and the last product will be the number in the decimal scale. 16. Change 3204 from the quinary to the decimal scale. Operation. LErplcanation. —Multiplying the left hand figure 3204 by 5, the given radix, evidently reduces it to the 5 next lower order; for in the quinary scale, 5.in 17 an inferior order make one in the next superior 5 order. For the same reason, multiplying this 85 sum by 5 again, reduces it to the next lower 5 order, &c. 429 Ans. 0as. This and the preceding operations are the same in principle, as reducing compound numbers from one denomination to another. 17. Change 1322232 from the quaternary to the decimal scale. Ans. 7854. 18. Change 2546571 from the octary to the decimal scale. 19. Change 34120521 from the senary to the decimal scale. 20. Change 145620314 from the septenary to the decimal scale. 21. Change 834107621 from the nonary to the decimal scale. 22. Change 403130021 from the quinary to the decimal scale. 23. Change 704400316 from the octary to the decimal scale. 24. Change 903124106 from the duodecimal to the decimal scale. ARTs. 163-165.] PROPERTIES OF NUMBERS. 9~ ANALYSIS OF COMPOSITE NUMBERS. 164.o Every composite number, it has been shown, may be resolved into prime factors. (Art. 161. Prop. 19.) Ex. 1. Resolve 210 into its prime factors. Operation. We first divide the given number by 2, which 2)210 is the least number that will divide it withb 3)105 out a remainder, and which is also a prime 5)35 number. (Prop. 20.) We next divide by 3, 7 then by 5. The several divisors and the last Ans. 2, 3, 5, and 7. quotient are the prime factors required. PnOOF.-2 X 3 X 5 X 7=210. Hence, 1650 To resolve a colp)osite number into its prime factors. Divide thle given number by the smallest number which will divide it without a remainder; then divide the quotient in the same way, and thus continue the operation till a quotient is obtained which can be divided by no number greater than 1. The several divisors with the last quotienzt, will be the prime fictors required. (Art. 161. Prop. 19.) Demoglstratzonn.; -Every division, of a number, it is plain, resolves it into two factors, viz: the divisor and dividend. (Art. 112.) But according to the rule, the divisors, in every case, are the smallest numbers that will divide the given number and t he successive quotients without a remainder; consequently they are all prime numnbers. (Art. 1-61. Prop. 20.) And since the division is continued till a quotient is obtained, which cannot be divided by any number greater Ihan 1, it follows that the last quotient must also be a Prime number; for, a Itrime number is one which cannot be exactly divided by any whole number except a tunit and itself. (Art. 160 Def. 4.) OBs. 1. Since the least divisor of every number is a p,'rime number, it is evident that a composite number may be resolved into its prime factors, by dividing it continually by anyl p?-iimne?;zumbe that will divide the given number and the quotients without a remainder. Hence, 2. A composite number can be divided by any of its prime factors without a remainder, and by the product of any two or more of them, but by nto other number. Thus, the prime factors of 42 are 2, 3, and 7. Now 42 can be diQUEST. —-165. How do you resolve a composite number into its prime factors? Obs. Will tbe same result be obtained, if we divide by any of its prilme factors? 98 GREATEST COMMON [SECT. VI. vided by 2, 3, and 7; also by 2X3, 2X7, 3X7, and 2XXX'~; but it can be divided by no other number. 2. Resolve 4 and 6 into their prime factor. Solution.-4=2X2; and 6=2X3. 3. Resolve 8 into its prime factors. Ans. 8=2 X 2 X 2. Resolve the following composite numbers into their prime factors: 4. 9. 22. 34. 40. 57. 58. 81. 5. 10. 23. 35. 41. 58. 59. 82. 6. 12. 24. 36. 42. 60. 60. 84.'7. 14. 25. 38. 43. 62. 61. 85. 8. 15. 26. 39, 44. 63. 62. 86. 9. 16. 27. 40. 45. 64. 63. 87. 10. 18. 28. 42. 46. 65. 64. 88. 11. 20. 29. 44. 47. 66. 65. 90. 12. 21. 30. 45. 48. 68, 66. 91. 13. 22. 31. 46. 49. 69. 67. 92. 14. 24. 32. 48. 50. 70. 68. 93. 15. 25, 33. 49. 51. 72. 69. 94. 16. 26. 34. 50. 52. 74. 70. 95. 17. 27. 35. 51. 53.'75. 71. 96. 18. 28. 36. 52. 54. 76. 72. 98. 19. 30. 37. 54. 55. 77. 73. 99. 20. 32. 38. 55. 56. 78. 74. 100. 21. 33. 39. 56. 5'7. 80. 75. 108, 76. Resolve 120 and 144 into their prime factors. 77. Resolve 180 and 420 into their prime factors. 78. Resolve 714 and 836 into their prime factors. 79. Resolve 574 and 2898 into their prime factors. 80. Resolve 11492 and 180 into their prime factors. 81. What are the prime factors of 650 and 1728? 82. What are the prime factors of 1492 and 8032? 83. What are the prime factors of 4604 and 16806? 84. What are the prime factors of 71640 and 20324? 85. What are the prime factors of 84705 and 65948? 86. What are the prime factors of 9235"2 and 8r, 378?9 ARTs. 166-168.] DIVISOn. 99 GREATEST COMMON DIVISOR. 1 6(6. A common divisor of two or more numbers, is a number which will divide each of them without a remainder. Thus 2 is a common divisor of 6, 8, 12, 16, 18, &c. 1 7. The greatest common divisor of two or more numbers, is the greatest number which will divide them without a remainder. Thus 6 is the greatest common divisor of 12, 18, 24, and 30. OBS. A common divisor is sometimes called a common measure. It will be seen that a common divisor of two or more numbers, is simply a factor which is coimon to those numbers, and the greatest common divisor is the greatest factor common to them. Hence, 16. To find a common divisor of two or more numbers. 2Resolve each number into two or morefactors, one of which shall be common to all the given numbers. Or, resolve the given numbers into their prime factors, then if the same factor isfound in each, it will be a common divisor. (Art. 165. Obs. 2.) OBS. If the given numbers have not a commonz factor, they cannot have a common divisor greater than a unit; consequently they are either prime qziulrbers, or are privme to each other. (Art. 160. Def. 3. Obs. 2.) Note.-The following facts may assist the learner in finding common divisors: 1. Any numbe- ending in 0, or an even number, as 2, 4, 6, &c., may be divided by 2. 2. Any number ending in 5 or 0, may be divided by 5. 3. Any number ending in 0, may be divided by 10. 4. When the two right hand figures are divisible by 4, the whole number may be divided by 4. 5. If the three right hand figures of any number are divisible by 8, the whole is divisible by 8. Ex. 1. Find a common divisor of 6, 15, and 21. Solution.-6 3 X 2; 15= 3 X 5; and 21= 3 X 7. The factor 3 is common to eiach of the given numbers, and is therefore a common divisor of them. QUEsT.-166. What is a common divisor of two or more numbers? 167, What is the greatest common divisor of two or more numbers? Obs. What is a common divisor sometimes called? 168. How do you find a common divisor of-two or more numbers Obs. I? two given numbers have not a common factor, what is true as to a-common divisor 5* 100 GCRI1ATEST COMMON [SECT. VI. 2. Find a coinmon divisor of 15, 18, 24, and 36. 3. Find a common divisor of 14, 28, 42, and 35. 4. Find a common divisor of 10, 35, 50, 75, and 60. 5. Find a common divisor of 82, 118, and 146. 6. Find a common divisor of 42 and 66. Ans. 2, 3, or 6. 169. It will be seen from the last example that two numbers may have more than one common divisor. - In many cases it is highly important to find the greatest divisor that will divide two or more given numbers without a remainder. 7. What is the greatest common divisor of 35 and 50? Operation. Dividing 50 by 35, the remainder is 15, 85)50(1 then dividing 35 (the preceding divisor) by 35 t5 (the last remainder) the remainder is 5' 15)35(2 finally, dividing 15 (the preceding divisor) by 30 5 (the last remainder) nothing remains; con5)15(3 sequently 5, the last divisor, is the greatest 15 common divisor. Hence, 1 70. To find the greatest common divisor of two numbers. Divide the greater number by the less; then divide the preceding divisor by the last remainder, and so on, till nothing remains..The last divisor will be the greatest common divisor. When there are more than two numbers given. First find the greatest common divisor of any two of them; then, that of the common divisor thus obtained and of another given number, and so on through all the given numbers. The last common divisor found, will be the one required. Demonstration.-Since 5 is a measure of the last dividend 15, in the preceding solution, it must therefore be a measure of the preceding dividend 35; because 35=2X-15+5; and 35 is oile of the given numbers. Now, since 5 measures 15 and 35, it must also measure their sum, viz: 35 -15, or 50, which Is the other given number. (Art. 161. Prop. 13.) In a similar manner it may be shown that the last divisor will, in all cases, be the g reatest common divisor. Note.-Numbers which have no common measlt,'e greater than 1, are said to be incommensurable. Thus 17 and 29 are incommensurable. QUEST. —170. Ilowl find the greatest common divisor of two numbers? Of inore than twol ARTS. 169-1 71.] DIVISOr. 01 8. What is the greatest common divisor of 285 and 465? 9. What is the greatest common divisor of 532 and 1274? 10. What is the greatest common divisor of 888 and 2775? 11. What is the greatest common divisor of 2145 and 3471? 12. What is the greatest common divisor of 1879 and 2425? 13. What is the greatest common divisor of 75, 125, and 160? Suggestion. —Find the greatest common divisor of 75 and 125, which is 25. Then that of 25 and 160. Ans. 5. 14. What is the greatest common divisor of 183, 3996, 108? 15. What is the greatest common divisor of 672, 1440, and 3472? 16. What is the greatest common divisor of 30, 42, and 66? Analysis.-By resolving the given num- Operation. bers into their prime factors, (Art. 165,) 30=2X 3 X b we find that the factors 2 and 3 are both 42=2 X 3 X 7 common divisors of them. But we have 66-2X3Xll seen that a composite number can be Now 2X3=6 Ans. divided by the product of any two or more of its prime factors; (Art. 165. Obs. 2;) consequently 30, 42, and 66 can'all be divided by 2 X 3; for 2 X 3 is the product of two prime factors common to each. And since they are the only factors common to the given numbers, their product must be the greatest common divisor of them. Hence, we deduce a 1 7 1. Second Method of finding the greatest common di'isor of two or more numbers. Resolve the given numbers into their prime factors, and the continued product of those factors which are common to each, will be the greatest common divisor. OBs. If the given numbers have but one common factor, that factor itself is the greatest commnon divisor. 17. What is the greatest common divisor of 105 and 165? 18. What is the greatest common divisor of 36, 60, and 108 19. What is the greatest common divisor of 108, 126, and 162? 20. What is the greatest common divisor of 140, 210, and 315? 21. What is the greatest common divisor of 24, 12, 54, and 60? 22. What is the greatest common divisor of 56, 84, 140, and 168? 102 IEAST CONMMON [SEC'r. VI. LEAST COMMON MULTIPLE. 17 2. One number is said to be a mnultiple of another, when the former can be divided by the latter without a remainder (Art. 160. Def. 8.) Hence, 173. A common multiple of two or more numbers, is a number which can be divided by eac. of them without a remainder, thus, 12 is a common multiple of 2, 3, and 4; 15 is a common multiple of 3 and 5, &c. OBs. A common multiple is dlways a composite number, of which each of the given numbers must be v.'zctor; otherwise it could not be divided by them. (Art. 165. Obs. 2.) 174. The continued product of two or more given numbers will always form a common multiple of those numbers. The same numbers may have an unlimited number of common multiples; for, multiplying their continued product by any number, will form a new common multiple. (Art. 161. Prop. 14.) 175. The least common multiple of two or more numbers, is the least number which can be divided by each of them without a remainder. Thus, 12 is the least common multiple of 4 and 6, for it is the least number which can be exactly divided by them. OBs. The least common multiple of two or more numbers, is evidently composed of all the prime factors of each of the given numbers repeated once, and only once. For, if it did not contain all the prime factors of any one of the given numbers, it could not be divided by that number. (Art. 165. Obs. 2.) On the other hand, if any prime factor is employed more times than it is repeated as a factor in some one of the given numbers, then it wou,ld not be the least common multiple. Ex. 1. What is the least common multiple of 10 and 15? Analysis.-10=2X5, and 15=3X5. The prime factors of the given numbers are 2, 5, 3, and 5. Now since the factor 5 occurs once in each number, we may therefore cancel it in one QUEST.-172. When is one number said to be a multiple) of a:nother? 173. What is a common multiple? 174. How may a common multiple of two or more r Dmbers be formed 1 How many common multiples may there be of ar y giver lumbers? 175. What is the least common multiple of two or more numbers? AnTs. 1 7'2- 176.] MULTIPLE. 103 instanlce, and the continued product of the remaining factors 2 X 3 X 5, or 30, will be the least common multiple. Operation. We first divide both the numbers by 5 5)10 " 15 in order to resolve them into prime fac2 " 3 tors. (Art. 175. Obs.) Thus, all the dif5 X 2 X 3=30 Ans. ferent factors of which the given numbers are composed, are found in the- divisor and quotients once, and only once. Therefore the product of the divisor and quotients.5 X 2 X 3, is the least common multiple required. Hence, 17G6. To find the least common multiple of two or more numbers. Write the given numbers in a line with two points between them. Divide by the smallest number which will divide any two or more of them without a remainder, and set the quotients and the undivided numbers in a line below. -Divide this line and set down the results as before; thus continue the operation till there are no two numbers which can be divided by any number greater than 1. The continued product of the divisors into the numbers in the last line, will be the least common multiple required. Ons. 1. We have seen that the least divisor of every number is aprimne number; hence, dividing by the smallest number which will divide two or more of the given numbers, is dividing them by a prime number. (Art. 161. Prop. 20.) The result will evidently be the same, if, instead of dividing by the smalless number, we divide the given numbers by any prime number, that will divide two or more of them, without a remainder. 2. The preceding operation, it will be seen, resolves the given numbers into their primeefactors, (Art. 165,) then multiplies all the different factors together, taking each factor as many times in the product, as are equal to the greatest qnumber of times it is found in either of the given numbers. 3. If the given numbers are prime numbers, or are prime to each other, the continued product of the numbers themselves will be their least common multiple. (Art. 168. Obhs.) Thus, the least common multiple of 5 and 7 is 35; of 8 and 9 is 72. QUEsT. —176. How is the least common multiple of two or more numbers found? Obs. If the given numbers are prime, or are primne to each other, what is the least comnon multiple of them? 176. a. Upotl what principle does this rule depend? Obs. Why do you divide by the smallest number that will divide two or more of the given numbers without a remainder? 104 LEAST COMMON |SECT. Vi, Ex. 2. What is the least common multiple of 6, 8, ancd 12? Ancalysis. —By resolving the given numbers Operation. into their prime factors, it will be seen that 2 6 =2 X 3 is found once as a factor in 6; twice in 12; and 8-2 X 2 X 2 three times in 8. It must therefore be taken 12 —2 X 2 X3 three times in the pproduct. Again, 3 is a fac- 2 X 2 X 2X3=24 tor of 6, and 12, consequently it must be taken only once in the product. (Art. 176. Obs. 2.) Thus, 2X2X2X3-24 _Ans. Ex. 3. What is the least common multiple of 12, 18. and 36? First Operation. Second Operation. Third Operation. 2)12' 18 s 36 9)12 11 18 11 36 12)12 " 18 "I 36 2) 6 9 "18 2)12 " 2 4 3) 1 18" 3 3) 3 " 9 9 2) 6 " 1 "2 1 "I 6 1 3) 1 " 3 " 3 3 " 1 " 1 And 12X3X 6=216. 1 " 1 " ow 9X2X2X 3-108. 2X2X3X3 —36 Ans. Explanation.-In the first operation, we divide by the smallest numbers which will divide any two or more of the given numbers without a remainder, and the product of the divisors, &c., is 36, which is the answer required. In the second and third operations, we divide by numbers that will divide two or more of the given numbers without a remainder, and in both cases, obtain erroneous answers. Note.-It will be seen from the second and third operations above, that "dividing by any number, which will divide two or more of the given numbers without a remainder," according to the rule given by some authors, does not always give the least common multiple of the numbers. 1 76. a. The reason of the preceding rule depends upon the principle that the least common multiple of any two or more numbers, is composed of all the prinie. factors of the given numbers, each taken as many times, as are equal to the greatest number of times it is found in either of the given numbers. (Art. 175. Ohs.) Note.-1. The reason for dividing by the smallest number, is because the divisor may otherwise be a composite number, (Art. 161. Prop. 20,) and have a factor common to some one of the quotients, or undivided numbers in the last line; consequently the continued product of them world be too large for ARTS. 176. a. 177.1 MULTIPLE. 105 the least common multiple. (Art. 175. Obs.) Thus, in t.' e second operation the divisor 9, is a composite number, containing the factor 3 common to the 3 in the quotient; consequently the product is three times too lare. In the tlhird operation the divisor 12, is a composite number, and contains the factor 6 common to the 6 in the quotient; therefore the product is six times too lange. 2. The object of arranging the given numbers in a line, is that all of them may be resolved into their prime factors at the same time; and also to present at a glance the factors which compose the least common multiple required. 4. Find the least common multiple of 6, 9, and 15. 5. Find the least common multiple of 8, 16, 18, and 24. 6. Find the least common multiple of 9, 15, 12, 6, and 5. 7. Find the least common multiple of 5, 10, 8, 18, and 15, 8. Find the least common multiple of 24, 16, 18, and 20. 9. Find the least common multiple of 36, 25, 60, 72, and 35. 10. Find the least common multiple of 42, 12, 84, and 72. 11. Find the least common multiple of 27, 54, 81, 14, and 63. 12. Find the least common multiple of 7, 11, 13, 3, and 5. 17 7. The process of finding the least common multiple may often be shortened, by canceling every number which will divide any other given number, without a remainder, and also those which will divide any other number in the same line. The least common multiple of the nztmbers that remain, will be the answier ~required. OBs. By attention and practice, the student will be able to discover, by inspection, the least common multiple of numbers, when they are not large. 13. Find the least common multiple of 4, 6, 10, 8, 12, and 15. Operation.' Since 4 and 6, will exactly di2)4 / 6 /" 10 " 8 " 12 " 15 vide S, and 12, we cancel them. 2) t " 4" 6 15 Again, since 5 in the second line 3) 2 " 3 " 15 will exactly divide 15 in the same 2' 1 " 5 line, we therefore cancel it, and Now, 2 X 2 X 3X2 X 5 =120 Anys. proceed with the remaining numbers as before. 14. Find the least common multiple of 9, 12, 72, 36, and 144. 15. Find the least common multiple of 8, 12 20, 24, and 25. 16. Find the least common multiple of 1, 2, 3, 4, 5, 6, 7, 8, 9. 106 COMMON LSECT. VJ. 17. Find the least common multiple of 63, 12, 84, and 7. 18. Find the least common multiple of 54, 81, 63, and 14. 19. Find the least common multiple of 72, 120, 180, 24, and 36. 1 7 7 a. The least common multiple of two or more numbers may also be found in the following manner. First find the greatest common divisor of two of the given numbers; by this divide one of these two numbers, and multiply the quotient by the other. Then perform a similar operation on the product iend another of the given numbers; thus continue the process until all of the given numbers have been employed, and the final result will be the least common multiple required. 20. What is the least common multiple of 24, 16, and 12? Solution.-By inspection, we find the greatest common divisor of 24 and 16, is 8. Now 24 -8=3; and 3X16-=48. Again, the greatest common divisor of 48 and 12, is 12. Now 48.12 _4; and 4X12=48. Ans. PRoo. —Resolving the given numbers into their prime factors, 24.-2X2X2X3; 16=2X2X2X2; and 12=2X2X3; (Art. 165;) consequently, 2X2X2X2X3 —48, the least common multiple. (Art. 175. Obs.) Ons. The reason of this rule depends upon the principle, that if the product of any two numbers be divided by anvy factor which is common to both, the qotiemt will be a common multiple of the two numbers. Thus, if 48, the product of 6 and 8, be divided by 2, a factor of both, the quotient 24, will be a multiple of each, since it may be regarded either as 8 multiplied by the quotient of 6 by the factor 2, or as 6 multiplied by the quotient of 8 by the same factor. Hence, it is obvious, that the greater the common measure is, the less will be the multiple; and, consequently, the greatest common measure will oduce the least common multiple. When the common multiple of the first two numbers is found, it is evident, at any number which is a common multiple of it and the third number, villI be a multiple of the first, second, and third numbers. 21. What is the least common multiple of 75, 120, and 300? 22. What is the least common multiple of 96, 144, and 720? 23. What is the least common multiple of 256, 512, and 1728? 24. What is the least common multiple of 375, 85 ), and 3 400? AnTS. 77. a,-181.] FRACTIONS. 107 SECTION VII. FRACTIONS. ART. 17 8. When a number or thing is divided into two eqail parts, one of those parts is called one half. If the number or thing is divided into three equal parts, one of the parts is called one third; if it is divided into four equal parts, one of the parts is called one fourth, or one quarter; and, universally, When a number or thing is divided into equal parts, the parts take their name from the number of parts into which the thing or number is divided. 17 9. The value of one of these equal parts manifestly depends upon the number of parts into which the given number or thing is divided. Thus, if an orange is successively divided into 2, 3, 4, 6, 6, &c., equal parts, the thirds will be less than the halves; the fourths, than the thirds; the fifths, than the fourths, &c. Ons. A half of any number is equ'al to as many units, as 2 is contained times in that number; a thi'rd of a number is equal to as many, as 3 is contained times in the given number; a fourth is equal to as many, as 4 is contained in the number, &c. 180. When a number or thing is divided into equal parts, these parts are called FRACTIONS. OBs. Fractions are used to express parts of a collection of things, as well as of a single thing; or parts of any number of units, as well as of one unit. Thus, we speak of ~ of six oranges; a of 75, &c. In this case the collectio, or pnumber to be divided into equal parts, is regarded as a whole. 181. Fractions are divided into two classes, Common and D)ecimal. For the illustration of Decimal Fractions, see Section IX. QVEST.-178. What is meant by one half? What is meant by one third? What is meant by a fourth? What is meant by fifths? By sixths? How many sevenths make a whole one? How many tenths? What is meant by twentieths? By hundreds? When a number or thing is divided into equal parts, from what do the parts take their name 1 179. Upon what does the value of one of these equal parts depend? 180. What are frao. tions 7 181. Into how many classes are fractions divided 3 108 FRACTIONS. [SECT. VII. q1 S,2. Common,Fractions are expressed by two numbers, one plated over the other, with a line between them. One half is written thus ~; one third, I-; one fourth, I; nine tenths, -; thirteen forty-fifths, j —, &c. The number below the line is called the denominator, and shows into how many parts the number or thing is divided. The number above the line is called the numerator, and shows how many parts are expressed by the fraction. Thus, in the frac tion -., the denominator 3, shows that the number is divided into three equal parts; the numerator 2, shows that two of those parts are expressed by the fraction. The denominator and numerator together are called the terms of the fraction. Oss. 1. The term fraction, is of Latin origin, and signifies brokeen, or sepa rated into parts. Hence, fractions are sometimes called broken numbers. 2. Common fractions are often called vulgar fractions. This term, however, is very properly falling into disuse. 3. The number below the line is called the denominator, because it gives the name or dCno1mivation to the fraction; as, halves, thirds, fifths, &c. The number above the line is called the numerator, because it numbers the parts, or shows how many parts are expressed by the fraction. I 83. A proper fraction is a fraction whose numerator is less than its denominator; as, -, 3, A. An improper fraction is one whose numerator is equal to, or greater than its denominator; as, 3, -. A mixed number is a whole number and a fraction expressed together; as, 4-, 251~,. A simple fraction is a fraction which has but one numerator and one denominator, and may be proper, or improper; as, 3-, -9. A co7mapound fraction is a fraction of a fraction; as, - of AL of -, L of l2 of -1 of -9. (QuasT.-182. Iow are common fractions expressed? What is the number below the lisla called? What does it show? What is the number above the line called? What does it show? What are the denominator and numerator, taken together, called? Obs. What is the meaning of the term fraction? What are common fractions sometimes called? Why is the lower number called the denominator? Why is the upper one ealled the numnerator? 183. What is a proper fraction An improper fraction? A Mixed number? A simple fraction? A compound fraction? ARTS. 182-188.] FRACTIONS. 109 A complex fraction is one which has a fraction in its numerator 2J 4 2or denominator, or in both; as, -, 5 8. 1 84. Fractions, it will be seen both from the definition and the mode of expressing them, arise from division, and may be treated as expressions of unexecuted division. The numnerator answers to the dividend, and the denominator to the divisor. (Arts. 25, 182.) Hence, 1 8 50 The value of a fraction is the quotient of the numerator divided by the denominator. Thus, the value of { is two; of A is one; of Ja- is one third, &c. Hence, 1 86. If the denominator remnains the same, mnulti2plying the numerator by any number, multiplies the value of the f:acetion b6 that vumber. For, since the numerator and denominator answe, to the dividend and divisor, multiplying the numerator is the sams as multiplying the dividend. But multiplying the dividend, we have seen, multiplies the quotient, (Art. 141,) which is the same as the value of the fraction. (Art. 185.) Thus, the value of ~=2; now, multiplying the numerator by 3, the fraction becomes.oA, whose value is 6, and is the same as 2 X 3. 1 8 7. Dividing the numerator by any number, divides the value of the fraction by that numyber. For, dividing the dividend, divides the quotient. (Art. 142.) Thus, — =2; now dividing the numerator by 2, the fraction becomes a, whose value is 1, and is the same as 2-.2. Hence, OBS. With a given denominator, the greater the numerator, the greater will ne the value of the fraction. 188. If the numerator remains the same, mzulti2plying thle denominator by any numzber, divides the value of the fraction by th/at nzumber. For, multiplying the divisor, we have seen, divides th QUEST.-What is a complex fraction? 184. From what do fractions arise? 185. VWhat is the value of a fraction? 186. What is the effect of multiplying the numerator, while the denominator remains the same? Explain the reason. 187. Who t is the effect of di viding the numerator? Obs. With a given denominator, what is the effect of increasing the nurnerator? 188. What is the effect of multiplying the denominator? 110 FRACTIONS.,[SECT. VII. quotient. (A_ t. 143.) Thus, -2-=4; now Tmultiplying the denominator by 2, the fraction becomes G-, whose value is 2, and is tlhe same as 4 2. I ) 9. D)ividing the denominator by any number, nmultiplics the value of the fraction by that number. For, dividing the divisor multiplies the quotient. (Art. 144.) Thus, t-i-4; now dividing the denominator by 2, the fiaction becomes 234-, whose value is 3, nd is the same as 4 X 2. Hence, Oss. With a given numerator, the greater the denominator, the less wnll oe the value of the fraction. 190. It is evident from the preceding articles, that multiplying the numerator by any number, has the same effect on the value of the fraction, as dividing the denominator by that number. (Arts. 186, 189.) And, Dividing dte numerator has the same effect, as multiplying the denominator. (Arts. 187, 188.) OBs. It will be observed, that multiplying or dividing the Inimerator of a fraction, has the same effect upon its value, as the same operation has upon a whole number; but, the effect of multiplying or dividing the denominator is exactly contr'ary to that of the same operation upon a whole number. 191. If the numerator and denominator are both multiplied or both divided by the same number, the value of the fraction will not be altered. (Art. 146.) Thus, -L==3; now if the numerator and denominator are both multiplied by 2, the fraction becomes 2 -, whose value is 3. If both terms are divided by 2, the frac-'ion becomes ~, whose value is 3; that is,:= —= —= 3. 192. Since the value of a fraction is the quotient of the numerator divided by the denominator, it follows, If the numerator and ldenominator are equal, the value is a unit or one. Thus, 7=1, -=1, &c. QUEST.-189. Winat is the effect of dividing the denominator? VWhy? Obs. With a given nunlerator, what is the effect of increasing the denominator? 190. What may be done to the denominator to produce the same effect on the value of the fraction, as multiplying the numerator by any given number? What, to produce the same effect as dividing the numerator by any given number? 191. What is the effect if the numerator and lenominator are both multiplied, or both divided by the same number? 192 When the uanmerator an d denominator are equal, what is the value of the fraction? ARTS. 189-194.] FRACTIONS. 111 If the numerator is greater than the denominator, the value is greater than one. Thus, -= 2, -=1. If the numerator is less than the denominator, the value is less than one. Thus, -=1 third of 1, -=4 fifths of 1. 193. Fractions may be added, subtracted, multiplied, and divided, as well as whole numbers. But, in order to perform these operations, it is often necessary to make certain changes in the. terms of the fractions. OBs. It is evident that any changes may be made in the terms of a fraction, which do not alter the quotient of the numerator divided by the denominator; for, if the quotient is not altered, the value remains the same. Thus, the terms of the fraction A may be changed into -, A, a8',- &c., without altering its value; for in each case the quotient of the numerator divided by the denominator is 2. Hence, for any given fraction, we may substitute any other fraction, which wih give the same qolzt/iemtg. REDUCTION OF FRACTIONS. 194. The process of changing the terms of a fraction into others, without altering its value, is called REDUCTION OF FRACTIONS. CASE I. jx. 1. Reduce f- to its lowest terms. First 0peration. Dividing both terms of the 2)l=~-: again, 5)-A —1 Ans. fraction by 2, it becomes -,0: again, dividing both by 5, we obtain -, whose terms are the lowest to which the given fraction can be reduced. Second Operation. If we divide both terms by 10, theil 10)I-a=~ Ans. greatest common divisor, (Art. 170,) the given fraction will be reduced to its lowest terms by a single division. Hence, QUEST. —WVhen the numerator is larger than the denominator, Nwhat? When smaller, what? Obs. What changes may be made in the terms of a fraction? 19-4. What is meant by reduction of fractions i 195. How is a fraction redatled to its lowest terms? 112 REDUCTION OF [SECT. VII. 1 9 5. To reduce a fraction to its lowest terms. Divide the numerator and denominator by any number whic will divide them both without a remainder; and thus continue the operation, till there is no number greater than 1 that will divide them exactly. Or, divide both the numerator and denominator by their greatest common divisor; the two quotients thence arising will be the lowest erms to which the given fraction can be reduced. (Art. 170.) OBs. 1. Since halves are larger than tvwentieths, it may be asked, how the fraction 6, can be said to be in lower terms than A-. It should be observed, the expression lowest term, has reference to the number of parts into which the unit or thing is divided, and not to the value or size of the parts. Thus, in x, thlere are fewer parts than in Ia in 4, there are fewer parts than in -A%-, &c. Hence, a fraction is said to be reduced to its lowest terms, when its numerator and denominator are expressed in the smallest numbers possible. 2. The value of a fraction is not altered by reducing it to its lowest terms; for, the numerator and denominator are both divided by the same number. 3. When the terns of the fraction are small, the former method will generally be found to be the shorter and more convenient; but when the terms are large, it is often difficult to determine whether the fraction is in its simplest form, without finding the greatest commlon d livisor of its terms. 2. Reduce — 9- to its lowest terms. AZns.. 3. RPeduce -f5 iL. Reduce -. 4. Reduce'r. 12. Reduce -1i2_-. 5. Reduce ~-. 13. Red-uce 7. 6. Reduce -3. 14. Reduce -. 7. Reduce 1-. 15. Reduce A_7_. 8. Reduce -7-. 16. Reduce ~_ 9. Reduce -~- 17. Reduce las.M 10. Reduce -t 18. Reduce --- CASE II. 19. Reduce -Z to a whole or mixed number. Analysis.-The object in this example, is to Operation. find a whole, or mixed number, whose value is 7)23 equal to the given fraction. Now, since 7 32 Ans, QVEST. —Obs. What is meant by the expression, lowest terms? When is a fraction said to be reduced to its lowest terms? Is the value of a fraction altereJ bhy reducinl it to its lowest terms? Why not. ARTS. 195-197.] FRACTIONTS. 113 sevenths mnake 1 uwhole one, 23 sevenths will make as many whole ones as 7 is contained times in 23. And 23-7=3~-. But the value of a fraction is the quotient of the numerator divided by the denominator. (Art. 185.) Hence, 196. To reduce an improper fraction to a whole, or mixed number. ~Divide the numerator by the denominator, and'he quotient Vill be the whole, o7 mixed number required.'20. Reduce 3 to a whole or mixed number. Ans. 6-. Reduce the following friactions to whole or mixed numbers: 21. Reduce 26. Reduce. 26. Redce 22. Reduce a. 27. Reduce 6 -. 23. Reduce ~. 28. Reduce -1a —5a, 24. Reduce ~. 29. Reduce -'25. Reduce -. 30. Reduce -2. C' A S III. 31. Reduce tlhe mixeditl nmll-er 027-5- to an imprloi,-' fl'ltic nio Analysis. —In 1 there are 5.fifths, and in 27 27 1 there are 27 times as many. Now 5X27-135, 5 and 2 fifths make 137 fifths. Hence, -,jz Ans 197. To reduce a mixed number to an improper fraction. Multiply the whole number by the denominator of the fraction, and to the product add the given numerator. The sum placed over' the given denominator, will form the impjroper fraction required. Oss. 1. Any whole number may be expressed in the form of a fraction without altering its value, by matking I thle denominator. 2. A whole number may also be reduced to a fraction of any denominator, by wmtttiplying the given number by the proposed denominatl;5 the product will be the numerator of the fraction required. QUvST.-196. IHow is an improper fraction redtced to a whole or mixed number 197. How reduce a mixed number to an improper fraction? Obs. How express a whole'number in the form of a fraction? TIow redsce it to a fraction of a given denominator I 114 REDUCTION OF [SECT. VII Thus, 25 may be expressed by A-, -O, or -~, &c., for 232=-r10- =QpQ, &c. So 121 2 o__ —3__, for the quotient of each of these numerators divided by its denominator, is 12. 32. Reduce 14~ to an improper fraction. Ans. 4. Reduce the following numbers to improper fractions: 33. Reduce 17-. 38. Reduce 856-t —. 34. Reduce 25-a. 39. Reduce 1304-. 35. Reduce 48`. 40. Reduce 4725*. 36. Reduce 70-,a. 41. Reduce 445 to tenths. 37. Reduce 11s5A2. 42. Reduce 672 to eighths. 43. Reduce 3830 to one hundred and fifteenths. 44. Reduce 5743 to six hundred and twenty-fifths. CASE IV. 45. Reduce - of -7 to a simple fraction. Ancalysis. —~a of 7 is 2 times as much as 1 third of j. Now - of 8 is 8X3' or -4; for, multiplying the denominator divides the value of the fraction. (Art. 188.) And 2 thirds is 2 times -2, or 2 which is equal to, or (Art. 195.) The answer is -d. OBs. This operation consists in simply multiplying the two numerators together and the two denominators. Hence, 198. To reduce compound fractions to simple ones. MuZltiply all the numerators together for a new numerator, and all the denominators toyether for a new denominator. Oas. 1. That a compound fraction may be expressed by a simple one, is evident from the fact that a part of a part, must be equal to some part of the whole. 2. The s'eason of the rule may be seen from the analysis of the preceding example. 46. Reduce - of - of 6 of {- to a simple fraction. Ans. 12-2, or.u 47, Reduce 4 of -1 of -j of -9- to a simple fraction, 48. Reduce - of * of - of * of -_z to a simple fraction. QUEST. —198. How are compound fractions reduceed to simple ones? ARTS. 198, 199.] FRACTIONS. 115 49. Reduce - of 3 of -a% of a% to a simple fraction. 50. Reduce I of - of - of - of 1- to a simple fraction. Analysis.-Since the product of Operation. the numerators is to be divided 1 O 5 4 5 by the product of the denomina- - of of of 7 of b 6tors, we may cancel the factors 2, 3, and 4, which are common to both; for, this is dividing the terms of the new fraction by the same number, (Art. 148,) and therefore does not alter its value. (Art. 191.) Multiplying the remaining factors together, we have G, which is the answer required. Hence, 199. To reduce compound fractions to simple ones by CANCELATION. Cancel all the factors which are common to the numerators and denominators; then multiply the remaining terms together as beforle. (Art. 198.) OGs. 1. The reason of this rule depends upon the fact that the numerator and denominator of the new fraction are, in effect, divided by the same numbers; for, canceligg a factor of a number divides the number by that factor. (Art. 148.) Consequently the value of the fraction is not altered. (Art. 191.) 2. This method not only s1tortens the operation of nultiplying, but at the same time reduces the answer to its lowest terms. A little practice will give the student great facility in its application. 51. Reduce { of 2 of 4 to a simple fraction. Operation. 3 First we cancel the 3 and 8 in the $ $o $ 3 numerator, then the 24 in the denominaof - of'-7 Ans. tor, which is equal to the factors 3 into 8. Finally, we cancel the 5 in the denominator and the factor 5 in the numerator 15, placing the other factor 3 above. We have 3 left in the numerator, and 7 in the denominator. Ans... 52. Reduce I- of A4 of - of -Lt to a simple fraction. 53. Reduce X- of 4- of - of -17- of 5 to a simple fraction. iQUPST.-199. How by cancelation? How does it appear that this method will give the true answer? Obs. Wvhat advantages does this method possess 3 6 116 rrEDUC'TON OF [SECT. VIL 54. R;dule 4 of -i of }4 of 4- of -Pq% to a simple fr wction. 55. Reduce 4 of ~ of i of - to a simple fraction. 56. Reduce 6 of o of 4 of -g to a simple fraction. 57. Reduce 4 of +1 of I - of 1 to a simple fraction. 58. Reduce 7 of -? of -b-f of f of - to a simple fraction. 59. Reduce ~ of - of 44 of -~ of 4 to a simple fraction. 60. Reduce 4- of 3- of of jr of of to a simple fraction. Note.-For reduction of complex fractions to simple ones, see Art. 239 CASE V. Ex. 61. Reduce - and - to a common denominator. Note.-Two or more fractions are said to have a common denominator, when they have the same denominator. Solution. —If both terms of the first fractioan -, are multiplied by the denominator of the second, it becomes -A; and if both terms of the second fraction -, are multiplied by the denominator of the first, it becomes -j. Thus the fractions -~% and g~ have a common denominator, and are respectively equal to the given fractions, viz: m-=~, and -a-=}. (Art. 191.) Hence, 200. To reduce fractions to a common denominator. Multiply each numerator into all the denominators except its own for a new numerator, and all the denominators together for a common denominator. 62. Reduce L, A, and - to a common denominator. Operation. 1X4X6=24 ) 3 X 3 X 6=54 the three numerators. 5X3X4=60 3 X 4 X 6=72 the common denominator. Ans..2, 4, and ~X. Ons. The reason that the process of reducing fractions to a common denom inator does not alter their value, is because the numerator and denominator of each of the given fractions, are multiplied by the same number.; and multiplyig QUEST.-J-Note. WThat is meant by a common denominator? 200. How are fractions ret dauced to a common denominator? Obs. Does the Process of reducing fractions to a corn mon denominator alter their value? Why not? AnTs. 200, 201.] FRACTIONS. 117 both the numerator and den(,minator of a fraction by the same number, does not alter its value. (Art. 191.) 63. Reduce,, -, I, and - to a common denominator. 64. Reduce -, -, a, and a to a common denominator. Reduce the following fractions to a common denominator: 65. Reduce A, j, 4, and -~. 69. Reduce.3, ~-, and 4J. 66. Reduce,, +, and 7. 70. Reduce 2, A-2, and H. 67. Reduce a, -, A-, and I-Z. 71. Reduce P-, _a, and -1-. 68. Reduce,,:-5, and i.'72. Reduce I, I, and -. CASE VI. 73. Reduce i, {, and ~ to the least common denominator. Analysis.-We first find the least Operation. common multiple of all the given de- 2)3 " 4 " 8 nominators, which is 24. (Art. 176.) 2)3 " 2 " 4 The next step is to reduce the given 3 " 1 "2 fractions to twenty-fourths without Now 2 X 2 X 3 X 2 =24, the altering their value. This may evi- least common denominator, dently be done by multiplying both terms of each fraction by such a number as will make its denominator 24. (Art. 191.) Thus 3, the denominator of the first fraction, is contained in 24, 8 times; now, multiplying both terms of the fraction - by 8, it becomes 8-. The denominator 4, is contained in 24,'6 times; hence, multiplying the second fraction - by 6, it becomes H-?. The denominator 8, is contained in 24, 3 times; and multiplying the third fraction A by 3, it becomes, 4-q. There.. fore z-, ~}, and k5- are the fractions required: Hence, 201[. To reduce fractions to their least common denominator. I. Find the least common multiple of all the denominators of the given fractions, and it will be the least common denominator. (Art. 176.) II. Divide the least common denominator by the denominator of each given fractirm, and multiply the quotient by the numerator; the products twill be the numerators of the fractions required. QUEsT. —201. Ilow are fractions reduced to the least common denominator? 118 REDUCTION OF [SEam. VIIL OBs. 1. This process, in effect, multiplies both the numerator r.nd denominator of the given fractions by the same number, and consequently does not alter their value, (Art. 191.) 2. The rule supposes each of the given fractions to be reduced to its lowest terms; otherwise, the least common multiple of their denominators may not be the least common denominator to which the given fractions are capable of being reduced. Thus, the fractions 4-, A, and -?,, when reduced to the least comnmon denominator as they stand, become -3, _j-6, and -9. But it is obvious that these fractions are not reduced to their least common denominator; for, they can be reduced to 1, 2, and A. Now, if the given fractions are reduced io the lowest terms, they become ~, ~, and a, and the least common mult'iple of their denominators, is also 4. (Art. 176.) 3. By a moment's reflection the student will often discover theleast common denominator of the given fractions, without going through the ordinary process of finding the least common multiple of their denominators. Take the fractions, t, and —; the least common denominator, it will be seen at a glance, is 4. Now if we multiply both terms of 1 by 2, it becomes a; and if we divide both terms of X by 3, or reduce it to its lowest terms, it becomes i. Thus the given fiactions are equal to 4, -, and 4, and are reduced to the lenst common delwminalor. 74. Reduce -, i, and I to the least common denominator. Operation. Now 2 X 2 X 3 X 2 = 24, the least com. denom. 2)4 " 6 " 8 Then 24- -4= 6, and 6X 3 = 18, the 1st num. 2)2 " 3 "4 24- 6=4, and 4X5=20, the 2d " 1 "3 " 2 24 —8=3, and 3X7-=21, the 3d " Ans. X JO-, and -H. 75. Reduce 4- and 4 to the least common denominator. Reduce the following fractions to the least common denominator:'7. X, -, i, and a, 84. 12, -,:, and —. 77. -, 6, and i. 85. i-AL,x1 q_, and -47o, 48Y 78. 1, A, -, and 11t 8. ~. -, z4S,:X, and -_a ~~4, 9, Si 17, 35'5 79.-4, -, A, and -X. 87. i,,, and A-. 80. i, 1,, and A-. 88. 2-84, 3, A7, and -'.8~ P-9,-Iand 4 88 367 15S 81. A,,, and 304. 89. A-, 1, 7, nd an. 82. ad,,, and ga. 9 -4, a:, A- and - 15. 83. 7, st 211 146 83. A, A, j, ad - 9nd11. A, and ^. Q.UEST.-Obs. Does this process alter the value of the given fiactions7 Why note What does this rule suppose respecting the given fractions? ARTS. 201, 202.1 FIRACTIONS. 119 ADDITION OF FRACTIONS. Ex. 1. A beggar meeting four persons, obtained a of a dollar from the first, 6 from the second, i from the third, and i from the fourth: how much did he receive from all? Solution.-Since the several donations are all in the same parts of a dollar, viz: sixths, it is plain they may be added together in the same manner as whole dollars, whole yards, &c. Thus, 1 sixth and 3 sixths are 4 sixths, and 4 are 8 sixths, and 5 are 13 sixths. Ans. 1-, or 21 dollars. Ex. 2. What is the sum of - and -? OCs. A difficulty here presents itself to the learner; for, it is evident, that 2 thir'ds and 3 fourths neither make 5 thirds, nor 5 fourths. (Art. 51.) This difficulty may be removed by reducing the given fractions to a common denominator. (Art. 200.) Thus, Operation. 3 X 3 = 9 the new numerators. 3 X 4=12, the common denominator. The fractions, when reduced, are - and -T; now 8 twelfths X 9 twelfths=1 7 twelfths. Ans. jj, or 1-~. 202. From these illustrations we deduce the following general RULE FOR ADDITION OF FRACTIONS. Reduce the fractions to a common denominator; add tkeir numerators, and place the sum over the common denominator. OBs. 1. Compound fractions must, of course, be reduced to simple ones, before attempting to reduce them to a common denominator. (Art. 198.) 2. Mixed numbers may be reduced to improper fractions, and then be added according to the rule; or, we may add the whole numbers and fractional parts separately, and then unite their sums. 3. In many instances the operation may be shortened by reducing the given fractions to the least com2mon denominator. (Art. 120) QuEsr. —202. How are fractions added? Obs. What must be done with compound fractions Hlow are mixed numbers added? -flow may the operation frequently be lhorteaed I 120 ADDITION OF [SECT. V1I. EXAMPLES. 3. What is the sum of 1, A, and i? Ans. 12-2. 4. What is the sum of ~, -, a, and A? 5. What is the sum of a-, A,, and? 6. What is the sum of A, A., +, and 1?'7. What is the sum of.a, -a, r, and -? 8. What is the sum of 9, i, -r, and A? 9. What is the sum of 5, -l', 7, and A? 10. What is the sum of i, A-, Az, and A? 11. What is the sum of 1, j, 1, 2, and ~? 12. What is the sum of a of 1, a of ~, and A? 13. What is the sum of ~ of -, - of -, and -7? 14. What is the sum of - of - of -z of -, and -1? 15. What is the sum of ~, + of 3, -a of 1, and A? 16. What is the sum of 4~, 84, 2L, 6~, and 9? 17. What is the sum of 41 of 6, - of 2, 3L, and 5? 18. What is the sum of A, 2a, -2 J,- and 9-? 5, 20, 30, 45I 19. What is the sum of 211, 35~, 3-2, and 2 of? 20. What is the sum of ~ of A, o2, 06, 1, and A? 21. ~What is the sum of ~ and -i%? Note.-It is obvious, if two fractions, each of whose numerators is 1, are re. duced to a common denominator, the new numerators will be the same as the given denominators. (Art. 200.) Thus, if ~ and - are reduced to a common denominator, the new numerators will be 12 and 8, the same as the given denominators. Now, the sum of the new numerators, placed over the product 128_ 20 of the denominators, will be the answer; (Art. 202;) that is 12x8-6 or 4, the answer required. Hence, 203. To find the stun of any two fractions whose numerators alre one. Add the denominators together, place this sum over their product, and the reszult will be the answer required. OBs. 1. The r'eason of this rule may be seen from the fact that the opera tion is the same as reducing the given fractions to a common denominator then adding their numerators. 2. When the numer'ato'rs of two factors are the samne, their sum may be found QUEST. —S3. How is the sum of any two fractions found whose numerators are 1 Obs. IIow find the sum of two fractions whose numerators are the same? ARTS. 203, 204.] FRACTIONS. 121 by multiplying the sum of the two denominatbrs by the common niumerator, and placing the result over the product of the given denominators Thus, the (4+-5)X3 9X3 27 - sum of and is equal to (44X5 4X520 or 2 22. What is the sum of -l~ and 17? Of A- and -1-? 23. What is the sum of -5 and 0? Of -T4- and -8? 24. What is the sum of -6I8 and ~~? Of _ t_ and 9 25. What is the sum of - and — ~? Of -1 and I-? 26, What is the sum of 2 and. -? Of -{ and 4- -? 27. What is the sum of En and a-? Of ~ and -~? 28. What is the sum of 5 and'? Yote. —The design in this and the following examples, is to incorporate the integers with the fractions, and express the answer fractionally. Solutionz. —5b —. (Art. 197. Obs. 2.) Now l3 -+=a- Ans. 204.- Hence, to add a whole number and a fiaction together. Reduce the whole number to a fraction of the same denominator as that of the given Ofraction; then add theigr numerators together. (Arts. 202, 197. Obs. 1, 2.) Note.-The process of incorporating a whole number with a fraction, is the same as that of reducing a mixed number to an improper fraction. (Art. 197.) 29. What is the suml of 45 and 2? 30. What is the sum of 320 atd -%? 31. What is the sum of 452 and -'4-ojT? 32. What is the sum of 635-'-1i-4227 — +-1625-? 33. What is the sum of 195-j- +600J- 5303' 160? 34. What is the sunm of 671-4+483J-'~84211-Z+4325~? 35. What is the sum of 590-LL+f100+~4005AZ+3020-_t~? 36. What is the sum of 2398 + 644-~ +16504-I+4500r? 37. What is the sum of 6563~-1-1000q-+1830-a+8396-l? 38. What is the sum of 356-t-+46~-{-1615-L-600 — 3212? 39, What is the sum of 41-1+105-t-3003-+241-3-472? 40. What is thE sum of 8672+- 163645+-1+1800a-t-66251-i-? 41. What is the sum of 26003-,~+19352- -92831~ +68693-? 42. What is the sum of 1925f6r —+45600O+-t- of o of 9? 43. What is the sum of A of 28-+6-+45-a-i+1 of 300? QuzST.-204. Hfow add a whole number andl a fraction? 122 SUBTRACTION OF Lb:ECT. V'I.'SUBTRACTION OF FRACTIONS. 205. Ex. 1. A man bought 7- of an acre of land, and afterwards sold -j3i of it: how much land had he left? Solution.-7 tenths from 9 tenths leave 2 tenths. Ans. e of an acre. 2. A laborer having received - of a dollar for a day's work spent A of a dollar for liquor: how much money had he left? Note.-The learner meets with the same difficulty here as in the second ex. ample of adding fractions; that is, he can no more subtract fifths from eighths, than he can add fifths to eighths; for, A offa dollar taken from m of a dollar will neither leave 4 fifths, nor 4 eighths. The fractions must therefore be reduced to a common denominator before the subtraction can be performed. Operation. 73X825=35 the numerators. (Art. 200.) 8X 5=40, the common denoniinator. The fractions become 4o- and u.a Now 3-a -2_L Ans. 206. From these illustrations we deduce the following general RULE FOR SUBTRACTION OF FRACTIONS. Reduce the given fractions to a common denominator; subtract the less numerator from the greater, and place the remainder over the common denozinator. OBs. Comnpoundfractionns must be reduced to simple ones, as in addition of fractions. (Art. 198.) EXAMPLES. 3. From a take A. Ans. of. 4. From -H take -%. 9. From -- take i. 5. From A5- take -~. 10. From - of - take ~ of 7. 6. From MH- take -. 11. From % of - take i of i 7. From 2x take -1 12. From - of 40 take a of 20. 8. From 7_ take —. 13. From X of -9 of A take - of. QUEST.-206. How is one fraction subtracted from another? 0/1s. Whlat is to be done with compound fractions.? a nrs. 205-208.] FRACTIONS. 123 207. Mixed numbers may be reduced to improper fractions, then to a common denominator, and be subtracted; or, the fractional part of the. less number may be taken from the fractional part of the greater, and the less whole number from the greater. 14. From 9-L take 7i. First Operation. Second Operation. 9-a4LZ 9+ 4 — 4 4 - 4 Ans. =-1-2, or 1-. Ans. 1i, or 11-. Note.-Since we cannot take 3 fourths from 1 fourth, we borrow a unit in the second operation and reduce it to fourtls, which added to the 1 fourth, make 5 fourths. Now 3 fourths from 5 fourths leave 2 fourths: 1 to carry to 7 makes 8, and 8 from 9 leaves 1. 15. From 25 — take 13-. 17. From 1'78 0- take 56a. 16. From 230~s- take 160G-o-. 18. From 7617-a5 take 482 —%%. 19. From 5 take I. Suggestion.-Since 3 thirds make a whole one, in 5 whole ones there are 15 thirds; now 2 thirds from 15 thirds leave 13 thirds. Ans. t3, or 4j. Hence, 208. To subtract a fraction from a whole number. Change the whole numnber to a fraction having the same denorrinator as tle fraction to be subtracted, and proceed as before. (Art. 197. Obs. 2.) OBS. If the fraction to be subtracted is a proper fraction, we may simply borrow a unit and take the fraction from this, remembering to diminish the whole number by 1. (Art. 69. Obs. 1.) 20. From 20 take n. Ails. 192. 21. From 135 take 9. 26. From 729 take 125.4 0 22. From 263 take 24-:j. 27. From 1000 take 25-ah. 23, From 168 take 30g. 28. From 563 take 5624-. 24. From 567 take 1004kt. 29. From 9263 take 999+. 25. From 634 take 342-V. 30. From 857 take 7858j. Quest. —207. How are mixed numbers subtracted? 208. How is a fraction subtracted from a whole number 1 6* 124 MULTIPLICATION OF L[SCT. VII. MULTIPLICATION OF FRACTIONS. 209. We have seen that multiplying by a whole number, is taking the multiplicand as many times as there are units in the multiplier. (Art. 82.) On the other hand, If the multiplier is only a part of a unit, it is plain we must take only a part of the multiplicand. That is, Multiplying by ~, is taking 1 half of the multiplicand once. Thus, 12 X =6. Multiplying by ~, is taking 1 third of the multiplicand once. Thus, 12 X -=4. Multiplying by, is taking 1 third of the multiplicand twicec Thus, 12Xs=8. Hence, 210. M2iultiplying by/a fraction is taklcing a certain PORTIOX of the nultiplicand as many times, as there are like portions of a unit in the multip9lier. OBs. If the multiplier is a Ornit or 1, the product is equal to the multiplicand; if the multiplier is grveatcer than a unit, the product is greater than the multiplicand; (Art. 82;) and if the multiplier is less than a unit, the product is less than the multiplicand. CASE I. 21 1. T uo multiply a fraction and a whole number together. Ex. 1. If I man drinks - of a barrel of cider in a month, how much will 5 men drink in the same time? Analysis.-Since 1 mana drinks - of a barrel, 5 men will drink,5 times as mueh; and 5 times 2 thirds are 10 thirds; that is, X 5=1&, or 3~. (Art. 196.) Ans. 3- barrels. Ex. 2. If a pound of tea costs - of a dollar, how much will 4 pounds cost? Sqlution. —~ X 4=-a; and 2=2s, or 2~t dolls. Ans. Or, since dividing the denominator of a fraction by any num bet, multiplies the value of the fraction by that number, (Art. 189,) QUEST.-209. What is meant by multiplying by a whole number 1 210. What is meant by multiplying by a fraction? Obs. If the multiplier is a unit or 1, what is the product, equal to? When the multiplier is greater than 1, how is the product, compared with the multipliand? When less, how? ARTS. 209-213.] FRACTIONS. 125 if we divide the denominator 8 by 4, the fraction will become A, which is equal to 2x, the same as before. Hence, 2 1 2. To multiply a fraction by a whole number. Iliultil2ly the numerator of the fraction by the whole number, and write the product over the denominator. Or, divide the denominator by the whole number, when this can be done without a remainder. (Art. 189.), OBS. 1. A fraction is multiplied into a number equal to its denominator by canceling the denominator. (Ax. 9.) Thus 7 X7=-4. 2. On the same principle, a fraction is multiplied into any factor in its de. nominator, by canceling that factor. (Art. 189.) Thus, - -X3= —. 3. Since multiplication is the repeated caddition of a number or quantity to itself, (Art. 80,) the student sometimes finds it difficult to account for the fact that the product of a number or quantity by a proper fraction, is always less than the number multiplied. This difficulty will at once be removed by refiecting that ndltiplyin, by a fraction is talin0g or repeeating a certain portion of the multiplicand as many times, as there are like portions of a unit in the Multiplier. (Art. 210.) EXAMPLES. 3. Multiply j1 by 15. Ans. -aL, or 10~. 4. Multiply z-7- by 8. 9. Multi ply A4Q by 165. 5. Multiply "- by 30. 10. Multiply 4a- by 100. 6. Multiply - by 27. 11. hM[ultiply -2zi by 530. 7. Multiply e~ by 45. 12. Maultiply 7-2 by 1000. 8. Multiply 4 — by 100. 13. Multiply:7zj by 831. 14. Multiply 12- by 8. Operation. 123 8 times a are -.~, which are equal to 5 and ~. 8 Set down the ~. 8 times 12 are 96, and 5 (which Ans. 1011. arose from the fraction) make 101. Hence, 213. To multiply a mixed number by a whole one. lM/ultiply the fractional paart and the whole numgber separately and unite the poroducts. QUEST.-212. How multiply a fraction by a whole numnber? Obs Hlow is a fraction!multiplied by a number equal to its denomilator? low by ally factot in its denominator * 213. Ilov is a misted number multiplied by a whole one? 126 MULTIPLICATION OF [SECT. VII 15. Multiply 45-1 by 10. Ans. 451-. 16. Multiply 813 by 9. 19. Multiply 127- by 35. 17. Multiply 31 iH by 20. 20. Multiply 48f-er by 47. 18. Multiply 148 — by 25 21. Multiply 250-lr- by 50. 214. 4Multiplying by a fraction, we have seen, is taking a certain portion of the multiplicand as many times, as there are like portions of a unit in the multiplier. Hence, To multiply by': D)ivide the multiplicand by 2. To multiply by': Divide the multiplicand by 3. To multiply by 1: Divide the multiplicand by 4, &c. To multiply by': Divide by 3, and multiply the quotient by 2. To multiply by -': Divide by 4, and multiply the quotient by 3. 2 1 5. Hence, to multiply a whole number by a fraction. Divide the multiplicand by the denominator, and multiply the quotient by the numerator. Or, multiply the given number by the numerator, and divide the product by the denominator. Obs. 1. When the given number cannot be divided by the denominator without a remainder, the latter method is generally preferred. 2. Since the product of any two numbers is the same, whichever is taken for the multiplier, (Art, 83,) the fraction may be taken for the multiplicand, and the whole number for the multiplier, when it is more convenient. 22. If 1 ton of hay costs 21 dollars, how much will -- of a ton cost? Operation. Analysis.-Since 1 ton costs 21 dollars, - of 4)21 a ton will cost A as much. Now, 1 fourth of 21 5L is -21; and a- of 21 is 3 times as much; but 3 21 21X3 63 4X3=- 2X3=4, or 153 dollars. Ans. 154 dolls. 23. Multiply 136 by l. Ans. 45 -. 24. Multiply 432 by -. 26. Multiply 360 by ]. 25. Multiply 635 by l. 2 7. Multiply 580 by i. QUEST.-215. How is a whole number multiplied I y a fraction? 216. How find a frao. itonal part of a number? A Rus. 21h-217.] FRAcrTION. 127 28. Multiply 672 by 3. 31. Multiply 660 by 7-. 29. Multiply 710 by 8. 32. Multiply 840 by A A. 30. Multiply 765 by X. 33. Multiply 975 by -4. 2 1 6. Since multiplying by a fraction is taking a certain portion of the multiplicand as many times, as there are like portions of a unit in the multiplier, it is plain, that the process of finding a fractional part of a number, is simply multiplying the number by the given fraction, and is therefore performed by the same rule. Thus, - of 12 dollars is the same as the product of 12 dollars, multiplied by i; and 12 X — 8 dollars. OBS. The process of finding a fractional part of a number, is often a source of confusion and perplexity to the learner. The difficulty arises from the erroneous impression that finding a fractional part, implies that the given number must be divided by the fraction, instead of being multiplied by it. 34. What is -7- of 457? Ans. 266 -i7. 35. What is J-A of 16245? 38. What is - of 5268? 36. What is- of 25000? 39. What is -22-i of 45260? 37. What is sir of 4261? 40. What is -iool of 452120? 41. Multiply 64 by 5. Operation. 2)64 We first multiply 64 by 5, then by ~, and the 51 sum of the products is 352. But multiplying by 320 - is taking one half of the multiplicand once. 32 (Arts. 82, 214.) Hence, _Ans. 352. 21 7. To multiply a whole by a mixed number. Multiply first by the integer, then by the fraction, and add the products together. (Art. 214.) 42. Multiply 83 by 7T. Ans. 597f. 43. Multiply 45 by 8~. 47. Multiply 225 by 30 -. 44. Multiply 72 by 10. 48. Multiply 342 by 20k. 45. Multiply 93 by 12I. 49. Multiply 432 by 35-. 46. Multiply 184 by 18g. 50. Multiply 685 by 42f. QIEST. —217. How is a whole number multiplied by a mixed number I 128 MULTIPLICATION OF [SECT. VII 51 Multiply 125 by 10-. 5. 56. Multiply 457 by 123. 52. Multiply 26 by 1029 57. Multiply 107 by 471-. 53. Multiply 256 by 17-19-. 58. Multiply 510 by 851-a 54. Multiply 196 by. - 59. Multiply 834 by 89 —L. 4. Mueltil y I9 b ~: 55. Multiply 341 by. 30-a. 60. Multly. 60. M ultiply 963 by 95. CASE II. 21 8. To multiply a fraction by a fraction. Ex. 1. A- man bought A of a bushel of wheat, at 7 of a dolla:per bushel' how much did he pay for it? Analysis.-Since 1 bushel costs I of a dollar, l of a bushel must- cost ~ of IS, which is -4 of a dollar; for, multiplying the denominator, divides the value of the fraction. (Art. 188.) Now, if l of a bushel costs -i0- of a dollar, - of a bushel will cost 4 times as much; and 4 times 7 are ~-, or -- dolls. (Art. 195.) Ans. -17 of a dollar,. Or, we may reason thus: since 1 bushel costs i of a dollar, a- of a bushel must cost { of - of a dollar. Now A of - is a compound fraction, whose value is found by multiplying the numera — tors together for a new numerator, and the denominators for a. new denominator. (Art. 198.) Solutioi.. —X-}= —, or 1-7 dollars, Ans. Hence, 21. 9 To multiply a fraction by a fraction. f~uiti2,ly the numerators together for a new numerator, and the denominators together for a new denominator. Ons. 1. It will be seen that the process of multiplying one fraction by another, is precisely the same as that of reducing compound fractions to simple ones. (Art. 198.) 2. The r'eason of this rule may be thus explained. Multiplying by a fraction is taking a certain part of the multiplicand as many times, as there are Iikhe parts of a unit in the multiplier. (Art. 210.) Now multiplying the denominator of the multiplicand by the denominator of the multiplier, gives the value of only one of the parts denoted by the given multiplier; (Art. 188;) we therefore multiply this new product by the numerator of the multiplier, to find the number of parts denoted by the given multiplier. (Art. 186.) QUEST.-219. How is a fraction multiplied by a fraction? Obs. To what is the pro"es of multiplying one fraction by another similar? &wrs. 218-220.] FRACTIONS. 129 2. Multiply ~ by. Ans. — 3& Multiply y -. 6. Multiply -f- by 4A. 4. Mul-tiply- ~ by.'7. Multiply 3L by 1s. 5. M!Wfultiply:La by o4. 8. Multiply 6-5 by 12~. 9. What is the product of A into A into ~ into 1 into i-? 10. What cost 6O yards of cloth, at 4~ dollars per yard? Analysis.-4- dollars=-, and 62 yards=-:-. (Art. 197.) Noiw 2- 9 or 30. (Art. 196.) Ans. 30 dollars. Hence, 220@ When the multiplier and multiplicand are both mized nunmbers, they should be reduced to inmproper fractions, and then bce multiplied according to the rule above. OBs. Mixed numbers may also be multiplied together, witloutt?'educing them to improper fractions. Take, for instance, the last example. We first multiply by 4, Oper'ation. the whole number. Thus, 4 times -3 are A; equal to 2 and -2; 61 set down the, and carry the 2. Next, 4 times 6 are 24, and 4f to carry are 26. We then multiply by 8, the fractional part. 2G63 Thus, i of 6 is 3; and - of 2 thirds is l. The sum of the two 3~ partial products is 30 dollars,' the samle as before. 30 dolls, i! 1. Multiply 63 by 2~7. 23. Multiply 246-i — by ~-5. s2. Multiply 8-i5- by 6~. 24. {Multiply 91 - by a of 4s. s13. Multiply 13~ by 177. 25. Multiply 1475 by z of 21.;44. Multiply 15- by 20~. 26.'Multiply 341 by - of 68. 115. Multiply 30~ by 44-~~. 27. Multiply 800 by 4 of 1000, 16. Multiply 63 -L by 50-. 28. Multiply 4 of 75 by - of 28.:i. Mlultiply 1'7- 3- by 254L. 29. Multiply 2~ by?- of 4 of 85. 18. Multiply 4795-E by 174-. 30. Multiply 7 of 2S by 4 of 61. 19. Multiply 61-1Z- by 32aJ. 31. Mu31 ltiply A of 10A- by 3 of 8-t. 20. Multiply 71-42- by 45-. 32. M ultiply ~ of 16 by -I of 9-0'. 21. Multiply 83 —9- by 614-5. 33. Multiply 7 of -% of 20 by 25-. 22. Multiply 96-4 by 724-. 34. 14Multiply n- of 651 by 4o-. 35. What cost 125~ bbls. of flour, at 73 dollars per barrel 36. What cost 2504 acres of land, at 251 dollars per acre? 37. If a man travels 404 miles per day, how far will he travel inu 135-4- days? Qrr sr. —2-0. When the multiplier andl multiplicand are mixed numbers, how proceed 130 MULTIPLICATION OF [SECT. VII. CONTRACTIONS IN MULTIPLICATION OF FRACTIONS. Ex.' 1. Multiply - by - and ftr and - and. Operation. Since the factors 3, 5 and 8 are' 1 7 7 common to the numerators and denomX X 2-22 inators, we may cancel therm; (Art. 191;) and then multiply the remainu, ing factors together, as in reduction of compound fiactions to sim.ple ones. (Art. 199.) Hence, 22 f. To multiply fractions by CANCELATION. Cancel all the Jfcctors commnon both to the numerators and denominsators; then multiply together the factors remaining in the ~numerators for a new numerator, and those remaining in the denonzinators jbr a new denominator, as in )reduction of comp2ound fractions. (Art. 199.) OBS. 1. The 1reason of this process may be seen from the fact that the produ'ct of the numerators is divided by the same nusmbees as that of the denominators, and therefore the vailte of the answer is not allteed. (Art. 191.) 2. Care lmust be taken that the factors canceled in the numerators are e.actly eqoal to those canceled in the denominators. 2. Multiply -3 by -- and {. Ans. -. 3. Multiply 3- by - -into. 7. Multiply. of - by -. 4. Mulliply ~ by 9y into 5. 8. Multiply 9 by j~ of -ro 5. lMultiply - by I into j-. 9. 9. ultiply I of 4 by -4-. 6. Mltiply y of. 1. ultiply -3 by - of 1. Multiply 3-a by -L- of 8. 11. Multiply - by X and n and- and -'. 12. Multiply - by - and -1-3 and -2 and 4. 13. Multiply j- by a and I — and -a- and -~. 14. Multiply 9- into I into A into -z into - into -. 15. MIultiply -1%- into -25 into I into - into A into l5. 16. Mulltiply iil into -r into into 8 — into 2A into A. 17. WAhat must a man pay for 3~- barrels of flour, when flcur il worth 6 dollars a barrel? QUEST. —221. IfIow are fractions multiplied by cancelation? Obs. How does it appean. that this process will give the true answer? What is necessary to be observe( with trgard to canceling factors? ARTS. 2-21-223.] FRACTIONS. 13] Analysis.-.31 bbls. is 1 of 10 Operation. bbls.; now since 1 bbl. costs 6 dolls. 6 price of 1 bblo dollars, 10 bbls. will cost 10 times 10 as much, or 60 dollars. But we 3)60 " of 10 bbls. wished to find the cost of only 3~ dolls. 20 " of 3~ bbls. barrels, which is ~ of 10 bbls. Therefore if we take ~ of the cost of 10 bbls., it will of course be the price of 3~ bbls. PROOF. —6 dolls. X 31=20 dolls., the same as before. NoIle.-In like manner, when the multiplier is 331, 3331, &c., if we multiply by 100, 1000, &c., I of the product will be the answer. Hence, 2 SS To multiply a whole number by 3L, 33~, 3331, &c. Annex as many ciphers to the multiplicand as theree are 3s in the integral part of the multiplier; then take ~ of the number thus o roduced, and the result will be the answer required. Os. 1. The reason of this contraction is evident fwom the principle that anne-xing a cipher to a number multiplies it by 10, annexing two ciphers multiplies it by 100, &c. (Art. 98.) But 31 is 1 of 10; 331 is - of 100, &c.; therefore annexing as many ciphers to the multiplicand, as thererare 3s in the inte-gral part of the multiplier, gives a product 3 times too large; consequently' of this product must be the tr'ue answer. 2. When the multiplicand is a mired number, and the multiplier is 31, 33k &C., it is evident we may multiply by 10, 100, &c., as the case may be, and i oS the number thus produced will be the answer required. 18. M~iultiply 158 by 33f. Ans. 5266-0. 19. Multiply 148 by 3~. 22. Multiply 297 by 333-3. 20. Multiply 256 by 33~. 23. Multiply 561L by 3~. 21. Multiply 1728 by 33. 24. Multiply 426~ by 33~. 2230 To multiply a whole number by 63, 66, 6662, &e. Annex as mzany ciphers to the nzultiplicand as there are 6s in the integral part of the nmultiplier; then take ~ of the number thus produced, and the result will be the answer required. OBs. The?reason of this contraction is manifest from the fact that 6i is } of 10; 6 of 100, &c. 25. What will 6~ tons of iron cost, at 75 dollars per ton? F.UEST. —222. tIow may a whole number be multiplied by 31, 331, &c. 1 223 Hoew may a, whole lnumber be multiplied by 61, 66-1, &c. 132 MULTIPLICATION OF [SECT. VIL Analysis. -6` tons is - of 10 Operation. tons. Now if 1 ton costs 75 dol- dolls. 75, price of 1 ton. lars, 10 tons will cost 10 times as 10 much, or 750 dollars; and - of 3)750 " of l0 tons", 750 dollars, (6=2- of 10,) are 250 500 dollars, whi:h is the answer 2 required. dolls. 500, " of 6a tons PROOF.-75 dolls. X 6= 500 dolls., the same as above. 26. Multiply 320 by 62. 28. Multiply 837 by 6.. 27. Multiply 277 by 66-. 29. Multiply 645 by 666.. 30. What will 12~ acres of land cost, at 46 dollars per acre i Analysis.-121 acres is ~ of 100 Op0eration.,cres; now since 1 acre costs 46 dol- dolls. 46, price of 1 A. lars, 100 acres will cost 100 times as 100 much, or 4600 dollars. But we wished 8)4600 " 100 jA. to find the cost of only 12-L acres, which dolls. 575 " 12 A. is ~ of 100 acres. Therefore ~ of the cost of 100 acres, will obviously be the cost of 12~ acres. PROOF.- 46 dolls. x 12 —575 dolls., the same as before. Note. In like manner, if the multiplier is 37-, 62b, or 87?, we may milltiFriy by 100, and f, f, or 8 of the product will be the answer. Hence, 22 4 To multiply a whole number by 121, 37-1, 62~, or 87~1-. Annex Lwo ciphers to the multizplicand, then take, a, a-, or;> of the number thus p2roduced, as the case may be, acnd the result will be the answer ~required.:Oes. The r'eanso. of this contraction may be seen from the fact that 121 is J, 371 is I-, 621 is -, and 871 is 8 of 100. 31. Multiply 275 by 37-~. 4Ans. 10312~. 32. Multiply 381 by 12~. 34. Multiply 643 by 62~. 33. Multiply 425 by 37~. 35. Multiply 748 by 871. 225. To multiply a whole number by 1-, 16-], 166-~, &c. Annex as many ciphers to the multi2plicand as there are integral figue,' in the nzcultplier, then ~ of the number thus produced will be the,product reg uired. AItTSa 224-226.] FRACTIONS. 133 O-Es. The rveason of this contraction is evident from the fact that 1I is 6 of 10; 166 is of 100 1 is of 1000, &c.'36~ Wlhat will 16. bales of Swiss muslin cost, at 735 dollars per bale? Solution.-Annexing two ciphers to 735 dolls., it becomes V31500 dolls.; and 73500- 6=12250 dolls. Ans, 370 Multiply 767 by 1.-. 39 Multiply 4:89 by i 6. 38, Multiply 245 by 16'. 40. Multiply 563 by 166. Note.-Specific rules might be added for multiplying by i-L, 11-, 11-, 8, 83}, 8331- 6, 6&c, but they will naturally be suggested to the inquisitive mind from the contractions already given. DIVISION OF FRACTIONS. CASE I. 2260 Dividing a fraction by a whole number. EX. 1. If 4 yards of calico cost - of a dollar', what will 1 yard cost? Analysis.-1 is 1 fourth of 4; therefore 1 yard will cost 1 fourth part as much as 4 yards. And 1 fourth of 8 ninths of a dollar, is 2 ninths. Ans. - of a dollar. Operation. We divide the numerator of the fraction by. 4, —:- Ans, 4, and the quotient 2, placed over the donomio nator, forms the answer required. 2. If 5 bushels of apples cost -iL of a dollar, what will 1 bushel cost? Operation. Since we cannot div'lae the numer 1 11 11 a. tor by the divisor 5, without a re. *-5 1SX5' 60or; ns mainder, we multiply the denomina, tor by it, which, in effect, divides the fraction. (Art. 188.) PROOF. ——, dolls. X 5=- dolls., the same as -above. Hence, 134 DIVISION OF [SECT. V[L] 2270 To divide a fraction by a whole number. Divide the numerator by the whole nulmber, when it can be don.w without a remaindeer; but when this cannot be clone, vmultitly th0 denominator by the whole number. 3. What is the quotient of - divided by 5? First Jethod. Second f~ethod. 15 3 15 ~ 15 15 3 5-20 Ans. 20 _5- 20X5-00, or 20 An. 4. Divide - by 9. 7. Divide 7 by 12. 5. Divide 2i by 7. 8. Divide 52- by 25. 6. Divide V7 by 10. 9. Divide';o by 29. CASE II. 228. Dividing a fraction by a friaction. 10. At * of a dollar a basket, how many baskets of peaches can you buy for - of a dlollar? Analysis.-Since * of a dollar will buty 1 basket, - of a dollar will buy as many baskets as - is contained times in -; and * is contained in -, 4 times. Ans: 4 baskets. 11. At - of a dollar per yard, how many yards of cloth can be boug'ht for * of a dollar? Ons. 1. Reasoning as before, 7 of a dollar will buy as many yards, as. of a dollar is contained in 7. But since the fractions have different denominators, it is plain we cannot divide one numerator by the other, as we did in the last example. This difficulty may be remedied by reducing the fractions to a common denominator. (Art. 200.) First Operation. -M and - reduced to a common denominator, become -2 and -4. (Art. 200.) Now 2-1 =-21; and _1 s Ans. 1-1s yards. OBs. 2. It will be perceived that no use is made of the covmmon denominator, after it is obtained. If, therefore, we invert t/te diviso2r, and then multiply tho two fractions together, we shall have the same result as before. Second Operation. iX3 (divisor inverted)=-2-, or 1-%5- yards, the same as above.* Q UEST.-227. I-low is a fraction divided by a whole number I IARTS. 227-230.] FRACTIONS. 135 22~9 Hence, to divide a fraction by a fraction. 1. If the given fractions have a common denominator, divide the numerator of the dividend by the numerator of the divisor. II. WThen the fractions have not a common denominator, invert ihe divisor, and p2roceed as in multi2plication of fractions. (Art. 219.) OBs. 1. When two fractions have a common denovminator, it is plain one Inumerator can be divided by the other,' as well as one whole number by ansthler; for, the parts of the two fractions are of the same denominvation. 2. When the fractions do not have a common denominator, the 9season that (verting the divisor and proceeding as in multiplication, will produce the tgqte a:?Inswe?, is because this process, in effect, reduces the two fractions to a coN1in0on denomivnaton, and then the numerator of the dividend is divided by the numerator of the divisor. Thus, reducing the two fractions to a common denominator, we multiply the numerator of the dividend by the denominator of She divisor, and the numerator of the divisor by the denominator of the divirend; (Art. 200;) and, then dividing the former product by the lattcr, we have the same combizlrtiton of thee sesrc e',lbe7rs as in the rule above, which will consequently produce the s(eae resvlt,. AWe do not smlulti l; y the two dlfenominators together for a commlon denomninator; for, inl dividing, no use is made of a comnmon denominator when found, thlerefore it is unnecessary to obtain it. (Art. 228. Obs. 2.) The object of inver/etiqg the divisor is simply for convezielce in multiplying. 3. Compoulnd fractions occurring in the divisor or dividend, must be re-:tuced to simple ones, and smixed numbers to improper fractions. 230@ The principle of dividing' a fraction by a fraction may:i:also be illustrated in the following manner. Thus, in the last example, Dividing' the dividend z by 2, the quo- Operation. sient is -1Lo (Art. 188.) But it is required Z — 2-i to divide it by 1 third of two consequently - X 3 = — the — z- is 3 times too small for the true And ~ — 1 1 — And, quotient; therefore multiplying -_z — by 3, will give the quotient required; and -7-X 31 —-, or 1-a. Aote.-By examination the learner will perceive that this process is precisely QUEsT.v229. How is one fraction divided br another when they have at common denominator? How, when they have not commot denomlinators? Obs. When the f'actions have a common denominator, how does it appear that dtividing any nummierator by the otlher wvill give the true answer? Wvhen the fiactions have not a common denorninator, how d)es it appear that inverting the divisor and proceeding as in multiplication will give the true answer? What is the object of invertintg the divisor? IIow proceed when the divis(o or dividend are compound fractions or mixed numbers? 136 DIVISION or [SECT. VII the same in effect as the preceding; for, in both cases the denominator of th& dividend is multiplied by the numerator of the divisor, and the numerator of -the dividend, by the denominator of the divisor. 12. Divide 3- of -3- by 23. Ans. L8, or -i. 13. Divide 8-9 by 3~. Ans. o, or 2z0-. 14,. Divide V4 by. 16. Divide 555 by 16a. 15. Divide. byT l-l 17. Divide 46a by 68~. 2310 The process of dividing fractions may often be conitracted bS- canceling equal factors in the divisor and dividend; (Art. i46;) or, after the divisor is inverted, by canceling factors, which are common to the numerators and denominators. (Art. 191.) 18. Divide -L of + of -2r by { of a of J. 0leration. For convenience we arrange the numera$ 1 tors, (which answer to dividends,) on the 74 right of a perpendicular line, and the de-!1 $ nominators, (which answer to divisors,) onr s 5 the left; then canceling the factors, 2, 3, 4,, and 7, which are common to both sides,, 1 7~ (Art. 151,) we multiply the remaining facll -1 A n2 5. tors in the numerators together, and those remaining' in the denominators, as in the rule above. Hence, 232. To divide fractions by CANCELATION. Hacving inverted the divisor, cancel all the factors commzon both to the numerators and denominactors, and the product of those re-?maining on the right of the line placed over the product of those remaining on the left, will be the answer rcequired. Ons. 1. Before arranging the terms of the divisor for cancelation, it is always necessary to invert them, or suppose them to be inverted. 2. The'reason of this contraction is evident from the principle, that if the, numerator and denominator of a fraction are both divided by the sacme is.b e-r, the,value of the fraction is not altered. (Arts. 148, 191.) 19. Divide 18' by 6-f. Answver 3. EUST.-2393, -low divide fractions by cancelation? -Iowr arrange the terms of tho' glven fractlions? Obs. What irmist be done to the (livisorbelfre arranging its terms. ItoW' does it appea.r that this contractiroen will give-the true answver? ARTS. 231 —234.] FRACTIONS. 137 20. Divide a- of - by of T 23. Divide of by 0 of q. 21. Divide a7 of L- by 6-. 24. Divide o of - of a by {. 22. Divide 153- by -9 of 4. 25. Diviide- of 7 by -- of 42. 26. Divide 4L- of -.. of - of ~13 by -2 of — a of of 5. CASE III. 23$0 o ]Dividing a vhole nzumber by a fraction. f2. How many pounds of tea, at -" of a dollar a pound, can be -iutght for:15 dollars? Analysis.-Since - of a dollar will buy 1 pound, 15 dollars will.buy as many pounds as -a is contained times in 15. Reducing the d ividend 15, to the form of a fraction, it becomes ih; (Art. 197. C bs. 1;) then. inverting the divisor and proceeding as before, we hal'e e -X-%=~-,- or 20. Anrs. 20 pounds. Or, ive may reason tiras: J- is contained in 15, as many times,there are l/obUts inl 15, viz: 60 times. But 3 fourths will be rntaiaed in 15, only a, tiitd as many times as 1 fourthll, and J3-8-320, the sa-ie riesltu as Lefore. Hence, 234o To divide a wAlYhole Ium ber by a firaction. Reduce ithe wiaole 2-,iib(.e lo zthe i? o cc ia actionz, ( Art. 1 07. )bs. 1,) and then proceed accordlizq to the rule Job dividing a fraction by a fraictionz. (Art. 229.) Or, nizultpl.y the wuhole numvzber by tize denominator, cand divide he prodzlcct by the nzumerator. OBS. 1. When the divisor is a nixzedi number, it must be reduced to an improper fraction; then proceed as above. Or, reducing the dividend to a fiaction having the sncne denrominartor, (Art. 197. Ohs. 2,) we may divide one numerator by the other. (Art. 2'29. I.) 2. If the divisor is a scit or 1, the quotient is ec/lan to the (lividend; if the divisor is g'crzcr than a unit, the quotient is less than the dividend; and if the divisor is less than a unit, the quotient is g?'eactr than the dividend. 28. HT-ow much cloth, at o3 dollars per yard, can you buy for 2.8 dollars? OQUEST. —-,34, -How is a whole number divided by a fraction? Obs. Tlow by a mixed:tillnlber? 138 DIVISION OF [SECT. VII. Operation. Since the divisor is a mixed number, 3~)28 we reduce it to halves; we also reduce 2 2 the dividend to the same denominator;" 7) 56 halves. (Art. 197. Obs. 2;) then divide one nu.i Ans. 8 yards. merator by the other. (Art. 229. I.) 29. Divide 75 by I. 32. Divide 145 by 12~. 30, Divide 96 by A. 33. Divide 237 by 25-. 31, Divide 120 by 103. 84. Divide 425 by 31s. CONTRACTIONS IN DIVISION OF FRACTIONS. 235. When the divisor is 3 3, 33, 333, &c. Multiply the dividend by 3, divide the product by 10, 100, o:r 1000, as the case may be, and the result zwill 1e the true quotienet. (Art. 131.) GOBS. The r'eason of this contraction will be understood fromi the principl that if the divisor and dividend are both multiplied by the sncaie number, t} quotient will not be aler'ed. (Art. 146.) Thus 31X3=10; 331X3=10l 3331X3=1000, &c. 35. At 3~ dollars per yard, how many yards of cloth can be bought for 561 dollars? Operation. We first multiply the dividend by'u dolls. 561 then divide the product by 10; for, mul-. 3 tiplying the divisor 3~ by 3, it becomes 10. 110)16813 (Art. 146.) Ans. 168-r3 yds. 36. Divide 687 by 33~. Ans. 220-'1 37. Divide 453 by 33~, 38. Divide 2783 by 3331. 236. When the divisor is 1 2, 163, 166-s, &C..Multiply the dividend by 6, and divide the product by 10, 100, or 1000, as the case may be. OBs. This contraction also depends upon the principle, that if the divisor and dividend are both multiplied.by the same number, the quotient will not b6 altered. (Art. 14G.) Thus, 1X6=10; 16lX6=100; 166 X6-1000, &c. AL&RTS. 235-239.] FRACTIONS. 139 39. What is the quotient of 725 divided by 16-? Solutio6n.-725X6=4350; and 4350- -100=43~ Ans. 40. Divide 367 by 1j. 42. Divide 849 by 16-2. 41. Divide 507 by 16-. 43. Divide 1124 by 166-". 237. When the divisor is 1-, 11k, 11.1-, &c..Multiply the dividend by 9, and divide the product by 10, 100, or 1000, as the case mzay be. OBs. This contraction depends upon the same principle as the preceding rThus, 1L-X9-=10; 11-X9 — 100; 111-X9=-1000, &c. 44. Divide 587 by 11x..Solution. —587 X 9 — 5283, and 5283 -- 100=- 52 -M3or Ans. 4i5. Divide 861 by 1L. Ans. 774n-. 46. Divide 4263 by 11d. 47. Divide 6037 by 111x. AOote. —Other methods of contraction might be added, but they will naturally su:igest themselves to the student, as he becomes familiar with the pnnciples of fractions. ~23~0 From the definition of complex fractions, and the manners of expressing them, it will be seen that they arise from di41 vis.on of fractions. (Art. 183.) Thus, the complex fraction, is the same as -.'a; for, the numerator, 4x-L, and the denominatos, -=A-; but the numerator of a fraction is a dividend, and the dehominator a divisor. (Art. 184.) Now, —.- o. which is a simple fraction. Hence, 239. To reduce a complex fraction to a simple one. Consider the denominator as a divisor, and proceed as in divzs} ion of fractions. (Arts. 229, 232.) OBs. The reason of this rule is evident from the fact that the denornmtzator if a fraction denotes a divisor, and the numeratol', a dividend; (Art. 184; lence the process required, is simply performzing the divi.lion which is ex..re:ssed by the given fraction. Q;UEST. —238. Fronm what do complex fractions arise 239. tIow reduce them to sin. le'ifractions? 7 140 conMPJ,Ex [SECT. 1Vl 48, Reduce 4-d- to a simple fraction. Solution.-4ba= -, and 7:=-L. (Art. 197.) 3 4 4 Now 2+9X# -, or _a Ans. Reduce the following complex fractions to simple odes: 8 5z 49. Reduce 8y 53. Reduce e a-5' 50. Reduce 6. 54. Reduce 2~ 51. Reduce 55. Reduce 7 9 9 52. Reduce - 56. Reduce 5 4 15 36 240. To multiply complex fractions together. First reduce the complex fractions to simple ones; (Art. 3f,,;) then arrange the terms, and cancel the coemmnon factors, as in nmq:ub tiplication of simple fractions. (Art. 219.) Oss. The terms of the complex fractions may be arranged for reducing theInm to simple ones, and for multiplication at the same time. 57. Multiply by -47 Operation. The numerator 3x=2 (Art. 197.) Place the 7 on the right hand and 2 on the left'of ATK5 the perpendicular line. The denominator 2:~ t ~ l =-1=2, which must be inverted; (Art. 239;,) 9 $ i. e. place the 12 on the right and the 5 on 915=~. Ans. ~fithe left of the line, 1 —a, and 41=-, both of which must be arranged in the same malnner as the terms of the multiplicand. Now, canceling the comnmon factors, we divide the product of those remaining on the right of the line by the product of those on the left, and the answer is 6. (Art. 219.) QUEST.-240. How are complex fractions multiplied together? 241. How is one coam, plex fraction divided by another? ARTS. 240-242.] FRACTIONS. 141 58. Multiply 2by 60. Multiply by into 9 up 4~ 2~1- 2~ 1~ 590. Multiply by 4. 61. Multiply 2~ by 2 into 8-a y 17~' 241. To divide one complex fraction by another. Reduce the complex fractions to simple ones, then proceed as in division of simple fractions. (Arts. 229, 239.) 4~ ~ 62. Divide 4- by. 14 4 4 -, 9 4 36 Q 1 4 4 Soolution. -2-X and:-X7-=-. (Art.239.) 2~ 2 9 1=' l3- 3 2'' 36 4 36 21'756 Now, X _ -o. Ans. 18 21 18 4 72 9X4 1X4 Or, since the given dividend=-2- and the divisor3X 7; 2X9 3X7' 9X4 3X'7 tthen x =the answer. (Art. 231.) 2X9 1X4 9X4 3X7 OX X3X7 21 But,.(Art. 232,) 2X 9 X —- 4=XX= -- or 10 Ans 63a Dividoe by l. 64. Divide 4 by.2 4 4 ~ APPLICATION OF FRACTIONS. 242. Ex. 1. A merchant bought 15 S yards of domestic flannel of one customer, 19} of another, 12~ of another, and 41-a3- of another: how many yards did he buy of all? 2..A grocer sold 161 lbs. of sugar to one customer, 112~ to anrother, and 33 to another: how many pounmds did he sell? 3. A clerk spent 26a dollars for a coat, 93 dollars for pants, 6& dollars for a vest,.51 dollars for a bat, and 6~ dollars for a'pair of boots: how much did his suit cost him? 4. A man having bought a bill of goods amounthig to 85-13- dolb >iars, handed the clerk a bank note of 100 dollars: how much:change ought he t'i receive back? 42 DIVISION OF [SECT. VII. 5. A lady went a shopping with 135 dollars in her purse' she paid 17 — dollars for silk, 35 dollars for trimmings, 371 dollars for a shawl, and 144 dollars for a muff: how much money had she left? 6. A man having 1563-s- dollars, spent 36543 dollars, and lost5624 dollars: how much had he left? 7. What will 563 sleep cost, at 24 dollars per head? 8. What cost 748 bcarrels of flour, at 7a dollars per barrel? 9. What cost 3783 yards of cloth, at 4 dollars per yard? 10. What cost 1121 -Ir- lbs. of tea, at 5 shillings per pound? 11. What cost 430 gallons of oil, at 14 dollar per gallon? 12. What cost 3 of an acre of land, -at 150 dollars per acre? 13. A man worth 25000 dollars, lost -2- of it by fire: what, was the amount of his loss? 14. A garrison had 856485 pounds of flour; after being blockaded 60 days, it was found that _2~ of it were consumed: how many pounds of flour were left? 15. At 174 dollars per ton, what cost 103L tons of hay? 16. How many bushels of corn will 115-3- acres produce, at 31 bushels per acre? 17. What cost 6754 tons of iron, at 454 dollars per ton? 18. If a ship sails 140I-ia miles per day, how far will she sail. in 491 days? 19. If a Railroad car should run 414 miles per hour, how far, would it go in 12 days, running 104 hours per day? 20. A young man having a patrimony of 12234 dollars, spent, 4 of it in dissipation: how much had he left? 21. At 4 of a dollar per yard, how many yards of satinet can be bought for 124 dollars? 22. How many pounds of tea, at a of a dollar a pound can you buy for 131 dollars? 23. How many gallons of molasses, at A of a dollar per er gallon can you buy for 235 dollars? 24. At 8 pence a pound, how many pounds of sugar can ycuL buy for 1634x pence? 25. At 5- pence a yard, how many yards of lace can be bought, for 279 pence? ART. 242.] FRACTIONS. 143'26. A dairy-man has 2294 pounds of butter whiclh he wishes to paclI in boxes containing 84- pounds each: how many boxes will it require? 27. A farmer wishes to put 384 bushels of apples into barrels, each containing 21 bushels: how many barrels will it require? 28. If 43 yards of cloth make a suit of clothes, how many suits will 1414 yards make? 29. One rod contains 51 yards: how many rods are there in 21.0 yards? 30. A merchant paid 204i dollars for 57 yards of cloth: how much was that per yard? 31. A grocer sold 50 barrels of flour for 311L dollars: what lid he get per barrel? 32. A merchant wishes to lay out 657- dollars for wheat, which worth 14 of a dollar a bushel: how much can he buy? 33. At 184 cents a dozen, how many dozen of eggs can you,,uy for 874 cents? 34. A grocer sold 154 pounds of coffee for 934 cents: how much was that a pound? 35. A shopkeeper sold 164 yards of satin for 163 —- shillings: aow much was that per yard? 36. Bought 19 sacks of wool for 2504} dollars: what was that per sack? 37. Paid 5754 dollars for 964 yards of cloth: what was the cost per yard? 38. Paid 15654 dollars for iron, valued at 37L4 dollars per ton: iiow many tons were bought? 39. Paid 13154 dollars for the transportation of 1286 barrels of pork: what was that per barrel? 40. Bought 3754 pounds of indigo for 6523 dollars: what was the cost per pound? 41. Paid 16794 dollars for 475 kegs of lard: how much was that per keg? 42. If an army consumes 563Z pounds of meat per day, how long will 150000 pounds supply it? 43. The cost of makling 254 miles of Railroad was 856035- dolljars: what was the cost per mile? 144 COMPOUND NUMBERS. [SECT. T SECTION VIII. COMPOUND NUMBERS. ART. 2430 Numbers which express things of the same kind or denomination, are called Siml2le lYumzbers. Thus, 3 oranges, 7 books, 12 chairs, &c., are simple numbers. Numbers which express things of different kinds or denominations, as the divisions of money, weight, and vmeasure, are called COMPOUND NUMBERS. Thus, 15 shillings 6 pence; 10 bushcl 3 pecks, &c., are compound numbers. Oss. The origin of Compound Numbers is ascribed to the wants and net sities of the earlier ages of the world. Their divisions and subdivisions generally irregular, and seem to have been suggested by the caprice, or the lit. ited business transactions of the rude ages of antiquity. It is much to be re: gretted, both on account of simplicity and their adaptation to scientific pur poses, that their different denominations were not graduated according to th' law of ibncrecase in the decigmal notation. Note.-Compound Numbers, by some authors, are called Denominai Numbers. FEDERAL MONEY. 244. Federal.Mioney is the currency of the United States The denominations are, Eagles, Dollars, Dimnes, Cents, and Mill 10 mills (m.) make 1 cent, marked ct. 10 cents cc 1 dime, " d. 10 (limes C" dollar, " doll. or $. 10 dollars'" 1 eagle, E. OBs. 1. Federal money was established by Congress, Aug. 8, 1786. It ia based upon the principles of the decimal notation. The law of icrease ar radix, is the same as that of simple numbers, and it is confessedly one of the most simple and comprehensive systems of currency in the civilized world. Previous to its adoption, English or sterling money was the principal currency "of the country. QUCEST.-243. What are simple numbers? What are compoundl numbers? 244. Whist s Federal money? Recite the table. Obs. When and by whom nwas it established ARTS, 243 —246.] COMPOUND NUMBERS. 145 2. The names of the coins or denominations less than a dollar, are significant of their value. The term dime, is derived from the French disme, which signifies teay; the terms cent and aill, are from the Latin centna and ille, the former of which signifies a hundred, and the latter a thousaand. Thus, 10 dimes, 100 cents, or 1000 mills, make 1 dollar. 3. The sign ($), which is prefixed to Federal money, is called the Dollar zar/c. It is said to be a ccntraction of " U. S., the initials of United States, which were originally prefixed to sums of money expressed in the Federal currency. At length the two letters were moulded or merged into a single character by dropping the curve of the U, and writing the S over it. Thus, the suml of seventy-five dollars, which was originally written " U. S. 75 dollars, is now written $75. 24-5. The national coins of the United States are of three kinds, viz: gold, silver, and copper. l. The golds coin are the eagle, the double eagle,* half eagle, guar ter eagle, and gold dollar?.* The eagle contains 258 grains of standardc gold; the half eagle and quarter eagle lile proporlionsl.t 2. The silver coins are the dollar, half dollar, quarter dollar, the dime, half dime, and three-cent-piece. The dollar contains 412- grains of standard silver; the otherslike proportions.t 3. The copper coins are the cent, and half cent. The cent contains 168 grains of puTre copper; the half cent, a like propoRition.jt /ills are not coined. OBs. The fineness of gold used for coin, jewelry, and other purposes, also the gold of commerce, is estimated by the number of parts of gold which it contains. Pure gold is commonly supposed to be divided into 24' equal parts, called carats. Hence, if it contains 10 parts of alloy, or some baser metal, it is said to be 14 carats fine; if 5 parts of alloy, 19 carats fine; and when absolutely pure, it is 24 carats fine. 24G6 The p2resent standard for both gold sand silver coin of the United States, by Act of Congress, 1837, is 900 parts of pure QUEsT.-245 Of how many Ilnds are the coins of the United States? What are they? What are the gold coins? The silver coins? The copper? Obs. Hlow is the fineness of gold estimated? Into howv many carats is ptre gold supposed to be divided? When it contains 10 parts of alloy, how fine is it said to be? 5 parts of alloy? 246. What is the present standard for the gold and silver coin of the United States? What is the alloy of gold coin?. What of silver coin? * Added by Act of Congress, 1849. t According to Act of Congress, 1837. 146 COMPOUND NUMBERS. [SECT. VIIT metal by weight to 100 parts of alloy. The alloy of gold coin is composed of silver and copper, the silver not to exceed the copper in weight. The alloy of silver coin is pure copper. Note.-The origiqnal standard for the gold coin of the United States by Act of Congress, 1792, was 22 parts of pure gold to 2 parts of alloy; the alloy consisting of 1 part silver and 1 part copper. The or iginal standard for the silver coin was 1489 parts of pure silver to 179 parts of alloy; the alloy being of pure copper. The eagle by the same act contained 270 grains of standard gold. The d./-,.a' contained 416 grains of standard silver. The cent contained 11 pennyweights, or 264 grains of pure copper. STERLING MONEY. 247. English or Sterlingy Ml]oney is the national currency of Great Britain. 4 farthings (qr. or far.) make I penny, marked d. 12 pence " 1 shilling, " s. 20 shillings " 1 pound, or sovereign, ~. 21 shillings 1 guinea. OBS. 1. It is customary, at the present day, to express farthings in fractions of a penny. Thus, 1 qr. is written 4 d.; 2 qrs. 4 d.; 3 qrs. 4 d. 2. The Pound Sterling is represented by a gold coin, called a Sovereiger,. According to Act of Congress, 1842, its value is 4 dolla6rs and 84 cents. Hence. the value of a shilling is 24- cents; that of a penny 2 cents, very nearly. 3. The letters A. s. d. and q. are the initials of the Latin words, libra, soli.. dls, denacrins, and quacedran~s, which respectively signifiy a pouand, shillin7g, penniy, andlfart/bing or quarter. The mark /, which is often placed betweerl shillings and pence, is a corruption of the longf. Note.-l. Sterling money is supposed by some to have received its name from the Easterllings, who it is said first coined it; others think it is so called to distinguish it from stocks, &c., whose value is nowminal. 2. The poutnd is so called, because in ancient times the silver for it weighed a pound Troy. A pound Troy of silver is now worth 66 shillings, or ~63, 6s. The GuiqLea is so called, because.the gold of whicl it was originally made, was brought from Guinea, on the coast of A2frica. 248. The following' denominations are fiequently met with, viz: the Groat: —4d.; the Crown=-s.; the Noble-= s. 8d.; QUEST..-247. W~hat is Sterling Money? Repeat the Tal le? Obs. Ic w are farthings usually expressed? low is a pounld sterling represented? W~hat is its A alue in dollats and cents? ARTS. 247-250.] COMPOUND NrUMBERS. 14" the Angel-lOs.; the Mark=13s. 4d.; the Pistole=16s. 10d. the Moidore:-27s. OBS. The present standard gold coin of Great Britain, consists of 22 Darts p2utre gold, and 2 parts of copper.* The weight of a Sovereign or ~, is 5 pwts., 3- -- grains. The standard silver coin consists of 37 parts of pure silver, and 3 parts of copper. The weight of a shilling is 3 pwts. 15 X-I grs. In copper coin, 24 pence weigh 1 pound avoirdupois. TROY WEIGHT. 249. Troy Weight is used in weighing. gold, silver, jewels, liquors, &c., and is generally adopted in philosophical experiments. 24 grains (gr.) make 1 pennyweight, marked pwt. 20 pennyweights " 1 ounce " oz. 12 ounces " 1 pound, cc lb. OBS. 1. The abbreviation oz., is derived from the Spanish oez'ca, which slgnifies an oe2nce. 2. The standar'd of Weights and Measures is different in different countries, and in different States. of the Union. In 1834, the Government of the United States adopted a uniform standard, for the use of the several Custom-houses and other purposes. 250 The standard Uniit of Weight adopted by the Governmeent, is the Troy Pouind of the United States Mint. It is equal to 22.T74422 cubic inches of distilled water, at its maximum density,+ the barometer standing at 30 inches, and is identical with the Imperial Troy pound of Great Britain, establisbed by Act of Parliament, in 182.0. OBs, The weights and meascres in present use, were derived from very imn. pResect and varciable standards. A gr'ain of wheat, taken from the middle of the ear or head, and being thoroughly dried,-was the original element of al. weights used in England, and was thence called a grain. At first, a weight (UEST.-249. In vwhat is Troy Weight used? Repeat the Table? Obs. Do all the S3tates have the same standard of weights and Tneasures? 250. What Is the standard Anit of Nweirght adopted by the Government of the United States?.Note. lYhen was Troy iWeight introduced into Europe 3 From what was its name derived. * Hind's Arithmetic; also, Hutton's Mathematics. T1 The maximnum density of water, according to Mr. Hassler, is at the temperature of 39'83 deg. Fahrenheit. t The Troy pound of the U.S. Jliint, is an exact copy, by Captain Kater, of the British Imperial Troy poulnd. Report of the Secretary of the Treasury, March 3, 1831.,7I 148 COMPOIUND NUMBERS. [SECT. VIII., equal to 32 gra ns, was called a peqnnywesghlt, from its being the weight of the silver penJy then in circulation. At a later period the pennyweight was divided into 24 equal parts instead of 32, which are still called grains, being the smallest weight now in common use. Note.-Troy Weight was formerly used in weighing articles of every kin-X. It was introduced into Europe from Cairo in Egypt, about the time of the Crusades, in the 12th century. Some suppose its name was derived from Tr7oyes, a city in France, which first adopted it; others think it was derived finom Troy-novant, the former name of London.* AVOIRDUPOIS WEIGHT. 2 5 1. Avoirdupois Weight is used in weighing groceries anta all coarse articles; as, sugar, tea, coffee, butter, cheese, flour, hay; &c., and all metals except gold and silver. 16 drams (dr.) make 1 ounce, marked oz. 16 ounces "' 1 pound, " lb. 25 pounds' 1 quarter, " qr. 4 quarters, or 100 lbs. " 1 hundred weight, " cwdt. 20 hundred weight "1 ton, " T. Note.-In weighing wool in England, 7 pounds make I clove; 2 cloves, 1 stone; 2 stone, 1 tod; f6 tods, 1 wey; 2 weys, 1 sack; 12 sacks, 1 last; 240 pounds, 1 pack. OBs. 1. Formerly it was the custom to allow 112 pounds for a hundred weight, and 28 pounds for a quarter; but this practice has become nearly cquite obsolete. In buying and selling all articles of commerce estimated by weight, the laws of most of the States as well as general usage, call 100' pounds a hundred weight, and 25 pounds a quarter. 2. Gross weight is the weight of goods with the boxes, casks, or bags whicf zontain them. Net weight is the weight of the goods only. 252. The Avoirdulpois Pound of the United States, is equal to 27.701554 cubic inches of distilled water, at the maximum density, and at 30 inches barometer.t It is determined from the Troy Pound, by the legal proportions of 5760 grains, which coi.QvesT. —51. In what is Avoirdupois Weight used? Repeat the Table? Cbs. -Iov many poundll were formerly allowed for a hundred weight? For a quarter? What is gross weighti. Net weight? 252. IHow is the Avoirdupois pound of the United States determined? * ind's Arithmetic, Art. 224. Also, North American Review, Vol XLV. Reports of Secretary of Treasury, March 3, 1832: uune 30, 1832. Also, Congressional Documents ot 1833. ,Ar~TS. 251-253.j COMPOUND NUMIBERS. 149 stitute the Troy pound, to 7000 graiils Troy, which constitute the Avoirdupois pound. That is, 5760 grains Troy make 1 pound Troy. 7000 grains " " I pound Avoirdupois. 4371 grains' 1 i ounce c 279j grains 1 " 1 dram " Ons. 1, The Br'itish ILmrperial Pound Avoirdupois is equal to 27-7274 cubic iches of distilled water, at the temperature of 620 Fahrenheit, when the barometer stands at 300. It is determined from the Imperial Troy pound, wIhich contains 5760 grains, while the former contains 7000 grains. 2. Since the Troy pound of the United States is identical with the Troy pound of England, the Avoirdupois pound of the former must be equal to that of the latter; for both bear the same ratio to the Troy pound. But the Engfish avoirdupois pound is said to contain 27.7274 cu. in. of distilled water, xyhile that of the IUnited States, according to Mr. Hassler, contains 27.701554 cu. in. This slight difference may be accounted for by the fact that the forser was measured at the temperature of 62~, while the latter was measured at~ its maximum density, which is 39.83 degrees. 3. The standar'd of seioht adopted by the State of New York, in 1827, is the ~:voidlJzl(ois p2oun(,d, whose magnitude is such that a cubic foot of distilled water, at the maximum density, in a vacuum, will weigh 621 pounds, or 1000 ounces. Note.-The term avoirdutpois, is thought by some to be derived from the French avoiq' duL poids, a phrase signifying to have weight. Others think it fis from avoir's, the ancient name of goods or chattels, and poids signifying viaeigt in the Norman dialect.* APOTHECARIES' WEIGHT. 253. Apothecaries' Weight is used by apothecaries and physicians in mnixing medicines. 20 grains (gr.) make 1 scruple, marked sc., or 3. 3 scruples - " 1 dram, " dr., or 3. 8 drams " 1 ounce, " oz., or S. 12 ounces " 1 pound, lb. Ons, 1. The pound and ounce in this weight are the same, as the nToY p.ound and ounce; the other denominations are different. 2. Drugs and medicines are bought and sold by avoir'dupois weight. QUEST.-253. In what is Apothecaries' Weight used? Recite the Table? Obs. To.rvhat are the apothecaries' ounce and pound equal? H(fw are drugs and medicines bought nald sold? * President John Quincy Adams on Weights and Measures; also, IHind's Arithmetic. 150 COMPOUND NUMBERS. [SECT. VJ1, LONG- MEASURE. 25L-o.Long Measure is used in measuring distances where length only is considered, without regard to breadth or depths, It is frequently called linear or lineal measure. 12 inches (in.) make 1 foot, marked ft. 3 feet " 1yard, it yd. 51 yards, or 16l feet " 1 rod, perch, or pole, r. orp. 40 rods " 1 furlong, " fiur. 8 furlongs, or 320 rods " 1mile, s4 M. 3 miles " 1 league, " 1. 60 geographical miles, or degee. or, 69x statute miles' ~ " 1 degree, " deg. or O0 360 degrees make a great circle, or the circumference of the earth. Note. —4 inches make 1 hand; 9 inches, 1 span; 18 inches, 1 cubit; 6 fees 1 fathom, In measuring roads and land, surveyors use a chain which is 4 rods long, arid which is divided into 100 links. Hence, 25 links make I rod, and 7-1 O2 0 inches make 1 link. This chain is commonly called Gsenter's CChain, from the namnie of its inventor. OBS. 1. The inch is commonly divided either into eighths or tenths; some-; times, however, it is divided into twelfths, which are called lines. Formerl3i the inch was divided into 3 barleycorns; but the barleycorn is not employed as; a measure at the present day. The term barleycorn, is derived from a grain of' barley, which was the original element of Linear Measure. 2. The terms rod, pole, and soerch; from the French peerche signifying a rod are each expressive of the instrument, which was originally used as a measure of this length. 25 5. The standard Unit of Length adopted by the Uniteod States, is the Yard of 3 feet, or 36 inches, and is identical with the British Imperial Yard. It is made of brass, at the temperature of 62~ Fahrenheit, from the scale of eighty-two inches prepared by Troughton, a celebrated English artist, for the survey of the Coast of the United States. OBs. 1. The Imper'ial stanrdcard yard of Grelat Britaiin is determined from tlm. sendulsm which vibrates seconds in a'vacuum, at the level of the sea, in, QUEST.-254. In what is Long Measure used? What is Long Measure sometimese called? Recite the Table? Obs. How are inches usually divided? What is the origin of the measure called barleycorn? Is this measure now used? 255. What is the standard unit of Length adopted by the United States? ARTS. 254-257.] COMPOUND NUMBERS. 151 Greenwich or London. This pendulum is divided into 391393 equal parts a d 360000 of these parts are declared, by act of Prclliament, to be the stand. mald yard, at the temperature of 620; consequently, since the yard is divided into 36 inches, it follows that the length of a pendulum vibratirng seconds, under these circumstances, is 39.1393 inches. The Ewngssls yard is said to have been originally determined by the length of the arm of Henry I. King of England. 2. The stanzdard of linear measure adopted by the,State of New York, is the pendlulnse which vibrates seconds, in a vacuum, at Columbia College, in the city of New York, which is in the latitude of 400 42', 43". The yard is declared to be _aa 2 0_ of this pendulum; hence, the length of the pendulum is 39.101688 inches, at the temperature of 320. Should the standard yard ever be lost, it could be recovered by resorting to the preceding experiment CLOTH MEASURE. 256. Cloth JMeasure is used in measuring cloth, lace, and all kinds of goods, which are bought and sold by the yard. 21 inches (in.) make 1 nail, marked na. 4 nails, or 9 in. " 1 quarter of a yard, " qr. 4 quarters " 1 yard, ". d. 3 quarters, or i of a yard " 1 Flemish ell, " F/. e. 5 quarters, or 1I yard " 1 English ell, " E. e. 6i quarters, or 1D yard " 1 French ell, " F. e. Oas. Cloth measure is a species of long measure. Cloth, laces, &c., are bought and sold by the linear? yard, without regard to their width. SQUARE MEASURE. 257. Square Measure is used in measuring surfaces, or things whose length and breadth are considered without regard to height or depth; as, land, flooring, plastering, &c. 144 square inches (sq. in.) make 1 square foot, marked sq. ft. 9 square feet " 1 square yard, " sq. yd. 301 square yards, or 272k square feet W " 1 sq. rod, perc., or pole, sq. r. 40 square rods I" 1 rood, " R. 4 roods, or 160 square rods " 1 acre, " A. 640 acres " 1 square mile, " l. QUEST.-256. In what is cloth measure used? Repeat the Table. Obs. Of what is cloth measure aspecies? What is the kind of yard by which cloths, laces, &c,are bought and sold X 257. In what is Square Measure used? Recite the Table. 152 COMPOUND NUMBERS. [SECT. VIIL Note.-16 square rods make 1 square chain; 10 square chains, or 100,000 square links, make an acre. Flooring, roofing, plastering, &c., are frequently estimated by the " square," which contains 100 square feet. A hide of land, which is spoken of by ancient writers, is 100 acres. OBS. 1. A sqncare is a figure which hasfour equal sides, and all its angles ri ht angles, as seen in the diagram. Hence, A Square Iiwlt is a square, whose sides are each a 9 sq. ft.=- sq. yd. wear inch in length. A Square Foot is a square, whose sides are each a lnear foot in length. A Square Yard is a square, whose sides are each a linear yard, or three linear feet in length, and contains 9 square feet, as represented in the adjacent figure. 2. Square measure is so called, because its measuring unit is a square. The' standard of sqnuare measure is derived from the standard linear measure. Hencn,. A unit of square measure is a square whose sides are respectively equal, in length, to the linear unit of the same name. CUBIC MEASURE. 25 S Cubic Yieasure is used in measuring solid bodies, or things which have length, breadth, and thickness; such as timber, stone, boxes of goods, the capacity of rooms, ships, &c. 1728 cubic inches (cb. in.) make 1 cubic foot, marked cq. ft.'27 cubic feet " 1 cubic yard, " cn. yd. 40 feet of round, or o " ton, orload, T. 50 ft. of hewn timnbe.r 42 cubic feet I 1 ton of shipping, " T. 16 cubic feel 1 foot of wood, or,, ft. a cord foot, 128 cubordic feet or " 1 cord, " C. A pile of wood 8 feet long, 4 feet wide, and 4 feet high, contains 1:)rdl For, 8X4X4=128. QUEST.-Obs. What is a square? What is a square inch? A square foot? A square yard. 258. In what is Cubic Measure used? Recite the Table. RTnTS. 258, 259.] COMPOUND NUMBERS. 15 Obs. 1. A Cube is a solid body bounded by six equal sides. It is often called a heuaedro~n. Hence, A Cubic Incah is a cube, each of whose sides s a square inch, as represented by the adjoin-'Ig figure. A Cubic Foot is a cube, each of whose sides is a square foot. 2. Cubic Measure is so called, because its measuring unit is a cube. It is often called solid measure. The standard of cubic measure is derived from the standard tndeaw measure. A unit of cubic measure, therefore, is a cube whose sides are respectively equal in length to the linear unit of the same name. 3. The cubic ton, sometimes called a toad, is chiefly used for estimating the cartage and transportation of timber. By a ton of round timber is meant, such a quantity of timber in its rough or natural state, as when hewn, will mnake 40 cubic feet, and is supposed to be equal in weight to 50 feet of hewn itimber. The cubic ton or load, is by no means an acc'urate or uniform, standard of estimating weight; for, different kinds of timber, are of very different degrees of density. But it is perhaps sufficiently accurate for the purposes to which it is applied. Note.-For an easy method of forming models of the Ctbe and other regular Solids, see Thomson's Legendre's Geometry, p. 222. WINE MEASURE. 259. Wine Measure is used in measuring wine, alcohol, nmolasses, oil, and all other liquids except beer, ale, and milk. 4 gills (gi.) make 1 pint, marked pt. 2 pints " 1 quart, " qt. 4 quarts " 1 gallon, " gal. 31t gallons " 1barrel,' bar. or bbl. 42 gallons 1 tierce " ir. 63 gallons, or 2 barrels "1 hogshead, hhld. 2 hogsheads " 1 pipe or butt, " pi. 2 pipes " t tun, " tun. OBs. 1. In England, 10 gallons make 1 anker; 18 gallons, 1 runlet; 3 tierces: or 84 gallons, 1 puncheon. 2. Liquids are generally bought and sold by the gallon or its subdivi ions, na QUST. —Obs. What is a cube? What is a cubic inch? A cubic foot? What is meant by a ton of round timber? 259. In what is Wine Measure used? Recite the Table. 154 COMPOUND NUMBnIiS. [SECT. VII. the quart, pint, &c. Cider and a few cheap artiles are bought and sold by the barrel. The capacities of cisterns, vats, &c., are sometimes estimated itr hogshteads, and the quotations or prices-current of oils in foreign markets, ari e usually made in l1,ns. But the lierce, and the'pipe or blttt are never used, a,':s such, in business transactions; their contents are given in gallons, quarts, &u. 260.o'The standard Unit of Liquid Measure adopted by the rrited States, is the WHine Gallon of 231 cubic inches, which is: equal to 58372.175,: grains of distilled water, at the maximum density, weighed in air at 30 inches barometer, or 8.33Q lbs. avoirdupois, elry nearly.* Oss. The Brilis/. imperial standarlad measaure of capacity, both for liqpuids and dry goods, is the im7erial gallon, which is equal to 10 pounds avoirdupois of distilled water, at 620 thermometer and 30 inches barometer, and contain~sh 277.274 cubic inches. It is equal to 1.2 gal. wine measure U. S. BEER MEASURE. 26 1 ~ Beeir Measure is used in measuring beer, ale, and milk,. 2 pints (PIs.) make 1 quart, marked qt. 4 quarts " 1 gallon, " gal. 36 gallons " I barrel, " ba,-. or bbl. I barrels, or 54 gallons" 1 hogshead, I" hId. OBS. 1. In England, 9 gallons make 1 firkin; 2 firkins, 1 kilderkin; 2 kil.derkins, 1 barrel. 2. The beer gallon contains 282 cubic inches, and is equal to 10.1799321 pounds avoirdupois of distilled water, at the maximum density. In many places, milk is measured by wine measure. DRY MEASURE. 262. Dry Measure is used in measuring grain, fruit, &c. 2 pints (pt1.) make 1 quart, marked qt. 8 quarts " I peck, " Pk. 4 pecks, or 32 qts. " 1 bushel, " bu. 8 bushels " 1 quarter, " qr. 32 bushels, or 4 qrs. " 1 chaldron, ". QUTEST.-260. What is the standard unit of Liquid Measure of the l-aited ataest How iany cubic itches in a wrine gallon? 261. Inwhat'is Beer Measure!sea? Rccito the Table. 262. In what is Dry Measure used' Repeat the Table. * Reports of the Secretary of the Treasury, March, 1831, and June, 1 832. At.so, lassletr on Weights and Measures. ARTs. 260-263.] COMPOUND NUMBERS. 155 OBs. In England flour is often sold by weight. A sack is equal to 280 lbs., and contains about five imperial bushels. The following denominations, are sometimes used, viz: 2 quarts make 1,pottle; 2 bushels, 1 strike; 2 strikes or 4 bu., 1 coom; 2 cooms or 8 bu., 1 quarter; 5 quarters, 1 wey or load; 2 loads, i last. In London 36 bushels of coal make a chaldron, but in New Castle 79f bushels are said to be allowed for a chaldron. But coal in England and in this country, is now usually bought and sold by weight. oVte. —Wine, Beer, and Dry Measures are often called capaci/fj measures and are evidently a species of cubic measure. 263. The standard Unit of Dry Measure adopted by the, United States, is the Winchester Bushel, which is equal to 77.027413 pounds avoirdupois of distilled water, at the max-.imunm density, weighed in air at 30 inches barometer, and contains.2150.4 cubic inches, nearly. The Winchester bushel is so called, because the standard measireir was formerly kept at Winchester, England. By statute, it is )an upright cylinder, 18- inches in diameter, and 8 inches deep. Oas. 1. The imperial bushel of Great Britain is equal to 80 lbs. avoirdupois,'f distilled water, at 620 Fahrenheit, and 30 inches barometer, and contains!'218.19'. cubic inches; consequently, it is equal to 1.032 bushel U. S., nearly. It is an upright cylinder, whose internal diameter is 18.789 inches, and its depth 8 inches. The use of heaped measu're was abolished by Act of Parliament, in 1835. 2. The standarced butshel of the State of New York, is equal to 80 pounds r.ivoirdupois of distilled water, at thb maximum density, at the mean pressure of the atmosphere, and contains 2218.192 cubic inches.* It is customary, at the present day, to determine capacity measures by the,wreight of distilled water which they contain. This is evidently more accu-iate than the former method of measurement by cubic inches. 3. In buying and selling grain, when no special agreement as to measurement or weight, is made by the parties, a bushel, in the State of New York, by Act of 1836, consists of 60 lbs. of wheat, 56 lbs. rye or Indian corn, 48 lbs. cf barley, and 32 lbs. of oats. There are sinilar statutes in most of the other States of the Union. This is this most impartial method by which the value of grain can be estimated. QuEsT.-2-63. What is the standard unit of Dry Measure adopted by the Governmenti * By the same Act it was declared, that the standard liquid gallon should be 8 lbs., and the standard dry gallon 10 lbs. avoirdupois of distilled water, at its maximum density luit tbis part of the statute was subsequently repealed, and the previous standard gallon a;n the office of the Secretary of State, was continued in use. ]56 COMPOUND NUMBERS. [SECT. VIII TIME. 264. Time is naturally divided into days and years; the forrper are caused by the revolution of the Earth on its axis, the~ latter by its revolution round the sun. 60 seconds (sec.) make 1 minute, marked wit. 60 minutes " 1 hour, " IA'r. 24 hours " 1 day, " d. 7 days " 1 week, " nqk. 4 weeks " 1 lunar month, " no. 12 calendar months, or 365 days and 6 hrs., (nearly,) civil year, yr. The following are the names of the 12 calendar months into which the civi or legal year is divided, with the number of days in each. January, written (Jan.) the first month, has 31 days. February, " (Feb.) " second " " 28 " March, " (Mar.) " third " " 31 " April, " (Apr.) " fonr?-th " (" 30 " MlIay, " (May) " fifth " " 31 c June, " (June) " sixSz " " 30 " July, " (July) " seventh " " 31 " August, " (Aug.) " eighth " " 31 c September, " (Sept.) " ninth "c " 30 6 October, " (Oct.) " tenth " " 31 " November, " (Nov.)'c eleventh " " 30 c December, " (Dec.) " twelfth, " 31 " The number of days in each month may be easily remembered from the folf. lowing lines: " Thirty days hath September, April, June, and November; February twenty-eight alone, All the rest have thirty-one; Except in Leap year, then is the time, When February has twenty-nine." OBs. 1. A Solar year is the exact time in which the earth revolves round the sun, and contains 365 days, 5 hours, 48 minutes, and 48 seconds. 2. Since the civil year contains 365 days and 6 hours, (nearly,) it is plain that in four years a whole day will be gained, and therefore every fourth yea; must have 366 days. This day was originally added to the year, by repeativng thd sixth of the Calenlds of HMarnch in the Roman calendar, which corresponds wvithi QUEST. —.64. How is time niaturally divided? Recite the Table. Obs What is a sotoloz bear t How is leap year occasioned? To which month is the odd day added. ARTS. 264-266.] COMPOUND NUMBERS~ 157 the 24th of Febrary in ours. It was called the intercelary day, from the Latin intercalo, to insert. The year in whch this day is added, is called Bissextile, from the Latin bis, twice, and sextilis, the sixth. It is also called " Leap Year," because it leaps over a day more than a common year. 3. The civil or legal year is often called the Julianb year, from Julits Caesar emperor of Rome, who adapted the calendar or register of the civil year to the supposed length of the solar year, oy adding 1 day to everyfoQurt/h year. 265o In process of time, as mathematical and astronomical science advanced, it was found that the length of a solar year was only 365 d. 5 hrs. 48min. 48 sec., or 11 min. 12 sec. less than 365-L days, which in 400 years amounted to about 3 days; consequently, the Julian calendar was behind the solar time. This error at the time of Pope Gregory XIII., amounted to 10 days, which he corrected in 1582 by suppressing 10 days in the month of October, the day after the 4th being called the 15th. Hence this calendar is sometimes called the Gregorian calendar. OBs. 1. This correction was not adopted in England till 1752, when the error amounted to 11 days. By Act of Par'liament, 11 days, after the 2d of September, -were therefore omitted; and. the civil year by the same Act, was made to commence on the 1st of January, instead of the 25th of March, as it had done previously. 2. Dates reckoned by the old method or Julian'calendar, are called Old Style; and those reckoned by the new method, are called New Style. To change any date from Old to New Style, we must add 11 days to it; and if the given date in Old Style, is between the Ist of January and the 25th of March, we must add 1 to the year in New Style. Russia still reckons dates according to Old Style. The difference now amounts to 12 days. 266o To ascertain whether a year is LEAP YEAR. Divide the given year by 4, and if there is no remainder, it is Leap year. The remainder, if any, shows how many years have Elap)sed since a Leap year occurred. Thus, dividing the year 1847 by 4, the remainder is 3; hence it is 3 years since the last leap year, and the ensuing year will be leap year. OBs. 1. To this rule there is an exception. For, we have seen, that a solzX year is 11 min. and 12 sec. less than a Julian year, which is 365, days, This error, in 400 years, amounts to about 3 days; consequently, if' 1 day is added U.EST. —266. 1iow (do you ascertain whether a year is leap year 1 158 COMPOUND NUMBERS. [SECT. VII1 everyfourtla year; that-is, if we have 100 leap years in 400 years, according to the Julian calendar, the reckoning would fall 3 days behind the solar time. Thus, reckoning from the commencement of the Christian era, when it was January Ist, 401 by the Julian time, it was January 4th by the solar time. 2. To remedy this error only 1 cenltennial year in jaulrs is regarded a leact year?; or, which is the same in effect, whenever the celtenseial year, or the nsLsmber expressing the century, is not divisible by 4, that year is not a leap year, while the other centennial years are. Thus, 17, 18, 19, denoting 1700, 1800, and 1900, are not divisible by 4, consequently they are not leap years, though according to the rule above they would be; on the other hand 16 and 20, denoting 1600 and 2000, are divisible by 4, and are therefore leap years. There is still a slight error, but it is so small that in 5000 years it scarcely amounts to a day. CIRCULAR MEASURE, OR MOTION. 267. Circular Ifeasure is applied to the divisions of the circle, and is used in reckoning latitude and longitude, and t1hmotion of the heavenly bodies. 60 seconds (1/) make 1 minute, marked 60 minutes " 1 degree, " o 30 degrees " 1 sign, " s. 12 signs, or 3600 " 1 circle, " C. This measure is often called Anguclar Measure, and is chiefly used by astronomers, navigators, and surveyors. 90~ OBs. 1. The circumference of every circleis divided or supposed to be divided, into 360 equal parts, called degrees, as in the subjoined figure. o o 2. Since a degree is -60- part of the o\ circumference of a circle, it is obvious that its length must depend on the size of the circle. 270~ Note.-The division of the circumference of the circle into:360 equal parts, took its on gin from the length of the year, which, (in roundl numbers) was supposed to contain 360 days, or 12 months of 30 days each. The 12 s'igscEs QUCST. -267. In what is Circular Measure used.? Repeat the Table. Obs. HIow is the idremnforence of every circle divided? On what does the length of a degree depend? ARTS. 267-269.] COMPOUND NUMBERS. 159 Correspon I to the 12 months. The term minuites, is from the Latin srninzurm,'which signifies a small part. The term seconds, is an abbreviated expression iQr second minutes, or minutes of the second order. 268. Since the earth turns on its axis from west to east once hin 24 hours, it evidently revolves 15~ per hour; or 10 in 4 minIltes, and 1' in 4 seconds of time. Hence, When the difference of longitude between two places is 1', the ifference in the time, or the hour of the day at these twoplaces, is 4 4econds; if the difference of longitude is 1~, the difference of time is 4 minutes; if 2~, the difference of time is 8 minutes, &dc. Thus, when it is noon at London, in Philadelphia, which is about'75~ west from London, it is only 7 o'clock, A. M. For, if the earth revolves 10 in 4 minutes, to revolve 75~, it will require 75 times as long, and 4X75=300 min., or 5 hours. OBs. 1. Since the earth revolves from west to east, it is manifest, that the timne is ear'lier as we go eastward, and later as we go westward.:i. This principle affords navigators and others a convenient and useful me-hod of ascertaining the diT2irence of time between two places, when the difference of their longitude is known; also, for ascertaining the diffeence of lo;gitude between two places, when the difference in their time is known. MISCELLANEOUS TABLE. 269. The following denominations not included in the preeding Tables, are frequently used. 12 units make 1 dozen, (doz.) 12 dozen, or 144 " 1 gross. 12 gross, or 1728 " 1 great gross. 20 units 1 1 score. 56 pounds 1 firkin of butter. 100 pounds " 1 quintal of fish. 30 gallons 1 bar. of fish in Mass. 200 lbs. of shad or salmon " 1 bar. in'N. Y. and Conn. 196 pounds " I bar. of flour. 200 pounds " I bar. of pork. 14 pounds of iron or lead " 1 stone. 211 stone l pig. 8 pigs 1 father.,Note.-Formerly it was customary to allow 112 lbs. for a quintal. i.UEST. —268. When the difference of longitude between two;laces is 1', what Is the d1tfference of titne? When 10, what is the difference of tinme 160 COMPOUND NUMBERS [SECT. VIIb,1 PAPER AND BOOKS. 270. The termsfolio, quarto, octavo, &o., applied to book*, denote the number of leaves into which a sheet of paper i;s folded. 24 sheets of'paper make 1 quire. 20 quires " 1 ream. 2 reams " 1 bundle. 5 bundles " 1 bale. A sheet folded in two leaves forms a folio. A sheet " "four leaves " a quarto, or 4to. A sheet " "eight leaves " an octavo, or 8vo. A sheet " " twelve leaves " a duodecimo, or 1dmo. A sheet " "eighteen leaves " an 18mo. A sheet " " thirty-six leaves " a 36mo. DIMENSIONS OF DIFFERENT KINDS OF ENGLISH PAPER. Names. Writing. Drawing. Printing. Pott, 151 by 124 in. 151 by 12J in. Small Post, 16k by 13- in. Fool's Cap, 164 by 13L in. 164 by 131 in. Crown, 20 by 15 in. 20 by 15 in. Demy, 20 by 154 in. 22 by 17 in. Medium, 221 by 17k in. 23 by 18 in. Royal, 24 by 191 in. 24 by 194 in. 26 by 20 in. Super Royal, 271 by 191 in. 274 by 194 in. Elephant, 28 by 23 in. Double Crown, 30 by 20 in. Imperial, 301 by 22 in. 304 by 22 in. Atlas, 34 by 26- in. Columbier, 34- by 234 in. Double Demy, 384 by 26 in. Double Elephant, 40 by 264 in. Antiquarian, 52 by 31 in. Double Atlas, 55 by 31- in. *Emperor, 68 by 48 in. Note.-Agnercan paper is usually rather larger than English paper of th0 same name. QUEsT. —270. What' do the terms, folio, quarto, &c., denote, when applied to books'? What is a folio. A quarto? An octavo? A duodecilo? An 18ro.? A 36mo. URTS. 270-273.] COMPOUND NUMBERS. 161 FRENCH MONEY, WEIGHTS, AND MEASURES. 27 1 The new system of Money, Weights, and Measures of France, adopted in 1795, was formed according to the decimat Notation. FRENCH MONEY. 272. The Franc is the unit money of the new system of'rench currency. It is a silver coin, consisting of ~- pure silver, and -#L of alloy. 10 centimes make I decime., 10 decimes " 1 franc. Note.-The value of a franc by Act of Congress in 1843, is $.186. The value of the avre tournois, the former unit of money, is $.185. FRENCH LINEAR MEASURE. 27 3. The standaird unit of the French Linear ATeasavre, is 1h1e Zlifetre. Its length, according to the 9mean of the several comparisons of Troughllto, Nicollet and Hassler, is equall to 39.38091 71 English, or United States inches. 10 metres make 1 decametre = 32.817431 U. S. feet. 10 decametres " 1 hectometre = 328.17431 " 10 hectometres " 1 kilometre 3281.7431 " 10 kilometres " myriametre. = 32817.431 " " Note.-l. The standa'rd by which the new F-ench measuLes of length are determined, is the quadranvt of a meridian of the earth, or the terrestrial aro from the equator to the pole, in the meridian of Paris. The ten- millionth: part of this arc is called a metre, which is equal to 39.378 U. S. in., nearly. 2. The metr-e is divided into 10 decimetres; the decimetlre into 10 centimetres; she centimetsc into 10 millimetres. 3. The denominations of the old system of linear measure were the toise,,i.oot, inch, line, and point. 12 points=1 line; 12 lines-= inch; 12 inu.=l oot; 6 ft.-1 toise. The old French foot was equal to 1.066 U. S. feet. 4. By a decree of 1812, the Toise, Aune, Foot, &c., are allowed to be used, hlaving the following ratios to the metre, viz: the toise-2 metres; the foot=,i metle; the inch —-', metre; the aune or ell=-l metre; the bushel. — hee., tolitre. 162 COMPOUND NUMBERS. [SECT. VIII. FRENCH SQUARE MEASURE. 274. The unit of French Superficial Measure, is the Are, whose sides are each a decametre in length; consequently, it con;,tins 100 square metres, or 119.6648496 U. S. sq. yds. 10 ares make 1 decare = 1196.648496 U. S. sq. yds. 10 decares " 1 hectare - 11966.48496 " 10 hectares " 1 kilare = 119664.8496 " " 10 kilares " 1 myriare - 1196648.496 " " Note.-The are is divided into 10 declares; the deczare into 10 centiares; the centzare into 10 milliares. FRENCH CUBIC MEASURE. 27 5. The unit of French Cubic Measure, is the Stere, whickh is a cubic metre, and is equal to 61074.1564445 cu. in. U. S 10 decisteres make 1 stere - 35.34384 cu. ft. U. S. 10 steres " decastere= 353.4384 " " FRENCH LIQUID AND DRY MEASURE. 276. The unit of French Liquid and Dry Mieasures, is called the Litre, which is a cubic decimetre, and is equal to 61.0741564445 cu. in. U. S., or 1.05756 qts. wine measure. 10 litres make I decalitre = 2.6439 gals. wine meas. 10 decalitres " l hectolitre — 26,439 " " " 10 hectolitres " 1 kilolitre = 264.39 " "' Note.-The litre is divided into 10 decilitres; the decilitre into 10 centilitres; the centilitre into 10 millilitres. FRENCH WEIG-HTS. 277. The unit of French Weigyhts, is the weight of a cubic centimetre of distilled water, at the maximum density, and is called the Gramme. It is equal to 15.433159 grains Troy. 10 grammes make 1 decagramme 154.331.59 g;s. Troy. 10 decagrammes " 1 hectogramme = 1543.3159 " 10 hectogrammes" 1 kilogramme = 15433.159 " " 10 kilogrammes, " myliagramme - 154331.59 " ARTS. 275-279.] FoREIGN WEIGHTS, ETC. 163 Note.-I. The gr?'amme is divided into 10 decigrammes; the decigamrnme into 10 centigrammes; the centigr'amme into 10 milligrammes.,). The denomination chiefly used in making out invoices of goods sold by iwreight, and in business transactions, is the kilogranL7tme, which is equal to 1000 kramnmes, or 2.21 lbs. avordupois, very nearly. 3. In the old system of French weight, the livre-poids=2 mnarcs; the marc;>28 onces; the once=8 gros; the gros=72 grains. The livre is equal to one-,half the kilogramme. FRENCH CIRCULAR MEASURE. 27 S0 The circle is divided into 400 equal ptirts, called grades, ailnd the quadrant into 100 grades. The grade is again divided into 100 equal parts, and each of these parts is subdivided into 100 other equal parts, according to the centesimal scale. Hence, The seconde -.00009 English deg. The minute =.009 " " The grade =.9 " " 2Note.-The names of the denominations lafrger than the unzt in the French Co mpound Numbers, are formed by prefixing to the name of the unit, the Gi,'eek words, deca,?lerco, kilo, and eyri7ia; those less than the q.nit, are formed hy prefixing to the name of the ullit, the Latin words, deci, celti, and milli. 279. F'oreign Weights and Measures compared with those of tie United States.*.4qmsterda, m.-100 lbs. (I centner)_108.923 lbs.t; 1 last-85.25 bu.; 1 ahm41 gals.; 1 foot Amlsterdam=111 - in.; 1 foot Antwerp=1ll in.; 1 ell Amsterdam=-2.26 ft.; 1 ell Brabant=2.3ft.; 1 eli Hague=2.28 ft. B.-ataxviac.-l picul=136 lbs.; 1 kann-.39 gal.; 1 ell=2.25 ft. Beng-al.-l haut=1.5 ft.; 1 guz=3 ft.; 1 coss or mile_=1.24 miles; 1 bazar maud=82.14 lbs.; 1 factory nmaud=74.66 lbs. Ben-coolen. — bahar=560 lbs.; 1 bamboo=l gal.; 1 coyang:=8 gals. Bombcay.-l maud=28 lbs.; I covid=1.5 ft.; 1 candy=25 bu. /lreseme.-l pound=l.1 lb.; 1 centner=116 lbs.; 1 last=80.7 bu.; 1 ft.=_-11 in. Canto7.-1 tael=lI oz.; 1 catty=l lbs.; 1 picul-1333 lbs.; 1 covid — 14 in. Dermnark.-100 lbs. (1 centner)=110.25 lbs.; 1 bbl. (toende)=3.95 bu.; 1. viertel=2.04 gals.; 1 foot Copenhagen, or Rhineland=l12 in.?lorenbce and Leglornw. —100 lbs. (1 cantaro)=74.86 lbs.; 1 mnoggio=16.59 bu.; 1 barile_=12.04 gals.; 1 palmo-91 in. * M'Culloch's Commercial Dictionary;' also Kelly's Universal Camanlist.? The poueils in this and the following comparisons are avoirdupois. S 164 FOREIGN vWET:rGL s,'tC. [SECT. VII. Genoa,.-100 I. s. (! peso grosso)-71i] l bs.; I peso sottile-=69.89 lbs,; 1 mina =3.43 bu.; 1 i ezz.arola —39.22 gcals.; 1 ialmo=9m — in..1r7abr1g.*'. — 1 foot — 1.3 in.; 1 ell=22.6 in. nearly; 1 ell Brabant=27.6 in. j 1 mile=4.68 miles 1 acss-1=-1 bu.; I last=S89.64 bu.; 1 ahlnm38 gals. JaCpaL.- catti —1.3 lbs; 1 picul-130 lbs.; 1 ichan=31 ft.; 1 inc or tetaim:y =61 ft.; 1 balec=1tf gals. MJadrlas. — covidl=1 ft.; i1 catty=13 lbs.; 1 picul-1331 lbs.; 1 maud-'; 25 lbs.; 1 candy=500 lbs.; 1 garee=140 bu. call ca.-1 foot=l101 in.; 100 lbs. (1 cantaro)=174.5 lbs.; 1 salma=8.22 bu,.lanillai.-l arroba=26 lbs.; 1 picul=:143 lbs.; 1 palmo=10.38. in. Naples. — cantaro grosso=19,6.5 lbs.; 1 cantaro piccolo=106 lbs.; 1 palmo=, 10. in.; 1 tomoo- 1.45 bu.; 1 carro=52.24 bu.; 1 carro of wine=264 gals. Nelherlcands.-I ell=3.28 ft.; 1 mudde=2.84 bu..; 1 lkan litre= —.11 pintsx 1 vat bectolitre=26.42 gals.; 1 pond kilogramllnme=2.21 lbs. Portingcal.-100 lbs.=101.19 lbs.; 1 arrob a-22.-26 lbs.; 1 quintal 89.05 lbs. 1 a.lmude:4.37 gals.; 1 alquiere=4 bu.; 1 mIoyo-=23.03 bu.; 1 last=70 bu. 1 pe or foot= 12- in.; I mile=l 1 mile. PPrnssia.-100 lbs=103.11 lbs.; 1 quintal (110 lbs.)=113.42 lbs.; 1 eimarl: 18.14 gal.; 1 scheffel —1.56 bu.; 1 foot,-1.03 ft.; 1 ell=2.19 ft.; I mile4.68 miles. Ro~me.-100 libras —74.77 lbs.; I rubbio=9.36 bu.; 1 barile 15.31 gals.; 1 fb.ot =11 in.; 1 canna=6~ ft.;'l mile=7T- fur. RPassia.-100 lbs.=90.26 lbs.; 1 berquit=361.04 ibs.; 40 lbs. (1 pood) 7 36 lbs.; I vedro=34 gals.; 1 chetwert-=5.95 bu.; 1 foot Petersbuirg=1.18 ft.-f; 1 foot Moscow-I.1 ft.; 1 arsheen:=2 ft.; 1 mile (verst)=5.3 fur. Sicilyl.-100 lbs. (libras)=70 lbs.; 1 cantaro ogosso:19P2.5 lbs.; I ceantahri. sottile.175 lbs.; 1 salma generale=7.85 bu.; 1 salina grossi:9.'77 bu,; I salma of wine=23.06 gals.; 1 palmoo=9. in.; 1 canna_=6, ft. Spbainv.-i arroba-25.36 lbs.; I quintal=101i.44 lbs.; 1 arroba of wine\ 41 gals.; 1 moyo —68 gals.; 1 fanega -=.Gbu.; 1 foot-11.128 in.; 1 vara-_ 2.78 ft.; 1 league (leagua)-4.3 nm., nearly. Sweden. —100 lbs, (victualie)=73.176 lbs.; 1 foot-11.69 in.; 1 elI= —1.95 ft.. 1 mile=G.64 il.; I kann —7.42 bu.; 1 last=75 bu.; 1 katin of wine=~ 69.09 gals. Snmyrna.-100 lbs. (1 quintal)=129.48 lbs.; I oke=2.83 lbs.; 1 quillotu 1.46 bu.; 1 quillot of wine=13.5 gals.; 1 pic=2,I ft. Triste. —100 lbs._-123.; lbs.; I stajo=21 biu.; 1 orna, or eimer=14.94 gals,; 1 ell (for silk) =2.1 ft.; 1 ell (for woollen)=2.2 ft.; 1 foot Austrian —1.037 ft.; 1 mile Austrian-4.6 m. 1,' nice. —100 lbs. (ipesso grosso) —105.18 lbs.; 1 peso sottile=64.42 lbs.; I stajo =2.27 bu.; 1 mocgio-9.08 bu.; 1 anifora=137 gals.; I foot=1.14 ft.; 1 brac. cio (for silk)=24.8 in.; 1 braccio (for woollen)=26.6 in. " New system of weights and measures adopted in 1843. In mneasuring tillber Enilish feet and inches are chlefly used throughout Russla. ART. 280.1 REDUCTIOI. 165 RE D U C T I O N. 280o Thle process of changing comnpound numbers from one denomination into another, without altering their value, is called REDUCTION. Ex. 1. Reduce ~5, 2s. Id. and 3 far. to farthings. Analysis.-Since in ~1 there are 20s., in ~5 there are 5 times as many, which is 100s., and 2, (the given shillings,) make 102s.,-,Again, since there are 12d. in is., in 102s. there are 102 times:s many, which is equal to 1224d., and 7 (the given pence) make i1231d. Finally, since in ld. there are 4 far., in 1231d. there are'2,231 times as many, or 4924 far., and 3, (the given far.,) make 4927 far. Ans. 4927 farthings. Operation. ~ s. d. far. We first reduce the given pounds to shil5 2 7 3..lings, by multiplying them by 20, because 20s. in.1. 20s. make ~1. (Art. 247.) We next re102 shillings. duce the shillings to pence, by multiply12d. in is. ing them by 12, because 12d. make is. Fi1'231 pence. nally, we reduce tble pence to farthings by 4 far in Id. multiplying them by 4, because 4 far. 4927 far. Ans. make id. 1Note.-t. In this example it is requmred to reduce higher denominations to ilower; as pounds to shillings, shillings to pence, &c. This is done by svz,eessive zuldtiplications. 2. In 4927 farthincs, how many pounds, shillings, and pence? Analysis.-Since 4 far. make ld., in 4927 farthings, there are is many pence as 4 is contained times in 4927, which is 1231d.,.and 3 far. over. Again, since 12d. make Is., in 1231d. there are Tas many shillings as 12 is contained times in 1231, which is I02s., and 7d. over. Finally, since 2Qs. make ~1, in 102s. there QIESTr. —280. What is Reduction? H ow are pounds reduced to shillings? Why mulitiply by 20? How are shillings reduced to pence? Why?'How pence to farthings? Whyv 166 REDUCTION. LSIECT. VIII: are as many pounds as 20 is contained times in 102, which:s ~5, and 2s. over. Ans. ~5, 2s. 7d. 3 far. Operation. We first reduce the given farthings: -4)4927 far. to pence, the next higher denomina4 12)1231d. 3 far. over. tion, by dividing them by 4, because! 20)102s. 7d. over. 4 far. make Id. (Art. 247.) Next we ~5, 2s. over. reduce the pence to shillings by di-' Ans. ~5, 2s. 7d. 3. far. viding them by 12, because 12d. make Is. Finally, we reduce-the shillings to pounds by dividing' them by 20, because 20s. make ~1l. The last quotient with the" several remainders, constitute the answer. Note.-2. The last example is exactly the reverse of the first; that is, lowe]r denominations are reduced to higher, which is done by successive divisions. 281] From the preceding illustrations we derive the folowing GENERAL RULE FOR REDUCTION. I. To reduce compound numbers to lower denominations. Multiply the highest denomination given, by that nzomber whicldh it takes of the next lower denomination to make ONE of this higher, to the product, add the number expressed in this lower denomina?tion in the given exanple. Proceed in thfis manner with eac/sh successive denomination, till you come to the one reguired. II. To reduce compound numbers to higher denominations. -Divide the given denomination by that number which it takes of: this denonzination to make ONE of the next higher'. Proceed in thisi mnanner wit/l each successive denomination, till you come to the one egquired. Thle last quotient, with the,everal remainders, will be the answer sought. 282. PROOF.-R- everse the operation; that is, reduce back tie answer to the original denominations, and if the result corresponciz with the numnbers given, the work is right. QUFrsT. —Iow are farthings reduced to pence. Why divide by 4? Ilow reduce pence to shillings? Why? How reduce shillings to pounds? Wliy? 281. HIow are compounid numbers reduced to lower denominations? Ilow to higher denominations? 282. IHo is Reduction proved? ARTs. 281, 282. | REDU ITION. 167 OBs. 1. Each remainder is of the same denomination as the dividend from which it arose. (Art. 113. Obs. 1.) 2. Reducing compound numbers to lower denominations may, with propriety, be called Reduction by Magltiplication,; reducing them to Aigher denominations, Reductionq by Division. The former is often called Reduction De%cealdinLg; the latter, Rcduction AscenBdiviOg. They mutually prove each other. EXAMPLES FOR PRACTICE. 1. In 136 rods and 2 yards, how many feet? Operation. Proof. r'ods. qyds. 3)2250 ft. 2)136 2 5,)750 yds. 5: yds. 1 r. 2 682 11)1500 68 136 r. 4 rem.=2 yards. 750 yds. Now 136 r. 2 yds. is the 3 ft. 1 yd. given number. 2250 ft. Ans. 2. In ~71, 13s. 6~d., how many farthings? 3. In ~90, 7s. 8d., how many farthings? 4. In ~295, 18s. 34d., how many farthings? 5. In 95 guineas, 17s. 94d., how many farthings? 6. How many pounds, shillings, &c., in 24651 farthings? 7. How many pounds, shillings, &c., in 415739 farthings? 8. How many guineas, &c., in 67256 pence? 9. In ~36, 4s., how many six-pences? 10. In ~75, 12s. 6d., how many three-pences? 11. Reduce 29 lbs. 7 oz. 3 pwts. to grains. 12. Reduce 37 lbs. 6 oz. to pennyweights. 13. Reduce 175 lbs. 4 oz. 5 pwts. 7 grs. to grains. 14. Reduce 12256 grs. to pounds, &c. 15. Reduce 42672 pwts. to pounds, &c. 16. In 15 cwt. 3 qrs. 21 lbs., how many pounds? 17. In 17 tons 12 cwt. 2 qrs., how many ounces? QUEST.-ObS. Of what denomination is each remainder? What may reducing compound numbers to lower denomlinations be called? To higher denominations? Which of the fundamental rules is employed by the former? Which by the latter 1 168 REDUCTION. [SECT. Vii. 18. In 52 tons 3 cwt., how many pounds? 19. In 140 tons, how many dranms? 20. In 16256 ounces, how many hundred weight, &c.? 21. In 267235 pounds, how many tons, &c.? 22. In 563728 drams, how many tons, pounds, &c.? 23. Reduce 95 pounds (apothecaries' weight) to drams. 24. Reduce 130 pounds to scruples. 25. Reduce 6237 drams (apothecaries' weight) to pounds, &t, 26. Reduce 25463 scruples to ounces, &c. 27. How many feet in 27 miles? 28. How many inches in 45 leagues? 29. How many yards in 3000 miles? 30, In 290375 feet, how many miles? 31. In 1875343 inches, how many leagues? 32. In 15 m. 5 fur. 31 1., how many rods? 33. In 1081080 inches, how many miles, &c.? 34. How many feet in the circumference of the earth? 35. How many nails in 160 yards? 36. How many quarters in 1000 English ells? 37. In 102345 nails, how many yards, &c.? 38. In 223267 nails, how many French ells? 39. In 634 yards, 3 qrs., how many nails? 40. In 28 hllds. 15 gals. wine measure, how many quarts? 41. In 5 pipes, 1 hhd., how many gallons? 42. In 3 tuns, 1 hhd. 10 gals., how many gills? 43. In 12256 pints, how many barrels, wine measure? 44. In 475262 gills, how many pipes, &c.? 45. In 50 hhds. 1 bbl. 10 gals., how many gills, wine measlaur 46. In 45 bbls., how many pints, beer measure? 47. How many barrels of beer in 25264 pints? 48. How many hogsheads of beer in 1362.56 quarts? 49. How many pints in 45 hhds. 10 gals. of beer? 50. In 15 bushels, 1 peck, how many quarts? 51. In 763 bushels, 3 pecks, how many quarts? 52. In 56 quarts, 5 bushels, how many pints? 53. In 45672 quarts, how many bushels, &c.? 54. In 260200 pints, how many quarts? ARTS. 282, 283.] REDUCTION. 169 55. Reduce 25 days, 6 hours to minutes. 56. Reduce 365 days, 6 hours to seconds. 57. Reduce 847125 minutes to weeks. 5S. Reduce 5623480 seconds to days. 59. How many seconds in a solar year? 60. How many seconds in 30 years, allowing 365 days 6 hours to a year? 61. How many years of Sabbaths are there in 70 years? 62. In 110 degrees, 2 0 minutes, how many seconds? 63, In 11 sigons, 45 degrees, how many secon ds? 64. In 7654314 seconds, how many degrees? 65. In 1000000000 minutes, how many signs? 66. Reduce 1728 sq. rods, 23 yds. 5 feet to feet. 67. Reduce 100 acres, 37 rods to square feet. 68. Reduce 832590 sq. rods to sq. inches. 69. Reduce 25363896 sq. feet to acres, &c. 70. In 150 cubic feet, how many inches?'71. In 97 yds. 15 ft., how many cubic inches?'72. In 49 cords, 23 feet, how many cubic inches? 73. In 84673 cubic inches, how many feet? 74. In 39216 cubic feet, how many cords?'75. In 65 tons of round timber, how many cubic inches?'76. In 458210:0 cubic inches, how many tons of hewn timber'? APPLICATIONS OF REDUCTION..83, To reduce Troy to Avoirdupois weight. First reduzce the given poutnzds, ounces, &c., to grains; then divide oy the number of grains in, a dramn, and the quotient will be the answer in dcrains. (Art. 25 2.) OBS. If the answer is required to be in pounds and a firaction of a rund, divide the grains by 7000. Ex. 1. In 175 pounds Troy, how many pounds avoirdupois? Solution.-175 X12X20 X 24-1008000 grs., and 1008000 grs. 271 —=36864 drams, or 144 lbs. avoirdupois. Ans. QuZsr.-283. How is Troy weight reduced to avoirnupois? 170 REDUCTION. [SECT. VIII. 2. In 700 lbs. Troy of silver, how many pounds avoirdupois? 3. In 840 lbs. 6 oz. 10 pwts., how many pounds, &c., avoirdupois? 4. An apothecary bought 1000 lbs. of opium by Troy weight, and sold it by avoirdupois: how many pounds did he lose? A. A merchant bought 1500 pounds of lead Troy weight, anidt sold it by avoirdupois: how many pounds did he lose? 284. To reduce Avoirdupois to Troy weight. First reduce the given pounds, ounces, &c., to drains, then multiply by the number of grcains in a dram, and the product will be the answer in grains. (Art. 252.) Ons. 1. When the given example contains pounds only, we may multiply them by 7000, and the product will be grains. 2. If the answer is required to be in poundls and a fraction of a pound, (l,vide the grains by 5760. 6. In 32 lbs. avoirdupois, how many pounds Troy? Solution.-32 X 16 X 16 X 27L= 224000 grs., and 224000 gr[s. =38 lbs. 10 oz. 13 pwts. 8 grs. Ans. 7. In 48 lbs. avoirdupois, how many pounds Troy? 8. A merchant bought 100 lbs. 10 oz. of tea avoir., and sold it by Troy weight: how many pounds did he gain? 9. A druggist bought 1260 lbs. of alum avoirdupois, and retailed it by Troy weight: how many more pounds did he sell' than he bought? 2$ 5. The area of a floor, a piece of land, or any surface which has four sides and four right-angles, is found by multiplgying its length and breadth e toether. lote 1. The area of a figure is the supeficial contents or space container4 within the line or lines, by which the figure is bounded. It is reckoned in squlaq'e inches, feet, yards, rods, &c. 2. A fiigure which has four sides and four right-anleos, like the following, diagram, is called a Reclctarle or P,/ralZlelogsras. QTEsT.-284. Ilonr is avoirdupois weight redilced to Troy? 285. How do you find, the area or superficial contl.ents of a surface hsaving four sides and four right-angles I 3N'ote. What is merant 1) ihec ternl area? HIow is it reckoned? What is a figure whiih has four sides and four right-angles called? ARTS. 284'286.] REDUCTION. 17. 10. How many square yards of carpeting will it take to cover a room, 4 yards long and 3 yards wide? Suggestion.-Let the given room be represented by the subjoined figure, the length of which is divided into 4 equal parts, and the breadth into 3 equal parts which. we will call linear yards. NXow it is plain that the room will contain as many square yards as there are squares in the given figure. But the number of squares in the figure is equal to the number of equal parts (linear yards) which its length contains, repeated as many times as there are equal parts (linear yards) in its breadth; that is, it is equal to 4X 3, or 12. Ans. 12 yds. 11. Howv many sq. feet in a floor, 20 feet long, 18 feet wide? 12. How many acres in a field, 50 rods long, 45 rods wide? 13. How many square yards in a ceiling, 35 feet long and 28 Ifeet wide? 14. How many acres in a farm, 420 rods long and 170 rods:wide? 15. What is the area of a square field, whose sides are 80 rods ~n length? 16. How many yards of carpeting, a yard wide, will it take to cover a floor i8 feet square. 17. How many yards of plastering are required to cover four sides of a room, 18 ft. long, 15 feet wide, and 9 ft. high? 18. How many square yards of shingling will cover both sides of a roof, whose rafters are 20 feet, and whose ridge pole is 25 feet long? 286. The cubical contents, or solidity of boxes of goods, piles of wood, &c., are found by multiplying the length, breadth, n(dI thickness toyether. 19. How many cubic feet in a box 5 feet long, 4 feet wide and 3 feet deep? Solution.-5 4=20, and 20 X 3=60. Ans. 60 cu. ft. QUEST. —06. I1ow are the cubical contents of a box of goods, a pile of wood, &c., found 8* 172 REDUCTION [SECT. VIII, 20. How many cubic feet in a block of granite, 65 in. long, 42 in. wide, and 36 in. thick? 21. How many cubic feet in a load of wood, 8 ft. long, 41 fti; high, and 3~ ft. wide.? 22. How many cords of wood in a pile, 46 ft. long, 16 ft. high and 14 A 5 feet wide 23. How many cubic feet in a vat, 12 ft. long, 12 ft. long, 8 ft. wide, and 7j ft. deep? 24. How many cubic feet in a bin, 12 ft. long, 9 ft. deep, and 7 ft. wide? 25. How many cubic yards in a cellar, 18 ft. long, 12 ft. wide, and 9 ft. deep? 26. How many cubic feet in a stick of timber, 2 ft. square, and 40 ft. long? 2'7. How many cubic feet in a cistern 15 ft. long, 12 ft. wide, and 10 ft. deep? 287, To reduce Cubic to Dry, or Liquid Measure. First reduce the given yards; feet, &c., to cubic inches; thet? divide by thie nucmber of cubic inches in a gallon, or bushel, as the case may be, and tile quotient will be the answer required (Arts. 260, 263.) 28. In 10752 cubic feet, how many bushels? Solutioz. - 10752 X 1728 = 18579456 cubic inches; and 18579456- - 2150 —=- 8640 bushels. 29. In 21504 cubic feet, how many bushels? 30. In 462 cubic feet, how many wine gallons? 31. In 1155 cubic feet and 33 inches, how many wine gallons? 32. In 846 cubic feet, how many beer gallons? 33. In 1128 cubic feet and 141 in., how many beer gallons? 34. How many bushels will a bin contain, which is 5 ft. long, 3 ft. wide, and 4 ft. deep? 35. How many bushels will a bin contain, which is 8 ft. long, 4~ ft. wide, and 3~ ft. deep? QUEsT-287. Ilow reduce cubic to dry, or liquid measure o ARTs. 287 —289.] REDUCTION. 173 3B. How many bushels will a bin contain, which is 14 ft. long, 10 ft. 8 in. wide, and 6 ft. 8 in. deep? 387. How many wine gallons in a cistern, which is 6 ft. long, 5 ft. wide, and 4 feet deep? 38. How many barrels of water (wine meas.) will a cistern hold, which is 20 ft. long, 15 ft. wide, and 10 ft. deep? 39. The distributing reservoir of the Croton Water Works in the City of New York, is 436 ft. square and 40 feet high: how many hogsheads of water will it hold? 288o To reduce Dry, or Liquid, to Cubic Aieasure. First find the number of bushels, if dry measure, or gallons, if liquid measure, in the given example; then multiply by the number of cubic inches in a gallon, or bushel; as the case may be, and M]ee product will be the answer required. (Art. 263.) 40. How many cubic feet in a bin, which contains 100 bushels? Solution. - 100X 2150-% = 215040, and 215040 -- 1728= l..24.-.j88, or 124- cubic feet. Ans. 41. How many cubic feet in a lime kiln, which holds 100.bushels? 42. How many cubic feet in the hold of a ship, which contains'1000 bushels of grain? 43. How many cubic feet in 1 hogshead, wine measure? 44. How many cubic feet in a cistern, which holds 50 barrels of water? 45. How many cubic feet in a vat, which contains 100 hogsheads wine measure? 289 To reduce Liquid to Dry Measure, or Dry to Liquid'MIeasure. First find the cubic inches in the given. exzanple; then divid0 them by the number of cubic inches in a gaallon, or bust'Gcl, as the casa may be, and the quotient wzill be the answer required. QURST. —288. H-Iow reduce dry, or liquid measure to cub Ic? 289. IIow reduce liquid to dry measure? How dry to liquid measure? 174 REDUCTION. [S ECT. YH.i 46. In 40 gallons wine measure, how many bushels? Solution.-40 X 231=9240 cu. in., and 9240 cu. in. - 2150-f=4ma bushels. Ans. 47. In 6 hogsheads, 16 gallons, how many bushels? 48. In 5 bushels, how many gallons wine measure? 49. In 3200 quarts dry measure, how many hogsheads wine measure? 290 To reduce Wine to Beer Measure, or Beer to Wine Measure. First find the number of cubic inches in the the given examzle; then divide them by the number of cubic inches whichl it takes toi make a gallon in the required measure. 50. In 94 wine gallons, how many beer gallons? Solution.-94 X 231==21714 cu. in., and 21714 cu. in.- 282~ 77 gallons. Ans. 51. In 1 hhd. wine measure, how many beer gallons? 52. A tavtern-keeper bought 4 hhds. of cider wine measure, aned retailed it by beer measure: how many gallons did he lose? 53. In 20 beer gallons, how many wine gallons? 54. A grocer bought 7238 gallons of milk beer measure, andi retailed it by wine measure: how many gallons did he gain? 55. A druggist bought 10000 gallons of alcohol beer measure, and sold it by wine measure: how many gallons did he gain? 56. A grocer bought 65 hhds. 29 gals. and 2 quarts of milk by beer measure, and sold it to his customers by wine measure: how many quarts more did he sell than he bought? 57. A liquor dealer bought 120 pipes of wine which his clerk retailed by beer measure how many gallons more did he buy than he sold? 291 o Since the earth revolves on its axis 1~ in four minutes, or 1' in 4 seconds of time; (Art. 268,) it is evident that longitude may be reduced to time. That is, multiplying degrees of longib tude by 4 reduces them to minutes of time, multiplying minutes of longitude by 4 reduces them to seconds of time, &c. QUEST.-290. 1How reduce wine to beer measure? HIIo oeer to wine measure I .ARis. 290-293.1 REDUCTION. 175 By reversing this process it is evident that time may be reduced 1'o lo lyitalde. Thus, dividingo seconds of time by 4, will reduce them to minutes of longitude; dividing minutes of time by 4, will Teduce them to degrees, &c. Hence, 21)2. To find the difference of time between two places from tle difference of their longitude. -Reduce the dfcerence of longitude to minutes; multiply them by 4, anld the product will be the difference of time in seconds, which'.cmay be reduced to hours and minutes. Ois. When the difference of longitude consists of degqrees only, we may multliply theln by 4, and the product will be the answer in mbinutes. 58. The difference of longitude between New York and Cinc/nnati is 10~ 2'': what is the difference in their time? Sol-ution. —10 and 26'=-626'; (Art. 281;) now 626'X4=!2504 seconds of time; and 2504 sec. — 60=41 min. 44 sec. Ans. 59. The difference of longitude between Albany and Boston is 2' 9': what is the difference in their time? 60. The difference of longitude between Albany and Detroit is 90 45' wharliLt is the difference in their time? 61. The difference of longitude between New Iavzen and New ()rleans is 17~ 10': what is the difference in their time? 62. The difference of longitude between Charleston, S. C. and'Mobile is 8~ 27': what is the difference in their time? 68. The difference of longitude between New York and Canton is 1817 3': vWhat is the difference in -their time? 293. To find the difference of loigitude between two places from the difference in their time. 2Reduce the given di2ferc)zce of time to seconzds; divide them by 4, qul1 the q/eotien t will be the dlitference of longitude in minutes, which may be recduced to degrees. (Art. 281.) OBs. Wthen there are no seconds in the difference of time, we may divide the minutes by 4, and the quotient will be the answer in degrees. QOUEs'r,* —2). Ilow find the difference of time between two places from their difference of longitudeC.'93. HIow findl the difference of longitude from the difference of time I 176 REEDUCTION. [SECT. VIII. 64. A C;hip sailed from Boston to Liverpool; on the fourth day the master took an obser-vi'tion of the sun at noon, iand found b5y his chroometer that it was 1. hr. 5 min. and 40 sec. earlier thar the Boston time: how many degrees east of Boston was the ship', Solution.-l hr. 5 m. 40 sec.= 3940 sec., (Art. 281,) and 3940, sec..4 — 985'. The ship had therefore sailed 985' east, which: is equal to 16~ 25'. Ans. 65. The difference of time between Albany and Buff-alo is 19 minutes: what is the difference of their longitude? 66. The difference of time between Richmond and New Orleans, is 51 min. 4 sec.: what is the difference of their longitude? 67. The difference of time between Boston and Cincinnati i.:s 53 min. 32 sec.: what is the difference of their longitude? COMPOUND NUMBERS REDUCED TO FRACTIONS. 294. That one concrete number may properly be said to be u part of another, the two numbers must necessarily express objects of the samnze kind, or objects which can be rleduzced to the same kind or denomination. Thus, 1 penny is -1 —- of a poucnd, but:1 penny cannot properly be said to be a part of a J/bot, or of a year; for, feet and years cannot be reduced to pence. So, 1 orange is -~of 5 oranges; but 1 orange cannot be said to be -1 of 5 apples, o r 5 pumpkins; for apples and pumpkins cannot be reduced tc, ql'lranges. Ex. 1. Reduce 2s. 7d. to the fraction of a pound. -Analysis.-The object in this example is to find what part of 1 pound, 2s. 7d. is equal to. To ascertain this, wve must reduce both the given numbers to the same denomination, viz: pence. NIow 2s. 7d.=31d., and ~1=240d. (Art. 281. I.) Tihe question, therefore, resolves itself into this: what part of 240 is 31? The anrswer is 4- o; consequently 2s. 7d. (31d.) is -2#6 of a pound. H3nce, QUIEST. —294. WThen can one concrete number be said to be a part cm't another? ARTS. 294, 296.1 REDUCTION. 177 295. To recdc' a compound number to a common fraction, of a higher denomination. First reduce the given compound number to the lowest denomina. tion mentioned for the numerator; then reduce a UNIT of the denomination of the r'equired fraction to the same denomination as the numerator, and the result will be the denominator. (Art. 281.) OBs. 1. The given number, and that of which it is said to be a Partl must, in all cases, be reduced to the same denomination. (Art. 294.) *2. When the given number contains but one denomination, it of coarse requires no reduction. If the given number contains a fraction, the denominator of the fraction is the lowest denomination mentioned. Thus, in 6Qs., the lowest denomination isfourt/ts of a shilling; in i-far., the lowest denomination isfif/his of a farthing. 2. Reduce ~- of a penny to the fraction of a pound. Solution.-Since sevenths of a penny is the lowest and only denomination given, we simply reduce ~1 to sevenths of a penny for the denominator. Now ~1 —240d., and 240d.X7=1680C,.4ns. L —g-o, or ~Z X. Hence, 296. To reduce a fraction of a lower denoulination to arn equivalent fraction of a higher denomination. -Reduce a unit of the denomination of the required fraction to she same denomination as the given fraction, and the result will be the denominator. Or, divide the given fraction by the same numbers as in reducing whole compound zumbenseq to higher denominations. (Art. 281. II.) Thus in the last example, 1-d. —12_ —-s., (Art. 227,) and -8s. — 20= 1 2.rf>,:L 8 o. Ans. OGs. When factors common to the numerator and denominator occur, the operation may be s7olrtenled by canceclngl those factors. (Art. 221.) 3. Reduce A of a penny to the fraction of a pound. Solution.-By the last article, --- the answer. 7X12X20 4 / 1 By Cancelation _ - 2- Ans. 7X12X20 7X12X 0,5 420 Q(IEST. —295. TIow is a compound number reduced to a coimmon fraction? 296. How is a fractiol of a lower denomination reduceN! to the fraction of a h',gher? 178 REDUCTION. [S:cT. VIIJ 4. Reduce 4-s. to the fraction of a pound. Ans. may be shown that the number of figures contained in any te2lminlace decimal,: is equal to the greatest number of times that either of the prime factors 2 or 5; is repeated in the denominator of the given fraction. The same reasoning will evidently hold true when the numerator is 2, 3, 4i, 5, &c., or any number greater than 1. In this case the decimal will be as: many times greater, as the numerator is greater than 1. 344. The number of figures in the period must always be one: less than there are units in the denominator; for, the number otf remainders different from each other which can arise from anwy operation in division, must necessarily be one less than the uniits in the divisor. For example, in dividing by 7, it is evident, the. only possible remainders are 1, 2, 3, 4, 5, and 6; and since in re-. ducing a common fraction to a decimal, a cipher is annexed to each remainder, there cannot be more than six different dividends', consequently, there cannot be more than six different figures inp the quotient. Thus, +=.142857,142857, &c. When the decimal is periodical or circulating, it is custom+uary to write the period but once, and put a dot, or accent over. the first and last figure of the period to denote its continuance. Thus,.461351355135, &c., is written.46135, and.633333, &c:,.63; Reduce the following fractions to circulating decimals: 31. 1. 36. i. 41. -. 46. A. 32. A. 37. +. 42. A-. 47. -. 33. i. 38. A. 43. I. 48.. 34. 2. 39. -. 44. a. 49. z. 35. A. 40. A4. -. 50. a. 51. How many decimal figures are required to express --? 52, Hofw many decimal figures are required to express -,-? 53 IHow many decimal figures are required to express -r i36? 54. How mlany decimal figures are necessary to express -j-6-? ARTS. 344-346.] DECIMALS. 215 55 How many decimal figures are necessary to express A? 56. How many decimal figures are necessary to express -— 6?4-? 57. Reduce -z%.to a decimal. 59. Reduce __6 to a decimal. 58. Reduce -,)g to a decimal. 60. Reduce ~3- to a decimal. NIVtoe.-For the method of finding the valute of periodical decimals, or of re, duticing them to Common Fractions, also of adding, subtracting, mnlltiplyin1gK and dividiwng them, see the next Section. CASE III. 345. Compound Numbers reduced to Decimals. Ex. Reduce 13s. 6d. to the decimal of a pound. Analysis. —13s. 6d.=162d., and ~1=240d. (Art. 281.) Now 1.62d. is ~]R of 240d. The question therefore resolves itself into this: reduce the fraction 6_~ to decimals. Ans. ~.675. Hence, 346. To reduce a compound number to the decimal of a;higher denomination. First reduce the given compound number to a common fraction; (Art. 295;) then reduce the commnon fraction to a decimal. (Art. 337.) 2. Reduce 4s. 9d. to the decimal of ~1. Ans. ~.237+. 3. Reduce 10s. 9d. to the decimal of ~1. 4. Reduce 16s. 6d. to the decimal of ~1. 5. Reduce 17s. 7d. to the decimal of ~1. 6. Reduce 5d. to the decimal of a shilling. 7. Reduce 6~d. to the decimal of a shilling. 8. Reduce 37 rods to the decimal of a mile. 9. Reduce S fur. X2 ocds to the decimal of a mile. 10. Reduce iL t-1AI. o0 sec. to the decimal of an hour. 11. Reduce 3 hrs. 3 min. to the decimal of a day. 12. Reduce 5 lbs. 4 6z. to the decimal of a cwt. 13. Reduce 7 oz. 8 drams to the decimal of a pound. 14. Reduce 3 pks. 4 qts. to the decimal of a bushel. 15, Reduce 4 qts. 1 pt. to the decimal of a peck. 16. Reduce 4 qts. I pt. to the decimal of a gallon. QuEST.-346. IIow is a'compound number reduced to the decimal of a higher denomr linatlon. 216 REDUCTION OF DECIMALS. [SECT. IX;. CASE IV. 342. -Decimal Compound Numbers reduced to whole ones. 1. Reduce o~.387 to shillings, pence and farthlings. O1pe?.ation. Multiply the given decimal by 20, becaus e ~.387 20s. make ~1, and point off as many figure;s 20 for decimals, as there are decimal places irl shil. 7.740 the multiplier and multiplicand. (Art. 330.:) 12 The product is in shillings and a decimal pence 8.880 of a shilling. Then multiply the decimal 4 of a shilling by 12, and point off as befar. 3.520 fore, &c. The numbers on the left of tlhke Ans. 7s. 8d. 3 fcar. decimal points, viz: 7s. 8d. 3 far., form tle answer. IIence, 348. To reduce a decimal compound number to whole nmimbers of lower denominations. _fiultiply the given decimal by that number which it takes of the next lower denomination to make ONE of this hiigher, as ina r'eductio;',, and point olf the product, as in multiplication of decimal fractionls. (Art. 330.) Proceed in this manner with the decimal figures cJ each szucceeding p2roduct, and the numbers on the left of the dccimnai point of the several products, will be the whole number required. 2. Reduce ~.725 to shillings, pence and farthings. 3, Reduce ~.1325 to shillings, &c. 4 Reduce.125s. to pence and farthings. 5. Reduce.825s. to pence and farthings. 6. Reduce.125 cwt. to pounds, &c. 7. Reduce.435 lbs. to ounces and drcams. 8. Reduce.275 miles to rods, &c. 9. Reduce.4245 rods to feet, &c, 10. Reduce.1824 hhds. to gallons, &c. I1. Reduce.4826 gal. to qts., &c. 12. Reduce.4258 day to hours, &c. 13. Reduce.845 hr. to minutes and seconds. QUEST.-348. How are decimal compound numbers reduc6d to whole ones of a low;e: denomination? AaTs. 347-353.1 CIRCULATING DECIMALS. 217 SECTION X. PERIODICAL OR CIRCULATING DECIMALS.* ART. 3 49. Decimals which consist of the same figures or set of figures repeated, are called PERIODICAL, oR CIRCULATING DECIMALS. (Art. 339.) 350. The repeating figures are called periods, or relpetends, If one figure only repeats, it is called a single period, or repetend; as.i1111, &c.;.33333, &c. When the same set of figures recurs at equal intervals, it is called a comzounzd:period, or repetend; as.01010101, &c.;.123123123, &c. 35 1o If other figures arise before the period commences, the decimal is said to be a mixed periodical; all others are called pure, or sim~ple periodicals. Thus.42631631, &c., is a mixed periodical; and.33333, &c., is a pure periodical decimal. Oss. 1. When a pure circulating decimal contains as many figures as there are units in the denominator less one, it is sometimes called a pefecjt period, or repetend. (Art. 344.) Thus, -. 142857, &c., and is a perfect period. 2. The decimal figures which precede the period, are often called the tersninate part of the fraction. 35 2. Circulating decimals are expressed by writing the period once with a dot over its first and last figure when compound; ancd when single by writing the repeating figure only once with a dot over it. Thus.46135135, &c., is written.46135 and.33, &c.,.3. 353. Similar periods are such as begin at the same place or distance from the decimal point; as.i and.3, or 2.34 and 3.76, &c..Dissimilar periods are such as begin at different places; as.123 and.42325. Similar and contermrinous periods are such as begin and end in the same places; as.232i and 1634. * Should Periodical Decimals be deemed too intrikate for younger classss, they can be omitted till review 218 PERIODICAL oR [SECT. X. REDIUCTION OF CIRCULATING DECIMALS. CASE I. —To redeuce pure circulating decincals to comzmon frac-a tions. 354. Tc investigate this problem, let us recur to the origino of circulating decimals, or the manner of obtaining them. F'or example, -=-.11111, &c., or.1; therefore the true value of.1111, &c., or.i, must be - from which it arose. For the same reason.22222, &c., or.2, must be twice as much or -; (Art. 186;>).33333, &c., or.3_-;.49=;.5=9, &c. Again, 9-=.010101, &c., or.01; consequently.010101, &c., oi-,oi=-l;.020202, c.,.or.030303, &e., or.03-o-r;.070707, &c., or.07 — -c, &c. So also --— l.001001001, &c., or.001; therefore.001001, 0&c., or.OOi: ——;.002 —-— e; &c.. In like manner j=.i42857; (Art. ~337;) and 1 42857 = -4-4-; for, multiplying thle numerator and denominator of -- byT 142857,, we have L-_-4-+-. (Art. 191.) So -4 is twice as much as -; -, thrleetimes as much, &c. Thus it will be seen that the value of a pure. periodical decimal is expressed by the common fraction whoseo numerator is the given period, and whose denominator is as many 9s as there are figures in the period. HIence, 355. To reduce a pure circulating decimal to a common fra tion. Macfake the yiven per2iod the nmzm-rator, a'acl the cenolominator will be as many 9s as t.here arse figtures in tlhe period. Ex. 1. Reduce.3 to a common fraction. Azns. -, or I. 2. Reduce.6 to a common fraction. Ans., or ]. 3. Reduce.18 to a common fraction. 4. Reduce.123 to a common fiaction. 5. Reduce.'290 to a common fraction. 6. Reduce.72) to a commiron fimcetion.'7. teduce.09 to a common firaction. 8. Reduce.045 to a common fiaction. 9. Reduce.142857 to a common fraction. 10. Reduce.076923 to a common fraction. AcrTs. 354-357.] CIRCULATING DECIMATS. 219 CASE II.-To seduce mixed circulating decimals to common fractions. 356. 11. Reduce.16 to a common fraction. Analysis.-Separating the mixed decimal into its terminate and periodical part, we have.16 =.1+.06. (Art. 320.) Now.l=-Ir; (Art. 312;) and.0d6 —=-; for, the pure period.6= —, (Art. 351,) and since the mixed period.06, begins in hundredths' place, its valtue is evidently only -J as much; but -&-10= —s. (Art. 227.) Therefore.16 = + —-r0. Now l and -, reduced to a common denominator and added together, make J —,, or A. Ans. OBS. In mixed circulating decimals, if the period begins in Aundredths' place i.t is evident from the preceding analysis that the value of the periodical part is Wonly -,T as much as it would be, if the period were pure or begun in tenths':place; when the period begins in thousandths' place, its value is only -i-i-v part ts much, &c. Thus.6-6-;.06= —.10= — -:.006- 6 —- 100=9 -, &c. 357. Hence, the denominator of the periodical part of a mixed circulating decimal, is always as many 9s as there ale figures in the period with as many ciphe'.rs annexed as there are decimals in the terminate part. 12. Reduce.8.567923 to a commopr fraction. Solution.-Reasoning as before.8567923' —&_ _-+ 6_ —z 0. Reducing these two fractions to the least common denominator, (Art. 261.) -o-f teX 99999=o-t r 9o whose denominator is the same as thlat of the other. Now 9 9 -L +-T-i999 - 990 99 — 99 0. Ans. Contraction. To multiply by 990999, annex as many 8500000 ciphers to the multiplicand as there are 85 9s in the multiplier, &c. (Art. 105.) This 8499915 1st Nu. gives the numerator of the first fraction 67923 2d Nu. or terminate part, to which add the nu8567838 merator of the second or periodical part, 9999900 Ans. and the sum will be the numerator of the answer. The denominator is the same as fhat of the second or periodical part. 10)* 220 PERIODICAL OR [SECT. X. Second 2Method. S567923 the given circulating decimal. 85 the terminate part which is subtracted 8567838 the numerator of the answer. 99999 0 nS. INote. —. The reason of this operation may be shown thus: 8567923_-. 8500000+67923. Now 8500000-85+67923 is equal to 8567923 —85. 2. It is evident that the required denominator is the same as that of the periodical part; (Art. 357;) for, the denominator of the periodical part is thbe least common multiple of the two denominators. Hence, 35S. To reduce a mixed circulating decimal to a common fraction. Change both the terminate and periodical part to common fractions separately, and their sam will be the answer required. Or, firom the given mixed peeriodical, subtract the terminate par t, and the remainder will be the numerator requtired. The denominzator is always as many 9s as there are fiqgures in the period with as many ciphers annexed as t/ere are decicmals in the termnicate part. PROOF.- Change the co)m)imoni ractionl backc to a decimal, and Ijf t)e result is the same se the givenl circelaCtin decimaCl, the work is righ't. 13. Reduce.138 to a common fiaction. Ans. -, or -. 14. Reduce.53 to a common fraction. 15. Reduce.5925 to a common fraction. 16. Reduce.583 to a common fraction. 17. Reduce.0227 to a commmon fraction. 18. Reduce.4745 to a common fraction. 19. Reduce.5925 to a common fraction. 20. Reduce.008497133 to a common fraction. CASE III. —Dissimilar periodicals reclduced to similar and canter — ininous ones. 359. In changing dissimilar periods, or repetends, to similar and conterminous ones, the following particulars require attentioil. 1, Any terminate decimal may be considered as interminarte by annexing ciphers continually to the numerator. Thtl s.46-.460000, &c.-=.460. Awrs. 358-361.] CIRCULATING DECIM.ALS. 221 2. Any pure periodical may be considered as mired, by taking the given period for the terminate part, and' making the given period the interminate part. Thus.46=.46+.0046, &c. 3. A single period may be regarded as a com2pound periodical. Thus.3 may become.33, or.333; so.63 may be made.6333, or.63333, &c. 4. A single period may also be made to begin at a lower order, regarding its higher orders-as terminate decimals. Thus.3 may be made.33, or.3333, &c. 5. Compound periods may also be made to begin at a lower order. Thus.36 uay be changed to.363, or.36363, &c.; or by extending the number of places.479 may be made.47979, or.4797979, &c.; or making both changes at once.532 may be changed to.5325325, &c. Hence, 360. To make any number of dissimilar periodical decimals similar. Move the points, so that each period shall begin at the same order as the period which has the most figures in its terminate part. 21, Change 6.814, 3.26, and.083 to similar and conterminous periods. Operation. Having made the given periods simi6.814= 6.8814814-8 i ar, the next step is to make them con3.2(6 —3.2 6262626 terminous. Now as one of the given.083 —0.08333333 periods contains 3 figures, another 2, and the othler 1, it is evrident the new periodical must contain a number of figures which is some multiple of the numwber of figures in th.e different periods; viz: 3, 2, and 1. But the least common multiple of 3, 2, and 1 is 6; therefore the new periods must at least contfain 6 figures. Hence, 361. To make any number of dissim')ilar periodical decimals, similar and conterminous. Fairst make the periods siimilar; (Art. 360;) thle extend the figures of each to cas many places, as there are nits in the least commrnon multiple of the uEMlE of goperiodtical flycres contained in each of the given decimals. (Art. 176.) 222 PERIODICAL ORn SECT. X. 22. Change 46.i18; 5.26, 63.423,.486, and 12.5, to similar and conterminous periodicals. Operation. 46.162=46.16216216 The numbers of periodical figures in 5.26 = 5.26262626 the given decimals are 3, 2, 3, and 1; 63.423=63.42342342 and the least common multiple of.486=- 0.48666666 them is 6. Therefore the new periods 12.5 =12.50000000 must each have 6 figures. 23. Make.27,.3, and.045 similar and conterminous. 24. Make 4.321, 6.4263, and.6 similar and conterminous. ADDITION OF CIRCULATING DECIMALS. Ex. 1. What is the sum of 17.23+41.24't6+8.61+-1.5+ 35.423? Operation. Dissimilar. Sim. & Conterminous. 17.23 = 17.2323232 First make the given decimals sim41.2476=41.2476476 ilar and conterminous. (Art. 361.) 8.61 = 8.61i6616 Then add the periodical parts as in 1.5 = 1.5000000 simple addition, and since there are 35.423 =35.4232323 6 figures in the period, divide their Ans. 104.0193648 sum by 909999; for this would be its denominator, if the sum of the periodicals were expressed by a common fraction. (Art. 355.) Setting down the remainder for the repeating decimals, carry the quotient 1 to the next column, and proceed as in addition of whole numbers. Hence, 362. We derive the following general RULE FOR ADDING CIRCULATING DECIMALS.,irst make the periods similar and conterminous, and nfin their surn as iv, Simple Addition. D)ivide this sum by as mcny 9s as there are figures in the period, set tfhe remainder under the figures added for the ]period of thle scum, carry the tquotient to the next; column, and proceed with the rest as in Simnple AJddition. ARTS. 362, 363.] CIRCULATING DECIMALS. 223 OBs. If the remainder has not so many figures as the period, ciphers must be prefixed to make up the deficiency. 2. What is the sum of 24.i32+2.2:3+85.24q+67.6'? 3. What is the sum of 328.1~6+81.23+5.624+61.6'? 4. What is the sum of 31.62-7.824+8.392+.027? 5. What is the sum of 462.34+60.82+71.164+.35? 6. What is the sum of 60.25-+.34~-6.43o+.45+-45.24? 7. What is the sum of 9.814+1.5+87.26+0.83+124.09? 8. What is the sum of 3.d+78.3i76-l-735.3-.375+.2'7+ 187.4? 9. What is the sum of 5391.357+72.38+187.2i+4.29a5+ 2] 7.8496 +42.176+.523 +58.30048? 10. What is the sum of.i62+ 134.b09+2.93+97.26+-3.769Q.36 +99.083+1.5+-.814? SUBTRACTION OF CIRCULATING DECIMALS. Ex. 1. From 52.86 take 8.37235. Operation. We first make the given decimals sil.i52.St =52.86868 lar and conterminous, then subtract as:n 8.37235= 8.37235 whole numbers. But since the period in 44.49632 the lower line is larger than that above it, we must borrow 1 from the next higher order. This will make the right hand figure of the remainder ofe less than if it was a terminate decimal. Hence, 363. We derive the following general RULE FOR SUBTRACTING CIRCULATING DECIMALS. Allake thee periods similar and conterminous, and subtrr(ct as in,lhole numbers. If the period in the lower line is larger titan that above it, diminish the right hand figure of thze remainder 7, 1. Ons. The rqeason for diminishing the right hand figure of the remainder by *1, if the period in the lower line is larger than that above it, may be explained thus: When the period in the lower line is larger than that above it, we must evitdently borrow 1 from the next higher order. Now if the given decimals were extended to a second period, in this period the lower number would also be 224 PERIODICAL OR [SECT. X. larger than that above it, and therefore we must borrow 1. But having bor. rowed 1 in the seo:nd period, we must also carry one to the next figure in the lower line, or, what is the same in effect, diminish the right hand figure of the remainder by 1. 2. From 85.62 take 13.76432. Ans. 71.86193. 3. From 476.32 take 84.7697. 4. From 3.8564 take.0382. B. From 46.123 take 41.3. 6. From 801.6 take 400.75. 7. From 4.7824 take.87. 8. From 1419.6 take 1200.9. 9. From.634852 take.02i. 10. From 8482.42i take 6031.035. MULTIPLICATION OF CIRCULATING DECIMALS Ex. 1. What is the product of.36 into.25? Operation. We first reduce the given periodi-.36 - -6~ —jr cals to common fractions; (Art. 357;).25 0 —s-+ —— =-z. then multiply them together. (Art. Now -ao X -=- -— 9%- 219.) Finally, we reduce the product But -% —o=.092 Ans. to a periodical decimal. Hence, 364. We derive the following general RULE FOR MULTIPLYING CIRCULATING DECIMALS. Fiirst redcce the given periodicals to common fractions, and multip/ly them) toyether as usual. (Art. 219.) Finally, reduce the prodzect to deciamls csnz it woill be the answer required. OBS. If the numerators and denominators have conmmon factors, the opera4 tion may be contrac!c(d by cancelir those factors before the multiplication is performeia. (Art. 221.) 2. Wha it is the prod-uct of 37.23 into.26?.Ans. 9.928. 3. What is the product of.123 into.6? 4. What is the product of.245 into 7.3? 5. What is the product of 24.6 into 15.7? 6. What is the product of 48.23 into 16.13? ARTS. 364, 365.] CIRCULATING DECIMALS. 225 7. What is the product of 8574.3 into 87.5? 8. What is the product of 3.973 into.8? 9. What is the product of 49640.54 into.70503? 10. What is the product of 7.72 into.297? DIVISION OF CIRCULATING DECIMALS. Ex. 1. Divide 234.6 by.7. Operation. We first reduce the divisor 234.6=-234-=-o and dividend to common frac9~/ —= tions; (Art. 357;) and divide Now 4.- = ---- eX — 636 one by the other; (Art. 229;) And 6-6-3=301.714285 Ans. then reduce the quotient to a decimal. (Art. 337.) Hence, 365. We derive the following general RULE FOR DIVIDING CIRCULATING DECIMALS. -Reduce the divisor and dividend to common fractions; divide one fraction by the other, and reduce the quotient to decimals. Ons. After the divisor is inverted, if the numerators and denominators have factors common to both, the operation may be contrtacted by cancelingl those factors. (Art. 232.) 2. Divide 319.28007i12 by 764.5. Ans. 0.4176325. 3. Divide 18.56 by.3. 4. Divide.6 by.123. 5. Divide 2.297 by.297. 6. Divide 750730.518 by 87.5. 7. Divide 42630.6 by 28421.3. 8. Divide 80000.27 by 20000.36. 9. Divide 24.08i by.386. 10. Divide.36 by.25. 226 REDUCTION OF [SECT. XI SECT1ON XI. FEDERAL MONEY. ART. 366. FEDERAL MONEY is the currency of the United States. Its denominations, we have seen, are Eagles, Dollcars,,Dimes, Cents, and Mills. (Art. 244.) 367. All accounts in the United States are required by law to be kept in dollars, cents, and mills. Eagles are expressed in dollars, and dimes in cents. Thus, instead of 8 eagles, wve say, 80 dollars; instead of 6 dimes and 7 cents, wie say, 67 cents, &c. 368. Federal Money is based upon the Decimal system of Notation. Its denominations increase and decrease from right to left and left to right in a tenfold ratio, like whole numbers and decimals. (Ait. 244. Obs. 1.) 369. The dollar is regarded as the unit; cents and mills are fractional parts of the dollar, and are distinguished from it by a decimal point or separatrix (.) in the same manner as common decintls are distinguished from whole numbers. (Art. 311.) _Dollars therefore occupy units' place of simple numbers; eagles, or tens of dollars, tens' place, &c. Dimes, or tenths of a dollar, occupy the place of tenths in decimals; cenzts, or hundredths of a dollar, the place of hundredths; mills, or thousandths of a dollar, the place of thousandths; tenths of a mill, or ten thousandths of a dollar, the place of ten thousandths, &c. OBs. 1. Since dimes in business transactions are expressed in cents, tsco places of decimals are assigned to cents. If therefore the nmnber of cents is less than 10, a cipher must always be placed oni the left haid of theme; for cents are h lndredtls of a dollar, and hundredths occupy the second decimal place. (Art. 313.) For example, 4 cents are written thus.04; 7 cents thus.07; &c. 2. Mills occupy the third place of decimals; for they are thousandths of l dollar. Consequently, when there are no cents in the given sum, tlvo cipher must be placed before the mills. Hence, QUEST. —366. What is Federal Money 367. In what are accounts kept in the U. S. How would you express 8 eagles? Hlow express 6 dimes and " cents? 368. Upon what iL Federal Money based? 369. What is regarded as the unit ir Federal Money? What art cents and mills? How are they distinguished from dollars? AnRs. 366-371.J FEDERAL MONEY. 227 374). To read any sum of Federal Money. Call all the fiygures on the left of the decimal point dollars; the.first t:wo fig uzres after the point, a)re cents; the third figure denotes mills; the other places on thie rig/ht are decimals of a mill. Thus, $3.25232 is read, 3 dollars, 25 cents, 2 mills, and 32 hundredths of a mill. Oss. Sometimes all the figures after the point are read as deciInals of a o~l. ar. Thus, $5.356 is read, " 5 and 35G thousandths dollars." Write the following sums in Federal money: 1. 70 dollars, and 8 cents. Alns. 870.08. 2. 150 dollars, 3 cents, and 5 mills. 3. 409 dollars, 40 cents, and 3 mills. 4. 200 dollars, 5 cents, and 2 mills. 5. 4050 dollars, 65 cents, and 3 mills. Note.-In business transactions, when dollars and cents are expressed together, the cents arc frequently written in the form of a common fraction. Thus, the sum of $75.45, is written 75%-00- dollars. REDUCTION OF FEDERAL MONEY. CASE I. Ex. 1. How many cents are there in 95 dollars? Solution.-Since in 1 dollar there are 100 cents, in 95 dollars there are 95 times as many. And 95X 100 = 9500. Ans. 9500 cents. 2. In 20 cents how many mills? Ans. 200 mills. Note.-To multiply by 10, 100, &c., we simply annex as many ciphers to ~ihe multiplicand, as there are ciphers in the multiplier. (Art. 99.) Hence, 37 1. To r'educe dollars to cents, annex two ciphers. To reduce dollars to mills, annex three ciphers. To reduce cents to mills, annex one ciphzer. Oss. To reduce dollars, cents, and mills, to mills, erase the sigtn of dollars eand the scpa'atlrix. Thus, s25.36 reduced to cents, becomes 2536 cents. Quest.-353. Ilow do you read Federal Money? Obhs. What other mode of reading ]Vederal Money is mentioned. 354. I-ow are dollars reduced to cents? Dollars to mills.enuts to mills? Obs. Dollars, cents, and mills, to mills? 228 ADDITION OF [SECT. XI 3. In $12 l ow many cents? Ans. 1200 cents. 4. I}n $460 how many cents? 5. In 895 how many mills? 6. In 90 cents how many mills? 7. Reduce $25.15 to cents. 8. Reduce $864.08 to cents. 9. Reduce $1265.05 to mills. 10. Reduce $4580.10 to mills. 11. Reduce $6886.258 to mills. 12. liaduce $85625.40 to mills. CASE I I. 13. In 6400 cents, how many dollars? Suggestion. — Since 100 cents make 1 dollar, 6400 cents will make as many dollars as 100 is contained times in 6400. And. 6400 — 100 —64. Ans. $064. 14. In 260 mills, how many cents? Ans. 26 cents. No/c. —To divide by 10, 100, &c., we simply cut cff as many figures from tht right of the dividend as there are ciphers in the divisor. (Art. 131.) Hence, 372. To reduce cents to dollars, ccut off two Jigures on th~ right. To reduce mills to dollars, cut off three,figures on the right. To reduce mills to cents, cut off one figure on the right. OBs. The figures cut off are cents and mills. 15. In 626 cents, how many dollars? Ans. $6.26. 16. In 1516 cents, how many dollars? 17. In 162 mills, how many cents? 18. In 1000 mills, how many dollars? 19, In 2360 mills, how many cents? 20. In 3280 mills, how many dollars? 21. Reduce 8500 cents to dollars. 22. Reduce 2345 cents to dollars, &c. 23. Reduce 92355 mills to dollars, &c. QUEST.-355. How are cents reduced to dollars? Mills to dollars? -Mills to cendx Obs. What are the figures cut offl ARTS. 372-374.] FEDERAL MONEY. 229 24. Reduce 150233 mills to dollars, &c. 25. Reduce 450341 cents to dollars, &c. 373. Since Federal Money is expressed according to the decimal system of notation, it is evident that it may be subjected to the same operations and treated in the same manner as Decimal Fractions. ADDITION OF FEDERAL MONEY. Ex. 1. A man bought a cloak for $35.375, a hat for $4.875, a pair of boots for $6.50, and a coat for g23.625: what did he pay for all? Operation. We write the dollars under dollars, cents $35.375 under cents, &c. Then add each column sepa4.875 rately, and point off as many figures for cents 6.50 and mills, in the amount, as there are places 23.625 of cents and mills in either of the given num$70.375 dAns. bers. Hence, 374. We derive the following general RULE FOR ADDING FEDERAL MONEY. Write dollars under dollars, cents under cents, &c., so that the same orders or denominations may stand under each other. Add each columon separately, and point of the amount as in addition of decimal fractions. (Art. 320.) OnS. If either of the given numbers have no cents expressed, it is customary to supply their place by ciphers. 2. What is the sum of $48.25, $95.60, $40.09, and $81.10? 3. What is the sum of $103.40, $68.253, $89.455, $140 02. and $180? 4. What is the sum of $136.255, $10.30, $248.50, $65.38, and $100.125? QUEST.-357. Hlow is Federal Money added? How point off the amount Obe. When muy of the given numbers have no cents expressed, how is their place supplied 1 230 SUBTRACTION OF [SECT. XI. 5. What is the sum of $170, $4040.02, $130, $250.10, ana $845.22? 6. What is the sum of $268.45, $800.05, $192.125,'80.625, and $90.25? 7. What is the sum of $1500.20, $1.050.07, *100.70, $95.025, $360.437, and $425? 8. What is the sum of $2600, $1927.404, $ 1603.40, $3304.17,$165.47, and $2600.08? 9. A man bought a load of hay for $1 9.675, a horse for $73.25, a yoke of oxen for $69.56, a cow for $17, and a calf for $5.80: what did he pay for all? 10. A lady gave $21.50 for a dress, $9.25 for a bonnet, $28.33 for a shawl, and $15.25 for a muff: what was her bill? 11. A jockey bought a span of horses for $276.87, and sold them so as to gain $73.45: how much did he sell them for? 12. A man gave $4925.68 for a farm, and sold it so as to gairs $1565.37: how much did he sell it for? 13. A man sold a sloop for $7623.87, which was $1141.25 less than cost: how much did it cost? 14. A man bought a block of stores for $15268, which was $1721 less than cost: what was the cost? 15. What is the sum of 134 dolls. 3 cts. 7 mills, 108 dolls. 6 cts. 8 mills, 90 dolls. 9 cts. 4 mills, and 46 dolls. 18 cts. 4 mills? 16. What is the sum of 61 dolls. 1 ct. 2 mills, 19 dolls. 11 cts. 4 mills, 140 dolls. and 80 dolls. 4 cts.? 17. What is the sum of 140 dolls. 10 cts., 69 dolls. 3 cts. 8 mills, 18 dolls. 7 cts., and 29 dolls. 5 mills? 18. What is the sum of 860 dolls. 8 cts., 298 dolls. 4 cts. 8 mills, 416 dolls., 280 dolls. 13 cts., and 91 dolls.? 19. What is the sum of 14209 dolls., 65241 dolls., 1050 dolls., 610 dolls. 7 cts., and 1000 dolls. 10 cts.? 20. What is the sum of 1625 dolls., 4025 dolls., 1863 dolls. 75 cts., 16000 dolls., and 48261 dolls.? 21. What is the sum of 8 thousand dolls., 2 hlundred and 60 dolls. 5 cts., 19 thousand dolls. 60 cts., 6 hundred dolls. 9 cts.? 22. What is the sum of 19 thousand dolls. 50 cts., 61 thousand dolls. 10 cts., 18 hundred dolls. 3 ct,.? ART. 375.] FEDERAL MONEY. 231 SUBTRACTION OF FEDERAL MONEY. Ex. 1. A merchant bought a quantity of molasses for $75.40, Aincl a box of sugar for $42.63: how much more did he pay for ine than the other? Operation. We write the less number under the greater, $75.40 placing dollars under dollars, &c., then subtract 42.63 and point off the answer, as in subtraction of $32.77 decimals. Hence, 37 5. We derive the following general RULE FOR SUBTRACTING FEDERAL MONEY. Wgrite tlhe less number ~under the ygreater, with dollars under doltars, cents under ceats, A&ic.; then subttr'act and point o(f the remainder as in subtractionz of oleciniacl fractionzs. (Art. 322.):Ons. If either of the given numbers have no cents expressed, it is customary tia supply their place by ciphers. 2. A man bought a horse for $75.50, and sold it for $87.63' 1how much did he make by his bargain? 3. If a man deposits $204.65 in a bank, and afterwards checks (out $119.83, how much will he have left? 4. A man owing $682.40, paid 8435.25: how much does ho ctill owe? 5. A man owing $982.68, paid all but $64.20: how much did he pay? 6. A merchant bought a quantity of goods for $833.63, and re. tailed them for $1016.85: how much did he make by the bargaint? 7. A merchant bought a lot of goods for $1265.82, and sold,them for $942.35: how much did he lose? 8. A grocer sold a lot of sugar for $635.20, and made thereby $261.38: how much did he pay for the sugar? 9. A man sold his farm for $1'2250.62, which was $1379.87 )more than it cost: how nmuch did it cost? QUEST.-358. Htow is Federal Money sbl)tracted l? T)ow point off the remainder? Obs When either of the given numbers have no cents, how is their place supplied 1 232 MULTIPLICATION OF [SECT. XI 10. From $10600.75 take $8901.26. 11. From $20206.85 take $10261.062. 12. From $61219.40 take $100.036. 13. From $19 take 1 cent and 9 mills. 14. From 89 dollars take 89 cents. 15. From 506 dolls. take 316 dolls. and 8 cts. 16. From 5 dolls. 7 mills take 2 dolls. 7 cts. 17. From 61 dolls. 6 cts. take 29 dolls. 4 mills. 18. From 11000 dolls. 10 cts. take 110 dolls. 3 cts, 19. From 100100 dolls. take 10110 dolls. 10 cts. MULTIPLICATION OF FEDERAL MONEY. 37 6. In multiplication of Federal Money, as well as in simple, numbers, the multiplier must always be considered an abstract number. (Art. 82. Obs. 2.) Ex. 1. What will 8 bbls. of flour cost, at $5.62 per bbl.? Analysis.-Since I bbl. costs $5.62, 8 bbls. will cost 8 times as much; and $5.62X 8 —$44.96 Ans. 2. What cost 21.7 bushels of apples, at 15 cts. per bushel? Operation. Reasoning as before, 21.7 bushels will cost, 21.7 21.7 times 15 cents. But in performing the.15 multiplication, it is more convenient to make. 1085 the.15 the multiplier, and the result will be 217 the same as if it was placed for the multipli$3.255 Ans. cand. (Art. 83.) Point off the product as before. Hence, 37 7. When the price of one article, one pound, one yard, &c., is given, to find the cost of any number of articles, pounds, yards, &c., Multiply the price of one article and the number of articles tagether, and point off the product as in multiplication of decimals. (Art. 324.) QUEST.-376. In Multiplication of Federal Money, what must one of the given factors> be ocnsidered? 377. When the price of one article, one pound, &c., is given, how is the cost Jf any numnber of articles found? Anwrs. 376-378.] FEDERAL MONLY. 233 3. What cost 17.6 yards of cloth, at $4.75 per yard? 4. Mlultiply $25.625 by 20.2. 37 8. From the preceding illustrations we derive the following,general RULE FOR MULTIPLYING FEDERAL MONEY. M'ultip)ly as in simp7nle numbers, cand point of the -product as ir?t. ltiplicatioz oJ decimzal fractions. (Art. 324.) OBS. 1, When the price, or the quantity contains a common fraction, the fraction may be changed to a decimal. (Art. 337.) 2. In business operations, when the mills in the answer are 5, or over, it is C;ustomary to call them a cent; when under 5, they are disregarded. 5. What cost 12~ yards of cotton, at 9- cts. per yard? Solution. —12- yards=- 12.5, and 9- cts. —.0925; now.0925 X 12 5-l1.15625. ins. 6. What cost 451 yards of satin, at 87~ cts. per yard? 7. What cost 1 6!) bbls. of pork, at 8S- per barrel? 8. What cost. 324a lbs. of sugar, at 12~ cts. a pound? 9. What cost 97 gals. of oil, at 87~ cts. perC gallon? 10. What cost 310 lbs. of tea, at 62~ cts. a pound? 11, What cost 23~ tons of hay, at $84 per ton? 12. What cost 45 bbls. of flour, at 8$7L per barrel? 13. At 15~ cts. per doz., what cost 1.3~ dozen of ecrs? 14. At 84 cts. per pound, what will 321 lbs. of pork come to? 15. At $64 per bbl., what will 145~ bbls. of flour cost? 16. At 22~ cts. per doz., what will a gross of buttons cost? 17. At 31~ cts. per doz., what cost 45 doz. skeins of silk? 18. At 17~1 cts. per yard, whllat cost 91~ yards of calico? 19. What cost 45 doz. plates, at 62~ cts. per doz.? 20. What cost 63 doz. pen-knives, at $3~ per d;.? 21. What cost 19 doz. silver spoons, at $.7~ per dozen? 22. What cost 1865~ bushels of wheat, at 81L per bushel? 23. What cost 2560:3 yds. of broadcloth, at,5- per yard? QUEST.-378. What is the rule for Multiplication of Federal rMoney? Obs. When tie nrice or quantity contains a conimmon fraction, what should be done with it? 234 DIVISION OF [SECI I. DIVISION OF FEDERAL MONEY. Ex. 1. A man bought 8 sheep for $42.24: what did n give apiece? Analysis.-If 8 sheep cost $42.24, 1 sheep will cost x of a 22.24 and $42.24.-8=$5.28. Ans. $5.28. PROOF.-If 1 sheep costs $5.25, 8 sheep will cost 8 times as much; and $5.28 X 8=$42.24. Hence, 379. When the number of articles, pounds, yards, &c., andi the cost of the whole are given, to find the price of one article, one pound, &c. Divide the whole cost by the whole number of articles, and pointg off the quotient as in division of decimal fractions. (Art. 33-0.) 2. A shoemaker sold 15 pair of boots for $67.50: how much did he get a pair? 3. A merchant sold 65 lbs. of sugar for $3.93: how much. was that a pound? 4. A man bought 6.5 yards of cloth for $20.345: how much was that per yard? 5. How many bbls. of flour, at $5.38 per bbl., can be boughtts for $34.97? Analysis.-Since $5.38 will buy I bbl., Operation. $34.97 will buy as many bbls. as $5.38 5.38)34.97(6.5 Ans are contained times in $34.97. We divide 32 28 as in simple numbers, and point off one de- 2 690 cimal figure in the quotient. 2 690 PRooF.-$5.38 X 6.5 =$34.97, the given amount. 380. Hence, when.the price of one article, pound, yard, &c., and the cost of the whole are given, to find the number of articles, &c. Divide the whole cost by the price of one, ancd point of the quolient as a'n division of decimals. QUEST.- 379. WVhen the number of articles, pound(s, &c., and the cost of the whole are given, how is the cost of one article found. 380. When the price, of one article, one pounld, &c., and the cost of the whole are given, how is the number of articles found? A.RTS. 379-381.1 FEDERAL MONEY. 235 6. How many coats, at $12.56, can be bought for $103.085? 7. How many times is $11.13 contained in 87.606? 8. A gentleman distributed $68 equally among 32 poor persons: how much did each receive? Op;eration. 32)$68(2.125 Ans. After dividing the $68 by 32, there is 64 a remainder of 4 dollars, which should be 4000 reduced to cents and mills, and then be 32 divided as before. (Art. 354.) The ciphers 80 thus annexed must be regarded as deci64 mals; consequently there will be three 160 decimal figures in the quotient. 160 38 1. From the preceding illustrations we derive the following ieneral RULE FOR DIVIDING FEDERAL MONEY. Divide cas in simple nzinmbeirs, and point oQf the quotient as nu division of decimal fractions. (Art. 330.) Os. 1. In dividing Federal Money, if the number of decimals in the divisor is the same as that in the dividend, the quotient will be a whvole number. (Art. 330. Obs. 1.) 2. When there are more decimals in the divisor than in the dividend, annex many ciphers to the dividend as are necessary to make its decimal places equclit to those in the divisor. The quotient thence arising will be a whole number. (Obs. 1.) 3. After all the figures of the dividend are divided, if there is a remainder, ciphers may be annexed to it, and the operation may be continued as in division of decimals. (Art. 330. Obs. 3.) The ciphers thus annexed must be re-.arded as decimal places of the dividend. 9. How many gallons of molasses, at 28 cts. per gallon, can,ou buy for $86.25? QUEST.-381. What is the rule for Division of Federal hMoney? Ohs. When there is a emnainder after all the figures of the dividiend are divided, how proceed! When there are',ore decimals in the divisor than in the dividend, how proceed? -i 1 236 DIVISION OF [SECT. XI. 10. HIow many yards of calico, at 131 cts. per yard, can be bought for $73.3 7? 11. How many doz. of eggs, at 91 cts. per doz., can be bought for $94.185? 12. At 184 cts. per doz., how many skeins of sewing silk can, be bought for Q67.50? 13. A man paid $72.25 for 20.5 yards of cloth: how much did he pay per yard? 14. A man paid $76.50 for 51 sheep: what was the price per head? 15. A man paid $150 for 24 pair of boots: how much was that a pair? 16. If you give $56.25 for 281 bbls. of flour, how much do you pay per barrel? 17. If a man gives $316.375 for 8741 yards of cloth, what i:s that per yard? 18. A grocer sold 9654 lbs. of sugar for $81.25: what did h!g get a pound? 19. The fare from Albany to Buffalo, a distance of 326 mliles', is $13.20: how much is it per mile? 20. The fare from Boston to Albany, a distance of 203 miles, is $5.50: how much is it per mile? 21. If a clerk's salary is $650 per annum, how much does lle receive per day? 22. If a man spends $563.38 a year, how much are his average expenses per day? 23. At 871 cts. per bushel, how many bushels of wheat can you buy for $1500? 24. How many tons of coal, at $6.625 per ton, can you buy for $752.36? 25. If a man's income is $100 per week, how much is it perhour? 26. At $14.50 per acre, how many acres of land can you bhi fo $3560? 27. At $151 apiece, how many cows can you buy for $7750? 28. At.375.75 apiece, how many carriages cdl be bought fo $56362.50? ART. 382.1 FEDERAL MONEY. 237 COUNTING ROOM EXERCISES. Ex. 1. What cost 320 yards of satinet, at $1.124 per yard? Analysis.-If the price were $1 per yard, the cloth would evi. dently cost as many dollars as there are yards. But $1.124 is equal to I and ~ dollars; hence, the cloth will cost - more dollars than there are yards; consequently, if we add to the number of yards, of itself, it will give the cost. Now l of 320-40, and 320+40=360. Ans. $360. PRoo.-$1.12LX320 —-$360, the same as before. Hence, 382. When the price of 1 article, 1 pound, &c., is $1.122 $1.25; $1.374; &c., to find the cost of any number of articles. To the given number of articles, add 4, 4, 4, &c., of itself, as the case may be, and the sum will be the cost required. OBs. When the price of 1 article, &c., is $2.121, f$2.25, $3.37~, &c., the operation may be contracted by multiplying the given number of articles by 2-, 2~, 3, &c., as the case may be. 2. What cost 640 bushels of wheat, at $1.25 per bushel? 3. What cost 372 pair of shoes, at $1.374 a pair? 4. What cost 480 bbls. of cider, at $1.624 a barrel? 5. What cost 520 yards of silk, at $1.50 per yard? 6. What cost 720 drums of figs, at 81.87~ per drum? 7. At $2.124 apiece, what will 480 sheep cost? 8. At $2.371 apiece, what will 364 vests cost? 9. At $3.25 per yard, what cost 744 yards of cloth? 10. At $4.621 apiece, what cost 960 hats? 11. At $5.124x a pair, what cost 278 pair of boots? 12. At $7.374 per lb., what will 365 lbs. of opium cost? 13. A collier sold 856 tons of coal, at $6.87L per ton: how much did it amount to? 14. At 19.624 per acre, what will 537 acres of land cost? 15. What cost 72 lbs. of flax, at $8.25 per hundred? Analysis. —72 pounds are -_Z%_- of 100 pounds; th:r-efore 72 8.25X72 pounds will cost 0- of $8.25; and -1do- of $8.25-=- 100-. 238 APPLICATIONS OF [SECT. XI. Operation. We multiply the price of 100 lbs. ($8.25) $8.25 by 72, the given number of pounds, and the 72 product $594.00, is the cost of 72 lbs. at 1650 $8.25 per pound. But the price is $8.25 per 5775 hundredl; consequently the product 8594.00 $5.9400 Ans. is 100 times too large, and must therefore be divided by 100, to give the true answer. But te divide by 100, we simply remove the decimal point two places towards the left. (Art. 331.) 16. What cost 367 bricks, at $4.45 per 1000? Oeration. Reasoning as before, 367 bricks will cost 4.45 -to~ of $4.45. We multiply the price of 1000 3 67 bricks by the given number of bricks, and di$1.63315 Ans. vide the product by 1000. (Art. 331.) Hence, 383. To find the cost of articles sold by the 100, or 1000. Multiply the given price by the given nunmber of articles; then if the price is for 100, divide the product by 100; but if the price is for 1000, divide it by 1000. (Art. 331.) 17. A farmer sold 563 lbs. of hay, at $1.12~ per hundred: how much did it come to? 18. What cost 1640 lbs. of beef, at $6.37~ per hundred? 19. What cost 2719 lbs. of fish, at $4.20 per hundred? 20. What is the freight on 3568 lbs. from New York to Buffalo, at $1.67 per hundred.? 21. What cost 6521 lbs. of cheese, at 7: cts. per hundred? 22. What cost 15214 lbs. of butter, at 12~ cts. per hundred? 23. At $6.25 per 1000, what cost 865 feet of spruce boards? 24. At $19.45 per 1000, what cost 2680 feet of pine boards? 25. At $67.33 per 1000, what cost 6500 feet of mahogany? 26. When ginger is $16.53 per cwt., what is it per pound? Analysis. —Since 100 lbs. cost $16.53, 1 lb. will cost _x_ of *16.53. But to divide by 100, we remove the decimal point two places to the left. (Art. 331.) Ans. $Q 1653. QuEsT.-383. How do you find the cost of articles sold, by the 100, or 1000 ARTS. 383-385.1 FEDERAL nMONEY 239 27. When pine boards are $21.63 per 1000, what are they per foot? Sol'ution.-Reasoning as before, 1 foot will cost 1 of $21.63. Now to divide by 1000, we remove the decimal point three places to the left. (Art. 331.) Ans. $.02163. Hence, 38 4. When the cost of 100, or 1000 articles, pounds, &c., -s given, the price of one is found by simply removing the decimal y.point in the given cost or dividend, as many places to the left as there are ciphers in the divisor. (Art. 331.) 28. Bought 1000 bricks for $7.20: what is that apiece? 29. If 1000 feet of hemlock boards cost $6.40, what will one foot cost? 30. Bought 42 cwt. of tobacco for $565.82: what is that per cwt.; and what per pound? 31. Bought 75 cwt. of butter for $966.38: what is that per cwt.; and what per pound? BILLS, ACCOUNTS, &C. 385. A Bill, in mercantile operations, is a paper containing a written statement of the items, and the price or amount of goods sold. 32. What is the cost of the several articles, and what the amount, of the following bill? NEW YORIK, IMay 21st, 1847. G. B. Grannis, Esq., Bought of M'arck H. -Newman & Co., 75 Thomson's Mental Arithmetic, at $.12~ - 50 s" Practical Arithmetic,".314 -- 36 Porter's Rhetorical Reader,'.62- -, 25 Willson's School History, ".46 - 30 M'Elligott's Young Analyzer, ".31 75 Thomson's Day's Algebra, ".50 50 " Legendre's Geometry,.47. - - Received Payment, Mark H. Newman & Co. 240 BILLS, ACCOUNTS, ETC. [SECT. Xl. (33.) PHILADELPIHIA, June 10th, 1847. [Ion. Horate Binney, Bought of Leverette & Griggs, 163 lbs. Butter, at $.14~ - - - 235 lbs. Coffee, ".08 - - - 86 lbs. Chocolate, ".11 - - - 685 lbs. Sugar, ".10 - - - 21 doz. Eggs, ".13 - - - 860 ]bs. Lard,.09 - - - What was the cost of the several articles, and what the amount. of his bill? (p4.) AlBA N Y, July 1st, 1847. _Messrs. Collins &c Brothers, To G. TV. Bunker, D)r. For 320 yds. Silk, at $1.12~ -' 256 " Broadcloth, " 3.621 - - - 1" 75 pair Cotton Hose, " 0.12 - - - it 100 " Silk " " 0.87 - - - " 15 doz. Gloves, " 0.62 - - - " 120 Straw Hats, " 1.87 - - - What was the cost of the several articles, and what the amount of his bill? (35.) ST. Louis, Aug. 25th, 1847. James Henry, Esq.,:To J. L. Iotffman & Co., Dr. For 15260 lbs. Pork, at $0.051 - - - " 72605 lbs. Cheese, " 0.08 - - - " 11.521 bu. Corn, " 0.50 - - " 1560 bbls. Flour, " 6.12~ - - - CREDIT. Bj 1150 lbs. Cotton, at $0.06~ - - " 8256 lbs. Sugar, " o.07 - " 6450 gals. Molasses, " 0.37~ - " Cash to balance account, - - What is the amount of cash requisite to balance the account? ARTs. 386, 387.4 PERCENTAGE. 241 SECTION X I I. PERCENTAGE. ART, 386. The terms Percen tage and Per Cent. signify a certain allowance on a U7ndred; that is, a certain 2art of a hundred, or simply hundrecdthts. Thus, the expression 6 per cent. signifies 6 hundredths, ( —,) 7 per cent., 7 hundredths, (yT,) &c., of the number, or sum of money under consideration. Nolc. —The terms Prcen'agcce and Per. Cent. are derived from the Latin per andt cetuZn, signifying by tie hu2zdr'ed.'387. We have seen that lzntdredthis are decimzal expressions, occupying the first two places of figures on the right of the decimal point. (Arts. 311, 314.) Now, since Percentaye and 2wer cent. signify lundredtls, it is manifest that they can bec expressed by decimals, as in the following PERCENTIAGE TABLE. 1. per cent. is written thus:.01 2 per cent... " ".02 3 per cent.., " " ".03 6 per cent....I" ".06 7 per cent... " " ".07 10 per cent.. " " ".10 12 per cent. " " "4.12 50 per cent. it" " "6.50 100 per cent.... " " " 1.00 103 per cent.... " " " 1.03 125 per cent., &c.. " " i " 1.25 J per cent., that is, i of I per cent. " " ".005 ~ per cent., that is, { of 1 per cent. (" " " 0025 4 per cent., that is, 1 of 1 per cent. t " ".0075 131 per cent... " ".13125 25- per cent.. t.' c.25375 OBS. 1. It will be seen from the preceding Table, that when the given per cent. is less than 10, a cipher must be prefixed to the figure expressing it, in the same manner as when the number of cents is less than 10. (Art. 369. Obs. 1.) QUEST. —86,. What do the terms percentage atnd per cent. signify? What is meant JY 8 per cent., 7 tper cent., &c., of any nunber, or suml I 242 PERCENTAGE. [SECT. XII, When the given per cent. is more than 100, it bnust plainly require a mixed number to express it. (Art. 315. Obs. 2.) 2. Parts of 1 per cent. may be expressed either by a commons fraction, or by decimals. Thus, the expression 17A per cent., is equivalent to.17625 per cent. 3. The first two decimal figures properly denote the pecr ccnt., for they are hulndredths; the other decimals being parts of hzandredt/;s, express parts of 1 per cent. EXAMPLES. 1 Write 1 per cent., 2 per cent., 4 per cent., 6 per cent., 7 per cent., 8 per cent.. in decimals. o. Write 11 per cent.; 12; 14; 15; 16; 23; 65; 93. 3. W rite Lper ct.; l; 1; 5; 5; A4; a; 1; A; A;'; -; 1;.2 4 5 5, 5 4 6, 2 8 3 1,6 4. Write 4~ per ct.; 6~; 7;' 9*; 12t; 16; 115; 400~. 5. An agent collected $700 for a merchant, and received.5 per cent. for his services: how much did he receive? Analysis.-Since 5 per cent. is the same as -{-O, the agent must have received -_-_ of $700. Now -amo of $700 is B —, which is equal to $7; and 5 hundredths is 5 times $7, or $35. Oyeraction. Since -Ta=.05, we multiply the given nunm$ 7 00 ber of dollars by.05, and it gives the answer in.05 cents, which we reduce to dollars by pointing $35.00 Ans. off 2 decimals. (Art. 372.) Hence, 388. To calculate percentage on any number, or sum of money..Aiultiply the given num76er or sum by the given per cent. exp2ressed decimally; and point off the product as in multiplication of decimal fractions. (Art. 324.) Oss. 1. It is important for the learner to observe, that the amount of money collected, is made the basis upon which the percentage is computed. That is, the agent is entitled to 3 dollars, as often as he collects 100 dollars, and'vot as often as he pays over 100 dollars, as is frequently supposed. For in the latter case he would receive only -Tv-g, instead of -T3-6- of the sum in question. This distinction is important, especially in calculating percentage on large sums. QeEST. —387. How nmay per centage or per cent. be expressed? Obs. When the given per cent. is less than t1, how is it written? When more than 100, how? 388. 1lcw is percentage calculated? Ohs. In collecting money, upon what basis is the per cent. calculated? If the per cent. contains a common fraction which cannot be expressed decs-. mally, how proceed? ART. 388 ] PERCENTAGE. 243 2. Hence, if the per cent. contains a common fraction which cannot be extpressed decimally, first multiply by the decimal, then by the common fraction of the given per cent., and point off the sum of their products as above. 6. What is 4- per cent. of $300? Solution.-Expressed decimally, 4* per cent.=.042; (Art. 387.'Obs. 2;) and 6300X.042=$12.60. Ans. 7. What is 3 per cent. of $256.25? 8. What is 2 per cent. of $437.63? 9. What is 21 per cent. of $138.432? 10. What is 6 per cent. of $I145.13? 11. What is 7 per cent. of $1630.10? 12. A man borrowed $150, and paid 7 per cent. for the use of it: how much did he pay? 13. A merchant bought goods amounting to $1825, and sold them so that he gained 12 per cent.: how much did he gain? 14. A constable collected $862.56, and charged 5 per cent. for hlis services: how much did he receive; and how much did he pay over? 15. What is 10 per cent. of $4020.50? 16. What is 8 per cent. of $1675? 17. What is 4* per cent. of $725? 18. What is 53 per cent. of $648.30? 109. What is 6* per cent. of $1000? 20. What is 7- per cent. of 82000? 21. What is 8- per cent. of $100.25? 22. A farmer having 1500 sheep, lost 25 per cent. of them: how many did he lose? 23. A merchant having $1960 on deposit, drew out 20 per cent. o'f it: how much had he left in the bank? 24. A merchant imported 1500 boxes of oranges, and 12* per clent. of them decayed: how many boxes did he lose; and how mrniny had he left? 25. What is * per cent of $1625? 26. What is 1 per cent. of:2526.40? 27. What is I per cent. of $42260.08? 28. What is 1 per cent. of $75000? 29. What is -3 per cent. of $100000? 11* 241 PERCENTAGE. [SECT. XII. 30. What is 1 per cent. of $45241.20? 31. What is j per cent. of $675264? 32. A merchant bought a stock of goods amounting to $4565:, and paid 31 per cent. for freight: what was the whole cost of his goods? 33. A man's salary is $2000 a year, and he lays up 371 pei cent. of it: how much does he spend? 34. A youth who inherited $20000, spent 40 per cent. of it ir, dissipation: how much had he left? 35. Two merchants embarked in business with $18250 capital apiece; one gained 20 per cent. and the other lost 20 per cent. the first year: what was then the amount of each man's property? 36. Two men invested $10000 apiece in stocks; one lost 8 per cent., the other 6 per cent.: what was the difference of their loss 2 37. What is the difference between 6 per cent. of $1040, and 7 per cent. of 8905? APPLICATIONS OF PERCENTAGE. 3S89. PERCE NW-AGE, or the qmethod of r'ck'oT1nfi by h'undredtlhs,. is applied to variovus calculations in the practical concerns of life. Among the most important of these are Commission, Brokerage, the Rise and Fall of Stocks, Interest, Discount, Insurance, Profi't and Loss, Duties, and Taxes. Its principles, therefore, should be thoroughly understood by every scholar. COMMISSION, BROKERAGE, AND STOCKS. 390. Commission is the per cezt. or sumn charged by agents for their services in buying and selling goods, or transacting other business. Ors. An Agent who buys and sells goods for another, is called a Cotmnis$. sion fMerchant, a Factor, or Correspondelt. 391. Brokerage is the ~er' cent. or sum charled by money dealiers, called 1Brokers, for negotiating Bills of Exchanye, and other monetary operations, and is of the same nature as Commission. QvUEST.-390. What is commission? Ohs. What is an agent who buys and sells goatas fox another usually called? AnTS. 389-395.] COMMISSION. 245 392. By the term Stocks, is meant the (Cquital of moneyed institutions, as incorporated Banks, AManufactories, Railroad and:insurance Companies; also, Government and State Bonds, &c. Ors. 1. Stocks are usually divided into portions Of $100 each, called shares; and the owners of these shares are called Stocl/holdcrs. 2. The association or company thus formed, is called a corporation; the in-.strument specifying the powers, rights, and privileges invested in the corporation, is called a clar'ter. 393. The original cost or valuation of a share is called its;-nominal, or par value; the sum, for which it can be sold, is its w"eal value. OBs. 1. The rtse or fall of Stocks is reckoned at a certain per cent. of its -par value. The term par is a Latin word, which signifies cqual, or a state oj qwuality. 2. When Stocks sell for their original cost or valuation, they are said to be act par; when they sell for more than cost, they are said to be above par, at a pr'cmiutm, or an advance; when they do not sell at cost, they are said to be b'elow par, or at a discount. 3. Persons who deal in Stocks are usually called Stlock Brokers, or Stock.Jobbers. 394. The comm7zission or ailowance made to factors and brokers, t.Also the rise and fall of stocks, are usually reckoned at a certain, 9y)ercentayz on the amounzt of money employed in the transaction, (or on the par value of the given shares. Hence, 395. To compute commiission, brokeragre, and the premium or isCount on stoclks..iftlui'jly thice iet'n_ Sltn by tIke gieon per centt. expr'essed in dIecienuals,, acd point o!2 the p2rod'uct uts in, Pc'cenct, qe. (Art;. 388.) Ocs. The.contlmission for the collection of bills, taxes, &c., also for the sale cr purchase of goods, varies firom;2- to 12 or 15 per cent., and should always e. reckloned on the oavolsl l,3 of mioncy c,//lc;((, o,.paid out, or employed in the. tri'ansaction. The brokerage for the sale or purchase of stocks, varies from 8 ts 4I per tent., reckoned on the par value of tle stoelk. ]UcEsT.-:'jl. Xh hatt is broker:ae? 392. W''.It is leant.', the terin stocks? Oba How are stocks usally lividedl' 393. Whst i. the par cvalu.e of stmicks? Wlhat the real value? Ohs. Whllat is the Ietrng of the terll par. Wxven are stocks tt par? When abovo par W~'hein below? 395. I!o do you coepute cominission, brokerage, &c,. 246 COMMISSION. [SECT. XII, EXAMPLES. 1. An auctioneer sold goods amounting to $463, at 3 per cent. commission: how much did he receive? ins. 813.89. 2. An agent bought goods amounting to $625.375: what is, his commission, at 2 per cent.? 3. What is the commission on $1682o25, at 31 per cent.? 4. What is the commission on $1463.18, at 5 per cent.? 5. What is the commission on $2560.07, at 4~ per cent.? 6. What is the commission on $10250, at 6 per cent.? 7. What is the commission on $8340.G60, at 7 per cent.? 8. What is the commission on $900.625, at 51 per cent.? 9. A commission merchant sold goods to the amount of $62 3 5, at 2~ per cent.: what was his commission? 10. An attorney collected a debt of $8265.17, and charged 7~ per cent. for his services: how much did he receive? 11. Bought $1108 worth of books, at 4 per cent. commission: what was the amount of commission? 12. A tax-gatherer collected $12250, for which he was entitled to 5~ per cent. commission: how much did lie receive? 13. Sold goods amounting to $1432.26: how much was the commission, at 4 per cent.? 14. A commission merchant sold a quantity of hardware amounting to $9240.71: how much would lihe receive, allowing 2L per cent. for selling, and 2 per cent. more for guaranteeing the payment? 15. An auctioneer sold carpeting amounting to $2136.63, and charged 21 per cent. for selling, and 23 per cent. for guaranteeing the payment: how much did the auctioneer receive; and how much did he remit the owner? 396. Commission merchants, agents, &c., generally keep an account with their employers, and as they make in;vestments or sales of goods, charge their commission on the amount invested, or the sum employed in the transaction. Sometimes, however, a specific amount is sent to an agent or broker, requesting him, after deducting his commfision, to lay out the balance in a certain manner. A rns. 396, 397.] IROICERfAGE. 24'7 10. A gentleman sent his agent $1500 to purchase a library: how much had he to lay out after deducting his commission at 5 per cent.; and what was his commission? Not~e.-The money actually laid out by the agent in books, is manifestly the proper basis on which to calculate his commission; for it would be unjust to charge commission on the sum he retains. (Art. 395. Obs.) Anclysis.-The money laid out is I of itself, and the commission is'-ri of this sum; consequently the money laid out added to the commission, must be -H-_ the whole amount. The question therefore resolves itself into this: $1500 is +-Q- of what sum? If $1500 is HE, _tro- must be 1500+ 105_1J,5L,', and~ Q I —Oio s X100=-1428.57, the sum laid out,. Now $1500-$1428.57=$71.43, the commission. PROOF. -$ 1428.57X.05=$71.43; and $1428.57+$71.43= $-1500, the amount sent. Hence, 397. To compute commission when it is to be deducted in advance from a given amount, and the balance is to be invested..Divide the given amount by 81 increased by the per cent. commission, and the quotient will be the iacrt to be invested. Subtract the part invested friom the given amount, and the remainder wuill be the commission. Ons. The commission may also be found by multiplying the sum invested by the given per cent. according to the preceding rule. (Art. 395.) 17. An agent received $21500 to lay out in provisions, after deducting 2 per cent. commission: what sum did he lay out? 18. A country merchant sent $3560 to his agent in the city, to purchase goods: after taking out his commission, at 9- per cent., how much remained to lay out? 19. Baring, Brothers & Co. sent their agents $800000 to buy flour: after deducting 5 per cent. commission, how much would be left to invest? 20. A broker negotiated a bill of exchange of $882531, at 5 per cent.: how much did lhe receive for his services? 21. What is the brokerage on $~94265, at 1- per cent.? 22. What is the brokerage on $6200, at 3- per cent.? 248 STOCKS. [SECT. XII 23. What is the brokerage on $8845.50, at { per cent.? 24. What is the brokerage on $2500, at a per cent.? 25. A broker made an investment of $21265, and chlarged 1-~ per cent.: what was the amount of his brokerage? 26. If you buy 20 shares of Western Railroad stock, at 7 per cent. advance, how much will your stock cost you? lzns. $2140. Nole.-The stock evidently cost its par value, which is $2000 ancl 7 per cent. besides. Now $2000X.07=$140.00; and $2000 +$140=$2140. 27. What cost 20 shares of bank stock, at 7 per cent. discount? Ans. $2000-$-140 =1860. 28. What cost 35 shares of New York and Erie Railroad stock, at 5~ per cent. premium? 29. A merchant bought 45 shares of Commercial Bank stock, at par, and afterwards sold them, at 50 per cent. discount: how much did he lose? 30. A man invested $8460 in the New England Manufacturing Co., and afterwards sold out at 4l per cent. advance: how mnuch dlid he sell his stock for? 31. Sold 64 shares of Hudson River Railroad stock, at 10~ per cent. premium: how much did they come to? 32. A man bought 35 shares of Utica and Syracuse Railroad stock, at par, and afterwards sold them at 1~ per cent. advance: how much did he get for them? 33. A man bought 15 shares of Albany and Schenectady Railroad stock, at 2 per cent. advance, and sold them at 10 per cent. disc.: how much did he sell them for; and how much did he lose? 334. Bought 71 shares in the Albany Gas Co. at 5~ per cent. premium: how much did they amount to? 35. A broker bought 48 shares of Michigan Railroad stock, at 14 per cent. discount, and sold them at 6 per cent. advance: how much did hle make by the operation? 36. If I employ a broker to buy mne 55 shares of R-Pailroad stock, which is 20 per cent. below par, and pay himlnL per cent. brokerage, how much will my stock cost me? 37. If my agent buys 78 shares of New York and Philadelphia Railroad stock, at 15 per cent. advance, and chargles nme - rer cent. brokerage, how much will my stock cost? ARTS. 398-401].] INTEREST. 249 INTEREST. 39S. INTEREST is the sum paid for the use of money bly the borrower to the lender. It is reckoned at a given per cent. per annurn; that is, so many dollars are paid for the use of $100 for one year; so many cents for 100 cents; so many pounds for ~100; &c. OBs. The student should be careful to notice the distinction between Conwissiorb and Interest. The Jbormer is reckoned at a certain per cent. without regard to time; (Art. 395;) the latter is reckoned at a certain per cent. for one year; consequently, for longer or shLorter periods than one year, lide proporstions of the percentage for one year are taken. The term per annczm, signifies for a year. 399. The money lent, or that for which interest is paid, is called the principyal. The per cent. paid per annum, is called the rate. The sum, of the principal and interest, is called the amount. Thus, if I borrow $100 for 1 year, and agree to pay 5 per cent. for the use of it, at the end of the year I must pay the lender $100, the sum which I borrowed, and $5 interest, making $105. The principal in this case, is $100; the interest $5; the rate 5 per cent.; and the amount $105. Oss. The term per asenlscz, is seldom expressed in connection with the rate per cent., but it is always understood; for the rate is the per cent. paid per alnnelc. (Art. 399.) 400. The rate of interest is usually established by law. It varies in different countries and in different parts of our own country. Oss. When no rate is mentioned, the rate established by the laws of the State in which the transaction takes place, is always understood to be the one intended by the parties. 40 1 Any rate of interest higher than the legal rate, is called,esury, and the person exacting it is liable to a heavy penalty. Any rate less than the legal rate may be taken, if th.. parties concerned so agree. CQUEST.-398. What is Interest? JIow is it reckoned? Obs. What is the dilterence between Commission and Interest? What is meant by the term per annum? 399. What is meant by the principal? The rate? The amount? 400. How is the rate istually determnined? Is it the same everywhere? Obs. When no rate is mentioned, what rate is understood? 401. What is any rate higher than the legal rate called? 250 INTEREST'. [SECT. XII. 402. The legal rates of interest, and the penalty for usury in the several States of the UTnion, are as follows: States. Legoal rates. Penalty for Usury. Maine, 6 per cent. Forfeit of the whole debt. N. Hampshire, 6 per cent. Forfeit of three times the usury. Velmont, 6 per cent. Recovery in action with costs. Massachusetts, 6 per cent. Forfeit of three times the usury. Rhode Island, 6 per cent. Forfeit of the usury and int. on the debt, Connecticu, 6 per cent. Forfeit of the whole debt. New York, 7 per cent. Forfeit of the whole debt. New Jersey, 6 per cent. Forfeit of the whole debt. Pennsylvania, 6 per cent. Forfeit of the whole debt. Delaware, 6 per cent. Forfeit of the whole debt. Maryland, 6 per cent. a Usurious contracts void. Virginia, 6 per cent. Forfeit of double the usury. N. Carolina, 6 per cent. Forfeit of double the usury. S. Carolina, 7 per cent. Forfeit of interest and usury with costs. Georgia, 8 per cent. Forfeit of three times the usury. Alabama, 8 per cent. Forfeit of interest and usury. Mississippi, 8 per cent. b Forfeit of usury and costs. Louisiana, 5 per cent. c Usurious contracts void. Tennessee, 6 per cent. Usurious contracts void. Kentucky, 6 per cent. Usury may be recovered with costs. Ohio, 6 per cent. Usurious contracts void. Indiana, 6 per cent. Forfeit of double the excess. Illinois, 6 per cent. d Forfeit of three times the usury, and int. due. Missouri, 6 per cent. e Forfeit of the usury, and the interest due. Michigan, 7 per cent. Forfeit of the usury, and one fourth the debt. Arkansas, 6 per cent. f Forfeit of usury. Florida, 8 per cent. Forfeit of interest and usury. Wisconsin, 7 per cent. g Forfeit of three times the usury. Iowa, 7 per cent. / Forfeit of three times the usury. Texas, 10 per cent. Usurious contracts void. Dist. Columbia, 6 per cent. Usurious contracts void. OBs. 1. On debts and judgments in favor of the United States, interest is computed at 6 per cent. 2. In Canada and Nova Scotia, the legal rate of interest is 6 per cent. in England and Fi'cance it is 5 per cent.; in Ireland 6 per cent. In Italy, about the commencement of the 13th century, it varied from 20 to 30 per cent. a On tobacco contracts 8 per cent. b By contract as high as 10 per cent. c Bank interest 6 per cent.; conventional as high as 10 per cent. d By agreement as high as 12 per cent. e By agreement as high as 10 per cent. f By agreement, any rate not exceeding 10 per cent. g By contract as high as 12 per cent. /h By agreement as high as 12 per cent AR 1 s. 402, 403] INTEREST. 251 403. Ex. 1. What is the interest of $30 for 1 year, at 6 per cent.? Acnalysis.-We have seen that 6 per cent. is -—; that is, $6 for $100, 6 cents for 100 cents, &c. (Art. 386.) Since therefore the interest of $1 (100 cents) Ifor year is 6 cents, the interest Af $30 for the same time must be 30 times as much; and $30X.06 =z$1.80. Ans. 01peration. We first multiply the principal by the $30'Prin. given rate per cent. expressed in decimals,.06 Rate. as in percentage, and point off as many de$1.80 Int. 1 yr. cimals in the product as there are decimal places in both factors. Ex. 2. What is the interest of $140.25 for 1 year, 1 month, and tO days, at 7 per cent.? What is the amount? Operation. $140.25 Prin. 1 month is -9~ of a year; there-.07 Rate. fore the interest for 1 month. is -a' 12)$9.8175 Int. 1 yr. of 1 year's interest. 10 days are - 3) 8181 ( 1 mo. of I month, consequently the interest 2727 6 10 d. for 10 days, is ~ of 1 month's inter$10.9083 Interest. est. The amount is found by add$140'25 Prin. added. ing the principal and interest to8151.1583 Amount. gether. Note. —1. In adding the principal and interest, care must be taken to add?dollars to dollars, cents to cents, &c. (Art. 374.) 2. When the rate per cent. is less than 10, a cipher must alwaysbe prefixed;to the figure denoting it. (Art. 387. Ohs. 1.) It is highly important that the.principal and the rate should both be writte~n correctly, in order to prevent mix-,takes in pointing off the product. Ex. 3. What is the interest of $250.80 for 4 years, at 5 per gent..? What is the amount? S+Zution.-$250.80X.05=$12.54, the interest for 1 year. Now $12.54X 4=$50.16, " " 4 years. And $250.80+$50.16=$300.96, the amount required. 252 INTEREST. [SECT. XII. 4 1.:o From the foregoing illustrations and principles we deduce the followyilng general RULE FOR COMPUTING INTEREST. 1. FOR ONE YEAR. iiultiply the princil2al by the given rate, and from the product point oq as many figures for decinmals, as therc are decial places in both factors. (Art. 324.) II. For TwO or MOR:E YEARS. Multiply the interest of 1 yeq by the given number of years. III. Fo(r MONTrIIS. Tcake such. a fractional pa2rt of 1 year's irterest, as is denoted by the given number of mzoths. IV. Foa DAYS. Tcake such a fractional part of one month's interest, as is denoted by the given numnber of days. The amouz-nt is found by addingL the principal and interest together. Ocs. 1. The rcason, of this rule is evident froml the consideration that the: given rate per CCent. Pe?nt p nerbarbn denotes ht'ndredt/ts. (Arts. 386, 398.) Now when the rate is 6 per cent. we multiply by.06, when 7 per cent. by.07, &c., and point off two figures in the product; consequently the result will be the same as to multiply by 7 —-%, &-_-_, &c. 2. In calculating interest, a month. whether it contains 30 or 31 days, or even but 28 or 29, as in the case of February, is assumed to be on e twelft/, of a year. Therefore, for 1 month we take l- of 1 year's interest; for 2 months, L; for 3 months, 4-; for 4 months, -; for 6 months, -; for 8 months, &, &c. Again, 30 days are commonly considered a month; consequently the interest for 1 day, or any number of days under 30, is so many tirt'ietihs of a month's interest. (Art. 303. Obs. 2.) Therefore, for I day we take -%J- of 1 month's interest; for 2 days, -I-; for 3 days, -2-L; for 5 days, 1; for 10 days, -, &c. This practice seems to have been originally adopted on account of its convenience. Though not strictly accurate, it is sanctioned by general usage. 3. Allowing 30 days to a month, and 12 months to a year, a year would contain only 360 days, which in point of fact is' —- or -13 less than an ordinary year. Fience, To find the interest for any number of (lays with entlire acc~,uracy, we must take so many 3065ths of 1 year's interest, as is deploted by the given number, of da.ys; or, find the interest for the days as above fromthis subtract 7-1 of QUEs'r.-4404. Ifow is interest codlputed for a year? lv w for any number of years * Hlow fIr months? E-Iow for days? How find the amount. Obs. In1 reckoning interesth what part of a year is a month consideredl? How many (lays are comlnonly considered a. wonth? Is this practice accurate? Aitr. 404.] INTEREST. 253 itself, and rthe remainder will be the exact interest. The laws of New York, and several other states, require this deduction to be made. In business, when the mills in the result are 5, or over, it is customary to add I to the cents; if under 5, to disregard them. EXAMPLES. 1. What is the interest of $423 for 1 yr., at 7 per cent.? 2. What is the interest of $240.31 for 3 yrs., at 6 per cent,? 3. What is the interest of $403.67 for 2 yrs., at 5 per cent.? 4. What is the interest of $640 for 1 yr., at 8 per cent.? 5. What is the interest of $430.45 for 2 yrs., at 7 per cent.? 6. What is the interest of $185.06 for 4 yrs., at 6 per cent.? 7. What is the interest of $864.80 for 5 yrs., at 4~ per cent.? 8. What is the interest of $763 for 4 months, at 7 per cent.? 9. What is the interest of $940.20 for 6 mo., at 6 per cent.? 10. What is the interest of $243.10 for 5 mo., at 8 per cent.? 11. What is the interest of $195.82 for 7 mo., at 6 per cent.? 12. What is the interest of $425.35 for 9 mo., at 6 per cent.? 13. At 7 per cent., what is the int. of $738 for 1 yr. and 2 mo.? 14. At 6 per cent., what is the int. of $894 for 1 yr. and 8 mo.? 15. At 7 per cent., what is the amount of $926 for 6 mo.? 16. At 7 per cent., what is the amt. of $648 for 2 mo. 15 d.? 17. At 6 per cent., what is the amt. of $1000 for 1 mo. 11 d.? 18. At 5 per cent., what is the amt. of $1565.45 for 3 mo.? 19. At 6 per cent., what is the amt. of $872 for 4 mo.? 20. What is the int. of $681 for 10 days, at 6 per cent.? 21. What is the int. of 8483.26 for 15 d., at 7 per cent.? 22. What is the int. of $569.40 for 20 d., at 6 per cent.? 23. What is the amt. of $95 for 1 yr. and 6 mo., at 5 per cent.? 24. What is the amt. of $148 for. 8 mo. 12 d., at 6 per cent,? 25. What is the amt. of $700 for 30 d., at 7 per cent.? 26. What is the int. of $340 for 60 d., at 51- per cent.? 27. What is the int. of 84685 for 90 d., at 6- per cent.? 28. What is the amt. of $3293 for 30 d., at 7 per cent.? 29. What is'the amt. of $5265 for 15 d., at 6 per cent.? 30. What is the int. of $8310 for 10 d., at 7 per cent.? 31. What is the int. of $50625 for 21 d., at 7 per cent.? 82. What is the amt. of $65256 for 4 mo., at 7 per cent.? 254 INTEREST. [SECT. XI1 SECOND METHOD OF COMlPUTING INTEREST. 40'5. There is another method of computing interest, which is very simple and convenient in its application, particularly wheni the interest is required for mzonths and days, at 6 per cent. 406, We have seen that for 1 year, the interest of A1 at C per cent. is 6 cents., or $.06; (Art. 404;) therefore, For I month, the interest of $1 is -92- of 6 cents, which is $.005; "2 months, " "is -2-, or - of 6 cents, " " 001;' 3 months, " is -3f t oxr of 6 cents, "'.015; 12 4 " 4 months, " "is 4-, or l of 6 cents, " ".025 months, " "is -sa, of 6 cents, " ".025; "6 months, is,6, or 1 of 6 cents, " ".03; Hence, The interest of $1 for 1 month, at 6 per cent., is 5 mnills; for every 2 months, it is 1 cent; and for any numzber of months, it is as Bmany cents, or hundredths of a dollar, as 2 is containeczc times in the given number of months. 40:7. Since the interest of $1 for 1 month (30 days) is 5 mills, or $.005, (Art. 406,) For 6 days ( of 30 das) the interest of 30 days) the interest of $1 is of 5 mills, or $.001; ~L~ VI V UUYU \5 V+ VV UIyCJUJ UIIV IVVIVUV V 5 "18 (lays (Ia of 30 dlays) " " is - of 5 mills, or.003; "3 days (- - of 30 days) " is O of 5 mills, or.0005; That is, the interest of 81 for every 6 days, is 1 mnill, or $.001; and for any number of days, it is as many mnills, or thousandths of a dollar, as 6 is contained times in the given number of days. 408. Hence, to find the interest of $1 for any number of days, at 6 per cent. )Divide the given number of days by 6, and set, the first quotient figyure in thousandths' place, when the days are 6, or mzore than 6, but in tent thousandths' place, when they are less than 6. OBs. For 60 days (2 mo.) the interest of $1 is 1 cent; (Art. 406,)j M nen, therefore, the number of days is 60 or over, the first quotient figure must occupy h&undredths' place. QuErsT. —408. I-low find the interest of $1 for any number of days, at 6 per cent. t kRTS. 405-409.] INTEREST. 255 Ex. 1. What is the interest of 8185 for 1 year, 6 months and 18 days, at 6 per cent.? Analysis.-The interest of $1 for 1 year is Operation. 3 cents; for 6 months it is 3 cents; and for $185 Prin. 1.8 days it is 3 mills. (Arts. 406, 407.) Now.093 Int. $1. 096+.03+.003=$.093. Since therefore the 555,interest of $1 for the given time is $.093, the 1665 wterest of $185 must be 185 times as much. $17.205 Ans. 409. Frmra these principles we may derive a SECOND RULE FOR COMPUTING INTEREST. I. To compute the interest on any sum, at 6 per cent. mlultiply the princoipal by the interest of $1 for the given time, aot 6 per! cent., cancl 2oint qof' the product as in multzillication of Cdecimals. (Art. 324.) II. rTo compute int. at any rate, greater or less than 6 per cent. First find the interest on the given sum at 6 per cent.; then acdd to this interest, or subtract from it, such a fractional plart of,itselJ; as the required rate exceeds or falls short of 6 per cent. bThe amoutnt is found by adding the princip)al cazd interest together as in the former method. (Art. 404.) Oss. 1. The amogunt may also be found by multiplying the given principal by the aeolunt of one dollar for the time.'. The r eason, of the first part of this rule, is manifest from the principle that;}le interest of 2 dollars for any given time and rate, mnust be twice as much as ithe interest of 1 dollar for the same time and rate; the interest of 50 dollars, ~O0 timnes as much as that of 1 dollar, &c. 3. When the required rate is 7 per cent., we first find the interest at 6 per cent., flthen add - of it to itself; if 5 per cent., subtract ~ of it from itself, &c., for the obvious reason, that 7 per cent. is once and 1 sixth, or -7 of 6 per cent.; 5 ier cent. is only A of 6 per cent., &c. 4. When the decimal denoting the int. of $1 for the days, is lonT, or is a r'epetild, it is more accurate to retain the common friaction. (Art. 387. Obs. 2.) 2. What is the interest of $746 for 4 months and 18 days, at 6 per cent.? Ans. 217.158. QuEssl.-409. What is the second mrethol of competing inlterest, at 6 per cent.? Whe oile rate per ceat. is greater or less thanl 6 per cent., how proceedl' 256 INTEREST. [SECT. XII. 3. What is the interest of $240 for 6 months and 12 days, at 7 pei cent.? Opere ion. $240 Prin. The interest of $1 for 6 mo. at 6.032 Int. of $1. per ct., is.03; for 12 d. it is.002; 480 and.03 -+.002=$.032. 720 The required rate is 1 per cent:. 6)$7.680=Int. at 6 per ct. more than 6 per cent.; we there-. 1.280=- of 6 per ct. fore find the interest at 6 per cent.,s Ans. $8.00 Int. at 7 per ct. and add } of it to itself. 4. What is the interest of $680 for 3 mo., at 5 per cent.? 5. What is the interest of $213.08 for 1 mo., at 6 per cent.'? 6. What is the interest of $859 for 1 yr. 2 mo., at 7 per cent.? 7. What is the interest of $768 for 1 yr. 7 mo., at 8 per cent.-? 8. What is the interest of $684 for 9 mo., at 6 per cent.? 9. At 7 per cent., what is the amount of $387 for 5 mo.? 10. At 4 per cent., what is the amt. of $1125 for 1 yr. 2 mo.? 9 11. At 6 per cent., what is the amt. of $1056 for 10 mo. 24 d.' 12. At 6 per cent., what is the int. of $1340 for 1 mo. 15 d.? 13. At 6 per cent., what is the int. of $815 for 2 mo. 21 d.? 14. At 8 per cent., what is the amt. of $961 for 4 mo. 10 d.? 15. What is the int. of $2345.10 for 6 mo., at' per cent.? 16. What is the int. of $1567.18 for 4 mo., at 7~ per cent.? 17. What is the int. of $3500 for 11 mo., at 10 per cent.? 18. What is the int. of $39.375 for 2 yrs., at 12~ per cent.? 19. What is the int. of $113.61 for 5 yrs., at 15 per cent.? 20. What is the int. of 81000 for 2 yrs., at 20 per cent.? 21. What is the int. of $1260.34 for 10 yrs., at 13 per cent.? 22. At 16 per cent., what is the int. of $15.0 for 6 years.? 23. At 30 per cent., what is the int. of $300 for 1 year.? 24. What is the amt. of $12645 for 10 d., at 6 per cent.? 25. What is the amt. of $16285 for 24 d., at 7 per cent.? 26. At 41 per cent., what is the int. of $10255 for 8 months 9? 27. At 51 per cent., what is the int. of $17371 for 3 months' 28. What is the amt. of $1 for 100 yrs., at 7 per cent.? 29. What is the amt. of 1 cent fol.100 yrs., at 6 per cent.? ARTS. 410, 411.] INTEREST. 257 410. Since the interest of $1 at 6 per cent. for 12 rmo. is 6 cents, (Ari. 406,) for 6 mo. it must be 3 cents; for 3 mo., 1L cents; for 2 mo., 1 cent; for 1 mo. or 30 d. J cent; for 15 d., -L cent; foT 20 d. 9 cent, &c. That is, the interest of $1 at 6 per cent. is as many cents as are equal to half the given number of months. * 1. H. lence, to compute interest at 6 per cent. by months. Mul6tip1y the princi2pal by half the number of months, and point O'g twvo 9more figures for clecimzals in the product than there are decimnal places in the mnulti2plicand. OBS. 1. When there are years and days, reduce the years to months, and the days to a common fraction of a month. Or, divide the dcys by 3, and annex the quotient to the months considered:~s h'undredths; half of the n2u2mber thus produced will be the decimal nultiplier 2. The latter method is the same as dividiing the days by 6, and setting the.ii st quotient figure in thoutsandth's place; for, we divide the days by 3 and 3,3 and 3X2=-6. (Arts. 407, 408.) 30. W ha/t is the int. of $460.384 for 8 mos. and 15 d., at 6 per ct.? Opemrction. $l460.3S84 We multiply by 4-, for, 8 monthls+15'4_ days=S8 months, and 8-1-2 =4. And 1841536 since there ale three decimals in the Imul115096 tiplicancld, we point off 5 in the product. $19.5G632 Ans. 31. What is thle interest of;$780 for 4 months, at 6 per cent.? 32. WIhat is the interest of $1406 for 3 nmo., at 6 per cent.? 33. Wlhat is the interest of 8109 for 2 mo., at 7 per cent.? 34. What is the interest of $119.45 for 8 mo., at 6 per cent.? 35, What is the interest of $618 for 1 yr. 3 mo., at 6 per cent.? 36. XWIhlat. is the interest of $861 for 2 yrs. 6 mo., at 6 per cent.? 37. What is the interest of $936.40 for 3 yrs., at 6 per cent.? 38. What is the interest of $4526 for 6 mo. 2 d., at O per cent.? 39, Wbhat is the interest of $88246 for 10 mo., at 7 per cent.? 40. TWh at is the interest of $31285 for 3 mo., at 5 per cent.? 41. What is the interest of 817500 for 1 yr. 3 mo., at 7 per ct.? 42. What is the amount of 8$286 for 8 mo. 15 d., at 6 per ct. 9 43. What is the amount of $15876 for 5 mo. 18 d., at 6 per ct.? 258 APPLICATIONS OF [SECT. XhI. 412. We have seen that the interest of $1 at 6 per cent. for any number of days is equal to as many mills, as 6 is contained, times in the given days. (Art. 407.) Hence, 413. To compute interest at 6 per cent. by d(ays..Multiply the princ'ipal by one sixth of the given number of clays, and point qf three more figures for decimals in the. product than there are decimal 2places in the principal. (Art. 411. Obs. 2.) Or, mutltiply the principal by the given number of days, dcivide the product by 6, and point oq the quotient as above. Oss. The product is in mills and parts of a mill. The object, therefore, of pointing off three more places for decimals in the product than there are deci; mnals in the principal, is to reduce it to dollars. (Art. 372.) 44. What is the interest of $976.22 for 33 days, at 6 per cent.?' Solution. — of 33 d. =5'; and,976.22X 5-=53d9.21 mills,? Pointing off 3 more decimals, we hlave $'.36921. Aens. 45. Whal is thle interest of 5 o30.30 for 24 days, at 6 per' cent.!, 46. What is the interest of $7085 for 63 d., a 6 per cent.? 47. What is the interest of 88126.21 for 8 d., at 6 per cent.? 48. What is the interest of $825681 for 93 d., at 6 per cent.? 49. What is the interest of $764.85 for 114 d., at 6 per cent.? APPLICATIONS OF INTEREST. 414. In the application of interest to business transactions, the following particulars deserve attention. 1. A promissory note is a writing which contains a promise of the payment of money or other property to another, at or before a time specified, in consideration of value received by the pfrolmiser or sealcse? of the note. Unless a note contains the words " value received," by some authorities it is deemed inuvatiscd; consequently these words should always be inserted. 2. The person who signs a note is called the makher, drawer, or give?- of the note. IThe person to whom a note is made payable, is called the pay/ee; the person who has the legal possession of a note, is called the holder of it. 3. A note which is made payable " to order," "or bearer," is said to be negos tiacbie; that is, the holder may sell or transfer it to whom he pleases, and it can, be collected by any one who has lawful possession of it. Notes without these words are not negotiable. (See Nos. 1, 2.) 4. If the holder of a negotiable note vhich is made payable to order wishes to sell or transfer it, the law requires him to endorse it, or write his name ron the back of it. The person to whom it is transferred, or the holder of it, is' ARTS. 412-415.] INTEREST. 259 then empowered to collect it of the drawer; if the drawer is qtnalbe, or refuses to pay it, then the endorser is responsible for its payment. (See No. 1.) 5. When a note is made payable to the beaser, the holder can sell or transfer it without endorsing it, or incurring the liability for its payment. Bank notes or bills are of this description. (See No. 2.) 6. When a note is made payable to any particular person without the words order or bearer it iH not negohtiable; for, it cannot be collected or sued except in the name of the person to whom it is made payable. (See No. 3.) 7. A note should always specify the time at which it is to be paid; but if no time is mentioned, the presumption is that it is intended to be paid on demand, and the giver must pay it when demanded. 8. According to custom and the statutes of most of the States, a note or draft is not presented for collection until thrce days after the time specified for its payment. These three days are called days of grace. Interest is therefore reckoned for:three days more than the time specified in the note. When the fiast day of grace comes on Sunday, or a national holiday, as the 4th of July,,&c.; it is customary to pay a note on the day previous. 9. If a note is not paid at maturitil or the time specified, it is necessary for fthe holder to notify the eendorser of the flact in a legal manner, as soon as cirirsumstances will admit; otherwise the responsibility of the endorser ceases. 10. Notes do not draw interest unless they contain the words " with interIest." But if a note is not paid when it becomes due, it then draws legal interest till paid, though no mlention is made of interest. (Art. 400. Obs.) 11. Notes which contain the words " qwit/, interest," though the rate is not mentioned, are entitled to the I/cgal rate established by the State in which the note is made. In writing notes therefore it is unnecessary to specify the rate, unless by agreement it is to be! less than the legal rate. 12. When a note is made pLiyable on a given day, and in a specified article of merchandlise, as grain, stock, &c., if the article specified is not tendered at the given time and place, thb holder can demand payment in money. Such notes, are iiot negotiable; nor is the drawer entitled to the days of grace. 13. When two or more persons jointly and severially give their note, it may be collected of either of them. (See No. 4.) 14. The sn1m7 for which a note is given, is called the pyrincipal, orface of te note; and should always be written out in words. 41 5. When it is required to compute the interest on a note, we must first find the time for which the note has been on interCest, by subtracting the carlier ftro h the later date; (Art. 303;) {then cast the interest on the faec,: of the note for the time, by either of the preceding methods. (brts. 404, 409.) OBS. In determining the time, the d.y on which a note is dated, and that ron which it becomes due should not bolh be reckoned; it is customary to exelude the former. 12 260 APPLICATIONS OF [SECT. XIL Ex. 1. What is the interest due on a note of $625 frcm FekL 2d, 1846, to June 20th, 1847, at 6 per cent.? Operation. $625 Prin. Yrs. mo. ds..083 Int. of $1, 1847 "6 "/ 20 1846 "2 " 2 5000 Time 1 " 4 " 18 $51.875 Ans. Compute the interest on the following notes: (No. 1.) $450. NEW YORK, June 3d, 1847O 2. Sixty days after date, I promise to pay George Baker, or. order, Four Hundred and Fifty Dollars, with interest, value rea.. ceived. ALEXANDER HAMILTON. (No. 2.) $630. BOSTON, Aug. 5th, 1847. 3. Thirty days after date, I promise to pay Messrs. Holmes & Homer, or bearer, Six Hundred and Thirty Dollars, with interest, value received. JAMES UNDERWOOD. (No. 3.) $850. PHILADELPHIA, Sept. 16th, 1847. 4. Four months after date, I promise to pay Horace Williams, Eight Hundred and Fifty Dollars, with interest, value received. JOI-IN C. ALLEN. (No. 4.) $1000. CINCINNATI, Oct. 3d, 1847. 5. For value received, we jointly and severally promise to pay to the order 6f Winm. D. Moore & Co., One Thousand Dollars, iri one year from date, with interest. JOSEPII HIENRY, SANDFORD ATWATER. 6. What is the interest on a note of $634 from -Jan. st, ] 846, to March 7th, 1847, at 6 per cent.? ARTS. 415, 416.] INTEREST. 261t 7. What is the interest on a note of $820 from April 16th, 1846, to Jan. 10th) 1847, at 6 per cent.? 8. What is the interest on a note of $615.44 from Oct. 1st, 1836, to June 13th, 1840, at 4 per cent.? 9. What is the, interest on a note of $1830.63 fiom Aug. 16th, 1841, to June 19th, 1842, at 7 per cent.? 10. What is the amount due on a note of 8520 from Sept. 2d, 1846, to March 14th, 1847, at 5 per cent,? 11. What is the amount due on a note of $25000 from Aug. 17th, 1845, to Jan. 17th, 1846, at 7 per cent? 12. What is the amount due on a note of $6200 from Feb. 3d, 1846, to Jan. 9th, 1847, at 6 per cent.? PARTIAL PAYMENTS. 416. When partial payments are made and endorsed upon Notes and Bonds, the rule for computing the interest adopted by the Supreme Court of the Uniited States, is the following. I. " The rule for casting interest, when partial payments have been made, is to apply the payment, in the first place, to the discharge of the interest then due. II. "I ff the payment exceeds the interest, the surplus goes towards discharging the principal, and the subsequent intesrest is to be computed on the balance of principal remaining due. II. "c If the payment be less than the interest, the surplus of interest must not be taken to augment the principal; but interest continues on the former principal until the period when the payments, taken together, exceed the interest due, and then the sugplus is to be applied towards discharging the principal; and interest is to be7 computed on the balance as aforesaid." Note.-The above rule is adopted by New lYork, Massachusetts, and most of the other States of the Union. It is given in the language of the distin. guished Chancellor I(ent.-Jolhnson's C/tancery Reports, Vol. I p. p. 17. QUzsrT.-416. What is the general method of casting interest on Notes and Bonds, when patial payments have been made? 262 APPLICATIONS OF [SECT. XII $965. NEw YORK, March 8th, 1843. 13. For value received, I promise to pay George 13. Granniss, or.order, Nine Hundred and Sixty-five Dollars, on demand, with interest at 7 per cent. HENRY BROWN. The following payments were endorsed on this note: Sept. 8th, 1843, received $75.30. June 18th, 1844, received $20.38. March 24th, 1845, received $80. What was due on taking up the note, Feb. 9th, 1846? Operation. Principal, -. - $965.00 Interest to first payment, Sept. 8th. (6 months,) 33.775 Amount due on note Sept. 8th, - - $998.775 Ist payment, (to be deducted from amount,) - 75.30 Balance due after 1st pay't., Sept. 8th, 1843, - $923.475 Interest on Balance to 2d pay't., June 18th, (9 mo. 10 d.,) 2d pay't., (being less than int. then due,) 20.38 Surplus int. unpaid June 18th, 1844, $29.898 Int. continued on Bal. from June 18th,.45 to March 24th, 1845, (9 mo. 6 d.,)559 Amount due March 24th, 1845, - - $1002.932 3d pay't., (being greater than the int. now due,) 80100 is to be deducted from the amount, Balance due March 24th, 1845, - - - $922.932 Int. on Bal. to Feb. 9th, (10 mo. 15 d.,) - - 56.529 B3al. due on taking up the note, Feb. 9th, 1846, $9793.46 $650. BOSTON, Jan. 1st, 1,4,'2, 14. For value received, I promise to pay John Lincoln, or order, Six Hundred and Fifty Dollars on demand, with interest at 6 per cent. GEonRE c TEWInS. Endorsed, Aug. 13th, 1842, $100. Endorsed, April 13th, 1843, $120. What was due on the note, Jan. 20th, 1844? ART. 4 7.] INTEREST. 263 $2460. PHILADELPIIA, April 10th, 1844. 45. Four months after date, I promise to pay James Buchanan, or order, Two Thousand Four Hundred and Sixty Dollars, with interest, at 6 per cent., value received. GEORGE WILLIAMS. Endorsed, Aug. 20th, 1845, $840. {" ~Dec. 26th, 1845, $400. " lMay 2d, 1846, $1000. HIow much was due Aug. 20th, 1846? $5000. NEW ORLEANS, May 1st, 1845. 16. Six months after date, I promise to pay John Fairfield, or order, Five Thousand Dollars, with interest at 5 per cent., value reo Zived. WILLIAM ADAMS. Endorsed, Oct. 1st, 1845, $700. Feb. 7th, 1846, $45. {" Sept. 13th, 1846, $480. What was the balance due Jan. 1st, 1847? CONNECTICUT RULE. 417. I. "Compute the interest on the principal to the time of the first payment; if that be one year or more from the time the interest commenced, add it to the principal, and deduct the payment from the sum total. If there be after payments made, compute the interest on the balance due to the next payment, and then deduct the payment as above; and in like manner, from one payment to another, till all the payments are absorbed; provided the time between one payment and another be one year or more." II. " If any payments be made before one year's interest has accrued, then compute the interest on the principal sum due on the obligation, for one year, add it to the principal, and compute the interest on the sum paid, from the time it was paid up to the end of the year; add it to the sum paid, and deduct that sum from the principal and interest added as above." III. " If a year extends beyond the time of payment, then find the amcunt of the principal remaining unpaid up to the time of settlement, likewise the amount of the endorsements from the time they were paid to the time of settlement, and deduct the sum of these several amounts from the amount of the principal." "If any payments be made of a less sum than the interest arisen at the time of such payrfient, no interest is to be computed, but only on the principal sum for any period."-Kirby's Reports. 264 APPLICATIONS OF [SECT. XII. THIRD RULE. 41 S. First find the amount of the given principal for the whole time; then find thx amto znt of eac/h payment frjon the time it was endorsed to the7w time of wettlement. Finally, subltact the amount of the several payme nts from the amount of the p~rincipal, and the remainder will be the sum due. Note.-It will be an excellent exercise for the pupil to cast the interest on the preceding notes by each of the above rules. 4 19 To compute Interest on Sterling Money. 17. What is the interest of ~241, 10s. 6d. for 1 year, at 6 pec cent.? Operation. ~241.525 Prin. We first reduce the 10s. 6d. to the.06 Rate. decimal of a pound, (Art. 346,) then ~14.49150 Int. 1 yr. multiply the principal by the rate, 20s.=~1. and point off the product as in Art. s. 9.83000 404. The 14 on the left of the deci12d.=ls. mal point, denotes pounds; the figd. 9.96000 ures on the right are decimals of a 4f.=1d. pound, and must be reduced to shilfar. 3.84000 lings, pence, and farthings. (Art. 348.) iAns. ~14, 9s. 94d. Hence, 41 9. a. To compute the interest on pounds, shillings, pence, and farthings. Reduce the given shillings, pence, and farthings to the decimal of a pound; (Art. 346;) then find the interest as on dollars and cents; finally, reduce the decimal figures in the answer to shillings, pence, and farthings. (Art. 348.) 18. What is the amount of ~156, 15s. for 1 year and 4 months, at 5 per cent.? Ans. ~167, 4s. 19. What is the int. of ~275, 12s. 6d. for 1 yr., at 7 per cent. 20. What is the int. of ~89, 7s. 6jd. for 2 yrs., at 5 per cent. 21. What is the int. of ~500 for 6 mo., at 5 per cent.? 22. What is the amt. of ~1825, 10s. for 8 mo., at 6 per cent.? 23. What is the amt. of ~2000 for 10 yrs., at 41 per cent.? QuEsT. —419. I1ow is interest computed on pounds, shillings, and pence.? &a-rs. 418-421.] INTEREST. 265 PROBLEMS IN INTEREST. 4200 It will be observed that there are fozur parts or terms connected with each of the preceding operations, viz: the principal, the rate per cent., the time, and the interest, or tle amount. These parts or terms have such a relation to each other, that if any three of them are given, the other may be found. The questions, therefore, which may arise in interest, are numerous; but they may be reduced to a few general principles, or Problems. OBs. A number or quantity is said to be given, when its value is stated, or may be easily inferred from the conditions of the question under consideration. Thus, when the principal and interest are known, the amount may be said to be given1 because it is merely the suw of the principal and interest. So, if the principal and the amount are known, the interest may be said to be given, because it is the dileence between the amount and the principal. PROBLEM2 I. 421. To o nd the INTEREST, the principal, rate per cent., and the time being given. This problem embraces all the preceding examples pertaining to Interest, and has already been illustrated. PROBLEM II. To find the RATE PEt CENT., the principal, the interest, and the time beinq given. Ex. 1. A man borrowed $80 for 5 years, and paid $36 for the use of it: what ~was the rate per cent.? Analiysis.-The interest of i80 at I per cent. for I year is 80 cents; (Art. 404;) consequently for 5 years it is 5 times as much, and $.80 X 5=84. Now since $4 is 1 per cent. on the principal for the given time, $36 must be ~ of 1 per cent., which is equal to 9 per cent. (Art. 196.) Or, we may reason thus: Since $4 is 1 per cent. on the principal for the given time, $36 must be as many per cent. as $4 is contained times in $36; and 836. $4=9. Ans. 9 per cent. QuEST. —420. How many terms are connected with each of the preceding examples? What are they. When three are given, can the fourth be found. Obs. When Is a num. bor or quantity said to be given. 266 APPLICATIONS OF [SECT. XII, PnOOF.-$80X.09=$7.20, the interest of $80 for 1 year at 9 per cent., and $7.20 X 5=$36.00, the interest for 5 years, which is equal to the sum paid. Hence, 422. To find the q'ate per cent. when the principal, interest, and time are given. Divide the given interest by the interest of the princi'pal at I per cent. for the given time, and the quotient will be the required per cent. Or, find the interest of the principal at 1 per cent. for the given time; then mance the interest thus found the denominator and the given interest the numerator of a common fraction; reduce thi?' fraction to a whole or mixed number, and the result woill be the per cent. required. (Art. 196.) 2. If I loan $500 for 2 years, and receive $50 interest, what is the rate per cent.? Ans. 5 per cent. 3. A man borrowed $620 for 8 months, and paid $24.80 for the use of it: what per cent. interest did he pay? 4. At what per cent. interest must $2350 be loaned, to gain $,57 in 4 months? 5. At what per cent. interest must $1.925 be loaned, to gain $154 in I year? 6. A man has $12000 from which he receives $900 interest annually: what per cent. is that? 7. A man deposited $2600 in a savings bank, and received $143 interest annually: what per cent. was that? 8. A man invested $4500 in the Bank of New York, and received a semi-annual dividend of $1st7.50: what per cent. was the dividend? 9. A man paid $16250 for a house, and rented it for $975 a year:- what per cent. did it pay? 10. A hotel which cost $250000, was rented for $12500 a year: what per cent. did it pay on the cost? 11. A capitalist invested'500000 in manufacturing, and received a semi-annual dividend of $12500:'what per cent. was his dividend? QUEsT. —421' When the principal, interest and tine are given, how is the rt te per et. found I ARTS. 422, 423.] INTEREST. 267 PROBLEM III. To find the PRINCIPAL, the interest, the rate per cent., and the time being given. 12. What sum must be put at interest, at 6 per cent., to gain $75 in 2 years? Analysis.-The interest of $1 for 2 years at 6 per cent., (the given time and rate,) is 12 cents. Now 12 cents interest is eo —o of its principal $1; consequently, $75, the given interest, must be 9~r of the principal required. The question therefore resolves itself into this: $75 is -~1% of what number of dollars? If'75 is %o1-<, Thlv is -Lt of $75, which is $61; and 1-0 —=$6 X 100, which is $625, the principal required. Or, we may reason thus: Since 12 Cents is the interest of 1 dollar for the given time and rate, 75 dollars must be the interest of as many dollars for the same time and rate, as 12 cents is conltained times in 75 dollars. And $75-+.12=625. Ans. $625. PooF. —$625X.06=$37.50, the interest for 1 year at the given per cent., and $37.50X2=$75, the given interest. Hence, 423. To find the princip2al, when the interest, rate per cent., and time are given. Divide the given interest by the interest of $1 for the given time and rate, expressed in decimals; and the quotient will be the priwczipal required. Or, gmake the interest of $1 for the given time and rate, the numer-:ator, and 100 the denominator of a common fraction; then divide the given interest by this fraction, and the quotient will be the principal requzired. (Art. 234.) 13. What sum must be put at 7 per cent. interest, to gain,63 in 6 months? 14. What sum must be put at 5 per cent. interest, to gain 890 in 4 months? 15. What sum must be invested in 6 per cent. stock, to gain i300 in 6 months? QUEST. —4'3. When the interest, rate per cent., and time are given, how is the princil'pal found? 12' 268 APPLICATIONS OF [SECT. XII. 16. What sum must be invested in 7 per cent. stock, to gain $560( in one year? 17. A man founded a professorship with a salary of $1000 a year: what sum must be invested at 7 per cent. to produce it? 18. What sum must be put at 6 per cent. interest to pay a salary of $1200 a year? 19. What sum must be invested in 5 per cent. stock to make a simi-annual dividend of $750? 20. A man bequeathed his wife $1250 a year: what sum muslt: bc invested at 6 per cent. interest to pay it? PROBLEM IV. To find the TIME, the principal, the interest, and the rate rper; cent. being given. 21. A man loaned $200 at 6 per cent., and received $42 interest: how long was it loaned? Analysis.-The interest of $200 at 6 per cent. for 1 year is $12. (Art. 404.) Now, since $12 interest requires the principal 1 year at the given per cent., $42 interest will require the same principal ~ of I year, which is equal to 3~ years. (Art. 196.) Or, we may reason thus: If 812 interest requires the use of the; given principal I year, $42 interest will require the same principal as many years as $12 is contained times in $42. And $42-.-$12-3.5. Ans. 3.5 years. Hence, 424. To find the time, when the principal, interest, and rate per cent. are given. Divide the given interest by the interest of the princip2al at the given rate for 1 year, and the quotient will be the time required. Or, mzake the given interest the numerator, and the interest of the vrincipal for 1 year at the given rate the denominator of a commoU, fraction; reduce this fraction to a whole or mixed number, and it. will be the time required. OBs. If the quotient contains a decimal of a year, it should be reduced to months and days. (Art. 348.) QuEs'r.-424. WVhen the principal, interest, and rate per cent. are given, how is the timne found? Obs. When the quotient contains a decimal of a year, what should be done with it T' AnT. 424 l INTEREST 269 22. A man loaned $765.50, at 6 per cent., and received $183.72 interest: how long was it loaned? 23. In what time will $850 gain $29.75, at 7 per cent. per annum? 24. A man received $136.75 for the use of $1820, which was 6 per cent. interest for the time: what was the time? 25. In what time will $6280 gain $471, at 5 per cent. interest? 26. How long will it take $100, at 5 per cent., to gain $100 interest; that is, to double itself? Operation. The interest of $100 for 1 year, at 5 per cent., $5)$100 is $5. (Art. 404.) 20 Ans. 20 years. PRooF.-$100 X.05 X 20=$100, the given principal. (Art. 404.) TABLE, lSowimng in what time any given princical will double itself at any rate, from 1 to 20 per cent. Simple I7zterest..Per cent. Years. Per cent. Years. Per cent. Ylears. Per cent. Years. 1 100 6 16.- 11 9Y-.- 16 61 2 50 7 147 12 8- 17 51 - 3 331 8 121 13 7 a 18 56 4 25 9 liI 14 7-l 19 56-N 5 20 10 10 15 6( 20 5 27. How long will it take $365 to double itself, at 6 percent.? 28. How long will it take ~$1181 to double itself, at 7 per cent.? 29. In what time will $2365.24 double itself at 7 per cent.? 30. In what time will ~$5640 double itself, at 10 per cent.? 31. How long will it take $10000 to gain $5000, at 6 per cent. interest? 32. A man hired $15000, at 7 per cent., and retained it till the principal and interest amounted to $25000: how long did he have it? 33. A man loaned his clerk $25000 to go into business, and agreed to let him have it, at 5 per ct., till it amounted to $60000: how long did he have it? 270 COMPOUND [SECT. XlI. COMPOUND INTEREST. 425. Compound Interest is the interest arising not only from the principal, but also from the interest itself, after it becomes due. Oss. Compound Interest is often called interest upon interest. When interest is paid on the principal only, it is called Simple Interest. Ex. 1. What is the compoand interest of $842 for 4 years, at 6 per cent.? Operation. $842.00 Principal. $842X.06=- 50.52 Int. for 1st year. 89'2.52 Amt. for l'year. $892.52 X.06= 53.55 Int. for 2d year. 946.07 Amt. for 2 years. $946.07 X.06= 56.76 Int. for 3d year. 1002.83 Amt. for 3 years. $1002,83X.06= 60.17 Int. for 4th year. 1063,00 Amt. for 4 years. 842.00 Prin. deducted. Ans. $221.00 Compound int. for 4 years. 426. Hence, to calcuJlate compound interest. Cast the interest on the given principal for 1 year, or the specified time, and add it to the princi2al; then cast the interest on this amount for the next year, or specified time, and adcid it to the princijpal as before. Proceed in this manner with each successive year of the proposed time. Finally, subtract the given principal from the last amount, and the remainder will be the compound interest. 2. What is the compound interest of $600 for 5 years, at 7 per cent.? Ans. $241.53. 3. What is the compound int. of $1260 for 5 yrs., at 7per cent.? 4. What is the amount of $1535 for 6 yrs., at 6 per cent. compound interest? 5. What is the amount of $4000 for 2 yrs., at 7 per cent., payable semi-annually? QuE ST.-426. How is compound interest calculated? ARTs. 425, 426.] INTEREST. 27] TAB LE, iUwfinlg tIhe amnoznt of $1, or ~L, at 3, 4, 5, 6, and 7 per cent. conmpaiemei in!er?'est, for any nznbe? of years, fron 1 to 40. Yrs. 3 per cent. 4 per cent. 5 perr cent. 1. 1.030,000 1.040,000 1.050,000 1.060,000 1.07,000 2. 1.060,900 1.081,600 1.102,500 1.123,600 1.14,490 3. 1.092,727 1.1-24,864 1.157,625 1.191,016 1.22,504 4. 1.125,509 1.169,859 1.215,506 1.262,477 1.31,079 5. 1.159,274 1.216,653 1.276,282 1.338,226 1.40,255 6. 1.194,052 1.265,319 1.340,096 1.418,519 1.50,073 7. 1.229,874 1.315,932 1.407,100 1.503,630 1.60,578 8. 1.266,770 1.368,569 1.477,455 1.593,848 1.71,818 9. 1.304,773 1.423,312 1.551,328 1.689,479 1.83,845 10. 1.343,916 1.480,244 1.628,895 1.790,848 1.96,715 11. 1.384,234 1.539,454 1.710,339 1.898,299 2.10,485 12. 1.425,761 1.601,032 1.795,856 2.012, 196 2.25,219 13. 1.468,534 1.665,074 1.885,649 2.132,928 2.40,984 14. 1.512,590 1.731,676 1.979,932 2.260,904 2.57,853 15. 1.557,967 1.800,944 2.078,928 2.396,558 2.75,903 16. 1.604,706 1.872,981 2.182,875 2.540,352 2.95,216 17. 1.652,848 1.947,900 2.292,018 2.692,773 3.15,881 138. 1.702,433 2.025,817 2.406,619 2.854,339 3.37,293 19. 1.753,506 2.106,849 2.526,950 3.025,600 3.61,652 20. 1.806,111 2.191,123 2.653,298 3.207,135 3.86,968 21. 1.860,295 2.278,768 2.785,963 3.399,564 4.14,056 22. 1.916,103 2.369,919 2.925,261 3.603,537 4.43,040 23. 1.973,587 2.464,716 3.071,524 3.819,750 4.74,052 24. 2.032,794 2.563,304 3.225,100 4.048,935 5.07,236 25. 2.093,778 2.665,836 3.386,355 4.291,871 5.42,743 26. 2.156,592 2.772,470 3.555,673 4.549,383 5.80,735 27. 2.221,289 2.883,369 3.733,456 4.822,346 6.21,386 28. 2.287,928 2.998,703 3.920,129 5.111,687 6.64,883 29. ~.356,566 3.118,651 4.116,136 5.418,388 7.11,425 30. 2 427,262 3.243,398 4.321,942 5.743,491 7.61,225 31..5Q00,080 3.373,133 4.538,039 6.088,101 8.14,571 32. 2.575,083 3.508,059 4.764,94, 6.453,386 8.71,527 33. 2.652,335 3.648,381 5.003,189 6.840,590 9.32.533 34. 2.731,905 3.794,316 5.253,348 7.251,025 9.97,811 35. 2.813,862 3.946,089 5.516,015 7.686.087 10.6,765 36. 2.898,278 4.103,933 5.791,816 8.147,252 11.4,239 37. 2.985,227 4.268,090 6.081,407 8.636,087 12.2,236 38. 3.074,783 4.438,813 6.385,477 9.154,252 13.0,792 39. 3.167,027 4 616,366 6.704,751 9.703,507 13.9,948 40. 3.262,038 4.801,021 7.039,989 10.285,72 14.9,744 292 DISCOUNT [SECT. XII, 427. To calculate compound interest by the preceding table. _Find the amount of $1 or ~1 for the given number of years by the table, multiply it by thle given principal, and the product will be tihe amnount required. Subtract the principal from the amount thus found, and the remainder will be the compound interest. 6. What is t1i compound interest of $500 for 15 years, at 6 per cent.? What is the amount? Operation. $2.396558 Amt. of $1 for 15 yrs. by Table. 500 The given principal. $1198.279000 Amt. required. $500 Principal to be subtracted. $698.279 Interest required. 7. What is the amount of $960 for 10 yrs., at 7 per et.? 8. What is the amount of $1000 for 9 yrs., at 5 per ct.? 9. What is the compound int. of $1460 for 12 yrs., at 4 per ct.?'' 10. What is the compound int. of $2500 for 15 yrs., at 6 per ct.? 11. What is the amount of $5000 for 20 yrs., at 6 per ct.? 12. What is the amount of $10000 for 40 yrs., at 7 per ct.? DISCOUNT. 42.8 DISCOUNT is the abatement or deduction made for the payimuent of money before it is due. For example, if I owe a man $100, payable in one year without interest, the present worth of the note is less than $100; for, if $100 were put at interest for 1 year, at 6 per cent., it would amount to $106; at 7 uer cent.,: to $107, &c. In consideration, therefore, of the present payment of the note, justice requires that he should make some 5batement from it. This abatement is called Discount. 429. The present worth of a debt payable at some future time without interest, is that sum which, being put- at legal interest,, will amnount to the debt, at the time it becomes due. QUEsT. —428. What is discount. 429 What is the present worth of a-debt, payable ai some futlre time, without interest? ARtTSo 427-430.3 DISCOUNT. 273 Ex. 1. What is the present worth of $756, payable in I year and 4 months, without interest, when money is worth 6 per cent. per annum? Analysis.-The amount, we have seen, is the sum of the principal and interest. (Art. 399.) Now the amount of $1 for 1 year and 4 months, at 6 per cent., is $1.08; (Art. 404;) that is, the amount is e+ of the principal $1. The question then resolves itself into this: $756 is etI of what principal? If $756 is Laof a certain sum, d-r is 18 of $756; now $756 - 108=$7, and -1u 4$a7 X 100, which is $700. Or, we may reason thus: Since $1.08 (amount) requires $1 principal for the given time,. $756 (amount) will require as many dollars as $1.08 is contained times in $756; and $756- $1.08$700, the same as before. PtOOF. —$700 X.08=$56, interest for I year and 4 months; and $700+56=$756, the sum whose present worth is required. Hence, 4-30o To find the present worth of any sum, payable at a future timne without interest..Fiirst find the amount of $1 for the time, at the given rate, as iz simple interest; then divide the given sum by this amount, and the quotient will be the present worth. (Art. 404.) TIe present worth subtracted fronz the debt, will give the true discount. Ons. This process is often classed among the Problems of Interest, in which the amount, (which answers to the given sum or debt,) the rate per cent., and the time are given, to find the princwipal, which answers to the present worth. 2. What is the present worth of $424.83, payable in 4 months, when money is worth 6 per cent.? What is the discount? Solution.-$424.83 — $1.02 $41 6.50, Present worth. And 8424.83 —$416.50=88.33, Discount. 3. What is the present worth of $1000, payable in 1 year, when the rate of interest is 7 per cent.? 4. What is the present worth of $1645, payable in 1 year and 6 Lmonths, when the rate of interest is 7 per cent.? QuEST. —430. How do you find the present worth of a det? HoW find the discount? 274 BANK DISCOUNT. [SECT. XIl, 5. What is the discount on a note for $2300, payable in (3 months, when the rate of interest is 8 per cent.? 6. What is the discount, at 6 per cent., on $4260, payable in 4 months? 7. What is the present worth of a note for $4800, due in 3 months, when the rate of interest is 6 per cent.? 8. What is the present worth of a draft for $0240, payable in 1 month, when the rate of interest is 6 per cent.? 9. A man sold his farm for $3915, payable in 2- years: what is the present worth of the debt, at 6 per cent. discount? 10. What is the present worth of a draft of 810000, payable at 30 days sight, w-hen interest is 6 per cent. per annum? 11. Wthat is the difference between the discount of $8000 for I year, and the interest of $8000 for 1 year, at 7 per cent.? BANK DISCOUNT. 43 t A Beentc, in commerce, is an institution established for the safe keeping and issue of money, for discounting notes, dealing il exchange, &c. OBS. 1. There are t/Ace kinds of banks, viz: banks of deposit, discount, and cir'culation. A banh f oqf dCeosit receives money to keep, subject to the order of the depositor. TIhis was the primary object of these institutions. A t/od!of discozott is one which loans money, or discounts ndtes, drafts, and bills of exchange. A ban/cf of circidlrtion issues bills, or nsotes of its own, which are redeemable in specie, at its place of business, and thus become a circulating medium of exchange. Banks of this country generally perforim the three-fold office of deposit, discount, and circulation. 2. Tbhe afltairs of a bank are managed by a boccrd of directors, chosen annually by the stockholders. (Art. 392. Ohs.) The directors appoint a yresident and cashier, who signll the bills, and transact the ordinary business of tble bank. A tcller is a clerk in a bank, who receives and pays the money on checks. A chick is an order for money, drawn on a banker, or t me casl ier, by a depositor, payable to the bearer. 3. Banks originated in Italy. The first one was established in Vence, ill 111, called the Bank of Venice. (Viuc'e. —431. rWhat is a bank? Obs. Of how nlaily khuls are baiks? ARTS. 431-433.] BANK DISCOUNT. 275 432. It is customary for Banks, in discounting a note or draft, to deduct in advance the legal interest on the given sum from the ti me it is discounted to the tinme it becomes due. Hence, Baauc di scount is the same as simple interest paid in advance. Thus, the bank discount on a note of' i106, payaTable in 1 year, at 6 per cei)t., is 06.36, while the true discount is butt-l6. (-Art. 430.) OBS. 1 Thle difference between bacnk discoCuhnt and Iore discolunt, is the interest of the true discount for the given time. On small sums for a short period his difference is trifling, but when the sum is large, and the time for which it is discounted is long, the difference is worthy of notice. 2. Taking legal interest in advance, according to the general rule of law, is swry?/..An exception is generally allowed, however, in favor of notes, drafts, &c., which are payable in less than aj year. The Safety Fund Banks of the State of New York, though the legal rate of interest is 7 per cent., are not allowed by their charters to take over 6 per vent. discount in advance on notes and drafts which mature within 63 days firom the time they are discounted.* Banks charge interest for the three days grace. C-A S E I. 12. What is the bank discount on a note for $850.20 for 6 months, at 6 per cent.? What is the presenut worth of the note? Opera' tion. $850.20 Principal..03 05 Int. $1 for 6 mo. 3 ds. grace. 4251 00 25 5060 $25.9311 00 Bank discount. And $850.20 — 25.93 —$824.27, Present worth. Hence, 433. To find the bank discount on a note or draft. Cast the interest on thre face of the note or draft for three days znore tlhan the specified time, and the result qwill be the discount. The discount subtracted frosn the face of the note, will gzve the 2oresent wortth or proceeds of a note discounrted at a bank. QjreST. —432. How do banks usually reckon discount? Wrhat then is bank discounit Ols. What is the difference between bank dliscount and true discount? Is this difference worth noticing Hlow is taking interest in advance generally regarded in law-? What erception to this rule is allowed? * Revised Statutes of New York, (3d edition,) Vol. I. p. 741. 276 BANK DISCOUNT. [SECT. XII. Note.-.nlltelest should be computed for the three days grace in each of the following examples. 14. What is the bank discount on a note for $465, payable in 6 months, at 6 per cent.? 15. What is the bank discount on a note for $972, payable in 4 months, at 5 per cent.? 16. What is the bank discount on a note for $1492, payable in 3 months, at 7 per cent.? 17. What is the bank discount on a draft of $628, payable at; 60 days sight, at 5 per cent.? 18. What is the present worth of $2135, payable in 8 months, att 7 per cent.? 19. What is the present worth of a note for 82790, payable in, 1 month, discounted at 6 per cent. at a bank? 20. What is the bank discount, at 51 per cent., on a draft of: $1747, payable at 90 days sight? 21. What is the bank discount, at 44 per cent., on a draft of $3143, payable in 4 months? 22. What is the bank discount on $5126.63, payable in 30 days, at 8 per cent.? 23. What is the bank discount on $3841.27, payable in 60 days, at 6G- per cent.? 24. What is the present worth of a note for $6721, payable in). 10 months, discouinted at 6 per cent. at a bank? 25. What is the present worth of a note for $1500, payable in 12 days, at 7 per cent. discount? 26. What is the bank discount on $10000, payable in 45 days, at 6 per cent.? 27. What is the bank discount on $25260, payable in 90 days, at 7 per cent.? 28. What is the difference between the true discount and bank discount on $5000 for 10 years, at 6 per cent.? CASE II 29. A man wishes to make a note payable in 1 year, at 6 per cent., the present worth of which, if discounted at a banVk, shall be just $200: for what sum must the note be made? &RT. 434.] BANK DISCOUNT. 277 Analysis.-The present worth of $1, payable in 1 year, at 6 per cent. discount, is 100 cts.-6 cts.=94 cts.; that is, the present worth is -tJL of the principal or sum discounted. The question then resolves itself into this: $200 (present worth) is -lox of what sum? Now, if $200 is xfi of a certain sum, dlv is -,4I of $200; and $200- 94=$2.12766, and t= —$82.12760X100, -which is $212.766. Ans. Or, we may reason thus: Since 94 cents present worth requires 1$, (100 cents) principal, or sum to be discounted for the given time,:$200 present worth will requil:e as many dollars, as 94 cents its contained times in $200; and $200 -.-.94=$212.766. PROOF.-$212.766X.06=$12.7659, the bank discount for 1 year; and $212.766- 12.7659-$200, the given sum. Hence, 434. To find what sum, payable in a specified time, will produce a given amount, when discounted at a bank, at a given "Per cent. Divide the given amount to be raised by the present worth of $1, for tile time, at the given rate of bank discount, and the quotient'will be the sum required to be discounted. 30. How large must I make a note payable in 6 months, to raise $400, when discounted at 7 per cent. bank discount? 31. What sum payable in 4 months must be discounted at a bank, at 5 per cent., to produce $950? 32. What sum payable in 60 days, will produce $1236, if discounted at a bank, at 8 per cent.? 33. For what sum must a note be drawn, payable in 34 days, Al the person, what are they called? When assessed upon the property, how are they almportioned? Obs. IIow is property divided? Whlat does real estate denote? What isto personal property? 455. When a tax is to be assessed, what is the first step? yaiTs. 454-456.] oF TAXES. 293 gOs. I. By the nscmber of polls is meant the number of taxable individuals, which usually includes every native or?atu'zalized fjreemanv over the age of 21, and under 70 years. In Massachusetts poll taxes are assessed upon every male inhabitant of the state, between the ages of 16 and 70 years, whether a citizen or an alien.*'2. -When any part or the whole of a tax is assessed upon the polls, each citizen is taxed a specific sum, without regard to the amount of property he possesses. Ex. 1. The tax assessed by a certain town is $990; its property, both personal and real, is valued at $28000, and it contains 300 polls, which are assessed 50 cents apiece. What per cent. is the tax; that is, how much is the tax on a dollar; and how much is a man's tax who pays for 3 polls, and whose property is valued at $1500? Solution. —Since 1 poll pays 50 cents, 300 polls must pay 300 tines 50 cents, which is $150. Now $990 —$150-$840, the sum to be assessed on the property. Now if $28000 is to pay $840, $1 must pay n-a0-r of $840; and $840 — 28000=$.03, or 3 per cent. Finally, the tax on $1500, the amount of the nan's property, at 3 per cent., is $1500X.03='$45; and 845 — $1.50 (3 polls)=$46.50, the man's tax. Hence, 456. To assess a State, County, or other tax. I. F'irst fiad the amount of tax on all the polls, if any, at the given rate, and seIubtract this sum from the whole tax to be assessed. T'ien ddividing the reemainder by the whole amount of taxable prop-'erty in the State, County, dc., the quotient will be the per cent. or tax on one dollar. II. Ji2idltiply the amoutnt of each man's pro-perty by the tax on one dollar, and the prohduct zwill be thle tax on his property. 111. _Add each manz's poll tax to the tax he p2ays o02 his properly, and the amount wvill be his whole ta,. Pnoacc.-T- Ahen a tax bill is m1~ade ouGt, add together the taxes of all,7e individutals in the toOwn, district, &c., cand if the amount is eutaGl to the whole tax assessed, the workl is right. QUEST.-..Obs. What is meant by the nnniber of polls? 456. TIow are taKes assesaOed When a tax bit is madtte out, how is its correctness proved? * Revised Statutes of Massachusetts. 294 ASSESSMENT [SECT. XII. 2. A certain corporation is taxed $537.50; the whole property of the corporation is valued at $35000, and there are 50 polls which are assessed 25 cents apiece. What per cent. is the tax; and how much is a man's tax, who pays for 2 polls, and whose property is valued at $4240. Op2eration. Multiply $.25 the tax on 1 poll, By 50 the number of polls. $12.50 Amount on polls. But $537.50-12.50=$525, the sum assessed on the corporation; and $525 — $35000=.015, the per cent. or tax on $1. Now $4240 X.015=$63.60, the tax on the man's property, And.25 X 2=.50, the tax for polls. Ans. $64.10, whole tax. 3. What is B's tax, who pays for 3 polls, and whose property is valued at $3560? 4. What is C's tax, who is assessed for 1 poll and $5350? 5. The city of New York levied a tax of 81945600; its taxable property was rated at $243200000: what per cent. was the tax? 6. What was A's tax, whose property was valued at $10000? 7. What was B's tax, who was assessed for $15240? 8. What was C's tax, who was assessed for $35460? 45,7. Having ascertained the expenditures of a State, County, Town, &c., it is necessary in assessing the tax, to take into con sideration the expense of collecting it. Collectors are paid a certain per cent. commission on the amount collected; (Art. 388. Obs. 1;) consequently, in determining the exact sum to be assessed, allowance must be made not only for the commission on the net amount to be raised; but also on the commission itself; for the commission is to be paid out of the money collected. 9. If the expenses of a town are $950, what sum must be assessed to raise this amount, with 5 per cent. commission for collecting it? ARTS. 157-459.1 OF TAXES. 295 Analysis.-Since the commission is 5 per cent. the net value of $1 assessment is 95 cents. Therefore, if 95 cents net, require $1 assessment, $950 net, will require as many dollars assessment, as 95 cents are contained times in $950; and $950+-$-.95 —1000. Ans. $1000. PRooF.-$1000 X.05=$50, the commission; and $1000-$50=$950, the net sum required. Hence, 4 5 8. To find what sum must be assessed, to raise a given net aliount. Suebtract the given per cent. commission from $1, and the remainder wvill be the net value of $1 assessment. Divide the net amount to be raised by the net value of $1 assessment, and the quotient will be the sum to be assessed. Oss. To meet the expense of collecting a tax, assessors not unfrequently calculate the commission at the given per cent. on the net amount to l)e raised, and add it to the tax bill. This method is wrong, and leads to erroneous results. Thus, on a tax of $1000, at 5 per cent. commission, the net amount is $2.50 too small; on $100000, the error is $250; on $1000000, it is $2500. 10. What sum must be assessed to raise a net amount of $8500. withl 4 per cent. commission for collection? 11. What sum must be assessed to raise $15400 net, allowing 4- per cent; commission for collection? 12. Allowing 5 per cent. for collection, what sum must be assessed to raise $16475 net? 13. Allowing 3- per cent. for collection, what sum must be assessed to raise $32860 net? FORMATION OF TAX BILLS. 459. In making out a tax bill for a Town, District, &c., having found the tax on $1, it is advisable to make a table, showing the amount of tax on any number of dollars from 1 to $10; then from $10 to $100; and from $100 to $1000. 14..A -township composed of 16 citizens, levies a tax of $5700; the tbwn contains 30 polls, which are assessed 50 cents each, and QUEST-4.58. Hlow find what suml must be assessed to raise a tax of a given.amount 296 ASSESSMENT [SECT. XII. its taxatble property is inventoried at $199500. What amount of tax must be raised to pay the debt and 5 per cent. commission for collection; and what is the tax on a dollar? Solution.-The sum to be raised is $6000; (Art. 458;) and the tax is 3 cents on a dollar. (Art. 456.) Now, since the tax on $1 is $.03, it is obvious that multiplying $.03 by 2 will be the tax on $2; multiplying it by 3, will be the tax on $3, &c., as seen in th following TABLE. $1 pays $.03 $10 pay $.30 $100 pay $3.00 2 I'.06 20 ".60 200 " 6'.00 3 ".09 30 ".90 300 " 9.00 4 ".12 40 " 1.20 400 " 12.00 5 ".15 50 " 1.50 500 " 15.00 6 ".18 60 " 1.80 600 " 18.00'7 ".21 70 " 2.10 700 " 21.00 8 ".24 80 " 2.40 800 " 24.00 9 ".27 90 " 2.70 900 " 27.00 10 ".30 100 " 3.00 1000 " 30.00 15. In the above assessment, what is A. B.'s tax, who is rated at $2256, and pays for 3 polls? Operation. %2000 pay $60.00 82256=2000+200+50+6 dollars. 200 " 6.00 Now if we add together the tax paid 50 " 1.50 on each of these sums, as found in the 6 ".18 table above, the amount will be the tax 3 polls " 1.50 on $2256. Amount, $69.18 A. B.'s tax therefore is $69.18. 16. What is G. A.'s tax, who is assessed for 2 polls, and;2400. 17. What is H. B.'s tax, who is assessed for 1 poll, and $3850? 18. What is W. C.'s tax, wiho is assessed for 3 polls, and $15000? 19. E. D. is assessed for $16024, and I poll: what is his tax? 20. J. F. is assessed for $10450, and 2 polls: what is his tax? 21. T. G. is assessed for 2b0680, and 3 polls: what is his tax? 22. WV. H. is assessed for'$17530, and I poll: what is his tax? ART. 460.] OF T1AXES. 297 23. Lo. J. is assessed for $8760, and 1 poll: what is his tax? 24. W. L. is assessed for $21000, and 2 polls: what is his tax? 25. J. K. is assessed for $6530, and 2 polls: what is his tax? 26. G. L. is assessed for $13480, and 1 poll: what is his tax? 27. F., M. is assessed for $12300, and 3 polls: what is his tax? 28. C. P. is assessed for $15240, and 2 polls: what is his tax? 29. J. S. is assessed for $16000, and 1 poll: what is his tax? 30. R. W. is assessed for $18000, and 2 polls: what is his tax? Note.-Rate Bills for schools are generally apportioned according to the number of days each scholar has attended. Hence, 4-60,, To make out Rate Bills for schools. First find the number of days attendance of all the scholar;s, and the whole amount of expenses, including teacher's salary, fuel, r epairs, &c. From the amount of expenses deduct the pIublic money, if any, then divide the remainder by the whole number of days attendance, and the quotient will be the rate _per day. Finally, multliply the rate per day by the number of days attendance of each man's children, and the product will be his tax. OaBs. In New York and some other states, the geleral principle is to include only the Teacher's Salary in the Rate Bill. (Revised Statutes. N. Y.) 31. A certain district paid $130 for teacher's salary, $34 for board, $19.42 for fuel, and $2.58 for repairs; the district dre-w $30 public money, and the whole number of days attendance was 2400: what was the rate per day; and how much was A's tax, who sent 115 days? Solw.tion. —Amount of expenses, $186- - 30-$156; and $156.:2400=-=.065, the rate per day. Now $.065X115-67.475, A's tax is therefore $7.475. 32. If the expenses of a district are $313.20, and the Vwhole attendance 3915 days, what is B's tax, who sends 167 days? 33. A district paid their teacher $115, and their fuel cost $21.10; it drew $38.50 public money, and the number of days attentlance was 1954: what was C's tax, who sent 69 days? 34. The expenses of a district wvere $215.20, and the number of days attendance 2150: what was D's tax, who sent 134 days? 298 ANALYSIS. LSECT. XII SECT10N XI!I. ANALYSIS. ART. 46 1. The term Analysis, in physical science, signifies tle resolving of a compound body into its elements, or component parts. ANALYSIS, in arithmetic, signifies the resolving of numlbers into the jftCtors of which they are composed, and the tracing of the relations which they bear to each other. (Art. 95. Obs. 2.) Oss. In the preceding sections the student has become acquainted with the method of anlyl/zing- particlcdar examples and combinations of numbers, and thence deducing g'eneral princciples anld r1aztles. But analysis may be applied with advantage not only to the development of mathematical truths, but also to the solsetion of a great variety of problems, both in arithmetic and practical life. Indeed, it is the method by which business men generally solve practical questions. A little practice will give the student great facility in its application. 462. No specific directions can be given for solving examples by analysis. None in fact are requisite. The judgmenzt, from the conditions of the question, will suggest the process. Hence,.Analysis may, withl propriety, be called the COMMON SENSE RULE. OBs. In solving questions analytically, it may be remarked in general, that we reason from the g-':en number to 1, then from 1 to the mneumber e'cqtired. Ex. 1. If 60 yards of cloth cost $240, what will 85 yards cost? Analytic solution.-Since 60 yds. cost $240, 1 yd. will cost -,'V of $240; and -Al- of $2i40 is $4. Now if 1 yd. costs $4, 85 yds. will cost 85 times as much; and $4 X 85=$340. Ans. Or, we may reason thus: 85 yds. are H- of 60 yds.; therefore 85 yds. will cost -8- of 8240, (the cost of 60 yds.) and — _ of $240 is $240X- -`$340, the same as before. (Arts. 210, 212.) OGs. 1. Other solutions of this example might be given; but our present object is to show how this and similar questions may be solved by analysis. The QOeEST.-461. What is meant by analysis in physical science? What in arithmetic'? To what mtay analysis be advantageously applied? 462. Can any particular rules be prescribed fo)r solving questions by analysis? How then will you know howl_') proceed I Obs. What is the operation of solving questions by analysis called? ARTS. 461, 462.] ANALYSIS. 299 former method is the simplest and most strictly analytic, though not so short as the latter. It contains two steps: Fir'st, we separate the given price of 60 yds. ($240) into 60 equal parts, to find the value of one part, or the cost of 1 yd., which is $4. Secoo(dI, we multiply the price of 1 yd. ($)4) by 85, the number of yds. whose cost is required, and the product is the answer sought. 2. This and similar questions are usually placed under the rule of Simple P'Loportioz1n, or the Riule of T/hree. 3. The operation of solving a question by analysis, is called an analyrtz oltion. In reciting the following examples, each one should be analyzed, ud the reason for every step given in full. 2. A man bought a horse, and paid $45 down, which was q of the price of it: what did he give for the horse? Analysis.-Since $45 is A of the price, the question resolves itself into this: $45 is -- of what sum? If $45 is a- of a certain sum, 1 is 1 of $45; and 1 of $45 is $9. Now if $9 is 1 seventh, 7 sevenths are 7 times as much; and $9X 7=$63. -Ans. $63. Pnoor. —17 of $63=$9, and 5 sevenths are 5 times as much, which is $45, the sum he paid down for the horse. N';te.-In solving examples of this kind, the learner is often perplexed in finding the value of 1, &c. This difficulty arises from supposing that if i of a certain number is 45, 1 of it must be I of 45. This mistake will be easily avoided by substituting in his mind the word parts for the given de-?nomninator. Thus, if 5 parts cost $45, 1 part will cost 1- of $45, which is $9. But this part is a seventit. Now if 1 seventh cost $9, then 7 sevenths will cost 7 times as much. 3. If 40 cords of wood cost $120, how much will 100 cords,ost? 4. Bought 35 tons of hay for $700: how much will 16 tons cost? 5. What cost 3'7 gallons of molasses, at 821 a hogshead? 6. What cost 1500 pounds of hay, at $14 per ton? 7. What cost 18 quarts of chestnuts, at $3 a bushel? 8. If 55 tons of hemp cost $660, what will 220 tons cost at the same rate? 9. If 165 bushels of apples cost $132, how much will 31 bushels cost? 30() ANALYSIS. [SECT. XIII. 10. It 72 bushels of peanuts cost $253.44, what will a pint cost at the same rate? 11. If 150 acres of land cost $7000, what will a square rod cost? 12. If 2 pipes of wine cost $315, what is that per gill? 13. A farmer bought a yoke of oxen, and paid $40 in work, which was - of the cost: what did they cost? 14. Bought a house, and paid $630 in goods, which was -,r of the price of it: what was the cost of the house? 15, A young man lost $256 by gambling, which was -a of all lie was worth: how much was he worth? 16. A man having $i1500, paid 3- of it for 112~ acres of land: how much did his land cost per acre? 17. If a stack of hay will keep 350 sheep 90 days, how long will it keep 525 sheep? 18. If 440 bbls. of flour will last 15 men 55 months, how long will the saime quantity last 28 men? 19. If 136 men can build a block of stores in 120 days, how long will it take 15 men to build it? 20. If E of a pound of tea cost 40 cents, what will j of a pound cost? 21. If - of a yard of broadcloth cost $2.50, how much will { of a yard cost? 22. Bought -~ar of a ton of hay for $3.42: how much will -?r of a ton cost? 23. Bought 1 of a hogshead of molasses for $38.19: how much will -2; of a hogshead cost? 24. If " of an acre of land cost $108, how much will ~- of an acre cost? 25. If - of a barrel of beef cost $6.48, how much will d of a barrel cost? 26. Paid $4200 for b- of a sloop: how much can I afolrd to sell -,i7 of the sloop for? 27. Sold 18~1 baskets of peaches for $34: how much would 65-,baskets come to? 28. If I pay $60.50 for building 20- rods of -wall, how much must I pay for 2153 rods? ART. 463.] ANALYSIS. 301 29. A ma.n can hoe a field of corn in 6 days, and a boy can hoe it in 9 days: how long will it take tahem both together to hoe it? 4lctlysis.-Since the man can hoe the field in 6 days, in 1 day he can hoe 1 of it; and since the boy can hoe it in 9 days, in 1 day he can hoe ~ of it; consequently in 1 day they can both hoe 6+9_1_ of the field. (Art. 202.) Now if -?8 of the field requires them both 1 day, -'-S of it will require them - of a day, and will require them 18 times as long, or 1-8 of a day, which is equal to S3 days. Ans. 30. If A can chop a cord of wood in 4 hours, and B in 6 hours, how long will it take them both to chop a cord? 31. A can dig a cellar in 6 days, B in 9 days, and C in 12 days: ~how long will it take all of them together to dig it? 32. A man bought 25 pounds of tea at 6s. a pound, and paid for it in corn at 4s. a bushel: how many bushels did it take? zlnalysis.-If 1 lb. of tea costs 6s., 25 lbs. will cost 25 times as much, which is 150s. Again, if 4s. will buy 1 bushel of corn, 150s. will buy as many bushels as 4s. is contained times in 150s.; a-nd 150s.. 4-=37- times. Ans. 3l7 bushels. 463. The last and similar examples are fiequently arranged tunder the rule of -Barter. Barter signifies an exchange of articles of commerce at prices agreed upon by the parties. OBS. Such examples are so easily solved by Analysis that a speccfic rule fct them is Lnnecessary. 33. A farmer bought 110 lbs. of sugar at 18 cents a pound, and paid for it in lard at 5~ cents a pound: how much lard did it take? 34. How much butter, at 12~ cents a pound, must be giqven for 250 lbs. of tea, at 75 cents a pound? 35. How many cords of wood, at $2~ per cord, must be given for 56 yds. of cloth, at $41 per yard? 36. How many pair of boots, at $4.50 a pair, must be given (or 50 tons of coal at $9 per ton? 302 ANALYSIS. [SECT. XIII. 37. A, B, and C, united in business; A put in $250; B, $270; and C, $340; they gained $258: what was each man's share of the gain? Analysis.-The whole sum invested is $250-+$270~+-340= $860. Now since $860 gain $258, it is plain $1 will gain — Jv of $258, which is 30 cents. And If $1 gains 30 cts. $250 will gain $250X.30=$75, A's share, " 1 " " $270 " $270X.30= 81, B's share, $"1 " " $340 " 8340X.30=102, C's share. Or, we may reason thus: Sin'ee the sum invested is $860, A's part of the investment is equal to 6 or 2; B's " ", or-; C's, or -'. Consequently, A must receive ~-2 of the whole gain 8258-=75; B " " " 258= 81; C it" " "U cc"it 258=102; PRooF. —The whole gain is $258. (Ax. 11.) 464. When two or more individuals associate themselves together for the purpose of carrying on a joint business, the union is called a partnership or copcgrtner.ship. Oss. The process by which examples like the last one are solved, is often called Fcltlmwship. 38. A andB join in a speculation; A advances $1500 and B $2500; they gain $1200: what was each one's share of the gain? 30. A, B, and C, entered into partnership; A furnished $3000, B $4000, and C $5000; they lost $1800: what was each one's share of the loss? 40. A's stock is $4200; B's $3600; and C's $5400; the whole gain is $2400: what is the gain of each? 41. A's stock is $7560; B's $8240; C's $9300; and D's $0200; the whole gain is $625: what is the share of each? 42. A bankrupt owes one of his creditors 8400; another $500; and a third $600; his property amounts to $1l000: how much can he pay on a dollar; and how much p~ill each of his creditors receive? OBs. The solution of this example is the same in principle as that of Ex. 37. ARTS. 464-466.] ANALYSIS. 303 465. Examples like the preceding are commonly arranged under the rule of Bankruptcy. Note, -A ban,. r-ypt is a person who is insolvent, or unable to pay his just debts. 43. A bankrupt owes $5000, and his property is worth $3500: how much can he pay on a dollar? 44. A -man died owing 816400, and his effects were sold foi t$4100: what per cent. did his estate pay? 45. If a man owes A $6240, B $8760, and C $9000, and has but $11500, how much will each creditor receive? 46. If I owe $48000, and have *property to the amount of $32000, what per cent. can I pay? 47. What per cent. can a man pay, whose liabilities are $120000, and whose assets are W45000? 48. What per cent. can a man pay, whose liabilities are $1500000O, and whose assets are $150000? 46Go It often hcalpelns in storms and other casualties at sea, that masters of vessels are obliged to throw portions of their cargo overboard, or sacrifice the ship and their crew. In such cases, the law requires that the loss shall be divided among the owners of the vessel and cargo, in proportion to the amount of each one's property at stake. The process of finding each man's loss, in such instances, is called Genzeral Average. OBS. Thie operation is the same as that in solving questions in bankruptcy and partnership. 49. A, B and C, freighted a ship from New York to Liverpool; A had on board 100 tons of iron, B 200 tons, and C 300 tons, in a storm 240 tons were thrown overboard: what was the loss of each? 50. A packet worth $36000 was loaded with a cargo valued at $65000. In a tempest the master threw overboard $25250 worth of goods wh:lat per cent. was the general average? 51. A steam ship being in distress, the master threw -1 of,Ihe cargo overboard; finding she still labored, he afterwards threw overboard - of what remained. The steamer was worth 304 ANALYSIS. [SECT. XIII. $120000, and the cargo $240000: what per cent. was the general average, and what would be a man's loss who owned 1 of the ship and car4go? 52. A man mixed 25 bushels of peas worth Os. a bushel, with 15 bushels of corn worth 4s. a bushel, and 20 bushels of oats worth 3s. a bushel: what was the mixture worth per bushel. Analysis.-25 bu. peas at 6s.=150s., value of the peas; 15 bu. corn at 4s.= 60s., " " corn; and 20 bu. oats at 3s.= 60s., " " oats. The mixture=60 bu. and 270s., value of whole mixture. Now if 60 bu. mixture are worth 270s., I bu. mixture is worth 10- of 270s.; and 270s.-. 60=41s. Ans. PROOF. —60 bu. at 4-1s.=270s., the value of the whole mixture. 4:67. The process of -finding the value of a compound or mixture of articles of different values, or of forming a compound which shall have a given value, is called Alligation. Alligation is usually divided into two kinds,.Medial and Alternate. OBS. 1. When the prices of the several articles and the number or quantity of each are given, the process of finding the valule of the mixture, as in the last example, is called A lligatiiozn Medial. 2. When the price of the mixture is given, together with the price of each article. the process of finding how much of the several articles must be taken to form the required mixture, is called Alligzatiomn Alterzate. Alligation Alternate embraces thtee varieties of examples, which are pointed out in the following notes. 53. If you mix 40 gallons of sperm oil worth 8s. per gallon, with 60 gallons of whale oil worth *3s. per gallon, what will the mixture be worth per gallon? 54. At what price per pound can a grocer afford to sell a mixture of 30 lbs. of tea worth 4s. a pound, and 40 lbs. worth 7's. a pound? 55. If 120 lbs. of butter at 10 cts. a pound are mixed with 24 lbs. at 8 cts. and 24 lbs. at 5 cts. a pound, what is the mixture worth? 56. A tobacconist had three kinds of tobacco, worth 15, 18, and 25 cents a pound: what is a mixture of 100 lbs. of each worth per pound? ART 467.1 ANALYSIS. 305 57. A liquor dealer mixed 200 gallons of alcohol worth 50 cts. a gallon, with 100 gallons of brandy worth $1.75 a gallon: what was the value of the mixture per gallon? 58. A grocer sells imperial tea at 10s. a pound, and hyson at 4s.: what part of each must he take to form a mixture which he can afford to sell at 6s. a pound?.Note 1. —It will be observed in this example that the price of the mzzxtre r,'. also the price of the several acticlcs or ivgcr(ldic~ts are given, o find what p/xt of each the mixture must con;ain. Analysis.-Since the imperial is worth 1Os. and the required mixture 6s., it is plain he would lose 4s. on every pound of inmperial which he puts in. And since the hyson is worth 4s.a pound and the. mixture 6s., he would gain 2s. on every pound of hyson he puts in. The question then is this: How much hyson must he put in to make up for the loss on 1 lb. of imperial? If 2s. profit require 1 lb. of lhyson, 4s. profit will require twice as much, or 2 lbs. IHe must therefore put in 2 lbs. of hyson to 1 lb. of imperial. PROOF-2 lbs. of byson, at 4s. a pound, are worth 8s., and 1 lb. of imperial is worth 10s. Now Ss.+10s.=l8s. And if 3 lbs. mixture are worth 18s., 1 lb. is worth ~ of 18S., which is 6s., the price of the mixture required. 59. A farmer has oats which are worth 20 cts. a bushel, rye 55 cts., and barley 60 cts., of which he wishes to make a mixture worth 50 cts. per bushel: what part of each must the mixture contain? innalysis.-The prices of the rye and barley must each be compared with the price of the oats. If 1 bu. oats gains 30 cts. in the mixture, it will take as many bu. of rye to balance it, as 5 cts. (the loss per bu.) are contained times in 30 cts., viz: 6 bu. Again, since 1 bu. oats gains 30 cts., it will take as many bushels of barley to balance it, as 10 cts. (the loss per bu.) are contained times;in 30 cts., viz: 3 bu. Hence, the mixture must contain 2 parts of oats, 6 parts rye, and 3 parts barley. 60. If ra man have four kinds oC sugar worth, 8, 9, 11, and 12 306 ANALYSIS. [SECT. XIII. cents a pound respectively, how much of each kind must he take to form a mixture worth 10 cents a pound? Note 2.-In examples like the preceding, we compare two kinds together, one of a higher and the other of a lower price than the required mixture; then compare the other two kinds in the same manner. In selecting the pairs to be compared together, it is necessary that the price of one article shall be above, and the other below the price of the mixture. Hence, when there are several articles to be mixed, some cheaper and others dearer than the mixture, a variety of answers may be obtained. Thus, if we compare the highest and lowest, then the other two, the mixture will contain 1 part at 8 cts.; 1 part at 9 cts.; 1 part at 11 cts.; and 1 part at 12 cts. Again, by comparing those at 8 and 11 cts., and those at 9 and 12 cts. together, we obtain for the mixture 1 part at 8 cts.; 2 parts at II cts.; 2 parts at 9 cts.; and 1 part at 12 cts. Other answers may be found by comparing the first with the third and fourth; and the second with the fourth, &c. 61. A goldsmith having gold 16, 18, 23, and 24 carats fine, wished to make a mixture 21 carats fine: what part of each must the mixture contain? 62. A farmer had 30 bu. of corn worth 6s. a bu., which he wished to mix with oats worth 3s. a bu., so that the mixture may be worth 4s. per bu.: how many bushels of oats must he use? NVote 3.-In this example, it will be perceived, that the price of the mixture, with the prices of the several articles and the quantity of one of them are given, to find how much of the other article the mixture muzst contain. Analysis.-Reasoning as above, we find that the mixture (without regard to the specified quantity of corn) in order to be worth 4s. per bu., must contain 2 bu. of oats to 1 bu. of corn. Hence, if 1 bu. of corn requires 2 bu. of oats to make a mixture of the required value, 30 bu. of corn will require 30 times as much; and 2 bu. X 30= 60 bu., the quantity of oats required. 63. A merchant wished to mix 100 gallons of oil worth 80 cts. per gallon, with two other kinds worth 30 cts. and 40 cts. per gallon, so that the mixture may be worth 60 cts. per gallon: how many gallons of each must it contain? 64. A merchant has Havana coffee at 12 cts. and Java at 18 cts. per pound, of which he wishes to make a mixture of 150 lbs., which he can sell at 16 cts. a pound: how numch of each must he use? ART. 468.] ANALYSIS. 307 Note 4.-In this example, the whole quantity to be mixed, the yrFice of the mixture, and the prices of the several articles are given, to fi:ld Acw vaouc of each must be taken. zAnalysis.-On 1 lb. of the Havana it is obvious he will gain 4 cts., and on 1 lb of the Java he will lose 2 cts.; therefore to balance the 4 cts. gain lie must put in 2 lbs. of Java; that is, the mixture must contain 1 part of Havana to 2 parts of Java. Now if 3 lbs. mixture require 1 lb. Havana, 150 lbs, mixture, (the quantity required,) will require as many pounds of Havana as 3 is contained times in 150, viz: 50 lbs. But the mixture contains twice as much Java as Havana, and 50 lbs X 2 —100 lbs. Ans. 50 lbs. Havana, and 100 lbs. Java. 65. It is required to mix 240 lbs. of different kinds of raisins, worth 8d., 12d., 18d., and 22d. a pound, so that the mixture may be worth 10d. a pound: how much of each must be taken? 66. If 10 horses consume 72C quarts of oats in 6 days, how long will it take 30 horses to consume 1728 quarts? Anclysis.-Since 10 horses will consume 720 qts. in 6 days, 1 horse will consume -,l) of 720 qts. in the same time; and -9,-r of'720 qts. is 72 qts. And if 1 horse will consume 72 qts. in 6 days, in 1 clay he will consume ~ of 72 qts., which is 12 qts. Again, if 12 qts. last 1 horse 1 day, 1728 qts. will last him as many days as 12 qts. are contained times in 1728 qts., viz: 144 days. Now if 1 horse will consume 1728 qts. in 144 days, 30 horses will consume them in -l- of the time; and 144 d. — 30=4a. Anls. 30 horses will consume 1728 qts. in 44 days. 468. This and similar examples are usually placed under the rule of CUonzound Proportion, or Double Rule of Tlhee. 67. If 15 horses consume 40 tons of hay in 30 weeks, how many horses will it require to consume 56 tons in 70 weeks-? 68. If 8 men can make 9 rods of wall in 12 dlays, how long wil it take 10 men to make 36 rods? 69. If 35 bbls. of water will last 950 men 7 months, how many men will 1464 b}bls. of water last 1 month? 14 308 ANALYSTS. [SlEC. XTIIJ 70. If 13908 men consume 732 bbls. of flour in 2 months, in how long time will 425 men consume 175 bbls.? 71. If the interest of $30 for 12 months is $2.10, how much is the interest of $1560 for 6 months? 72. If the interest of $750 for 8 months is $28, how much is the interest of $16425 for 6 months? 73. A man being asked how much money he bad, replied that -, -a, and 8 of it made $980: what amount did he have? Analysis.-It is plain that +-~-.=-__. (Art. 202.) The question then resolves itself into this: 8980 are 9} of what sum.? Now if $980 are eH of a certain sum, ~-4 is -4T of $980; and $980.-.49 —$20, and 24 is $20 X 24-$480. Ans. PnOOFa. —{ of $480=$320; -A of 8480=$3600; and { of $480 -$300. Now $320-+$360+$300=8980, according to the conditions of the question. 469. This and similar examples are placed under the rule of Position. The shortest and easiest method of solving them is by Analysis. 74. A sailor having spent l of his money for his outfit, deposited i of it in a savings bank, and had $50 left: how much had he at first? 75. A man laid out - of his money for a house, ~ for furniture, and had $1500 left; how much had he at first? 76. A man lost - of his money in gambling, - in betting, and spent - in drinking; he had $l,259 left: how nmch had he at first? 77. What number is that and -a- of which is 102?'78. What number is that, -l, l, and r- of which is 450? 79. What number is that - and -L of which being added to itself, the sum will be 164? 80. What number is that - of which exceeds — of it by 18? 81. A post stands 40 feet above water, ~ in the -vater, and A in the ground:' what is the length of the post? 82. What will 376 yds. of muslin cost, at 2s. and 6d. per yd.? Analysis.-2s. 6d.=-~~. Now if 1 yd. costs ~L;, 376 yds. wili cost 376 times as much; and ~ X 370 c=C47. Ans. Airs. 469, 470. 1 ANALYSIS. 309 83. If 1 yard of silk costs 50 cents, what will 250 yards cost? Analysis.-50 cts.=$1. Now if 1 yd. costs $-, 256 yds. will cost 256 times as much; and $lX256-$128. Ans. 470. Examples like the preceding, in which the price of a single article is an aliquot part of a dollar, &c., are usually classed under the rule of Practice. Practice is defined by a late English author to be " an abridged mnethod of performing operations in the rule of proportion by means of aliquot Qparts; and it is chiefly employed in computing the prices of commodities." Oas. After giving several tables of aliquot parts in money,.weight, and measure, the same author proceeds to divide his subject into twelve subdivisions or cases, and gives a specfic irule for each case, to be committed to memory by the pupil. It is believed, however, that so maznny specific rules are worse than useless. They have a tendency to prevent the exercise of thought and reason, while they tax the time and memory of the student with a multiplicity of particular directions for the solution of a class of examples, which his commonl sense, if permitted to be exercised, will solve more expeditiously by Analysis. TABLE OF ALIQUOT PARTS OF $1, ~1, AND 1S. Parts of a Dollar. Parts of a Pound sterling. Parts of a Shilling sterling. 50 cts. =- 1 Os. = I 6 pence-= shil. 331 ts.= $ Os. 8d.=X-~ 4 pence = shil. 25 cts. $-L 5s. = -- 3 pence=4 shil. 20 cts.=:$sx 4s. = X~* 2 pence =- shil. 16- cts. = $- 3s. 4d.= - 1~ pence =- shil. 1 2-1 cts.=8 $ 2s. 6d.= — 1 penny=-,ly- slril. 10 cts.- -1 2s. =~ —- ~ penny= -I shil. 81 cts.-M- i s. 8d.=~Ch 7 pence =-s. +-1-Ls, 61 cts. — $ Is1. =~I - 8 pence =-s.-' +s. 5 cts. -2 1 l ls.= +~ - - 9 pence -=s.~+s..iotc.-.Is the price itself is not an aliquot part of $1, or A1, &c., it may sometimes be divided illto such parts as will be aliquot parts of $1, ~1, &c., or which;will be aliquot parts of each other. Thus, 872 cts. is not an al'quot part of $1 but of 871 cts, -50- 25 —-1 cts. Now 501 cts. Now 50 2 cts.; =$; and 12. cts.a=$-. Or thus: 50 cts.:$L, 25 cts. —r of 50 cts., and 121 cts.=_ oQf'25 cts. 310 ANALYS1S. [SECT. XIII. 84. What will 680 bu. of wheat cost, at 87~ cts. per bushel? Analysis. —It is plain, if the price were $1 per bu., the cost of 680 bu. would be $680. Hence, Were the price 50 cts. the cost would be ~ of $680, whicL is $340... c 25 cts. " " 1 of 6080, which is $170 " <" 12~ cts. " " l of -680, which is S 85 But since the price is 50+25+12~ cents, the cost lmust be $595 Or, thus: $1 X 680=$680, the cost at 81 per bushel. At 50 cts., or $L, it will be - of $680, or $340 "25 cts., ~ of 50 cts., " " of $340, or $170 " 12~ cts., of 25 cts.," " of $170, or $ 85 Therefore the whole cost is $595. Ans. 85. What cost 478 yards of cashmere, at 50 cts. per yard? 86. What cost 1560 lbs. of tea, at 75 cts. per pound? 87. What cost 2400 gals. of molasses, at 37~ cts. per gal.? 88. What ost cost 1800 yds. of satinet, at 62 cts. per yard? 89. At 25 cts. per bushel, what cost 1470 bu. of oats? 90. At 33~ cts. a pound, what cost 1326 lbs. of ginger? 91. At 6- cts. per roll, what cost 3216 rolls of tape? 92. At 8L cts. per pound, what cost 4200 lbs. of lard? 93. At 12L cts. per dozen, what cost 1920 doz. of eggs? 94. At 163 cts. a pound, what cost 4524 lbs. of figs? 95. At 66]- cts. per yard, what cost 1620 yds. of sarcenet? 96. What cost cost 840 bu. of rye, at R$ per bushel? 97. What cost 690 yds. of cloth, at 6s. Sd. per yard? Analysis.-At ~1 per yard the cost would be ~690. But 6s. 8d. is ~~; therefore the cost must be ~ of ~690, which is ~230. Ans. 98. What cost 360 gals. of wine, at 16s. per gallon? Analysis.-16s. =10s. -5s. +Is. Now 1Os.=~1; 5s.=~~; ls.=~ of 5s. If the price were ~1 per gal., the cost of 360 gals. would be ~36O0. At 1Os., ~~, it will be ~ of ~360, or ~180 " 5s., ~ of 10s., " 1 of ~180, or ~ 90 " ls., of S.,, " of~ 0, or ~ 18 Therefore the whole cost is ~288.,Ans. ART'. 471. ANALYSIS. 311 90. What cost 1240 yds. of flannel, at 3s. 4d. per yard? 100. What cost 2128 lbs. of spice, at 2s. 6d. per pounld? 101. What cost 5250 yds. of lace, at 6d. per yard? 102 What cost cost 56480 yds. of tape, at lLd. per yard? 47 1. Notwithstanding the law requires accounts to be kept in Federal Money, goods are frequently sold at prices stated in t]he denominations of the old state currencies. When the price per yard, pound, &c., stated in those currencies, is an adiquot part of a dollar, the answer may be easily obtained in Federal MIoney. TABLE OF ALIQUOT PARTS IN DIFFERENT STATE CURRENCIES. Parts of a Dollar, Parts of a Dollar, Parts of a Shilling, New York Currency. New England Currency. N. E. and N. Y. Currency. 4 shil. $, 3 shil. -- 6 pence = shil. 2s. Sd.=$-L 2 shil. =$: 4 pence = shil. 2 shil. =8$ Is. 6d.=$~ 3 pence =4 shil. is. 4d.=8 1 shil. =$-1 2 pence =- shil. I shil. _ - 4s.=$ -— 8- 1 pence =4 shil. 5Ss.-$ ~+8~ 5s.-=$+$L 1 penny=-z shil. N2ote.-l. In N. Y. currency 8s. make $1; in N. E. currency 6s. make $1. From example 103 to 119 inclusive, the prices are given in N. Y. currency; from example 120 to 132 inclusive, they are given in N. E. currency. For the mode of reducing the difierent State currencies to each other and to Federal Money, see Section XVII. 103. At Is. 4d. per yard, what cost 726 yds. of cambric? Analysis.-If.the e price WTere $1 per yard, the cost would be $1X726=$726. But Is. 4d.=-8; therefore the.cost must be X of $726, which is $121. Ans. 104. What cost 896 bu. of wheat at 6s. per bushel? Analysis.-6s.=4s.+2s. Now 4s.=$4-; and 2s.=1 of 4s. At $1 a bushel the cost would be $896. At 4s, $, it will be 4 of $896, or $448 2 m,~Of 4s., " "' of $448, or $224 Therefore the whole cost is $672. Ans. Or, thus: 6s-$a-; therefore the number of bu. minus 4 of itself, -will be the cost, and 896-224 (4 of 896)=672. Ans. $672. 912 ANALYSIS. [SECT. XIII. 105. What cost 752 yds. of balzorine, at 2s. 8cl. per yard? 106. What cost 1232 yds. of calico, at is. 6d. per yard? 107. What cost 763 lbs. of pepper, at is. 3d. a pound? 108. What cost 1116 bu. of apples, at is. 4dcl. per bushel? 109. What cost 1920 yds. of shirting, at is. 2d. per yard? 110. At 6s. a basket, what will 1560 baskets of peaches cost? 111. At 5s. 4d. a pound, what will 1200 lbs. of tea come to? Note.-2. Since 5s. 4d. is 1 less than $1, it is plain 1200-400=$800. Ais. 112. At 7s. per yard, what will 432 yds. of crape cost? 113. At 6s. 8d. a pound, what cost 972 lbs. of nutmegs? 114. At 2s,. 8d, a pair, what cost 864 pair of cotton hose? 115. At ld. a yard, how much will 2800 yds. of tape come to? 116. What cost 1628 yds. of flannel, at 4s. per yard? 117. WVhat cost 2560 bu. of oats, at 2s. per bushel? 118. What cost 9600 lbs. of wool, at 2s. 6d. a pound? 119. What cost 3200 lbs. of sugar, at 6d.'per pound? 120. What cost 600 yds. of damask, at 5s., N. E. cur'., per yard? Nole.-3. 5s. N. E. cur. is - less than $1; hence, 600-100-=$500. AMs. 121. What cost 2500 bu. of potatoes, at is. 6d. per bushel? 122. What cost 1440 yds. of gingham, at 2s. per yard? 123. How much will 4848 chickens cost, at Is. apiece,? 124. How much will 1680 slates cost, at Is. Gd. apiece? 125. How much will 920 turkeys cost, at 4s. 6d. apiece? 126. What cost 4860 lbs. of butter, at ls. 1d. per pound? 127. WThat cost 1260 melons, at 8d. apiece? 128. What cost 2340 lbs. of tea, at 4s. a pound? 129. What cost 040 bu. of peas, at 4s. Gd. per bushel? 130. What cost 720 pair of gloves, at 5s. 3d. a pair? 131. What cost 360 bushels of corn, at 3s. per bushel? 132, What cost 7686 lbs. of butter, at is. per pound? 133, What cost 960 yds. of silk, at 5s. per yard? 134. What will 75 lbs. of butter cost, at $16.80 per cwt.? 135. What will 125 lbs. of wool cost, at 836 per hundred? 136. What will 15 cwt. of hemp cost at $60 per ton? 137. What will 2500 lbs. of iron cost, at $72 per ton? 138. What cost 1' acre of land, at $120 per acre? ARTS. 472-475.] RATIO. 313 SECTION XIV.o RATIO AND PROPORTIONi, ARnT 47 2 In comparing numbers or quantities with each other, we may inquire, either hotw muca greater one of the numnbers or quantities is than the other; or howv many times one of' them contains the other. In finding the answer to either of these inquiries, we discover what is called the relation between the two numbers or quantities. 47 3, The relation between the two quantities thus compared, is of two kinds: First, that which is expressed by their difference. Second, that which is expressed by the quotienzt of the one divideci by the other. 47 4o RATIO is that relation between two numbers or quantities, which is expressed by the guotient of the one divided by the other. Thus, the ratio of 6 to 2 is 6 -- 2, or 3; for 3 is the quotient of 6 divided byT 2. Ons. The relation between two numbers or quantities denoted by their difference, is sometimes called 7'wilAnetical?'crlaio; while that denoted by the quotient of the one divcdeld by the other, is called gcometdicat ratdio. Thus 4 is the arithmetical ratio of 8 to 4; and ti 2 is the geometrical ratio of 8 to 4. But as the term ar'ihmctical'C atio is merely a substitute for the word differeAce, the term difference, in the succeeding pages, is used in its stead; and when the word r'atio simply is used, it signifies that which is denoted by the quotient, of the one divided by the other, as in the article above, 475. The two given numbers thus compared, when spchen of together, are called a couplet; when spoken of separately, they are called the ternms of the ratio. The first termf is the antecedent; and the last, the consequent. QiUESTr. —472. In how many ways are numbers or quantities compared? 474. What is ratio? 475. What are the two given numbers called w lien spoken of together?'When spoket of separately? 314 RATIOo [SECT. XI 476. Ratio is expressed in two ways: First, in the form of a fraction, making the antecedent the tumcn'rator, and the consequent the denominator. Thus, the ratio of 8 to 4 is written -8; the ratio of 12 to 3, A, &c. Second, by placing two points or a colon (:) between the numbers compared. Thus, the ratio of 8 to 4 is written 8: 4; the ratio of 12 to 3, is 12: 3, &c. The expressions -, and 8: 4, are of the same import, and one may be exchanged for the other, at pleasure. Oss. 1. The sign (:) used to denote'ratio, is derived from the sign of division, (~) the horizontal line being omitted. The English mathematicians put the antecedent for the numerator, and the consequent for the denominator as above; but the French put the consequent for the numerator and the antecedent for the denominator. The English method appears to be equally simple, while it is confessedly the most in accordance with reason. 2. In order that concrete numbers may have a'ratio to each other, they must necessarily express objects so far of the same nature, that one can be properly said to be eqral to, or greater, or less than the other. (Art. 294.) Thus a foot has a ratio to a yard; for one is three times as long as the other; but a foot has not properly a ratio to an hour, for one cannot be said to be longer or shorter than the other. 4770 A direct ratio is that which arises from dividing the antecedent by the consequent; as 6- 2. (Art. 474.) 47 S. An inverse, or reciprocal ratio, is the ratio of the reciprocals of two nunbers. (Art. 160. Def. 10.) Thus, the direct ratio of 9 to 3, is 9: 3, or -; the reciprocal ratio is': J, or 4_ i —=; (Art. 229;) that is, the consequent 3, is divided by the antecedent 9. Note.-The term inverse, signifies inverted. Hence, An inverse, or reci2procal ratio is expressed by inverting tThe fraction which expresses the direct ratio; or wheen the notation is b6 points, by inverting the order of the terms. Thus, 8 is to 4, inversely, as 4 to 8. QUrST.-476. In how many ways is ratio expressed? The first? The second? Ob. Which of the terms do the English malthematicians put for the numerator? Which do the French? In order that concrete numbers jmay have a ratio to each'other, what kind of objects lmust they express? 477. What is a direct ratio' 478. What is an inverse ed reciprocal ratio? Etow is a reciprocal ratio expressed by a fractioa? Blow by points. ARTS. 476-481.] RATIO. 315 47 9o A simp2le ratio is a ratio which has but one antecedent and one consequent, and may be either direct or inverse; as 9: 3, or. 480. A complound ratio is the ratio of the products of the corresponding terms of two or more simple ratios. Thus, The simple ratio of 9: 3 is 3; And " c" of 8: 4is2; The ratio compounded of these is 72: 12=6. OBs. 1. A compound ratio is of the same nature as any other ratio. The term is used to denote the origin of the ratio in particular cases. 2. The compound ratio is equal to the product of the simple ratios. Ex. 1. What is the ratio of 27 to 9? Ans. 3. 2. What is the ratio of 8 to 32? Ans. -. Required the ratio of the following numbers: 3. 14 to 7. 13. 324 to 81. 23. 63 lbs. to 9 oz. 4. 36 to 9. 14. 802 to 99. 24. 68 yds. to 17 yds. 5. 54 to 6. 15. 9 to 45. 25. 40 yds. to 20 ft. 6. 108 to 18. 16. 17to 68. 26. 60 miles to 4 fur. 7. 144 to 24. 17. 13 to 52. 27. 45 bu. to 3 pks. 8. 156 to 17. 18. 27 to 135. 28. 6 gals. to 1 hhd. 9. 261 to 29. 19. 53 to 212. 29. 3 qts. to 20 gal. 10. 567 to 63. 20. 47 to 329. 30. ~1 to 15s. 11. 405 to 45. 21. 18 lbs. to 6 lbs. 31. 15s. to ~3. 12. 576 to 64. 22. 28 lbs. to 4 lbs. 32. ~10 to 10d. 4$8 o From the definition of ratio and the mode of expressing it in the form of a fraction, it is obvious that the ratio of two numbers is the same as the value of a fraction whose numerator and denominator are respectively equal to the antecedent and consequent of the given couplet; for, each is the quotient of the numerator divided by the denominator. (Arts. 474, 185.) Ors. From the principles of fractions already established, we may, therefore, deduce the following truths respecting ratios. QE-ST.-47?9. What is a simple ratio? 480. What is a compound ratio? Obs Does it differ in its nature from other ratios? What is the term used to denote? 14:* 316 RATIO. [SECT. XIV. 482. To multiply the antecedent of a couplet by any number, multiplies the ratio by that number; and to divide the antecedent, divides the ratio: for, multiplying the numerator, multiplies the value of the fraction by that number, and dividing the numerator, divides the value. (Arts. 186, 187.) Thus, the ratio of 16: 4 is 4; The ratio of 16 X 2: 4 is 8, which equals 4 X 2; And " 16 - 2: 4 is 2, which equals 4 - 2. OBS. With a given consequent the greater the antecedent, the greater the ratio; and on the other hand, the greater the ratio, the greater the antecelent. (Art. 187. Obs.) 4-83. To multiply the consequent of a coupglet by any number, divides the ratio by thzat number; and to divide the consequent, mnultiplies the ratio: for, multiplying the denominator, divides the value of the fraction by that number, and dividing the denominator, multiplies the value. (Arts. 188, 189.) Thus, the ratio of 16: 4 is 4; The " 16 4 X 2 is 2, whicll equals 4 - 2 And " 16: 4 2 is 8, which equals 4X2. OGs. With a given antecedent, the greater the conseq1tnt, the less the ratio; and the greater the ratio, the less the consequent. (Art. 189. Obs.) 48 4. To C nmutiepiy or divide both the antecedent and consequent of a couplet by the same number, does not alter the ratio: for, multiplying or dividing both the numerator and denominator by the same number, does not alter the value of the fraction. (Art. 191.) Thus, the ratio of 12: 4 is 3; The " 12X2:4X2 is 3; And " 12 -- 2: 4~- 2 is 3. 485. If the two numbers compared are equal, the ratio is a tunit or 1, and is called a ratio of equality. Thus, the ratio of 6X2: 12 is 1; for the value of i-=21. (Art. 196.) QUEST.-482. What is the effect of multiplying the antecedent of a couplet by any num3er? Of dividing the antecedent? 483. MWhat isthe effect of multillying the consequent by any number? Of dividing the consequent? Why? 484. ha-t.is the effect of multiplying or divi ding both the antecedent and consequent by the sane nuinber? Why 485. When the two numbers compared are equal, what is the ratio? What is it called? ARTS. 482-488.1 RATIO. 317 4~86 If the antecedent of a couplet is greater than the consequent, the ratio is yreater than a unit, and is called a ratio of greater inequality. Thus, the ratio of 12: 4 is 3; for the value of'-a=3. (Art. 196.) 4870 If the antecedent is less than the consequent, the ratio is less than a unit, and is called a ratio of less inequality. Thus, the ratio of 3: 6 is -, or ~; for 3:-. (Art. 195.) OBs. I. The direct ratio of two fractions which have a common nemeratmr, is the same as the reciprocal ratio of their denlominators. Thus, the ratio of.-: - is the same as 1-:, or 8: 4. 2. The ratio of two fractions which have a common denominator, is the same as the ratio of their nuvzLeralors. Thus, the ratio of:' is the same as that of 8: 4, viz: 2. Hence, 488. The ratio of any two fractions may be expressed in whole numbers, by reducing them to a common denominator, and then using the numerators for the terms of tlle ratio. (Art. 484.) Thus, the ratio of - to - is the same as -r: -%-', or 6: 2. 33. What is the direct ratio of 4 to 12, expressed in the lowest terms? Ans. -.1 34. What is the inverse ratio of 4 to 12? Ans. -i — I-=3, 35. What is the direct ratio of 64 to 8? Of 9 to 63? 36. What is the direct ratio of 84 to 21? Of 256 to 32? 37. What is the inverse ratio of 4 to 16 Of 28 to 7? 38. What is the inverse ratio of 42 to 6? Of 8 to 72? 39. Which is the greater, the ratio of 63 to 9, or that of 7 2 to 8? 410. Which is the greater, the ratio of 86 to 240, or that of 45 to 72? 41. Which is the greater, the ratio of 120 to 85, or that of 240 to 1io0? 42. Which is the greatesr, t-he ratio of 624 to 416, or that of 936 to 560? 43. Is the ratio of 5X 6 to 24, a ratio of greater, or less inequality? QUnEST. —— 48. When tile antecedent is greater thean the conseqltent, whet is the ratio calletd? 187. If the anltecedetnt is less than the coscequent, what is the -ratio called 488. How,:ay tlhe ratio of two fNCtioj3s jie eixpressed in whole numbers? 318 RATIOc [SECT. XIV 44. Is the ratio of 6 X 9 to 7 X 8, a ratio of greater, or less inequality? 45. Is the ratio of 2X4X 10 to 4 X 32 a rltiO of greater, or le.ss inequality? 46. What is the ratio compounded of the ratios of 5 to 3, and 12 to 4? 47. What is the ratio compounded of 8: 10, and 20: 16? 48. What is the ratio compounded of 3: 8, and 10: 5? 49. What is the ratio compounded of 18: 20, and 30: 40? 50. What is the ratio compounded of 35: 40, and 60:75, and 21 to 19? 51. What is the ratio compounded of 60: 40, and 12: 24, and 25: 30? 489. In a series of ratios, if the consequent of each preceding couplet is the antecedent of the following one, the ratio of the first antecedent to the last consequent, is equal to that compounded of all the intervening ratios. Thus, in the series of ratios 3: 4 4: 7 7.: 10 the ratio of 3 to 16, is equal to that which is compounded of the ratios of 3: 4, of 4: 7, and 7: 16; for, the compound 3X4X 7 3 ratio is 4X7X16i-, or 3: 16. 490. If to or from the terms of any couplet, tewo other num hers having the same r'atio be added or subtracted, the sumns or re mainders will also have the same ratio. (Thomson's Legendre, B. III., Prop. 1, 2.) Thus, the ratio of 12: 3 is the same as that cf 20: 5. And the ratio of the sugm of the antecedents 12+20 to the sulm of the consequents 3 -+5, is the same as the ratio of either couplet. That is, t12+20 12 -20 12+20:3+5::12: 3=20:5, or 3+ 5 —- = 4. So also the ratio of the diference of the antecedents, to the dif. ference of the conseguents, is the same. That is, 20-12 12 2020-12: 5 —3:: 12: 3=20: 5, or -_ 3z-~ — = ArTS. 489-494.] PaROPOTION. 31.9 49 M. If in several couplets the ratios are equal, the sumn of all thA antecedents has the samne ratio to the sum of all the conse. quents, which any one of the antecedents has to its consequezt. ( 12: 4-=3 Thus, the ratio of 1 5: 5 3 18: 6 —3 Therefore the ratio of (12 15 -+ 18): (4 + 5 + 6)=3. OBs. 1. A ratio of gr'eater inequality is diminished by adding the same num, h6er to both terms. Thus, the ratio of 8: 2, is 4; and the ratio of 8+-L:'2 4 is 2. 2. A ratio of less inequality is increased by adding the same lbutmber to bothb trie terms. Thus, the ratio of 2: 8 is a, and the ratio of 2+16: 8+16 is a. PROP ORTION. 492. PRnPORTION is an equality of ratios. Thus, the two ratios 6: 3 and 4: 2 form a proportion; for ~-A, the ratio of each being 2. OBS. The terms of the two couplets, that is, the numbers of which the pro@. portion is composed, are called proportionals. 493, Proportion may be expressed in two ways. First, by the sign of equality (=) placed between the two ratios. Second, by four points (::) placed between the two ratios. Thus, each of the expressions, 1 2: 6=4: 2, and 12: 6:: 4: 2, is a proportion, one being equivalent to the other. The latter expression is read, "the ratio of 12 to 6 equals the ratio of 4 to 2," or simply, "12 is to 6 as 4 is to 2." OBS.!, ie sign (::) is said to be derived from the sign of equality, the fozr poins being merely the extrelmities of the lines. 494. The number of terms in a proportion must at least be four, for the equality is between the ratios of two cotuplets, and eahll couplet must have an antecedent and a consequient. (Art. 476.) There may, however, be a pi-oportion formed from three nzwimbers, for one of the. numbers may be repeated so as to form two QUEST.-492. What is Proportion? 493. I-ow many ways is proportion expressed? What is the first? The second? 494. flow mlany terms must there be in a proportion 1 Why? Call a proportion be formed of three numubers? Howl 320 PREOPORTJIO N. [SECT. XIV terms. Thus, the numbers 8, 4, and 2, are proportional; for the ratio-of 8: 4=4: 2. It will be seen that 4 is the consequent in the first couplet, and the antecedent in the last. It is therefore a mean pro2portional between 8 and 2. OBS. 1. In this case, the number repeated is called the meddle lerm or mean proportional between the other two numbers. The last term is called a third proportional to the other two numbers. Thus 2 is a third proportional to 8 and 4. 2. Care must be taken not to confound propotion with?ratzo. (Arts. 474, 492.) In a simple ratio there are but two terms, an antecedent and a consequent; whereas in a proportion there must at least befoner terms, or two cov~p)lets. Again, one ratio may be greater' or less than another; the ratio of 9 to 3 is greater than the ratio of 8 to 4, and less than that of 18 to 2. One pyro)prtion, on the other hand, cannot be greater or less than another; for erquality does not admit of degrees. 495o The first and last terms of a proportion are called the extremes; the other two, the mneans. OBs. Homologous terms are either the two antecedents, or the two consequents. Analogous terms are the antecedent and consequent of the same couplet. 4-96. Direct proportion is an equality between two direct ratios. Thus, 12: 4:: 9: 3 is a direct proportion. OBs. In a direct proportion, the first term has the same ratio to the second, as the third has to the fourth. 497. Inverse or?reciprocal proportion is an equality between a direct and a reciprocal ratio. Thus, 8:4::1: 1; or 8 is to 4, reciprocally, as 3 is to 6. OBS. In a reciprocal or inverse proportion, the first term hlas the same ratio to the second, as the fourth has to the third. 498. If four nunmbers are propor1tional, the product of the extremes is equal to the product of thie mea6ns. Thus, 8 4:: 6: 3 is a proportion; folr -n =, (Art. 492,) an d 8 X3=X 0. Q.UEST.-Obs. WVhat is the nullmber repeated called t VWhat is the last terni called in such a case? W~hait is the difference between proportion and ratio' 495. Which terms are the extremes? Which the means? Obs. What are homologous terems s Analogous terms. 496. What is direct proportion? Obs. In direct proportion what ratio has the first term. to the second? 497. What is inverse proportion? Obs.- What ratio has the first term to the second in this case? 498. If four numbers are proportioilal, what is the prAduct of the extremes equal to? ARTS. 495-501.] PROPORTION. 321 Again, 12: 6:: -: is a proportion. (Art. 496.) And 12XL=-6X. Ons. 1. The truth of this proposition may also be illustrated in the following manner: The numbers 2: 3:: 6: 9 are obviously proportional. (Art. 492.) For, 3_- = (Art. 195.) Now, Multiplying each ratio by 27, (the product of the denominators,) The proportion becomes2X27 6X7 (At. 2. Ax. 6 Dividing both the numerator and the denominator of the first couplet by 3, (Art. 191,) or canceling the denominator 3 and the same factor in 27, (Art. 221,) also canceling the 9, and the same factor in 27, we have 2X9-6X3. But 2 and 9 are the extremes of the given proportion, and 3 and 6 are the means; hence, the product of the extremes is equal to the product of the means. 2. Conversely, if the product of the extremes is equal to the product of the means, the four numbers are proportional; and if the products are not equal, the numbers are not proportional. 499. Proportion, in arithmetic, is usually divided into Simple and Compound. SIMPLE PROPORTION. ~500. SIMPLE PROPORTION is an equality between two simple ratios. It may be either direct or inverse. (Arts. 479, 406, 497.) The most important application of simple proportion is the solution of that class of examples in which three terms are given, to find a fcurth. 501. We have seen that, if four numbers are in proportion, the product of the extremes is equal to the product of the means. (Art. 498.) Hence, If the product of the means is divided by one of the extremes, the quotient will be the other extreme; and if the product of the extremes is divided by one of the means, the quotient will be the Q.UTr. —-Obs. If the product of the extremes is equal to the product of the means, what ti true of the four numubers If the products are not equal, what is true of them? 499 How is prolportion usually divided? 500. What is simple proportion? What is the most importaut application of it? 501. If the product of the inmeans is divided by one of the extremes, what will the quotient be? If the prodlct of the extremles is divided by one of the means, what will the quotient be? 322'ROPORIvloN [SECT. XIV other mean. For, if the product of two fitctors is divided by one of them, the quotient will be the other factor. (Art. 156.) Take the proportion 8: 4:: 6: 3. Now the product 8X3 3 4= 6, one of the means; So the product 8 X3 6= 4, the other mean. Again, the product 4 X 6 8-3, one of the extremes; And the product 4 X 6 3 8, the other extreme. 5 02~ 2 f, therefore, any three terms of a porop)ortion are tiv,,en, tlae fiuzrtl may be fouznd by dividing the product of two of them by the otiher term. OBS. Simiple Proportion is often called the Rti e of Three, from the circumstance that three tcrms are given to find a fourth. In the older arithmetics, it is also called the Golden Rule. But the fact that these names convey no idea of the nature or object of the rule, seems to be a strong objection to their use, not to say a sufficient reason for discarding them. Ex. 1. If the product of the means is 84, and one of the extremes is 7, what is the other extreme, or term of the proportion? 2. If the product of the means is 54, and one of the extremes is 18, what is the other extreme? 3. If the product of the means is 720, and one of the extremes is 45, what is the other extreme? 4. If the product of the means is 639, and one of the extremes is 213, what is the other extrele? 5. If the first three terms of a proportion are 8, 12, and 16, Nhat is the fourth term? Solution.- 2 X 1 6 = 192, and 192 -8-24, the fourth term, or number required; that is, 8: 12::16: 24. 6. It is required to find the fourth term of the proportion, the first three terms of which are 36, 30, and 24. 7. Required the fourth term of the proportion, the first three terms of which are 15, 27, and 31. 8. Required the fourth term of the proportion whose first three terms are 45, 60, and 90. QJEST. —Obs. What is simple proportion often called? Do these terms convey ail Mea of the nature or object ot the rule? ARTs. 502, 503.] PROPORTION. 323 9. If 8 yds. of broadcloth cost $96, how much will 20 yds. cost at the same rate? Solution. —It is plain that 8 yds. has the same ratio to 20 yds. as the cost of 8 yds., viz: $906, has to the cost of 20 yds. That is, 8 yds.: 20 yds.:0::to the cost of 20 yds. Now 96 X20 -$1 920; anid ~1920 8=r,240. -lns. 10. If 35 men will consume a certain quantity of flour in 20 lays, how long will it take 50 men to consume it? Note. — Since the answer is days, we put the given days for the third term, Then, as the flour will not last 50 men so long as it will 35 men, we put the smaller number of men for the second term, and the larger fbr the first. Operation IMen. Men. I)ays. 50: 35: 20: to the number of days required. 20 Multiply the second and third terms to50) 700 gether, and divide the product by the first 14 days. Ans. term, as in the last example. PROOF.-50 X 14=35 X 20. (Art. 498.) 503. From the preceding illustrations and principles, we deduce the following general RULE FOR SIMPLE PROPORTION. I. Place that number for the third term, which is of the sagme kind as the answoer or number required. II. Then, if by the nature of the question the answer must be greater than the third term, place the greater of the other two num6ers for the second termn; but if it is to be less, place the less of the other two numbers for the second term, and the other for the first. III. Finally, multiply the second and third terms together, divide the p2roduct by the first, and the quotient will be the answer in the same denomination as the third term. PRooF.-Multiply the first tetrm and the answer together, and if the product is equcal to the product of the second and -third terms, the worh is, right. (Art. 500.) QUEST. —503. In arranging the terms in simple proportion, which number is put for the third term'! How arrange the other two numbers? Having stated the question how is the ans wer found. Of what denomination is the answer -low is simple proportion proved X 324 SIMPLE [SECT. XIV. Demonstrationl. -If four numbers are proportional, we have seen that the product of the mecCans is equal to the product of the extremes; (Art. 498;) therefore the prd ict of the seconad and third terms must be equal to that of the first and fourth. But if the product of two factors is divided by one of them, the quotient will be the other; (Art. 156;) consequently, when the first three terms of a proportion are given, the product of the secon7d and tAird terms divided by the first, must give the fon?'th term2 or answver. The object of placing that number, which is of the same kind as the answer, for the thAird term, instead of the second, as is sometimes done, is twofold: 1st, it avoids the necessity of the Ruile of TThree Inverse; 2d, the third term, ir many cases, has no raclio to the first; consequently it is inconsistent with tha principles of proportion to put it fbr the second term. Thus, in the ninth example, if we put:$`96 for the second term, it would read, 8 yds.: $96:: 20 yds.: $240, the answer. But a /'yard can have no ratio to a dollar; for one -annot be said to be,g-reater nor less than the other. (Art. 476. Obs. 2.) OBS. 1. If the first and second terms are compound numbers, reduce them to the lowest denomination mentioned in either, before the multiplication or division is performed. When the third term contains different denominations, it must also be reduced to the lowest denomination mentioned in it. 2. The process of arranging the terms of a question for solution, or putting it into the fbrrmn of a proportion, is called strctisqg the questions. 3. Questions in Simple Proportion, we have seen, may be solved by Analysis. After solving the following examples by proportion, it will be an ex-:ellent exercise for the student to solve them by analysis. (Art. 462. Obs. 2.) 11. If 16 barrels of flour cost $112, what will 129 barrels cost? 12. If 40 acres of land cost $540, what will 97 acres cost? 13. If 641 sheep cost $1923, what will 75 sheep cost? 14. At the rate of 155 miles in 12 days, how far can a man travel in 60 days? 15. How much hay, at $17.50 per ton, can you buy for $350? 16. If $45 buy 63 lbs. of tea, how much will $1540 buy? 17. If 90 lbs. of pepper are worth 72 lbs. of ginger, how many lbs. of ginger are 64 lbs. of pepper worth? 18. A bankrupt compromised with his creditors, at 64 cts. on a dollar; how much will be received on a debt of $2563.50 19. An emigrant has a draft for ~1460 sterling: how much is it vw)rth, allowing $4.84 to a pound? QUEST.-Obs. [f the first and second ternls contain different denominati..ns, how proceed? When thie tb ird terln contains different denominations, what is to be done? What Is meant by stating a question? ART. 504.] PROPORT'ION. 325 SIMPLE PROPORTION BY CANCELATION. 20. If 72 tons of coal cost $648, how mucli will 9 tons cost? Operation. Having stated the question as beTions. To. Doll6s. fore, we perceive the factor 9 is com48 Ans. mnon to the first two terms, and 8:1 ~ e648:~ $1 1 A therefore may be canceled. (Art. Now $648 -8-1,A. 151A) 151.) >X 648 Or thus, 7 —2 the answer. (Art. 503.) 9X648 Q X 648 But - =$8 S81, the same as before. HI-ence, 504. When the first term has factors common to either of the other two terms. Cancel the factors which are common, then proceed according to the r'ule above. (Arts. 151, 221.) PnooF.-Place the answer in the denominator, or on the left of the peaTendicdalr line, as the case mcay be, and if the factors of the divisor exactly cancel those of the dividenzd, the worik is right. OBs. 1. The question should be stated, before attempting to cancel the coinmon factors. When the terms are of different denominations, the reduction of them may sometimes be shortened by Cavncelation. 2. Instead of points, it may sometimes be more convenient to place a perpendicular line between the first and second terms, as in division of fractions. (Art. 231.) In this case the third term should be placed under the second, with the sign of proportion (: ) beforeit to denote its origin, and its relation') the fourth term or the answer. 3. It will be perceived that cancelation is applicable in Simple Proportion ti all those examples, whose first term has one or more factors common& to either of the ollier terms. 21. If 24 yds. of cloth cost $63, what will 32 yds. cost? Operation. g4 yds. $i0 yds., 40 When arranged in this way, the $ l: $l', 21 question is read, 24 yds. is to 320 Ans. 121 X 40=$840. yds., as $63 is to the answer required. 22. If 20 bu. of oats cost ~1, how much will 2 quails cost? 826 SIMPLE [SECT. XIV 23. If 12 bbls. of flour cost 88S, what will 108 barrels cost 9 24. If 30 cows cost $480, what will 173 cows cost? 25. If a man can travel 240 miles in 16 days, how far can he travel in 29 days? 26. If 48 men can build a ship in 84 days, how long would it take 16 men to build it? 27. If ~ of a ton of hay costs X~, what will { of a ton cost? tonl. ton. ~ Solution. —: i: ~: Ans. Now, sX-}X —~2 Ans. Hence, 505e If the terms in a proportion are fractional, the question is stated, and the answer obtained in the same manner as if they were whole numbers. Ons. When the first and second terms are fractions, we may reduce them to a common denominator, and then employ the numerators only; for the ratio of two firactions which have a common.'denominator, is the same as the ratio of their numerators. (Art. 487. Obs. 2.) 28. If - of a cord of wood cost $1.35, what will - of a cord cost? 29. If -I of a yard of berege cost 6 shillings, what will - of a yard cost? 30. If -~- of a yard of sarcenet cost - of a dollar, what will 38 yds. cost? 31. If -3- of a pound of chocolate cost ~ of a dollar, what will 25] pounds cost? 32. What will 165 melons cost, at - of a dollar for 5 melons? 33. A man had 420 acres of land which he wished to divide among his three sons A, B, and C, in proportion to the numbers'7, 5, and 3: how much land would each receive? Solution.-Since the several parts are to be proportional to the numbers 7, 5, and 3, the sum of which is 15, it is evident that the sum of all the given numbers is to any one of them, as the whole quantity to be divided to the part corresponding to the number used as the second term. That is 15'7::420A. to A's share, which is 190 acres; Also 15: 5:: 420A. to B's " " " 140 acres; And 15: 3:: 420A. to C's " " "84 acres. PnROOF.-196+-140n+84-420A. the given number. (Ax. 11.) ARTS 505, 506.] PROPORTION. 327,506. Hence, to divide a given number or quantity into parts which shall be proportional to any given numbers. Place the whole nzumber or quantity to be dividedfor the third term, the sum of the given numrbersfor the first term, and each of the given nunmbers respectively for the second; then multiply ard divide as before. (Art. 503.) 34, A farmer wishes to mix 100 bushels of provender of oats and corn in the ratio of 3 to 7: how many bushels of each must he put in? 35. Bell metal is composed of 3 parts of copper, and 1 of tin: how much of each ingredient will be used in making a bell which weiglhs 2567 pounds? 36. Gunpowder is composed of 76 parts of nitre, 14 of charcoal, and 10 of sulphur: how much of each of these ingredients will it take to make a ton of powder? 37. If 40.12 lbs. of suigar are worth 85.13, how much can be bought for 8125.375? 38. The Vice-President's salary is $5000 a year: if his daily expenses are $10, how much can he lay up? 39. If -rx lb. of snuff cost ~Xa-, what will 150 lbs. cost? 40. If -o off a of a sloop cost 1 500, what will the whole cost? 41. If - of - of an acre of land on Broadway is worth 88200, how much is I of - of an acre worth? 42. A man bought - of a vessel and sold 5 of what he bought for 88240, which was just the cost of it: what was the whole vessel worth? 43. How many times will the fore wheel of a carriage which is 7 ft. 6 in. in circumference turn round in going 100 miles? 44. How many times will the hind wheel of a carriage 9 ft. 2 in. in circumference, turn round in going the same distance? 45. There are two numbers which are to each other as 12 to 34, the smaller of which is 75: what is the larger? 46. What two numbers are those which are to each other as 5 to 6, the greater of which is 240? 47. If two numbers are as 8 to 12, and the less is 320, what is the greater? 328 PROPORTION. [SECer. XIV. 48. There are two flocks of sheep which are to each other as 15 to 20, and the greater contains 500: how many does the less contain? 49. An express traveling 60 miles a day had been dispatched 5 days, when a second.was sent after him traveling 75 miles a day: how long will it take the latter to overtake the former? 50. A. fox has 150 rods the start of a hound, but the hound yluns 8 rods while the fox runs 5 rods: how far must the hound un before he catches the fox? 51. A stack of hay will keep a cow 20 weeks, and a horse 15 weeks: how long will it keep them both? 52. A traveler divided 80s. among 4 beggars in such a manner, that as often as the first received 10s., the second received's., the third 8s., and the fourth 7s.: what did each receive? 53. Pure water is composed of oxygen and hydrogen in the ratic of 8 to 1 by weight; what is the weight of each in a cubic foot of water, or 1000 ounces avoirdupois? COMPOUND PROPORTION. 507 COMPOUND PROPORTION iS an equallity between a conm5pound ratio alnd a simple one. (Arts. 4'79, 480.) ntohus,:3: 12: 3, is a compound proportion. That is, 6X4: 3X2:: 12: 3; for, 0X4X3-3X2X12. OBS. Compound proportion is chiefly applied to the solution of examinles which would require twio or mzore statements in simple proportion. It is sometimes called Double Ruzle of Three. Ex. 1. If 8 men can reap 32 acres in 6 days, how many acres, canl 12 men reap in 15 days? Suggestion.-When stated in the form of a compouend Fpoortiong, the question will stand thus: 8m. ~ 12m. a 6d. 125d.: 32 A.: to tile answer. That is, the product of the antecedents 8 X 6, has the stlme QUEST. —507. What is compound proportion? Obs. To what-is it ciiefl- epplied I What is it sometimes called? A RTS. 507, 508.] PROPORTION. 329 ratio te the product of the consequents 12 X 15, as 32 has to the answer; o, simply, 8 into 6: 12 into 15:: 32: to the answer. Op2eration. The product of the numbers 32 X 12 X 1-5 5760, standing in the 2d and 3d places And 8X 6 —48. divided by the product of those Now 5760 -48: 120. standing in the first place, will Ans. 120 acres. give the answer. lAote.-The learner will observe that it is not the ratio of 8 to 12 alone, -ior that of 6 to 15, whLich is equal to the ratio of 32 to the answer, as it is sometimes stated; but it is the ratio compovrlded of 8 to 12, and 6 to 15, which is equal to the ratio of' 32 to the answer. Thus, 8X6: 12X 15:: 32: 120, the answer. A compound proportion when stated as above, is read, " the ratio of 8X6 is to 12X15 as 32 is to the answer." 2. If 6 men can earn ~42 in 60 days working 8 hours per day, how much can 10 men earn in 84 days working 12 hours a day? 02Opeation. State the question, then Cm.: 10m.. GOd.: 8-ld. i:: ~42: t~o mns mllultiply and divide, as 10X84 X 12 42=423360; and 6 X 0 X 8-2880. Now 423360 -2880-147. Ans. ~147. 50S. From the foregoing illustrations we derive the following general RULE FOR COMPOUND PROPORTION. I. Place that number whlichl is of the same kind as the answer required for the thir d ternm. II. Thenz takce the other numbers in pairs, or two of a kinzd, and a?rrange them as in simple proportion. (Art. 503.) III. dFinally, multi2)1ly together all the second and tliird terms, divide the 1result by the pr.oduct of the first termns, and the quotient iwill be thefouzrth e ter or answver reqtuired. QUEs'r.-508. In stating a question in coinpound proportion, which number do you put for the third term Tlows arrange the other numbers? tHaving stated the question,.how i. the answer founii d1 330 COMPOUND [SECT. XIV. PROOF. — fultiply the answer into all of the first termns or antecedents of the first couplets, and if the product is equal to the continued pTroduct of all the second and third terms, the workl is right. (Art. 498.) OBS. 1. Among the given lnumbers there is but one which is of the same kind as the answer. This is sometimes called the odd term, and is always to be placed for the tfhirdt, term. 2. If the antecedent and consequent of any couplet are compound numbers, they must be reduced to the lowest denomination mentioned in either, before the multiplication is performed. When the thir-d term contains different denominations, it must also be reduced to the lowest mentioned in it. 3. Questions in Compound Proportion may be solved by Analysis; also by Simple Proportion, by making two or more separate statements. 3. If 12 horses can plough 11 acres in 5 days, how many horses can plough 33 acres in 18 days? 4. If a man walking 12 hours a day, can travel 250 miles in 10 days, how long will it take himn to travel 400 miles, if he walks but 10 hours a day? 5. If 40 gallons of water will last 20 persons 5 days, how many gallons will 9 persons drink in a year? 6. If 16 laborers can earn ~15, 12s. in 18 days, how many laborers will it take to earn ~35, 2s. in 24 days? COMPOUND PROPORTION BY CANCELATION. 7. If a person can make 60 rods of wall in 45 days, working 12 hours a day, how many rods can he make in 72 days, working 8 hours a day? Statement. 45d. 72d } Rods., 12hrs. 8hrs..: 60: to the answer. That is, 6,2 4 2XX2- - 64 rods. Ans. Hence, 45X12 40 XYX QuniST.-How are questions in compound proportion proved? Obs. Among the given numbers, how many are of the same kind as the answer? Can questions in compound proportion be solved in any other way? ART. 509.] PROPORTION. 331 509. When the first terms have factors common to thle second or third terms. Cancel the factors which are common, then divide the product of those remaining in the second and third termis by the product of those remaining i.n the i2rsi, and the queotient wvill be t!he anszver. PROOF. —Place the answer in the denominator, or on the left of the peypendcicular line, and i)f the jftctors of the divisor and dividend exactly cancel each other, the work is right. OBs. 1. Instead of placing points between the antecedents and consequents cf the left hand couplets of the proportion, it is sometimes more convenient to put a perpendicular line between them, as in division of fractions. (Art. 232.) This will bring all the terms whose product is to be divided on the right of the line, and those whose product is to form the divisor, on the left. In this case the third term should be placed below the second terms, with the sign of proportion (::) before it, to show its origin, and its?relation to the answer. 2. It will be observed that Cancelaotion can be applied in Compound Proportion to all those examples whosefir'st terms have factors common to the second terms, or to tbe tLi?'rd term. 8. If 24 men can saw 90 cords of wood in 6 days, when the days are 9 hours long, how many cords can 8 men saw in 36 days, when they are 12 hours long?,Operation. $, bm. I 0m. Od. }d., 2 ihrs. 12.hrs. " 0c., lo Ans. 2X12 X 10 —240 cords. 9. If 6 men can make 120 pair of boots in 20 days, working 8 hours a day, how long will it take 12 men to make 360 pair, working 10 hours a day? 10. If 12 men can build a wall 30 ft. long, 6 ft. high, and ft. thick, in 18 days, how long will it take 36 me;n to build on 360 ft. long, 8 ft. high, and 6 ft. thick. 11. If a horse can travel 120 niles in 4 days when the day are 8 hours long, how faro can lie travel in 30 days when the days are 10 hovurs long? QUEST.-509. When the first termsnhave factors common to the secoid or third terms, how proceed? 15 332 CONJOINotD [SECT. XIV. 12. If $25() gain $30 in 2 years, what will be the interest of $750 for 5 years? 13. What will be the interest of $500 for 4 years, if $600 will gain $42 in 1 year? 14. If $300 gain $1.4.40 in 8 months, what will $4800 gain in 32 months? 15. If a family of 8 persons spend $200 in 9 months, how much will 18 persons spend in 12 months? 16. If 15 men, working 12 hours a day, can hoe 60 acres in 20 days, how long will it take 30 boys, working 10 hours a day, to hoe 96 acres, 6 men being equal to 10 boys? CONJOINED PROPORTION. 5 1 0. When each antecedent of a compound ratio is equal in value to its consequent, the proportion is called Conjoined Proportion. OBs. Conjoined Proportion is often called the chtain raule. It is chiefly used in comparing the coins, weights and measures of two countries, through the imedium of those of other countries, and in the higher operations of exchange. The odd term is sometimes called the demzand. 17. If 20 lbs. United States make 12 lbs. in Spain; and 15 lbs. Spain 20 lbs. in Denmark; and 40 lbs. Denmark 60 lbs. in Russia: how many pounds in Russia are equal to 100 lbs. U. S.? Operation. Arrange the given terms in 20 lbs. U. S.-12 lbs. Spain pairs, making the first term the 15 lbs. Spain= 20 lbs. Den. cantecedent, and its equal the 40 lbs. Den. -60 lbs. Rus. consequent; then since it is How many lbs. R.=- 100 lbs. U. S. required to find how many of the last kind are equal to a given number (100 lbs.) of the first, place the odd term or demand under the consequents. Then, 20X15X40: 12X20X60:: 100: Ans. That is 0 12 Cancel the -factors common 1$.20 to both sides, and the produtt g0, 40 6~, 4 of those remaining- on the right 0:: 0o, 10 divided by the product of those Ans. 12X10=120 lbs. on the left, is the answer. ARTS. 510, 511.1 PROPORTION. 33:5 1 1. From these illustrations we derive the following RUILE FOR CONJOINED. PROPORTION. I. Taking the terms in pairs, place the first term on the left of the sign of equality or a- pependicular line for the antecedent, anzd its equal on the right for the consequent, and so on. Then, if the answer is to be of the same kind as the first term, place the odd term under the antecedents; but if not, place it under the consequents. II. Cancel thefactors common to both sides, and if the odd term falls under the consequents, divide the product of the factors remaining on the right by the'product of those on the left, and the quotient will be the answer; but if the odd term falls under the antecedents, divide the product of the factors r'emaining on the left by the product of those on the right, and the quotient will be the answer. PROOF.-Reverse the operation, taking the consequentsfor the antecedents, and the answer for the odd term, and if the result thus obtained is the same cas the odd term in the qiven question, the work is right. Ons. In arranging the terms, it should be observed that the first antecedelnt and the last consequent will always be of the same kind. 18. If 100 lbs. United States, make 95 lbs. Italian; and 19 lbh, Italian, 25 lbs. in Persia; how many pounds in the U. S. are equal to 50 lbs. in Persia? Ans. 40 lbs. 19. If 10 yds. at New York make 9 yds. at Athens; and 90 yds. at Athens, 112 yds. at Canton; how many yds. at Canton are equal to 50 yds. at New York? 20. If 50 yds. of cloth in Boston are worth 45 bbls. of flour in Philadelphia; and 90 bbls. of flour in Philadelphia 127 bales of cotton in New Orleans; how many bales of cotton at New Orleans are worth 100 yds. of cloth in Boston? 21. If $18 U. S. are worth 8 ducats at Frankfort; 12 ducate at Franrkfort 9 pistoles at Geneva; and 50 pistoles at Geneva, t2 rupees at Bombay: how many rupees at Bombay are equal to $100 United States? 3314 DUODECIMALS. [SECT. X V. SECTION XV DUODECIMALS. ART. 5 12. DUODECIMALS are a species of compound numblers, lhe denomtinations of which increase and decrease uniformly in a twelvefold ratio. The denominations are feet, inches or primes, seconds, thi'rds, fourths, fifths, &c. Note.-The term dzuodecimal is derived from the Latin numeral ditodecim, which signifies twelve. TAB L E. 12 fourths (fI) make 1 third, marked "' 12 thirds " 1 second, " U 12 seconds " 1 inch or prime, "9 in. or / 12 inches or primes " 1 foot, " ft. Hence 1' ='-_L of 1 foot. 1 o' —-- i o of 1 in., or - of 1- o f 1 t. —T of 1 ft. I",'= —~ of 1", or a-t of ~- of,-of 1 ft. —r~-,- of 1 ft. OBs. The accents used to distinguish the different denominations below feet, are called indices. 5 1 3. Duodecimals may be added and subtracted in the same manner as the other compound numbers. (Arts. 300, 302.) MULTIPLICATION OF DUODECIMALS. 5 1 4. Duodecimals are principally applied to the measurement of suifaces and solids. (Arts. 285, 286.) Ex. 1. How many square feet are there in a board 12 ft. 7 in. long, and 4 ft.: 3 in. wide? QUTEST.-512. What are duodecirnals? What are the denominations? i.ote. WVihat is the meaning of the term duodecimal? Repeat the Table. Obs. What are the accents called, which are usedl to distinguish the different denominations? 513. Hobw ar duodecimals added and subtracted? 514. To what are duosdecimnals chiefly applied? ARtTS. 512-515.1 DUODEECIMALS, 335 Operation. W'e first multiply each denomination of the 12 ft.'' multiplicand by the feet in the multiplier, be1 ft. 3/ 4 ft. 3' ginning at the right hand. Thus, 4 times 7' 50 ft. 4' are 28', equal to 2 ft. and 4'. Set the 4' 3 ft. 1' 9" under inches, and carry the 2 feet to the next 53 ft. 9,, product. 4 times 12 ft. are 48 ft. and 2 to carry make 50 ft. Again, since 3':= —3- of a ft. and 7'=-7 of a ft., 3' into 7' is -1- of a ft.=21", or 1' and 9' VWrite the 9"' one place to the right of inches, and carry the 1' to the next product. Then 3' or -3 of a ft. multiplied into 12 ft.="LA of a ft., or 36', and 1' to carry malke 37'; but 37'=3 ft. and 1'. Now adding the partial products, the sum is 53 ft. 5/ 9"1. OBs. It will be seen from this operation, that feet multiplied into feet, produce feet; feet into inches, produce inches; inches into inches, produce seconds, &c. That is, the product of any two factors has as many accents as the factors themselves have. Hence, 5 a To find the denomination of the product of any two factors in duodecimals. Add the indices of the two factors together, and the szum will be the index of their pr'oduct. Thus, feet into feet, produce feet; feet into inches, produce inches; feet into seconds, produce seconds; feet into thirds, produce thirds; &ec. Inches infto inches, produce seconds; inches into seconds, produce thirds; inlches into fourths, produce fJfths, &dc. Seconds into seconds, producefourths; seconds into thirde, produce fifths; seconds into sixgts, produce eighths, &c. Thirds into tlirds, produce sixths; thirds into fifths, produce eighths; thirds into sevenths, produce tenths, &'c. Fourths into fourths, produce eighths; fourths into eighths, pro. duce twelfths, &c. Note.-The foo', is considered the unit and has no index. QUEST.-515. HUqV find the denoinin tion of the product in duodecimals? What do'fet into feet produce? Feet into inches? Feet into seconds? What do inches into inches produce? Inches into thirdls? Inches into fourths? Seconds into seconds? Seconds into thirds? Secontls into eighths? Thirds into thirds? Thirds into sixths? 336 DUODECIMALS. [SECT. XV. 5 1 6. From these illustrations we derive the following RULE FOR MULTIPLICATION OF DUODECIMALS. I. Place the several terms of the multiplier under the correspazcd. ing terms of the multi'plicand. II. Miultiply each term of the multiplicand by each term of the multiplier separately, beginning wzith the lowest denomination in the multiplicand, and the highest in the multiplier, acd write the Jfrsi figure of each partial product one place to the right of that of the preceding product, under its corresponding denomination. (Art. 515.) III. Finally, add the several partial products together, carrying I for every 12 both in multiplying and adding, and the sum will be the answer required. OBS. 1. It is sometimes asked whether the inches in duodecimals, are linear, square, or cubic. The answer is, they are neither. An Znclh is I twelfth of a foot. Hence, in measuring surfaces an inch is -jt of a sqlcare fbot; that is, a surface 1 foot wide and 1 inch blong. In measuring solids, an inch denotes Y.of a cubic foot. In measuring lumber, these inches are commonly called carpenter's inlches. 2. Mechanics, also surveyors of wood and lumber, in taking dimensions of their work, lumber, &c., often call the inches afr.rctional part of a foot, and then find the contents in feet and afra'ection of a foot. Sometimes inches are regarded as decimnals of a foot. 3. We have seen that one of the factors in multiplication, is always to be considered an abstract number. (Art. 82. Obs. 2.) How then, can feet be multiplied by feet, inches by inches, &c. It should be observed, that when one geometrical quantity is multiplied by another, some particular extent is to be considered the unLit. It is immaterial what this extent is, provided it remains the same in the different parts of the same calculation. Thus, if one of the factors is one foot and the other half a foot, the former being 12 in., and the latter 6 in., the product is 72 in. Though it would be nonsense to say that a given length is repeated as ofteqn as anowther is long, yet there is no impropriety in saying that one is repeatled as nanzy timncs as there are feet or inches in another. 4. On the principles of duodecimals, it has been supposed that pounds shillings, pence, and farthings can be multiplied by pounds, shillings, pence, and farthings. But it may be asked, what denomination- shillings multiplied by pence, or pence by farthings, will produce 1 It is absurd to say that 2s. and 6d. is r'epealed 2s. and 6d. times. QUEST. —516. What is the rule for multiplication of duodecinals? Obs. What kind of inches are those spoken of in measuring surftces by duodecimals? In measuring solids I ART. 516.] DUODECIIALSo. 337 Ex. 2. How many square feet are there in a piece of marble 9 ft. 7 in. 2"' long, and 3 ft. 4 in. 71" wide? Note.-It is not absolutely necessary to begin to multiply by the highest denomination of the multiplier, or to place the lower denomination to the right of the multiplicand. The result will be the same if we begin with the lowest denomination of the multiplier, and place the first figure of each partial product under the figure by which we multiply. Comm?/onez telod. Second X.iethod. 9o 7ft. 7 2" ft'.' 2' 3 ft. 4' 7" 3 ft. 4' 7" 28 ft. 9' 6" 5' 7" 2"' 2/"" 3 ft. 2' 4" 8"' 3 ft. 2' 4" 8"' 5' 7"t 2"' 1/2" 28 ft. 9' 6".aAns. 32 ft. 5' 51 10"' 2"". Ans. 32 ft. 5' 5" 101/ 211"o 3. How mlany squale feet are there in a board 15 ft. 7 in. long, and 1 ft. 10 in. -wvide? 4. How many cubic feet in a stick of timber 15 ft. 3 in. long, 2 ft. 4 in. wide, and 1 f. 8 in. thick? 5. How niany cubic feet in a block of granite 18 ft. 5 in. long, 4 ft. 2. in. wide, and 3 ft. 6 in. thick? 6. HI-ow many square feet in a stock of 10 boards, 15 ft. 8 in. long, and 1 ft. 6 in. wide? 7. How many square feet in a stock of 15 boards, 20 ft, 3 in. long, and 2 ft. 5 in, wide? 8. Multiply 16 ft. 3' 4" by 6 ft. 5' 8" 10"'. 9. iMultiply 20 ft. 4' 8" 5,"' by 7 ft. 6' 9"1 4'. 10. Multiply 18 ft. 0' 5', 10"' by 4- ft 8' 7" 9"' 11. Multiply 50 ft. 6' 0' 2"' 6"'" by 3 ft. 10' 5",1 12. How mnany cords in a pile of wood 50 ft. 6 in. long, 8 ft. 3 in. wide, and 7 ft. 4 in. high? 13. If a cistern is 30 ft. 10 in. long', 12 ft. 3 in. wide, and 10 ft. 2 in. deep, how many cubic feet will it contain? 14. What will it cost to plaster a room 20 ft. 6 in. long, 18 ft. wide, and 10 ft. high, at 12L cts. per square yard-? 15. IHow many bricks 8 in. long, 4 in. wide, and 2 in. thick, will make a wall 50 ft. long, 10 ft. high, and 2 ft. 6 in.i-thick' 338 EQUATION [SECT. X Vl SECTION XVI. EQUATION OF PAYM IENTS. ART 5 1 EQUATION OF PAYMENTS is the process of findirng the equalized or average time when two or more payments due at different times, may be made at once, without loss to either party. OBS. The equalized or aver'age time for the payment of several debts, due at different times, is often ca.lled the nzean time. 5 ~ From principles already explained, it is manifest, when the rate is fixed, the interest depends both upon the princia ld a the time. (Art. 404.) Thus, if a given principal produces a certain interest in a given time, Double that principal will produce twice that interest; Haylf that principal will'produce half that interest; &c. In double that time the same principal will produce twice tha.t interest; In half that time, half that interest; &c. 5 1 9o Hence, it is evident that any given principal will prem duce the same interest in any given time, as One half that principal will produce in double that time; One ticird that principal wi.ll ". " thrice that time; Twice that principal will " " half that time; Thrice that principal will 6 6"a third of that time; &c. For example, at any given per cent. The int. of 82 for 1 year, is the same as the int, of $1 for 2 prs.; The int. of $3 for 1 year,'" "' 1 for 3 yrs.; &e, The int. of 84 for 1 mo.'. " $1 for 4 mos.; The ilt. of $, for 1 mo. " " " $1 for 5 mos.; &c QuesT.-517. What is eqnation of paylents? OWs. What is the average time for the payment of several debts sometimes called? 518. ~WAhen th6 ra,e is f'Xed, upon wlh does the interest depend? ARTS. 5,17-521.] OF PAYMENTS. 339 5 20~. a7e interest, therefore, of any given principal for I yea?, or 1 month, &dc., is the same, as the interest of 1 dollar for as many years, or months, as there are dollars in the given principacl. Ex. 1. Suppose you owe a man $15, and are to pay him $5 in 10 nmonths, and $10 in 4 months, at what time may both payments be made without loss to either party? A1nalysis.-Since the interest of $5 for 1 month is the same as the interest of $1 for 5 months, (Art. 519,) the interest of $5 for 10. months must be equal to the interest of $1 for 10 times 5 months. And 5 mo. X 10-=50 mo. In like manner the interest of $10 for 4 months is equal to the interest of $1 for 4 times 10 months; and 10 mo. X 4 —40 months. Now 50 months added to 40 months make 90 months; that is, you are entitled to the use of $1 for 90 months. But $1 is X of $15, consequently you are entitled to the use of $15, -ly of 90 months, and 90- -15=6. Ans. 6 months. Prloof. The interest of $5 at 6 per cent. for 10 mo. is $5 X.05-$.25 The interest of 810 "." " 4 mo. is $10X.02=.20:Sum of both $.45 The interest of $15 at 6 per cent. for 6 mo. is 15X.03=-$.45, 521., From these principles we derive the following general RULE FOR EQUATION OF PAYMENTS. First mvultily each debt by the time beibre it becomes due; them divide the sum of the products thus obtained by the sumn of the debts, and the quotient wvill be the average timze requirecd. OBs. 1. If one of the debts is paid dowvn, its product will be nothing; but in finding the sutm of the debts, this payment must be added with the others. 2. When there are months and days, the months must be reduced to days, or the days to the fractional part of a month. 3. This rule is based upon the supposition that discount and ivterest paid i ada1vcc are eqtual. But this is not exactly true; consequently, the rule, though in general use, is not strictly accurate. (Art. 432. Obs. 1.) QUEST. —521. What is the rule for equation of payments'? 15' 310 PARTNERSHIP. [SECT. X VI. 2. If you owe a man $60, payable in 4 months, $120 payable in 6 months, and $180 payable in 3 months, at what time may you justly pay the whole at once? Operation. 8 60X4 = 8240, the same as $1 for 240 months. (Art. 520.) $120X6 = $720, " " " $1 for 720 " $180X3 = 8540, " " 8" $1 for.540 " $360 debts. $1500, sum of products. Now 1500 -- 360=4- months. Ans. 3. A merchant bought one lot of goods for $1000 on 5 months; another for $1000 on 4 months; another for $1500 on 8 months: what is the average time of all the payments? 4. If a man has one debt of $150, due in 3 months; anothei of $200, due in 41 months; another of $500, due in 7 — months: what is the aiverage time of the whole? 5. A man bought a house for $3500, and agreed to pay $500 down, and the balance in 6 equal annual instalments: at what time may he pay the whole? 6. If you owe one bill of $175, due in 30 days; another of $81, due in 60 days; another of $120, due in 65 days, and another of $200, due in 90 days: when may you pay the whole at once? PARTNERSHIP. 522. PARTNERSHIP is the associating of two or mnore indzividuags toyether for the transaction of business. (Art. 464.) The persons thus associated are called partners; and the association itself, a compccany or firm. The money employed is called the capital or stock; and the profit or loss to be shared among the partners, the dividend. CtASE I. — Vhen stock is emnployed an equal length of time. Ex. 1. A and B formed a partnership; A furnished $600 capital, and B $900; they gained $300: what was each partner's share of the gain? QUEST.-522O. What is partnership? What are the persons thus associated called? What is the association itself called? What is the money emIployed-called? What the profit or loss? ARTS. 522, 523.] PARTNERSHIP. 341 Analysis.-Since the whole stock is $600+$900=$1500, A's part of it was a%- —, and B's part was No-w=U. NOW since A put in - of the stock, he must have -- of the gain; and $300 X)-=$120. For the same reason B must have - of the gain; and $300 X==$180. Or, we may reason thus: As the whole stock is to the whole gain or loss, so is each man's particular stock to his share of the gain or loss. That is, $1500 $300:: $600: A's gain; or $120. And $1500: $300:: 900: B's gain; or $180. PROoF.-$! 2 0 + 818 0-3 00, the whole gain. (Art. 21. Ax. 11.) 523. Hence, to find each partner's share of the gain or loss, when the stock of each is employed for the same time. Mi ultiply each mnan's stock by the whole gain or loss; divide the product by the whole stock, and the quotient will be his share of the gain or loss. Or, make each man's stock tem nrumerator, and the whole stock the denominator of a common fraction; multiply the gain or loss by the fraction which expresses each nman's share of the stock, and Ehe product will be his share of the gain or loss. PROOF. —Add the several shares of the gain or loss together. and if the soum is equal to the whole gain or loss, the work is right. (Art. 21. Ax. 11.) OBs. 1. The preceding case is often called Single Fellowship. But since a 7Carntlers7Lip is necessarily composed of tho or nmore individuals, it is somewhat difficult to see the propriety of calling it sinqgle. 2. This rule is applicable to questions in Bankruptcy, and all other operations in which there is to be a division of property in specifiedproporytions. (Arts, 465, 466.) 2. A, B, and C formed a partnership; A put in $12l00 of the capital, B $1600, and C $'2000; they gained $960: what was each map's share of the gain? QuEs'r. —.23. FIow is each man's share of the gain or loss found', when the stock of each is einlployed for the same tise.? 1-low is the operation proved? Ohs. VWhet is t sometimes ca.lled? To what is this rule applicable, 842 GENERAL [SECT X'V 3. A, B, and C entered into partnership; A furnished $2350, B $3200, and C $1820; they lost $3860: what was eacad man's share of the loss? 4. A 1tankrupt owes A $2400, B $4600, C $6800, and D 89000; his whole effects are worth $11200: how much will each creditor receive? 5. A, B, C, and D, engaged in an adventure; A put in 8170, B $160, C $140, and D $130; they made $3000: what was each man's share? CASE II.- When the stocks are employed unequal lengths of time. 6. A and B formed a partnership; A put in $900 for 4 months, and B put in $400 for 12 months; they gained $763: what was each man's share of the gain? Note.-It is obvious that the gain of each depends both upon the capital he furnished, and the time it was employed. (Art. 518.) Analysis.-Since A's capital 0900, was employed 4 months, his share of the gain is the same as if le had put in $3600 for I month; (Art. 519;) for 8900X4-$3600. Also B's capital $400, being employed 12 months, his share of the gain is the same as if he had put in $4800 for 1 month; for ~$400X 12= $4800. The sum of $3600 and $4800 is $8400. Therefore, A's share of the gain must be -4 = -. B's A" " " " -$ 7. Now $763X3=$327, A's share. And 87`63X= $436, B's share. Hence, 524. To find each partner's share of the gain or loss, when the stock of each is employed unequal le2ngths of time..Multiply each partner's stock by thle time it is employed; mace each man's product the numerator, and the sum of the p)roducts the denominator of a common fraction; then mulliplsy t/he whzole gain or loss by eaclh man's fractional share of the stock, and the product xill be his share of tfie gain or loss. Oss. This case is often called Comop0ond or Double Fellowship. QUIZST. —524. XWhen the stoclr of each partner is employed unequa.l lellgths vf time, how is each man's sllhare found? Obs. What is this case sometimes cllted? ARTs. 524-527.] AVER AGE. 343 7. The firm of X, Y, and Z lost $4500; X had $3200 employed for 6 months, Y $2400 for 7 months, and Z $1800 for 9 nmonths: what was each partner's loss? 8. A, B, and C hired a pasture for $60; A put in 15 oxen for 20 days, B 17 oxen for 160 days, and C 22 oxen for 10 days: what rent ought each man to pay? 9. In a certain adventure A put in $12000 for 4 months, then adding 8000 he continued the whole 2 months longer; B put n $25000, and after 3 mlonths took out $10000, and continued the rest for 3 months longer; C put in $35000 for 2 months, then withdrawing - of his stock, continued the remainder 4 months long'er; they gained $15000: what was the share of each? GENERAL AVERAGE. 525. The term Genercll Average, in commerce, signifies the apportionment of certain losses among the different interests corncerned, when a part of the cargo, furniture, &c., of a ship has been voluntarily sacrificed to preserve the rest. (Art. 466.) The property thus sacrificed is called the jettison. 5 2G Losses thus incurred are charged to the shldp, the cow'go, and the frei/ht, pro rata; or according to the value of each. The contributory intterests are to be freed from all charges upon them before the average is made. OBS. 1. In estimating the freight, in New York, one-half, but in most ports one-tMhird is deducted from the gross amount, for seLaen's wages, pilota:ge, and otletr small charges. 2. In the valuation of masts, spars, cables, rigging, &c., of the ship, it is customary to deduct a thbir'd from the cost of replacing them; thus calling the old, two-thirds the value of the new, in making the average. 3. The cargo is valued at the price it would bring at its destined port, after the storage and other necessary charges are deducted. The property sacrificed must be taken into the account as well as that which is saved. 527o General Average may be calculated both by Analysis and P]arnerheisp. (Arts. 464, 522.) Ocs. 1. Losses arising from the ordinary wear and tear, or from a sacrifice, made for the safety of the ship only, or a particular part of the cargo, must be borne by thle individuals who own the property lost, and hot by gen ral OveragP e. 344 EXCHAINGE OF [SECT. XVI. 2. General average is not allowed, unless the peril was imminent, and the sacrifice indispensable for the safety of the ship and crew. 10. The ship Minerva from London to New York, had on board a cargo valued at $75000, of which A owned $30000; B $27000; and C $18000; the gross amount of freight and passage money was $11040. The ship was worth $40000, and the owner paid $520 for insurance on her. Being overtaken by a severe tempest, the master threw $18000 worth of A's goods overboard, and cut away her mainmast and anchors; finally, he brought her into port, where it cost $2796.75 to repair the injury: what was the loss of each owner of the ship and cargo? Operation. Ship valued at o e $40000.00 Less premium for insurance 520.00 $39480.00 Cargo worth... 75000.00 Freight and passage money. $11040.00 Less one-half for wages of crew 5520.00 5520.00 Amount of contributory interests $120000.00 Goods thrown overboard valued at $18000.00 Cost new masts, spars, &c. $2796.75 Less one third for wear of old. 932.25 1864.50 Commissions on repairs., 15.13 Port duties and incidentals. 120.37 Azmouvnt of loss $20000.00 Now $20000X30000 $120000=$5000, loss of A. $20000 X 27000o L$120000= —$450 0 " B. $20000x 18000~o $120000=83000 " C. $20000X394480$120000 3948 1650 " Ship. $20000X 5520 — $120000=$ 920 " Freight. PRooF.-Whole loss (Ax. 11.) $20000, the same as above. Nole,-We may also find what per cent. the loss is; then multiply each contributory interest by the per cent. Thus, since $120000 lose hi 0000, $1 will lose,__l___ of $20000; and 20000+ $120000-=.16-; that is, the loss is 16per cent. Now $30000X.16 —=$5000, A's share of the loss. The loss of the others may be found iqr a similar manner. ARPTS. 528-532.] CURRENCIES. 345 EXCHANGE OF CURRENCIES. 528. The term currency, signifies money, or the circulatiirg medium, of trade. 529, The intrinsic value of the coins of different nations, depends upon their weight and the purity of the metal of which they are made. (Art. 245. Obs.) Oss. For the present standard weight and purity of the coins of the United States, see Arts. 245, 246. For that of British coin, see Art. 248. Obs. 530. The relative value of foreign coins is determined by the laws of the country and commercial usage. O:s. The legal value of a pound sterling in this country has been different at different times. By act of Congress, 1799, it was fixed at $4.4-4. In 1832 its value was raised by the same authority to $4.80; and in 1842, to $4.84. 53 1. The process of changing money from the denominations of one country to its equivalent value in the denominations of another country, is called Exchange of Uurrencies. CASE T.-Reduction of Sterling to Federal Money. Ex. 1. Change ~60 sterling to Federal money. Solution.-Since ~1 is worth $4.84, ~60 are worth 60 times as much, and $4.84 X 60=$290.40. Ans. 2. Change ~8, 7s. 6d. to Federal money. Operation. We first reduce the 7s. 6d. to the decimal $4.84 of a pound; (Art. 346;) then multiply $4.84, 8.375 and ~8.375 together, and point off the prod$40.535 Ans. uct as in multiplication of decimals. Hence, 532. To reduce Sterling to Federal )Money. Miultigly the legal value of one pound, $4.84, by the given nunzmber of poounds, point qff the product as in multiplication of decinzals, and it will be the answer required. (Art. 324.) If the example contains shillings, pence, and farthincs, they.?lust be reduced to the decimal of a p2ound. QSrT. —528. What is meant by currency? 529. On what does the ir trinsic value of the coins of differelnt countries depend 1 530. SIow is the relative value of foreign coins determinei(d Obs. What is the value of a poundc sterling? 531. What is ime nt by ex, change of currencies? 532. How is Sterling money reduced to Federal? 346 EXCHIANGE OF [SECT. XVI. OBs. 1. The reason, of this rule is obvious from the principle that ~5 are worth 5 times as nmuch as ~1, &c. 2.'The rule usually given for reducing Sterllinz to Federal Money, is to reduce the shillings, pence, and farthings to the decimal of a pound, and placing it on the right of the given pounds, divide the whole s an by 9 —. This rule is bas.ed on the law of 1799, whichl fixed the value of a pound at $4.44-, and that of a dollar at 4s. Gd. But $4.44;-4 is 9 per cent. of itself, or 40 cents less than $4.84, which is the present e c~'ta value of a pound; consequently, the result or answer obtained by it, must be 9 per cent. too small. A dollar is now equal to 49.6d. very nearly, instead of 54d. as formerly. 5 3 3. From the preceding rule it is plain that Guineas, Francs )Doubloor zs, and all foreign coins, may be reduced to Federa/l money by multiplying the legal valzue of onze by the given number. Change the following sums of Sterling to Federal money: 3. 3~850, 10s. 8. ~1000, 4s. 6d. 13. 350173, 12s. 6-d. 4.. ~175, I5so 9. ~1600, 8s. 7-~d. 14. ~53262, 13s. 81d. 5. ~8s5, 13s. 6d. 10. ~12531, 10s. 4d. 1. 15. ~76387, 15s. 7-3d. 6. 1~200, 7s. 6d. 11. ~43116, 9s. 10d. 16. ~c58762, 18s. 91d. 7. ~L421, 16s. 4d. 12. ~68318, 10s. 3,?d. 17. ~1000000. CAsE 1I. —]edzuction of.Federal to Sterling Money. 18. Changge $40.535 to sterlinig money. So/gut ion. Since $4.84 are worth ~1, $40.535 are worth as many pounds as $,4.84 are contained times in $40.535; and $410.535 -4.84=8.375; that is ~8.375. Now reducilng the decimal.375 to shillings and pence, (Art. 348,) we have ~8, 7s. 6d. for the answer. Hence, 53 *4 To reduce Federal to Sterling money. Divide the given sumn b7y 84.84, (the value of ~1,) and point oql the quotient as in divigioi of decigmals. The figures on the left hancl of the decinzal point will be pournds; those on the rig ht, decimazcls of a pFound, whiclrh sntst be reduced to shillings, pence, acnd fartlhings. (Art. 348.) Oas. Federal money may be reduced to Guineas, Francs, or any foreign coin, by dividili, the givcnl sltm by the value of one gineaC, oe franc, &c. QU:ST.-Os. I-ovw may foreign coins be reduced to Federal money 534. tlow it- Fed eral noney reduced to Sterling? ARTS. 533-536.] CURRENCIES. 347 Change the. following sums of Federal to Sterling money: 19. $396.88. 23. $2160.50. 27. $25265, 20. 435.60. 24. 975.66. 2. 41470. 02. 876.25. 25. 4275.10. o9. 502S3. 22. 1265.33. 26. 5300.75. 30. 1600'00. 5303 Previous to the adoption of Federal money in 1 7'6, accounts in the United States were kept in pounds, shillings. plence, and farothings. Ons. At the time Federal money was adopted, the colontial czqrencyl or bills of credit issued by the colonies, had more or less depreciated in value: that is, a colonial pound was worth less than a pound Sterling; a colonial shilling, than a shilling Sterling, &c. This depreciation being greater in some colonies than in others, gave rise to the di'Crent vCntes of the Stale currecncies. In N. E. cur., Va., Ky., Tenn., Ia., Ill., Miss., Missou., 6s. or ~f-1-=31. In N. Y. cur., N. C., Ohio, and Mich., - - 8s. or:=$ 1. In Penn. cur., New Jer., Del., and Md., 7s. 6d. (7as.) or ~ —-41. In Georgia cur., and South Carolina, 4s. 8d. (4is.) or ~ s-z =$1. In Canada cur., and Nova Scotia, - - - 5s. or ~-_=1. Ala., La., Ark., and Florida use Federal Money exclusively. CASE III. — eduction o.f ederal Mfoney to State currencies. 31. Reduce 863.25 to New England currency. Solution.-Since $1 contains 6s. N. E. cur., $G3.25 contains 63.25 times as many; and 6s.X63.25=379.50s. Now 379+- 20 =~18, 19s., and.5s.X12-=6d. (Art. 348.) Alns. ~18. 19s. 6d. 53. Hence, to reduce Federal money to State currencies. Multi2ly the given sum by the number of shillings which, in the reguired currency, makce $1, acnd the product zill be the answver in shillings, and decimnals of a shilling. The shillings should be reduced to pounds, and the decimals to pence and fartthingso (Art. 34 8.) 32. Reduce $4450 to New England currency. 33. Reduce $567.50 to New York currency. 34-. Peduce $840.10 to Pennsylvania currency. 35.:iReduce $1500 to Canada currency. QUiST. —535. Previous to the adoption of Federal money, in what were-accounts kIp1 $36. Itow is Federal inonev reduced to the State currencies? 348 FOREIGN MONEYS [SECT. XVI CASR IV. —leduction of State currencies to -Federal Money. 36. tleduce ~23, 12s. Gd. N. E. currency, to Federal money. Solution.-~23, 12s. 6d.=472.5s. (Art. 348.) Now since 6s. N. E. cur. miake $1, 472.Ss. will make as many dollars as 6s. is contained times in 472.5s.; and 472.5s.-6s._78.75. Ans. *758.75. 537. Hence, to reduce State currencies to Federal money. Rieduce the pounds to shillings, and the given pence and fcarthing8 to the decimal of a shilling; then divide this sum by the number of Shilling/s which, in the give.n currency, make $1, and the quotient will be the answer in dollars and cents. Ons. One state currency may be reduced to another by first reducing the given currency to FedeTal money, then to the currency required. 37. Reduce ~160, 5s. N. E. currency, to Federal money. 38. Reduce ~245, 13s. 6d. N. Y. culrency, to Federal money. 39. Reduce ~369, 15s, V7d. Penn. currency, to Federal money. 40. Reduce X~1800, Georgia currency, to Federal money. 41. Reduce ~5000, Canada currency, to Federal money. FOREIGN COINS AND MONEYS OF ACCOUNT. 538. The denominations of money, in which the laws of a country require accounts to be kept, ale called Jifoneys of account. They are generally represented by a coin of the same name; sometimes, however, they are merely nominal, like mills in Federal money. (Art. 245.) 539o Foreign Mlboneys of Account, wuith the p1ar value of the unit established by commercial usage, expressed in Federal MIoney.* Au1stria. —60 kreutzers — florin; I florin, (silver) is equal to $0.485 Beltgium.-100 cents- 1 guilder or florin; t guilder, (silver).40 The coinage of Belgium in 1832, was made similar to that of France. Bencoolten.-8 satellers — soocoo; 4 soocoos=l dollar or rial, - 1.10 Brazil,-1000 rees=l1 milree=$.828. The silver coin, 1200 rees.994 Bremen.-5 schwares-1 grote; 72 grotes-1 rix dollar, (silver).787 3ritisA Iedirc.-12 pice-1 anna; 16 annas=l Co. rupee, (silver).445 The current (silver) rupee of Bengal, Bombay and Madras, is worth.444 Qussa. —537 I-How are the several State currencies reduced to Federal Money. *'CUlloch's Comlmercial Dictionary; Kelly's Universal Cawmnbit. AITS. 537-539.] OF ACCOUNT. 349 Blenos Ayres.-S rials=l dollar currency, (fluctuating) - $0.93 Canton. —10 cash-= candarine; 10can.=l mace; 10 miace1 tael 1.48 The cash, which is made of copper and lead, is said to be the only money coined in.China. Cape of Good Hop2e.-6 stivers=l schilling; 8 schillingsl1 rix dollar.313 Ceylon.-4 pice=1 finam; 12 fanams=l rix dollar -.40 Ciiba.-S rials plate=l dollar; 1 dollar - 1.00 Colombia.* —8 rials —1 dollar; 1 dollar, (variable) mean value - 1.00 Cl/ili. — rials —l dollar; 1 dollar, (silver) - - - 1.00 Denzmzcak.-l2 pfenings=I skilling, 16 skillingsl mar-c; 6 mnarcs= I rigsbank or lix dollar, (silver) - - - - -.52 Eg?/ypt.-3 aspers=1 para; 40 paras=1 piastre, (silver) - -.048 France and Gs'eat B itaine.-See Tables. (Arts. 247, 272.) Greece.-100 lepta=1 drachm; 1 drachdrachm, (silver) -.166 IIollaad. —100 cents=l florin or guilder; 1 florin, (silver) -.40 HaCmba'rg.-12 pfenings-1 schilling or sol; 16 schillings=1 marc Lubs; 3 marcs=l rix dollar. The current marc, (silver)-$.28; marc banco.35 The term Lubs, signifies money of Lubec. The ma'rc cqsereTncy is the common coin; the marc banco is based upon certificates of deposit of bullion and jewelry in the bank of Hamburg. Invoices and accounts are sometimes made out in pozLnds, sc/lilling's, and pence, Flemish, whose subdivisions are like sterling money; the pound Flemish —7 mares banco. apczan.-10 candarines-1 mace; 10 mace=l tael -.75 Java.-100 cents=l florin; 1 florin, as in Netherlands - -.40 Also 5 doits=1 stiver; 2 stivers=1 dubbel; 3 dub.=-1 schilling; 4 schillings:=1 florin -.40 Mallta.-20 granit=l taro; 12 tari=1 scudo; 21 scudi=1 pezza 1.00 Mclauritiz s.-In public accounts 100 cents=l dollar -.968 In mercantile accounts 20 sols=l livre; 10 livres=l dollar. Man1,illac.-34 maravedis1 rial; 8 rials-l dollar, (Spanish) - 1.00 Milavn.-12 denari=1l soldo; 20 soldi=l lirat -.20 2Mexico.-8 rials=l dollar; 1 dollar - - - 1.00 M.lnle Video. —100 centesimos=1 rial; 8 rials=1 dollar - -.833 Naples.-10 grani=l1 carlino; 10 carlini=l1 ducat, (silver) -.80 Ncel/liccl s.-Accounts are kept throughout the kingdom in florins or gullders, and cents, as adopted in 1815. See Holland. Ne2w,Sol/b WCales.-Accounts are kept in sterling money. Noe'vaeay.:t-l 20 skillings=1 rix dollar specie, (silver) - 1.06 Paptal Slzates.-10 bajocchi=l paolo; 10 paoli=l1 scudco or crown 1.00 Pe1. —8 rials=l -dollar, (silver) - 1.00 Venezuela, New Grenada, and Ecuador. t Grani is the plural of grano, tari of taro, scudi of scude lire of lira, pezze of pezza. t Norway has no national gold coin 350 FOREIGN COINS. [SECT. XV1. Portugtal.-400 rees-1 cruzado; 1000 rees-I= milree or crown - $1.12 Priissma.-12 pfenings=-1 grosch, (silver) 30 groscllen=1 thaler or dol..69 Rqtssia.* —100 copeclks=l rouble, (silver) - - -.8 Sae'disia.-100 centesimi=1 lira; 1 liraa=1 fianc, Frehch -.186 Sivedesn.-~12 rundstycksl skilling; 48 skillings=i 1ix dol., specie 1.06 Sicily. —20 grani=l taro; 30 tari=l oncia, (gold) - - 2.40 Saiz.-2 maravedis=:l quinto; 16 quintos=l rial of oid plate -.10 20 rials vellon=l1 Spanish dollar - - - - 1.00 The rial of old plate is not a coin; but it is the denomination in which invoices and exchanges are generally computed. St. Dombr$i'eo.-100 centimes=l1 dollar; 1 dollar - - -.33{ Titscasy.-12 denari di pezza-= soldo di pezza; 2 soldi di pezza= — pezza of 8 rials; 1 pezza, (silver) - - - -.90 T,,rlkey.-3 aspers=1 para; 40 paras=1 piastre, (fluctuating) -.05 Vesice.-100 centesimi=1 lira; 1 lira-1 franc, French - -.181 Formerly accounts were kept in ducats, lire, &c. 12 denari=-1 soldo; 20 soldi=-1 lira piccola; 6~ lire piccole-l ducat current; 8 lire pic. —1 ducat effective. The value of the lira piccola is.09f West Indies, British.-Accounts are kept in pounds, shillings, pence and farthings, of the samne relative value as in England. The value of the pound varies very much in the different islands, and is in all cases less than the pound sterling. 5 40. T/le following coins and moneys of account leave been madcle current in the United States, by act of Congyress, at tie 2rates annexed. f Pound sterling of Gt. Britain, 84.84 Rix Dollar of Bremen,. $0.78Pound of Canada, Nova Scotia, Specie Dollar of Denmark, 1.05 Do. New Brunswick and New- Do. Sweden and Norway,. 1.06 foundland,. 4.00 Rouble, silver, of Russia,..75 Franc of France and Belgium,.186 Florin of Austria,.485 Livre Tournois of France,.185 Lira of Lombardo, Venetian Florin of Netherlands,..40 kingdom,,,.16 Do. Southern States Germany,.40 Lira of Tuscany,...16 Guilder of Netherlands,..40 Do. of Sardinia,..186 Real Vellon of Spain,..05 Ducat of Naples,.. 80 Do. Plate of Spain,...10 Ounce of Sicily, o 2.40 Milree of Portugal,.. 1.12 Leghorn Livres,.,.16 Do. Azores,....83k Tael of China,. 1,.48,Marc Banco of Hamburg,..35 Rupee, Company,. 445 Thaler or Rix Dollar, Prussia, Do. of British Incdia,.. 445 and North. States Germany,.69 Pagoda of India, 1.84 * Previous to 1840, accounlts were kept in paper roubles, 3.1 of wvhich made a silvet rouble. t Laws of United States. AR'rs. 540-543.] EXCHANGE. 351 5 41. Foreign gold and silver coins, at thle sates established by the Custom tHouses oandc commercial usage.* Guinea, English, (gold) $5.00 Leghorn Dollar, (S.) $0.90 Crown, 4" (silver) 1.12 Scuda of Malta, (s.).40 Shilling piece," (s.).23 Doubloon, Mexico, (g.) 15.60 Bank token, " (s.).25 Livre of Neufchatel, (S..26t Florin of Basle, (s.).41 Half Joe, Portugal, (g') 8.53 Mloidore, Brazil, (,.) 4.80 Florin, Prussia, (s.).22Livre of Catalonia. (s.).53k Imperial, Russia, (g.) 7.83 Florence Livre, (s.).15 Rix Dollar, Rhenish, (s.).604 Louis d'or, French, (g.) 4.56 Rix Dollar of Saxony, (s.).69 Crown, " (s.) 1.06 Pistole, Spanish, (g.) 3.97 40 Francs, " (g.) 7.66 Rial " (s.).12l 5 Francs, " (s.).93 Cross Pistareen, (s.).16 Geneva Livre, (s.).21 Other Pistareens, (s.).18 10 Thalers, German (R.) 7.80 Swiss Livre, (s.).27 10 Pauls, Italy, (s.).97 Crown of Tuscany, (s.) 1.05 Jamaica Pound, oa7cialcdl, 3 00 Turkish Piastre, (s.).05 Note.-The true method of estimating the value of foreign coins, is by their wcigr't and pa-rils/y. EiXC HANGE. 5 42. EXCHANGE, in commerce, signifies the receiving or paying of money in one place, for an equal sum in another, by diraft or bill of.Exchange. OBs. 1. A -Pill of Eaxc/aage is a written order, addressed to a person, directing him to pay at a specified time, a certain sum of money to another person, or to his order. 2. The person who signs the bill is called the draower or zak7cer'; the person in whose favor it is drawn, the bityer or'egritter; the person on whom it is drawn, the drawvee, and after he has accepted it, the accepter; the person to whom the money is directed to be paid, the payee; and the person who has legal possession of it, the bolder-. 3. On the reception of a bill of exchange, it should be immediately pre. sented to the dralzwee for his accepltance. 543. The acceptcawce of a bill or draft is a promise to pay it at ncaturity~l or the specified timne. The common method of acceptQUEST.-542. What is meant by exchange? Obs. What is a bill of exchange? Wha is the drawer of a bill? The irawee? The payee? The holder? 543. What is mneast, by the acceptance of a Iill? What is the comanion m ethol of acce)pting a bill 3 * See Manual of Gold and Silver Coins by Eckfeldt & D1 Bois' Ogden on the Tariff of J14; Taylor's Gold and Silver Coin Examiner 352 EXCHANGE. [SECT. XVI. ing a bill, is for the drawee to write his na?me under the word accelted, across the bill, either on its face or back. The drawee is not responsible for its payment, until he has accepted it. OBs. 1. If the payee wishes to sell or transfer a bill of exchange, it is necessary for hhin to endorse it, or write his name on the back of it. 2. If the endorser directs the bill to be paid to a particular person, it is called a special endorsement, and the person named, is called the encdorsee. If the endorser simply writes his name upon the back of the bill, the endorsenent is said to be blank. When the endorsement is bl6alk, or when a bill is drawn payable to the bearer, it may be transferred from one to another at pleasure, and the drawee is bound to pay it to the holder at maturity. If the drawee or accepter of a bill fail to pay it, the endorsers are responsible for it. 54:40. When acce-ptance or payment of a bill is refused, the holder should duly notify the endorsers and drawer of the fact by a legal protest, otherwise they will not be responsible for its payluent. OBs. 1. A protest is a formal declaration in writing, made by a civil officer termed a wtolary public, at the request of the holder of a bill, for its non-acceptasce, or non-payJment. 2. When a bill is returned protested for non-acceptance, the drawer must pay it immediately, though the specified time has not arrived, otherwise he is liable to prosecution. 3. The timze specified for the payment of a bill is a matter of agreement between the parties at the time it is negotiated. Some are p)ayable at sihht, others in a certain nucszber of dlays or moalbhls after sighlt, or after date. When payable after sight or date, the day on which they are presented is not reckoned. When the time is expressed in months, they are always understood to mean calendar months. Hence, if a bill payable in one month is dated the 25th of January, it will be due on the 25th of February. And if it is dated he 28th, 29th, 30th, or 31st of January, it will be due on the last day of February. It is customary to allow three days g'race on bills of exchange. 5 4-5. Bills of exchange are usually divided into inland and foreign bills. When the drawer and dcrawee both reside in the same country, they are termed inland bills or dsrafts; when they reside in different countries, foreign bills. Ors. In negotiating foreign bills, it is customary to draw three of the samer date and amon'0nt, which are called the Fir-st, Second and -Third of Exchanzge; and collectively, a Set of Exchange. These are sent by different shipe or QUcEST. —5-4. When the acceptance or payment of a bill is refuscdi',what should be dcne? Obs. What is a protest? 545. fIow are blills of exechange divided? Obs WVhat is meant by a set of exchange 1 ARTS. 544-547.] EXCHIANGE. 353 conveyances, and when the first that arrives, is accepted or paid, the otncrs become void. The object of this arrangement is to avoid delays, which might arise from accidents, miscarriage, &c. FORM OF A FOREIGN BILL OF EXCHIANGE. Exchange ~1000. BOSTON, Oct. 3d, 1847. At ninety days sight of this first of Exchange, (the second and third of the same date antd tenor unlpaid,) pay George Lewis, Esq., or order, One Thousand Pounds sterling, with or without farther advice. JOHN W. ADAMS. To Tflessrs. ROTIISCHILD & Co. Brokeirs, London. FORM OF AN INLAND BILL OR DRAFT. $2500. NEW YORK, Sept. 27th, 1847. Thirty days after sight, pay to the order of Messrs. Newman & Co., Twenty-five Hundred Dollars, value received, and charge the same to MACY & WOODBURY. gTo MeCssres. D. BIA1 R & Co. lcrc/antlls, N -wt Orleans.!54 o The term par of exchang-e, denotes the slandalrd by which the comparative wiorith of the money of different countries is estimated. It is either intrinsic or comnmercial. The intrinsic par is the real value of the money of different countries, determined by the zveight and purity of their coin. The commnercial par is a nominal value, fixed by law or commercial usage, by which the worth of the money of different countries is estimated. OCs. 1. The intrinsic par remains the same, so long as the staCndard coilns of each country are of the same mnetals, and of the same zweig$ht and pur'itsl; but in case the standard coins are of difclrent metals, the intrinsic par must vary, as the comparative values of the metals vary. 2. The commercial par is conventional, and may at any time be changed. by law or custom. 5v47o By the ternl comssse of exchange is meant the current.price which is paid in one place for bills of a given amount drawn on another place. Ors. 1. The course of exchange is seldom stlationaryJ or at 1pr. It variea QUs'rT.546. What is meant by par of exchane? Inlrinsie par? Comliercial par 7 354 EXCHANGE. [SE CT. XVI according to the circumstances of trade. When the balance of trade is against a country, that is when the exports are less than the imports, bills on the foreign country will be above par, for the reason that there will be a greater demand for them to pay the balance due abroad. On the other hand, when the balance of trade is in favor of a country, fbreign bills will be below P? a,, for the reason that fewer will be required. 2. It should be remarked that the course of exchange can never exceed very much the intrinsic par vcalue; for it is plain that coin or bailio~n instead of bills will be remitted, whenever the course of exchange is such that the ex pense of insuring and transporting it from the debtor to the creditor country, is less than the pre'1imLsu for bills, and the exchange will soon sink to par. 54o. Rates of exchange on Great _Britain are commonly reckoned at a certain per cent., on the old commercial pcar, instead of the new par. OBS. 1. According to the old par, the value of a pound sterling is $4.44~, as fixed by act of Congress. in 1799. According to the newv pcar it is $4.84. The inltinsic value of a ~ ster.,or sovereigcll, according to assays at the U. S. slint, is 4.8t;1. The nezv parz is the value fixed by tile government in 18412, anud is used in calculating dulties, when the invoice is in sterling mloney. 2. The old par is?iqse per cent. less than the nLew pear or legal value; conse-!uently the rate of exchange must reach the nominal premium of 9 per cent before it is at par according to the nsew stanclard. Table of _Excha1nge showing the value of ~1 Sterling frosn 1 to to 12L yer cent. pfresniunm on the old par of $4.44~. Old Par $4.444 5~ per ct. $4.689 8 per ct. $4.800 9 — perct. $4.878 1 per ct. 4.489 6 " 4.711 8~ " 4.811 10 " 4.889 2 " 4.533 6 4.733 8 " 4.822 160 " 4.911 3 " 4.578 7 " 4.756 8- c " 4.833 11 " 4.933 4 4.622 71 " 4.767 9 New Par 4.844 111J " 4.956 4c"c 4.644 7L " 4.77819- perct. 4.856112 " 4.978 5 c" 4.667 77 " 4.789 9 " 4.8(s7 12~" 5.000 Ovte.-1. When exchange is 10 per cent. advance or over, on the old par, it will cause a shipiment of specie to.England; for the freight, interest and insurance will not amount to so much as the premium. When the premium is less than 9 per cent. English funds are, in reality, below their intrinsic par. 2. Tie practice of quoting rates of exchange at the old par, is calculated to lead persons unacquainted with the subject into serious amerccastile mistiskes, and to degrcade our seational cruresncy by making it apearc to foreign nations to be so much below pac'. ARTS. 548, 549.] EXCHANGE. 355 Ex. 1. A merchant negotiated a bill of exchange on London for ~500, 10s., at 8 per cent. premium on the old par: how much did he pay for the bill? Solution. —~500, 10s. —~500.5. (Art. 346.) Now $4.44- X 500.5=$2224.44- at the old par value. Then $2224.44-X.08= 177.95A the premium. The sum paid $2402.40 Ans. Or, the val. of ~1 by table, $4.80X500.5=$2402.40. Ans. 2. A merchant negotiated a bill on Liverpool for ~1000, at 1 per cent. discount from the new par: what did he pay for it? 3. What will a bill cost, on England, for ~5265, 13s. 6d., at 8~ per cent. advance on the old par? 4. How much is a bill worth on France for 1500 francs, at 2 per cent. above par, which is $.186 per franc? 5. What will a bill cost on Paris for 56245 francs, exchange being 5 francs and 54 centimes to the dollar? 6. What cost a bill of exchange on Hamburg for 2000 mares banco, at 1 per cent. above par, which is 35 cts. per mare? 7. What cost a bill of exchange on St. Petersburg for 2560 roubles, at 2 per cent. discount, the par being 75 cts. per rouble? 8. What cost an inland bill of exchange at Boston, on New Orleansi for 815265.85, at 1 per cent. advance? 9. What cost a draft at Albany, on Mobile, for $20260, at 2 per cent. premium? 10. What cost a draft at St. Louis, onl New York, for $35678, at 2- per cent. premium? ARBITRATION OF EXCHANGIE. 549. Arbitration of -Exchange is the method of finding the exchange between two countries through the medium of that of other countries. OBs. 1. 1. When there is but one intervening country, the operation is termed strmyle zarbitration, when more than one, it is termed compyourd arbitration. 2. Problems in Arbitration of exchange are usually solved by conjoined prod portion. (Art. 511.) Care must be taken to reduce all the quantities which are of the same kind, to. the same denomination. 16 356 ALLIGATION [SECT. EC XVI. Ex. 1. If the exchange of New York on London is 8 per cent. advance on old par, or $4.80 for ~1 sterling, and that of Amsterdam on London is 12 florins for ~1, what is the arbitrated ex-,hange of New York on Amsterdam; that is, how many florins are equal to $1 U. S.? Ans. $1 =2-1 florins. 2. A merchant in Baltimore wishes to remit 1200 mares banco to Hamburg, and the exchange of Baltimore on Hamburg is 35 cents for 1 marc. He finds the exchange of Baltimore on Paris is 18 cents for 1 franc; that of Paris on London, is 25 francs for ~1 sterling; that of London on Lisbon, is 180 pence for 3 mrilrees; that of Lisbon on Hamburg, is 5 milrees for 18 marcs banco. How much will he gain by the circuitous exchange? Ains. Direct Ex. $420; circuitous Ex. $375: Gain $45. 3. A man in England owes a man in Portugal ~420; the direct exchange from London to Lisbon is 70d. for 1 milree; but the exchange between London and Amsterdam is 48 florins for ~1 sterling; between Amsterdam'and Paris it is 16 florins for 3 francs; and between Paris and Lisbon it is 6 francs for 2 m1il. rees. Is it better for the man in Portugal to have a direct remittance from London to Lisbon, or a circuitous one through Amsterdam and Paris? ALLIGATION. 550. Alligation is the method of findings the value of a compound or mixture of articles of different values, or of forming a compound which shall have a given value. (Art. 467.) OBs. 1. The term dlligctioln is derived fromn the Latin aZligo, which signifies to bind or tie together. It had its origin in the manner of connecting the numbers together by a curve line in the solution of a certain class of examples. 2. Alligation is usually divided into Afedial and -Alternate. (Art. 467. Obs.) Note.-For a new mlethod of Alligation Alternate, see Key, p. 72. MEDIAL ALLIGATION. 5 51. Medial- Alligation is the process of finding the mean pyrice of a mixture of two or more ingredients or articles of different values. Note.-The term medial is derived from the Latin melilns, signifying a mean or average. ARTS. 550-554. ] ALLIGATION. 357 552. To find the mean value of a mixture, when the quantity and the price of each of the ing'redients are given. -Divide the whole cost of the ingredients by the whole quanttity mixed, and the quotient will be the mnean price of the mixture. PROOF. — ultiply the whole mnixture by the mean price, and if the product is equal to the whole cost, the workl is right. Ex. 1. A grocer mixed 10 lbs. of tea worth 5s. a pound, with 18 lbs. worth 3s. a pound, and 2-0 lbs. worth 2s. a pound: what is the mixture worth per pound? Solution.- 10 lbs. at 5s.-50s. 18 lbs. at 3s.=54s. 20 lbs. at 2s.=40s. Whole quantity 48 lbs. and 144s. whole cost of mixture. Now 144s.- 48 —3. Ans. 3s. a pound. 2. A drover bought S70 lambs at 75 cts. apiece, and 290 sheep at $`31.25 apiece: what is the amean price of the lot per head?. A grocer mixed 12 gals. of wine at 4s. 10d. per gal., with 21 gals. at 5s. 3d., and 29L gals. at 5s. Sd.: what is a gallon of the mixture worth? ALTERNATE ALLIGATION. 5,533. Alternate Alligation is the process of finding what quanlltity of any number of ingredients, whose prices are given, will form a mixture of a yiven meanc price. Note —The term altcernte is derived from the Latin allernaltus, signifying by lturns, and in its present application, refers to the connection of the prices which are less than the ecr7, price, with those which are greater. Alternate alligation embraces three varieties of examples. CASE I. 5 54. To find the quantity of each ingredient, when its price and that of the required mixture are given. I. ti'rite the prices of the engredients under each other, beginning with the least; then connect, with a curve line, each jprice which is less thase that of the mzixture wit/ one or snore of those that are greater; and each greater price with one or more af those that are less. 858 ALLIGATION. [SLcT. XVI. II. Write the dcli rence between the price of the mixture and that of each of the ingredients o2pposite the price with which they are connected. If only one dic'erence stands acsgainst any price, it will denote the quanlity to be taken of that price; but if there ar'e more than one, their sum will be the quantity. OEs. It is immaterial in what manner we select the pairs of ingredients, provided the price of one of the ingredients is less and the other greater than the smean price of the mixture required. PROOF.-Find the value of all the ingredients at their given lrices; if this is equal to the value of the whole mixture at the given price, the work is right. 4. A man'mixed four kinds of oil, worth 8s., 9s., Ils., and 12s. per gal.; the mixture was worth 1Os. per gal.: required, the quantity of each. lst Ans. 2d Ans. 3d Ans. K ) 2I. 8 2g 10 9 r2 =2g. II-{ IgI g. 11 2 - 11 2 = 12 2 g. 12 1 g. 1 2 2+1=3 g. OBS. 1. It is manifest that other answers may be obtained by connecting the prices in a different manner. 2. It is also manifest, if we multiply or divide the answers already obtained by any number, the results will fulfil the conditions of the question; consequently the number of answers is unlimited. 5. A goldsmith has gold of 18, 20, 22, and 24 carats fine: how much may be taken of each to form a mixture 21 carats fine? CASE II. 555. When the quantity of one of the ingredients and the mean price of the mixture are given. Find the diTference between the price of each ingredient and the meatn p-ice qof the required mixtubre, as before; then by proportion, As the difference of that ingredient whose quantity is given, is to each particular difference, so is the quantity givciz to the qu.antity required of each ingredient. ARTS. 555, 556.] ALLIGATION. 359 6. How many pounds of sugar at 10, and 15 cents a pound, must be mixed with 20 lbs. at 9 cents, so that the mixture may be worth 12 cents a pound? Solution.-Connecting the prices as directed, the differences between them and the mean, are 3 cts., 3 cts. and 5 cts. Then 3 cts.: 3 cts.:: 20 lbs.: to the lbs. at 9 cts. Also 3 cts.: 5 cts.: 20 lbs.: " " 10 cts. Ans. 20 lbs. at 9 cts., and 33- lbs. at 10 cts. 7. How much gold of 16, 18, and 22 carats fine must be mixed with 10 oz. 24 carats fine, that the mixture may be 20 carats fine? 8. How much wool at 20, 30, and 24 cts. a pound must be mixed with 95 lbs. at 50 cts. to form a mixture worth 40 cts. a pound? CASE III. 5560. When the quantity to be mixed and the mean price of the required mixture are given. Fiind the dii erence between the price of each ingredient and the mean price of the required mixture, as before; then by plrolportioln, As the sum of the diiferences is to each pcrticular dizference, so is the whole quantity to be mnixed, to the quantity required of each ingredient. 9. A grocer hlas raisins worth 8, 10, and 16 cents a pound: how many of each kind may be taken to form a mixture of 112 lbs. worth 12 cents a pound? Solution.-The sum of the differences between the prices of the ingredients, and the mean price, 6 cts.+4 cts.+4 cts.=14 cts. Then 14 cts.: 6 cts.:: 112 lbs.:-to the lbs. at 16 cts. And 14 cts.: 4 cts.:: i12 lbs.: to the lbs. at 8 and I0 cts. Ans. 48'lbs. at 16 cts., 32 lbs. at 10 cts., and 32 lbs. -t:, 8cts. 10. How much wine at 15, 17, 18, and 22 shillings per gallon, may be mixed to form a mixture of 320 gals. worth 20 shillings per gallon? 11. How much water must be mixed with wine worth 9s. per gal. to fill a pipe, so that the mixture may be worth 7s. per -gal.? 360 INVOLUTION. [SECT. XVII. SECTION XV I. IN VO LUTI ON. Anrt'. 5 57. When any number or quantity is mzultiplied into itsel/; the jg2rod'tlct is cillIed a 2owe7r. Thus, 5 X 5 —25; 3 X 3 X 3 27; 2X2X2X2 12-16; the products 25, 27, and 16 are powers. The oriein-l number, that is, the number which being multi. plied into itself, produces a power, is called the root of all the powers of that number, because they are derived fom it;. 55 8. Po~wers are divided into dig?'rent orders; as the first, secolnd, third, fburthi, fi/ power, &c.'i'lThey take their name from thle nulimber of times the given number is used as a factor, in producing the given power. Ores. 1. The first power of a number is said to be the number itself. Strictly spealking, it is not a power, but a r'oot. (Art. 557.) 3 yards. 2. The second power of a number is also called the sqluare; (Art. 257. Obs. I;) for, if the side of a square is 3 yards, then the product of 3X3-9 yards, wil be the area of the given square. (Art. 285.) But 3X3-9 is also the seco~nd power of 3; hence, it >. is called the squayre. 3. The diacgo~nal of a square is a line connecting two of its opposite corners. 3X3=9 yards. 3 feet. 3. The, ithid power of a number is also called the cube; (Art. 25S. Obs. 1;) for, if the side of a cube's 3 feet, then the product of 3X3X3X327 feee, vili be the solidity of the given cube. (Art. I 281;.) But X3X3X3=27 is also the third power e - of 3f,;-ldrnce it is called the cube. Leg. VII. 11. Sch. C - 4. The jovrth power of a number is called the biq7oadralte. 3)X3( 3-'-J fieet. ~QUEST. —-)57..What is a power? 558. How are powers dividled? —From what do they take their nalne? Obs. Whlat is said to be the first power? What is the secoand powo called? The third? The fourth? ARrs. 557-56 1. INVOLUTION. 361 5 5 90 Powers are denoted by a small figure placed above the given number at the right hand. This figure is called the index or exponent. It shows how many times the given number is employed as a factor to produce the required power. Thus, The index of the first power is 1; but this is commonly omitted; for, (2)'-2. The index of the second power is 2; The index of the third power is 3; The index of the fifth power is 5; &c. That is, 21'2, the first power of 2; 2-'2 X 2, the square, or second poswer of 2; 2' —2X2X2, the cube, or third power of 2; 24- 2 X 2 X 2 X 2, the biquadrate, or fourth power of 2; 25=2X2 X 2 X 2, the fifth power of 2; 26=2 X 2 X 2 X 2 X 2 X 2, the sixth power of 2; &c. Ex. 1. Express the square of 17, and the cube of 19. Ans. 172, 190 Express the given powers of the following numbers: 2. The square of 54. 7. The 2d power of 299. 3. The cube of 43. 8. The 4th power of 785. 4. The square of 87. 9. The 5th power of 228. 5. The biquadrate of 91. 10. The 8th power of 693. 6. The 3d power of 416. 11. The 32d power of 999. 5t3I0 T7Ae process of finding a power of a given number by multiplying it into itself, is called INVOLUTION. 561. Hence, to involve a number to any required power..i dtCi 71 y the given number into itself, till it is taken as a factor, as many times as there are units in the index: of the power to which the number is to be xraised. (Art. 55S.) OBs. 1. The nun6mber of?gulltijlictlions in raising a number to any given power, is one less than —the index of the required power. Thus, 32=3X3; the 3 is taken twice as a factor, but there is but one multiplication. Q(UZST. —55t. How are powers denoted? WVhat is this figure called? What does it shovw? VWiWat is the index. of the first power? Of the second? The third Fifth I 560. What is involution? 561. Hiow is a nuttmber involved to any required Vpbwe:r 362 INVOLUTION. [SECT. XVII 2. A Fraction is raised to a power by multiplying it into itself. Thus, the square of i is 2 a iMixed numbers should be reduced to improper fractions, or the common fraction to a decimal. They may however be involved without reducing them. (Art. 220, Obs.) 3. The process of raising a number to a high power, may often be contraclted by multiplying together powers already found. The index of the power thus found, is equal to the sun of the indices of the powers multiplied together. Thus, 2X2=4; and 44= —2X2X2X2, or 21. So 32X33=3X3X3X3X3, or 35; and 54X53 —57. 12. What is the square of 23? Cbommon Operations. AIznalytic Operation. 23 23=2 tens or 20+3 units. 23 23=2 tells or 20+3 units. 69 60+9 46 400 + 60 529 Ans. And 400+120+9-529. Ans. It will be seen from this operation that the square of 20+3 contains the square of the first part, viz: 20X 20= 400, added to twice the product of the two, parts, viz: 20X 3 +20X 3=120, added to the square of the last part, viz: 3 X 3= 9. Hence, 5620 The square of the sumn of two numbers is equal to the square of the first part, added to twice the product of the two'parts, and the square of the last part. OBs. 1. The product of any two factors cannot have more figures than both factors, nor but one less than both. For example, take 9, the greatest number which can be expressed by one figure. (Art. 34.) And (9)2, or 9X9=81, has two figures, the same number which both factors have. 99 is the greatest number which can be expressed by two figures; (Art. 34;) and (99)2, or 99X 99-9801, has four figures, the same as both factors have. Again, I is the smallest number expressed by one figure, and (1)2, or 1XI =1, has but one figure less than both factors. 10 is the smallest number which can be expressed by two figures; and (10)2, or 10X10=100, has one figure less than both factors. Hence, QvEsr. —Obs. How many multiplications are there in raising a number to a given power? How is a fraction involved? A mixed nunlber? 562. What. is the square of the sum of two numbers equal to? Ohs. How Inany figures are there in the product of any two factors? How imany figures will the square of Ia number contain? The cube? ARTS. 562-564.] EVOLUTION. 363 2. A squgare cann7ot Rhave more figures thpan, double thle number of the root or first power, nor but one less. 3. A cube cannot have mzore figures than triple thle number of the r'oot or Jfist nweIr, 10o? but two less. 4. All powers of 1 are the same, viz: 1; for, 1XIX1X1, &c. =1. 13. What is the square or second power of 123? 14. The cube of 135? 23. The cube of.012? 15. The square of 2880? 24. The square of.00125? 16. The 4th power of 10? 25. The square of -? 17. The 5th power of 5? 26. The cube of A? 18. The 7th power of 6? 27. The square of i-? 19. The 6th power of 7? 28. The cube of -3%i 9-? 20. The 8th power of 4? 29. The square of 4~? 21. The 9th power of 9? 30. The square of 7i? 22. The souare of 2.5? 31. The square of 38?- 9 EVOLUTION. 563. If we resolve 25 into two equal factors, viz: 5 and 5, each of these equal factors is called a root of 25. So if we resolve 27 into three eqtcal fectorts, viz: 3, 3, and 3, each factor is called a root of 27; if we resolve 16 into four equal factors, viz: 2, 2, 2, and 2, each factor is called a 9root of 16. And, universally, when a number is resolved into any number of equal factors, each of those factors is said to be a root of that number. Hence, 564. A root of a number is a factor, which, being mult2plied into itself a certain number of times, will produce that number. Oms. Roots, as well as powers, are divided into djifeirent orders. Thus, when a number is resolved;nto twoo eqital factors, each of these factors is.called the second or square root; when resolved into three equal factors, each of these factors is called the t/hir-d or cube root, &c. Hence, Tlhe naame of tAe root expresses the nuzlmber of equal factors into wqlhich th e given numbcr is to be resolved. Roots. I P 2 3 1 4 5 6' 1 7 1 8 i 9 1 101 11 12 Squares. 1 4| 9j161 251 36 491 64 81 100K I l-7'2 144 Cubes. 1 i 8 127 164' 125 j 216 1 343 l512 1729 1 ]000/ 1331 I 172Si QUEST.-Obs. What are all powers of I? 564. What is a root of a number? Obs. What does the name of the root express? 16' 1364 EVOLUTION. [SECT. XVII 5665 The process of resolvhin/ n~um~bers into equal factors zs called EvoLUTION, or the Extractionz of Rloots. OBs. 1. Evolution is the opposile of involution. (Art. 560.) One is finding a power of a number by multiplying it into itself; the other is finding a r'oot by resolving a number into cqseal factO?'s. Powers and 7'oots are therefore correlatigve terms. If one number is a power of another, the latter is a s'oot of the former. Thus, 27 is the cube of 3; and 3 is the cube root of 27. 2. The learner will be careful to observe, that In sulbttactione, a number is resolved into twoo par'ts; In division, a number is resolved into two facto7's; In evolutiosr, a number is resolved into equzalfactors. 566. Roots are expressed in two ways; one by the radical sigqn (/) placed before a number; the other by a fjactional index placed above the number on the right hand. Thus, V4, or 42 denotes the square or 2d root of 4; V27, or 27 a denotes the cube 4 1 or 3d root of 27; V16, or 164 denotes the 4th root of 16. OBs. 1. The figure placed over the radical sign, denotes the?oot, or the number of eqlrc/ factlors into which the given number is to be resolved. The figure for the sqarcl root is usually omitted, and simply the radical sign / is placed before the given number. Thus the square root of'25 is written,/25. 2. lVhen a root is expressed by a frafctional index, the dclcroomeisator, like the figure over the radical sign, denotes the s'oot of the given number. Thus, (25)2 denotes the square root of 25; (27)3 denotes the cube root of 27. 3. A fractionzal index whose walracrtor is greater thllan 1, is sometimes used. In such cases the deloolriraztor denotes the soot, and the cesaerator7 the polcer of the given number. Thus, 83 denotes the square of the cLube root of 8, or the cbe root of the sqcas'e of 8, each of which is 4. 4. The radical sign ~/, is derived from the letter s, the initial of the Latin radcix, a root. i. Express the cube root of 74. Ans. V74, or 743 2. The square root of 119. 5. The square root of. 3. The 4th root of 231. 6. The cube root of 4. 4. The 9th root of 685. 7. The 4th'root of -. 8. Express the 3d power of tlie 4th root of 6. - s. 6'. 9. Express the 2d power of the 3d root of 81. QUJEST.-565. WMhat is evolution? Obs. Of what is it the opposite? Into what iare painbers resolved in subtraction? In division? In evolution?5 6,6.' Ioorv many: ways are roots expressed? WVhat are they t O~ls. What tloes the figure over the radical sign denote?' What the denoi-llnator of the fractional lhdex'. ARTS. 565-570.] SQUARES AND CUBES. 365 567. A number which can be resolved into equal factors, or whose root can be exactly extracted, is called a perfect power, and its root is called a rational number. Thus, 16, 25, 27, &c., are perfect powers, and their roots 4, 5, 3, are rational numbers. 568. A number, which cannot be resolved into equal factors, or whose root cannot be exactly extracted, is called an imperfect power; and its root is called a Surd, or irrational number. Thus, 15, 17, 45, &c., are imperfect powers, and their roots 3.8+; 4.1 +; 6.7 +, &c., are surds, for their roots cannot be exactly extracted. OBs. A number may be a perfect power of one degree and an imperfect power of another degree. Thus, 16 is a perfect power of the second degree, but an imperfect power of the third degree; that is, it is a perfect square but not a perfect cube. Indeed numbers are seldom perfect powers of more than one degree. 16 is a perfect power of the 2d and 4th degrees; 64 is a perfect power of the 2d, 3d and 6th degrees. 56 9s Every roo t, as well as power of 1, is 1. (Art. 562. Obs. 4.) T.hus, (1)2, (1)3, (1)6, and Vi, v1, VI, &c., are all equal. PROPERTIES OF SQUARES AND CUBES. 5 7 @, The properties of numbers in general, have already been given. The following pertain to square and cubic numbers. i. The product of any two or more square numbers, is a square; and the product of any two or more cubic numbers, is a cube. Thus 22X3 =36, the square of 6; and 2 X3 =216, is the cube of 6. 2. If a square number is divided by a square, the quotient will be a square. Thus, 144+. 9=16, which is the square of 4. 3. If a square number is either multiplied or divided by a number that is not a square, neither the product nor quotiezt will be a square. 4. If you double the number of times a number is taken as a factor, it will not produce double the product, but the square of it. Thus, 3X3-9, and 3X3 X3X3=8l, and not 18. 5. The product of two differclnt prime numbers cannot be a square. 6. The product of no two different numbers, which are prime to eacrl other, will muake a square, unless each of those numbers is a square. 7. The square and cube of an even number are even?; and the square annd cube of an odd number are odd. (Art. 161. Prop. 6, 10.) Hence, QUEST. —567. WVhat is a perfect power? What is a rational nubtler? 568. An imper. fect power? A surd? Obs. Are numlbers ever perfect powers of one degree and imperfect powters of another degree? 569. What are all roots and powers of I 1 366 SQUARES AND, CUJBES. [SECT. XVI1, 8. The square or cube root of an even number, is even; and the square or cube root of an odd number, is odd. 9. Every square number necessarily ends with one of these figures, 1, 4, 5, 6, 9; or with an even numnber of ciphers preceded by one of these figures. 10. No number is a square that ends in 2, 3, 7, or 8. 11. A cubic number may end in any of the natural numbers, 1, 2, 3, 4, 5, 6 7, 8, 9, or 0. 12. All the powers of any number, ending in 5, will also end in 5; and if a number ends in 6, all its powers will end in 6. 13. Every square number is divisible by 3, and also by 4, or becomes so when dimntished by cn~ity. Thus, 4, 9, 16, 25, &c., are all divisible by 3, and by 4, or become so when diminished by 1. 14. Every square number is divisible also by 5, or becomes so when increased or diminished by unity. Thus, 36-1, and 49+-1, are divisible by 5. 15. Any even square number is divisible by 4. 16. An odd square number, divided by 4, leaves a remainder of 1. 17. Every odd square number, decreased by unity, is divisible by 8. 18. Every number is either a square, or is divisible into two, or three, or Jowln squares. Thus 30 is equal to 25+4+1; 33=16+16+1; 63=49+9+4 —1. 19. The product of the suem and differeince of two numbers, is equal to the difference of their squares. Thus, (5+3)X(5-3): _16; also 52-32=16 20. If two numbers are such, that their squarles, when added together, form a squar'e, the product of these two numbers is divisible by 6. Thus, 3 and 4, the sum of whose squares, 9 —16i=25, is a square number, and their product 12, is divisible by 6. Hence, 21. To find two numbers, the sltm of whose squares shall be a square number. Talke anyJ two numbers and mbnltiply thevm together; t/e (double of thcir produIt will be oTbe of the nse.mbers souglbt, and the' diftrenee of th.eir sqtuar'es will be the other. Thus, take any two numbers, as 2 and 3; the double of their product is 12, and the difference of their squares is 5; now 122+52=-169, the square of 13. 22. When two numbers are such, that the difference -of their squares is a square number; the ssesa and difference of these numbers are themselves square numbers, or the double of square numbers. Thus, 8 and 10 give for the difference of their squares 36; and 18, the sum of these numbers, is the double of 9, which is a square number, and 2, their difference, is the double of 1, which is also a square number. 23. If two numbers, the difference of which is 2, be multiplied together, their product increased by unity, will be the square of the intermlediate number. 24. The sum or de'erzece of two numbers, will measure the differelce of their squares. 25. The sum of two numbers, differing by unity, is equal to the dier~ece of their squares. 26. The sum of two numbers will measure the sunt of their cubes; rand the difference of two numbers will measure the differenece of their cubes. ART, 571.] SQUARE ROOT. 367 27. If a square measures a square, or a cube a cube, the'root will also incas. uore the root. 28. If one number is prime to another, its square, cube, &c., will also be prime to it. 29. The difference between an integral cube and its root, is always divisir ble by 6. 30. If any series of numbers beginning from 1, be in continued geometrical proportion, the 3d, 5th, 7th, &c., will be squares;* the 4th, 7th, 10th, &c., cubes; and the 7th will be both a square and a cube. Thus, in the series, 1, 2, 4, 8, 16, 32, 64, &c., the 3d, 5th, and 7th terms are squares; the 4th and 7th are cubes; and the 7th is both a square and a cube. EXTRACTION OF THE SQUARE ROOT. 5 7 1. To extract the SQUARE ROOT, is to resolve a given number into two equal factors; or, to.find a number which being multiplied into itself, will produce the given number. (Art, 564. Obs.) Ex. 1. What is the square root of 36? Solution.-Resolving the given number into two equal factors, we have 36-6 X 6. Atns. The square root of 36 is 6. 2. What is the length of one side of a square field which contains 529 square rods? Operation. Since we may not see what the root of 529 529(23 is at once, we will separate it into two periods 4 by placing a point over the 9 and another over 43)129 the 5. Now the greatest square of 5, the left 129 hand period, is 4, the root of which is 2. Placing the 2 on the right of the number, we sub. tract its square from the period 5, and to the right of the re. mainder bring down the next period. We then double the 2, the part of the root already found, and, placing it on the left of the dividend for a partial divisor, we perceive it is contained in the dividend, omitting its right hand figure, 3 times. Placing the 3 on the right of the root, also on the right of the partial divisr, we multiply the divisor thus completed by 3, and subtract the prodtct from the dividend. The answer- is 23 rods. QUEST.-571, What is it to extract the square root of a number 3 368 saUARE ROOT. [SECT. XVII Note,-Since the root is to contain 2 figures, the 2 stands in tens place, hence the first part of the root found is properly 20; which being doubled, gives 40 for the divisor. For convenience we omit the cipher on the right; and to colrmpensate for this, we omit the right hand figure of' the dividend. This is the same as dividing both the divisor and dividend by 10, and therefore does not alter the quotient. (Art. 146.) 5 7 2. Hence, we derive the following general RULE FOR EXTRACTING THE SQUARE ROOT. I. Separate the given number into periods of two figures each, by p2lacing a point over the units figures, then over every second figure towards the left in whole numbers, and over every second figure towards the right in decimals. II. Find the greatest square number in the first or left hand period, and place its root on the right of the number for the first figure in the root. Subtract the squate of this figure of the root,o'0m the period under consideration; and to the right of the re-.mainder bring down the next period for a dividend. III. Double the root just found and place it on the left of the dividend for a partial divisor; find how many times it is contained in the dividend, omitting its right hand figure; place the quotie~nt on the right of the root, also on the right of the cpartial divisor; multiply the divisor thus completed by the figure last placed in the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. IV. Double the root already found for a new partial divisor, divide, &c., as before, and thus continue the operation till the root of all the periods is extracted. If there is a remainder after all the periods are broug/ht down, the operation may be continued by annexing periods of cipher~s. PROOF.-Multip2ly the root into itself; and if the product. is equal to the given number, the worlc is right. (Art. 564.) 573 Dcmonstlration.-Take any number as that in the last example; then separating it into parts, 529 —500+29. Now the greatest square in 500 is 400, the root of which is 20, with a remainder of 100; consequenily, the first part of GQrIEST.-572. What is the first step in extracting lhe sq(tare root? T4le.secondi? Thirdl Fourth? WShen there is a rerliinder, hoxe proceetl? Ilow is the square'root provedl ARTS. 572-574.1 SQUARE ROOT. 369 the root must be 20, and the true remlnainder is 100+29, or 129. And since there are three figures in the given number, there must be two figures in the root; (Art. 562. Obs. 2;) but the square of the sum of two numbers, is equal to the square of the first part adled to twice the product of the two parts and the sqear'e of the last part; it follows therefore that the remainder 1.29, must be twice the product of' 20 into the part of the root still to be found, together with the square of that part. (Art. 562.) Now dividing 129 by 40 the double of 20, the quotient is 3, which being added to 40 makes 43; finally, multiplying 43 by 3, the product is 129, which is manifestly twice the product of 20 into 3, together with the square of 3. In the same manner the operation may be proved in every case. (For illustration of this rule by geometrical figures, see Practical Arithmetic, p. 318.) I. The season for separating the given numbers into per'iods of two fiqugres each, is that a squlacre number can not have.ore figques than doeuble the number of figures in the root, nor but one less. It also shows how nnase/nyfg'r'es the root will contain, and thus enables us to find pCeat of it at a time. (Art. 562. Obs. 2.) 2. The reason for doubling that part of the root already found for a divisor, is because the remainder is double the product of the first part of the root into the second part, together with the sqycare of the second part. 3. In dividing, the right haend fJigu~re of the dividendlis onmitted, because the cipher on the right of the divisor being omitted, the quotient would be 10 times too laroge for the next figure in the root. (Arts. 130, 146.) 4. The last figure of the root is placed on the light of the divisor simply for convenience in multiplying it into itself. OBS. 1. The product of the divisor completed into the figure last placed in the root, cannot exceed the dividend. Hence, in finding the figure to be placed in the root, some allowance must be made for ca'rr'ying, when the product of this figure into itself exceeds 9. 2. If the right hand period of decimals is deficient, it must be completed by annexing a cipher to it. 3. There will always be as many decimal figures in the root, as there are periods of decimals in the given number. 57 4 The square root of a common fraction is found by ex.. tracting the root of the numerator and denominator. A mixed number should be reduced to an improper fiaction. When either the numerator or denominator of a common fiaction is not a pesfect square, the fraction may be reduced to a decimal, and the acproxioimate root be found as above. QUEST. —573. Dest. Why do we separate the given number into periods of two figures each? Why double the root thus found for a divhsor? Why otmit the right hand figure of the dividend? Why place the last figure of the root on the right of the divisor? Obs. fHow many decimal figures will there be in the root? 574. How is the-square root of a conmmon fraction found? Of a mixed number? 370 SQUARE ROOT. LSECT. XVII. Required the square root of the following numbers: 3. 2601. 10. 27889 17. 566.44. 24. _iA 4. 5329. 11. 961. 18. 7.3441. 25. -a-. 5. 784. 12. 97. 19..81796. 26. i. 6, 87. 13. 7. 20. 1169.64. 27. 17A. 7, 4761. 14. 190. 21. 627264. 28. 794~. 8. 7056. 15. 43681. 22. 3.172181. 29. 207 -T 9. 9801. 16. 47089. 23. 10342656. 30. 34967ih. 31. What is the square root of 152399025? 32. What is the square root of 119550669121? 33. What is the square root of 964..5192360241? 57 5. When the root is to be extracted to many figures, the operation may be contracted in the following manner. First find half, or one more than half the numnber of figures reguired in the root; then having found the next true divisor, cut of its right hand.figure, and divide the remainder by it; place the quotient in the root, and continue the operation as in contraction of division of decimals. (Art. 333.) 34. Required the square root of 365 to eleven figures in the root. Ans. 19.104973174. 35. Required the square root of 2 to twelve figures. 36. Required the square root of 3 to seventeen figures. APPLICATIONS OF THE SQUARE ROOT. 576. A triangle is a figure which has three sides and three angles. When one of the sides of a triangle is p)e)rpend'icular to another side, the angle between them is called a rifght-anyle. 577. A right-angled triangle is a triangle which has a right-angle. / The side opposite the right-angle is called the hypothenuse, and the other O i0 / two sides, the base and perpendicular. / The triangle ABC is right-angled at B, and the side AC is the hypothenuse. A Base. B ARTs. 575-580.] SQUARE ROOT 371 57 8. The square described on the /hypothenuse of a rightangled triangle, is equal to the sum of the squares described on the other two sides. (Thomson's Legendre, B. IV. 11, Euc. I. 47.) Thle truqth of this principle may be seen friom the following geometrical illustr'ation. Thus, Let the base AB of the rightangled triangle ABC be 4 feet,'he perpendicular AC, be 3 feet; hen will the squares described on the base AB, and the perpendicular AC, contain as many square feet as the square described on the hypothenuse BC. Now (4)2+(3)2=25 sq. ft.; and the square described on BC also contains 25 sq. ft. Hence, the square described on the hypothen~bse of any righ/t-angled triangle, is equal to the sum of the squar'es described on the other two sides. OBs. Since the square of the hypothenuse BC, is 25, it follows that tle V25, or 5, must be the hypothenuse itself. Hence, 57 9. When the base and perpendicular are given, to find the hypothenuse. Add the square of the base to the square of the perpendicular, and the square root of the sunm will be the hypothenuse. Thus, in the right-angled triangle ABC, if the base is 4 and the perpendicular 3, then (4)2 +-(3)2 =25, and 25 = 5, the hypothenuse. 580. When the hypothenuse and base are given, to find the perpendicular. From the square of the hypothenuse subtract the square of the base, and the square root of the remainder will be the perpendicular. QUCST. —576. What is a triangle? What is a right-angle? 577. What is a rightangled triangle? What is the side opposite th, right-angle called' What are the other two sides called? 578. What is the square described on the hypothenuse equal toe 579. When the base and perpendicular are given, how is the hypothenuse found I 580. Whet the hypothenuse and base are given, how is the perpendicular found? 372 S(BUARE ROOT. [SECT. XVII. Thus, if the hvpothenuse is 5 and the base 4, then (5)2-(4)2 -9, and V9=3, the perpendicular. 5 8 1. When the hypothenuse and the perpendicular are given, to find the base. From the square of the hypothenuse subtract the s/quare of the perpendicular, and the square root of the remainder will be the base. Thus, if the hypothenuse is 5 and the perpendicular 3, then (5)2 _(3)2=16, and V16=4, the base. Ouis. 1. From the preceding principles it is manifest that the Carec of a square may be found by dividing the square of its hypothenuse by'2. (Arts. 285, 578.) 2. The areas of all similar' figurfes are to each other as the squab'es of their /lontoloe,,ols sides, or their like dimensions. (Leg. IV. 25, 27. V. 10.) Hence, The s.1_n of the areas of equilateral or other similar tirianqgles, also of simrilar poltygons, circles and semicircles described on the base and perpendicular of a right-angled triangle, is equal to the area of a similar figure described on the hypothenuse.:3. The squ.agre of a simple ratio is called a duplicate ratio; the cube of a simple ratio, a triplicate ratio. The ratio of the square roots of two numbers is called a sub-duplicate ratio; that of the cube roots, a sub-tripl2icate ratio. Ex. 1. If a street is 28 feet wide, and the height of a tower is 96 feet, how long must a rope be to reach from the top of the to wer to the opposite side of the street? Solution.-(96)2 +(28)2=10000, and /10000=100 ft. Ans. 2. A ladder 40 feet long being placed at the opposite side of a street 24 feet wide, just reached the top of a house: how high was the house? 3. Two ships, one sailing 7 miles, the other 12 miles an hour, spoke each other at sea; one was going due east, the other due south: how far apart were they at the expiration of 12 hours? 4. What is the length of the side of a square farm which contains 360 acres; and how far apart are its opposite corners? 582. A mean proportional between two numbers is equal Io the square root of their product. (Arts. 494, 498. Obs. 2.) QUErsT. —i"".o When the hypothenuse and perpendicular are given,. how is the base found X ARTS. 581-584.] SQUARE ROOT, 373 5. Find a mean proportional between 2 tmd 8. Solultion. —8X2=16; and V16=4. Ans. Find a mean proportional between the following numbers: 6. 4 and 25. 10. 121 and 36. 14. - and I-. 7. 9 and 36. 11. 196 and 144. 15. - and -%a. 8. 16 and 81. 12. 2.56 and 49. 16. and and'9 64 and 25. 13. 6.25 and 729. 17 9 and.5 58 3. To find the side of a square equal in area to any given surface. Extract the square root of the given area, and it will be the side of the square soaght. OBS. When it is required to find the dimensions of a rectangular field, equal in area to a given surface, and whose length is double, triple, or quadruple, &c., of its breadth, the square root of h, ~, ~, of the given surface, will be the widItA; and this being doubled, tripled, or quadrupled, as the case may be, will be the length. 18. What is the side of a square equal in area to a rectangular field 81 rods long, and 49 rods wide? 19. What is the side of a square equal in area to a triangular field which contains 160 acres? 20. What is the side of a square equal in area to a circular feld which contains 640 acres? 21. What are the length and breadth of a rectangular field which contains 480 acres, and whose length is triple its breadth? 22. A general arranged 10952 soldiers, so that the number in rank was double the file: how many were there in each? 5 4. When the sum of two numbers and the difference of their squares are given, to find the numbers. )Divide the dif'erence of their squarces by the sunm of the numbers, and the quotient will be their difference; then proceed as in A.rt. 155. 23. The sum of two numbers is 42, and the difference of their squares is 756: what are the numbers? Ans: 12 and 30. 24. The sum of two numbers is 65, and the difference of their squares is 975: what are the numbers? 374 CUBE ROOT. [SECT. XVII 585. When the difference of two numbers and the difference of their squares are given, to find the numbers. Divide the dilerence of the squares by the difference of the numbers, and the quotient will be their sum; then proceed as in Art. 155. 24. The difference of two numbers is 29, and the difference of their squares is 1885: what are the numbers? EXTRACTION OF THE CUBE ROOT. 586. To extract the cube root, is to resolve a given number into three equal factors; or, to find a number which being multiplied into itself twice, will produce the given number. (Art. 564.) Ex. 1. What is the cube root of 64? Solution.-Resolving the given numbers into three equal factors, we have 64=4X4X4. Ans. 4. 2. What is the cube root of 12167? Operation. -Wre first separate the Col.. Col. II. 12167(23 given number into two pe1st term 2 4X2= 8 riods, by placing a point 2d " 4 1200 divisor,) 4167 over the units' figure, then 3d " 63 1389X3 — 4167 overthousands. This shows us that tlie root must have two figures, (Art. 562. Obs. 3,) and thus enables us to find part of it at a time. Beginning with the left hand period, we find the greatest cube of 12 is 8, the root of which is 2. Place the 2 on the right of the given number for the first part of the root, and also in Col. I. on the left of the number. Multiplying the 2 into itself, write the product 4 in Col. II.; and multiplying 4 by 2, subtract its product from the period, and to the right of the remainder bring down the next period for a dividend. Then adding 2, the first figuie of the root, to the first term of Col. I., and multiplying the sum by 2, we add the product 8 to the 1st term of Col. II., -and to this sum QUEST. —86. What is it to extract the cube root. ARTS. 585-587.] CUBE ROOT. 375 annex two ciphers, for a divisor; also add 2, the first figure of the root, to the 2d term of Col. I. Finding the divisor is contained in the dividend 3 times, we place the 3 in the root, also on the right of the 3d term of Col. I. Then multiply the 3d term thus increased, by 3, the figure last placed in the root, and add the product to the divisor. Finally, we multiply this sum by 3, and subtract the product from the dividend. Ans. 23. 587. Hence, we derive the following general RULE FOR EXTRACTING THE CUBE ROOT. I. Separate the given number into periods of three figures each, placing a point over units, then over every third figure towards the left in whole numbers, and over every third figure towards the right in decimals. II. Find the greatest cube in the first period on the left hand; place its root on the right of the number for the first figure of the root, and also in Col. I. on the left of the number. Then multiplying this figure into itself, set the product for the first term in Col. II.; and multiplying this term by the same figure again, subtract this product from the period, and to the remainder bring down the next period for a dividend. III. Adding the figure placed in the root to the first term in Col. I., multiply the sum by the same figure, add the product to the first term in Col. IT., and to this sum annex two ciphers, for a divisor; also add the figure of the root to the second term of Col. I. IV. Find how many times the divisor is contained in the dividend, and place the result in the root, and also on the right of the third term of Col. I..Next multiply the third term thus increased by the fiyure last placed in the root, and add the product to the divisor; then multiply this sum by the same fJgure, and subtract the product from the dividend. To the remainder bring down the next peeriod for a new dividend. V. WFind a new divisor in the same manner that the last divisor was found, then divide, &c., as before; thus continue the operation till the root of all the periods is found. QrEST.-587. What is the first step in extracting the cube root? The second — Thirdl Fourth? Fifth 1 How is the cube root proved. 376 CUBE ROOT. [SECT. XV1I PRooF.. — Mfulti)ly the root into itself twice, and if the last prod. uct is equal to the given number, the work is right. OBs. 1. When there is a remainder, periods of ci2hers may be added, as in square root. 2. If the right hand period of decimals is deficient. this deficiency must be supplied by ciphers. The root must contain as many decimals as there are periods of decimals in the given number. 5~~. Demonstration.-This rule depends upon the principle that the cube f the sLum of two numbers is equal to the cube of the first part, added to. times the square of the first part into the last part, added to 3 times the first part into the square of the last, added to the cube of the last part. Take any nunlber, as 23; we have 23=20-3. Then (23)3=(20)3+(3X202X3)(3X20X32)+33 —12167. Or, (23)3=8000+-3600+-540+-27=12167. After subtracting the greatest cube from the left hand period, it is plain the remainder must contain 3 times the square of the first part of the root into the last part, &c. Hence, if we divide the remainder by 3 times the square of the first part of the root, the quotient will be the last part. But it will be seen that the divisor is 3 times the square of the first part of the root, consequently the quotient must be the last part of the root required. 1. The r'cason for separating the given number into periods of tIrce figurel' s, is that the culbe of a number can not have vwre figures than tlrip]le the number of figures in the root, nor but two less. It also shows how many figures the?root will contain, and thus enables us to find part of it at a time. (Art. 562. Obs. 3.) 2. The reason for annexing 2 ciphers to the divisor is (manifestly) because the first figure of the root, of which the divisor is 3 times the square, stands in tens' place 3. Placing the last figure of the root on the right of the 3d term in Col. I., then multiplying it by this figure, and adding the product to the divisor, and this sum being multiplied by the figure last placed in the root, the product will evidently be 3 times the square of the first part of the root into the last part, together with 3 times the first into the square of the last part, and the cube of the last part. In a similar manner the operation may be illustrated in all other cases. Note.-The preceding method of extracting the cube root was discovered by the late Mr. Horner of Bath, England, and is often called Horner's Metho 1. (For the common method, and its demonstration by cubical blocks, see P1 actical Arithlmetic, p. 325). QuiEST. —Obs. WVhen there is a remainder, how proceed? When the right hand period of decimals is deficient, what must be done? How many decimals must the root eoumtain I 588. Why separate the given number into periods of three figures'? Why annex two ciphers to the right of the divisor? ARTS. 588-590.] CUBE -ROOi. 377 589. The cube root of a common fraction is found by extracting the root of its numerator and denominator, or by first reducing it to a decimal. A mixed number should be reduced to an improper fraction, or the fractional part to a decimal. 3. Required the cube root of 78314.6. Operation. Col. I. Col:. 7T. 87314.606(42.78.4 1st term 4 16X4 - 64 2d " 8 4800, 1st divisor ) 14314 3d " 122 5044X2 = 10088 4th " 124 529200, 2d divisor) 4226600 5th " 1267 538069X7 - 3766483 6th " 1274 54698700, 3d divisor) 460117000 7th " 12818 54801244X8 - 438409952.590. When the root is required to many places of decimals, the operation may be contracted in the following manner. First find one more thanz ehacf the number of decimal figures required. For a new divisor, take as nmany figures plus one on the left of the last term in Col. II. as remain to be fobund in the root; and for a dividend retain one more figure on the left of the remainder than the divisor has; then proceed as in the contraction of division of decimals. (Art. 333.) Required the cube root of the following numbers: 4. 91125. 8. 10218313. 12. 37. 16. I -. 5. 140608. 9. 11543.176. 13 6 17. 7.5 9 6. 571787. 10. 20.570824., 14. 376. 18. 44-. 7. 2515456. 11..241804367. 15. 575. 19. 49-2. 20. Whlat is the cube root of 2 to eight decimals? 21. What is the cube root of -e%- to eleven decimals? 22. What is the side of a- cubical mound which contains 314432 solid feet? Note. -Similar solids are to each other as the cubes of their homologous sides, or like dimensions. (Leg. VII. 20, VIII. 11, Cor.) Hence, QuesT.-589. How find the cube root of a common fraction? Of a mixed number I 378 HIGHER ROOTS. [SECT,. XVIL 591. To find the side of a cube whose solidity shall be double, triple, &c., that of a cube whose side is given. Cube the given side, multiply it by the given proportion, and the cube root of the product will be thle side of tle cube required. 23. Whatt is the side of a cubical bin which contains 8 times as many solid feet as one whose side is 4 feet? Ans. 8 ft. 24. What is the side of a cubical block which contains 4 times as many solid yards as one whose side is 6 feet? 25. If a ball 6 inches in diameter weighs 32 lbs., what is the weight of a ball whose diameter is 3 inches? 26. If a globe 4 ft. in diameter weighs 900 lbs., what is the weight of a globe 3 ft. in diameter? 592. To find two mean proportionals between two given numbers. Divide the greater number by the less, and extract the cube root of the quotient. Multiply the root thus found by the least of the given numbers, and the product will be the least proportional sought; then multiply the least mean proportional by the same root, and this product will be the greater mean proportional required. Find two mean proportionals between the following numbers: 27. 8 and 216. 29. 12 and 1500. 31. 71 and 15396. 28. 64 and 512. 30. 40 and 2560. 32. 83 and 60507. EXTRACTION OF ROOTS OF HIGHER ORDERS. 593. When the index denoting the root to be extracted is a composite number. First extract the root denoted by one of the prime factors of the given index; then of this root extract the root denoted by anotler prime factor, and so on. Thus, For the 4th root, extract the square root twice. For the 6th root, extract the cube root of the square root. For the 8th root, extract the square root three times. For the 27th root, extract the cube root three times.1. What is the 4th root of 81? Ans. 3. ARTS. 591-594.] HITs ER RE noOS. 379 2. What is the 8th root of 256? 3. The 4th root of 65536? 4.'The 4th root of 19987173376? 5. The 6th root of 46656? 6. The 6th root of 308915776? 7. The 8th root of 390625? 8. The 9th root of 40353607? 9. The 18th root of 387420489? 10. The 27th root of 134217728? 594O When the index denotillg tile r'oot is not a comnposite number, iwNe have the following general RULE FOR EXTRACTING ALL ROOTS. 1. Point of the nzumber into periods of as many fi'gur2es each,, as ther'e are tunits in the given index, comzmencing wilth the zunits figyure. Ii. Find the first 2figzle of the root, and stublract its vower ftomn the left hand period; then to thae right of the remainder bring downu the first figure inz the next 2period for a divideld. I 1[. Involve the root to the power next i2'ferior to that of the incdet, of the qrequir ed root, an2d mcl?zltily it by the inc/lex itself; for a divisor. IV. _Find hoow mvt'ny i9mes tthe divisor is containzec in tie dividend, and the guoStient will be the qneht figZ6ure of the r'oot. V. Involve the whole root to the power? denzoted by the index oqf the reqzuireed root, and subtract it fr'om thle lwo left hancd periods of,~he. givenz number. VI. Fi'nally, bring down the first fisgure of the next perliocd to the remainder, for a new dividend, ancd ifind a new divisor as befor'e. Thus vroceed till the whole root is ext'racted.'OBs. 1. The'reason of this rule may be illust;ated in the same manner as that for the extraction of the Square and Cube Roots.'. The proof of all roots is by -involzution. 3. Any r'oot whlateveer may be extracted by an extesioit, of the principle applied to the extraction of the cube root. In 1this general application of tbi principle, the given number must be divided into periods, each consisting of as many figures as there are units in the index of the requied root, and the number of columns employed wi!i be one ess than there are units in the given index. The operation then proceeds exactly as in the extraction of the, cube root; and if there be a remainder, a like contraction is admissible. 17 380 HIGHER' ROOTS. [SECT. XV11 11. Required the 5th root of 35184372088832. Operation. 35184372088832(512 Ans 3125 54X5=3125) 3934 515 =345025251 514X5=33826005) 68184698 5125=35184372088832 12. Required the 5th root of 95-hr. 13. Required the 7th root of 2103580000000000000. Note.-The preceding method in most of the practical cases, gives perhaps as easy solutions, as the nature of the case will admit. But when roots of a very high order are required, the process may be shortened by the following * APPROXIMATE RULE. 595. Call the index of the given power n; and find by trial a number nearly equal to the required root, and call it the assumed root. Raise the assumed root to the power whose index is n. Then, As n 1 times this power, added to n-1 times the given number, is to n-1 times the same power added to n-+ l times the given number, so is the assumed root to the true root nearly. The number thus found may be employed as a new assumed root, and the operation repeated to find a result still nearer the true root. 14. Required the 365th root of 1.06. Solution. —Take 1 for the assumed root, the 365th power of which is 1; and n being 365, we have n+1-=366, and n-1364. Then proceed in the following manner: 1X 366=366 1X364=364 1.06X364= 385.84 1.06X366-387.96 As 751.84 751.96:: 1: Ans. Ans, 1.000159b. 15. The 7th root of 2? 17. The 12th root of 1.05? 16. The 9th root of 2? 18. The 100th root, of 100? * Hutton's Mathematical Tracts; also Bonnycastle's Arithmetic. ARTS. 595-400.] PROGRESSION. 38] SECTION XVIII. PROGRESSION. ART. 596. When there is a series of numbers such, that the'atios of the first to the second, of the second to the third, &c., are all equal, the numbers are said to be in Continued Proportion, or Progression. Progression is commonly divided into arithmetical and geometrical. Note.-The terms arithmetical and geometrical are used simply to distinguisl the different kinds of progression. They both belong equally to arithmetic and geometry. ARITHMETICAL PROGRESSION. 5 97. Numbers which increase or decrease by a common di'erence, are in arithmetical progression. (Art. 474. Obs.) OBS. 1. Arithmetical progression is sometimes called progr-ession by difference, or equidifferent series. 2. When the numbers increase, the series is called ascending; as, 3, 5, 7, 9, 11, &c. When they decrease, the series is called descending; as, 11, 9, 7, 5, &c. 598. When four numbers are in arithmetical progression the sum of the extremes is equal to the sum of the means. Thus, if 5 —3=9 —-7, then will 5+7=- 3+9. Again, if three numbers are in arithmetical progression, the suni of the extremes is double the mean. Thus, if 9-6=6-3, then will 9 + 3 = 66. 599. In any arithmetical progression, the sum of the two extremes is equal to the suam of any other two terms equally distant fiom the extremes, or equal to double the middle termn, when the number of terms is odd. Thus, in the series 1, 3, 5, 7, 9, it is obvious that 1+9 =3+7=5+5. 600. In an ascending series, each succeeding term is found by adding the common difference to the preceding term. Thus, if the first term is 3, and the common difference 2, the selies is 3, 5, 7, 9, 11, 13, 15, 19, 17, 21, &c. :352 A,CRT Vr TlcIATI, S aCT. XVII I. In a descending series, each succeedting term is fomnd by subtiact:ing the common dlifference from the preceding' term. Tirus, if the first term is 15, and the common difference 2, the series i3 15, 13, 11, 9, 7, &c. G@io. In arithmetical progression there are five parts to be considered, viz' the first ter)m, the last termsn, the nu21ber of ter'ms, the comzihzon difl'erence, aznd thce su11 of all the ter?'ms. These parts have such a relation to each otherl, that if any three of them arce given, the other t2zo may be easily found. 602o If the sum of the two extremes of an arithmletical progression is multiplied by the number of the terms, the produtt will be double the sum of all the terms in the series. Take the series 2, 4, 6, 8, 10, 12. The same inverted 12, 10, 8, 6, 4, 2. r'he sums of the terms are 14, 14, 14, 14, 14, 14. Thus, the sum of anl the terms in the double series, is equal to the sum of the extremes repeated as many times as there are terms; that is, the sum of the double series is equal to 12- -2 multiplied by 6. But this is twice the sum of the single series. Hence, 6@0g3e To find the sum2c, of all tlhe terms, when the extremzes and the number of terms are given. Miultiply half the sum?, of the extremgzes by the number of terms, aznd tlze product will be the s7m, of the givenz series. OBS. The reason of this process is manifest from the preceding illustration. Ex. 1. The extremes of a series are 3 and 25, and the number of terms is 12: what is the sum of all the terms? Anzs. 108. 2. What is the sum of the natural series of numbers, 1, 2, 3, 4, 5, &c., up to 100? 3. How many strokes does a common clock striike in 1'2 hourls O@4o To find the commzonz d;icflrezce, when. the extremes and the nuzz7berZ of terms are given. Divide tle diebrence of thle extremzes by the: numZber of terms less 1, and tfhe quotient wzill be the co70mmonz cdi'erC ce. re eqeZdredC Oss. The trut1h of this rule is manifest from Art. 603. AI TS. 601-607.1 PIOGREISSION. 383 4. The extremes are 5 and 56, and the number of terms is 18: what is t1he common difference? Ans. 3. 5. If the extremes are 3 and 300, and the number of terms 10, what is the common difference? 05.* To find the nzzmber of terms, Awhen the extremes and o?2o2),202 cdicferenzce are given. Divide.le the dffereznce of the extremnes by the common, cli/'erence, and the qzotient i?creased by 1 will be the zzumnber of terms. Oms. The t'Lnt/c of this principle is manifest from the manner in which the successive terms of a. series are formed. (Art. 600.) 6. If -the extremes are 6 and 470, and the common difference is 8, what is the number of terms? Ans. 59. 7. If the extremes are 500 and 70, and the common difference is 10, what is the number of terms? {O0. When the sumz of the series, the number of terms, and one of the extremes are given, to find the other extreene. Divide twice the sumn of the series by the number of terms, and from tIze quotieczt take the given extremze. Orns. The lreason of this rule is manifest from Art. 602. S. If the sum of a series is 576, the number of terms 24, and the first term 1, whlat is the last term? Ans. 47. 9. If the sum of a series is 1275, the number of terms 50, and the greater extreme 474%, what is the less extreme? 7o To find aczny given, term, wlen the first term and the common difference are given. itflltiply the common difference by one less than the nzumber of terms eqzuired; then if the series be ascendizng, adcd the product to the first terem; buz t;f it be descending, subtract it. OBs. The reason of this rule may be seen from the manner in which the succeeding terms of a selies are forjned. (Art. 600.) 10. If the first term of an ascending series is 7, and the common difference 3, what is the 41st term? Ans. 127. 11. If- the first term of a descending series is 100, and the common lifference 1i-, whllat is the 54th1 term.? 384 GECMETRICAAL [SECT. XVIII. 12. If the first lerm of an ascending series is 7, and the common difference 5, what is the 100th term? 608. To find any given number of arithmetical means, when the extremes are given. Subtract the less extreme from the greater, and divide the remainder by i more than the number of means required; the quotient qwill be the common difference, which being continually added to the less extreme, or subtracted from the greater extreme, will give the mean terms required, One mean term may be found by taking!half the sums of the extremes. (Art. 598.) oBs. This rule depends upon the same principle as that in Art. 604. 13. Required 3 arithmetical means between 7 and 35. 14. Required 6 arithmetical means between 1 and 99. GEOMETRICAL PROGRESSION. 609. Numbers which increase by a common multiplier, oi decrease by a common divisor, are in Geometrical Progression. The numbers 4, 8, 16, 32, 64, &c., are in geometrical progression; and if each preceding term is multiplied by 2, the product will be the succeeding term; thus, 4X2=8; 8 X 2=16, &c. Again, if the order of this series be inverted, the proportion will still be preserved and the common multiplier become a common divisor. Thus, in the series 64, 32, 16, 8, &c., 64+ 2=32; 32+. 2=16, &c. Note.-If the first term and ratio are the same, the progression is simply a series of powers; as 2; 2X2; 2x2X2; 2X2X2X2, &c. Oss. 1. Geometrical Progression is geometrical proportiotn continued. It is therefore sometimes called continual proportionals, or progression by quolients. If the series increases it is called ascending; if it decreases, descendling, 2. The numbers which form the series, are called the terms of the progression. The common.multiplier, or divisor, is called the ratio. For most purposes, however, it will be more simple to consider the ratio as always a multzplier,. either integ'ral or friactional. Thus,' in the series 64, 32, 16, &c., the ratio is either 2 considered as a divisor, or I considered as a multiplier. 3. In Geometrical as well as in Arithlmetical progression, there are five parts to be considered, viz: the first term, the last term, the number of terms, the ratw, and the sum of all the terms. These parts have such a relation to each other, that if any three of them are given, the other twbo may be easily found. AIRTS. 608-611.] PRO aRESSION. 385 6 10. To find the last term, when the first term, the ratio, and the number of terms are given. Alrultiply the first term into that fpower of the ratio whose index is 1 less than the uztmber of terms, and the product will be the last terms required. OBs. 1. The reason of this process may be seen by adverting to the manner in which each successive term is formed. (Art. 609.) Thus, in the series 4, 8, 16, 32, &c., the 2d term 8-4X2; 16=4X2X2, or 4X22; 32=4X23, &c. 2. It will be seen that the several amounts in compound interest, form a geometrical series of which the principal is the 1st term; the amount of $1 for 1 year the ratio; and the number of years-+-i the number of terms. Hencne the required amount of compound interest may be found in the same way as the last term of a geometrical series. 1. If the first term of a geometrical progression is 2, and the ratio 4, what is the 5th term? Ans. 512. 2. The first term is 64, and the ratio i: what is the 5th term? 3. The first term is 2, and the ratio 3: what is the 8th term? 4. The first term is 7, and the ratio 5: what is the 10th term? 5. A farmer hired a man for a year, agreeing to give him $1 for the 1st month, $2 for the 2d, 84 for the 3d, and so on, doubling his wages each month: how much did he give the last month? 6. What is the amount of $250, at 6 per cent., for 5 years compound int.? Of $500, at 7 per ct., for 6 years? Of $1000, at 5 per ct., for 10 years? 61 1, To find the sum of the series, when the ratio and the extremes are given. 1Multiply the greatest term into the ratio, from the product subtract the least term, and divide the remainder by the ratio less 1. O0s. 1. When the first term, the ratio, and the number of terms are given, to find the sum of the series we must first find the last term, then proceed as above. 2. The sum of an infil ite series whose terms decrease by a common divisor, may be found by wulttiptying the greatest term into the ratio, and divzding the product by the ratio less 1. The least term being infinitely small, is of no comparative value, and is therefore neglected. 7. What is the sum of the series, whose extremes are 5 and 1215, and the ratio 3? Ans. 1820. 8. The extremes of a series are 1 and 512, and tlhe, ratio 2' what is the sum of the series? 386 ANNUITIES. [SEcTlv XVIl 9. The extreamtes of a series are 102 14 and 15274~j, and the ratio is 1J: what is the sum of tlhe series? 10. A merchant hired a clerkli for a year, and agreed to pay hlim 1 mill the 1st month; 1 cent the 2d; 10 cents the 3d, and so on, increasing in a tenfold ratio for each successive month: what was the amount of his wages? 11. What is the sum of the infinite series 1 +~ —~ -+ —, &c.; that Is, the descending series whose first term is 1 and the ratio 2? Alns. 2. 12. What is thle sum of the infinite series 1 -+ - ~,-~-5- - &c, 6 1 2o To find the ratio, when the extregmes and znnzuber of terms are given. Divide the gr~eater extreme by the less, and extract that 9root of the quotient whose index is 1 less than the number of terms. 13. The extremes of a series are 3 and 192, and the number of terms 7 what is the ratio? Ans. 2. 14. What is the ratio of a series of 5 terms, whose extremes are 7 and 567? Note.-Other formulas in arithmetical and geometrical progrression might be added, but they involve principles with which the student is supposed as yet to be unacquainted. For a fuller discussion of the subject, see Thomson' Day's Algebra. A N N U I T I E S. 6 3. The term annuity properly signifies a sulm of money payable annually, for a certain length of time; or forever. OBs. 1. Payments made semi-annually, quarterly, monthly, &c., are al1.) called annuities. Annuities therefore embrace pensions, salaries, rents, &c. 2. When annuities remain unpaid after they are due, they are said to be forborne, or ine arrecvrs. The scem of the annuities in arrears, added to thle inbterest due on each, is called the acmocznt. The presentz,o-t/A of an annuity is the sum, which being put at interest, wll exactly pay the annuity. 3. Wlhen an annuity does not commence till a given time-has elapsed, it is called.an annuity i reversion,; when it continues fonever, a pcrpetseity. 4. In finding the amzougnt of annuities in arrvears, it is customary to reckon compound interest on each annuity from the time it is due to the time of payment. The'process therefore is the same as findlrag the sum of' ai aICccnding geometrical series. (Art. 611.) Hence, AR s, 612-615.] ANNUITIES. 387 G I 4o To find the amount of an annuity in arrears. _cakce the annuity the first term of a geometrical series, the amnount of'1 for 1 year the ratio, and the. given number of years the number of terms; thenz fnd the sum of thle series, and it will be the amount required. (Arts. 610, 611.) Gas. When the payments are not yearly, for the amount of $1 for 1 year, use its amount for the time between the payments; and instead of the number of /ye.srs, use the number of payments that have been omitted, and proceed as before. 1. What is the amount of an annuity of $100 which has not been paid for 3 years, at 6 per cent. compound interestS SolzLtiou.-10OOX(i.06)2=112.36; and (112,36X1.06)-100 —.06= $318.36, TABLE, sowwing t/le amount of annuity of $1, or ~1, at 5, 6, and 7 per cent. for anzy umber?' of year's from 1 to 20. Yrs. 5 per ct. 6 per ct. /7 per et. Yrs. 5 per ct. 6 per et. 7 per ct. 1 1.00000 1.00000 1.00000 1 14.20678 14.97164 15.7836 2 2.05000 2,06000 2.07000 12 15.91712 16.86994 17.8884 3 3.15250 3.18360 3.21490 13 17.71298 18.88213 20.1406 4 4.31012 4.37461 4.43994 14 19.59863 21.01506 22.5504 5 5.52563 5.63709 5.75073 15 21.57856 23.27596 25.1290 6 6.80191 6,97532 7.15329 16 23.65749 25.67252 27.8880 7 8.14-201 8.39383 8.65402 1 7 25.84036 28.21287 30.8402 8 9.54911 9.89746 10.2598 18 28.13238 30.90565 33.9990 9 11.02656 11.49131 11.9799 19 30.53900 33.75999 37.3789 10 12.57789 13.18079 13.8164 20 33.06595 36.78559 40.9954 Note.-Multiply the given annuity by the amt. of $1, for the given numlber of years found in the Table, and the product will be the amount required. 3. What will an annual rent of $75 amount to in 9 years, at 5 per cent.' 4. What is the amount of $200 forborne for 9 years, at 6 per cent.. 5. What is the amount of $350 forborne for 10 years, at 7 per cent.. 6. What is the amount of $1000 forborne for 20 years, at 6 per cent.. Gt 5[ To finld the present worth of an annuity. Find the amount of $1 anzuity for the given time as before; then? divide thiis amount by the amgount of 81 at comp2ound interest 3or the samze time, nultiplyy the quotient by the given annuity, and the product voill be the prcescnzt worth. Iff the c annuity is a perpetuity, or to convtinze forever, mnultiply it by 100, divide the —product by the given grate, and the quotient will be the present value lrequired. Ors. For the amount of'.$ at compound interest, see Table, p. 271. 7. What is the present worth of an annuity of $40 to continue 5_years, at 5 per cent; (ompound interest? Ans. $ L73.178. I?* 388 PERMUTATIONS. LSECT. XVIII. 8. What is the present worth of an annuity of $80 to continue forever, at 6 per cent.? 6 1 6. To find the present worth of an annuity in reversion. Find the present worth of the annuity from the present time till its termination; also find its present worth for the time before it commences; the difference between these two results will be the prescnt worth required. I3, What is the present worth of $79.625 at 5 per cent., to commence in jyears and continue 6 years. Ans. $332.50. PERMUTATIONS AND COMBINATIONS. 6 1 7. By Permutations is meant the- changes which may be made in the arrangement of any given number of things. The term combinations, denotes the taking of a less number of things out of a greater, without regard to their order or position. 6 1 8. To find how many permutations or changes may be made in the arrangement of any given number of things. Multiply together all the terms of the natural series of numbers from? 1 up to the given number, and the product will be the answer. 1. How many changes may be rung on 5 bells'I Ans. 120. 2. How many different ways may a class of 8 pupils be arranged. 3. How many different ways may a family of 9 children be seated 3 4. How many ways may the letters in the word ar'itlhetic, be arranged? 5. A club of 12 persons agreed to dine with a landlord as long as he could seat them differently at the table: how long did their engagement last? 6 1 9. To find how many combinations may be made out of any given number of different things by taking a given number of them at a time. Take the series of numbers, beginning at the number of things given, and decreasing by 1 till the number of terms is equal to the number of things taken at a time; the product of all the terms will be the anszoer required. 6. How-many different words can be formed of 9 letters, taking 3 at a time 3 Solntgion —9X8X7-504. Ans. 504 words. 7. How many numbers can be expressed by the 9 digits, taking 5 at a time 7. 2. How many words of 6 letters each can be formed out of the 26 letters of the alphabet, on the supposition that consonants will form a word 3 ARTS. 616-625.] MENSURATION. 389 SECTION XIX. APPLICATION OF ARITHMETIC TO GEOMETRY, 620. In the preceding sections abstract numbers have been applied to concrete substances, or to objects in general, considered arithmetically. On the same principle, geometrical magnitudes may be compared or measured by means of the numbers representing their dimensions. (Arts. 7, 516. Obs. 3.) OBS. The measurement of magnitudes is commonly called ntesuratwr. MENSURATION OF SURFACES. 621. In the measurement of surfaces, it is customary to assume a square as the measuring unit, whose side is a linear unit of the same name. (Leg. IV. 4. Sch. Art. 257. Obs. 2.) Note.-For the demonstration of the following principles, see references. 622. To find the area of a parallelogram, also of a square. Multiply the length by the breadth. (Art. 285, Leg. IV. 5.) OBs. When the area and one side of a rectangle are given, the other side is rund by dividing the area by the given side. (Art. 156.) 1. How many acres in a field 240 rods long, and 180 rods wide. 2. How many acres in a square field the length of whose side is 340 rods X 3. If the diagonal of a square is 100 rods, what is its area! 4. A rectangular farm of 320 acres, is i a mile wide: what is its length? 623. To find the area of a rhombus. (Leg. I. Def. 18I. V. 5.) MAultiply the length by the altitude or perpendicular height. 5. Find the area of a rhombus whose length is 20 ft., and its altitude 18 ft. 624. To find the area of a trapezium. (Leg. IV. 7.).Multiply half the sum of the parallel sides by the altitude. 6. Find the area of a trapezium the lengths of whose parallel sides are 27 ft. and 31 ft., and whose altitude is 15 ft. 625. To find the area of a triangle. (Leg. IV. 6.) Multiply the base by half the altitude or perpendicular height. 7. Find the area of a triangle whose lbase is 50 r't., nd its -altitude 44 ft. 390( ns ENSU I ATU -'. I ) l f'Srr. XIX. 626. To find the area of ta triangle, the three sides beincg given. From kcsdf tlhe sem i of three t si6des subtr'act, CCL side 9respectively; then mzulttiplp together harcif the sz6r acrd thfe three remaindefs, ancd extract the sqcuare r'oot of t/7e p1rodtuct. 9. What is the area of a triangle whose sides are'20, 30, and 40 ft. 10. How many acres in a triangle whose sides are, each 40 rods. 2 7r'To find the circumference of a circle from its diameter. Izultrpgly the diaemeter by 3.141509. (Leg. V. 11. Sch.) iNote. —The circrnrfeorenzce of a circle is a curve line, all the points of which are equally distant from a point within, called the cenItire. The ditameler of ta circle is a straight line which passes through the centre, and is terminated on both sides by the circumference. The r'adius or semi-diametler is a straight line drawn from the centre to the circumference, 11. What is the circumference of a circle, whose dianmeter is 20 ft.. 12. What is the circumference of a circle, whose dialleter is 45 rsods 628. To find thle diameter of a circle from its cicumference. Divide tfhe circuslnferernce by 3.14159. oas. The diaczretcer of a circle may also be found by dividing the areca by 7854, and extracting the square root of the quotient. 13. W7hat is the diameter of a circle, whose circumference is 314.159 ft. I G290 To filnd the area of a circle. (Leg. V. 11.) ifulGtti3ply Iceif ti.re circumsferenrce by half thJe diescreter; or, mZultlipy tihe szucare of the diacgeter by tize decizmal.17854. 15. What is the area of a circle, whose diameter is 50 rods 1 16. Find the area of a circle 200 ft. in diameter, and 628.318 ft. in circumn. 6330~ To find the side of the greatest squr'e that can be inscribed in a circle of a giiven diameter. Divide the squcare of tlre givez dicametetr by 2, and extract the square soot of the quotienzt. (Art. 581. Obs. 1.) 17. The diameter of a round tabrle is 4 ft.; what is the side of the greatest square table which can be made from it 3 63 1. To find the side of the greatest equilateral triangle that can be inscribed in a circle of a (iven diameter. Multiply - the givenz diameter by 1.73205. (Leg. V. 4. Sch.) 18. Required the side of an equilateral triangle inscribed int rile of 20 ft, diameter, A nis. 626-637.] MENSURATION. 391 MEASUREMENT OF SOLID.S. 6320 In the measuerment of solids it is customary to assume a cube as the gmeasuring unit, whose sides are squares of the same naime. (Art. 258. Obs. 2.) 633, To find the solidity of bodies whose sides are perpendi cular to each other. _lf'ltyiply the length, breadth, and thickness together. (Art. 286.) OBs. When the conte~nts of a solid body and two of its sides are given, the other side is found by dividing the contents by the product of the two given sides. (Art. 159.) 1. What are the contents of a stick of timber 4 ft. square, and 851 ft. long 3 2. What is the capacity of a cubical vessel, 14 ft. 8 in. deep? 6 3 Q4 To find the solidity of a prism. lufltiply the acrea of the base by the height. (Leg. VII. 12.) Ons. This rule is applicable to all prisms, triangular, quadrangular, pentaganal, &c., also to all pdiacelolpipedons, whether rectangulai' or oblique. 3. Finhd the solidity of a prism 461 ft. high, whose base is 71 ft. square? G35. To find the lateral surface of a right prism. JMueltiply the length by the perimeter of its base. (Leg. YII. 5.) OBS. If we add the areas of both ends to the lateral surface, the sum will be the whole surface -of tile prism. 4. What is the surface of a triangular prism, whose sides are each 3 ft., and its length 12 ft.' 6360 To find the solidity of a pyramid and cone. pffultiply the area of the base by _ of the height. (Leg. VI[. 18.) 5. What is the solidity of a pyramid 100 ft. high, whose base is 40 ft. square X 6. What is the solidity of a cone 150 ft. high, whose base is 15 ft. in diameter 3 i3370 r To find the lateral or convex surface of a regular pyramid, or cone. (Leg. VII. 16, VIII. 3.) Mfultiply the perimeter of the base by ~ the slcant-heig/ht. 7. What is the lateral surface of a regular pyramid, whose slant-height is, 15 ft., and base is 30 ft. square? 8. What is the convex surface of a right cc ne, whose slant-height is 94 f and the perimeter of its base 37 ft. 7 392 MENSURATION. [SECT. XIX, 63 S To find the solidity of a frustum of a pyramid and cone. To the sum of the areas of the two ends, add the square root of the product of these areas; then multiply this sum by J of the perpendicular height. (Leg. VII. 19. Sch., VIII, 6.) 9. If the two ends of the frustum of a pyramid are 3 ft and 2 ft. square, and the height is 12 ft., what is its solidity? 6 3"9. The convex surface of a frustum of a pyramid and cone is found by multiplying half the sum of the circumferences of the two ends by the slant-height. (Leg. VII. 17, VIII. 5.) 10. If the circumferences of the two ends of the frustum of a cone are 18 ft and 14 ft., and its slant-height 11 ft., what is its convex surface 1 640. To find the solidity of a cylinder. Multiply the area of the base by the height. (Leg. VIII. 2.) 11. Find the solidity of a cylinder 10 ft. in diameter, and 35 ft. high. 12. Find the solidity of a cylinder 100 ft. in circumference, and 150 ft. high. 641. To find the convex surface of a cylinder. Mfultiply the circumference of the base by the height. (Leg. VIII. 1.) 13. Find the convex surface of a cylinder 5 yds. in diameter, and 5 yds. long. 642. To find the convex surface of a sphere or globe. Multiply the circumference by the diameter. (Leg. VIII. 9.) 14. What is the surface of a globe 18 inches in diameter! 15. If the diameter of the moon is 2162 miles, what is its surface. 643. To find the solidity of a sphere or globe. Multiply the surface by -1 of the diameter. (Leg. VIII. 11.) 16. Find the solidity of a globe 15 inches in diameter. 17. The diameter of the moon is 2162 miles: what is its solidity' MEASUREMENT OF LUMBER. 644. The area of a board is found by multiplying the length into tohe twa oreadth. (Arts. 622, 623.) The solid contents of hewn or square timber are found by multiplying the,engtA into the mean breadth and depth. The solid contents of round timber are found by multiplying the lengtA by i the mean girt o' circumtference. Obs. 1. The mean breadth of a tapering board is found by measuring if in the middle,,r by taking j the sum of the breadths of the two ends. ARTS. 638-647.1 ME, NSURATION., 393 2. The mean dimensions of square and round timber are found in a similar manner. 3. The method for finding the solidity of round timber makes an allowance of about - for waste in hewing. (Arts. 640, 258. Obs. 3.) 18. Find the area of a board 12 ft. long, and the ends 14 in. and 12 in. wide. 19. Find the solidity of a joist 16 ft. long, the ends being 8 in. and 4 in. sq. 20. Find the solidity of a log 50 ft. long, the circumferences of the ends being 6 ft. and 4 ft. GAUGING OF CASKS. 645. The process of finding the contents or capacities of casks and other vessels is called GAUGING. 646. The contents of casks are found by multiplying the square of the mean diameter into the length; then this product multiplicd by.0034 will give the wine gallons, and multiplied by.0028 will give the beer gallons. Ois. The mean diameter of a cask is found by adding to the head diameter.7 of the difference between the head and bung diameters when the staves are very much curved; or by adding.5 when very little curved; and by adding.55 when they are of a medinm curve. 21. How many wine gallons in a cask but little curved, whose length is 45 in., its bung diameter 40 in., and its head diameter 36 in.' 22. How many beer gallons in a cask much curved, whose length is 64 in., its bung diameter 52 in., and head diameter 46 in.' TONNAGE OF VESSELS. 64[J. Government Rule.-I. If the vessel be double-decked, take the length from the fore part of the main stern to the after part of the stern-post, above the upper deck; then the breadth at the broadest part above the main wales, half of which breadth shall be accounted the depth of such vessel; from the length deduct three-fifths of the breadth, multiply the remainder by the breadth and the product by the depth; divide the last product by 95, and tile quotient shall be deemed the true tonnage of the vessel. II. If the vessel be-single-decked, take the length and breadth as above directed, deduct from the length three-fifths of the breadth, and take the depth from the under side of the deck plank to the ceiling in the hold, then multiply and divide as before, and the quotient shall be deemed the tonnage. 2arpenter's Rule.-The continued product of the length of the keel, the.breadth at the main beam, and the depth of the hold in feet, divided by 95 will give the tonnage of a si'ngle-decked vessel. For a double.decker, instead of the depth of the hold, take half the breadth at the beam. 23. What is the government tonnage of a dcable-decker, whose engtl is 150 ft., the breadth 35 ft., and the depth 25 ft.. 24. What is the carpenter's tonnage of the same vessel. 394 MLCIHANICAL POWEitS. [SECT. XIX. MIECHANICAL POWERS. G40 l The lM~ecaznical pozuers are six, viz: the lever, the w7heel and axle, the pulley, the inclined plaze, the screw, and the wedge.,b49. When the power and weight act perpen'edicularly to the arms of a straighlt lever, the power is to the weight, as the distance from the fulcrum to the weight is to the disagznce from the fulcrum to the power. 1. If the power is 100 lbs., the long arm 10 ft., and the short arm 2 ft., what weight can be ra:sed. 2. The arms of a lever are 15 ft. and 4 ft., and the weight raised 500 lbs.: what is the wower.? 650. When a weight is sustained by a lever resting on two props, T7'he tlong ara: the sholwt cnar: the weight supported by the long arm: the weight supported by the short arm. Hence, The whoble lengthe: shoort arm?'' whole weiglht: weight ons s. a. (Leg. III. 16.) 3. A and B carry 256 lbs. suspended upon a pole 5 ft. from A and 3 ft. from B: how many pounds does each carry. 4. A and B carry 90 lbs. upon a lever 12 ft. long: where must it be placed that!B may carry 3 of it. 6i11, The wheel and axle operate on the same principle as the lever; the semi-diameter of the wzeel answers to the long arm, and the semi-diameter of the axle to the short arm. 5. If the diameter of a wheel is 6 ft., and that of the axle 1 ft., what weight vill 100 lbs. raise. 6. A wheel is 8 ft. diameter, an axle 1i ft.: what weight will 200 lbs. raise X,o* In the application of mzovable pulleys, Tlte PowER: the WEIGHT:: 1.: twice the NUMBER of plleys. 7. What weight can a power of 200 lbs. raise with 4 movable pulleys. 8. What power with 8 pulleys will raise a pillar of granite weighing 10 tons',63, Thde perpesedicuslaar leiglt of acn inclined plane is to its length, as the power to the weigt/R.: 9. What power will draw a train of cars weighing 100000 pounds up an inclined plane which rises 60 ft. to a mile. 664. The screw acts upon the principle of the inclined plane. Hence, Thte distance between the threcad7s is to the ciqrcumference of a circle described by the power, as the power is to the weightt. 10. Whatt weight can be raised by a power of 1000 lbs. applied to a screw whose threads are 1 inch apart, at the end of a lever 12 ft. long 2 635. The p1oweq- applied to the head of a wedge is to the..eight, as half the thickness.of the head is to the length of its side. In the use of the wedge, not less than half the power is lost by fi:ction agoainst the si les. MISCELLANEOUS z. XAitPLESI 395 MISCELLA1NEOUS EXAMPLES. 1. The sum of two numbers is 980, and their difference 62: what are the numlbers 3 T,'he pro'duct of two numbers is 441 0, and one is (;3: what is the other I t)3. What number multiplied by 28 —, will produce 145 3 4. What number multiplied by 64, will be equal to 74 multiplied by 5- 3 5. If an army of 24000 men have 5B20000 lbs. of bread, how long will it last llem, allowing each man 14 lbs. per day! 6(. What is the interest of $5256 for 60 days, at 7 per cent. 3 7. What is the amount of $16;230 for 4 months, at 64 per cent. 3 8. What is the bank discount on $1200 for 90 days, at 6 per cent 3 9. For what sum must a note be made, payable in 4 months, the proceeds of which shall be $1800, discounted at a bank at 7 per cent.? 10. A capitalist sent a broker $25000 to invest in cotton, after deducting his commission of 241 per cent.: what amount of cotton ought he to receive 3 11. A merchant bought 500 yards of cloth for $1800: how must he retail it by the yard to gain 25 per cent. 3 12. A man bought 640 bbls. of beef for $5000, and sold it at a loss of 12 per cent.: how much did he get a barrel? 13. If a man buys 1000 geographies, at 374 cents apiece, and retails them at 50 cents, what per cent. will he mnake. 14. A grocer bought 180 boxes of lemons for $360, and sold them at 10 per cent. less than cost: what did lIe lose 3 15. How many dollars, each weighing 4124 grs., can be made from 16 lbs. 5 oz. of silver 3 16. How many eagles, weiglling 258 grs. apiece, will 21 lbs. 10 oz. nmake 17. How long a thread can be spun from 1 ton of flax, allowing 5 oz. will make 100 rods of thread 3 18. How many revolutions will the hind wheel of a carriage 5 ft. 6 in. in circumlference, make in 2 miles 4 furlongs 3 19. 1-How mlany revolutions will the fore wheel of a carriage 4 ft. 7 in. in circumsference, make in the same distance 3 20. Bought 1500 doz. buttons for $187.50: what was that per gross! 21. A man paid $132 for 40 bbls. of cider: what is that a quart 3 22. A man paid $150 for 10 rods of land, what, was that per acre 3 23. A man having $2500, laid out.z of it in flour, at $5 per barrel: h6w nmany barrels did he buy 3 24. The commander of an exploring expedition found that 4 of his provisions were exhausted in 28 months: how much longer would they fast 3 25, What cost 15- lbs. of cheese, at $84 per hundred 3 26. I-low many yards of carpeting - yd:; wide will it take to cover a floor 18 ft. long and 15 ft. wide! 27. If -x yard of calico cost -Zs., what will. of an ell English cost. 28. Iow lollg will 4683256 lbs. of beef last an army of 8245 s?.ddiers, allcwing each man 1 h. l)or day 3 396 Mrl;SCELLANEOUS EXAM3PLES. 29. I-low long would the same quantity of beef last the army, if reinforced ay 2500 men, allowing each man 14 lb. per day' 30. Bought J of a pipe of wine for $126: what was that per gallon? 31. If a man can walk 17 miles in 5 hours, 12 minutes, 31 seconds, how far can he walk in 3 hours, 40 minutes, 36 seconds? 32. If a man traveling 14 hours per day, performs half his journey in 9 days, how long will it take him to go the other half traveling 10 hours a day? 33. If you lend a man $700 for 90 days, how long ought he to lend you $1200 to requite the favor? 34. A milkman's measure was deficient half a gill to a gallon: how much did he cheat his customers in selling 8720 gallons? 35. If 85 yds. of calico cost $10.20, what will 1500 ydls. cost? 36. If 1650 lbs. of sugar cost $206.25, what will 87 lbs. cost? 37. If 1424 gals. of oil cost $106,2, what will 210 gals. cost? 38. If wind moves 21 miles per hour, how long is it in moving from the pole to the equator, a distance of 6214 miles? 39. if -a- of a barrel of flour costs ai of a dollar, what will i of a bbl. cost? 40. If 3 of a ton of chalk cost ~}, what will -z of a ton cost' 4 7' 5 41. If A of a bushel of wheat cost $~, how much will A of a bushel cost? 5 6'42. If i of a ship cost $16000, what will.-? of her cost?.43. If 16- bbls. of mackerel cost $65-, what will 481 bbls. cost? 44. If 28La gals. of oil cost $31.25; how much will $250 buy? 45. At $31 for 40 doz. eggs, what will 460k doz. cost? 46. The President's salary is $25000 per annum: how much can he spend per day, and lay up $10000 of it? 47. If one person lies in bed 9 hours per day, and another 6 hours, how much time will the one gain over the other in 20 years? 48. A cistern has three faucets; the 1st will empty it in 10 min., the 2d in 20 min., the 3d in 30 min,: in what time will they all empty it? 49. A man and his wife drink a barrel of beer in 30 days, and the man alone can drink it in 40 days: how long will it last the wife? 50. A teacher being asked how many scholars he had, replied I study Arithmectic, J study Latin, -o- study Algebra, -,- study Geometry, and 24 study French: how many scholars had he 1 51. A man having spent i and ~ of of his money, had ~48R left: how much had he at first? 52. A man bequeathed ~ of his property to his wife, - to his son, 1 to his daughter, and the remainder, which was $1500, to the Bible Society: what did his whole property amount to? 53. What is that-number - of which exceeds A of it by 45? 54. Pewter is composed of 112 parts of tin, 15 of lead, and 6 of brass: how much will it take- of each ingredient to make 5650 pounds of pewter X 55. Two travelers start at the same time from Boston anl Washington to meet each other; one goes 5 miles an hour, the other 7 miles; the whole distance is 436 miles: how far will each travel? MISCELLANEOUS EXAMPLES. 397 56. A grocer divided a barrel of flour into 2 parts, so that the smaller con. tained I as much as the other: how many pounds were there in each. 57. A, B, and C, build a ship together; A advanced $1000, B $12000, and C $13000; they gain $5000: what is the gain of each X 58. A, B, and C, entered into partnership; A furnished $600, B and C together $1800; they gained $960, of which B took $280: how much did A and C gain; and B and C put in respectively? 59. The liabilities of a bankrupt are $63240, and his assets $12648: what per cent. can he pay? 60. A bankrupt compromises with his creditors for 37J per cent.: h( w muca will he pay on a claim of $3656? 61. How much will he pay on a debt of $12680.375? 62. A owns - and B - of a ship; A's share is worth $10000 more thlar B's: what is the value of the ship X 63. A man gave his oldest son ~ of his i roperty less $50; to the second, he gave 1; and to the youngest he gave the remainder, which was 1 less $10: what was the amount of his property 64. A man and boy together can frame a house in 9 days; the man can frame it alone in 12 days: how long will it take the boy to frame it? t65. A cistern has a receiving and a discharging pipe; when both are running it takes 18 hours to fill it; if the latter is closed it requires 15 hours to fill it: if the former is closed, how long will it take the latter to empty it? 66. Four men, A, B, C, and D, spent ~255, and agreed that A should pay i; B ~; C t; and D D: how much must each pay? 67. A, B, and C, formed a joint stock of ~820, and gained. ~640, in the division of which A received ~5 as often as B did ~7, and C ~8: how much did each put in and receive? 68. A, B, and C, gained a certain sum, of which A and B received $640, B and C $880, and A and C $800: what was the gain of each 69. What number is that 5 and J of which being multiplied together, will produce the number itself? 70. A club spent ~2, 12s. id.; on settling, each paid as many pence as there were individuals in the party: how many were there in the party? 71. The sum of two numbers is 120, and the difference of their squares is 4800: what are the numbers? 72. The difference of two numbers is 53, and the difference of the squares is 10759: what are the numbers? 73. The diagonal of a square is 80 ft.: what is its side? 74. The diagonal of a square field is 120 rods: what is its ares? 75. Find the side of the greatest square beam which can be hewn`:om log 5 ft. in diameter? 76. The mainmast of a ship is 95 ft. long, the diameter of the base is 3} ft., that of the top 21 ft.: what is its solidity? 77. A man wished to tie his horse by a rope so that he could feed on just an acre of ground now long must the rope be? 398 MISCELLANEOUS EXAMPLES. 78. What is the area of a circle I mile in circumference' 79. If the diameter of the sun is 887000 miles, what is its surface! 80. If the diameter of Jupiter is 86255 miles, what is its solidity' 81. A conical stack of hay is 20 ft. high, and its base 15 ft. in diameter i what is its weight, allowing 5 lbs. to a cubic foot 1 82. How many bushels will a cubical bin contain whose side is 9 ft.' 83. How many hogsheads will a cylindrical cistern 10 ft. deep and 6 fRt. diameter contain' 84. How far from the end of a stick of timber 30 ft. long, of equal size from end to end, must a lever be placed, so that 3 men, 2 at the lever, and 1 at the end of the stick, may each carry ] of its weight? 85. How many different ways may a class of 26 scholars be arranged? 86. If 100 eggs are placed in a straight line a rod apart, how many miles must a person travel to bring them one by one to a basket placed a rod from the first egg' 87. What is the sum of the series 1, 1,, 2, 2~, 3, &c., to 50 terms' 88. A blacksmith agreed to shoe a horse for 1 mill for the first nail in his shoe, 2 mills for the second nail, and so on: the shoes contained 32 nails: how much did he receive' 89. Said a mule to an ass, if I take one of your bags, I shall have twice as many as you, and if I give you one of mine, we shall have an equal number: with how many bags was. each loaded? 90. What number taken from the square of 48 will leave 16 times 54' 91. Divide $1000 between A, B, and C, and give A $120 more than C, and C $95 more than B. 92. A persob being asked the hour of the day, said, that the time past noon was i of the time till midnight: what was the hour? 93. A, B, and C, can trench a meadow in 12 days; B, C, and D, in 14 days; C, D, and A, in 15 days; and D, B, and A, in 18 days. In what time would it be done by all of them together, and by each of them singly? 94. Suppose A, B, and C, to start from the same point, and to travel in the same direction, round an island 73 miles in compass, A at the rate of 6, B of 10, and C of 16 miles per day: in what time will they be next together 2 95. At what time between 12 and 1 o'clock do the hour and minute hands of a common clock or watch point in directions exactly opposite' 96. In how many years will the error of the Julian Calendar involve the loss of a day' 97. A man's desk was robbed 3 nights in succession; the first night half the number of dollars were taken and half a dollar more; the second, half the remainder was taken and half a dollar miore; the third night, half of what was then left and Hralf- a dollar more, when he found he had $50 left: howy much had he at first' THE END. ANSWERS TO EXAM'PLES. NOTE.-At the urgent request of several distinguished Teachers, who ht received Thomson's Higher Arithmetic with favor, the publishers have issu an edition of it, containing the answers in the end of the book, It is hoj that pupils, who may use this edition, will have sufficient regard to their o, improvement, never to consult the answer till they have made a strenulous a persevering effort to solve the problem themselves. N. B.-The work without the answers is published as heretofole. ADDITION.-A'Ls. 5 9- 1 o Ex. ANS. Ex. Airs. Ex. ANs. 1. $5445. 17. 288011295. 35. 9429190. 2. 41757 bushels. 18. 14303433. 36. 11178170. 3. 11596 pounds. 19. 100611775. 37. 10306156. 4. $31551. 20. 1805851434. 388. 10662291. 5. $5583. 21. 337351. 39, 40. Given. 6. 65440 sq. miles. 22. 7221. 41. 214. 7. 102451 sq. nm. 23. 4251988. 42. *253. 8. 528524 sq. m. 24. 3795. 43. 276. 9. 666327 sq. m. 25. 73464390. 44. 19443. 10. 1362742 sq. m. 26. 33604444. 45. 20714. 11. 233890. 27. 15821984. 46. 2476372. 12. 828463. 28. 97059404. 47. 8132085946. 13. 990240. 29. 1038220930. 48. $107109740. 14. 96181521. 32. $570805. 49. 2069857 tons. 15. 127215713. 33. 6460458 yards. 50. $57981492. 16. 869754587. 34. 6657039 pounds. 51.. Given. SUBTRACTION.- ART. 7 6. I. $7095. i0. $12280043. 19. 5313439. 2. 28984 bushels. 11. $23563746. 20. 543679. 3. $30954. 12. 430143 tons. 21. 20079S-84. 4. $46025. 13. 149237. 22. 45103074. 5.5 8000000 miles. 14. 3393329. 23-. 06729549. 6. $6327597.. 15. 54399581. 24. 72820280. 7. $26176670. 16. 8825431. 25. 55301760. 8. $1644737. 17. 4001722. 26. 80200180. 9, $7977899. 18.- 2601900..27. 95658143. 400 ANSWERS. [PAGES 46-60. SUBTRACTION CONTINUED.-ART. 7 76 Ex. ANS. Ex. ANS. EX. ANs. 28. 9000001. 39. 85807025. 50. 925. 29. 09899999. 40. 1598. 51. 1511. 30. 83128433..41. 4004. 52. 41845. 31. 40592424. 42. 1384. $46900, W. 32. 55352005. 43. 14061. $69450, H. 33. 19957466. 44. 12494. 54. $2410 lost. 34. 77919201. 45. 11547. 55. 171825. 35. 70051563. 46. 3295. 56. $1674737. 30. 53201371. 47. 1606. 57. $97. 37. 25311703. 48. 3707. 58. $3893. 38. 86282745. 49. 2664. 59. Given. MULTIPLICATION.-ART. 93. 1. $24795. 15. 3931476. 29. 239968374861. 2. $36099. 16. 415143630. 30. 449148410434. 3. $56700. 17. 31884470. 31. 289975559744. 4. 90520 miles. 18. 8468670. 32. 294144537440. 5. 74175 pounds. 19. 43506216. 33. 335834314400. 6. 372500 days. 20. 11847672. 3i4. 18834782688. 7. 960000 rods. 21. 57380625. 35. 109588282650. 8. 20835. 22. 11050155200. 36. 654638320927. 9. 21576. 23. 12810000. 37. 396890151372. 10. 68198. 24. 48288058. 38. 554270292192. 11. 176400. 25. 3473567604. 39. 2985984. 12. 1554768. 26. 88789980848. 40. 57111104051. 13. 5497800. 27. 9313702853. 41. 60435595442394 14. 1674918. 28. 67226401140. 42. 87112343040000 CONTRACTIONS IN MULTIPLICATION.-ARTs. 97 —18S. 8. $1776. 20. 312046700000. 29. 96000 pounds. 9. $5760. 21. 52690078000000 30. 359400000. 10. $8100. 22. 6890634570000- 31. 143759940000. 11. 5782 s. 000. 32. 287-08635000000 12. 23808 miles. 23. 494603050600- 34.. 123240000. 13. $11736. 000000. 35. 2309760000. 14. 1984' s. 24. 87831206507- 36. 26366200000. 15. $32256. 000000000. 37. 144447000000. 17. 46500 bushels. 25. 678560051090- 39. 31276000000. 18. 365000 days. 000000000. 40. 3747BG0000000, 19. 1534860000. 28. 18750 pounds. 41. 18054680000000 PAGEs 61 —74.] AN S WE nRS. 40 CONTRACTIONS IN MULTIPLICATION CONTINUED. Ex. ANS. Ex. ANm. E. ANs. 42. 664726500000- 74. 4140. 99. 180600000. 000. 75. 27936. 100. 2722946304. 43. 1075635900000- 76. 154250. 101. 2172069918. 000. 77. 11348400. 102. 7225. 45. 45514. 78. 34639552. 103. 65536. 46. 68476. 79. 2685942. 104. 104650. 47. 400624. 80. 2801960. 105. 12744790. 48. 907002. 81. 72156000. 106. 31049291000. 50. 132525. 82. 1680000000. 107. 2732116062240 51. 307664. 83. 2000000000. 108. 222310980000. 52. 2333616. 84. 43644865. 109. 2006685774553. 5691627. 85. 81708550. 896. 55. 474309. 86. 401939564. 110. 1256700743298 56. 6027966. 87. 476413195. 111. 37968867755.' 57. 1293699. 88. 62220780. 112. 39073118478. 58. 4629537. 89. 637049231. 113. 1021288493520 63. 54530. 90. 406101366. 114. 1421400000000 64. 72819. 91. 42261696. 115. 6030240000065. 346896. 92. 504159579. 000. 66. 6624403632. 93. 6724232757. 116. 9130020300067. 17651712450. 94. 7306359. 000. 68. 21983532672. 95. 21760506. 117. 68004000000071. 625. 96. 39429936. 000. 72. 2916. 97. 2283344802. 118. 400000000073. 5184. 98. 650633256. 000000. DIVISION, -ART. 127. 1. 4.5 bu. 13. 5697-ff. 25. 3679. 35. 8264512. 85 bbls. 14. 38238-5. 26. 4500. - -. 3. $68 ]. 15. 4166wA. 27. 50830ofti. 36. 13878054. $3. 16. 21276-4. 28. 630. 64 2. 5. $682-. 17. 121522-9. 29. 235. 37. 900009009 6. $73972sj. 18. 198. 1 -. 30. 648. 9009Ti T. 7. 20-1+ days.119. 873. 31. 26710J-6-. 38. 900090008. 173-14 ds. 20.,-48. 32. 563. -9000ooo 9. 2773-32. 21. 4827-4. 33. 8826211- 39. 9000090010. 1139.41. 22. 874a, O O9ihr 11, 1443-5a2. 23. 108. 34. 2343440212. 1489". 24. 45. 7a. 402 A N S WE rS a C.PAGES 75-96. CONTRACTIONS IN DIVISION.-ARTS. 129-1-39, 1O.. ANSS. X. ANS. X. N. E. AS. x ANS. 1, 2. Given. 18. 36. 37. 15. 55. 2283781{43 1 a. 19. 68. 3. 56.. 41501 4. 4. 672. 20. 363-2 39. 17. 57. 478676J0. 5. 460. 21. 75. 40. 30. 58. 59207. 6. 205. - 23. 1207. 41. 250 days. 59. 1826896. 7 1265. 24. 1690. 42. 950 years. 60. 138791w3. 8. 20; 34; 56d. 25. 65121 43. $lOI-v. 61. 65964142. 9. $650;7650; 26. 8654. 44. $2852. 62. 6162-v~. $43200. 27. 83~45 45. $39-i. 63. 158314 2. 10. 267, and 28. 7644. 46. $24 7-8. 64. -214-u7o-. 50000 R. 29. 774. 47. $11-4ar 65. 413414~. 11. 144, and 30. 1424A. 48. $a4- 31-. 66. 396624L. 360791 R. 31. 94. 49. $2194a4. 67. 1658-2-4. 12. 5823, and 32. 10944. 50. 184,. 68. 7405-121 - 67180309R 33. 169344. 51. 13529~-4. 69. 4362 —z-l. 14. 105 b. 34. 3795 —. 52. 1246649. 70. 3186-9t# 15. 184 bbls. 35. 67. 53. 124544 —. 71. 974644o i6. 19744. 36. 203-Th%-. 54. 134469133 72. 920 u 7u 80 CANCELATION.-ARTs. 50_9 2I~]. 2. 45. 1 4. 65. 6, 7. Given. 9. 3. 3. 63. 15. -73. 8.6. 10. 3. APPLICATIONS OF TIlE FUNDAMENTAL RULES. ARTS. 152 —A59. 1. Given.'11. 79. years; 19. Given. 27. 632. 2. 255 acres. 94 yrs.. 20. 48 beggars. 28. 974. 3. 925 bu. 12. $5104 cair. 21. 20 flocks. 29. 7124; 5516 4. Given. S,345Ii hor..22. Given. 30. 13.000; 5. $190. 14. 65 years. 23. 20 years. 12264. 6. 1125 sleep. 15. 175 rods. 24. 10 months. 31. 21151 s. $2240. 117. 1 878 25 2. 18. 42. 20975. 9. $3436. 18. 1033062. 203. 1062. 32. 786. PROPERTIES OF NU.MBERS.-ARTs. 1629 63o 1 —9. Given.. 11-3. 2024122. 18. 707961. 22. 1614386.. 10, 20212331. 14. 1522365. 19. 1036993. 23. 118620366. 11. 2350147. 15. Given. 20. 9753020. -24, 387909012. 1331124. 16, 17, Given. 21. 360913096. 582. PAGES 97, 98.] A N SW E S. 403 ANALYSIS OF COMPOSITE NUMBERS.-ART, 165. Ex. ANS. Ex. ANS. Ex. ANS. 4. 9=3X3. 28. 2, 3, and 7. 52. 2, and 37. 5. 2, and 5. 29. 2, 2, and 11. 53. 3, 5, and 5. 6. 2, 2, and 3. 30. 3, 3, and 5. 54. 2, 2, and 19. 7. 2, and 7. 31. 2, and 23. 55. 7, and 11. 8. 3, and 5. 32. 2, 2, 2, 2, and 3. 56. 2, 3, and 13. 9. 2, 2, 2, and 2. 33. 7, and 7. 57. 2, 2, 2, 2, and 5, 10. 2, 3, and 3. 34. 2, 5, and 5. 58. 3, 3, 3, and 3. 11. 2, 2, and 5. 35. 3, and 17. 59. 2, and 41. 12. 3, and 7. 36. 2, 2, and 13. 60. 2, 2, 3, and 7, 13. 2, and 11. 37. 2, 3, 3, and 3. 61. 5, and 17. 14. 2, 2, 2, and 3. 38. 5, and 11.' 62. 2, and 43. 15. 5, and 5. 39. 2, 2, 2, and 7. 63. 3, and 29. 16. 2, and 13. 40. 3, and 19. 64. 2, 2, 2, and 11. 17. 3, 3, and 3. 41. 2, and 29. 65. 2, 3, 3, and 5, 18. 2, 2, and 7. 42. 2, 2, 3, and 5. 66. 7, and 13. 19. 2, 3, and 5. 43. 2, and 31. 67. 2, 2, and 23. 20. 2; 2, 2, 2, and 2. 44. 3, 3, and 7. 68. 3, and 31. 21. 3, and 11. 45. 2, 2, 2, 2, 2, and2 69. 2, and 47. 22. 2, and 17. 46. 5, and 13. 70. 5, and 19. 23. 5, and 7. 47. 2, 3, and 11. 71. 2, 2, 2, 2, 2, and3 24. 2, 2, 3, and 3. 48. 2, 2, and 17. 72. 2, 7, and 7. 25. 2, and 19. 49. 3, anld 23. 73. 3, 3, and 11. 26. 3, and 13. 50. 2, 5, and 7. 74. 2, 2, 5 and 5. 27. 2, 2, 2, and 5. 51. 2, 2, 2, 3, and 3. 75. 2, 2, 3, 3, and 3. 76. 120_2X2X2X3X5. 82. 1492=2X2X373. 144=2X2X2X2X3X3 8032=2X2X 2X2X2X 77. 180=2X2X3X3X 5. 251. 420=2X2X3X5X7. 83. 4604-=2X2X1151. 78. 714=2X3X7X17. 16806-2 X 3X2801. 836-2XX2X11 X19. 84. 71640-2X2X2X3X3X5 79 574 —2X 7X41. X 199. 2898-2X3X3X7X23. 20780- -2X)X5 X l039. 80. 11492=2X2X13X13X17 85. 84570 —2X3X5X2819. 980 —-2X2X5X7X 7. 65480-2X2X 2 X 5X1637. 81. 650=2X5X5X13. 86. 92352=2X2X2X2X2X2 1728=2X2X2X2X2X2 X3X 13X37. X3X3,3. 81660=2 X X 3 X 5X1361. 18 404 X N s T,E Rt S. [p AGES 99 —118 GREATEST COMiMON DIVISOR. —ARTS. 16~ —][.7t Ex. ANS. EX. ANS. Ex. ANS. Ex. ANS. 1. Given. 6, 7. Given. 12. 1. 18. 12 2. 3. 8. 15. 13. Given. 19. 18. 3. 7. 9. 14. 14.3. 20. 35. 4. 5. 10. 111. 15. 16. 21. 6. 5.2. 11. 39. 17. 15. 22. 28. LEAST COMMON MULTIPLE.-ARTS. 1t6, 17E7e 1 —3. Given. 8. 720. 14. 144. 19. 360. 4. 90. 9. 12600. 15. 600. 21. 600. 5. 144. 10. 504. 16. 2520. 22. 1440. 6. 180. 11. 1134. 17. 252. 23. 13824. 7. 360 12. 15015. 18. 1134. 24. 51000. REDUCTION OF FRACTIONS.-Awr's. I 95P-201 1, 2. Given. 15. 3. 30. 10a 7- - 4L7. T-x. 3.. 16. 3. 3.1 48. 4 7 4.. 1 7. 1c7. 34, z7. 49. __t~_ 7., 3 325~ 5.. 18. 4 35. 52. 5. 5 4 9,7 163~ 6. A 21. 9. 36. JI(1- ~ 3. -SL )7. m. 22. 5. 37. -3. 54. -r~~~~~. 9.~~i 4 0 8. 29~ 5r ~~~~~~~~~~~~38. -k 1 4 _Q_5 8. -. 23. 3-3-. 38 0 o 55. 9. 24. 91. 39. 3x- 56 33~~~ 24. 9L ~~~~ 3 3 5. 10. -i. 25. 1. 40.a 5 B2* 57. -Il?. 11. 1. o6. 60. 1. 1o. 8. -. 12. 2. 27. 21. 42. 53;. 59. -5 13. -'~ 28. 52. 43. -4,h1'~. 60. S 14.2. 60-. ~ 44. 5 61 62 Given 9 o. 4. 6}a.62 63. "SQ* 2 i a. a0Q 75 1. -_!;_.4 420 4 20 420 42' 4 1 4 3 540 -,_. -3 * - 76 -.; 2~ *. _l 6z. 54U~ v540 540 ~ 540' * 36~ 36 5 6 36' 65. xQ. ~o. 1z3:a *. o~18 * 29 2 0270 270 2 27 0 ~ 24 4 24 24' 66. 2izo; -2~m o L0,5- 22 85. -L~ -40-. -ca. 2; 2205 512215 20 72- 7~ 7 9 7267. 4 0zfa. o AA-o. A _7. 64. E' 7 6 * 0 7 6 7 6 7 7 7 7 6 756' 72 268. s~-~~~7s, ~-,~a~2. _~oo.axo.~,t_ 475 8.0,: _n..aa Xo__ 32 9 a0** 3A 58. ff 5 77'5 5 7 7 5 5 7 7 5 3 6 3 6 86 3 6 6gQ9Q. 3_2o__o. ~ai _._a_ 8 Q1 Ea *:a __._G. ao_ 6 92400 ~ 92400 ~ 92400 8 "' "~ 40 c_ ~ 4 ~ 0 _ a _OAo A_5a~. _.4 r, ar,_2o0-_o 8') a. 5 7. A, 5A 65.o 11L_ LLI -Lin i 0; {% o,; { o. 70. 6 5 6 0 0 i 6 5 U O ) 6 5 O U U - 6 0 1 6 U 6 U 6 O. 2- - -0- ~ *' 2 —_ o.65 QQ S30o 0 ~'"~ ~ a-.. 1.5; 25 0, 5 6 2 5 o0 5 6 2 5 0 * 4 2, 4 2, 4-2, 4 2' 72. of. a4 _L3o.,- _ L. 8 48 _7. -4'';l 6 0 5 0 0 6 7 5 0 U * 6 0, 6 0 6 G 6 oG''3, 74, Given. 85. -%u%%, -o0, 4 -285o; -M.608 PAGES 118 —l7.1 ANSW ER S. 405 REDUCTION O)F FRACTIONS CONTINUED. —ART. 21. Ex. ANS. Ex. ANs. 86. 7- 76767 8.; 5 5 89- 6 76755 1260 67 7f6725* 90. 7,; 71; tt5; 7. 87. 63~oo, 0 7as~o _. ~l~ 91 78~ a_ 2 U z ao 8 7. 6300 6300 u, 5 4 0; 540 54 5 4 0 88. 18444; p44 4i; t~' 44. ADDITION OF FRACTIONS.-ARTS. 202-2@04 1 —3. Given. 13. 1-35. 23. -G,6; 3. 34. 3-< 7 a 9,Q 4. 3H. 14.. 24. OF A CNS-A 35 _ - 9_0. 5. 2j. 15. 23 21. 1-. 252. 7; 4-36. 64690 6. Given.16. 2-3k. 26. 1614... 37. 7. 4.~=3-i. 18. 3715-~. 3 305 3-. 49. 15282. 7. 27-a-. 17. 1 1 9+. 27. 13,2-; -. 38. 279.la 9 8. —. 18. 18. 29. A209 3. 4 39.' 02 9. 2T. 1 197618. 30. 350. 10. 3,-?. 20. 15240. 31. 5415. 1 4 z t3-. 11. 5-. 22. 35, or -4-r; 32. 37-5t. 42. -672 k8. 12. 2z-2. _ l0_,or 33. 1 43. a8Ga. 43. SUBTRACTION OF FRACTIONS.-ARTs. 206-208..~ -~. 10. 1t.az 107. 1213-a-o Z 25. 291z. 5. 6w. 11. 18. 278. 26. 40342. 6. 12. 023. 12. 1235. 21.. 27. 94-. 71. 13. 1 3-i 4 6t. 20. 238920. 28.. 8. MS. 15. 12-i. 23. i173-. 29. 8263a. 9.?60 -. 16. 6945V-. 24. 466. 30. 71*. MIULTIPLICATION OF FRACTIONS,. —CASE I.-ARTS. 21l —lY. 1 —3. Given. 17. 633L. 32. 5141. 48. 6897. 4. --— 3-. 18. 3715. 33. 305-n 0. 49. 15282. 5. 4-12. 19. 44489. 35. 10396-. 50. 2931 S. 6. 189-. 20. 2264-,Af, 36. 17460'63. 51. 1280. 7. 37-. 21. 12519}. 37. 366[-2g. 52. 279. 8. 48 93. 24. 108. 38. 2067 -S. 53. 4496 9. 89-&%. 25. 127. 39. 35650-f-. 54, 8113. 10. 66f2. 26. 240. 40. 23555-1-a 55. 10413-f. 11. 167T-,. 27.-435. 43. 375. 56. 5-672 2. 12. 776B68Y. 28. 560. 44. 738. 57.- 508618. 13. 662 za1 29. 621-. 45. 1178. 58. 43452. 14, 15. Given. 30. 701$. 46. 3450. 59. 7-4290-.. 16. 735 31. 3 i. 76. 47. 6795. 60. 92280H. 406 AN SW ER S. [PAGES 28- 142. MULTIPLICATION OF FRACTIONS CONTINUED.-CASE II. ARTS. 2199 220. gx. ANs. Ex. ANS. Ex. ANS. EX. A'Ns. 1 2. Given. 11. 17. 20. 3232#3243. 29. 99_. 3. A = Z=-. 12. 534. 21. 5143-14v37-. 30. 784. 4. -13=i. 13. 242-2. 22. 6998,f. 31. 47-:,%. 5.-~jk. 14. 329. 23. 167444. 32. 86. 6. 4. 15. 13624. 24. 53W4o-. 33. 4014. 7..szz. 16. 319841. 25. 24091a. 34. 1474-m6-. 8. 75. 17. 451,-H. 26. 466B-. 35. $9724. 9. 4. - 18. 8344S 7. 27. 300000. 36. $6323,ju. 10. Given. 19. 2001j-f. 28. 700. 37. 5501-4m. CONTRACTIONS IN MULTIPLICATION OF FRACTIONS, ARTS. 221-225. 3. A. 11. 4. 21. 57600. 32. 47621. 4.. 12. -4. 22. 99000. 33. 159374. 5. 4. 13.. 23. 3. 871. 34. 401871. 6. -. 14. 1-. 24. 14220. 35. 65450. 7. 4. 15. -4. 26. 21334. 37. 12784. 8. a. 16. #. 27. 184664. 38. 40834. 9. 4. 19. 4931. 28. 5580. 39. 8150. 10. 26. 20. 85334. 29. 430000. 40. 938333. DIVISION OF FRACTIONS.-ARTs. 226 —-241. 1 —3. Given. 18, 19. Given. 33. 9-zi. 51. -. 4. 20. 0. 34. 13444. 52. 4. 5. 21. 37. 3. 3. 53. 311. 6. 22. 234. 38. 8 —-4. 54.. 7. -4. 23. 404. 40. 220o. 55, 1-. 8. -oo-=-. 24. 4. 41. 302-. 56. 1. 9, TH%. 25. 4. 142. 504o 7 58. 164. 10 —1.3. Given. 26. 3-,. 43. 6-95. 59. ~. 14. 1r-. 29. 135. 46. 383-3L,j-. 60. 33. 15. 2+6. 30. 168. 47. 54-jaoa%. 61. 5s. 16. 342. 31. 11 ~7. 49. 2-h. -3. 4. 71. 244. 32. 116 7. 50. 44. 64. -. APP.LICATION OF FRACTIONS.-ART. 242. 1. 88)-' yds. 4. 7.144-4. 7 8$1548~. 10. 5606-A s. 2. 162- lbs. 5..02-4. 8. $55164. 11. $483-. 3. 8 544.. 4635-4. 9. $1515. 1Q. $100. 6 6 3~~~~~~~~~~~~~~~~~~~~~~~~~~ PAGES 143 —170.] A N S W i R S. 407 APPLICATION OF FRACTIONS CONTINUED.-ART. 24:2. Ex ANS Ex. ANs. Ex ANS. Ex. ANs. 13. $16111. 21. 165~ yds. -29. 38-,- rods. 37. $5 -5-. 14. 404552 b 22 218lbs. 30. 821 30. 38. 42_7q tons. 15. $1806. 23. 6264 gals. 31. $6-. 39. $1W-~. 16. 3612T bu. 24. 20 — lbs. 32. 5844 bu. 40. $1i4l-. 17. $30968-. 25.-53 - yds. 33. 44 doz. 41. $3.9-_. 18. 6939-1- m. 26. 27 boxes. 34. 6-32- cts. 42. 266T4-z- d 19. 5229 m. 27. 153a bbls. 35. 94-4-4 s. 43. $33798-i-. 20. $9175-. 28. 291{ suits. 36. $139-2-,-. REDUCTION.-ART. 282. 2. 68810 far. 28. 8553600 in. 52. 28992 pts. 3. 86768 far. 29. 5280000 yds. 53. 1427 bu. 1 pk. 4. 284079 far. 30. 54 m. 7 fur. 38 r. 54. 130100 qts. 5. 96615 far. 2 yds. 2 ft. 55. 36360 min. 6. ~25, 13s. 6d. 3 far. 31. 9 1. 2 mn. 4 fur. 31 r. 56. 31557600 sec. 7. ~433, Is. 2d. 3 far. l yds. 2 ft. 7 in. 57. 84wks. 6hrs. 45min 8. 266 guin. 18s. 8d. 32. 5031 rods. 58. 65d. 2h. 4m. 40 see 9. 1448 sixpences. 33. 17 m. 20 r. 59. 31556928 sec. 10. 6050 threepences. 34. 132105600 ft. 60. 946728000 sec. 11. 170472 grs. 35. 2560 na. 61. 10 yrs. 12. 9000 pwts. 36. 5000 qrs. 62. 397200". 13. 1010047 grs. 37. 6396 yds.2qrs. 1 na 63. 1350000." 14. 2 lbs. 1 oz. 10 pwts. 38. 9302F. e. 4qrs. 3na 64. 2126o, 11', 54". 16 grs. 39. 10156 na. 65. 555555s. 16o, 40'. 15. 1771bs. 9 oz. 12pwts 40. 7116 qts. 66. 470660 sq. ft. 16. 1596 lbs. 41. 693 gals. 67. 4366073i sq. ft. 17. 564000 oz. 42. 26528 gi. 68. 32640858360 sq. in. 18. 104300 lbs. 43. 48 bar. 20 gals. 69. 582 A. 1 R. 3 r. 19. 71680000 drs. 44. 117 pi. 1 hhd. 46 g. 269k- sq. ft. 20. 10 cwt. 16 lbs. 3 qts. 1 pt. 2 gi. 70. 259200 cu. in. 21. 133 T. 12 cwt. 35lbs 45. 102128 gi. 71. 4551552 cu. in. 22. 1 T. 202 lbs. I oz. 46. 12960 pts. 72. 10877760 cu. in. 23. 9120 drs. 47. 87 bar. 26 gals. 73. 49 cu. ft. 1 cu. in. 24. 37440 sc. 48. 630 hhds. 44 gals, 74. 306 C. 48 cu. ft. 25. 64 lbs. 11 oz. 5 drs. 49. 19520 pts. 75, 4492800 cu. in. 26. 88 lbs. 4oz. 7 drs.2sc 50. 488 qts. 76. 52 T. 40 cu. ft. 27. 142560 ft. 51. 24440 qts. 180cu. in. APPLICATIONS OF REDUCTION.-ARTs. 2S3-2c-~. 1. Given. 4. 1717 lbs. Troy, or 7. 58 lbs. 4 oz. Troy. 2. 576 lbs. avoir. 1454l-`- lbs.avoir. 8. 21-jZa3- lbs. Troy, 3. 691 lbs. 10 oz. 5. 265.bs. Troy, or 9. 271 lbs 3 oz. 5-A3- drams. 2 181 -5- lbs. avoir. 10. Given. L08 A E R S. LPAGES 171-180 APPLICATIONS O{F REDUCTION CONTINUED.-ARTS. 2-2-93. Ex. ANS. EX. ANS. Ex. ANS.!1. 360 sq. ft. 30. 3456 wine gals. 49. 14 hhds. 48iU g. 12. 14 A. 10 sq. rds. 31. 8640~ w. gals. 51. 51- _ beer gals. 13. 108 sq. y. 8 sq. ft. 32. 5184 beer gals. 52. 45-2 wine gals. 14. 446 A. 1 R. 33. 6912 b. gals. 2 qts 53. 24a_ w. gals. 15. 40 A. 34. 80-A5- bu. 54. 1598 w. gals. 16. 36 sq. yds. 35. 100 bu. 55. 2207f7 w. gals. 17. 66 sq. yds. 36. 800 bu. 56. 3125 — qts. 18. 111~ sq. yds. 137. 897ai~ w. gals. 57. 2734'' gals. 20. 56- Cu. ft. 38. 7122a bar. 59. 8 min. 36 see. 21. 126 cu. ft. 39. 902S6753L hhds. 60. 39 min. 22. 86 C. 2 cu. ft. 41. 6229 cu. ft. 61. 1 hr. 8 m. 40 see. 23. 748 cu. ft. 42. 1244~ cu. ft. 62. 33 min. 48 sec. 24. 756 cu. ft. 43. 827 cu. ft. 63. 12 h. 28 m. 12 s. 25. 72 cu. yds. 44. 210~4 cu. ft. 64. Given. 26. 160 cu. ft. 45. 842-ia- cu. ft. 65. 4~ 45'. 27. 1800 cu. ft. 47. 42-1%5' bu. 66. 12~ 46'. 29. 17280 bu. 48. 46 6i gals. 67. 13~ 23'. COMPOUND NUM'IBERS REDUCED TO FRACTIONS.-ARr. 296. 1-4. Given. 12. 7 yd. 20. U 6#4~v. 29. 3:6. ~ 14.' 10- A. 23. - 31. 8. jE lb. Troy. 16.: gal. 2 33. - 9. -6%- lb. Troy 17. l hhd. 26. -.i 34. -1-T-. 10. 3-5-lb. avoir. 18. - 7. d. 27. -1SUS4- - 35. 1. 11. a-_ T. 19. 9 -- hr. 28. 36. _-_z. FRACTIONAL COMPOUND NUMBERS REDUCED TO WHIOIE, NUMBERS OF LOWER DENOMINATIONS.-ARTS. 2979 298. 3. 17s. Gd. F13. 2 qts.1 pt. 1-' gi. 25. 6171 hrs. 4. 7d. far. 14. 55 gals. 1 pt. 26. 2688 min. 5. 5 oz.2 p.20-t. 16. 3 pks. lqt. lpts. 27. 8-,8,- na. 6. 12 pwts. 12 grs. 7. 46 min. 40 see. 38. 17 — qts. 7. 10 oz. 10 drs 1 2hrs. 36 min. 3 29..-174i qts. S. 571lbs. 2oz. 41drs. 19. 22 se. 330. - 4 2% oz. 9 12501bs. 120. 17' S31. 31. 66 pwts. 10 2 ft. 4 in. 2 2 d. 32. _7- 2 r. 11. 6 t. 2 in. h. -- oz. 33. -- sq. ft. 12. 177 r.12 ft.10 in 24. 1 4.., 70. PAGES 182-190.1 AN S WE R S. 409 COMPOUND ADDITION.-ART. 300. Ex. ANS. Ex. ANS. Ex. ANs. 3. F106, 3s. id. 10. 1091. 2 m. 6 fur. lft. 16. 240 gals. 4. ~188, 13s. ~d. 11. 114 yds. 3 qrs. 17. 181 hhds. 59 gals 5. 9 T. 8 cwt. 17 lbs. 12. 387 yds. I qr. 1 pt. 1 gi. 6. 45T. 4 cwt. 57 lbs. 13. 138 A. 114 sq. r. 18. 115 w. 15 h. 25 m. 2 oz. 80 sq. ft. 19. 322 bu. 1 pk. 5 qts. 7. 107 lbs. 7 o. 8 p. 1 g. 14. 468 A. 1 R. 33 sq. r. 20. 135 qrs. 3 bu. 3 pks 8. 330 lbs. 2 o. 3 p. 5 g. 15. 43 sq. yds. 5 sq. ft. 2 qts. 9. 4 fur. 13 r. 13 ft. 3 in. 125 sq. in. COMPOUND SUBTRACTION.-ARTs. 3029 303. 1. Given. 9. 35 bu. 2 pks. 6 qts. 19. 25~ 3' 15" 2. ~9, 2s. 8d. 3 qrs. 10. 19 qrs. 6 bu. 2 pks. 20. 350 3' 30' 3. ~60, 4s. 7d. 3 qrs. 11. 55 yds. 2 qrs. 3 na. 21. 100 26'. 4. ~499, 13s. 4d. 2 qrs. 12. 44 yds. 1 qr. 3 na. 22. 54 yrs. 2 mos. 2 wks 5. 8 cwt. 1 qr. 6 lbs. 13. 6 gals. 2 qts.'1 pt. 6 d. 2 hrs. 45 min. 6 s 10 oz. 14. 48 lhhd. 46 g. 2 qts. 23. Given. 6. 24 T. 1 cwt. 71 l56. 15. 85 A. 119 r. 24. 67 yrs. 9mos. 22 d, 7. 19 m. 289 -. 2 ft. 16. 235 A. 48 r. 25. 8. 11. 1 m. 7 fur. 10 r. 17. 56 C. 90 cu. ft. 26. 1 yr. mos. 11 d. 12}- ft. 18. 339 cu. ft. 26 in. 27. 3 yrs. 9 mos. 22 d. COMPOUND MULTITPLICATION.-ART. 305. 1, 2. Given. 14. 2044 1. 1 m. 4 fur. 23. 158910 -13' 30" 3. ~247, 6s. id. 30 r. 24. 2040 10'. 4. ~24, 9d. 15. 8962 bu. 16 qts. 25. 4581 bu. 8 qts. 5. 17 T. 55 lbs. 16. 2968 qrs. 5 bu. 2 pks. 26. 2453 bu. 4 qts. 6. 403 T. 17 cwl. 55 lbs 6 qts. 127. ~5, 16s. 10~d. 7. 689 lbs. 8 oz. 16pwts 17. 7821 A. 20 r. 28. ~679, 3s. 4d. 8. Glbs. 10 oz. 10 pwts 18. 25172 A. 1 R. 3 r 29.;297. 9. 3039 hhds. 39 gals. 19. 24645 cu. ft. 930 30. ~507, 16s. 3d. 1 qt. 1 pt. cl. i1'. 31. 36 C. 74 cu. ft. 10. 5668 pi. 32 gals. 20. 96350 C. 50 cu. ft. 944 in. 11. 2358 yds. 21. 12783 d. 11 h. 28 m. 32. 865 lbs. 12 oz. 12. 5375 yds. 22. 1199 yrs. 9 mos. 33. 25418 lbs. 12 oz. 13. 14778 m. 1 fur. 32 r. 3 wks. Id. 34. 8662 gals. 2 qts. COMPOUND DIVISION.-ART. 37,. 1-3. Given. 8. ~4, 17s. 3d. — qr. 4, 51 lbs. 3 oz,. 10 pwts. 9. 10 yds. 3 qxs. 1 na. 153 grs. 10. 9 yds. 1 qr. H- na. 5. 31 bu. 14~- qts. 11. 83 in. 2 fur. 26 r. 11 ft, 6. 25 bu. 1J pts. 12. 214 m. 2 fur. 27 r. 7. ~20, is. 6d. 4 ft. -} in. 410 AN SW ER S. [PAGES 196-201 COMPOUND DIVISION CONTINUED.-ART. 307O Ex. ANS. Ex. ANS. 13. 1 gal. 2 qts. 1 pt. 11-'} gi. 18. Is. 17. 52/ 214".. x 41 ~ 14. 44 hhds. 29 gals. I pt. 91- gi. 19 9 C. 84 ft. 1016-l in. 15. 24 d. 8 hrs. 42 lmin. 40 sec. 20. 6 C. 92 ft. 8506-j- in. 16. 10 yrs. 35 d. I hr. 13 min. 21. 6s. 10~d. 11-1' sec. 22. 7s. lId. 3-1 qrs. 17. 10 48' 411'". 23. 10s. lid. 2-9 qrs. ADDITION OF DECIMALS. —ART, 320. 1, 2. Given. 9. 857.005. 16. 2.471092. 3. 428.1739. 10. 1097.84143. 17. 0.0711824. 4. 103.8523. 11. 1408.25559. 18. 0.3532637. 5. 14.747274. 12. 127.05034. 19. 0.807711. 6. 60.149. 13. 33.3182746. 20. 0.1627165. 7. 332.1249. 14. 15674.1613. 21. 0.996052. 8. 501.15998. 15. 1.807. 22. 0.329773. SUBTRACTION OF DECIMALS. —ART. 322. 1, 2. Given. 13. 2.291. 24. 0.000999. 3. 1427.633782. 14. 9.9999999. 25. 699.93. 4. 20.987651. 15. 8.000001. 26. 28999.908. 5. 72.5193401. 16. 4635.5346. 27. 255999999.744. 6. 81.16877. 17. 541.787. 28. 0.414. 7. 0.066721522. 18. 46.43606. 29. 0.0041. 8. 0.01. 19. 0.0000999. 30. 0.000000000999, 9. 9.999999. 20. 0.000396. 31. 0.002873789. 10. 64.0317753. 21. 31.99968. 32. 0.062156. 11. 24680.12377. 22. 44.99955. 33. 0.71699. 12. 24.75. 23. 98.99999901. 34. 0.0000843174. MULTIPLICATION OF DECIIALS. —ART. 324. 1. 681.45 ft. 13. 36.740232. 25. 0.00164389993. 2. 25020 miles. 14. 919.82036. 26. 160.86701632806. 3. 2055.375 gals. 15. 0.000000072. 27. 0.06288405909156. 4. 136.125 nails. 16. 0.00105175. 28. 2.5067823. 5. 788.0125 sq. yds. -17. 390.657556. 29. 64.327106105314. 6. 43560 sq. ft. 18. 275.230594. 30. 0.-0000118260069. 7. 2465.375 sq. rods. 19. 148.64244532. 31. 11027.-140199543710 8. 0.250325. 20. 73.25771882. 32. 94167471.8696549. 18.93978. 21. 52.17977576. 039. 10. 14.78091. 22. 0.0306002448. 33..0000('00667654211. 0.613836. 23. 4701.169144360. 672; 12. 0.0320016. 24. 536o660(075952. PAGES 201-213.] ANSWERS. 411 CONTRACTIONS IN MULTIPLICATION OF DECIMALS. ARTS. 325-327. Ex. ANS. Ex. ANS. Ex. ANS. 1. Given. 9. 75000. 17-20. Given. 2. 429302.13401. 10. 6.5. 21. 0.09484. 3. 106723.50123. 11. 48. 22. 1.262643. 4. 608340.17.. 12. 2480. 23. 0.0769. 5. 304672.14067. 13. 381. 24. 0.0389254. 6. 44632140.32. 14. 65.04. 25. 0.00876. 7. 2134567.82106. 15. 834000. 26. 0.002516. 8. 500. 16. 10. 27. 0.001789. DIVISION OF DECIMALS. ART. 330. 1-3. Given. 12. 79098.8235-f. 21. 83671000. 4. 13 boxes. 13. 0.6344+. 22. 255.1210+. 6. 8 suits. 14. 1210.2344-. 23. 0.000005. 6. 4.98347+days. 15. 0.03. 24. 60.2589. 7. 82.9997floads. 16. 134.8805+. 25. 211.076. 8. 27.7173+days. 17. 59.4060+. 26. 400000. 9. 150.25 bales. 18. 24.8266-f-. 27. 60000000. 10. 5.9291+. 19. 4320.67. 28. 4000000. 11. 6.632. 20. 0.02. 29. 311.487360-+-. CONTRACTIONS IN DIVISION OF DECIMALS.-ARTs. 331-33 1, 2. Given. 7. 0.000012300456. 12. 1.611. 3. 67234.567. 8. 0.0000020076346. 13. 0.04026. 4. 103.42306. 9. Given. 14. 0.0954776. 5. 0.42643621. 10. 0.1274. 15. 2.0208. 6. 6.72300045. 11. 0.09471. 16. 0.980439. DECIMALS REDUCED TO COMMON FRACTIONS.-ART. 335. 1, 2. Given. 6. o 10. 250 14. -2. 3. 0. 7. -02.711. 26 15. Zi6n5a25 4. 2 8. 05. 12. 0. 8o 27. nt 5. fXzou. 9. ~%. 13. -qAL6%-. 17. 2-o0 0. COIMMON FRACTIONS REDUCED TO DECIMALS. ARTS. 337 —3I44 1-3. Given. 110. 0.6. 172. Terminate0.65. 26. Terminate. 4. 0.5. 11. 0.8. 18. 0.75. 7. Inteminat 5. -05.. 2. 0.5. 19. 0.875. 6. 0.5. 13. -0.125. 20, 21. Given. 28. Terminate. 7. 0.75. 14. 0.25. 22. Terminate. 31. 0.,3. 8. 0.2. 15. 0.375. 23. Terminate. 32. 0.6. 9. 0.4. 16. 0.5. 24. Interminato. 33. —0.16. 18* 412 A N S WV E R S. [PAGES 214 —224 COMMON FRACTIONS REDUCED TO DECIMALS CONTINUED. Ex. ANS. Ex. ANs. Ex. ANS. Ex. ANS. 34. 0.3. 41. 0.714285. 48. 0.6. 56. 0.004882835. 0. 42. 2..857142. 49. 0.7. 125. 36. 0.83. 43. O.i. 50. 0.8. 57. 0.583. 37. 0.i42857. 44. 0.2. 51. 0.1875. 58. 0.076923. 38. 0.285714. 45. 0.3. 52. 0.076923. 59. 0.0104895. 39.,0.42857i. 46. 0.4. 53. 0.024. 60. 0.4683544340. 0.571428. 47. 0.. 55. 0.0112.379. COMPOUND NUMBERS REDUCED TO DECIMALS.-ART. 346. 1, 2. Given. 6. 0.41 6 s. 10. 0.2583 hr. 14. 0.875 bu. 3. ~0.5375. 7. 0.5416 s. 11. 0.127083 d. 15. 0.5625 pk. 4. ~0.825. 8. 0.115625 m. 12. 0.0525 cwt. 16. 1.125 gals. 5. ~0.87916. 9. 0.25625 m. 13. 0.46875 lb. DECIMAL COMPOUND NUMBERS REDUCED TO WHOLE ONES.-ART. 34S, 2. 14s. 6d. 7. 6 oz. 15.36 drs. 11. 1 qt. 1 pt. 3.4432 gi, 3. 2s. 7d. 3.2 qrs. 8. 88 rods. 12. 10 h. 13 m. 9.12 sec, 4. id. 2 qrs. 9. 7 ft 0.51 in. 13. 50 min. 42 sec. 5. 9d. 3.6 qrs. 10. 11 gals. 1 qt. 1 pt. 6. 12 lbs. 8 oz. 3.7184 gills. REDUCTION OF CIRCULATING DECIMALS. —ARIS. 355 —61. 1, 2. Given. 8. -"59, or -,T. 17. -. 0.333. 3., or.. 18. 0.045. 4. 1, or-'33. 10. -i3.v 139. 495O 24. 4.3213. 5. 2_z, or al 14. 7% 20. at9 715 6 3 or1101 4 " 9$6.4263. 6. or 15. 9A-}A, J3- or. 1_o — 7. 9-, or -11- 16. -7A 23. 0.2776000. ADDITION OF CIRCULATING L)ECIMALS. —ART. 362. 2. 179.2745563. 5. 594.691 8. 1380.0648193. 3. 476.65i29. 6. 112.7224. 9. 5974.103i,. 4. 47.86683. 7. 223.5107744. 10. 339.626177443 SUBTRACTION OF CIRCULATING DECIMALS.-ART. 363. 1, 2. Given. 5. 4.789. 8. 218.60. 3. 391.5524. 6. 400.915. 9. 0.613640731. 4. 3.81824, 7. 3.9046. 10. 245La:386. PAGES 225-235.] A:N S E R s. 413 MULTIPLICATION OF CIRCULATING. DECIMALS.-ART. 364. Ex. ANS. Ex. ANS. Ex ANS. 1, 2. Given. 5. 389.185. 8. 31.9i1. 3. 0.082. 6. 778.14. 9. 34998.4199003. 4. 1.8. 7. 75o730.518. Il0. 2.297. DIVISION OF CIRCULATING DECIMALS.-ART. 36i5. 1, 2. Given. 5. 7.72. 9. 62.3238341963. 55.69. 6. 8574.3. 891i. 4. 5.41463. 7. 3.506493. 10. 1.42292490118. 3.145. 85770750988. ADDITION OF FEDERAL MONEY.-ART. 374. 1. Given. 7. $3531.432. 13. $8765.12. 18. $1945.258 2. $265.04. 8. $12200.524 14. $16989. 19. $82110.17. *3. $581.128. 9. $185.285. 15. $378.383. 20. $71774.75. 4. $560.56. 10. $74.33. 16. $300.166. 21. $27860.74. 5. $1195.34. 11. $350.32. 17. $256.213. 22. $81800.63. 6. $1431.50. 12. $6491.05. SUBTRACTION OF FEDERAL MONEY.-ART. 375. I. Given. 6. $183.22. 11. $9947.788. 16. $2.937. 2. $12.13. 7. $323.47. 12. $61119.364 17. $32.056. 3. $84.82. 8,. $373.82. 13. $18.981. 18. $10890.07. 4. $247.15. 9. $10870.75. 14. $88.11. 19. $89989.90. 5. $918.48. 10. $1699.49. 15. 8189.92. MULTIPLICATION OF FEDERAL MONEY.-ARTs. 377, 37S3. 3. $83.60. 9. $84.875. 14. $2.84375. 19. $28.125. 4. $517.625. 10. $193.75. 15. $909.375. 20. $220.50. 6. $39.59375. 11. $205.625. 16. $2.70. 21. $142.50. 7. $1440.75. 12. $326.25. 17. $14.0625. 22. $2331.875. 8. $40.59375. 13. $2.0925. 18. $15.78375. 23. $14084.125E DIVISION OF FEDERAL M'ONEY.-ARTs, 379-3S1. 1. Given. 5. Given. 10. 543.518+ yds. 2. $4.50. 6. 8.207 coats. 11. 991;421+ doz. 3. $0.06. 7. 7.871+ times. 12. 360 skeins. 4. $3.13. 9. 308.035+ gals. *13. $3.52-+ 114 AN SWE S. LPAGEs 236-248 DIVISION OF FEDERAL. MO)NEY CONTINUED.-ART. I]1. Ex. ANs. Ex. As. Ex. ANS. l4. $1.50. 19. $0.04049+. 24. 113.56377- tons l5. 86.25. 20. $0.02709~. 25. $0.595238+ l6. $1.973-. 21. $i.78008+. 26. 245.517+ acres. 17. $3.615+- 22. $1.5435+. 27. 500 cows. 18. $0.084+. 23. 1714.285+ bu. 28. 150 carriages. APPLICATIONS OF FEDERAL MONEY.-ARTS. ~32-ae& 1. Given. 13. $5885. 28.,P0.0072. 2. $800. 14. $10538.625. 29. $0.0064. 3. $511.50. 15, 16. Given. 30. $13.4719+4. $780. 17. $6.33375. per cwt.; 5. $780. 18. $104.55. $0.1347196. $1350. 19. $114.198. per lb. 7. $1020. 20. $59.5859. 31. $12.88506 CWt. 8. $864.50. 21. $505.3775. 80.1288506 lb. 9. $2418. 22. $1901.75. 32. $129.625. 10. $4440. 23. $5.40625. 33. 208.838. 11. $1424.75. 24. $52.126. 34. 81734.875. 12. $2691.875. 25. $437.645. 35. ~13703.78. PERCENTAGE.-ART. 38 S 5, 6. Given. 14. $43.13rec'd. 22. 375 sheep. 30. $90.4824. 7. $7.6875. $819.43 paid. 23. 9$1568. 31. $844.08. 8. $8.7526. 15. $402.05. 24. 187.5 lost; 32. $4724.775. 9. $3.4608. 16. $134. 1312.5 saved. 33. $1250. 10. $8.7078. 17. $32.625. 25. $8.125. 34. $12000. 11. $114.1070. 18. $34.03575. 26. $6.316. 35. $21900; 1st; 12. $10.50. 19. $62.50. 27. $84.52016. $14600, 2d 13. $219. 20. $146.666+. 28. $250. 36. $200. 21. $8.771875. 29. $750. 37. $0.95. APPLICATIONS OF PFRCENTAGE.-ARTS. 395-9-7. 1. Given. 10. $619.887. 19. $761904.761. 301 $8840.70. 2. $12.507. 11. $44.32. 20. $4126.55. 31. $7072. 3. $58.878. 12. $673.75. 21. $1413.975. 32, $3552.50. 4. $73.159. -13.' $57.29. 22. $46.50. 33. $1350 rec'd, 5. $115.2'03. 14. $415.831. 23. $22.113. $180 lost. 6. $615. 15. $106.831 A. 24. $9.375. 34. $7490.50. 7. $583.842. $2029.798 0. 25. $318.975. 35.; $960. 8. $52.834. 17. $21078.431.' 28. $3692.50. 36. $4427.50. 9. $155.875. 18. 1$3439.613. 29. $2-250. 37. $9028.50, PAGES 251-270.] A N S WV r S. 415 INTEREST.-ART. 404. Ex. ANS. Ex. ANS Ex. ANS. Ex ANS. 1. $29.61. 10. $8.103. 19. $889.44. 28. $3312.209. 2. $43.255. 11. $6.853. 20. $1.135. 29. $5278.162. 3. $40.367. 12. $19.14. 21. $1.409., 30. $16.158. 4. $51.20. 13. $60.27. 22. $1.898. 31. $206.718, at 5: $60.263. 14. $89.40. 23. $102.125. 360 days; 6. $44.414. 15. $958.41. 24. $154.216. $203.886, at 7. $194.58. 16. $657.45. 25. $704.083. 365 days. 8. $17.803. 17. $1006.833. 26. $2.975. 32. $66778.64 9. $28.206. 18. $1585.018. 27. $76.131. SECOND METHOD.-ARTs. 4.09-413. 4. $8.50. 15. $82.078. 26. $307.65. 38. $137.288. 5. $1.065. 16. $39.179. 27. $227.994. 39. $481.016. 6. $70.151. 17. $320.833. 28. $8. 40. $391.062. 7. $97.28. 18. $9.8437. 29. $0.07. 41. $1531.25. 8. $-30.78. 19. $85.207. 31. $15.60. 42. $3425.655. 9. $398.287. 20. $400. 32. $21.09. 43. $16320.528. 10. $1177.50. 21. $1638.442. 33. $1.272. 45. $2.145. 11. $1113.024. 22. $144. 34. $4.778. 46. $74.392. 12. $10.05. 23. $90. 35. $46.35. 47. $10.835. 13. $11.0025. 24. $12666.075. 36. $1-29.15. 48. $398.055. 14. $988.761. 25. $16360.996. 37. $168.552. 49. $14.532. APPLICATIONS OF INTEREST.-ARTS. 41. -419. 2. $5.25. 7. $36.08. 12. $6547.20. 20. ~8, 18s. 9 d. 3. $3.15. 8. $91.085. 14. $499.034. 21. ~12, 10s. 4. $17. 9. $107.854+. 15. $498.595. 22. ~1898, 10s. 5. $60. 10. $533.867. 16. $4149.689. 4 44d. 6. $45.014. 11. $25729.166-j 19. ~19, 5s. 10d 23. ~2900. PROBLEMS IN INTEREST.-ARTs. 4L22-4L24. 1, 2. Given. 10. 5 per cent. 19. $30000. 28. 14 y. 3 mo. 3. 6 per cent. 11. 5 per cent. 20. $20833~. 13 d. nearly. 4. 6 per cent. 13. $1800. 22. 4 years. 29. 14 y. 3 mo. 5. 8 per cent. 14. $5400. 23. 6 months. 13 d. nearly. 6. 79 per cent. 15. $10000. 24. 1 y. 3 mos. 30. 10 years. 7. 5~ per cent. 16. $8000. 1 d. nearly.. 31. 8 y. 4 mo. 8. 7 per cent, 17. $14285.7143. 25. 1 y. 6 mo. 32. 9 y. 6 mo. 8 d, 9. 6 per cent. 18. $20000. 27. 16 y. 8 mo. 33. 28 years. COMPOUND INTEREST.-ARTS. 426,.27. 1, 2. Given. 5. $4590.09. 8. $1551.328. 111. $16035.675. 3. $507.213. 6. Given. 9. $877.506. 12 $149744. 4. $2177.426..7. $1888.464. 10. $3491.395. 416 ANSWE RS. LrAG ES 272-295 DISCOUNT,-ARn, 430. Ex. ANS. Ex. ANS. Ex. ANS. Ex. ANS. 1, 2. Given. 5. $88.461+. 8. $6208.955+. 10. $9950.248+. 3. $934.579+. 6. $83.52+. 9. $3404.347+-. 11. $36.636. 4. $1488.687+. 7. $4729.064-+-. BANK DISCOUNT.-ARTS. 433, 434. 12, 13. Given. 20. $24.822. 27. $456.785. 34. $3821.883. 14. $14.1825. 21. $48.3237. 28. $1126.523. 35. $4355.102. 15. $16.605. 22. $37.595. 29. Given. 36. $63717.884. 16. $26.98. 23. $43.694. 30. $414.507. 37. $10416.666. 17. $5.495. 24. $6381.59. 31. $966.101. 38. $51194.539. 18. $2034.1213. 25. $1495.625. 32. $1252.70. 39. $46638.655. 19. $2774.655. 26. $80. 33. $2514.247. 40. $8301.342. INSURANCE.-AaRT. 4l37-442. 1. Given. 8. $1875. 16. 1 per cent. 25. $8365.482. 2. $20.70. 9. $487.50. 17. 14 per cent. 26. $13876.288. 3. $94.20. 10. $243.125. 19. $52000. 27. $27027.027 4. $63.75. 11. $192.78. 20. $65600. 29. $48.60. 5. $104. 12. $3375. 21. $65000. 30. $373.75. 6. $70.50. 14. 25 per cent. 22. $57333~. 31. $10000, ins. 7. $900. 15. 2] per cent. 23. $3416]. $12250,prem, PROFIT AND LOSS.-ARTS. 444-447. 1-3. Given. 10, 11. Given. 19. 15- per cent 127. $2622.222. 4. $218. 12. $156.804. 20. 100 per cent. 28. $2736. 5. $680. 13. $4238.50. 21. 201 — -A- per ct. 29. $13043.478. 6. $935.25. 14. $5926.85. 22. 22-]- per cent. 30. $6317.391. 7. $1366.75. 15. $29504.875. 23, 24. Given. 31. $17806.122. 8, C$68730.28. 17. 23-#3- per ct. 25. $460.869. 32. $42654.028. 9. $12500 lost. 18. 4L per cent. 26. $205.882. 33. $42160. DUTIES. —AnTs. 451 -453. 1. Given. 7. $3784. 13. $717.40. 19. $12642.40. 2. $370.80, 8. $345.744. 14. $492. 20. $2807.10. 3. $163.20. 9. $679.14. 15. $1051.71. 21. $11172.30 4. $1323. 10. $1882.406. 16. $715.75. 22. $17328.75. 5, $546. 11. Given. 17. $1230. 23. $15770.70. 6. $1235.22., 12. $248. 18. $15884.75. ASSESSMENT OF TAXES.-ARTs. 4z6, 457 1, 2. Given. 5. - of 1 per cent., or 7. $121.9g, B's tax, 3. $54.15, B's tax. 8 mills on $1. 8. $283.68, C's tax. 4. $80.50, C's tax,. 6. $80, A's tax.. 10. $8854.166. PAGES 296-310. AN S W ER S. 417 ASSESSMENT OF TAXES CONTINUED. —ARTS. 4599 460. Ex. ANS. Ex. ANS. Ex. ANS. 11. $16125.654. 20. $314.-50, J. F's. 27. $370.50, F. M's. 12. $17342.105. 21. $621.90, T. G's. 28. $458.20, C. P's. 13. $34051.815. 22. $526.40, W. H's. 29. $480.50, J. S's. 16. $73, G. A's. 23. $263.30, L. J's. 30. $541, R. W's. 17. $116, H. B's. 24. $631.00, W. L's. 32. $13.36. 18. $451.50, W. C's. 25. $196.90, J. K's. 33. $3.45. ]9. $481.22, E. D's. 26. $404.90, G. L's. 34. $13.40. ANALYSIS.-ARTs. 462 —47 1, 2. Given. 12. $0.039-96-. 22. $7.98. 32. Given. 3. $300. L3. $64. 23. $6.03. 33. 360 lbs. 4. $320. 14. $1080. 24. $160. 34. 1500 lbs. 5. $12.33-. 15. $480. 25. $3.70f. 35. 95.2 cords. 6. $10.50. 16. $8. 26. $3430. 36. 100 pair. 7. $1.683. 17. 60 days. 27. $119.918. 38. $450,. A's. 8. $2640. 18. 292 mos. 28. $636.479. $750, B's. 9. $24.80. 19. 1088 days. 29. Given. 39. $450, A's. 10. $0.055. 20. $0.56. 30. 2~ hours. $600, B's. 11. $0.29~. 21. $3. 31. 2- days. $750, C's. 40. $763.63-1a, A's. 49. 40 tons. A's. 65. 188~ lbs. at 8d. $654.54?r, B's. 80 tons, B's. 17~ lbs. " 12d. $981.81-r1, C's. 120 tons, C's. 17}L lbs. " 18d. 41. $150.951,- A's. 50. 25 per cent. 17~ lbs. " 22d. $164.53 31, B's. 51. 331 per cent. 67. 9 horses. $185.70-A-o-. C's. $30000, loss, 68. 38- days. $123.80-3%-, D's. 53. 5s. per gal. 69. 278160 men, 42. 66~ cts. on $1. 54. 5-s.. per lb. 70. 15~1- month~ $266.66a, 1st. 55. 9 cts. per lb. 71. $54.60. $333.33-, 2d. 56. 19~ cts. per lb. 72. $459.90. $400.000, 3d. 57. 91- cts. per gal. 74. $600. 43. 70 cts. on $1. 58-60. Given. 75. $3600 44. 25 per cent. 61. 1 part 16 car. 76. $630. 45. $2990.00, A's. 1 " 18 car. 77. 72. $4197.50, B's. 2 " 23 car. 78. 360. $4312.50, C's. 1 " 24 car. 79. 120. 46. 66~ per cent. -63. 100 gals. at 80cts. 80. 240. 47 374 per cent. 40 " 30cts. 81. 68~ feet. 48 10 per cent. 40 " 40cts. 85. $23(.. 418 ANSW E RS. [PAGES 311-328. ANALYSIS CONTINUED.-ART. 47;1. Ex. AN. Ex. ANs. Ex. ANS. Ex. ANS. 86. 1170. 10070. 100. 266. 114. $288. 127. $140. 87. $900. 101. ~131- 115. $43*. 128. $1560. 88. 1125. 102. ~353. 116. $814. 129. $180. 89. $367.50 105. $250~-. 117. $640. 130. $630. 90. $442. 106. $231. 118. $3000. 131. $180. 91. $201. 107. $119- 1119. 9. 200. 132. $1281. 92. $350. 108. $186. 121. $625. 133. $800. 93. $240. 109. $280. 122. $480. 134. 812.60. 94. $754. 110. 81170. 123. $808. 135.,845. 95. $1080. 111. Given. 124. $420. 136. $45. 96. $630. 112. 8378. 125. $690. 137. $90. 99. ~2064. 113. $810. 126. $877~. 138. $150. RATIO.-ARTs. 48O —zL ~. 1, 2. Given. 14. 8-~-. 26. 120. 40. 45 to 72. 3. 2. 15. _. 27. 60. 41. Equal. 4. 4. 16. 4. 28..2. 42. 936 to 56C. 5. 9. 17. 4. 29. -i-. 43. G. inequality 6. 6. 18. -. 30. 13. 44. L. inequality 7. 6. 19. 4. 31. J. 45. Equality. 8. 8. 20.'. 32. 240. 46. 60: 12=5. 9. 9. 21. 3. 35. 8; 4. 47. 1. 10. 9. 22. 7. 36. 4; 8. 48. 4. 11. 9. 23. 112 avoir. 37. 4; 4. 49. -)-. 12. 9, 24. 4. 38. 1; 9. 50. _4_-. 13. 4 25. 6. 39. 72 to 8. 51. 8. SIMPLE PROPORTION.-ARTs. 502 —506. 1. 12. 17. 51*1 lbs. 35. 19205lbs.cop.147. 480. 2. 3. 18. $1640.64. 6414 lbs. tin. 48. 375 sheep. 3. 16. 19. $7066.40. 36. 1520 lbs. n. 49. 20 days. 4. 3. 22. 3 far. 280 lbs. c. 50. 400 rods. 5. Given. 23. $792. 200 lbs. sul. 51. 84 weeks. 6. 20. 24. $2768. 37. 980.5155 lbs. 52. ~1, 3s. 6d 7. 55,. 25. 435 miles. 38. $1350. 1 1 - far. 1st. 8, 120. 26. 252 days. 39. ~45. ~1, is. 2d. 9, 10. Given 28. $2.70. 40. $3375. q-P far. 2nd. 11. $903. 29. 3s. 3d. 2A3- q. 41. $2562.50. ~0, 18s. 9d. 12. $1309.50. 30. $3.15. 42. $16480. 3q1- far. 3rd. 13. $225. 31. $8655. 43. 70400 times. ~0, 16s. 5d. 14. 775 miles. 32. $26.40. 44. 57600 imes. 21T 7far. 4th. 15. 20 tons. 3.4. 30 bu. oats; 45. 170. 63. 888-4 oz. ox. 16. 2156 lbs. 70 bu. corn. 46. 20{0 1 11 oz. hy. PAGES 329 —355.1 ANS W ERS. 419 COMPOUND PROPORTION. —ARTs. 5OS-511, Ex. ANS. Ex. ANS. Ex. ANS. EX. ANS. 3. 10 horses. 9. 24 days. 13. $1-40. 17, 18. Given. 4. 19L- (lays. 10. 144 days. 14. $768. 19. 56 yds. Can. 5. 1314 gals. 11. 1125 miles. 15. $600. 20. 127 b. N. O. 6. 27 laborers. 112. $225. 16. 32 days. 21. 16 rupees. DUODECIMALS.-ART. 5 6. 1, 2. Given. 8. 105 ft. 5'4"5"'5"' 11. 195 ft.4' 1"3"' 8" 3. 28 sq. ft. 6' 10". 4""'. 0...""'.. 6""". 4. 59 cu. ft. 3' 8". 9. 154 ft. 3' 1" 5"' 4"". I12. 23 C. 111 ft. 3'. 5. 268 cu. ft. 6' 11". 6""' 8""". 13. 3840 ft. 0' 5". 6. 235 sq. ft. 10. 85 ft. 1'611" 0"' 5"" 14. $15.819-. 7. 734 sq. ft. 0' 9". 2""'f 6""". 15. 33750 bricks EQUATION OF PAYMENTS.-ART. 521. 3. 6 months. I 4. 6 months. 1 5. 3 years. 1 6. 62 days. PARTNERSHIP.-ART. 523. 1. Given. $2259.649, B's. $1448.276, Y's. 2. $240, A's gain. $3340.351, C's. $1396.552, Z's. $320, B's gain. $4421.053, D's. 8. $22.486, A's. $400, C's gain. 5. $850, A's. $21.024, B's. 3. $274.217- —, A's. $800, B's. $16.490, C's. $373.40V —; B's. $700, C's. 9. $3492.06, A's. $212.37-aa —, C's. $650, D's. $4761.91, B's, 4. $1178.947, A's. 7. $1655.172, X's. $6746.03, C's. EXCHANGE OF CURRENCIES.-ARTS. 533-537. 3. $4116.42. 15. $369716.8&4+. 28. ~8568, 3s. 7ld. 4. $850.63. 16. 284412.622-. 29. 29. 10384, 18s. 4d. 5. $414.667. 17. $4840000. 30. ~20661, 3s. lid. 6. $969.815. 19. i~82. 32. ~135. 7. $2041.59+-. 20. ~90. 33. ~227. 8. $4841.089+. 21. ~181, 10 ~d. 34. ~315, 9d. 9. $7746.082-. 22. ~261, 8s. 73d. 35. ~375. t0. $60652.55-. 23. ~446, 7s. 8~d. 37. $534.166. 11. $208683.819-, 24. ~201, 11s. 7,-d. 38. $614.1875. 12. $330661.605-4-. 25. ~883, 5s. 8~d. 39. $986.083. 13. $242840.369+. 26. ~1095, 3s. ll-d. 40. $7714.285. 14. $257791.397-t-. 27. ~5220, 9-d. 41. $20000. EXCHANGE.-ART. 54S. 2. $4791.60. 5. $10152.527+-.. $10418.509. 3. $25391.084-. 6. $707. 9. $20665.20. 4. $284.58. 7. $1881.60. 10. $36480.755. 420 A N S WER S. [PAGES 356-378 ARBITRATION OF EXCHANGE.-ART, 549. Ex. ANS. Ex. ANS. Ex. ANS. 1. 2j florins. 2. $45 gain. 3. 180 milrees, circu, ALLIGATION.-ARTS. 552 —556. 2. $0.87%. 7. 10 oz. 16 car. fine. 10. 40 gals. at 15s. 3. 5s. 4d. 1-] qr. 5 oz. 18 " 40 gals. at 17s. 5, 3 grs. at 18 car. fine. 5 oz. 22 " 40 gals. at 18s. I gr. " 20 " 8. 133 lbs. at 20 cts. 200 gals. at 22 s. 1 gr. " 22 " 95 lbs. at 30 cts. 11. 28 gals. water; 3 grs," 24 " 190 lbs. at 54 cts. 98 gals. wine. INVOLUTION.-ART. 562. 1', Given. 17. 3125. 22. 6.25. 27. 1. 13. 15129, 18. 279936. 23..000001728. 28. — 1zn 14, 2460375. 19. 117649. 24..0000015625. 29. 20k. 15. 8294400. 20. 65536. 25. t 30. 543-4-. 16. 10000. 21. 387420489. 26. A. 1480 36. SQUARE ROOT.-AaTs. 5749 575. 3. 51. 112 9.848+. 21. 792. 30. 186.995144. 73. 13. 2.6457 —. 22. 1.7810-. 31. 12345. 5. 28. 14. 13.78404+. 23. 3216. 32. 345761. 6. 9.327+. 15. 209. 24. -. 33. 31.05671. 7. 69. 16. 217. 25. 4.. 34. 19.104973174,1 8. 84. 17. 23.8. 26..79056+. 35. 1.414213569. 99. 18. 2.71. 27. 4.1683+. 237+-. 10. 167. 19..9044+. 28. 28.181. 36. 1.732050807 11. 31. 20. 34.2. 29. 14.4116+. 5688772. APPLICATIONS OF THE SQUARE ROOT. -ARTs. 581-585. 1. Given. 7. 18. 14.. 20. 320 rods. 2. 32 feet. 8. 36. 15. A. 21. 480, length; 3. 166.709+m. 9. 40. 16. --. 160, breadth. 4. 240rds. side. 10. 66. 17 7- 22. 148 in rank; 339.4112 r. d. 11. 168. 6 8' 74 in file. 5. Given. 12. 11.2. 18. 63 rods. 24 25 and 40. 6. 10. 13. 67.5. 19. 160 rods. 25. 18 and 47, EXTRACTION OF THE CUBE ROOT.-ARTs. 590-92. 4, 45. 11. 0.623. 18. 3.5463+. — [26. 379-1- lbs. 5. 52. 12. 3.332222+. 19 32. 27. 24 and 72. 6. 83. 13. 1.817121+. 20. 1.25992104.' 28. 128 and 256. 7. 136. 14. 7.217652-. 21..64365958974; 29. 60 and 300. 8. 217. 15. 8.315517+. 2.2. 68 ft. 30. 160 and 640, 9 22.6 16. J. 24. 3.1748+yds. 31.. 426 and 2556. 10. 2.74'. 17.. 25. 4 lbs..32. 747 and 6723. PAGES 379-393.] A N S W E Rt s. 421 ROOTS OF HIGHER ORDERS.-ARTs. 693-05. Ex. ANS. Ex. ANS. Ex. ANS. Ex. ANS. -'. ANS. 2. 2..5. 6. 8. 7. 12. 2.4872+. 16. 1.080059. 3, 16. 6. 26. 9. 3. 13. 414.5+. 17. 1.004074. 4. 376. 7. 5. 10. 2. 15. 1.104089. 18. 1.047128. ARITHMETICAL PROGRESSION.-ARrTs. 603-60~. 1. Given. 4. Given. 9. 34. 113. 14, 21, & 28 2. 5050. 5. 33. 11. 334. 14. 15, 29,43,57, 3. 78 stroles. 7. 44. 12. 502. 71, & 85. GEOMETRICAL PROGRESSION. —ARTrs 6] O - s. 2. 4. $750.3651759245, 8. 1023 3. 4374. aimt. of $500. 9. 43774:4. 4. 13671875. $1628.894614622- ]0. $111111111.111. 5. $2048. 37890625, amt. 12. 14. 6. $334.5563944, amt. of $1000. 14. 3. of $250. ANNUITIES.-ARTs. 614, 615. 1, 2. Given. I 4. $2298.262. I 6. 36785.59. 8. $1333.333. 3: $826.992. 5. $4835.74. 7 Given. 9. Gi-etn,. PERMUTATIONS AND COMBINATIONS.-ARTs. 61S, 619. 2. 40320 ways.. I 4. 3628800 ways. I 7. 15120 numbers. 3. 362880 ways. 5. 479001600 days. 8. 165765600 words. MENSURATION OF SURFACES.-ARTS. 622-63:L1. 1. 270 acres. 7. 1100 sq. ft. 13. 100 ft. 2. 7224 acres. 9. 290.4737 sq. ft. 15. 12 A. 43.49375 r. 3. 314 acres. 10. 4 A. 52.82 rods. 16. 31415.9 sq. ft. 4. 320 rods, or 1 m. 11. 62.8318 ft. 17. 2 ft. 9.94 in. 5. 360 sq. ft. 12. 141.37155 rods. 18. 17.3205 ft. 6. 435 sq. ft. MENSURATION OF SOLIDS; —ARTs. 633-647. 1. 1364 cu. ft. 11. 2748.89125 cu. ft. 18. 13 sq. ft. 2,. 3154 ft. 11' 6" 8"'. 12. 119366.25 cu. ft. 19. 4 cu. ft. 3. 2615 cu.ft. 1080 in. 13. 78 yds. 4 ft. 123- 20. 624 cu. ft. 4. 115 ft. 114.368 in..1-128 in. 21. 220 gals. 3 qts. 1 pt 5 533331 cu. ft. 14. 7 sq. ft. 9.87516 in. 1.824 gi. 6. 8835.75. Cu. ft. - 15. 14684558.20796 22. 451 -gals. 2 qts. 7. 900 sq. ft. sq. miles. 0.729344 pt. 8. 1739 sq. ft. 16. 1767.14437 cu. in. 23. S31.71526i-tons. 9. 76 cu. ft. 17. 5291335807.60158 24. 967.10521-4 tons. 10. 176 sq. ft. CU. m. 422 A NS WE RS. [PAGES 394-398 MECHANICAL POWERS.-ARTs. 648-655. Ex. ANS. Ex. ANS. Ex. ANS. Ex. ANS. 1. 500 lbs. 160 lbs. B. 5. 600 lbs. -8. 1250 lbs. 2. 133 - lbs. 4. 4 ft. fiom A.; 6. 1066-} lbs. 9. 1136.3636 lb. 3. 96 lbs. A.; 8 ft. from B. 7. 1600 lbs. 10. 904777.92 lb. MISCELLANEOUS EXAMPLES. 1. 459 less. 34. 136 g. 1 q. $440, C's g. 79, 247170562. 521 greater 35. $180. 7700, B's s. 2710 s. m. 2. 70. 36. $1 0.875. $1100,C'ss. 80. 336009143. 5 -or. 37. $156.615. 59. 20per cent. 2264006.24. 6-,z-. 38. 94 d. 3 h. 60. $1371. 3104 c. m. 5. 20 days. 38mn. 10- 6s. 61. "4755.141. 81. 5890.5 lbs. 6. $61.32. 39. $2. 62. $32000. 82. 585.80357b 7.,$16581.65. 40. ~1. 63. t$360. 83. 39.401 hld. 8. $18.60. 41. $~J. 64. 36 days. 84. 7~ ft. 9. $1843.003. 42. $4800. 65. 90 hours. 85. 40329146110. $24390.243 43. $197.759. 66. ~51, A's. 12660563511. $4.50. 44. 228 gals. ~34. B's 584000000. 12. $6.875. 45. $40.298f-. ~68, C's 86. 31 n. 180r 13. 331 per ct. 46. $41.095. ~102, D's. 87. 6621. 14. $36. 47. 2 y. 182~ d. 67. ~160, A's. 88. $429496715. $2291j. 48. 5 —nL min. ~224, B's..295. 16. 49874- E. 49. 120 days. ~256, C's. 89. 5 bags, A. 17. 2000 miles. 50. 120 schol. ~205, A's. 7 bags, M. 18. 2400 times. 51. ~292. X~287, B's. 90. 1440. 19. 2880 times. 52. $6000. ~328, C's. 91. $230, B's. 20. $1.50perg. 53. 600. 68. $520, D's. $325, C's. 21. 21 cts. 54. 5600lbs. t. $ $280, A's. $445, A's. 22. $2400. 750 lbs. 1. $360, B's. 92. 5 o'clock, 23. 4371 bbls. 300 lbs. b. 69. 20. 20 min. 24. 21 months. 55. 254~ miles. 70. 25 persons. 93. 1034H d. all 25. $1.328. 56. 78* lbs. 71. 40 and 80. 47Tfif d. A 26. 40 yds. 117a- lbs. 72. 75 and 128. 38-O d. B. 27. is. 32L4 qrs. 57. $192.307-13 73. 56.5685 ft.j 27 —3P — d. C. 28. 37-a21-z* d. A's gain. 74. 7200 rods. 111-ja d. D. 29. 34A374- d. $2307.692 75. 3.535519ft! 94. 36~ days. 30. $1.6Q. A, B's g. 76. 677.73475f. 05. 12 o'clock. 31. 12 miles. $2500.000, 77. 7.13645 r. 32-r min. 32. 12] days. C's g. 78. 50 A. 3 R. 96. 128* yrs. 33. 52- days. 58. $240, A'sg. 28.7399+r. 97. $407.