"ErfMC> lt\0, V \ %| \, 1 -^rp 1 <,_; . ■ . ^____________ %:M/MU^ f2 Ji fCtUi'i^tci & i The gift of Mrs* R. H. Allen, ' 2004 Tenth Avenue, Oakland . / JJ* Cf. n IB THE NATIONAL ARITHMETIC, ON THE INDUCTIVE SYSTEM, COMBINING THE ANALYTIC AND SYNTHETIC METHODS; FORMING A COMPLETE COURSE OF HIGHER ARITHMETIC. Bt benjamin greenleaf, am. AUTHOR OF THE "COMMON SCHOOL ARITHMETIC," " ALGEBRA," ETC NEW ELECTROTYPE EDITION, WITH ADDITIONS AND IMPROVEMENTS. BOSTON: PUBLISHED BY ROBERT S. DAYIS & CO. NEW YORK : D. APPLETON & CO., AND MASON BROTHERS. PHILADELPHIA : J. B. LIPPINCOTT AND COMPANY. CHICAGO : KEEN AND LEE. 1858. Entered according to Act of Congress, in the year 1847, by BENJAMIN GREENLEAF, In the Clerk's Office of the District Court of the District of Massachusetts. Entered according to Act of Congress, in the year 1857, by BENJAMIN GREENLEAF, In the Clerk's Office of the District Court of the District of Massachusetts. GBEENLEAF'S SERIES OF MATHEMATICS. 1. PRIMARY ARITHMETIC ; Ok, MENTAL ARITHMETIC, upon the Inductive Plan ; designed for Primary Schools. Improved electrotype edition. 72 pp. 2. INTELLECTUAL ARITHMETIC | Ok, HIGHER MENTAL ARITHMETIC, upon the Inductive Plan ; designed for Common Schools and Academies. Improved edition. 3. COMMON SCHOOL ARITHMETIC ; Or, INTRODUCTION TO THE NATIONAL ARITHMETIC. Improved stereotype edition. 324 pp. 4. HIGHER ARITHMETIC ; Or, THE NATIONAL ARITHMETIC, being a com- plete course of Higher Arithmetic, for advanced scholars in Common Schools and Acade- mies. New electrotype edition, with additions and improvements. 444 pp. 5. PRACTICAL TREATISE ON ALGEBRA, for Academies and High Schools, and for advanced Students in Common Schools. Improved stereotype edition. 360 pp. 6. ELEMENTS OF GEOMETRY, for Academies and High Schools, and for Advanced Students in Common Schools. [In preparation, and will soon be published.] COMPLETE KEYS TO THE INTRODUCTION, AND NATIONAL ARITHME- TIC, AND THE PRACTICAL TREATISE ON ALGEBRA, containing Solutions and Explanations, for Teachers only. In 3 volumes. EDUCATIOU LIBH# O 3 Two editions of the National Arithmetic, and also of the Common School Arith- metic, one containing the answers to the examples, and the other without them, are pnb- lished. Teachers are requested to state in their orders which edition they prefer. ELECTROTYPED AND PRINTED BY METCALE & CO., CAMBRIDGE, MASS. PREFACE. The National Arithmetic was first presented to the American pub- lic in 1835. The generous favor with which it was received assured the author that he had not misunderstood the wants of the public in the department of arithmetical instruction, and that his labors had, to a considerable extent, supplied those wants. During the ten years following, increased attention was given to the subject of popular education, and great improvements were made in methods of imparting knowledge. Accomplished teachers soon began to demand a work on Arithmetic, which should embody the numerous improvements which had enriched that science. In re- sponse to a demand so reasonable, the author was induced, in 1847, to prepare a revised and enlarged edition of the National Arithmetic. Aided by important suggestions from eminent teachers, and directly assisted by gentlemen intimately acquainted with arithmetical sci- ence, he was enabled to produce a work which, up to the present time, has been steadily increasing in public favor. The last ten years have formed a period of unprecedented activity in all that relates to the interests of education. The numerous Arith- metics which, within this period, have become candidates for popular patronage, afford ample evidence that the department of knowledge to which they relate has meanwhile received its share of attention. Vigorous emulation among authors and publishers has produced thor- ough investigation, careful preparation, and valuable results. The author of this work, wishing, if possible, to keep pace with the rapid march of improvement, has again thoroughly revised, rewritten, and considerably enlarged it. The results of a long experience as a mathematical instructor, and the suggestions of many distinguished teachers of the present day, are embodied in this volume. In preparing this as well as the former editions of his National Arithmetic, the author has regarded the end to be sought in the study of Arithmetic as twofold, — a practical knowledge of numbers, and the discipline of the mind. With reference to the former, he has endeav- ored to present methods which are brief, accurate, and especially adapted to the wants of business life ; with reference to the latter, he has aimed to give a clear and logical analysis of every operation, from the simplest to the most involved. The author adheres to his opinion long since advanced, in relation M577021 iv PREFACE. to the value of rules in an arithmetical treatise. It is not an easy- thing for the experienced teacher to express in the most concise and accurate language the method of solving a problem. Much less can such an expression be given by the untrained scholar. Now, as precision in thought is essentially aided by precision in language, it is deemed expedient to furnish the scholar with rules which shall state in the fewest and clearest words the results of previous logical induc- tions. Moreover, when an intricate reasoning process may have been forgotten and cannot readily be recalled, the brief form of words im- pressed upon the memory in one's youth may oftentimes enable him in after life to perform an important mathematical operation in which he must otherwise have failed. It will be observed, that, while the author has expressed in rules his modes of operating, he has in every case first given the analysis upon which each rule is based. The author flatters himself that the present edition of the National Arithmetic embraces many improvements on former editions. He has endeavored to present clearer definitions, more rigid analyses, and briefer and more accurate rules. While almost every topic included in earlier editions has been treated in a more elaborate and compre- hensive manner, this volume comprises a large amount of new matter, which it is believed will be found useful in business. On comparing this with preceding editions, teachers will find exten- sive additions and improvements under the heads of Numeration, Addition, and the other fundamental rules, Properties of Numbers, Fractions, Ratio, Percentage, Notes and Banking, Roots, etc. Among the new material will be discovered methods of finding the greatest common divisor and the least common multiple of fractions, of re- ducing fractions to a common numerator, of contracting the operations in the multiplication and division of decimal fractions, of reducing continued fractions, of averaging accounts, of alligating, of extracting roots to any degree, and of reducing numbers from one system of no- tation to another. Especial attention is invited to the section on averaging accounts, — a subject rarely taught in schools, though of great importance in the counting-room, — to the manner of treating the roots, and to the many new problems which will be found in all parts of the book. In closing these remarks, the author desires to tender his hearty thanks to many teachers who have favored him with valuable sug- gestions; and to acknowledge in an especial manner his indebted- ness to Mr. H. B. Maglathlin, who has been constantly associated with him in making this revision, and to whose accurate scholarship and sound judgment the value of the work is largely due. Bradford, Mass., May 30, 1857. CONTENTS. Page INTRODUCTION 7 DEFINITIONS 13 SIGNS 14 AXIOMS 15 NOTATION 16 Roman Notation 16 Arabic Notation 18 NUMERATION 20 French Numeration ... 20 English Numeration ... 23 ADDITION ........ 24 SUBTRACTION 30 MULTIPLICATION 35 DIVISION 44 GENERAL PRINCIPLES AND APPLICATIONS 53 Cancellation 54 Contractions in Multiplication 56 Contractions in Division . . 62 PROBLEMS 66 UNITED STATES MONEY. . 70 Reduction of United States Money 72 Addition of United States Money 73 Subtraction of United States Money 74 Multiplication of United States Money 75 Division of United States Money . . • 76 General Principles and Ap- plications 78 Analysis by Aliquot Parts . 78 1* Page Bills 83 Ledger Accounts .... 87 COMPOUND NUMBERS. . . 88 Reduction of CompoundNum- bers 88 Addition of Compound Num- bers Ill Subtraction of Compound Numbers. .*.... 116 Multiplication of Compound Numbers 119 Division of Compound Num- bers 123 PRINCIPLES AND APPLICA- TIONS 126 Difference between Dates . . 126 Difference of Latitude . . . 129 Difference of Longitude . . 130 Longitude and Time . . . 132 PROPERTIES OF NUMBERS . 137 Definitions 137 Prime Numbers 139 Factoring 141 Divisibility of Numbers . . 142 Greatest Common Divisor . 146 Least Common Multiple . .149 COMMON FRACTIONS . . . 154 Reduction of Common Frac- tions 155 A Common Denominator . . 159 Addition of Common Frac- tions 161 Subtraction of Common Frac- tions 164 Multiplication of Common Fractions 169 VI CONTENTS. Division of Common Frac- tions 173 Complex Fractions . . . .177 Denominate Fractions . . .185 DECIMAL FRACTIONS ... 197 Notation and Numeration of Decimals 198 Addition of Decimals . . . 200 Subtraction of Decimals . . 202 Multiplication of Decimals . 203 Division of Decimals . . . 206 Reduction of Decimals . . 210 CIRCULATING DECIMALS . 217 Reduction of Repetends . . 219 Transformation of Repetends 222 Addition of Circulating Deci- mals 224 Subtraction of Circulating Decimals 225 Multiplication of Circulating Decimals 225 Division of Circulating Deci- mals 226 CONTINUED FRACTIONS . . 227 RATIO 229 Reduction and Comparison of Ratios 232 Analysis by Ratio .... 234 PROPORTION 237 Simple Proportion . . . .238 Compound Proportion . . . 244 Conjoined Proportion . . . 248 PERCENTAGE 251 Interest 259 Promissory Notes . . . .272 Partial Payments .... 273 Compound Interest . . . .279 Discount and Present Worth 286 Banking 289 Stocks 295 Brokerage and Commission . 299 Account of Sales . . . . . 302 Profit and Loss 303 PARTNERSHIP, OR COMPA- NY BUSINESS 308 Bankruptcy 313 Taxes 314 General Average 818 EQUATION OF PAYMENTS . 320 Averaging of Accounts . . 324 Accounts of Storage . . . 332 INSURANCE 335 Fire and Marine Insurance . 335 Life Insurance 337 CUSTOM-HOUSE BUSINESS . 341 COINS AND CURRENCIES . 345 Reduction of Currencies . . 347 EXCHANGE 349 Inland Bills 353 Foreign Bills 354 Arbitration of Exchanges . .358 ALLIGATION 363 Alligation Medial .... 363 Alligation Alternate . . . 364 INVOLUTION 369 EVOLUTION 371 Extraction of the Square Root 373 Extraction of the Cube Root 378 Extraction of any Root . .384 APPLICATIONS OF POWERS AND ROOTS 388 PROGRESSION, OR SERIES . 395 Arithmetical Progression . . 395 Geometrical Progression . . 400 ANNUITIES 405 PERMUTATIONS AND COM- BINATIONS 411 ANALYSIS BY POSITION. . 413 SCALES OF NOTATION . . 417 DUODECIMALS 420 Addition and Subtraction of Duodecimals 421 Multiplication of Duodecimals 421 Division of Duodecimals . . 423 MENSURATION 425 Definitions 425 Mensuration of Surfaces . . 425 Mensuration of Solids . . .433 Mensuration of Lumber . .437 Gauging of Casks .... 439 Tonnage of Vessels .... 439 INTRODUCTION. HISTORY OF ARITHMETIC. It is difficult to determine who was the inventor of Arithmetic, or in what age or among what people it originated. In ordinary history, we find the origin of the science attributed by some to the Greeks, by some to the Chaldeans, by some to the Phoenicians, by Josephus to Abraham, and by many to the Egyptians. The opinion, however, which modern investigations have rendered most probable, is, that Arithmetic, properly so called, is of Indian origin, — that is, that the science received its first definite form, and became the germ of mod- ern Arithmetic, in the regions of the East. It is evident, from the nature of the case, that some knowledge of numbers and of the art of calculation was necessary to men in the earliest periods of society, since without this they could not have per- formed the simplest business transactions, even such as are incidental to an almost savage state. The question, therefore, as to the invention of Arithmetic, deserves to be considered only as it respects the origin of the science as we now have it, and which, as all scholars admit, has reached a surprising degree of perfection. In this sense the honor of the invention must be awarded to the Hindoos. The history of the various methods of Notation, or the different means by which numbers have been expressed by signs or characters, is one of much interest to the advanced and curious scholar ; but the brevity of this sketch allows us barely to touch upon it here. Among the ancient - nations which possessed the art of writing, it was a natu- ral and common device to employ letters to denote what we express by our numeral figures. Accordingly we find, that, with the Hebrews and Greeks, the first letter of their respective alphabets was used for 1, the second for 2, and so on to the number 10, — the latter, how- ever, inserting one new character to denote the number 6, and evi- dently in order that their notation might coincide with that of the Hebrews, the sixth letter of the Hebrew alphabet having no corre- sponding one in the Greek. 8 INTRODUCTION. The Komans, as is well known, employed the letters of their alpha- bet as numerals. Thus I denotes 1 ; Y, 5 ; X, 10 ; L, 50 ; C, 100 ; D, 500 ; and M, 1,000. The intermediate numbers were expressed by a repetition of these letters in various combinations ; as, II for 2 ; VI for 6 ; XV for 15 ; IV for 4 ; IX for 9, &c. They frequently ex- pressed any number of thousands by the letter or letters denoting so many units, with a line drawn above ; thus, V, 5,000 ; VI, 6,000 ; X, 10,000 ; L, 50,000 ; C, 100,000; M, 1,000,000. In the classification of numbers, as well as in the manner of ex- pressing them, there has been a great diversity of practice. While we adopt the decimal scale and reckon by tens, the aborigines of Mexico, according to Humboldt, and some of the early nations of Europe, adopted the vicenary, reckoning by twenties ; some of the Indian tribes, and several of the African tribes, use the quinary, reckoning by fives ; and the Chinese for more than 4,000 years have used the Unary, reckoning by twos. The adoption of one or another of these scales has been so general, that they have been re- garded as natural, and accounted for by referring them to a common and natural cause. The reason for assuming the binary scale prob- ably lay in the use of the two hands, which were employed as counters in computing ; that for employing the quinary, in a similar use of the five fingers on either hand ; while the decimal and vicenary scales had respect, the former to the ten fingers on the two hands, and the latter to the ten fingers combined with the ten toes on the naked feet, which were as familiar to the sight of a rude, uncivilized people as their fingers. -»- It is an interesting circumstance, that in the common name of our numeral figures, digits (digiti) or fingers, we preserve a memento of the reason why ten characters and our present decimal scale of numeration were originally adopted to express all numbers, even of the highest order. It is now almost universally admitted, that our present numeral characters, and the method of estimating their value in a tenfold ratio from right to left, have decided advantages over all other systems, both of notation and numeration, that have ever been adopted. There have been those, as Leibnitz and De Lagni, who have advo- cated the binary scale ; a few, with Claudius Ptolemy, have claimed advantages for the sexagenary scale, or that by sixty ; and there are those who think that a duodecimal scale, and the use of twelve numeral figures instead of ten, would afford increased facility for rapid and ex- tensive calculations ; but most mathematicians are satisfied with the present number of numerals and the scale of numeration, which have attained an adoption all but universal. It was long supposed, that for our modern Arithmetic the world is indebted to the Arabians. But this, as we have seen, is not the HISTORY OF ARITHMETIC. 9 case. The Hindoos at least communicated a knowledge of it to the Arabians, and, as we are not able to trace it beyond the former people, they must have the honor of its invention. They do not, however, claim this honor, but refer it to the Divinity, declaring that the inven- tion of nine figures, with device of place, is to be ascribed to the benefi- cent Creator of the universe. But though the invention of modern Arithmetic is to be ascribed to the Hindoos, the honor of introducing it into Europe belongs unques- tionably to the Arabians. It was they who took the torch from the East and passed it along to the West. The precise period, however, at which this was done, it is not easy to determine. It is evident, that our numeral characters and our method of computing by them were in use among the Arabians about the beginning of the eighth century, when they invaded Spain, and it is probable that a knowledge of them was soon afterwards communicated to the inhabitants of Spain, and gradually to those of the other European countries. It is said, that the celebrated Gerbert, afterward Pope Sylvester II., returning to France from Spain, where he had been to acquire a knowledge of the Arabic or Indian notation, about the year 970, in- troduced it among the French. About the middle of the eleventh century it is supposed to have been introduced into England by John of Basingstoke, Archdeacon of Leicester. The Arabic characters, having been first used by astronomers, be- came circulated over Europe in their almanacs ; but do not seem to have secured general adoption in Europe earlier than the twelfth or thirteenth century. The science of Arithmetic, like all other sciences, was very limited and imperfect at the beginning, and the successive steps by which it has reached its present extension and perfection have been taken at long intervals and among different nations. It has been developed by the necessities of business, by the strong love of certain minds for mathematical science and numerical calculation, and by the call for its higher offices by other sciences, especially that of Astronomy. In its progress, we find that the Arabians discovered the method of proof by casting out the 9's, and that the Italians early adopted the practice of separating numbers into periods of six figures, for the purpose of enu- meration. To facilitate the process of multiplication, this latter people also introduced, probably from the writings of Boethius, the Multipli- cation Table of Pythagoras. The invention of the Decimal Fraction was a great step in the ad- vancement of arithmetical science, and the honor of it has generally been given to John Muller, commonly called Regiomontanus, about the year 1464. It appears, however, that Stevinus, in 1582, wrote 10 INTRODUCTION. the first express treatise on the subject. The credit of first using the decimal point, by which the invention became permanently available, is given by Dr. Peacock to Napier, the inventor of Logarithms ; but De Morgan says, that it was used by Richard Witt as early as 1613, while it is not shown that Napier used it before 1617. Circulating Decimals received but little attention till the time of Dr. Wallis, the author of the Arithmetic of Infinites. Dr. Wallis died at Oxford, in 1703. The greatest improvement which the art of computation ever re- ceived was the invention of Logarithms, the honor of which is un- questionably due to Baron Napier, of Scotland, about the end of the sixteenth or the commencement of the seventeenth century. The oldest treatises on Arithmetic now known are the 7th, 8th, 9th, and 10th books of Euclid's Elements, in which he treats of pro- portion and of prime and composite numbers. These books are not contained in the common editions of the great geometer, but are found in the edition by Dr. Barrow, the predecessor of Sir Isaac Newton in the mathematical chair at Cambridge. Euclid flourished about 300 B. C. A century later, Eratosthenes invented a method, which is known as his " sieve," for separating prime numbers from others. The next writer on Arithmetic mentioned in history is Nico- machus, the Pythagorean, who wrote a treatise relating chiefly to the distinctions and divisions of numbers into classes, as plain, solid, triangular, &c. He is supposed to have lived near the Christian era. About the middle of the fourth century lived Diophantus, a cele- brated mathematician, who, besides being the first known author on the subject of Algebra, composed thirteen books on Arithmetic, six of which are still extant. The next writer of note is Boethius, the Roman, who, however, copied most of his work from Nicomachus. He lived at the begin- ning of the sixth century, and is the author of tljp well-known work on the Consolation of Philosophy. The next writer of eminence on the subject is Jordanus, of Namur, who wrote a treatise about the year 1200, which was published by Joannes Faber Stapulensis in the fifteenth century, soon after the invention of printing. The author of the first printed treatise on Arithmetic was Pacioli, or, as he is more frequently called, Lucas de Burgo, an Italian monk, who in 1484 published his great work entitled Summa de Arithmetica^ &c, in which our present numerals appear under very nearly their modern form. In 1522, Bishop Tonstall published a work on the Art of Computa- tion, in the Dedication of which he says, that he was induced to study HISTORY OF ARITHMETIC. 11 Arithmetic to protect himself from the frauds of money-changers and stewards, who took advantage of the ignorance of their employers. In his preparation for this work, he professes to have read all the books which had been published on this subject, adding, also, that there was hardly any nation which did not possess such books. About the year 1540, Robert Eecord, Doctor in Physic, printed the first edition of his famous Arithmetic, which was afterward aug- mented by John Dee, and subsequently by John Mellis, and which did much to advance the science and practice of Arithmetic in Eng- land in its early stages. This work, which is now quite a curiosity, effectually destroys the claim to originality in some things of which authors much more modern have obtained the credit. In it we find the celebrated case of a will, which we have in the Miscellaneous Questions of Webber and Kinne, and which, altered in language and the time of making the testament, is the 2nd Miscellaneous Question in the present work. This question is, by his own confession, older than Record, and is said to have been famous since the days of Lucas de Burgo. In Record it occurs under the " Rule of Fellowship." Record was the author of the first treatise on Algebra in the English language. In 1556, a complete work on Practical Arithmetic was published by Nicolas Tartaglia, an Italian, and one of the most eminent mathema- ticians of his time. From the time of Record and Tartaglia, works on Arithmetic have been too numerous to mention in an ordinary history of the science. De Morgan, in his recent work (Arithmetical Books), has given the names of a large number, with brief observations upon them, and to this the inquisitive student is referred for further information in re- gard both to writers and books on this subject since the invention of Printing. It is remarkable that De Morgan knew next to nothing of any American works on Arithmetic. He mentions the " Ameri- can Accountant" by William Milns, New York, 1797, and gives the name of Pike (probably Nicholas Pike) among the names of which he had heard in connection with the subject. Of the compilation of Webber and the original work of Walsh, he seems to have been en- tirely ignorant. The various signs or symbols, which are now so generally used to abridge arithmetical as well as algebraical operations, were introduced gradually, as necessity or convenience taught their importance. The earliest writer on Algebra after the invention of printing was Lucas de Burgo, above mentioned, and he uses p for plus and m for minus, and indicates the powers by the first two letters, in which he is fol- lowed by several of his successors. After this, Steifel, a German, who in 1544 published a work entitled Arithmetica Integra, added 12 INTRODUCTION. considerably to the use of signs, and, according to Dr. Hutton, was the first who employed -f- and — to denote addition and subtraction. To denote the root of a quantity he also used our present sign t Jl x originally r, the initial of the word radix, root. The sign ==, to denote equality, was introduced by Record, the above-named English mathe- matician, and for this reason, as he says, that " noe 2 thynges can be moar equalle," namely, than two parallel lines. It is a curious cir- cumstance that this same symbol was first used to denote subtraction. It was also employed in this sense by Albert Girarde, who lived a little later than Record. Girarde dispensed with the vinculum em- ployed by Steifel, as in 3+4, and substituted the parenthesis (3 + 4), now so generally adopted. The first use of the St. Andrew's cross, X, to signify multiplication, is attributed to William Oughtred, an Englishman, who in 1631 published a work entitled Clavis Mathe- matical, or Key of Mathematics. It was intended to notice several other works, ancient and modern, but the length to which this sketch has already extended forbids it. We had thought of alluding to the ancient philosophic Arithmetic, and the elevated ideas which many of the early philosophers had of the science and properties of numbers. But a word must here suffice. Arithmetic, according to the followers of Plato, was not to be studied " with gross and vulgar views, but in such a manner as might enable men to attain to the contemplation of numbers ; not for the purpose of dealing with merchants and tavern-keepers, but for the improve- ment of the mind, considering it as the path which leads to the knowl- edge of truth and reality." These transcendentalists considered per- fect numbers, compared with those which are deficient or superabun- dant, as the images of the virtues, which, they allege, are equally remote from excess and defect, constituting a mean between them ; as in the case of true courage, which, they say, lies midway between audacity and cowardice, and of liberality, which is a mean between profusion and avarice. In other respects, also, they regard this anal- ogy as remarkable ; perfect numbers, like the virtues, are " few in number and generated in a constant order ; while superabundant and deficient numbers are, like vices, infinite in number, disposable in no regular series, and generated according to no certain and invariable law." We conclude this brief sketch in the earnest hope that the noble science of numbers may erelong find some devoted friend, who shall collect, arrange, and bring within the reach of ordinary students, much more fully than we have done, the scattered details of its long- neglected history. ARITHMETIC. DEFINITIONS. Article 1, Quantity is anything that can be increased, diminished, or measured ; as time, weight, lines, surfaces, and solids. 2. A unit is a single thing or quantity regarded as a whole. 3 1 An abstract unit is one that has no reference to any particular thing or quantity. 4. A concrete unit is one that has reference to some par- ticular thing or quantity. 5. A number is an expression of quantity, representing either a unit or a collection of units. 6. An abstract number is a number whose unit is abstract ; as, one, six, nine. 7. A concrete or denominate number is a number whose unit is concrete ; as, one dollar, six pounds, nine men. 8. A simple number is a unit, or a collection of units, either abstract or concrete, of a single kind or denomination ; as, 1, 15, 1 book, 13 dollars. 9t The unit of measure of any quantity is one of the same kind with that by which the quantity is measured or com- pared ; as, in the abstract number, six, the abstract unit is that of measure or comparison ; and in six pounds, the concrete unit, one pound, is that of measure or comparison. 10, Arithmetic is the science of numbers and the art of computing by them. It treats of the properties and relations N 2 14 SIGNS. of numbers, and teaches the methods of applying the prin- ciples of the science to practical purposes. 11. An axiom is a self-evident truth. 12. A problem is a question proposed for solution, or something to be done. 13. An operation is the process of finding, from given quantities, others that are required. 14. A sign is a symbol employed to indicate the relations of quantities, or operations to be performed upon them. 15. A rule is a direction for performing an operation. 16. An example is a particular application of a general principle or rule. 17. The principal or fundamental processes of arithmetic are Notation and Numeration, Addition, Subtraction, Multipli- cation, and Division. . SIGNS. 18. The sign of equality, two short horizontal lines, =, is read equal, or equal to, and denotes that the quantities be- tween which it is placed are equal to each other. Thus, 12 inches = 1 foot, signifies that 12 inches are equal to 1 foot. 19. The sign of addition, an erect cross, -\-, is read plus, and, or added to, and denotes that the quantities between which it is placed are to be added together. Thus, 8 — |— 6 signifies that 6 is to be added to 8. 20. The sign of subtraction, a short horizontal line, — , is read minus, or less, and denotes that the quantity on the right of it is to be subtracted from the quantity on the left. Thus, 8 — 6 signifies that 6 is to be subtracted from 8. 21. The sign of multiplication, an inclined cross, X? is read times, or midtiplied by, and denotes that the quantities between which it is placed are to be multiplied together. Thus, 7x6 signifies that 7 is to be multiplied by 6. 22* The sign of division, a horizontal line between two dots, --, is read divided by, and denotes that the quantity on the left of it is to be divided by that on the right. Thus, 42-7-6 signifies that 42 is to be divided by 6. 23. The sign of aggregation, a parenthesis, ( ), includ- AXIOMS. 15 ing several numbers, or a vinculum, , drawn over them, indicates that the value of the expression is to be used as a single number. Thus, (17 -)- 3) X 5, indicates that the sum of 17 and 3 , or 20, is to be multiplied by 5; and 12 +(9 — 3)-r-2, indicates that the difference between 9 and 3 divided by 2, or 3, is to be added to 12. AXIOMS. 24. Arithmetic, in common with other branches of the mathematics, is based upon axioms, few in number, and uni- versally admitted to be so clearly true, that no process of reasoning can make them plainer ; as, 1. If the same quantity, or equal quantities, be added to equal quantities, the sums will be equal. 2. If the same quantity, or equal quantities, be subtracted from equal quantities, the remainders will be equal. 3. If the same quantity, or equal quantities, be added to unequal quantities, the sums will be unequal. 4. If the same quantity, or equal quantities, be subtracted from unequal quantities, the remainders will be unequal. 5. If equal quantities be multiplied by the same quantity, or equal quantities, the products will be equal. 6. If equal quantities be divided by the same quantity, or equal quantities, the quotients will be equal. 7. If the same quantity be both added to and subtracted from another, the value of the latter will not be changed. 8. If a quantity be both multiplied and divided by the same quantity, its value will not be changed. 9. If two quantities be equally increased or diminished, their difference will not be changed. 10. Quantities which are equal to the same quantity are equal to each other. 11. Quantities which are like parts of equal quantities are equal to each other. 12. The whole of a quantity is greater than any of its parts. 13. The whole of a quantity is equal to the sum of all its parts. 16 NOTATION. NOTATION AND NUMERATION. NOTATION. 25. Notation is the process of representing numbers by letters, figures, or other symbols. The common methods of expressing numbers are three : by words, written or spoken ; by letters, called the Roman meth- od ; and by figures, called the Arabic method. 26. In common language, we express numbers by the terms one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eigh- teen, nineteen, twenty, twenty-one, etc., giving a distinct name to each unit as far as ten, when we begin a second ten, and pass on to twenty ; a third ten, and pass on to thirty ; and so on to forty, fifty, sixty, seventy, eighty, and ninety. Proceeding thus we reach ten tens, which we call one hundred, when we begin a second hundred, and pass to two hundred ; a third hundred, and pass to three hundred; and so on as far as ten hundred, which we call one thousand. A thousand thousand we call one mil- lion ; a thousand million, one billion ; a thousand billion, one trillion ; and so on with numbers still higher. Note 1. — The term eleven is a contraction of one left after ten ; and twelve, of two left after ten. Thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen, are derived from three and ten, four and ten, five and ten, etc. Twen- ty, thirty, forty, fifty, sixty, seventy, eighty, and ninety are contractions of two tens, three tens, four tens, etc. Note 2. — Billion is a contraction of the Latin his, twice, and million ; and trillion, of the Latin tres, three, and million. In like manner from the Latin numerals, quatuor, four ; quinque, five ; sex, six; septem, seven; octo, eight; novem, nine ; decern, ten; undecim, eleven ; duodecim, twelve ; tredecim, thirteen ; quatuordecim, fourteen ; quindecim, fifteen ; sexdecim, sixteen ; septendecim, seventeen ; octodecim, eighteen ; novemdecim, nineteen ; viginti, twenty, — are formed quadrillions, quintillions, sextillions, septillions, octillions, nonillions, decilr lions, undecillions, duodecillions, etc. Roman Notation. 27. The Roman Notation, so called from its having originated with the ancient Romans, employs in expressing numbers seven capital letters, viz. : NOTATION. 17 one, V, five, X, ten, fifty, C, one hundred, five hundred, M. one thousand. All intervening and succeeding numbers are expressed by use of these letters, either in repetitions or combinations. By a letter being written after another denoting equal or less value, the sum of their values is represented ; as, II represents two ; VI, six. By writing a letter denoting a less value before a let- ter denoting a greater, their difference of value is represented ; as, IV represents four ; XL, forty. A dash ( — ) placed over a letter increases the value denoted by the letter a thousand times; as, V represents five thousand; IV, four thousand. Table. I denotes I one. XXX de note s thirty. II a two. XL H forty. ni a three. L a fifty. IV u four. LX Ci sixty. V a five. LXX a seventy. VI a six. LXXX a eighty. vri a seven. XC a ninety. VIII u eight . C a one hundred. IX a nine. CI a one hundred and one X a ten. CC it two hundred. XI a eleven. CCC a three hundred. XII a twelve. CCCC a four hundred. XIII a thirteen. D a live hundred. XIV u fourteen. DC a six hundred. XV u fifteen. DCC a seven hundred. XVI a sixteen. DCCC a eight hundred. XVII a seventeen. DCCCC a nine hundred. XVIII a eighteen. M i( one thousand. XIX a nineteen. MD a fifteen hundred. XX a twenty. MM a two thousand. XXI u twenty-one. XIX a nineteen thousand. xxn a twenty-two. M a one million. XXIII U twenty-three. MM a two million. Note 1. — The Roman method of Notation is now but little used, except in numbering sections, chapters, and other divisions of books ; and for indicating the hours on the face of clocks, watches, or dials. 2* 18 NOTATION. Note 2. — Formerly CIO was used to represent one thousand, and the pre- fixing of a C and the annexing of a increased the number denoted ten times ; thus, CCIOO represented ten thousand, and CCCIOOO, one hundred thou- sand. Exercises. Represent the following numbers by letters : — 1. Forty-nine. Ans. XLIX. 2. Ninety-seven. 3. One hundred and eighty-eight. 4. Two hundred and nineteen. 5. Six hundred and sixty-three. 6. One thousand five hundred and six. 7. One thousand eight hundred and fifty-seven. 8. Four thousand four hundred and forty-four. 9. Eleven thousand nine hundred and eleven. 10. One hundred fifty thousand and fifty. 11. One million twenty thousand and twenty. 12. Three million one hundred thousand. Arabic Notation. 28. Arabic Notation, so called from its having been made known through the Arabs, employs in expressing num- bers ten characters or figures, viz. : 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. one, two, three, four, iive, six, seven, eight, nine, cipher. The first nine are sometimes called digits, and the cipher, naught or zero. 29. The place of a figure is its particular position with regard to other figures ; as in 61 (sixty-one) counting from the right, the 1 occupies the first place and the 6 the second place, and so on for any other like arrangement of figures. 30. The digits have been denominated significant figures, because each of itself has a positive value, always repre- senting so many units, or ones, as .its name indicates. But the size or value of the units represented by a figure differs with the place occupied by the figure. Thus in 235 (two hundred and thirty-five), each of the figures, without regard to its place, expresses units, or ones ; but these units or ones differ in value. The 5 occupying the first place represents NOTATION. 19 5 single units ; the 3 occupying the second place represents 3 tens, or 3 units each ten times the size or value of a unit of the first place ; and the 2 occupying the third place represents 2 hundreds, or 2 units each one hundred times the size or value of a unit of the first place ; the value of any figure being always increased tenfold by each removal of it one place to the left. 31. The cipher becomes significant when connected with other figures, by filling a place that otherwise would be va- cant ; as in 10 (ten) where it occupies the vacant place of units, and in 102 (one hundred and two) where it fills the va- cant place of tens. 32. The simple value of a figure is the value its unit has when the figure stands alone ; or, in a collection, when standing in the right-hand place. Thus 2 alone, or in 32 (thirty-two), expresses a simple value of two single units or ones. 33. The local value of a figure is the value its unit has when the figure is removed from the right-hand place, and de- pends upon the place the figure occupies. Thus, in 44 (forty- four), the 4 in the first place has the local value of 4 units, and the 4 in the second place has the local value of 4 tens, or forty. 34. The successive places occupied by figures are often called orders. Thus a figure in the first or units' place is called a figure of the first order, or of the order of units ; a figure in the second place is a figure of the second order, or of the order of tens ; in the third place, of the third order, or of the order of hundreds ; and so on, each figure next to the left belonging to a distinct order, the unit of which is tenfold the size or value of a unit of the order at the right. Exercises. 1. "Write three units of the first order. 2. Write H\e units of the first order. 3. Write eight units of the second order, with seven of the first. 4. Write two units of the third order, with none of the sec- ond, and one of the first. 5. Write seven units of the fourth order, with two of the third, none of the second, and none of the first. 20 NUMERATION. 6. Write one unit of the fifth order, with none of the four lower orders. 7. Write six units of the sixth order, five of the fifth, four of the fourth, three of the third, one of the second, and two of the first. 8. Write one unit of the eighth order, with none of the seven lower orders. 9. Write nine units of the ninth order, with six of each of the eight lower orders. 10. Write two units of the twelfth order, with none of the eleventh, none of the tenth, one of the ninth, five of the eighth, nine of the seventh, none of the sixth, none of the fifth, none of the fourth, three of the third, none of the second, and three of the first. 11. Write three units of the fifteenth order, with none of the fourteenth, none of the thirteenth, none of the twelfth, one of the eleventh, seven of the tenth, five of the ninth, one of the eighth, none of the seventh, none of the sixth, five of the fifth, three of the fourth, two of the third, two of the second, and seven of the first. 12. Write four units of the twenty-fifth order, with three of the twenty-fourth, two of the twenty-third, none of the twenty- second, none of the twenty-first, none of the twentieth, none of the nineteenth, none of the eighteenth, none of the seventeenth, five of the sixteenth, and none of the fifteen lower orders. NUMERATION. 35. Numeration is the process of reading numbers when expressed by figures. 36* There are two methods of numeration ; the French and the English. French Numeration. 37. The French method of numeration is that in gen- eral use on the continent of Europe and in the United States. Beginning at the right, figures occupying more than three places being separated into as many groups as possible of three figures each, called periods, it gives a disthict name to each period. 00 X o NUMERATION. 21 FRENCH NUMERATION TABLE. If. 4 § f| ga ?g id g§ ^1 J o?a cs fi| wg a§ h| -3*>§ -3»J -§C?.2 .fcft- ,gM ,gS ^H-g ^ ij03 £0-3 £ °Jj £ ° O £ ° « r£ ° S r% ° ^ £ 3 5 3 3 §'3 3 § S3 S3 g^S 3 §S 3 §3 53 §,§ S3 § 'fl WH^ KHO 1 WHO' WHH WHW WHg WHH WHP 7 8 9, 12 3, 4 5 6, 7 8 9, 12 3, 4 5 6, 7 8 9, 12 3. 8th Period. 7th Period. 6th Period. 5th Period. 4th Period. 3d Period. 2d Period. 1st Period. Sextil- duintil- Quadril- Trillions. Billions. Millions. Thousands. Units, lions. lions. lions. The value of the number represented in the table is, seven hundred eighty-nine sextillions, one hundred twenty-three quin- tillions, four hundred fifty-six quadrillions, seven hundred eigh- ty-nine trillions, one hundred twenty-three billions, four hundred fifty-six millions, seven hundred eighty-nine thousands, one hun- dred twenty-three. 38. The unit of the first period, or right-hand group, is 1 ; of the second, 1 thousand ; of the third, 1 million ; of the fourth, 1 billion ; of the fifth, 1 trillion ; of the sixth, 1 quad- rillion ; of the seventh, 1 quintillion ; of the eighth, 1 sextil- lion, etc. The periods above sextillions, in their order, are, Septillions, Octillions, Nonillions, Decillions, Undecillions, Duodecillions, Tredecillions, Quatuordecillions, Quindecillions, Sexdecillions, Septendecillions, Octodecillions, Novemdecillions, Vigintillions, etc. Note. —The idea of number is the latest and most difficult to form. Before the mind can arrive at such an abstract conception, it must be familiar with that process of classification by which we successively ascend from individ- uals to species, from species to genera, from genera to orders. The savage is lost in his attempts at numeration, and significantly expresses his inability to proceed by holding up his expanded fingers or pointing to the hair of his head. It is, indeed, difficult for any mind to form an adequate idea of the larger num- bers. To count a million, at the rate of one in a second, would require upward of twenty-three days of twelve hours each. A billion is equal to a million a thousand times repeated, or a number so great, as to exceed all the seconds of time that would elapse in thirty-two years. 22 NUMERATION. 39. To read numbers represented by figures according to the French method ; — Begin at the right hand, and point off the figures into as many periods as possible of three places each. Then, commencing at the left hand, read the figures of each period, adding the name of each period excepting that of units. Exercises. Read orally, or write in words, the numbers represented by the following figures, according to the French method : — 1. 31620 7. 21623417 13. 3256171894 2. 55216 8. 101315125 14. 5162251912 3. 29156 9. 876223399 15. 13141421211 4. 179213 10. 12345625 16. 457863497639 5. 918512 11. 21177792 17. 434378783434 6. 1219615 12. 6665551161 18. 7852463767445 19. 407000010703801 21. 478127815016666060707 20. 2000700078 01000 22. 800800800800800800800800 23. 127081061071081010009007007 24.407144140070060700007101800808 40. To write numbers in figures according to the French method ; — Begin at the left hand, and write in each successive order the figure belonging to it. If any intervening order would otherwise be vacant, fill ihe place by a cipher. Exercises. Represent by figures, and read, the following numbers, ac- cording to the French method : — 1. Twenty-nine. 2. Four hundred and seven. 3. Twenty-three thousand and seven. 4. Five millions and twenty-seven. 5. Seven millions, two hundred five thousand and five. 6. Two billions, two hundred seven millions, six hundred four thousand and nine. 7. One hundred five billions, nine hundred nine millions, three hundred eight thousand two hundred and one. NUMERATION. 23 8. Nine quintillions, eight billions and forty-six. 9. Fifteen quintillions, thirty-one millions and seventeen. 10. Five hundred seven septillions, two hundred three tril- lions, fifty-seven millions and eighteen. 11. Nine nonillions, forty-seven trillions, seven billions, two millions, three hundred ninety-two. 12. Fifteen duodecillions, ten trillions, one hundred twenty- seven billions, twenty-six millions, three hundred twenty thou- sand four hundred twenty-six. English Numeration. 41. The English method of numeration is that generally used in Great Britain, and in the British Provinces. It di- vides numbers into periods of six figures, and gives a distinct name to each. ENGLISH NUMERATION TABLE.- 2 a Si O Cm m O ^ m 85 O H S^ a ^H o ^ Cm '^ m3§* M* » O fl e3 .2 O Cm a © O ££ Cm H O Cm . O Is ro ^m cq Sw g «M . O S^ 22.2 f« 5 r » J3 £ jB ^ s r S B HWHH WHHMHpq OOOO K £ O Q 7 1 CQ h li tM F=< ji «M H OCm . o 'S ^ Is £.2 *S M CQ o ^ § » sS 8 J H | "I , "^ CQ 3^ » .2 g a o g fl^g 39883 2, 56387 1, 351615, 12356 1 4th Period. Trillions. 3d Period. Billions. y 2d Period. Millions. 1st Period. Units. The value of the figures in the above table, expressed in words according to the English method, is, Three hundred ninety-eight thousand, eight hundred thirty-two trillions; five hundred sixty-three thousand, eight hundred seventy-one bil- lions ; three hundred fifty-one thousand, six hundred fifteen millions ; one hundred twenty-three thousand five hundred sixty-one. 24 NUMERATION. 42. To read numbers represented by figures according to the English method ; — Begin at the right hand, and point off the figures into periods of six places each. Then, commencing at the left hand, read the figures of each period, adding the name of each period except that of units. Exercises. Read orally, or write in words, the numbers represented by the following figures, according to the English method : — 1. 23457896 2. 325487691 3. 1678912161 4. 98765421910311 5. 5632411132321300012 6. 6961771889133201443345567 43. To write numbers in figures according to the English method ; — Begin at the left hand, and write in each successive order the figure belonging to it. If any intervening order would otherwise be vacant, fill the place by a cipher. Exercises. Represent by figures, and read, the following numbers, ac- cording to the English method : — 1. Thirty-two million three hundred. 2. Seven billion seventeen thousand. 3. Five hundred sixty thousand one hundred two million, nine hundred twenty-nine thousand four hundred eleven. 4. One trillion, seven hundred forty-eight thousand nine hun- dred fifty-five billion. ADDITION 44. Addition is the process of collecting two or more numbers into one sum. The result thus obtained is called the amount. When the numbers added are all of the same denomination, as all dollars, all days, the process is termed addition of simple numbers. ADDITION. 25 45. To add simple numbers. Ex. 1. A man has three farms ; the first contains 378 acres, the second 586 acres, and the third 168 acres. How many acres are there in the three farms. Ans. 1132. operation. Having arranged the numbers so that all the units Acres. f the sa«me order shall stand in the same column, 3 7 8 we first add the column of units ; thus, 8 and 6 are 5 8 6 14, and 8 are 22 units, = 2 tens and 2 units. We 1 6 § write the two units under the column of units, and carry or add the 2 tens to the column of tens ; thus, Ans. 113 2 2 added to 6 make 8, and 8 are 16, and 7 are 23 tens, = to 2 hundred and 3 tens. We write the 3 tens under the column of tens, and add the 2 hundred to the column of hundreds; thus, 2 added to one make 3, and 5 are 8, and 3 are 11 hundred, = 1 thousand and 1 hundred. We write the 1 hundred under the column of hundreds ; and there being no other column to be added, we set down the 1 thousand in the thousands' place, and find the amount of the several numbers to be 1132. In practice, it is better not to name each figure added, but only the results, thus, 8, 14, 22 units, — 2 tens and 2 units, etc. Rule. — Write the numbers so that all the figures of the same order shall stand in the same column. Add, upward, all the figures in the column of units, and, if the amount be less than ten, write it underneath. But if the amount be ten or more, write down the unit figure only, and add in the figure denoting the ten or tens ivith the next column. Proceed in this way ivith each column, until all are added, observing to write under the last column its ivhole amount. 46. First Method of Proof — Begin at the top and add the columns downward in the same manner as they were be- fore added upward; and if the two sums agree, the work is presumed to be right. The reason of this proof is, that, by adding downward, the order of the figures is inverted ; and, therefore, any error made in the first addition would probably be detected in the second. Note. — This method of proof is generally used in business. 47. Second Method of Proof, — Separate the numbers to be added into two parts, by drawing a horizontal line be- tween them. Add the numbers below the line, and set down their sum. Then add this sum and the number or numbers above the line together ; and if their sum is equal to the first amount, the work is presumed to be right. N 3 26 ADDITION. The reason of this proof depends on the principle, that the sum of all the parts into which any number is separated is equal to the whole. (Art. 24, Ax. 13.) Examples. 2. OPERATION. Dollars. 765 381 872 315 2. OPERATION AND PROOF. Dollars. 765 3. 3. OPERATION. OPERATION AND PROOF. Tons. Tons. 126 126 381 872 315 384 876 243 384 876 243 Ans. 2 3 3 3 First ain't 2 3 3 3 Ans. 16 2 9 First am't 16 2 9 1568 Ans. 2 33 3 1503 Ans. 16 2 9 4. Barrels. 123 456 789 341 5. Pounds. 678 901 278 633 6. Acres. 456 789 127 815 7. Cents. 789 987 123 321 8. Eagles. 456 781 197 715 9. Rods. 781 175 564 337 1709 2490 2187 2220 2149 1857 11. 12. 13. 14. 15. Ounces. Inches. Rods. Furlongs. Cords. 7891 3256 6789 1234 4567 3245 7890 1234 5678 8912 6 7 89 1234 5678 9012 3456 1234 5678 6543 3456 7891 5678 7801 1234 7891 4567 10. Poles. 889 776 432 876 2973 16. Feet. 4561 7890 7658 8888 9199 24837 25859 21478 27271 29393 38196 17. Hogsheads. 1789 6 5 43 2177 8915 6781 4325 18. Furlongs. 6781 1371 8715 6371 1234 7171 19. Miles. 7890 1070 4437 6789 5378 1234 20. Dollars. 1785 5678 9137 8171 1888 1919 21. Casks. 7895 5678 7186 5176 4321 4127 22. Pence. 4371 1699 1098 8816 6171 7185 ADDITION. 27 23. Shillings. 78956 321 67 41328 45678 13853 71667 28. Acres. 789516 377895 378567 832156 7 8 9 5 67 8 13 13 8 24. Tons. 12345 87655 34517 65483 79061 20939 25. Miles. 34567 78901 32199 17188 88888 12345 29. Roods. 451237 813715 679919 787651 6 37171 813785 26. Trees. 76717 77776 67890 71444 47474 16175 30. Pole&r 1234567 8901234 5678901 3456789 5432115 7177444 27. Loads. 56789 12345 67819 34567 71888 33197 31. Yards. 789123 456789 987654 357913 245678 999999 32. What is the sum of 15 -f 26 + 18 -f 91 ? Ans. 150. 33. What is the sum of 6789 -f 5832 + 4671 -f 8907 ? Ans. 26199. 34. Required the sum of 76 + 48 + 59 -f 81. Ans. 264. 35. Required the sum of 123456 + 789012 -f 345678 + 901234 + 567890 -f 987654 + 321032 -f 765437. 36. Required the sum of 876543 + 789112 + 345678 + 965887 + 445566 + 788743 + 399378 + 456789. 37. Required the sum of 789012 -f 345678 + 901234 + 789037 -f 891133 -f- 477666 + 557788 + 888878. 38. Required the sum of 987654 -f 456112 + 222333 + 456789 + 987654 -f 321178 + 123456 -f 789561. 39. Required the sum of 678953 -f 467631 -f 117777 + 888888 + 444444 + 667679 -f 998889 + 671236. 40. Required the sum of 783256 + 7128 + 39 + 432815 -|_ 99 + 67851 + 125 -f 641236 + 801 + 4328. 41. Required the sum of 12004 -f- 32 -f- 1 -f- 7836 + 100 _(_ 46 + 3 + 6176 + 32 + 91876. 42. Add together 763, 4663, 37, 49763, 6178, and 671. Ans. 62075. 43. Add together 15, 7896, 1, 13, 106, 113, 156, 100, 2201. 28 ADDITION. 44. A butcher sold to A 369 lbs. of beef, to B 1G0 lbs., to C 861 lbs., to D 901 lbs., to E 71 lbs., and to F 8716 lbs. ; what did they all receive? Ans. 11087 lbs. 45. A owes to one creditor 596 dollars, to another 3961, to another 581, to another 6116, to another 469, to another 506, to another 69381, and to another 1261. What does he owe them all? Ans. $82871. 46. If a boy earn 17 cents a day, how much will he earn in 7 days? v Ans. 119 cts. 47. If a man's wages be 19 dollars per month, what are they per year ? Ans. $ 228. 48. If a boy receive a present every New Year's day of 1783 dollars, how much money will he possess when he is 21 years old? Ans. $37443. 48. Method of adding two or more columns at one operation. Ex. 1. A merchant paid for cloth 219 dollars, for flour 416 dollars, for hardware 711 dollars, and for rent 93 dollars. How much did he pay in all ? Ans. $ 1439. operation. Beginnillg with the number last written down, Dollars. ^ a jj ^ units an( j teng . thus> 93 an( j X _ 94^ 219 and 10 = 104, and 6 = 110, and 10 = 120, and 416 9 = 129, and 10 = 139. Of this amount we write 7 11 the 9 units and 3 tens under the columns added ; 9 3 and add in the 1 hundred with the column of ■ — hundreds ; thus, 1 (carried) and 7 = 8, and 4 = Ans. 14 3 9 12, and 2 = 14 hundred, of which we write the 4 hundred under the column of hundreds and t\\Q 1 thousand in the thousands' place ; and find the whole amount to be 1439. In like manner may be added more than two columns at one operation. Note. — The examples that follow can be performed as the above, or by the common method, or by both. 2. How many were the members of Congress in 1856, there being 2 Senators from each State, and Maine sending 6 Repre- sentatives, New Hampshire 3, Massachusetts 11, Rhode Island 2, Connecticut 4, Vermont 3, New York 33, New Jersey 5, Pennsylvania 25, Delaware 1, Maryland 6, Virginia 13, North Carolina 8, South Carolina 6, Georgia 8, Alabama 7, Mississippi 5, Louisiana 4, Tennessee 10, Kentucky 10, Ohio 21, Indiana 11, Illinois 9, Wisconsin 3, Iowa 2, Missouri 7, Arkansas 2, Michigan 4, Florida 1, Texas 2, California 2. Ans. 296. ADDITION. 29 3. Water-mills were invented in the year 555 after Christ ; windmills 744 years after water-mills ; pumps 126 years after windmills ; printing 15 years after pumps ; watches 37 years after printing ; the spinning-wheel 53 years after watches ; the steam-engine 119 years after the spinning-wheel; the fire- engine 14 years after the steam-engine ; the spinning-frame 98 years after the fire-engine ; and the electro-magnetic telegraph 71 years after the spinning-frame. What year was the electro- magnetic telegraph invented? Ans. 1832. 4. During the American Revolutionary war, the British lost at the battle of Lexington 273 men, at Bunker Hill 1054 men, at Long Island 400 men, at White Plains 300 men, at Fort Washington 1000 men, at Trenton 1020 men, at Princeton 400 men, at Hubbardton 200 men, at Bennington 800 men, at Brandywine 500 men, at Stillwater 600 men, at German- town 500 men, at Saratoga 400 men, by Burgoyne's sur- render 5791 men, at Fort Mercer 500 men, at Monmouth 400 men, at Rhode Island 260 men, at Brier Creek 23 men, at Stony Point 600 men, at Savannah 130 men, at Camden 325 men, at King's Mountain 1150 men, at Cowpens 800 men, at Guilford Court-House 600 men, at Hobkirk's Hill 250 men, at Eutaw Springs 700 men, at Yorktown 7000 men. What was the entire loss? Ans. 25976 men. 5. Required the number of square miles in the following States, there being in Maine 35,000, in New Hampshire 8,030, in Vermont 8,000, in Massachusetts 7,250, in Rhode Island 1,200, in Connecticut 4,750, in New York 46,000, in New Jersey 6,851, in Pennsylvania 47,000, in Ohio 39,964, in Michigan 56,243, in Indiana 33,809, in Illinois 55,405, in Wisconsin 53,924, in Iowa 50,914, and in California 188,982. Ans. 643,322. 6. Required the number of square miles in the following States, there being in Delaware 2,120, in Maryland 11,000, in Virginia 61,352, in North Carolina 45,500, in South Caro- lina 28,000, in Georgia 58,000, in Alabama 50,722, in Flor- ida 59,268, in Mississippi 47,147, in Tennessee 44,000, in Kentucky 37,680, in Missouri 67,380, in Arkansas 52,198, in Louisiana 46,431, and in Texas 325,520. Ans. 936,318. 7. According to the census of 1850, Maine had 583,169 3* 30 SUBTRACTION. inhabitants, New Hampshire 317,976, Massachusetts 994,514, Rhode Island 147,545, Connecticut 370,792, Vermont 314,120, New York 3,097,394, New Jersey 489,555, Pennsylvania 2,311,786, Delaware 91,532, Maryland 583,034, District of Columbia 51,687, Virginia 1,421,661, North Carolina 869,039, South Carolina 668,507, Georgia 906,185, Florida 87,445, Alabama 771,623, Mississippi 606,526, Lousiana 517,762, Texas 212,592, Arkansas 209,897, Tennessee 1,002,717, Missouri 682,044, Kentucky 982,405, Ohio 1,980,329, In- diana 988,416, Illinois 851,470, Michigan 397,654, Wisconsin 305,391, Iowa 192,214, California 92,597, and the Territories 92,298. What was the whole number of inhabitants ? Ans. 23,191,876. SUBTRACTION. 49. Subtraction is the process of taking one number from another to find the difference. When the two numbers are unequal, the larger is called the Minuend and the less number the Subtrahend ; and when the numbers are equal, either is the Minuend, and the other is the Subtrahend, The result, or the number found by the subtrac- tion, is called the Difference, or Remainder. When the numbers are of the same denomination, the pro- cess is termed Subtraction of Simple Numbers* 50. To subtract simple numbers. Ex. 1. From 935 take 673. . Ans. 262. operation. We first take the 3 units from the 5 Minuend 9 3 5 units, and find the difference to be 2 units, Subtrahend 6 7 3 wn i c ^ we write under the figure subtracted. We then proceed to take the 7 tens from Remainder 2 6 2 the 3 tens above it ; but we here find a difficulty, since the 7 is greater than the 3, and cannot be subtracted from it. We therefore add 10 tens to the 3 tens, which makes 13 tens, and then subtract the 7 from 13, and 6 tens remain, which we write below. Then, to compensate for the 10 tens, equal to 1 hundred, added to the 3 tens in the minuend, we add 1 hundred to the 6 hundred of the subtrahend, which makes SUBTRACTION. 31 7 hundreds, and subtract the 7 hundreds from 9 hundreds, and 2 hundreds remain. By adding the 10 tens to the minuend and the 1 hundred to the subtrahend, the two numbers being equally increased, their difference is not changed. (Art. 24, Ax. 9.) The remainder is 262. Note. — The addition of 10 to the minuend is sometimes called borrowing 10, and the addition of 1 to the subtrahend is called carrying 1. Rule. — Place the less number under the greater, so that units of the same order shall stand in the same column. Commencing at the right hand, subtract each figure of the subtrahend from the figure above it. If any figure of the subtrahend is larger than the figure above it in the minuend, add 10 to that figure of the minuend before subtracting, and then add 1 to the next figure of the subtrahend. 51 • First Method of Proof . — Add the remainder and the subtrahend together, and their sum will be equal to the min- uend, if the work is right. This method of proof depends on the principle, that the greater of any two numbers is equal to the less added to the difference between them. 52. Second Method of Proof. — Subtract the remainder or difference from the minuend, and the result will be like the subtrahend, if the work is right. This method of proof depends on the principle, that the smaller of any two numbers is equal to the remainder obtained by subtracting their difference from the greater. Examples. M* 2. 2. 3. 3. OPERATION. OPERATION AND PROOF. OPERATION. OPERATION AND PROOF. Minuend 4 6 9 469 788 788 Subtrahend 18 3 183 369 369 Remainder 2 8 6 286 419 419 Min. 469 Sub. 369 4. 5. 6. 7. Miles. Gallons. Minutes. Pecks. From 7 6 5 4 7116 6178 4567 Take 19 7 8 1997 17 69 1978 32 SUBTRACTION. 8. Barrels. From 7 6 5 116 Take 7 16 6 6 9 12. Hogsheads. From 6110 Take 19 9999 16. Hoods. From 10 2 Take 9 8 7 6 1 9. Degrees. 56789 10091 13. Bushels. 617853 190909 17. Acres. 511799 419109 10. Furlongs. 56781 39109 14. Yards. 7111111 909009 18. Poles. 610000 166666 From Take From Take From Take 20. Dollars. 10000000 9099019 21. Eagles. 99999999 1000919 11. Tons. 71678 18819 15. Pounds. 999000 199919 19. Cords. 789111 171670 22. Guineas. 888888 99999 23. Seconds. 100200300400500 90807060504039 25. Months. 61567101 91678 26. Days. 1000000 1 24. w Hours. 600700800900 19181891718 5 27. "Weeks. 10000000 9999999 28. What is the value of 6767851 — 81715 ? 29. What is the value of 761619161 — 916781 ? 30. What is the value of 31671675 — 361784? 31. What is the value of 16781321 — 100716? 32. What is the value of 1002007000 — 5971621 ? 33. Sir Isaac Newton was born in the year 1642, and he died in 1727 ; how old was he at the time of his de- cease? Ans. 85 years. 34. Gunpowder was invented in the year 1330 ; how long was this before the invention of printing, which was in 1440? Ans. 110 years. SUBTRACTION. 33 35. The mariner's compass was invented in Europe in the year 1302 ; how long was this before the discovery of America by Columbus, which happened in 1492 ? Ans. 190 years. 36. What number is that, to which if 6956 be added, the sum will be one million? Ans. 993044. 37. A man bought an estate for seventeen thousand five hundred and sixty-five dollars, and sold it for twenty-nine thousand three hundred and seventy-five dollars. Did he gain or lose, and how much? Ans. Gained $ 11810. 38. Bonaparte was declared emperor in 1804 ; how many years since ? 39. The union of the government of England and Scotland was in the year 1603; how long was it from this period to 1776, the time of the declaration of the independence of the United States? Ans. 173 years. 40. Jerusalem was taken and destroyed by Titus in the year 70 ; how long was it from this period to the time of the first Crusade, which was in the year 1096? Ans. 1026 years. 41. From the Creation to the Deluge was 1656 years; thence to the founding of Rome 1595 years; thence to the death of Charlemagne, which took place 814 years after Christ, 1567 years. In what year of the world was Christ born? Ans. 4004. 42. A gentleman 83 years old has two sons ; the age of the older son added to his makes -128 years, and the age of the younger son is equal to the difference between the age of the father and that of the older son. How old is each of his sons ? Ans. The older, 45 years ; the younger, 38 years. 43. During the year 1810, there were manufactured in the United States one hundred and forty-six thousand nine hun- dred and seventy-four yards of cotton cloth; and during the year 1855, five hundred and twenty million yards. What was the increase? Ans. 519,853,026 yards. 53. Method of subtracting, when there are two or more subtrahends. Ex. 1. From a pile of wheat containing 657 btrshels, A is to have 141 bushels, B 244 bushels, C 134 bushels, and D the remainder. How many bushels is D to have ? Ans. 138. 34 SUBTRACTION. FIRST OPERATION. SECOND OPERATION. Ill the first OpOTcl- Busheis. Bushels, tion, the several sub- Minuend 6 5 7 Minuend 6 5 7 trahends are added for "TTT s i a -i a single subtrahend to 141 \ 1 4 ] be taken from the 2 4 4 -^244 m i nu end. In the sec- 13 4 ^13 4 ond, the subtrahends Subtrahend ilo Remainder 188 Z%%%£$. Remainder 13 8 at } on i ^ us : beginning with units, 4 and 4 and 1 = 9, which from 17 units leaves 8 units; passing to tens, 1 (carried) and 3 and 4 and 4 = 12 tens; reserving the left-hand figure to add in with the figures of the subtrahends in the next column, the right>hand figure, 2 tens, which we subtract from the 5 tens of the minuend, and have left 3 tens ; and, passing to hun- dreds, we add in the left-hand figure 1 , reserved from the 1 2 tens, which with the other figures 1 and 2 and 1 = 5 hundreds, which, taken from 6 hundreds, leaves 1 hundred; and 138 is the answer sought. 2. John Drew has a yearly income of 2,500 dollars; his family expenses are 1,300 dollars, his expenditures in improv- ing his estate 450 dollars, and his contributions to several worthy objects 225 dollars. What remains to lay up or in- vest? 3. A speculator bought four village lots ; for the first he paid 620 dollars; for the second, 416 dollars; for the third, 350 dollars ; for the fourth, 225 dollars ; and sold the whole for 2,000 dollars. What did he gain ? 4. Daniel White, dying, left property to the amount of 27,563 dollars, of which his wife received 9,188 dollars, each of his two daughters, 4,594 dollars, and his only son the bal- ance. What did his son receive? Ans. 9,187 dollars. 5. The United States contain 2,983,153 square miles, of S which the Atlantic slope includes 967,576, the Pacific slope 778,266, and the Mississippi Valley the remainder. How many square miles does the Mississippi Valley contain ? Ans. 1,237,311. 6. The British North American Provinces contain 3,125,401 square miles; of which 147,832 square miles belong to Canada West ; 201,989 to Canada East ; 27,700 to New Brunswick ; 18,746 to Nova Scotia; 2,134 to Prince Edward's Island; 57,000 to Newfoundland ; 170,000 to Labrador ; and the re- MULTIPLICATION. 35 mainder to the Hudson's Bay Territory. What number of square miles belong to the Hudson's Bay Territory? Ans. 2,500,000. 7. James Howe has property to the amount of 63,450 dollars, and owes in all three debts ; one of 1000 dollars, another of 350 dollars, and another of 12,468 dollars. How much has he after paying his debts ? 8. The entire coinage of the mint of the United States, in- cluding the coinage of its branches, from 1792 to 1856, amounted in value to $498,197,382, of which $396,895,574 was gold, $ 100,729,602 silver, and the remainder of the amount cop- per. What was the value of the copper coinage ? Ans. $ 572,206. MULTIPLICATION. 54. Multiplication is the process of taking one number as many times as there are units in another number. In multiplication three terms are employed, called the Multi- plicand, the Multiplier, and the Product. The multiplicand is the number to be multiplied or taken. The midtiplier is the number by which we multiply, and denotes the number of times the multiplicand is to be taken. The product is the result, or number produced by the multi- plication. The multiplicand and multiplier together are called factors, from the product being made or produced by them. When the multiplicand consists of a single denomination, the process is termed multiplication of simple numbers. In the following table, the invention of Pythagoras, may be found all the elementary products necessary in performing any operation in multiplication, since the multiplication of numbers, however large, depends upon the product of one digit by an- other. The products, therefore, of each digit by any other, should be thoroughly committed to memory. Considerable more of the table, even, may be memorized with fully com- pensating results. 86 MULTIPLICATION. MULTIPLICATION TABLE. m CM o P P 71 o 17 — P P 71 m 71 71 o in 71 K3 5 P P CO 17 71 eg P ■ 7 CO O CO o o O CN P m O >n o m o o cn m b- o m m in in co m CN CO 01 00 xf 71 CO OS P 71 oo to 71 2 to 1— 1 71 ? 71 to 71 CO X. 71 71 C7 CO CO CO O CO' CO' Xj CO XJ c CM CO CO o O t* 00 CN CO oo o cn in r^ rf o m m in o o CO CO CO OS CO OS m go C7 to 1- P 71 O CO 71 CO i.7 71 CO I- 71 OS OS 71 71 71 C7 00 CO C7 ■— 1 OS CO i-h CO O CO CO 00 CO OS CN O cm m m m m m m CN to CO GO 00 o 71 •<* .7 00 OS CN 71 t 71 71 to X 71 00 S CO C7 71 i7 CO C7 § 2 CO "* O CM ^ CO 00 CO 00 O CN m in o m m CM 71 w (O 3 m o to 71 GO OS 2 P 71 CO 71 71 »-7 71 C7 I- 71 OS 71 .7 to CO CO 1-7 CO 00 OS t^. OS CO CO O ^-> CN t* 71 CO CO -T 00 o ■<* in m CN m O CM P - -o do 71 -+ t 71 71 71 71 71 p C7 CO 3 CO CO CO O CN c P to C 00 o m 2 x o co t 9) -r CO 77 71 .7 - I- O OS OS P 71 X 71 71 I- ^! 71 to to 71 i7 Xj 71 5 C7 CO 71 C7 CN —h -* CO co co O OS 00 OS co co X <-* t^. CO co m m 00 m I- OS O 71 to 71 eg 71 71 1- 71 71 C7 CN ,««* co co CO t^ co co OS C7 rf ^ 71 eg m t-^ 1.7 00 OP is 71 a at — CO eg eg i7 O 1- X g 71 71 71 X C7 71 i7 71 71 t- 71 OS tc 71 CO CO O CN CO CO o «^ t* in CO CO C7 at C7 00 ~ ~r m CN co 71 CO B0 to n at 71 -r I- OS g 71 71 71 -7 71 i7 71 I- 71 00 O CN CO CM CO co co i.7 C7 CO C7- 00 CO O m p cc <* Q -o n t- OS ■ 7 o VI .7 C7 .7 1.7 to C7 X .7 o 71 • 7 71 71 a 71 1.7 17 71 o »n t^ 00 CN CN 8 2 CO CO C7 .7 3 P CO CO m CO -t 00 71 7i m t^ X - 71 -f i7 -X X o. 71 71 71 C7 71 m to CM CN 00 OS CN CN g C7 71 CO CO 3 m CO CO CO 71 eg oi ■a 17 "O X t>. 01 — - c CO •-o »7 o. to 71 X • 7 OS as 71 71 71 CO 't CN CN O CO CO t- CN CN CO X 71 OS O 71 2 CO m CN CO CN 71 to C7 <* CO i - X at c M :r 4 .7 to X OS fl s a — i7 CN CN CO 71 71 00 00 71 s CO i— i I— < 71 71 3 • ■7 10 1- x 00 35 OS P — 71 71 C7 5 i7 17 ■O CO 1- I- X oo o OS o — CN O l-N 71 CO CN CN 9 71 CO 1.7 71 CO 71 m CN o as O eg * i.7 to t^. H OS 1 71 C7 -7 '7 CO j: CO OS § 2 CN CN 71 V, P 71 m CN OS 00 71 to cr -t in C7 CO 71 t> X o. OS O- X, I - CO 71 17 C7 C7 1.7 CN F-. co t^ O OS CO CO 00 OS o 71 CO 01 m CN CN 00 CO ct -r «f i-7 •x> I' x X OS r — 71 00 71 C7 ^ !m CO CO 1- 00 z: O CN I-*. -t 71 x 71 >.7 eg 71 rr oT «o~ i.7 c7 CO 1- X OS X o. i7 3 71 oT CO CO CN CO I to 00 CO in" CO 71 30 eg eg <* * L-7 to CO l^ i- X 81 " 6 O "- CN CN CO C7 — o" m m 2 <—< P 71 >7 71 >7 C7 i7 u7 17 CO ■ 7 CO o" I- i7 I- a So o o io 00 OS ' OS o m O P o in* D 71 in • ob~ ~cb o" — 71 71 CO 71 71 Tn" CN For example, suppose we wish to find the product of 7 by 5 ; we look for 7 at the top of the table, and for 5 at the left hand, and where the lines intersect is 35, the number sought ; or, we may look for the 7 at the left hand, and the 5 at the top, and find, where the lines intersect, the same result. MULTIPLICATION. 37 55. The repeated addition of a number to itself is equiva- lent to a multiplication of that number. Thus, 7 —J— 7 — j— 7 — J- 7 is equivalent to 7 X 4, the sum of the former and the product of the latter being the same. Hence multiplication has some- times been called a concise method of addition, 56. The product must be of the same kind or denomination as the multiplicand, since the taking of a quantity any number of times does not alter its nature. Thus : 5, an abstract number, X 3 = 15, an abstract number ; and 9 yards X 7 = 63 yards. 57. The multiplier must always be considered as an abstract number. Thus, in finding the cost of 4 boohs at 9 dollars each, we cannot multiply books and dollars together, which would be absurd, but we can, by regarding the 4 as an abstract number, take the 9 dollars, or cost of 1 book 4 times, and the product, 36 dollars, will be the result required. 58. The product of two factors will be the same, whichever is taken as the multiplier. Thus, 8x6= 6x8 = 48; and the cost of 5 hats at 2 dollars each gives the same product as 2 hats at 5 dollars each. Also, the product of any number of factors is the same, in whatever order they are multiplied. Thus, 2X3X5 = 3X5X2 = 5X2X3 = 30. 59. A Composite number is a number produced by multi- plying together two or more numbers greater than 1. Thus, 10 is a composite number, since it is the product of 2 X 5 ; and 18 is a composite number, since it is the product of 2 X 3 X 3. 60. To multiply simple numbers. Ex. 1. Let it be required to multiply 1538 by 9. Ans. 13842. operation. Having written the multiplier, 9, un- Multiplicand 15 3 8 der the unit figure of the multiplicand, Multiplier 9 we multiply the 8 units by the 9, ob- -r> j taining 72 units = 7 tens and 2 units. Product 13 8 4 2 We wr ; te an(1 4 Products 1 4 3 12 hundreds times 2156 = 86 24 hundreds; (8624 the sum of which partial products = -p , .. - a Q Q 911988, or the total product required. 1 roduct J 1 1 J 8 8 In the opera tion the right-hand figure of each partial product is written di- rectly under its multiplier, that units of the same order may stand in the same column, for convenience in adding. Rule. — Write the multiplier under the multiplicand, arranging units under units, tens under tens, jrc, Multiply each figure of the multiplicand by each figure of the midtiplier, beginning with the right-hand figure, writing the right-hand figure of each product underneath, and adding the left-hand figure or figures, if any, to the next succeeding product. If the multiplier consists of more than one figure, the right-hand figure of each partial product must be placed directly under the figure of the multiplier that produces it. The sum of the partial products will be the whole product required. Note. — When there are ciphers between the significant figures of the multiplier, pass over them in the operation, and multiply by the significant figures only, remembering to set the first figure of the product directly under the figure of the multiplier that produces it. 61 . First Method of Proof. — Multiply the multiplier by the multiplicand, and, if the result is like the first product, the work is supposed to be right. (Art. 58.) 62. Second Method of Proof \ — Divide the product by the multiplier, and, if the work is right, the quotient will be like the multiplicand. Note. — This is the common mode of proof in business ; but, as it antici- pates the principles of division, it cannot be employed without a previous knowledge of that process. 63. Third Method of Proof. — Begin at the left hand of the MULTIPLICATION. 39 multiplicand, and add together its successive figures toward the right, till the sum obtained equals or exceeds the number nine. If it equals it, drop the nine, and begin to add again at this point, and proceed till you obtain a sum equal to, or greater than, nine. If it exceeds nine, drop the nine as before, and carry the excess to the next figure, and then continue the ad- dition as before. Proceed in this way, till you have added all the figures in the multiplicand and rejected all the nines con- tained in it, and write the final excess at the right hand of the multiplicand. Proceed in the same manner with the multiplier, and write the final excess under that of the multiplicand. Multiply these excesses together, and place the excess of nines in their product at the right. Then proceed to find the excess of nines in the product obtained by the original operation ; and, if the work is right, the excess thus found will be equal to the excess contained in the product of the above excesses of the multiplicand and mul- tiplier. Note. — This method of proof, though perhaps sufficiently sure for common purposes, is not always a test of the correctness of an operation. If two or more figures in the work should be transposed, or the value of one figure be just as much too great as another is too small, or if a nine be set down in the place of a cipher, or the contrary, the excess of nines will be the same, and still the work may not be correct. Such a balance of errors will not, however, be likely to occur. Examples. 3. Multiply 7325 by 3612. Ans. 26457900. OPERATION. PROOF BY MULTIPLICATION. Multiplicand 7 3 2 5 Multiplicand 3 6 12 Multiplier 3 612 Multiplier 7 3 2 5 14650 18 6 7325 7224 43950 10836 21975 25284 Product 26457900 Product 26457900 4. Kequired the product of 82967 by 652. Ans. 54094484. 40 MULTIPLICATION. Multiplicand Multiplier Product OPERATION. 82967 652 165934 414835 497802 54094484 PROOF BY THE NINES. 5 excess. 4 excess. 201 2 excess. 2 excess. 5. 789123 4 9. 678954 24 6. 1234567 5 10. 616783 36 989898 2 11. 789563 57 8. 3 7 8 9 5 8 8 8 12. 789567 98 13. . 892001 329 14. 230442 701 15. 425016 645 16. 5061029 3408 17. What will 365 acres of land cost at 73 dollars per acre? Ans. $26645. 18. What will 97 tons of iron cost at 57 dollars a ton? Ans. $5529. 19. What will 397 yards of cloth cost at 7 dollars per yard? Ans. $2779. 20. What will 569 hogsheads of molasses cost at 37 dollars per hogshead? Ans. $21053. 21. If a man travel 37 miles in one day, how far will he travel in 365 days? Ans. 13505 miles. 22. If a vessel sails 169 miles in one day, how far will she sail in 144 days? 23. What will 698 barrels of flour cost at 7 dollars a bar- rel? 24. What will 376 lbs. of sugar cost at 13 cents a pound ? Ans. 4888 cts. 25. What will 97 lbs. of tea cost at 93 cents a pound? Ans. 9021 cts. MULTIPLICATION. 41 26. If a regiment of soldiers consists of 1128 men, how- many men are there in an army of 53 regiments ? Ans. 59784. 27. What is the product of 75432 X 47. Ans. 3545304. 28. What is the product of 76785316 X 7615. Ans. 584720181340. 29. What is the product of 67853 X 8765. Ans. 594731545. 30. What is the product of 3812345 X 31243. Ans. 119109094835. 31. What is the product of 40670007 X 10002. Ans. 406781410014. 32. What is the product of 31235678 X 10203. Ans. 318697622634. 33. What is the product of 76786321 X 3007. Ans. 230896467247. 34. What is the product of 6176777 X 22222. Ans. 137260338494. 35. What is the product of 7060504 X 30204. Ans. 213255462816. 36. Multiply 88888 by 4444. Ans. 395018272. 37. Multiply 7008005 by 10008. Ans. 70136114040. 38. Multiply 987648 by 481007. Ans. 475065601536. 39. Multiply 101010101 by 202020202. Ans. 204060808060402. 40. Multiply 304050607 by 3011101. Ans. 915527086788307. 41. Multiply 908007004 by 500123. Ans. 454115186861492. 42. Multiply 2003007001 by 6007023. Ans. 12032109124168023. 43. Multiply 9000006 by 9000006. Ans. 81000108000036. 44. A full-grown elm will, it is computed, yearly, on an average, produce three hundred twenty-nine thousand three hundred and seventy-five seeds. How many seeds will three such trees produce in fifty-three years. Ans. 52370625. • 45. John Alden can plant 3 plats of corn, containing 11 rows of 67 hills each, in 1 day, and Loring Blanchard can 4* 42 MULTIPLICATION. plant twice as much in the same time. How many hills can Blanchard plant in a month of 26 working days ? Ans. 114972. 46. If the multiplicand be three hundred and seventy-five millions two hundred and ninety-six thousand three hundred and twenty-one, and the multiplier seventy-nine thousand and twenty-four, what will be the product ? Ans. 29657416470704. 64. When the multiplier is a composite number. Ex. 1. What cost 35 acres of land at 316 dollars an acre? Ans. 11060 dollars. operation. The factors of 35 are 3 16 dollars, cost of 1 acre. 7 an d 5. Now if we 7 multiply the price of one 921 2 dollirs cost of 7 acres a ^ re by 7 ' we 3 et the cost 1 1 1 I dollars, cost 01 / acres. of 7 ^ reg . and? the ^ hy , ~ multiplying the cost of 7 110 6 dollars, cost of 35 acres. ? cres . J>y & e fac t° r 5 > {t is evident, we obtain the cost of 5 times 7 acres, or 35 acres. Hence, when the multiplier is a composite number, we may Multiply the multiplicand by one of the factors of the multiplier, and the product thus obtained multiply by another, and so on until each of the factors has been used as a multiplier : and the last product will be the one sought. Examples. 2. Multiply 3121 by 81, using its factors. 3. What will 63 horses cost at 175 dollars each? 4. A certain house contains 87 windows, and each win- dow has 32 panes of glass. How many panes in the whole house? Ans. 2784. 5. What is the product of 47134987 by 56? Ans. 2639559272. 6. If a garrison consume 6231 pounds of bread in 1 day, how many pounds will the same consume in 144 days ? Ans. 897264 pounds. C5. When there are ciphers on the right in the multiplier or multiplicand, or both. MULTIPLICATION. 43 OPERATION. Ex. 1. In 1 yard there are 36 inches; how many inches in 10 yards? In 100 yards? Ans. 360, 3600. We annex one cipher to the multiplicand to multiply it by 10, and two ciphers to multiply it by 100 ; since annexing one cipher removes each figure of the multiplicand one place to the left, and thus increases it Multiplicand Multiplier Product Or thus : 36 10 360 3 6 0, 36 100 3T00 3 600. 10 times ; annexing two ciphers removes each figure two places to the left, and increases it 100 times ; and so on, each additional cipher hav- ing the effect to increase its value 10 times (Art. 30). 2. What will 700 bales of cotton cost at 40 dollars per bale ? Ans. 28000 dollars. Multiplicand Multiplier Product OPERATION. 700 40 28000 The multiplicand we resolve into the factors 7 and 100, and the multiplier into the factors 4 and 10. Now, it is evident (Art. 58), that, if these several factors be multiplied together, they will produce the same product as the origi- nal factors, 700 and 40. Thus 7X4 = 28, and 28 X 100 = 2800, and 2800 X 10 = 28000, the same result as in the operation. Hence, when there are ciphers, one or more, on the right of the multiplier, or multiplicand, or both, we may, for the required pro- duct, Multiply the significant figures together, and to their product annex as many ciphers as there are on the right in both multiplicand and mul- tiplier. Examples. 3. Multiply 1819 by 10. 4. Multiply 4106 by 100. 5. Multiply 10000 by 7000. 6. Multiply 123000 by 78000. 7. Multiply 70000 by 10000. 8. Multiply 900900 by 70070. Ans. 70000000. Ans. 9594000000. Ans. 700000000. Ans. 63126063000. 9. What must be the distance sailed by a steamship, whose average rate is 310 miles a day, in making a voyage from New York to Liverpool, in 12 days? 10. The annual salary of a member of Congress being 3,000 dollars, how much do 296 members receive ? Ans. 888,000 dollars. 44 DIVISION. 11. The salary of the President of the United States is 25,000 dollars a year; how much will it amount to in 82 years? Ans. 2,050,000 dollars. 12. The earth is 95,000,000 of miles from the sun, and the planet Neptune is 30 times as far. How far is Neptune from the Sun ? Ans. 2,850,000,000 miles. DIVISION 66. Division is the process of finding how many times one number is contained in another ; or the process of separating a number into a proposed number of equal parts. In division there are three principal terms : the Dividend, the Divisor, and the Quotient. The dividend is the number to be divided. The divisor is the number by which we divide The quotient is the result, or number produced by the di- vision, and denotes the number of times the divisor is contained in the dividend, or one of the equal parts into which the divi- dend is divided. When the dividend does not contain the divisor an exact number of times, the excess is called a remainder, and may be regarded as a fourth term in the division. When the dividend consists of a single denomination, the process is termed Division of Simple Numbers. 67. Division is frequently indicated by writing the dividend above a short horizontal line and the divisor below ; thus, §; The expression J = 3 is read, 6 divided by 2 is equal to 3. Another method of indicating division, is by a curved line placed between the divisor and dividend. Thus, the expression 6) 12 shows that 12 is to be divided by G. 68. When a number is divided into two equal parts, one of the parts is called one half ; when divided into three equal parts, one of the parts is called one third, two of the parts two thirds ; when divided into four equal parts, one of the DIVISION. 45 parts is called one fourth, two of the parts two fourths, three of the parts three fourths ; etc. Such equal parts are called fractions, since they are frac- tured or broken numbers. They are expressed by figures, in a form of division ; thus, one half is written £ ; one third, \ ; two thirds, § ; one fourth, £ ; two fourths, f ; three fourths, £ ; and may also be read, one divided by two, one divided by three, and so on. In any fraction, expressed in the manner now ex- plained, the number above the line is called the numerator, and that below the line its denominator. Thus, in \, 1 is the nu- merator, and 2 the denominator. 69. When the divisor and dividend are of the same kind or denomination, the quotient will denote the number of times the divisor is contained in the dividend. Thus, to find how many pencils at 6 cents each can be bought for 24 cents, we inquire how many times 6 cents are contained in 24 cents, which are 4 times. Hence, 4 pencils, at 6 cents each, can be bought for 24 cents. 70 • When the divisor and dividend are of different kinds or denominations, the divisor will denote the number of equal parts into which the dividend is divided, and the quotient will denote the number of units in each part, and will be of the same kind or denomination as the dividend. Thus, to find the cost of 1 pencil, when 4 pencils cost 24 cents, we separate or divide the 24 cents into 4 equal parts, of which one part is 4 cents. Hence, 1 pencil costs 6 cents, when 4 pencils cost 24 cents. 71 # The remainder will always be of the same kind or de- nomination as the dividend, since it is a part of the dividend. 72. Division is the reverse of multiplication. The dividend answers to the product, and the divisor and quotient to the factors, of multiplication. In multiplication, two factors are given, to find their product ; and in division, one of two factors and their product are given, to find the other factor. 73. To divide simple numbers. Ex. 1. How many yards of cloth, at 4 dollars a yard, can be bought for 948 dollars ? Ans. 237 yards. 46 DIVISION. operation. ^\y e first inquire how many Divisor 4)948 Dividend, times 4, the divisor, is contained o 7 r\ -• a. in 9, the first left-hand figure of 1 6 i quotient. . the dlvidendj wbich is ] mn( ireds, and find it contained 2 times, and 1 hundred remaining. We write the 2 directly under 9, its dividend, for the hundreds' figure of the quotient. To 4, the next figure of the dividend, which is tens, we regard as prefixed the 1 hundred that was remaining, which equals 10 tens, and thus form 14 tens, in which we find the divisor 4 to be contained 3 times, and 2 tens remaining. We write the 3 for the tens' figure in the quotient, and the 2 tens that were remaining, equal 20 units, we regard as prefixed to 8, the last figure of the dividend, which is units, in which the divisor 4 is contained 7 times. Writing the 7 for the units' figure of the quotient, we have 237 as the entire quotient, equal the number of yards of cloth at 4 dollars a yard that can be bought for 948 dollars. 2. How many times does 3979 contain 17 ? Ans. 234^ times. opEnxTioN. We say, 17 in 39, Dividend. 2 times. The 2 we Divisor 17)3979(234 T U Quotient. ™ rite in the ( l uo " J 34 V TT ^ tient. 17X2 = 34, which we write 5 7 under the 39. 39 5 1 — 34 = 5, to which bringing down the 6 9 next figure of the 6 8 dividend, we form — ~ -o . , 57. 17 in 57, 3 1 Remainder. ^ ^ The 3 \ ye write in the quo- tient. 17 X 3 = 51, which we write under the 57. 57 — 51 = 6, to which bringing down the next figure of the dividend, we form 69. 17 in 69, 4 times. The 4 we write in the quotient. 17 X 4 = 68, which we write under the 69. 69 — 68 = 1, a remainder, or a part of the dividend left undivided. 1 divided by 17 — * A (Art. 68). The i f we write in the quotient, and obtain as the answer required 234^. In this illustration, to render the explanation the more concise, the naming of the denominations of the figures has been omitted. When, as in the operation preceding the last, results only are written down, the method is called short division ; and when, as in the last operation, the work is written out at length, it is called long division. The principle is the same in both cases. Hence the general Rule. — Beginning at the left, find how many times the divisor is contained in the fewest figures of the dividend that will contain it, for the first quotient figure. DIVISION. 47 Multiply the divisor by this quotient figure, and subtract the product from the figures of the dividend used. With the remainder, if any, unite the next figure of the dividend. Find how many times the divisor is contained in the number thus formed, and write the figure denoting the result at the right of the former quotient figure. Thus proceed until all the figures of the dividend are divided. Note 1. — The proper remainder is in all cases less than the divisor. If, in the course of the operation, it is at any time found to be larger than the divisor, it will show that there is an error in the work, and that the quotient figure should be increased. Note 2. — If at any time the divisor, multiplied by the quotient figure, produces a product larger than the part of the dividend used, it shows that the quotient figure is too large, and must be diminished. Note 3. — It will often happen that, when a figure of the dividend is taken, the number will not contain the divisor ; and, in that case, a cipher must be placed in the quotient, and another figure of the dividend taken, and so on, until the number is large enough to contain the divisor. Note 4. — If there be a remainder after dividing the last figure of the dividend, write it with the divisor underneath, with a line between them, at the right of the quotient. 74. First Method of Proof. — Multiply the divisor by the quotient, and to the product add the remainder, if any, and if the work be right, the sum thus obtained will be equal to the dividend. Note. — This method follows from division being the reverse of multiplica- tion. (Art. 72.) 75. Second Method of Proof. — Find the excess of nines in the divisor, quotient, and remainder. Multiply the excess of nines in the divisor and quotient together, and to the product add the excess of nines in the remainder. If the excess of nines in this sum equal the excess of nines in the dividend, the work may be supposed to be right. 76. Third Method of Proof. — Add together the remainder, if any, and all the products that have been produced by multi- plying the divisor by the several quotient figures, and the result will be like the dividend, if the work be right. 77. Fourth Method of Proofs — Subtract the remainder, if any, from the dividend, and divide the difference by the quotient. The result will be like the original divisor, if the work be right. Note. — The first method of proof (Art. 74) is usually most convenient, and is most commonly employed. 48 DIVISION. Examples. 3. Divide 9184 by 7. Ans. 1312. OPERATION. Divisor 7)9184 Dividend. 1312 Quotient. 4. Divide 18988 by 759. OPERATION. PROOF BY MULTIPLICATION. 1312 Quotient. 7 Divisor. 9184 Dividend. Ans. 25^. PROOF BY THB NINES. Dividend. Divisor = 3 excess. Divisor 759)18988(25 Quotient. Quotient == 7 excess. 1518 Remainder = 4 excess. 3808 3795 (3 X 7) + 4 = 7 excess. 1 3 Remainder. Dividend = 7 excess. 5. Divide 147856 by 97. Ans. 1524|f OPERATION. Dividend. Divisor 97)147856(1524 Quotient. +9 7 508 +485 235 +19 4 PROOF BY ADDITION. 97 416 +3 8 8 Remainder. 14785 6 Dividend. Ans. 297. PROOF BY DIVISION. Dividend. Dividend. Divisor 285)84645(297 Quotient. 297)84645(285 Divisor 570 594 +2 8 Remainder. 6. Divide 84645 by 285. OPERATION. 2764 2565 1995 1995 2524 2376 1485 1485 DIVISION. 49 7. 3) 67856336 22618778 § 10. 5)334652 8. 7) 178985 11. 8)96723578 9. 11)1667789 12. 9)186731 13. 14. 17) 678916( 35)9106013( x6. Divide 671678953 by 6. 17. Divide 166336711 by 7. 18. Divide 161331793 by 8. 19. Divide 161677678 by 9. 20. Divide 363895678 by 11. 21. Divide 164378956 by 12. 22. Divide 78950077 by 3. 23. Divide 678956671 by 4. 24. Divide 667788976 by 5. 25. Divide 777777777 by 6. 26. Divide 888888888 by 7. 27. Divide 789636 by 46. 28. Divide 7967848 by 52. 29. Divide 16785675 by 61. 30. Divide 675753 by 39. 31. Divide 5678911 by 82. 32. Divide 6716394 by 94. 33. Divide 1167861 by 135. 34. Divide 7861783 by 87. 35. Divide 1678567 by 365. 36. Divide 87635163 by 387. 37. Divide 34567890 by 6789. 15. 91) 6210011 ( Quotients. 111946492 23762387 20166474 17964186 33081425 13698246 153227 275175 71451 8650 90365 4598 226447 5091 38. What is the value of 213255467083 -f- 30204 ? 39. What is the value of 395020613 -=- 4444? 40. What is the value of 7207276639 -f- 9009 ? 41. What is the value of 454115186870257 — 500123 42. How many barrels of flour, at 9 dollars a barrel, bought for 18621 dollars? N 5 Rem. 1. 2. 1. 4. 3. 4. 1. 3. 1. 3. 6. 44. 1. Ill, 28. 297. 174. 5091. 4267. 2341. 4567. ? 8765. can be 50 DIVISION. 43. How much sugar at 15 dollars a hundred may be bought for 405 dollars? 44. A tailor has 938 yards of broadcloth ; how many cloaks can be made of the cloth, if it require 7 yards to make one cloak ? 45. What number multiplied by 1728 will produce 1705536 ? Ans. 987. 46. A. Hartmann has sold his wagon to J. Herr for 85 dol- lars. He is to receive his pay in wood at 5 dollars a cord. How many cords will it require to pay for the wagon ? Ans. 17 cords. 47. The Bible contains 31,173 verses; how many must be read each day, that the *book may be read through in a year of 365 days? Ans. 85^|§ verses. 48. A train on the Liverpool Railroad runs at the rate of 65 miles an hour ; how long would it take at that velocity to pass round the earth, the distance being about 25,000 miles ? Ans. 384 T 8 7j hours. 49. A gentleman possessing an estate of 66,144 dollars, be- queathed one fourth to his wife, and the remainder was divided among his 4 children ; what was the share of each ? Ans. 12,402 dollars. 50. If the dividend is 6756785 and the quotient 193051, what is the divisor? Ans. 35. 51. A's age multiplied by 17, or B's age multiplied by 19, is equal to 1292 years, and the sum of their ages is equal to C's age multiplied by 3. What is the age of each ? Ans. A's 76 years ; B's 68 years ; C's 48 years. 78. When the divisor is a composite number. Ex. 1. A farmer boughfc 21 horses for 2625 dollars; how many dollars did each cost? Ans. 125 dollars. operation. The factors of 21 are 3)2625 dolls., cost of 21 horses. 3 and 7. Now, if we rr \ q n r -, 11 . r rr -i divide the 2625 dollars, 7 ) 875 dolls., cost of 7 horses. the cost of 21 horgeSj h > 12 5 dolls., cost of 1 horse. 3, we obtain 875 dollars, the cost of 7 horses, since there are 7 times 3 in 21. Then, dividing the 875 dollars, the cost of 7 horses, by 7, we obtain 125 dollars, the cost of 1 horse. DIVISION. 51 2. Divide 3515 by 42. Ans. 83£f. OPERATION. FINDING THE TRUE REMAINDER. 2)3515 4 X 3 X 2 = 24, 1st Product. 3 ) 17 57, 1, 1st Rem. 2x2= 4, 2d Product. 7)585, 2, 2d Rem. 1, 1st Remainder. 8 3, 4, 3d Rem. 29, true Remainder. Or, (4 X 3 X 2) + (2 X 2) + 1 = 29, true Rem. Using as divisors 2, 3, and 7, the factors of 42, we obtain for remainders, 1, 2, and 4. The first remainder, 1 , is a unit of the given dividend, since it is a part of it (Art. 71). The second remainder, 2, is units of the quo- tient 1 75 7, whose units are 2 times as great as those of the given dividend. The third remainder, 4, is units of the quotient 585, whose units are 3 times as great as those of the quotient 1757, of which the units are 2 times as great as those of the given dividend. Now, these remainders must be all of the same units as the given dividend, to constitute the whole or true remainder. We therefore multiply the third remainder by 3 and 2, the divisors used in producing the quotient of which it is a part ; and the second remainder by 2, the divisor used in producing the quotient of which it is a part ; and the products with the first remainder added together give 29, the whole or true remainder sought. Hence, as shown by these illustrations, when the divisor is a composite number, we may Divide the dividend by one of the factors, and the quotient thus found by another, and thus proceed till each of the factors has been made a divisor. The last quotient will be the quotient required. If there be remainders, multiply each remainder, except the first, by all the divisors preceding the one which produced it; and the first remain- der being added to the sum of the products, the amount will be the true remainder. Note. — There will be but one product to add to the first remainder when there are only two divisors and two remainders. Examples. 3. Divide 7704 by 24 = 4 X 6. Ans. 321. 4. Divide 8317 by 27 = 3 X 9. 5. Divide 3116 by 81 = 9 X 9. 6. Divide 61387 by 121 = 11 X l£ Ans. 507 T 4 f T . 7. Divide 19917 by 144 = 12 X 12. Ans. 138^. 8. Divide 91746 by 336 = 6 X 7 X 8. Ans. 273^- 9. At 45 dollars an acre, a farm of how many acres can be bought for 5464 dollars ? Ans. 121£f acres. 52 DIVISION. 79. When the divisor contains one or more ciphers at the right hand. Ex. 1. If 10 men receive 792 dollars for a job of work, what will be each man's share of it ? Ans. 79-£$ dollars. operation. To multiply by 10, we annex one 1 | ) 7 9 | 2 cipher, which removes the figures one r\ & L n c\ o -r> place to the left, and thus increases Quotient 7 9, 2 Rem. i h e irva i ue io times (Art. 65). Now, Or thus : 7 9 | 2. it is obvious, that, if we reverse the process, and cut off the right-hand figure by a line, we remove the remaining figures one place to the right, and consequently diminish the value of each 10 times, and thus, divide the whole number by 10. The figures on the left of the line are the quotient, and the one on the right is the remainder, which may be written over the divisor and annexed to the quotient Hence each man's share is 79f . 2. How many years will it take a man at a yearly salary of 700 dollars to earn 3664 dollars? Ans. 5f§-$- years. operation. The divisor, 700, may 1|00)36|64 be resolved into the factors ' _ - . . , _, 7 and 100. We first divide 7 ) 3 6, 6 4, 1st Kern. by thc factor 100? by cut _ 5, 1, 2d Rem. **$ ofr two figures at the right, and get 36 for the Or thus : 7|00)36|64 quotient, and 64 for a rc- mainder. We then divide 5, 1 G 4. the quotient, 36, by the other factor, 7, and obtain 5 for a quotient and 1 for a remainder. The last remainder, 1, being multiplied by the divisor, 100, and 64, the first remainder, added, we obtain 164 for the true remainder (Art. 77); and for the answer required, 5}oo years. Hence, when the divisor contains one or more ciphers at the right, we may, to perform the division, Cut off the ciphers from the right of the divisor, and the same number of figures from the right of the dividend ; and then divide the remaining figures of the dividend by the remaining figures of the divisor. Note. — When, by the operation, there is a remainder, to it must be an- nexed the figures cut off from the dividend to form the true remainder. Should there be no last remainder, then the significant figures, if any, cut off from the dividend, will form the true remainder. Examples. Quotient. Kern. 3. Divide 123456789 by 10. 12345678 9. 4. Divide 987654300 by 100. Quotients. Rem. 5 2100. 11 91853. 3 137851. 37 411111. GENERAL PRINCIPLES AND APPLICATIONS. 53 5. Divide 32100 by 6000. 6. Divide 3678953 by 326100. 7. Divide 1637851 by 500000. 8. Divide 41111111 by 1100000. 9. Divide 89765432156 by 1000000. 10. The entire annual loss to the United States in conse- quence of intemperance has been estimated to be about 98,400,000 dollars. How many schools at a yearly expense of 600 dollars would that sum support ? Ans. 164,000 schools. 11. The late war with Russia was carried on at a cost of 600,000,000 dollars to Great Britain. Allowing that country to have a population of 28,000,000, what was the cost to each individual ? 12. If light moves at the rate of 192,000 miles in a second, how long is it in passing from the sun to the earth, a distance of 95,000,000 of miles. Ans. 494|f § seconds. GENERAL PRINCIPLES AND APPLICATIONS. 80 1 In division, the value of the quotient depends upon the divisor and dividend. 81. If the dividend be multiplied, or the divisor divided, by any number, the quotient is multiplied by the same number. Thus, if the dividend be 20 and the divisor 4, the quotient will be 5 ; but if the dividend be multiplied by any number, as 2, and the divisor remain unchanged, the quotient will be 2 times as large as before, or 10; as (20 X 2) ~ 4 = 10; and if the divisor be divided by the 2, and the dividend remain un- changed, the quotient will be, likewise, 2 times as large, or 10 ; as 20 -*• (4 -f- 2)=~ 10. 82. If the dividend be divided, or the divisor multiplied, by any member, the quotient is divided by the same number. Thus, if the dividend be 32 and the divisor 8, the quotient will be 4. But if the dividend be divided by any number, as 2, and the 5* 54 CANCELLATION. divisor remain unchanged, the quotient will be only half as large as before, or 2 ; as (32 -H 2) -r- 8=2; and if the divisor be multiplied by 2, and the dividend remain unchanged, the quotient will be, likewise, only half as large, or 2 ; as 32 -f- (8 X 2) = 2. 83. If the dividend and divisor be both multiplied, or both divided, by the same number, the quotient will not be changed. Thus, if the dividend be 16 and the divisor 4, the quotient will be 4. Now, if we multiply the dividend and divisor by some number, as 2, their relative values are not changed, and we obtain 32 and 8 respectively, and 32 -r- 8 = 4, the same as the original quotient. Also, if we divide the dividend and quotient by some number, as 2, their relative values are not changed, and we obtain 16 and 2 respectively, and 16 -s- 2 = 8, the same quotient as before. 84. If a factor in any number is rejected or cancelled, the number is divided by that factor. Thus, if 24 is the dividend and 6 the divisor, the quotient will be 4. Now, since the di- visor and quotient are the two factors which, being multiplied together, produce the dividend (Art. 72), it follows, if we reject or cancel the factor 6, the remaining 4 is the quotient ; and, by the operation, the dividend 24 has been divided by 4. CANCELLATION. 85. Cancellation is the method of abbreviating arithmet- ical operations by rejecting any factor or factors common to the divisor and dividend. Ex. 1. Sold 19 thousand shingles at 4 dollars a thousand, and received pay in wood at 4 dollars a cord ; how many cords of wood was received? Ans. 19 cords. operation. Having indicated by signs Dividend ^ X 19 . the multiplication and di- . ~Z =s 1™ Quotient, vision required by the ques- Divisor £ t j on ^ ^ en ^ s i nce dividing both dividend and divisor by the same number will not change the quotient (Art. 83), we divide them by the common factor 4, by cancelling it in both, and obtain 25 for the quotient. CANCELLATION. 55 2. Divide the product of 15, 3, 28, and 13, by the product of 7, 30, and 4. * OPERATION. Dividend j $ X 3 X ft X 1 3 _ 3 9 Divisor *X£0Xrf~ " * T = 19 * Qu0tient ' 2 The product of the 7 and 4 in the divisor equals the 28 in the divi- dend ; we therefore cancel all these numbers. Finding 15 in the dividend to be a factor of 30 in the divisor, we cancel both of the numbers, and use the remaining factor 2 in place of the 30. There now being no factor common to both dividend and divisor uncancelled, we multiply together the remaining factors in the dividend, and divide the product by the remaining factor in the divisor, and obtain the quotient 19J. Rule. — Cancel the factor or factors common to the dividend and divisor, and then divide the product of the factors remaining in the dividend by the product of those remaining in the divisor. Note. — 1. In arranging the numbers for cancellation, the dividend may be written above the divisor, with a horizontal line between them, as in division (Art. 67); or, as some prefer, the dividend may be written on the right of the divisor, with a vertical line between them. Note. — 2. Cancelling a factor does not leave 0, but the quotient 1, to take its place, since rejecting a factor is the same as dividing by that factor (Art. 84). Therefore, for every factor cancelled, either in the dividend or divisor, the factor 1 remains. Examples. 3. Multiply 24 by 16, and divide the product by 12. Ans. 32. 4. Divide 48 by 16, and multiply the quotient by 8. Ans. 24. 5. Divide the product of 7, 10, 12, and 5, by the product of 14, 18, and 6. 6. If 15 be multiplied by 7, 27, and 40, and the product divided by 54 multiplied by 14, 10, and 2, what will be the result ? Ans. 7£. 7. Divide the product of 13, 15, 20, and 5, by the product of 26, 10, 2, and 3. Ans. 121. 8. Divide the product of 28, 27, 21, 15, and 18, by the pro- duct of 7, 54, 7, 3, and 9. 9. How many pounds of butter at 28 cents a pound will be required to pay for 56 pounds of sugar at 11 cents a pound ? Ans. 22 pounds. 56 CONTRACTIONS IN MULTIPLICATION. 10. A. Holmes sold 14 boxes of soap, each containing 24 pounds, at 9 cents a pound, and received for pay 63 barrels of ashes, each containing 3 bushels. What was allowed a bushel for the ashes ? Ans. 1 6 cents. 11. M. Gardner sold 5 piles of brick, each containing 12 thousand, at 7 dollars a thousand, and was paid in wood, 3 ranges, at 4 dollars a cord. How many cords in each range ? Ans. 35 cords. 12. A merchant exchanged 8 cases of shoes, each containing 60 pairs, at 75 cents a pair, for a certain number of casks of molasses, each containing 90 gallons, at 40 cents a gallon. How many casks did he get ? Ans. 10 casks. CONTRACTIONS IN MULTIPLICATION. 86 • A contraction is the process of shortening any oper- ation. 87. When the multiplier is 13, 14, etc., or 1 with a sig- nificant figure annexed. Ex. 1. Multiply 3126 by 14. Ans. 43764. first operation. second operation. I n the first operation, 3126X14 3126 we multiply the multi- 12 5 4 14 plicand by the 4 units . _ „ . . , _ •, of the multiplier, and 4 3 7 6 4 Ans. 4 3 7 6 4 Ans. write tlie product un _ der the multiplicand one place to the right. To this partial product we add the multiplicand, since, as it stands, it represents the product of the multiplicand by the 1 ten of the multiplier ; and obtain 43764, the answer required. In the second operation, we add in the multiplicand taken as the product by the 1 ten of the multiplier, as we multiply by the 4 units ; thus, 6 X 4 = 24, of which we write down the 4 and carry the 2. 2X4 = 8,4-2 (carried) + 6 (from the multiplicand) = 16, of which we write down the 6 and carry the 1. lX4 = 4,-j-l-|- 2=7, which we write down. 3X4 = 1 2, -J-l =13, of which we write down the 3 and carry the 1. 3 — [— 1 = 4, which we write down ; and have as the entire result 43764 as before. Rule. — Write the product by the units' figure of the mvltiplier under the midtijrficand, one place to the right, and add them together. Or, Multiply each figure of the multiplicand "by the units 1 figure of the multiplier, and, after the units' place, add in the preceding figure of the midtiplicand. CONTRACTIONS IN MULTIPLICATION. 57 Examples. 2. Multiply 63013 by 17. Ans. 1071221. 3. Multiply 79245 by 19. 4. Multiply 32067812 by 16. Ans. 513084992. 88. When the multiplier is 101, 102, etc., or 1 with one or more ciphers and a significant figure annexed. Ex. 1. Multiply 8107 by 103. Ans. 835021. operation. We multiply by 3, the units' figure of the 8107X103 multiplicand, and write the product under 2 4 3 2 1 the multiplicand, two places to the right, o o g n 9 i a so *^ afc tne multiplicand, as it stands over o o Z 1 Ans. thig p ar tial product, will represent the pro- duct of the multiplicand by the 1 hundred of the multiplier ; and, adding these, we obtain 835021, the result re- quired. For the reason given, if the multiplier had been such as to have contained one more intervening cipher, we should have written the product by the units' figure three places to the right, and so on, one place farther to the right for every additional intervening cipher. Rule. — Write the product by the units 7 figure of the multiplier under the multiplicand, as many places to the right as there are in the multiplier intervening ciphers plus 1 ; and add them together. Examples. 2. Multiply 6651 by 108. Ans. 718308. 3. Multiply 111223 by 104. Ans. 11567192. 4. Multiply 2042 by 1009. 89 # When the multiplier is 21, 31, etc., or 1 with a signifi- cant figure prefixed. Ex. 1. Multiply 3113 by 41. Ans. 127633. operation. \y e multiply by 4, the tens' figure of the 3 113X^1 multiplier, and write the product under the 12 4 5 2 multiplicand, one place to the left, so that 9 _ the multiplicand, as it stands over this par- 1 L 4 b o o Ans. tial product, will represent the product of the multiplicand by the 1 unit of the multi- plier ; and, adding these, obtain the result required. Rule. — Write the product by the tens 9 figure of the multiplier under the multiplicand, one place to the left, and add them together. 58 CONTRACTIONS IN MULTIPLICATION. Examples. 2. Multiply 13317 by 51. Ans. 679167. 3. Multiply 71389 by 21. 4. Multiply 12062 by 91. Ans. 1097642. 90. When the multiplier is 201, 301, etc., or 1 with one or more ciphers and a significant figure prefixed. Ex. 1. Multiply 14118 by 601. Ans. 8484918. operation. We multiply by 6, the hundreds' 14118 X 601. figure of the multiplier, and write the 8 4 7 8 product under the multiplicand, two places to the left, so that the multipli- ed 4 o 4 J 1 o Ans. cand, as it stands over this partial pro- duct, will represent the product of the multiplicand by the 1 unit of the multiplier ; and adding these we have the answer required. Rule. — Write the product by the hundreds' figure of the multiplier under the multiplicand, as many places to the left as there are in the multiplier intervening ciphers plus 1 ; and add them together. Examples. 2. Multiply 8360 by 7001. Ans. 58528360. 3. Multiply 10613 by 801. Ans. 8501013. 4. Multiply 91603 by 2001. 91. When the multiplier or multiplicand has a fraction annexed. Ex. 1. Multiply 426 by 7£. By 8§. Ans. 3124 ; 3692. OPERATION. OPERATION. 426 426 7* 8f 2 9 8 2 = Product by 7. 3408 = Product by 8. 14 2= Product by £. 284 = Product by §. 3124 = Product by 7 J. 3 6 9 2 = Product by 8f. In multiplying 426 by 7^, we first obtain the product of 426 by 7, and then the product of 426 by |, and, adding these two partial pro- ducts together, have 3124, the product by 7£. In multiplying by £, we take one third of the multiplicand, by dividing it by 3 ; thus, 426 x ^ = 426 -7- 3 = 142. In multiplying 426 by 8|, we proceed as in the other case, except in obtaining the product by the fraction ; CONTRACTIONS IN MULTIPLICATION. 59 we take two thirds of the multiplicand, by taking one third of it 2 times ; thus, 426 X t = (426 -f- 3) x 2 = 284. If the fraction had been annexed to the multiplicand instead of the multiplier, we then would have found the product of the frac- tion and multiplier, for a partial product, in like manner as above. Rule. — Multiply the fractional part and the whole number sepa- rately, and add the products. Examples. 2. Multiply 915 by 22§. Aiis. 20496. 3. Multiply 1224| by 18. Ans. 22034f 4. If 69 £ miles make 1 degree, how many miles are 180 degrees ? Ans. 12450 miles. 92. When the multiplier is a convenient part of a number of tens, hundreds, or thousands. Note. — The following are some of the convenient parts often occurring as multipliers; 2£ =% of 10; 3£ = £ of 10; 12£ = £ of 100; 16§ = i of 100; 25 = £ of 100; 33£ = £ of 100; 125 =£ of 1000; 166| = £of 1000; 250 = £ of 1000; 333^ = J of 1000. Ex. 1. Multiply 785643 by 25. Ans. 19641075. operation. W e multiply by 1 00, by annex- 4)78564300 ing two ciphers to the multipli- i n n a i r\ n z -n i cand (Art. 65), and obtain a pro- 19 6 410 7 5 Product. duct | times £ Urge a? it s J uld be, since 25, the multiplier, is only one fourth part of 100; we therefore take one fourth part of that product, by dividing by 4, for the true product. Upon the same principle, if the multiplier had been 3£, we should have annexed one cipher and divided by 3, or, if the multiplier had been 125, we should have annexed three ciphers and divided by 8. Hence the following Rule. — Multiply by the number of tens, hundreds, or thousands, of which the multiplier is a part, and, of the product thus found, take the same part. Examples. 2. Multiply 68056 by 12£. Ans. 850700. 3. Multiply 17924 by 2£. 4. Multiply 192378 by 16§. Ans. 3206300. 5. Multiply 12345678 by 125. Ans. 1543209750. 6. How much can be earned in one year of 313 working days, at 3£ dollars a day ? 60 CONTRACTIONS IN MULTIPLICATION. 7. What will a farm containing 534 acres cost, at 33£ dollars an acre ? 8. In a certain large field there are 771 rows of corn, each containing 250 hills ; how many hills in all ? Ans. 192750 hills. 9. From a port in Louisiana there were exported, in a given time, 9168 boxes of sugar, averaging 166§ pounds to a box; required the whole number of pounds. Ans. 1528000 pounds. 10. An agent has bought for the army, at different times, in the aggregate, 1993 horses, at an average price of 125 dollars each. How much did he pay for them in all ? Ans. 249125 dollars. 11. How many are 333£ times 28044 ? Ans. 9348000. 93. When a part of the multiplier is a factor of another part. Ex. 1. Multiply 3263 by 568. Ans. 1853384. operation. ^ "We regard the multiplier 3 2 6 3 Multiplicand. as separated into two parts, 5 6 8 Multiplier. 56 tens and 8 units, or 560 _. _ . _ . .~ , . B . -(- 8 ; of which the smaller 2 6 10 4= Product by 8 units. part is evidently a factor of 18 2 7 2 8 = Product by 56 tens, the larger, since the 56 tens, l85lT^l = Product by 568. ^^^ft 8 units, obtaining the product for that part of the multiplier. Now, as this product is the same as that by the factor 8 of the other part of the multiplier, we multiply it by 7 tens, obtaining the product of the multiplicand by 8 X 7 tens or 56 tens. These products of the parts, 560 and 8, added together, give the true product by 568. Rule. — Multiply first by the smaller part of the multiplier; and then that partial product by a factor, or factors, of a larger part ; and soon with all the parts. The sum of the several partial products ivill be the product required. Note. — Care must be taken in writing down the partial products to have the units of the different orders stand in their proper places for adding. Examples. 2. Multiply 112345678 by 288144486. Ans. 32371787641631508. CONTRACTIONS IN MULTIPLICATION. 61 OPERATION. 112345 6 78 Multiplicand. 288144486 Multiplier. 674074068 = Product by 6 units. 5392592544 = 1st product X 8 tens for product by 48 tens. 1 (\ 1 77777fi^2 2( * P roduct x 3 thousands for product by 144 32355555264 = 3d *fiS2£* millions for product by m 32~3 71787641631508 = Product by 288144486. 3. Multiply 61370913 by 96488. Ans. 5921556653544. 4. Multiply 8649347864 by 1325769612. Ans. 11467042561708308768. 94. When the multiplier is any number of nines. Ex. 1. Multiply 87654 by 999. Ans. 87566346. t- operation. By annexing three ci- 87654000 = 87654X1000 phers to the multiplicand 87654 = 87654X 1 we . take & 1000 times, or — - „ ~ ~ ^ . n r, - n k . ^ ~ ~ 1 time more than is re- 87566346 = 87654X 999 quired by the given mul- tiplier. We therefore from this result subtract the multiplicand taken once, and thus obtain the product by 999. In like manner we may multiply by any number of nines. Hence the Rule. — Annex as many ciphers to the multiplicand as there are nines in the multiplier, and from the number thus produced subtract the given multiplicand. Note. — To multiply by any number of threes, find the product for the same number of nines, by the rule, and take one third of it by dividing by 3 ; and to multiply by any number of sixes, take twice the product of the same number of threes, by multiplying it by 2. 2. Multiply 7777777 by 9999. Ans. 77769992223. 3. Multiply 416231 by 99999. 4. Multiply 987654 by 333333. Ans. 329217670782. 5. Multiply 876543 by mm. Ans. 58435615638. 6. Multiply 999999 by 9999. Ans. 9998990001. 7. Multiply 32567895 by 3333. 8. Multiply 66666 by 66666. Ans. 4444355556. 9. Multiply 912345678 by 99. Ans. 90322222122. 10. Multiply 1234567 by 9999. Ans. 12344435433. 11. Multiply 98123452 by 999999. Ans. 98123353876548. N 6 62 CONTRACTIONS IN DIVISION. CONTRACTIONS IN DIVISION. 95. When the divisor is a convenient part of a number of tens, hundreds, or thousands. Ex. 1. Divide 19641075 by 25. Ans. 785643. operation. By multiplying both divisor and 1964107 5 dividend by 4, which does not change 4 the relation of the one to the other WfttAiftl v A - . (Art. 83), the divisor becomes 100, 785643100 Quotient. which ena bles us to perform the di- vision by simply cutting off two figures at the right of the dividend (Art. 79). In like manner, if the divisor had been 3£, by taking both it and the dividend 3 times as large, we could have performed the division by simply cutting off one figure at the right of the dividend ; or, if the divisor had been 125, by taking both it and the dividend 8 times as large, we could have performed the division by simply cutting off three figures at the right of the dividend. Hence the Rule. — Multiply both divisor and dividend by that number which ?/•/// change the divisor to a number of tens, hundreds, or thousands, and then divide. Examples. 2. Divide 89630 by 31. Ans. 26889. 3. Divide 123450 by 16§. 4. Divide 18621 by 12£. Ans. 1489-^. 5. Divide 317121 by 2J. Ans. 126848 T \. 6. Divide 876735 by 33£. Ans. 26302 T ^. 7. Divide 123456 by 125. 8. Divide 61678500 by 250. Ans. 246714. 9. J. Cushing bought a number of horses, at 166f dollars each, for $ 9500 ; required the number bought. Ans. 57 horses. 10. A company has received 12000 dollars from the sale of piano-fortes, at an average price of 333^ dollars each ; required the number sold. 11. How many shares of the Illinois Central Railroad, at 125 dollars each, can be bought for 150000 dollars ? Ans. 1200 shares. 12. How many cows, at 33£ dollars each, may be purchased for 333£ dollars. CONTRACTIONS IN DIVISION. 63 13. How many books, at 2£ dollars each, may be purchased for 120 dollars ? Ans. 48 books. 14. A certain magazine contains 616350 pounds of powder, in kegs of 25 pounds each ; required the number of kegs. Ans. 24654 kegs. 96. When the divisor is any number of nines. Ex. 1. Divide 316234 by 99. Ans. 3l94f|. operation. vy e fi rs t divide by 100, or 99 -f- 1, by cut- 3 16 2 3 4 ting off two figures at the right of the dividend, 3 19 6 obtaining for the first partial quotient 3162 12 7 an d a remainder 34. Since the divisor used 2 g was 1 larger than the given divisor, 99, the quotient obtained denotes that an excess of 3 1 9 4 f §- Ans. 3162, or a number equal to the quotient itself, must be added to the 34 for the true remainder, 34 _|_ 3162 = 3196, which exceeding the divisor 99, we write it for a second dividend; and dividing by 100, or 99 -f- 1, as before, we ob- tain for the second partial quotient 31, and a remainder 96. To the remainder 96 we add 31, the excess denoted by the last quotient, and obtain 1 2 7 for a third dividend, which being divided by 1 00, or 99 -j- 1 gives the third partial quotient 1, and a remainder 27. To the remainder 27 adding 1, the excess denoted by the last quotient, we have for the true remainder 28, which is ||. The sum of the partial quotients with the final remainder annexed gives 3194||, the quotient required. The above process is based upon the principle that 10 = 9 -f- 1 ; 100 = 99 -f- 1 ; 1000 = 999 + 1, etc. ; consequently 20 = (2 X 9) + 2, and 37 = (3 X 9) -f- 3 -f 7 ; 200 == (2 X 99) -f 2 ; 3859 = (38 X 99) + 38 + 59 ; 15987 = (15 X 999) -f 15 -f 987, etc. Hence* 316234 = 316200 -j- 34 = (3162 X 99) -f- 3162 -f 34; 3162 -f 34 == 3196 == 3100 -f- 96 = (31 X 99) + 31 -f- 96 ; 31 -f- 96 = 127 = 100 + 27= (1 X 99) +1 + 27; 1 -f 27 = 28 = jj; and 3162 — [— 31 — |— 1 — }— 11 == 3194||, the answer, as before obtained. By like process, and upon the same principle, may the quotient be found for any number of nines. Rule. — Add 1 to the given divisor, and, by it thus increased, di- vide by Gutting off figures at the right of the dividend. To the figures cut off on the right add those on the left for a true remainder, of which, if it equal or exceed the given divisor, make a second dividend, and divide as before. Proceed thus till there shall be no remainder as large as the given divisor ; and the sum of the several quotients, with the last remain- der, if any, will be the answer required. Note. — When the last remainder is the same as the given divisor, it must be cancelled, and 1 written as a partial quotient. 64 CONTRACTIONS IN DIVISION. Examples. 2. Divide 341 by 9. 3. Divide 123332544 by 999. OPERATION. OPERATION. 544 34 3 1 123332 5 123 8 1 876 000 3 7 f Ans. 12 3 4 5 6 Ans. 4. Divide 12332655 by 999. Ans. 12345. 5. Divide 987551235 by 9999. 6. Divide 9123456779876543211 by 999999999. Ans. 9123456789, 97. Abbreviated method of long division. Ex. 1. Divide 34634 by 134. Ans. 258 T %. operation. The operation we perform 134)34634(258 ^& Ans. thus : finding the first quotient „ Q cy figure to be 2, we say 4 X 2 = ' ° 6 8; 16 — 8 = 8; 3X2 = 6, 113 4 and 1 carried = 7 ; 14 — 7 = 7 ; 1X2 = 2, and 1 car- 6 2 ried = 3 ; 3 — 3 = 0. We now bring down 3, and find the next quotient figure to be 5, then, 4X5 = 20; 3 — = 3; 3 X 5 = 15, and 2 carried = 17; 8 — 7=1; 1x5 = 5, and 1 carried = 6 ; 7 — 6=1. We next bring down 4, and find the next quotient figure to be 8 ; then, 4X8 = 32 ;4 — 2=2; 3X8 = 2 I, and 3 carried = 27 ; 13 — 7 = 6 ; 1 X 8 = 8, and 2 -f 1 car- ried = 11; 1 — 1 = 0; 1 — 1 = 0. Hence, this method is that of ordinary long division (Art. 73), abridged by subtracting each figure of the product of the divisor and a quotient figure, as it is obtained, and writing down onbj the remainders. Examples. 2. Divide 39006 by 44. Ans. 886f |. 3. A certain orchard contains 1088 trees in 34 rows ; how many trees in each row ? Ans. 32 trees. 4. A speculator sold 191 mules at a gain of 5157 dollars; what was the gain on each ? 5. The population of Massachusetts, in 1855, was 1,133,123 ; how many would that be to each of its 7750 square miles of surface ? Ans. 146|f § g. 8# CONTRACTIONS IN DIVISION. 65 98. When the divisor or dividend has a fraction annexed. Ex. 1. Divide 3692 by 8|. Ans. 426. OPERATION. 3 6 9 2 We bring the divisor and divi- 3 dend to the same fractional parts — — as those denoted by the given 26)11076(426 Ans. fraction, which are thirds, by mul- 10 4 tiplying them both by 3, the num- 7TZ ber of thirds in 1. We thus change r 9 the divisor to 26 thirds, and the ^ dividend to 3692 thirds, without 15 6 changing the quotient (Art. 83) ; i k a and dividing obtain 426, the an- swer required. Rule. — Reduce the divisor and dividend to the same fractional parts as are denoted by the given fraction, and then divide as in whole numbers. Note. — In case there should be a remainder after the division, its true value may be found as in Art. 76. Examples. 2. Divide 13120 by 9f Ans. 1418£f 3. Divide 76672f by 36. Ans. 2129f ||. 4. Divide 2090£ by 205. Ans. 10£f£. 5. If 16J feet make a rod, how many rods are there in 10626 feet ? 6. If 69£ degrees make a mile, how many degrees in 12450 miles ? Ans. 180 degrees. 7. How many barrels of pork, at 17f dollars a barrel, may be bought for 559 1|- dollars ? 8. A merchant has sold 2667 yards of silk in dress patterns of 10^- yards each ; required the number of patterns. Ans. 254 patterns. 9. How many plats, each of 272J square feet, are equal in extent to 136125 square feet ? Ans. 500 plats. 10. How many days will 119 pounds of bread last a man, if he consume 2^ pounds per day ? Ans. 49 days. 11. How many casks of 31£ gallons each will be required to hold 12968J gallons of cider? Ans. 415 casks. G* 66 PROBLEMS. PKOBLEMS, FOUNDED UPON THE FUNDAMENTAL RULES. 99. The following problems are founded upon the general principles of addition, subtraction, multiplication, and division, the fundamental operations of arithmetic, which have already been explained. 1. The parts of a number being given, to find the number. — Add the parts together (Art. 47). 2. The sum of two numbers and one of the numbers beinjr given, to find the other number. — From the sum subtract the given number (Art. 50). 3. The difference between two numbers and the larger num- ber being given, to find the smaller. — From the larger number subtract the difference (Art. 52). 4. The difference between two numbers and the smaller number being given, to find the larger. — Add the smaller number and the difference together (Art. 51). 5. The sum and the difference of two numbers being given, to find the numbers. — From the sum subtract the difference and divide the remainder by 2, for the smaller number ; add the difference to the smaller number, for the larger (Art. 52). Note. — In like manner, when the sum and differences are given, may be found any number of required numbers. After subtracting the differences from the sum, if there are 3 required numbers, divide by 3 for the smaller number; if 4, divide by 4, and so on. 6. The product of two numbers, and one of the numbers being given, to find the other numbers. — Divide the product by the given number (Art. 62). 7. The product of three numbers, and two of the numbers being given, to find the other number. — Divide the given pro- duct by the product of the two given numbers (Art. 72). 8. The dividend and quotient being given, to find the di- visor. — Divide the dividend by the quotient (Art. 77). 9. The divisor and quotient being given, to find the divi- dend. — Multiply the divisor and quotient together (Art. 74). PROBLEMS. 67 Examples. 1. A carpenter has contracted to build one house for 2763 dollars, another for 4650 dollars, and a third for 8950 dollars. How much is he to receive for them all ? Ans. 1 6363 dollars. 2. N. Chandler has invested in railroad stock and a small farm 929 dollars. If the amount invested in the stock was 279 dollars, how much did the farm cost him ? Ans. 650 dollars. 3. Mount Black, the highest peak of the Blue Ridge, is 6476 feet high, which is 242 feet higher than Mount Washington, the highest peak of the White Mountains. What is the height of Mount Washington ? 4. The city of Mexico, in 1519, was taken by Cortes, and, 328 years after, by General Scott. In what year did it yield to Scott ? Ans. 1847. 5. Two travellers, A and B, meeting on a journey, found they had both travelled 1963 miles, and that A had travelled 199 miles more than B. What distance had each travelled ? Ans. A, 1081 miles ; B, 882 miles. 6. A father gave his three sons 4698 dollars, of which James received 250 dollars more than George, and Edwin 410 dollars more than George. What sum did each receive ? Ans. George $1346; James $1596; Edwin $1756. 7. There was paid for 217 chests of tea 8463 dollars. How much was that a chest ? 8. How many weeks will 684 bushels of oats last 19 horses, each horse consuming 3 bushels a week ? 9. On dividing 3808 dollars among a certain number of men, it was found that the share of each was 224 dollars. Eequired the number of men. Ans. 17 men. 10. A certain missionary society divides its income among 99 missions, giving to each an average of 575 dollars. What is its income ? Ans. 56925 dollars. 11. The product of 96, 22, and one other number is 63360. What is the other number ? Ans. 30. 12. The divisor being 13 and the quotient 1101, what is the dividend? Ans. 14313. 68 MISCELLANEOUS EXAMPLES. MISCELLANEOUS EXAMPLES. 1. What is the distance by railroad from Boston to Galena, it being from Boston to Albany 200 miles, from Albany to Niagara Falls 305, from Niagara Falls to Detroit 230, from Detroit to Chicago 282, from Chicago to Galena 171 ? 2. Sold J. Weimer my best horse for 175 dollars, my second- best chaise for 87 dollars, and a good harness for 31 dollars. He has paid me in cash 38 dollars, and has given me an order on S. Lantz for 12 dollars. How many dollars remain due? 3. Bought 97 barrels of molasses at $ 5 a barrel. Gave 17 barrels to support the poor, and the remainder was sold at $ 8 a barrel. Did I gain or lose, and how much ? Ans. $ 155 gain. 4. It requires 1728 cubic inches to make one cubic foot; required the number of cubic inches in 3787 cubic feet. 5. If a garrison of 987 men are supplied with 175686 pounds of beef, how much will there be for each man ? Ans. 178 lbs. 6. Albert Peyton sold off from his farm 120 acres, gave his son 80 acres, and had remaining 1 60 acres ; what number of acres did his farm contain before he disposed of any portion of it? Ans. 360 acres. 7. The annual revenue of a gentleman being $ 8395, how much per day is that equivalent to, there being 365 days in a year? Ans. $23. 8. What is the difference between half a dozen dozen, and six dozen dozen ? Ans. 792. 9. Bought of F. Johnson 8 barrels of flour at $ 7 per barrel, and 3 hundred-weight of sugar at $ 8 per hundred. What was the amount of his bill ? 10. George Adams bought an equal number of cows and oxen for 3952 dollars. For the cows he paid 31 dollars each, and for the oxen 45 dollars each. How many of each kind did he buy ? Ans. 52. 11. If a certain quantity of provisions will sustain 13 men 4 days, how long would it sustain 1 man ? Ans. 52 days. 12. The Globe Manufacturing Company has a capital of 250,000 dollars, divided into 500 shares. How much is each share ? Ans. 500 dollars. MISCELLANEOUS EXAMPLES. 69 13. Purchased a farm of 500 acres for $17,876. I sold 127 acres of it at $ 47 an acre, 212 acres at $ 96 an acre, and the remainder at $ 37 an acre. What did I gain by my bar- gain ? Ans. $ 14,402. 14. There is a certain island 18 miles in circuit, which A and B undertake to travel round, both starting from the same point and going round in the same direction. When A has travelled 17 miles and B 7 miles, how far apart are they ? Ans. 8 miles. 15. If 15 men can reap a certain field in 5 days, how long will it take 1 man to reap the same ? Ans. 75 days. 16. What number is that, which being multiplied by 24, the product divided by 10, the quotient multiplied by 2, 32 subtracted from the product, the remainder divided by 4, and 8 subtracted from the quotient, the remainder shall be 2 ? Ans. 15. 17. From 126 + (16 + 4) X 2 take (48 -4- 2) + (34 X 6) -r- (17 — 5). Ans. 125. 18. There are in the library of a certain school 683 books, which number will give 23 books to each pupil, and 16 books over ; what is the number of pupils ? 19. R. Howland in making a journey, after walking 12 miles and travelling 40 miles by stage, went by steamboat 5 times the distance he had travelled by stage, and by cars 6 times as far as he had walked, and 7 miles besides. What was the length of his journey? Ans. 331 miles. 20. At an election A and B were candidates for the same office, and the whole number of votes cast for them was 9891, of which B received 1211 majority. What number of votes did each receive? Ans. A, 4340 ; B, 5551. 21. The product of 3 numbers is 4080 ; one of the numbers is 15, and another 16. What is the third number ? 22. How many tons of coal at 6 dollars a ton will be re- quired to pay for 17 yards of cloth at 4 dollars a yard and 32 bushels of wheat at 2 dollars a bushel ? Ans. 22 tons. 23. James Cooper has manufactured in 4 years 5608 pairs of shoes, making each successive year 100 pairs more than the year before ; how many pairs did he manufacture each year? Ans. 1252; 1352; 1452; 1552. 70 UNITED STATES MONEY. 24. How many years will it take a young man earning 45 dollars, and spending 85 dollars of it, every month, if he have already 620 dollars, to lay up enough more to pay for a house costing 1100 dollars? Ans. 4 years. 25. If the product of the divisor by the quotient be 19782, and the remainder 31, what is the dividend? Ans. 19813. 26. John Franklin purchased railroad stock to the amount of 4473 dollars, and sold a part of it for 1885 dollars, obtain- ing 65 dollars a share, which was at a loss of 6 dollars on every share sold ; but the stock advancing much in value, he was able to dispose of the balance so as to gain by the whole operation 812 dollars. What did he get a share for the bal- ance of the stock ? Ans. 100 dollars. UNITED STATES MONEY. Art. 100t United States Money is the legal currency of the United States. It was established by Congress in 1786. Its denominations and their relative values are shown in the following TABLE. 10 Mills make 1 Cent, marked ct. 10 Cents u 1 Dime, u d. 10 Dimes u 1 Dollar, a S- 10 Dollars u 1 Eagle, U E. Mills. Cents. 10 = 1 Dimes. 100 = 10 = 1 Dollars. 1000 = 100 = 10 1 Eagle. 10000 = 1000 = 100 10 = 1 In accounts and ordinary business transactions, the only denominations mentioned are dollars and cents, eagles being expressed as dollars, dimes as cents, and mills as a fraction of a cent. The dollar is the integer, or unit of measure of United States money ; and therefore dimes are tenths of a dollar ; cents, hundredths ; and mills, thousandths. UNITED STATES MONEY. Yl 101 • Parts of a dollar, or any quantity thus expressed, in tenths, hundredths, etc., are termed decimals, or fractions, whose denominator (Art. 68) is 1, with one or more ciphers annexed. They are usually expressed by simply writing the numerator (Art. 68) with a point ( . ) before it, called the deci- mal point, or separatrix ; the first place at the right of the point being tenths ; the second place, hundredths ; the third place, thousandths ; the fourth place, ten-thousandths ; and so on. Thus, T V is written .1 ; TT ^ is written .01 ; To -W * s writ- ten .001 ; ijjfau is written .0001, etc. 102t In writing dollars and cents together, the decimal point or separatrix is placed between the dollars and the cents or decimal part ; and since cents occupy two places, the place of dimes and of cents, when the number of cents is less than 10, a cipher must be written before them, in the place of dimes. Thus, $30,375 is read, thirty dollars thirty-seven cents five mills, or thirty dollars and three hundred seventy-five thou- sandths of a dollar ; $ 12.05 is read, twelve dollars five cents, or twelve dollars and five hundredths of a dollar, etc. 103 1 The denominations of United States Money increasing from right to left, and decreasing from left to right, in the same manner as do the units of the several orders, as simple whole numbers, they may therefore be added, subtracted, multiplied, and divided according to the same rules. 104. The coins of the United States are of gold, silver, and copper. The gold coins are the double eagle, eagle, half-eagle, quar- ter-eagle, three-dollars, and dollar. The silver coins are the dollar, half-dollar, quarter-dollar, dime, half-dime, and three-cent-piece. The copper coins are the cent and half-cent. Note. — All the gold and silver coins of the United States are now made of one purity, 9 parts of pure metal and 1 part alloy. The alloy for the silver is copper, and that for the gold 1 part copper and 1 part silver. The cent and half-cent, old coinage, are of pure copper; the cent, new coinage, is made 88 parts copper and 12 parts nickel. The standard weight of the eagle, as fixed by present laws, is 258 grains Troy, and the gold coins in proportion according to their values. The weight of the silver dollar is 412 \ grains; half-dollar, 192 grains; quarter-dollar, 9G 72 UNITED STATES MONEY. grains ; dime 38| grains ; half-dime, 19| grains ; three-cent piece, 11^ grains ; the cent, old coinage, 168 grains; the cent, new coinage, 72 grains. The weight of the silver dollar, it will be seen, is greater in proportion to its value than the other silver coins. This is owing to their standard weight hav- ing been reduced, while that of the dollar remained unchanged ; but since this reduction of weight of the smaller silver coins, no more of the silver dollar ap- pear to have been coined. In circulation, the gold dollar, which of late years has been extensively coined, has almost entirely taken the place of the silver dollar. The symbol $, or dollar sign, represents, probably, the letter U written upon an S, denoting U. S. (United States). REDUCTION OF UNITED STATES MONEY. 105. Reduction of United States Money is changing the units of one of its denominations to the units of another, either of a higher or lower denomination, without altering their value. 106. To reduce units from a higher denomination to a lower. Ex. 1. Reduce 58 dollars to cents and mills. Ans. 5800 cents ; 58000 mills. OPERATION. 5 8 dollars. \ q Wc multiply the 58 dollars by 100 to reduce them to cents, be- 5 8 00 cents. cause 100 cents make 1 dollar; 1 and multiply the cents by 10 to reduce them to mills, because 10 5 8 mills make 1 cent. Hence, Or thus : 5 8 mills. To reduce dollars to cents, annex two ciphers ; to reduce dollars to mills, annex three ciphers; and to reduce cents to mills, annex one cipher. Note. — Dollars, cents, and mills, expressed by a single number, are re- duced to mills by merely removing the separating point; and dollars and cents, by annexing one cipher and removing the separatrix. 107. To reduce units from a lower denomination to a higher. Ex. 1. Reduce 58000 mills to cents and to dollars. Ans. 5800 cents; 58 dollars. operation. "We divide the mills by 10 to re- 10) 58000 mills. duce them to cents, because 10 mills i a a \ k a a a * make 1 cent; and the cents by 100 1 )5800 cents. to reduce th ' em to dollarg) ^ cause 5 8 dollars. 100 cents make 1 dollar. Hence, UNITED STATES MONEY. 73 To reduce mills to cents, point off one figure on the right ; to reduce cents to dollars, point off two figures ; and to reduce mills to dollars, point off three figures. Examples. 2. Reduce $765 to cents. 3. Change 726 mills to cents. Ans. 72^ cents. 4. How many dollars are 329 cents? Ans. $3.29. 5. Change 12345 mills to dollars. Ans. $ 12.345. 6. Reduce $ 123.56 to mills. Ans. 123560 mills. 7. Reduce 2 eagles, 2 dollars, and 2 dimes to cents. Ans. 2220 cts. ADDITION OF UNITED STATES MONEY. 108. Ex. 1. Add together 17 dollars 13 cents 5 mills ; 8 dollars 4 cents 6 mills ; 63 dollars 20 cents 3 mills ; and 29 dollars 87 cents 5 mills. Ans. $ 118.259. We write units of the same denomina- tion in the same column, and add as in addition of simple numbers (Art. 45), and separate the dollars from the cents in the answer by the decimal point. OPERATION. $ cts. m. 1 7.1 3 5 8.0 4 6 6 3.2 3 .29.875 Ans. 1 1 8.2 5 9 Rule. — Write dollars, cents, and mills, so that units of the same denomination shall stand in the same column. Add as in addition of simple numbers, and place the decimal point directly under that above. Proof. — The proof is the same as in addition of simple numbers. Examples. 2. 3. 4. 5. $ cts. m. $ cts. m. $ cts. m. $ cts. m. 3 7 5.875 78.193 171.013 861.073 6 7 1.1 2 7 1 8.0 1 4 3 8 2.0 9 4 5 1 6.7 1 6 3 8 7.143 91.038 999.900 344.673 1 8 4.1 8 9 1 6.8 1 7 1 5 5.0 6 8 6 1 7.8 1 4 14 7.758 81.476 48.153 169.973 6 3.0 7 2 4 3.18 4 4 9.6 19 8 10.426 18 29.16 4 N 7 74 UNITED STATES MONEY. 6. Add the following sums, $18,165, $701.63, $151,161, $ 375.089, and $ 471.017. Ans. $ 1717.062. 7. Bought a horse for eighty-seven dollars nine cents, a pair of oxen for sixty-five dollars twenty cents, and six gallons of molasses for two dollars six cents five mills ; what was the amount of my bill ? Ans. $ 154.355. 8. Sold a calf for three dollars eight cents, a bushel of corn for ninety-seven cents five mills, and three bushels of rye for three dollars five cents ; what was the amount received ? Ans. $ 7.105. SUBTRACTION OF UNITED STATES MONEY. 109. Ex. 1. From 106 dollars 7 cents 7 mills, take 92 dollars 83 cents 8 mills. Ans. $ 13.239. operation. "We write the less number under the $ cts. m. greater, place mills under mills, cents under 1 6.0 7 7 cents, and dollars under dollars, and subtract 9 2.8 3 8 as in subtraction of simple numbers (Art. a o qq 50), and separate the dollars from the cents Ans. 1 6,1 6 J j n ^ e answer by the decimal point. Rule. — Write the several denominations of the subtrahend under the corresponding ones of the minuend. Subtract as in subtraction of simple numbers, and place the decimal point directly under that above. Proof, — The proof is the same as in subtraction of simple numbers. Examples. 2. 3. 4. 5. | cts. m. 8 7 1.1 6 1 8 9.9 1 8 | cts. m. 4 7 8.4 7 7 1 9 9.9 9 1 $ cts. m. 1 6 7.1 6 3 9 8.0 9 7 $ cts. m. 1 6 3.1 6 7 9.0 9 8 7 8 1.2 4 3 6. Bought a farm for $1728.90, and sold it for $3786.98; what did I gain by my bargain ? 7. Gave $79.25 for a horse, and $106,875 for a chaise, and sold them both for $ 200 ; what did I gain ? 8. Bought a farm for $8967, and sold it for nine thousand eight hundred seventy-six dollars seventy-five cents ; what did I gain ? Ans. $ 909.75. UNITED STATES MONEY. 75 9. Bought a barrel of flour for $ 7.50, three bushels of rye for $ 2.75, and three cords of wood at $ 5.25 a cord ; I sold the flour for $ 6.18, the rye for $ 3.00, and the wood for $ 6.75 a cord ; what was gained by the bargain ? Ans. $ 3.43. 10. A young lady went a " shopping." Her father gave her a twenty-dollar bill. She purchased a dress for $ 8.16, a muff for $ 3.19, a pair of gloves for $ 1.12, a pair of shoes for $ 1.90, a fan for $ 0.19, and a bonnet for $ 3.08 ; how much money did she return to her father ? Ans. $ 2.36. MULTIPLICATION OF UNITED STATES MONEY. 110. Ex. 1. What will 365 barrels of flour cost at $5.75 a barrel ? Ans. $ 2098.75. OPERATION. $ 5.7 5 We multiply as in multiplication of sim- 3 6 5 pie numbers, and, obtaining the product in o o 7 k cents, the lowest denomination of the multi- o a n n plicand, we reduce them to dollars by point- ^ 4 „ v ing off two places at the right for cents 1725 (Art. 107). $2 9 8.7 5 Ans. Rule. — Multiply as in multiplication of simple numbers. The product will be in the lowest denomination in the multiplicand, which must be pointed off as in reduction of United States money. Proof. — The proof is the same as in multiplication of simple numbers. Examples. 2. What will 126 pounds of butter cost at 13 cents a pound? 3. What will 63 pounds of tea cost at 93 cents a pound ? 4. What will 43 tons of hay cost at 13 dollars 75 cents a ton? Ans. $591.25. 5. If 1 pound of pork is worth 7 cents 3 mills, what are 46 pounds worth? Ans. $3,358. 6. If 1 hundred of beef cost 3 dollars 28 cents, what are 76 hundred worth ? 7. What will 96 thousand feet of boards cost at 11 dollars 67 cents a thousand? Ans. $ 1120.32. 8. If a barrel of cider be sold for 2 dollars 12 cents, what will be the value of 169 barrels ? Ans. $ 358.28. 76 UNITED STATES MONEY. 9. What will be the value of a hogshead of wine, containing 63 gallons, at 1 dollar 63 cents a gallon? Ans. $ 102.69. 10. Sold a sack of hops, weighing 396 pounds, at 11 cents 3 mills a pound ; to what did it amount ? Ans. $ 44.748. 11. Sold 19 cords of wood, at $5.75 a cord; to what did it amount ? 12. Sold 169 tons of timber, at $4.68 a ton; what did I receive ? 13. Sold a hogshead of sugar, weighing 465 pounds ; to what did it amount, at 14 cents a pound? Ans. $ 65.10. 14. What will be the amount of 789 pounds of leather at 18 cents a pound? Ans. $ 142.02. 15. What will be the expense of 846 pounds of sheet-lead at 5 cents 7 mills a pound ? Ans. $ 48.222. 16. When potash is sold for $ 132.55 a ton, what will be the price of 369 tons ? Ans. $ 48910.95. 17. What will 365 pounds of beeswax cost at 18 cents 4 mills a pound ? Ans. $ 67.16. 18. If 1 pound of tallow cost 7 cents 3 mills, what are 968 pounds worth ? Ans. $ 70.664. DIVISION OF UNITED STATES MONEY. 111. Ex. 1. If 78 barrels of fish cost $ 303.42, what will 1 barrel cost ? Ans. $ 3.89. OPERATION. 78) $3 3.4 2 ( $3.8 9 2 3 4 We divide as in division of sim- n Q * pl e numbers, and point off as many places for cents in the quotient as 6 2 4 there are for cents in the divi- 7 2 dend - 702 Rule. — Divide as in division of simple numbers. The quotient will be in the lowest denomination of the dividend, which must be pointed off as in reduction of United States money. Note. — When the dividend consists of dollars only, and is either smaller than the divisor or not divisible by it without a remainder, reduce it to a lower denomination by annexing two or three ciphers, as the case may require, and the quotient will be cents or mills accordingly. When it is required to find the number of times one sum of money is con- tained in another, both sums, if of different denominations, must be reduced to the same, before dividing. 3. How many yards of cloth at 9 cents a yard can be bought for $ 28.89 ? Ans. 321 yards. OPERATION. 9)2889 Ans. 3 2 1 yards. UNITED STATES MONEY. 77 Proof. — The proof is the same as in division of simple numbers. Examples. 2. Bought 11 chests of tea for $234.30. What did I give a chest ? Ans. $ 21.30. OPERATION. 11) $ 2 3 4.3 Ans. $2 1.3 4. When 19 bushels of fine salt can be bought for $30.87£, what costs one bushel ? 5. At 12 cents a pound, how many pounds of sugar can be bought for $51? 6. If 78 barrels of fish cost $303.42, what will one barrel cost? 7. Bought 42 bushels of pears for $ 73.50 ; what cost one bushel? Ans. $1.75. 8. How many yards of broadcloth at $ 2.75 a yard can be bought for $ 904.75 ? Ans. 329 yards. 9. When rye is sold at the rate of 628 bushels for $ 471.00, what is that a bushel ? Ans. $ 0.75. 10. If it should cost $1460 to support a family 365 days, what would be the expense a day ? Ans. $ 4. 11. How many gallons of oil at $ 1.62 J- a gallon can be bought for $ 234 ? 12. If 1624 pounds of pork cost $97.44, what cost one pound? Ans. $0.06. 13. If 47 thousand shingles cost $ 176.25, what is the cost per thousand ? Ans. $ 3.75. 14. Bought 148 tons of plaster of Paris for $ 337.44; what was it per ton ? Ans. $ 2.28. 15. At $37.75 an acre, how many acres of land may be bought for $ 1774.25 ? Ans. 47 acres. 16. At 67 cents a pound, how many chests of tea, each weigh- ing 59 pounds, can be bought for $ 672.01 ? Ans. 17 chests. 17. How many bushels of wheat at $ 1.25 a bushel can be bought for $ 863.75 ? Ans. 691 bushels. 7* 78 UNITED STATES MONEY. 18. Sold 169 tons of timber for $790.92; what cost one ton? 19. "What cost one pound of leather, if 789 pounds cost $ 142.02 ? 20. If 369 tons of potash cost $48910.95, what will be the price of one ton ? Ans. $ 132.55. 21. Bought 47 hogsheads of salt, each hogshead containing 7 bushels, for $ 368.48 ; what cost one bushel ? Ans. $ 1.12. GENEKAL PRINCIPLES AND APPLICATIONS. 112. Analysis is an examination of a question by re- solving it into its parts, in order to consider them separately, and thus, by reasoning from the nature of the question, to render each step in the solution plain and intelligible. 113. An aliquot part of any number or quantity is such a part as will exactly divide that number or quantity. Thus, 3 is an aliquot part of 6 ; and 4, an aliquot part of 16. 114. The method of performing operations by means of aliquot parts is called Practice, or Analysis by Aliquot Parts. 115. In business calculations frequent use is made of the following Aliquot Parts of a Dollar. 50 cents = £ of $1. 33£ cents = £ of $ 1. 25 cents = £ of $1. 20 cents = 4 of $ I . 12i cents = £ of $ 1, 10 cents = T \j- of $ 1. 6£ cents = T \ of $ 1. 5 cents = J$ of $ 1. Practical Questions by Analysis. 116. The price of a unit of any quantity being given, to find the cost of that quantity. Ex. 1. If 1 barrel of flour cost $7, what will 125 barrels cost? Analysis. Since 1 barrel costs $ 7, 125 barrels will cost 125 times as much : $ 7 X 125 = $ 875. GENERAL PRINCIPLES AND APPLICATIONS. 79 2. If 1 pound of beef cost 13 cents, what will 914 pounds cost? Ans. $118.82. 3. If 1 yard of calico cost 23 cents, what will 31L yards cost ? 4. When 44 cents are paid for 1 pound of tea, how much must be paid for 15 chests of tea, each containing 47 pounds ? 5. At 39 J cents a pound, how much must be paid for 9 bales of wool, each containing 317 pounds ? Ans. $ 1126.93J. 6. If the cost of building 1 mile of railroad be $41315, what will be that of 113 miles ? Ans. $ 4668595. 117# The price of a unit of any quantity being a given aliquot part of a dollar, to find the cost of that quantity. 7. At 12£ cents a yard, what will 208 yards of calico cost? Ans. $26. Analysis. Since, at $ 1 a yard, the cost would be as many dollars as there are yards, the cost at 12 J cents, or J of $ 1, will be one eighth as many dollars as there are yards, or as many dollars as 8 is contained times in 208 ; 208 -7-8 = 26. Therefore 208 yards cost $26. That is, we take such a part of the number denoting the quantity as the price is of 1 dollar, 8. What will be the cost of 362 pounds of feathers at 33£ cents a pound ? Ans. $ 120.66§. 9. At 50 cents a bushel, what will 7 loads of potatoes cost, each containing 30 bushels ? Ans. $ 105. 10. At 6£ cents a pound, how much must be paid for 1163 pounds of shingle nails ? 11. What will be the cost of 96 yards of broadcloth at $4.33^ a yard? Ans. $416. Analysis. Since, at $ 1 a yard, the cost would be as many dollars as there are yards, the cost at $ 4.33^, or $ 4£, will be 4£ times as much : $ 4£ X 96 = $ 416. 12. If the cost of 1 pair of boots be $3.16|, what will be the cost of 20 cases, each case containing 60 pairs ? 13. At $ 2.25 each, what will 150 hats cost ? Ans. $ 337.50. 14. What will 98 dozen knives cost at $ 5.12^ a dozen ? Ans. $ 502.50. 80 UNITED STATES MONEY. 15. If James Spooner spends uselessly every clay for cigars G£ cents, how much does he thus spend in 360 days ? 16. At $ 2.20 a bushel, what will 15 car-loads of wheat, each containing 212 bushels, cost ? Ans. $ 6996. 118. The price of any article sold by the 100, or 1000, with the quantity, being given, to find the cost of that quantity. 17. What will 732 fishes cost at $ 5 a hundred? Ans. $36.60. Analysis. Since 100 fishes cost $ 5, 1 fish will cost -j-^ of $ 5, or $ T fo ; and if 1 fish cost $ T fo, 732 fishes will cost 732 times as much: $ 5 X 732 = $3660; $3660 ~- 100 = $36.60. That is, we multiply the quantity and price together ; and cut off two figures at the right (Art. 79). 18. How much must be paid for 950 apple-trees at 20 dol- lars a hundred ? 19. What will 7235 arithmetics cost at $ 45 a hundred ? Ans. $ 3255.75. 20. What will it cost to excavate 19875 solid feet of earth, at the rate of 52 cents per hundred solid feet? Ans. $ 103.35. 21. What is the value of 1765 feet of timber at $ 3 a hun- dred? Ans. $52.95. 22. What is the value of 2355 cedar rails at $5.50 per hundred ? 23. What will 3100 bricks cost at $ 6 a thousand? Ans. $ 18.60. Analysis. Since 1000 bricks cost $ 6, 1 brick will cost tttW of $ 6 > or $ tcjW 5 and > if * brick cost $ tthttd 3*00 w m cost 3100 times as much : $ 6 X 3100 = $ 18600 ; $ 18600 ~- 1000 = $ 18.60. That is, we multi-ply the quantity and price together, and cut off three figures at the right (Art. 79). 24. At $ 142.50 a thousand, what will 6150 chestnut posts cost? Ans. $876,375. 25. How much must be paid for 15750 feet of boards at $ 30 a thousand ? 26. At $ 2.25 per thousand, how much must be paid for 3550 laths ? Ans. $ 7.98*. 27. I have bought of L. T. Bobbins 363 feet of plank at GENERAL PRINCIPLES AND APPLICATIONS. 81 $ 2.50 per hundred ; 3150 pickets at $ 14 per thousand ; and 350 feet of boards at $ 20 per thousand. What is the amount of the whole ? Ans. $ 60.17£. 28. What must be paid for 92350 railroad ties, at $ 135 per thousand ? 119. The price of any article sold by the ton of 2000 pounds, with the quantity, being given, to find the cost of that quantity. 29. At $ 8 a ton, what will 2550 pounds of coal cost ? Ans. $ 10.20. Analysis. Since 2000 pounds cost $ 8, 1000 pounds will cost one half of $ 8, or $ 4 ; and, if 1000 pounds cost $ 4, 1 pound will cost T xjW °f $ 4, or $ yxny^, and 2550 pounds will cost 2550 times as much : $ 4 X 2550 = $ 10200 ; $ 10200 ~- 1000 == $ 10.20. That is, we multiply the number denoting the quantity and half the price together \ and cut off three figures at the right (Art. 79). 30. How much must be paid for 3760 pounds of hay, when the price is $ 15 a ton ? Ans. $ 28.20. 31. What is the value of 19 casks of plaster of Paris, each weighing 500 pounds, at $ 9 a ton ? 32. How much must be paid for transporting 5163 pounds of freight, at $ 3.50 per ton ? 33. At $ 46 a ton, what will 33550 pounds of railroad iron cost? ,% Ans. $771.65. 34. G. Reed has bought 30 bags of superphosphate of lime, each bag containing 125 pounds, at $45 per ton; and 7500 pounds of Peruvian guano, at $ 53 per ton. What is the cost of the whole ? Ans. $ 283.125. 35. I have bought 6350 pounds of anthracite coal at $7.12J per ton, and 3560 pounds of Pictou coal at $7.25 per ton. What is the cost of the whole ? Ans. $ 35.526J. 120* The price of any quantity, and the quantity, being given, to find the price of a unit of that quantity. 36. If 125 barrels of flour cost $875, what will 1 barrel cost ? Ans. $ 7. Analysis. Since 125 barrels cost $875, 1 barrel will cost 82 UNITED STATES MONEY. as many dollars as 125 is contained times in 875 ; $ 875 -~ 125 = $7. 37. If 914 pounds of beef cost $ 118.82, what will 1 pound cost? Ans. $0.13. 38. If 96 yards of broadcloth can be bought for $400, for what can 1 yard be bought ? Ans. $ 4.1 6f. 39. If $ 510 are paid for 120 pairs of boots, what is the cost of 1 pair ? What is the cost of 17 pairs ? 40. A merchant paid $ 270 for 300 casks of lime ; what did he pay a cask ? What did he pay for 33 casks ? Ans. $ 29.70. 41. When boards are $25 per thousand, what are they per foot ? What per hundred ? Ans. $ 2.50. 42. If 3 pieces of calico, each containing 30 yards, cost $ 19.80, what does 1 yard cost ? Ans. $ 0.22. 43. When 10 firkins of butter, each containing 5G pounds, can be bought for $ 112, what is the price per pound ? Ans. $0.20. 121. The price of any quantity, and the price of a unit of that quantity, being given, to find the quantity. 44. At $ 7 per barrel, how many barrels of flour can be purchased for $875 ? Ans. 125 barrels. Analysis. Since 1 barrel will cost $ 7, as many barrels can be purchased with $ 875 as 7 i3 contained times in 875 ; 875 -f- 7 = 125 ; therefore, there can be purchased 125 barrels. 45. At $ 3 a yard, how many yards of broadcloth can I purchase with $456 ? Ans. 152 yards. 46. How many cords of wood, at $ 4.25 each, can be bought for $ 357 ? 47. At 33^- cents a pound, how many pounds of feathers can be bought for $ 120.66§? Ans. 362 pounds. 48. At 50 cents a bushel, how many barrels of apples, each containing 3 bushels, can be purchased for $ 40.50 ? Ans. 27 barrels. 49. Expended for hemp $11600, at $232 per ton. How many tons did I buy ? Ans. 50 tons. 50. How many boxes of raisins, at $ 3.50 per box, can be bought for $ 756 ? Ans. 216 boxes. BILLS. 83 BILLS. 122. A Bill is a paper, given by merchants, containing a statement of goods sold, and their prices. An invoice is a bill of merchandise shipped or forwarded to a purchaser, or selling agent. The date of a bill is the time of the transaction. The bill is against the party owing, and in favor of the party who is to receive the amount due. A bill is receipted, when the receiving of the amount due is acknowledged by the party in whose favor it is, as in bill 1. A clerk, or any other authorized person, may, in his stead, receipt for him, as in bill 2. When the items of a bill have been rendered at different dates, the several times may be given at the left hand, as in bill 4. When the bill is in the form of an account, containing items of debt and credit in its settlement, it is required to find the difference due, or balance, as in bill 4. Note. — A due-bill is a ■written acknowledgment that a debt is due, and is often given in making settlements to remove cause of subsequent disputes as to a claim. A bill is sometimes settled or receipted by a due-bill, as in bill 4. What is the cost of each article in, and the amount due of, each of the following bills ? (1-) Be )ston, July 4, ! Mr* James Dow, Bought of Dennis Shai 17 yds. Flannel, at $0.45 19 " Shalloon, a .37 16 " Blue Camlet, a .46 13 " Silk Vesting, a .87 9 " Cambric Muslin, a .63 25 " Bombazine, a .56 17 " Ticking, ft .31 19 " Striped Jean, Received payment, u .16 ']? Dennis Sharp. 61.33 84 UNITED STATES MONEY. (2.) Chicago, Jan. 1, 1857. •. Samuel Smith, Bought of David Johnson, 13 lbs. Tea, at $0.98 16 " Coffee, .15 36 " Sugar, " .13 47 " Cheese, " .09 12 " Pepper, '.« .19 7 " Ginger, *• " .17 13 " Chocolate, " .61 $ Received payment, David Johnson, by J. Keen (3.) Philadelphia, Sept 5, 1856. Messrs. Wilson, Niles, & Co., Bought of Peck & Bliss, 2 doz. National Arithmetics, at % 6.00 3 " Coin. Sch. Arithmetics, " 5.00 5 " Primary Arithmetics, " 17 Parker's Exercises in Composition, " L80 .25 13 Maglathliris National Speaker, " 19 Lever etC s Ccesar, " .60 .50 3 Greenleafs Algebra, " 7 Sillimaris Chemistry, " 15 6V?^s £71 aSI History, " 5 imp. Quarto Bibles, " 3 Webster's Quarto Dictionaries, " 5 Felton's Greek Readers, " .60 1.12* .22 15.00 4.50 1.50 1 Kane's Arctic Explorations, " 4.50 <2 171 Received payment, Peck & Bliss. BILLS. 85 (4.) New Orleans, Jan. 1, 1856. Mr. Albert Crawford, To Baldwin, Sherwood, & Bice, Dr. 1855. Jan. 2. .For 17 yefo. , Broadcloth, at $5.25 " 15. " 29 " Gassimere, u 1.62 Feb. 3. " 60 " Bleached Shirting, a .17 "• 7. u 49 « Ticking, a .27 " 15. « 18 " Blue Cloth, a 3.19 June 17. « 27 " Habit do. a 2.75 Aug. 3. " 75 " Flannel, u .61 Sept. 19. " 36 " Plaid Prints, a .75 Dec. 2. u 49 « Brown Sheeting, a • .18 4b Q70 on Cr. Jan. 28. i?^ cash, $83.00 Feb. 19. " 3 if. ^oar^, at $ 30.00 May 9. " 7 M. > Shingles, a 4.00 June 17. " 4 efa^s' Zafor, u 2.00 Ji% 3. " 5 u a a 1.75 Oct. 7. " 7 C. Timber, a 2.25 Nov. 4. " Cash, 60.00 Dec. 29. " Draft, 45.00 $338.50 $ 34.40 Bal. due B., S., $ R., Jan. 5. Settled by due-bill, Baldwin, Sherwood, & Rice at (5.) Mr. Benjamin Treat, 37 chests Green Tea, 41 « Black do. 40 " Imperial Tea, 13 crates Liverpool Ware, Received payment, N 8 St. Louis, May 1, 1856. Bought of John True, $25.50 16.17 97.75 169.37 John True. 86 UNITED STATES MONEY. (6.) Portland, May 16, 1857. Mr. J. C. Porter, Bought of Willard & Hale, 17 bbl. Canada Flour, a* $ 8.25 50 lbs. Duponfs Eagle 140 " Sheet Zinc, \ Gunpowder, .50 .081 120 " Prussian Blue, ■ " .63 * Received payment, Willard & Hale. (7.) New York, July 11, 1856. Mr. John Cummings, Bought o/*Lord & Secomb, 97 bbl Genesee Flour, at $ 6.25 167 " Philadelphia 87 " Baltimore do. a a 5.95 6.07 196 " Richmond do. (f 5.75 275 " Howard St. do. a 7.25 69 bu. Rye, 136 " Virginia Corn, a a 1.16 .67 68 " North River do. a .76 169 " Wheat, a 1.37 76 tons Lehigh Coal, 89 " Iron, a a 9.67 69.70 49 Grindstones, u 3.47 39 Pitchforks, 197 ifa£es, a 1.61 .17 86 .floes, a .69 78 Shovels, r< 1.17 187 Spades, 91 Ploughs, 83 Harrows, a a .85 11.61 17.15 47 Handsaws, IK 3.16 35 Mill-saws, a 18.15 47 ctttf. A$fee£, a 9.47 57 " Zeatf, E. T. Lowe, 6.83 ^ 17 315 3° Received payment, far Lord & Secomb. LEDGER ACCOUNTS. 87 LEDGER ACCOUNTS. 123. The principal book of accounts among merchants is called a ledger. In it are brought together scattered items of accounts, often making long columns. As a rapid way of finding the amounts, accountants generally add more than one column at a single operation (Art. 48). The examples below may be added both by the usual method and by that which is more rapid. 1. 2. 3. 4. $ cts. $ cts. I cts. 8 cts. 3.3 3 9 1.16 1 3 4.6 2 1 2 0.0 3 9.10 4 0.4 4 5 1 1.4 2 6 3.17 16.3 7 6 0.6 6 2 7.3 4 1 0.0 41.3 9 1.3 1 3 3.3 5 7 8 1.1 4 3 1.7 4.0 6 0.4 1 5 1.6 1 5 7.71 6.3 4 1 4 9.3 1 3 1.2 1 3.6 7.19 9 2 1.3 1 1 5.5 2 5.0 2 8.3 7 3 9.16 3 6.0 6 .5 81.0 1 1 0.0 5 17.7 5 .13 6.0 1 4.0 3 8 5.2 5 1.5 5 15.0 3 6 3.4 13.0 7.9 1 4 3.0 5 9.0 10.0 1.19 10.0 3 11.12 1 3 5.7 5 9.0 9 5.0 5 0.10 1 1 1.1 5 13.4 61.18 17.9 1 1 6 3.0 5 3 7.6 5 1 1 2.1 5 11.13 7 1 2.1 5 4 5.7 5 1 0.6 1 4 5.9 2 3 1.15 5.2 5 1 6 0.0 17.7 5 9 5.2 5 7.7 5 3 9.4 1 13.8 5 2 5.17 5.3 1 2 2.3 9 1 1.0 1 7 7.2 5 8.13 9 2.7 7 6 2.19 1 2 3.1 4.2 7 3 3.0 1 9 5.17 1 1 6.5 1 2.6 4 6 4.3 2 1 9.1 2 2 3.0 6 11.12 5 9.11 1 1 1.0 9 6 6.7 5 41.2 1 17.8 9 1 9 5.2 5 1 2.1 5 3 3.7 9 7 6 1.1 5 6 1 3.2 4 1 6.3 1 4 7.3 1 8 2 3.5 4 1 8.0 9 6 1 2.1 9 10.0 2 3 8.4 7 3 3 4.0 5 7 1 1 9.8 1 88 KEDUCTION OF COMPOUND NUMBERS. COMPOUND NUMBERS. 124. A Compound Number is a collection of concrete units of different denominations ; as, 5 pounds and 6 ounces ; 4 feet and 5 inches. 125. A scale expresses the law of relation between the different units of a number. The different units of simple numbers have a uniform tenfold increase from lower to higher orders, and a like decrease from higher to lower orders. They, therefore, are said to have a uniform scale. In compound numbers, the names of different measuring units (Art. 9) are included in the expression of a single quantity, so that the relation of the units of one order to those of another is that of a varying scale ; as in the expression of pounds, shillings, and pence, it is 4, 12, and 20. REDUCTION OF COMPOUND NUMBERS. 126. Reduction is the process of changing numbers from one denomination to another, without altering their values. It is of two kinds, Reduction Descending and Reduction As- cending. Reduction Descending is changing numbers of a higher de- nomination to a lower denomination ; as pounds to shillings, &c. It is performed by multiplication. Reduction Ascending is changing numbers of a lower denom- ination to a higher denomination j as farthings to pence, &c. It is the reverse of Reduction Descending, and is performed by division. ENGLISH MONEY. 127. English or Sterling Money is the currency of Eng- land. Table. 4 Farthings (qr. or far.) make 1 Penny, d. 12 Pence " 1 Shilling, s. 20 Shillings " 1 Pound, £. REDUCTION OP COMPOUND NUMBERS. 89 far. d. 4 = 1 s. 48 = 12 = 1 £. 960 = 240 = 20 = 1 Note 1. — The symbol £. stands for the Latin word libra, signifying a pound; s. for solidus, a shilling; d. for denarius, a penny; qr. for quadrans, a quarter. Note 2. — Farthings are sometimes expressed in a fraction of a penny; thus, 1 far. = i d.; 2 far. = £ d.; 3 far. = | d. Note 3. — The term sterling is probably from Easterling, the popular name of certain early German traders in England, whose money was noted for the purity of its quality. Note 4. — The English coins consist of the five-sovereign piece, the double- sovereign, the sovereign, and the half-sovereign, made of gold; the crown, the half-crown, the shilling, the six-pence, the four-pence, the three-pence, the two- pence, the one-and-a-half-pence, and the penny, made of silver ; the penny, the half-penny, the farthing, and the half-farthing, made of copper. The sovereign represents the pound sterling, whose legal value in United States money is $4.84. The value of the English guinea is 21 shillings ster- ling. The guinea, the five-guinea, the half-guinea, the quarter-guinea, and the seven-shilling piece, are no longer coined. The guinea is so called because the gold of which the first guineas were made was brought from Guinea, in Africa. The English gold coins are now made of 11 parts of pure gold, and 1 part of copper, or some other alloy ; and the silver coin, of 37 parts of pure silver, and 3 parts of copper. The present standard weight of the sovereign is 223i|l grains Troy; the crown, 440^f grains ; the copper penny, 291| grains. 128. To change numbers expressed in one or more de- nominations to their equivalents in one or more other denom- inations. Ex. 1. In 48£. 12s. 7d. 2far. how many farthings? operation. "VVe multiply the 48 by 20, be- 4 8 £. 1 2 s. 7 d. 2 far. cause 20 shillings make 1 pound, and to this product we add the 12 shillings in the question, and obtain shillings. 972 shillings. We then multiply by 12, because 12 pence make 1 shilling, and to the product we add P the 7 pence, and obtain 11671 pence. Again, we multiply by 4, Ans. 4 6 6 8 6 farthings. because 4 farthings make 1 penny, and to this product we add the 2 farthings, and obtain 46686 farthings, the answer sought. 8* 20 972 12 11671 4 90 REDUCTION OP COMPOUND NUMBERS. Ex. 2. In 46686 farthings how many pounds ? operation. We divide by 4, because 4 far- 4 ) 4 6 6 8 6 far. things make 1 penny, and the re- i o \ -i t nr, t i o e sm1 t is 11671 pence, and 2 farthings 1 I )Hb71 d. Z tar. rem aining. We then divide by 12 20) 972 s. 7 d. because 12 pence make 1 shilling, and the result is 972 shillings, and 4 8 £. 1 2 s. 7 pence remaining. Lastly, we di- Ans. 48£. 12s. 7d. 2 far. T» h r . 20 > because 20 shillings make 1 pound, and the result is 48 pounds, and 12 shillings remaining. By annexing to the last quotient the several remainders, we obtain 48£. 12s. 7d. 2far. as the required result. From these illustrations, for the two kinds of reduction, we deduce the following Rule. — For Reduction Descending. Multiply the highest denomination given by the number of units required of the next lower denomination to make one in the denomination multiplied. To this product add the corresponding denomination of the multiplicand, if there be any. Proceed in this way, till the reduction is brought to the denomination required. For Reduction Ascending. Divide the lowest denomination given by the number of units required of that denomination to make one of the next higher. The quotient thus obtained divide as beforehand so proceed until it is brought to the denomination required. The last quotient, with the several remainders, if there be any, annexed, will be the answer. Examples. 3. In 127£. 15s. 8d. how many farthings ? 4. In 122672 farthings how many pounds ? 5. How many farthings in 28£. 19s. lid. 3 far. ? 6. How many pounds in 27839 farthings ? 7. In 37 8£. how many pence ? 8. In 90720 pence how many pounds ? 9. Reduce 967 guineas to pounds. 10. Reduce 1015£. 7s. to guineas. AVOIRDUPOIS WEIGHT. 129. Avoirdupois or Commercial Weight is used in weigh- ing almost every kind of goods, and all metals except gold and silver. REDUCTION OP COMPOUND NUMBERS. 91 Table. 16 Drams (dr.) 16 Ounces 25 Pounds 4 Quarters 20 HundredWeight 4r. oz. 16 = 1 256 = 16 make 1 Ounce, oz. " 1 Pound, lb. " 1 Quarter, qr. " 1 Hundred Weight, cwt. M 1 Ton, T. lb. 1 qr. 6400 = 400 = 25 = 1 cwt. 25600 == 1600 = 100 = 4 = 1 T. 51200 = 32000 = 2000 = 80 = 20 =* 1 Note 1. — The oz. stands for onza, the Spanish for ounce, and in cwt. the c stands for centum, the Latin for one hundred, and wt for weight. Note 2. — The laws of most of the States, and common practice at the present time, make 25 pounds a quarter, as given in the table. But formerly, 28 pounds were allowed to make a quarter, 112 pounds a hundred, and 2240 pounds a ton, as is still the standard of the United States government in col- lecting duties at the custom-houses. Note 3. — The term avoirdupois is from the French avoir dupoid, signifying to have weight. Note 4. — The standard avoirdupois pound of the United States is the weight, taken in the air, of 27 I 7 ° 1 5 cubic inches of distilled water, at its max- imum density, or when at a temperature of 39^ degrees Fahrenheit, the ba- rometer being at 30 inches. It is the same as the Imperial pound avoirdupois of Great Britain, which is the weight of 27^§j cubic inches of distilled water at the temperature of 62 degrees. Examples. 1. In 165T. 13cwt. 3qr. 191b. 14oz. how many ounces ? 2. In 5302318 ounces how many tons ? 3. If a load of hay weigh 3T. 16cwt. 2qr. 181b., required the weight in ounces. 4. In 122688 ounces how many tons ? 5. Required the number of drams in 2T. 17cwt. 3qr. 161b. 15oz. 13dr. 6. In 1482749 drams how many tons ? 7. What is the value of 7T. 17cwt. at 7 cents per pound? Ans. $ 1099.00. 8. What will 19cwt. 3qr. 201b. of sugar cost at 9 cents per pound? Ans. $179.55. 92 REDUCTION OP COMPOUND NUMBERS. TROY OR MINT WEIGHT. 130. Troy or Mint Weight is the weight used in weighing gold, silver, jewels, and liquors ; and in philosophical experi- ments. Table. 24 Grains (gr.) 20 Pennyweights 12 Ounces make a U 1 Pennyweight, 1 Ounce, 1 Pound, pwt. oz. lb. gr. 24 — pwt. 1 oz. 480 — 20 = 1 lb. 5760 = 240 = 12 = 1 Note 1. — Troy weight was introduced into Europe from Cairo in Egypt, in the 12th century, and was first adopted in Troyes, a city in France, where great fairs were held, whence it may have had its name. Note 2. — A grain or corn of .wheat, gathered out of the middle of the ear, was the origin of all the weights used in England. Of these grains, 32, well dried, were to make one pennyweight. But in later times it was thought sufficient to divide the same pennyweight into 24 equal parts, still called grains, being the least weight now in use, from which the rest are computed. Note 3. — Diamonds and other precious stones are weighed by what is called Diamond Weight, of which 16 parts make 1 grain ; 4 grains, 1 carat. 1 grain Diamond Weight is equal to f grains Troy, and 1 carat to 3} grains Troy. In weighing pearls, the pennyweight is divided into 30 grains instead of 24, so that 1 pearl grain is equal to | grains Troy. The carat as a weight must not be confounded with the assay carat, a term whose use is to indicate a proportional part of a weight, as in expressing the fineness of gold, each carat means a twenty-fourth part of the entire mass used. Thus, pure gold is termed 24 carat gold, and gold that is not pure is termed 18 carat gold, 20 carat gold, &c, as its mass may be 18 twenty-fourths, 20 twenty-fourths, &c. pure gold. Each assay carat is subdivided into 4 assay grains, and each assay grain into 4 assay quarters. Note 4. — The Troy pound, the standard unit of weight adopted by the United States Mint, is the same as the Imperial Troy pound of Great Britain, and is equal to the weight, taken in the air, of 22^^ cubic inches of distilled water, at its maximum density, the barometer being at 30 inches. Examples. 1. How many grains in 281b. lloz. 12pwt. 15gr. Troy? 2. In 166863 grains Troy how many pounds ? 3. If a silver pitcher weigh 31b. 10oz., what is its weight in grains ? 4. How many pounds Troy in 22080 grains ? REDUCTION OF COMPOUND NUMBERS. 93 5. What is the value of 731b. lloz. of standard silver at $ 0.062 per pennyweight ? 6. How many pounds of standard silver can be purchased for $ 1099.88, at the rate of $0,062 per pennyweight? 7. A Californian has 571b. 7oz. of pure gold. What is its value at $ 20.593± per ounce ? Ans. $ 14229.901f 8. What is the value of a mass of standard gold weighing 191b. 6oz. 16pwt. at 93 cents per pennyweight? Ans. $4367.28. 9. I have a lump of pure silver weighing 131b. 9oz. What is its value at $ 1.385 & per ounce ? Ans. $ 228.640 J. APOTHECARIES' WEIGHT. 131 • Apothecaries' Weight is used in mixing medical pre- scriptions. Table. 20 Grains (gr make 1 Scruple, sc. or £). 3 Scruples u 1 Dram, dr. or 5 • 8 Drams ii 1 Ounce, oz. or § . 12 Ounces a 1 Pound, lb. or ft,. gr. sc. 20 = 1 dr. 60 = 3 =s 1 oz. 480 = 24 = 8 = 1 lb. 5760 = 288 = 96 = 12 = 1 Note 1. — In this weight the pound, ounce, and grain are the same as in Troy Weight. Note 2. — Medicines are usually bought and sold by Avoirdupois Weight. Note 3. — In estimating the weight of fluids, 45 drops, or a common tea- spoonful, make about 1 fluid dram ; 2 common table-spoonfuls, about 1 fluid ounce; a wineglassful, about 1^ fluid ounces; and a common teacupful, about 4 fluid ounces. Examples. 1. In 23ft) 9 § 05 29 13gr. how many grains? 2. In 136853 grains how many pounds ? 3. How many scruples in 231b. ? 4. How many pounds in 6624 scruples? 5. In 47tb OS 05 19 19gr. how many grains? 6. In 270759 grains how many pounds ? 94 REDUCTION OF COMPOUND NUMBERS. 7. A physician bought 1 pound of ipecacuanha for $ 1.80, and retailed it out in doses of 5 grains, at 12£ cents each. How much did he get for it over the cost ? Ans. $ 142.20. Avoirdupois, Troy, and Apothecaries' Weight Com- pared. 132. The relative value of the pound, and its subdivisions, of the several weights, in Troy grains, and in denominations of each other, is shown in the following Table. 1 lb. Av. = 7000 gr. Tr. = lib. 2oz. llpwt. 16gr. Tr. 1 lb. Tr. or Ap. — 57G0 " — 13oz. 2{i|dr. Av. 1 oz. Tr. or Ap. = 480 " — loz. lftdr. Av. 1 oz. Av. = 43 7 J " = 18pwt. 5jgr. Tr. 1 dr. Ap. = 60 " = 2 T 3 7 4 5 dr. Av. 1 dr. Av. = 27J£ " — lpwt. SUgr. Tr. 1 pwt. Tr. = 24 " = Iffdr. Av. 1 sc. Ap. = 20 " = }f|dr. Av. 1 gr. Tr. or Ap. = 1 " = g 3 f 5 dr. Av. Note. — To change a quantity from one weight to its equivalent in another weight, reduce the yivcn quantity to Troy grains, and Hienjind their value in de- nominations of die weight required. Examples. 1. Change 131b. 6oz. Avoirdupois weight to Troy weight. 2. Change 161b. 3oz. lpwt. lgr. Troy weight to Avoirdupois weight. 3. Change 31b. 8oz. lOpwt. to drams of Apothecaries' weight. Ans. 356dr. 4. Change 356 drams Apothecaries' weight to Troy weight. Ans. 31b. 8oz. lOpwt. 5. An apothecary bought by Avoirdupois weight 21b. 8oz. of quinine at $ 2.40 per ounce, which he retailed at 20 cents a scruple. What was his gain on the whole ? 6. If I should buy by Avoirdupois weight 121b. of opium at 37^ cents per ounce, and sell it by Troy weight at 40 cents per ounce, should I gain or lose by so doing ? Ans. Lose $ 2. REDUCTION OF COMPOUND NUMBERS. 95 LINEAR OR LONG MEASURE. 133. Linear or Long Measure is used in measuring dis- tances in any direction. Table. 12 Inches (in.) make 1 Foot, ft. 3 Feet it 1 Yard, yd. 5J Yards, or 16$ Feet, u 1 Rod, or Pole, rd. 40 Rods u 1 Furlong, fur. 8 Furlongs, or 320 Rods, a 1 Mile, m. 3 Miles a 1 League, lea. 69J Miles (nearly) a 1 Degree on the equator, deg. or °. 360 Degrees a 1 Great Circle of the Earth. in. ft. 12 = 1 yd. 36 = 3 mm 1 rd. 198 = 16|- = &h = 1 fur. 7920 = 660 = 220 = 40 = 1 m . 63360 = 5280 = 1760 = 320 = 8 = 1 Note 1. — 12 lines make 1 inch; 4 inches, 1 hand; 6 feet, 1 fathom; 120 fathoms, 1 cable-length; 7 J cable-lengths, 1 mile; i of a degree of the circum- ference of the earth, 1 knot, or geographical mile, equal to l|i statute miles. Note 2. — The yard adopted by the United States government as the standard unit of linear measure is the same as the imperial yard of Great Britain, which, as compared with a pendulum vibrating seconds in the latitude of London, the pendulum moving in a vacuum, at the level of the sea, and at the temperature of 62° Fahrenheit, should bear the proportion of 36 to 39^3 inches. A metre, the unit of linear measure, as established by the French government, is equal to about 39^ English inches. Note 3. — The English statute mile is the same as that of the United States, but that of other countries differs in value from it; as the German short mile is equal to 6857 yards, or about 3j 9 English miles; the German long mile, to 10125 yards, or about 5| English miles; the Prussian mile, to 8237 yards, or about 4j T g English miles; the Spanish common league, to 7416 yards, or about 4| English miles; the Spanish judicial league, to 4635 yards, or about 2| English miles. . Note 4. — A degree of longitude is ^ of any circle of latitude. As the circles of latitude diminish in length, the degrees of longitude vary in length under different parallels of latitude. Thus, under the equator, the length of a degree of longitude is about 69§ statute miles; at 25° of latitude 62 T 7 S miles; at 40° of latitude, 53 miles; at 42° of latitude, 5l£ miles; at 49° of latitude, 45£ miles; at 60°, 34/ § miles. 96 REDUCTION OF COMPOUND NUMBERS. Examples. 1. In 96deg. 56m. 7fur. 32rd. 12ft. 6in. how many inches ? 2. In 424320486 inches how many degrees ? 3. How many feet in 79 miles ? 4. Required the miles in 417120 feet. 5. How many inches in 396 furlongs ? 6. Required the furlongs in 3136320 inches ? 7. How many inches from Haverhill to Boston, the distance being 30 miles ? 8. Required the miles in 1900800 inches. CLOTH MEASURE. 134. Cloth Measure is used in measuring cloth, ribbons, lace, and other articles sold by the yard or ell. Table. 24 Inches (in ) make 1 Nail, na. 4 Nails K 1 Quarter of a yard, qr. 4 Quarters u 1 Yard, yd. 3 Quarters it 1 Ell Flemish, E. F. 5 Quarters U 1 EU English, E. E. in. na. 2i = 1 qr. 9 — 4 — 1 E. F. 27 = 12 = 3 = 1 yd. 36 = 16 = 4 = H = 1 E.E. 45 = 20 = 5 = i| = li = 1 Note 1. — The Ell French is 6 quarters; the Ell Scotch, 4qr. l|in. Note 2. — Cloth measure is a species of linear measure, and the yard and inch are the same in both. Examples. 1. In 17yd. 3qr. 2na. how many nails ? 2. In 286 nails how many yards ? 3. In 365yd. lqr. 3na. how many nails ? 4. In 5847 nails how many yards ? 5. In 71E. E. 4qr. how many nails ? 6. In 1436 nails how many ells English ? 7. What cost 47yd. 3qr. of silk velvet at $ 1.25 per quarter ? 8. A merchant bought a roll of cloth containing 31£E. E. and paid for it at the rate of $ 3 per yard. What did it cost him? Ans. $117. REDUCTION OP COMPOUND NUMBERS. 97 135. kinds. SUEFACE OR SQUARE MEASURE. Square Measure is used in measuring surfaces of all Table. Square inches (sq. in.) make Square feet " 144 9 30£ Square yards 40 Square rods 4 Roods 640 Acres in. 144 = 1596 = 39204 = 1568160 = 6272640 = Square foot, Square yard, Square rod or pole, Rood, Acre, Square mile, ft. yd. P- R. A. S.M. ft. 1 9 = 272i = 10890 = 43560 = yd- 1 30£ = 1210 = 4840 = P- 1 40 = 160 = R. 1 4 4014489600 = 27878400 = 3097600 = 102400 = 2560 A. 1 S.M. 640 = 1 Square foot. Note. — A square is a figure bounded by four equal lines, perpendicular to each other. When the four lines are each 1 foot in length, the space enclosed is 1 square foot; when 1 yard in length, 1 square yard; when 1 rod in length, 1 square rod; and so for any other dimension. 3ft. = 1yd. In this diagram the large square repre- sents a square yard, and each of the smaller squares within it represents one square foot. Now, since therfc are three rows of small squares, and three square feet in each row, there will be 3 times 3 = 9 sq. ft. in the large square. But the large square is 3ft. in length and 3ft. in breadth ; hence, To find the contents of a square, multiply its length by its breadth. Examples. 1. In 57A. 3R. 27p. 21yd. 8ft. 57in. how many square inches ? 2. In 363331893 square inches how many acres? How many square feet in 25 acres ? How many acres in 1089000 square feet? How many square rods in 365 square miles ? How many square miles in 37376000 square rods ? How many acres in 12345678 square inches? Ans. 1A. 3R. 34p. 27yd. 4ft. 54in. N 9 3, 4. 5. 0. 7. 98 REDUCTION OF COMPOUND NUMBERS. 8. Bought 39 A. 2R. 16p. of land for $3.75 per square rod, and sold the same for $0.25 per square foot. What did I gain by my bargain ? Ans. $ 407,484.00. SURVEYORS' MEASURE. 136. This measure is used by surveyors in measuring land, roads, &c. Table. 7jfc Inches (in.) make 1 Link 25 Links u 1 Pole, 00 Links, 4 Poles, or 66 Feet U 1 Chain, 10 Chains u 1 Furlong, 8 Furlongs, or 80 Chains, u 1 Mile, Inches. Link. 7 92 _ l Tole. 198 = 25 = 1 Chain. 792 = 100 = 4 = 1 Furlong. 7920 = 1000 = 40 = 10 = 1 1. p- ch. fur. m. Mile. 63360 = 8000 = 320 — 80 = 8 == 1 Note 1. — Guntcr's chain, in length 4 poles, or 66 feet, and divided into 100 links, is that mostly used in ordinary land surveys ; but in locating roads, and like public works, an engineer's chain is usually 100 feet in length, containing 120 links, each 10 inches long. Note 2. — A section of government lands is 1 square mile, or 640 acres. An acre, as a square piece of land, will measure on each side about 209 feet or 70 paces. 625 square links make 1 square rod or pole ; 16 square rods make 1 square chain, and 10 square chains make 1 acre. Note 3. — A rod or pole is sometimes called a perch, and each of the names given to this measure is expressive of the instrument by which it was formerly measured. Examples. 1. How many links in 46m. 3fur. och. 251. ? 2. In 371525 links how many miles ? 3. In 97m. Ofur. how many links ? 4. In 776000 links how many miles ? 5. The extent of a certain farm is found, by survey, to be 1377 square chains (Note 2). How many acres does it con- tain ? Ans. 137A. 2E. 32p. 6. What will be the cost of a field measuring 2,126,250 square links, at $ 80 per acre ? Ans. $ 1701.00. REDUCTION OF COMPOUND NUMBERS. 99 CUBIC OR SOLID MEASURE. 137. Cubic or Solid Measure is used in measuring such bodies or things as have length, breadth, and thickness; as timber, stone, &c. Table. 1728 Cubic inches (cu. in.) make 1 Cubic foot, cu. ft. 27 " feet u 1 " yard, cu. yd. 40 " feet (4 1 Ton, T. 16 " feet ft 1 Cord foot, eft. 8 Cord feet, or ") 128 Cubic feet, } a 1 Cord of wood, C. in. 1728 — ft. 1 yd. 46656 = 27 = 1 T. 69120 = 40 = m = 1 c. 221184 = 128 = 4f = S\ = 1 Note 1. — A pile of wood 8ft. in length, 4ft. in breadth, and 4ft. in height, contains a cord. Also, one ton of timber, as usually surveyed, contains 50j 9 Q 2 cubic or solid feet. Sawed timber, joists, plank, and scantlings are now generally bought and sold by what is called board measure. Note 2. — A cube is a solid bounded by six square and equal sides. If the cube is 1 foot long, 1 foot wide, and 1 foot high, it is called a cubic or solid foot. If the cube is S feet long, 3 feet wide, and 3 feet thick, it is called a cubic or solid yard. Now, since each side of a cubic yard, as represented in the diagram, contains 9 sq. ft. of surface (Art. 93), it is plain, if a block be cut off from one side, one foot thick, it can be divided into 9 solid blocks, with sides 1 foot in length, breadth, and thickness, and therefore will contain 9 solid feet ; and since the whole block or cube is three feet thick, it must contain 3 times 9 = 27 feet; for 3ft. x 3ft. X 3ft. = 27 solid feet. Hence, To find the contents of a cubic or solid body, multiply its length, breadth, and thickness together. Note 3. — A cubic foot of distilled water at the maximum density, at the level of the sea, and the barometer at 30 inches, is equal in weight to 62£lb. or lOOOoz. avoirdupois. ^1 /- / *s •> ^ 4\ •ff-/^ / „rHBIIIII d Bll llli II 111 lliiiillii CO 111 ill ill ipr 3 ft. = 1 yd. 100 REDUCTION OF COMPOUND NUMBERS. Note 4. — A cubic foot of lead weighs 708|lb. ; of brass, 534|lb. ; of cop- per, 5551b.; of wrought-iron, 486flb.; of cast-iron. 450£lb.; of marble, 1711b.; of granite, 1651b.; of clay, 1301b.; of common soil, 1241b.; of bricks, 1241b.; of sand, 951b.; of sea-water, 64^1^; of oak wood, 551b.; of Anthracite coal, 541b.;' of Bituminous coal, 501b.; of red-pine wood, 421b. ; and of white-pine wood, 301b. Examples. 1. In 29 cords of wood how many solid inches? 2. In 6414336 cubic inches how many cords? 3. In 19 tons of timber how many solid inches ? 4. How many tons of timber in 1313280 cubic inches ? 5. How many cubic feet of wood in 128 cords? 6. How many cords of wood in 16384 cubic feet ? 7. How many cubic feet in a pile of wood, 40 feet long, 4 feet wide, and 7 feet high ? 8. How many cords of wood in 8650 cubic feet ? Ans. 67 cords, 74 cubic feet. 9. How many cubic feet in a granite block, 17 feet long, 11 feet wide, and 9 feet high ? Ans. 1683 cubic feet. LIQUID OR WINE MEASURE. 138. Liquid or Wine Measure is used in measuring all kinds of liquids, except, in some places, beer, ale, porter, and milk. Table. 4 Gills (g i) 2 Pints 4 Quarts 63 Gallons 2 Hogsheads 2 Pipes gi- pt. 4 = 1 8 = 2 32 = 8 2016 = 504 4032 = 1008 8064 = 2016 take K a u 1 Pint, 1 Quart, 1 Gallon, 1 Hogshead, pt. qt. gal. hhd. u u 1 Pipe, or 1 Tun, Butt, pi. tun. qt. 1 gal. 4 = 1 hhd. 252 = 63 = : 1 pi. 504 = 126 = : 2 = 1 tun. 1008 = 252 = 4 == 2 = 1 EEDUCTION OF COMPOUND NUMBERS. 101 Note 1. — Bylaws of Massachusetts, 32 gallons make 1 barrel. In some States 3l£ gallons, and in others from 28 to 32 gallons, make 1 barrel. 42 gal- lons make 1 tierce, and 2 tierces make 1 puncheon. Note 2. — The term hogshead is often applied to any large cask that may contain from 50 to 120 gallons, or more. Note 3. — The Standard Unit of Liquid Measure adopted by the govern- ment of the United States is the Winchester Wine Gallon, which contains 231 cubic inches, and is of a capacity to hold 8f^jlb. Avoirdupois of distilled water, at its maximum density, weighed in air, the barometer being at 30 inches. It has the name Winchester, from its standard having been formerly kept at Winchester, England. The Imperial Gallon, now adopted in Great Britain, contains 277^^ cubic inches; so that 6 Winchester gallons make about 5 Imperial gallons. Note 4. — 1 gallon of alcohol weighs 71b. ; of camphene, 7|lb. ; of proof spirits, 7]Jlb. ; of spirits of turpentine, 7j g lb. ; of sperm oil, 7|lb. ; of olive oil, 7£lb. ; of linseed oil, 7f lb. ; and of molasses, llflb. Note 5. — The fluid measure of apothecaries, used by them in measuring liquids of medical prescriptions, divides the gallon (marked Cong.) into 8 pints (0.); the pints into 16 fluid ounces (f^ ); the fluid ounces into 8 fluid drams (f5 ); and the fluid drams into 60 minims (m) or drops. The abbrevi- ation Cong, stands for congiarium, the Latin for gallon, and the O. is the initial of octans, the Latin for an eighth, the pint being an eighth of a gallon. Examples. 1. In 57T. 3hhd. 50gal. 3qt. how many pints ? 2. In 116830 pints how many tuns? 3. Reduce 96hhd. 47gal. 2qt. to gills. 4. How many hogsheads in 195056 gills ? 5. What cost 40 hogsheads of wine at $ 0.37^ per pint ? 6. How much may be gained by buying 2 hogsheads of molasses, at 40 cents a gallon, and selling it at 12 cents a quart? Ans. $10.08. BEER MEASURE. 139. Beer Measure is used in measuring beer, ale, porter, and milk. Table. 2 Pints (pt.) 4 Quarts 54 Gallons make (4 u 1 Quart, 1 Gallon, 1 Hogshead, qt. gal. hhd. pt. 2 8 432 qt. = 1 = 4 = 216 9* gal. = 54 = hhd. 1 102 REDUCTION OF COMPOUND NUMBERS. Note 1. — The gallon of beer measure contains 282 cubic inches; and has been usually reckoned, 36 gallons equal 1 barrel; 2 hogsheads, 05/IO8 gallons, 1 butt; 2 butts, or 216 gallons, 1 tun. Note 2. — Beer Measure is becoming obsolete. Milk and malt liquors, at the present time, are bought and sold, very generally, by wine or liquid measure. Examples. 1. How many pints in 46khd. 49gal.? 2. In 20264 pints how many hogsheads ? 3. In 368hhd. how many pints ? 4. In 158976 pints how many hogsheads ? 5. At 29 cents per gallon, what cost 76 hogsheads of ale ? Ans. $1190.16. 6. How much may be obtained by selling 47hhd. 36gal. of lager-bier at 5 cents a quart? Ans. $514.80. DRY MEASURE. 140. This measure is used in measuring grain, fruit, salt, &c. Table. 2 Pints (pt.) 8 Quarts 4 Pecks make lit u 1 Quart, 1 Peck 1 Bushel, qt. pk bu. pts. 8 gal. 1 P k. 16 = 2 = 1 bu. 64 — 8 = 4 = 1 Note 1. — The Standard Unit of Dry Measure adopted by the United States government is the Winchester bushel, which is in form a cylinder, 18£ inches in diameter, and 8 inches deep, containing 2150^ cubic inches. The Stand- ard Imperial bushel of Great Britain contains 22181^ cubic inches, so that 32 Imperial bushels equal about 33 Winchester bushels. The gallon in Dry Measure contains 268| cubic inches. Note 2. — Of wheat a standard bushel is 601b. ; of shelled corn, 561b. ; of corn on the cob, 701b. ; of rye, 561b. ; of barley, 481b. ; of buckwheat in Pa., Ky., &c, 521b. ; of buckwheat in Mass., 481b. ; of oats in Ohio, 111., Mass., &c, 321b.; of oats in Ky., 33£lb.; of oats in Me., 301b.; of oats in Pa., 241b.; of clover-seed, 601b. ; of flax-seed, 561b. ; of Timothy-seed, 451b. ; of bran, 201b. ; of beans, 601b. ; of onions, in Pa., Ky., &c, 571b. ; of) onions in Mass., 52ib. ; of salt in Ky., 561b.; of salt in 111., 501b.; of dried apples in Pa., 221b.; of dried apples in 111., 241b. ; of dried peaches in Pa., 331b. ; of dried peaches in 111., 321b.; of stove coal in 111., 801b.; of bituminous coal in the Western States, 761b. ; and of hard-wood charcoal, 301b. The weight by law, of a few REDUCTION OF COMPOUND NUMBERS. 103 of the articles named, to a bushel, is not uniform in all the States, and there- fore may vary slightly from the above, in a few States not mentioned. Note 3. — In some places it is customary, in measuring coal, potatoes, and like articles, to " heap " the bushel, as it is called, and in that case 5 even pecks are about equal to 1 "heaped bushel." The " coal bushel," as estab- lished by laws of Massachusetts, Ohio, and some other States, is of greater capa- city than the Winchester bushel. In some parts of the United States a chaldron, a measure of coal, consists of 36 bushels ; and in other parts of the country it consists of 32 bushels, or of 4 quarters, each consisting of 8 bushels. The quarter, however, in England is 8 Imperial bushels, a measure of grain equal to 5601b., or one quarter of a ton of 22401bs. Examples. 1. How many pints in 35bu. 3pk. ? 2. In 2288 pints how many bushels ? 3. In 676 chaldrons, of 36 bushels each, how many pecks? 4. How many chaldrons, of 36 bushels each, in 97344 pecks ? 5. A grocer purchased 50 bushels of potatoes, by "heaped" measure, at 60 cents a bushel, and sold the same, by " even " measure, at 15 cents a peck ; did he gain or lose by the oper- ation ? Ans. Gain $ 7.50. 6. If I purchase by weight, in Pennsylvania, 96 bushels of oats, at 42 cents a bushel, and sell the same by weight, in Ohio, at 45 cents a bushel, shall I gain or lose by so doing ? Ans. Lose $ 7.92. Dry, Liquid, and Beer Measures Compared. 141 • The relative value of the gallon and its subdivisions, of the several measures, in cubic inches, and in denominations of each other, are shown in the following Table. cu. in. 1 gal. B. M. = 282 = lgal. lpt. 3£gi. L. M. = lgal. «pt. D. M. 1 gal. D. M. = 268§ == lgal. lpt. lifgi. L. M. = 3qt. ligpt. B. M. 1 gal. L. M. = 231 = 3qt. £pt. D. M. = 3qt. |«pt. B. M. 1 qt. B. M. == 10h = lqt. Opt. lffgi. L. M. = lqt. ggt. D. M. 1 qt. D. M. = 67a = lqt. IJJpt. L. M. = lpt. 3i|Jgi. B. M. 1 qt. L. M. = 57| = l||pt. D. M. = lfjpt. B. M. 1 pt. B. M. = 35.i = l||pt. L. M. mm 1^-pt. I). M. 1 pt. D. M. = 333 ^ ipt. ggj, L. M. = 3||igi. B. M. 1 pt. L. M. = 28 j = Jjpt. D. M. = iipt. B. M. 1 gi. L. M. = lJ- 2 = &pt D. M. == $#t. B. M. 104 REDUCTION OP COMPOUND NUMBERS. Note 1. — By the table, it is evident that each of the measures of capacity- is a species of cubic measure; and to change cubic measure, expressed in cubic inches, to any denomination of either dry, liquid, or beer measure, divide by the number of cubic inches required to make a unit of the proposed denominar tion. Thus, to reduce 14ft. 294in. cubic measure to gallons of liquid measure: 14 cu. ft. 294 cu. in. = 24486 cu. in. ; 24486 cu. in. -f 231 = 106 gallons, liquid measure. Note 2. — To change a quantity from one measure of capacity to its equiv- alent in another, reduce the given quantity to cubic inches, and then find their value in denominations of the proposed measure. Examples. 1. Change 4hhd. 15gal. beer measure to liquid measure ? 2. Change 4hhd. 30gal. liquid measure to beer measure ? 3. If a milkman buy milk at 4 cents per quart, beer meas- ure, and sell the same at 6 cents per quart, liquid measure, what will he gain in disposing of 2820 gallons ? Ans. $ 375.02ff 4. If 2538 gallons of milk have been purchased by liquid measure, at 4 cents per quart, and the same has been sold by beer measure at G cents per quart, what has been the gain? Ans. $92.88. 5. A merchant bought 385 bushels of seed peas at $ 4.00 per bushel, dry measure. lie sold the same at 20 cents per quart, liquid measure. "What did he gain by the pur- chase? Ans. $1327.20. 6. J. Day bought 1000 bushels of corn at $ 1.05 per bushel, dry measure, and sold the same at $ 1.12 per bushel, liquid measure. Did he gain or lose by the operation, and how much ? 7. My hogshead contains 30 cubic feet. How many more gallons of dry measure will it contain, than of beer meas- ure? 8. Bought of my neighbor, John Smith, 365 gallons of milk, at 5 cents per quart ; but by mistake he measured it in his liquid measure. How much did I lose ? MEASUEE OF TIME. 142. This measure is applied to the various divisions and subdivisions into which time is divided. REDUCTION OF COMPOUND NUMBERS. 105 Table. 60 Seconds (sec.) make 1 Minute, m, 60 Minutes " 1 Hour, h. 24 Hours " iDay, d. 7 Days " 1 Week, w. 365£ Days, or 52 weeks 1% days, " 1 Julian Year, y- 12 Calendar Months (mo.) " 1 Year. sec. m. 60 == 1 h. 3600 = 60 = 1 d. 86400 = 1440 = 24 = 1 w. 604800 = 10080 = 168 = = 7=1 3 31557600 = 525960 = 8766 = = 365i = jr- Note 1. — The true Solar or Tropical Year is the time measured from the sun's leaving either equinox or solstice to its return to the same again, and is 365d. 5h. 48m. 49j 7 sec. The Julian Year, so called from the calendar instituted by Julius Caesar, contains 365^ days, as a medium; three years in succession containing 365 days, and the fourth year 366 days ; which, as compared with the true solar year, produces an average yearly error of 11m. lOj^sec, or a difference that would amount to 1 whole day in about 120 years. The Gregorian Year, or that instituted by Pope Gregory XIII., in the year 1582, and which is now the Civil or Legal Year in use among most nations of the earth, contains, like the Julian year, 365 days for three years in succession, and 366 days for the fourth, excepting the last year of the odd centuries. The Gregorian year is so nearly correct as to err only 1 day in 3866 years, a difference so little as hardly to be worth taking into account. The manner of reckoning time according to the Julian Calendar is termed Old Style, and that according to the reformed calendar of Gregory, New Style. England did not adopt the new style till 1752, when., according to an act of Parliament, the difference between the two styles, which then amounted to 11 days, was removed, by the day following the 2d of September of that year being accounted the 14th day. The difference now between old and new style is 12 days. A Common Year is one of 365 days, and a Leap or Bissextile Year is one of 366 days. Any year, excepting the last year of the odd centuries, that can be divided by 4 without a remainder, is Leap Year. A Sidereal Year is the time in which the earth revolves round the sun, and is 365d. 6h. 9m. 9 6_sec. Note 2. — The names of the 12 calendar months, composing the civil year, are January, February, March, April, May, June, July, August, September, October, November, December, and the number of days in each may be readily remembered by the following lines : — " Thirty days hath September, April, June, and November ; And all the rest have thirty-one, Save February, which alone Hath twenty -eight ; and this, in fine, One year in four hath twenty-nine." 106 REDUCTION OF COMPOUND NUMBERS. TABLE Showing the Number of Days from any Day of one Month to the same Day of any other Month in the same Year. from any DAY OF to the same day of next Jan. Feb. Mar. Apr. May. June. July. Aug. Sept. Oct. Nov. 304 Dec. 834 January 365 31 59 90 120 151 181 212 243 273 February 334 365 28 59 89 120 150 181 212 242 273 303 March 306 337 365 31 61 92 122 153 184 214 245 275 April 275 306 334 365 30 61 91 122 153 183 214 244 May 245 267 304 335 365 31 61 92 123 153 184 214 June 214 245 273 304 334 365 30 61 92 122 153 183 July 184 215 243 274 304 335 365 31 62 92 123 153 August 153 184 212 243 273 304 334 365 31 61 92 122 September 122 153 181 212 242 273 303 334 365 30 61 91 October 92 123 151 182 212 243 273 304 335 365 31 61 November 61 92 120 151 181 212 242 273 304 334 365 30 December 31 62 90 121 151 182 212 243 274 304 335 365 For example, suppose we wish to find the number of days from April 4th to November 4th, we look for April in the left-hand vertical column, and for No- vember at the top, and where the lines intersect is 214, the number sought. Again, if we wish the number of days from June 10th to September 16th, we find the difference between June 10th and September 10th to be 92 days, and add 6 clays for the excess of the 16th over the 10th of September, and so we have 98 days as the exact difference. If the end of February be included between the points of time, a day must be added in leap year. "When the time includes more than one year, there must be added 365 days for each year. Examples. 1. How many seconds in a solar year ? 2. In 31556929 seconds how many days ? 3. How many seconds from the deluge, which took place 2348 years before Christ, to the year 1856, a year being 365£ days? Aiis. 132636592800. 4. In 74726807872 seconds how many solar years ? Ans. 2368 years. 5. How long did the last war with England continue ; it having commenced June 18th, 1812, and ended February 17th, 1815 ? Ans. 974 days = 2 years 244 days. CIRCULAR ANGULAR MEASURE. 143. Circular Angular Measure is applied to the measure- ment of circles and angles, and is used in reckoning latitude and longitude, and the revolutions of the planets round the sun. REDUCTION OP COMPOUND NUMBERS. 107 60 Seconds (") make 1 Minute, '. GO Minutes u 1 Degree, °. 30 Degrees a 1 Sign, S. 12 Signs, or 360 Degrees, The Circle of the Zodiac, C. 60 — / 1 o 3600 = 60 mm 1 S. 108000 = 1800 mm 30 = 1 C . 1296000 = 21600 = 360 = 12 = 1 Note 1. — A Circle is a plane figure bounded by a curve line, all parts of which are equally distinct from a point called its centre. The Circumference of a circle is the line which bounds it, as shown by the diagram. An Arc of a circle is any part of its cir- cumference; as AB. A Radius of a circle is a straight line drawn from its centre to its circumference ; as CA, CB, or CD. Every circle, large or small, is supposed to be divided into 360 equal parts, called degrees. A Quadrant is one fourth of a circle, or an arc of 90° ; as AB. An Angle, as ACB, is the inclination or opening of two lines which meet at a point, as C. The point is the vertex of the angle. If a circle be drawn around the vertex of an angle as a centre, the two sides of the angle, as radii of the circle, will include an arc, which is the measure of the angle ; as the arc AD = 120° is the measure of the angle ACD, and AB == 90©, the meas- ure of the angle ACB; hence the one is called an angle of 120°, and the other an angle of 90°. Note 2. — As the earth turns on its axis from west to east every 24 hours, the sun appears to pass from east to west §5 of 360o of longitude every hour, or 15° of longitude in 1 hbur's time, or 1° in 4 minutes of time, and 1' in 4 seconds of time ; so that, for instance, when it is noon at any place, it is 1 hour earlier for every 15° of longitude westward, and 1 hour later for every 15° of longitude eastward. Thus, Boston being 71° 4' west of Greenwich, and San Francisco 51° 17' west of Boston, when it is noon at Boston, it is 4h. 44m. 16sec. past noon at Greenwich, and wanting 3h. 25m. 8sec. of noon at San Francisco. Examples. 1. How many minutes in 27S. 27° 43'? 2. In 50263 minutes how many signs ? 3. How many seconds in 44S. 18° 57' 23"? 4. How many signs in 4820243"? 5. How many seconds in 360° ? 6. How many degrees in 1296000"? 108 REDUCTION OF COMPOUND NUMBERS. MISCELLANEOUS TABLE. 144. The following denominations, frequently used, are not embraced in the preceding tables. 12 units ] nake 1 dozen. 12 dozen u 1 gross. 1 2 gross II 1 great gross, 20 units U 1 score. 24 sheets of Paper u 1 quire. 20 quires a 1 ream. 2 reams ll 1 bundle. 5 bundles u 1 bale. 14 pounds of Iron or Lead U 1 stone. 21£ stone K ipig- 8 pigs II 1 fother 18 inches a 1 cubit. 24| cubic feet of Stone it, 1 perch. 60 pairs of Shoes M 1 case. 25 pounds of Powder II 1 keg. 56 pounds of Butter It 1 firkin. 100 pounds of Fish <( 1 quintal. 196 pounds of Flour II 1 barrel. 200 pounds of Beef M 1 barrel. 200 pounds of Pork u 1 barrel. 200 pounds of Shad or Salmon. , in N. Y. & Ct. n 1 barrel. 220 pounds of Fish, in Md. u 1 barrel. 256 pounds of Soap II 1 barrel. 300 pounds of Hydraulic Cement u 1 barrel. 30 gallons of Fish, in Mass. M 1 barrel. 5 bushels of Shelled Corn, in Southern States 1 " 1 barrel. 8 bushels of Salt U 1 hogshead. Note 1. — A sheet of paper folded into 2 leaves forms a folio; a sheet folded into 4 leaves, a quarto; a sheet folded into 8 leaves, an octavo; a sheet folded into 12 leaves, a 12mo; a sheet folded into 18 leaves, an 18mo; a sheet folded into 24 leaves, a 24mo. Note 2. — Of shoemaker's measure, No. 1 of small size is 4£ inches in length ; and No. 1, large size, is 8££ inches in length ; and each succeeding number of either size is J of an inch additional length. Examples. i. j naies 4 bundles 1 ream 10 quires of paper how any sheets ? Ans. 23760. 2. In 23760 sheets of paper how many bales ? 3. In 10 fothers 6 pigs 8 stones of iron how many pounds ? 4. In 25998 pounds of iron how many fothers ? 1. In 4 bales many sheets ? REDUCTION OF COMPOUND NUMBERS. 109 5. At 23 cents a pound, what will 12 firkins of butter, cost? 6. If $ 22 is paid for a barrel of pork, how much is that by the pound? Ans. 11 cents. 7. How much must be paid for 302 hogsheads of salt at $ 0.30 a bushel ? Ans. $ 724.80. 8. At $ 4 per quintal, how many pounds of fish may be bought for $ 50.24 ? Ans. 1256 pounds. 9. If the wholesale price of one writing-book be 4 J cents, what will be the cost of a great-gross of writing-books ? 10. A dairyman sells 2 firkins of butter at 20 cents a pound, and takes in pay half a barrel of flour at 5 cents a pound, and the balance in cash. How much cash does he receive ? Ans. $ 17.50. MISCELLANEOUS EXAMPLES IN EEDUCTION. 1. In 57£. 15s. how many half-pence ? . Ans. 27720. 2. In 591b. 13pwt. 15gr. how many grains ? 3. In 340167 grains how many pounds ? 4. How many ells English in 761 yards ? Ans. 608 E. E. 4qr. 5. How many yards in 61 ells Flemish ? Ans. 45yd. 3qr. 6. How many bottles, that contain 3 pints each, will it take to hold a hogshead of wine ? Ans. 168. 7. How many steps, of 2ft. 8in. each, will a man take in walking from Bradford to Newburyport, the distance being fifteen miles ? Ans. 29700. 8. How many spoons, each weighing 2oz. 12pwt., can be made from 51b. 2oz. 8pwt. of silver ? Ans. 24. 9. How many times will the wheel of a coach revolve, whose circumference is 14ft. 9in., in passing from Boston to Washing- ton, the distance being 436 miles ? Ans. 156073 T 3 T 9 T . 10. I have a field of corn, consisting of 123 rows, and each row contains 78 hills, and each hill has 4 ears of corn ; now if it takes 8 ears of corn to make a quart, how many bushels does the field contain ? Ans. 149bu. 3pk. 5qt. Opt. 11. If it take 5yd. 2qr. 3na. to make a suit of clothes, how many suits can be made from 182 yards ? 12. A goldsmith wishes to make a number of rings, each N 10 # 110 REDUCTION OP COMPOUND NUMBERS. weighing 5pwt. 10gr., from 31b. loz. 2pwt. 2gr. of gold; how many will there be ? Ans. 137. 13. How many shingles will it take to cover the roof of a building, whifth is 60 feet long and 56 feet wide, allowing each shingle to be 4 inches wide and 18 inches long, and to lie one third to the weather ? Ans. 20160. 14. There is a house 56 feet long, and each of the two sides of the roof is 25 feet wide ; how many shingles will it take to cover it, if it require 6 shingles to cover a square foot ? Ans. 16800. 15. If a man can travel 22m. 3fur. 17rd. a day, how long would it take him to walk round the globe, the distance being about 25000 miles ? Ans. 1114fff f days. 16. If a family consume 71b. lOoz. of sugar in a week, how long would lOcwt. 3qr. 161b. last them ? Ans. 143^ 5 T weeks. 17. What will ^ hogsheads of wine cost, at 9 cents a quart? 18. What will 15 hogsheads of beer cost, at 3 cents a pint? Ans. $194.40. 19. What will 73 bushels of meal cost, at 2 cents a quart? Ans. $46.72. 20. A merchant has 29 bales of cotton cloth ; each bale contains 57 yards ; what is the value of the whole at 15 cents a yard ? Ans. $ 247.95. 21. There is a certain pile of wood 120 feet long, 4£ feet high, and 4 feet wide ; what is its value at $ 4.00 per cord? Ans. $67.50. 22. How much must be paid, at twenty cents a square yard, for plastering overhead in a room which is 33 feet long and 18 feet wide ? 23. An apothecary, in compounding 20 boxes of pills, each box containing 25 pills, used 6 grains of aloes, 5 grains of rhubarb, and 4 grains of calomel in each pill. What was the entire quantity used ? Ans. 7500 grains. 24. Purchased a cargo of molasses, consisting of 87 hogs- heads ; wTiat is the value of it at 33 cents a gallon ? Ans. $ 1808.73. 25. If a cubic foot of white-oak wood weighs 880 ounces, and a cubic foot of white-pine wood weighs 480 ounces, how much will a load weigh, which is composed of half a cord of white-oak, anil of a cord of white-pine ? Ans. 73601bs. ADDITION OF COMPOUND NUMBERS. Ill 26. How many chests of tea, weighing 24 pounds, at 43 cents a pound, can be bought for $ 1548 ? 27. Joseph Eldredge received $ 10, as a birthday present, from his father, on every 29th day of February, from 1837 to 1857. How much less than $ 200 did he receive, in all ? Ans. $ 150. 28. If 25f grains of standard gold be worth $1, how many pounds avoirdupois of standard gold will be worth $ 1,000,000 ? Ans. 3685f pounds. 29. A merchant, who had bought 188 gallons of molasses, at 40 cents a gallon, intended to have it sold at the rate of 50 cents a gallon ; but his shop-boy retailed half of the quan- tity at 12^ cents a quart, beer measure, when, finding he had made a blunder, he sold the balance at 14 cents a quart, wine measure, thereby expecting to exactly make up for the mis- take. How much less d^l the whole bring than was intended ? Ans. $ 2.86. ADDITION OF COMPOUND NUMBERS. 145i Addition of Compound Numbers is the process of finding the amount of two or more compound numbers. Ex. 1. Required the amount of 31£. 17s. 9d. 2far.; 16£. 16s. 6d. lfar.; 16£. lis. lid. lfar.; 19£. 19s. 9d. 3far. ; 61£. 17s. Id. 2far. Ans. 147£. 3s. 2d. lfar. Having written units of the same denomination in the same column, we find the sum of farthings in the right- hand column to be 9 farthings = 2d. lfar. We write the lfar. under the column of farthings, and carry the 2d. to the column of pence ; the sum of which is 38d. = 3s. 2d. We write the 2d. under the column of pence, and carry the 3s. to the column of shillings ; the sum of which is 83s. = 4£. 3s. Having written the 3s. under the column of shillings, we carry the 4£. to the column of pounds, and find the entire amount sought to be 14 7£. 3s. 2d. lfar. The same result can be arrived at by reducing the numbers as they are* added in their respective columns. Thus, beginning with far- things, we can add, in this way : 2far. -f- 3far. = 5far. = Id. lfar. ; £. OPERATION. s. d. far. 31 17 9 2 16 16 6 1 16 11 11 1 19 19 9 3 61 17 1 2 Ans. 14 7 3 2 1 112 ADDITION OP COMPOUND NUMBERS. § and lfar. =- Id. 2far., and lfar. = Id. 3far., and 2far. = 2d. lfar. Writing the lfar. under the column of farthings, we carry the 2d. to the column of pence ; and add, 2d. (carried) -J- 1 = 3d., and 9d. = 12d. = Is., and lid. = Is. lid., and 6d. = 2s. 5d., and 9d. = 3s. 2d. Writing the 2d. under the column of pence, we carry the 3s. to the column of shillings ; and add, 3s. (carried) -[- 17s. = 20s. = l£., and 19s. = l£. 19s., and lis. = 2£. 10s., and 16s. = 3£. 6s., and 17s. = 4£. 3s. Writing the 3s. under the column of shillings, we carry the 4£. to the column of pounds, and so find the whole amount to be, as before, 14 7£. 3s. 2d. lfar. The last method of operation may be rendered more concise, as it should always be in practice, by merely naming results as the adding is performed (Art. 45). From the illustrations given, it is evident that the adding of com- pound numbers is like that of simple numbers, except in carrying ; which difference holds also in subtracting, multiplying, and dividing compound numbers. Rule. — Write all the given numbers, so that units of the same de- nomination may stand in the same column. Add as in addition of simple numbers p and carry, from column to column, one for as many units as it takes of the denomination added to make a unit of the denomination next higher. Proof. — The proof is the same as in addition of simple numbers. Examples. 2. 3. Ton. cwt. qr. lb. oz. dr. cwt. qr. lb. oz. dr. 61193171515 612111114 63 133161111 163151511 51 12 3 17 7 6 41 3 13 9 9 61161111212 38 2111010 13 13 3 12 13 15 42 1 9 8 13 71182131414 313171112 324 15 2 4, 15 12 9 5. lb. oz. pwt gr. lb. oz. pwt. gr. 16 11 19 23 123 9 7 13 31 10 18 16 98 11 17 14 63 9 12 15 49 7 13 21 17 8 13 12 13 10 10 20 61 7 12 16 ' 47 9 19 23 17 6 17 22 51 5 15 15 209 7 15 ADDITION OP COMPOUND NUMBERS. 113 6. 7. lb 5 5 B gr. lb § 5 3 gr. 27 1 1 7 2 19 37 9 6 1 18 16 10 6 1 13 14 4 4 2 11 41 9 3 2 16 61 6 3 2 6 38 10 5 2 14 41 4 7 2 16 41 4 4 1 11 39 8 4 1 12 16 6 6 2 6 51 11 7 2 19 183 6 3 8. 1 19 9. eg.. m. fur. rd. ft. in. m. fur. rd. j rd. ft. 3 17 69 7 39 16 11 69 7 31 5 2 11 6162 31712 9 1661641 6 1616 61613 10 617 32 32 10 48 19 3 15 15 9 73 3 16 4 2 9 17 33 58 6 35 5 33 19 14 7 9 9 19 4 14 1 1 75 5 25 5 2 8 7 195 55^ 1 i-4 24 i 7 195 55 5 24 1 1 Note. — As half a mile is equal to 4 furlongs, we add them to the 1 fur- long, which makes 5 furlongs. And as half a foot is equal to 6 inches, we add them to the 7 inches, which makes 13 inches ; and these are equal to 1 foot 1 inch. 10. Add together 37yd. 3qr. 3na. 2in. ; 61yd. 3qr. Ina. lin.; 13yd. 2qr. 2na. 2in.; 32yd. lqr. Ina. lin.; 61yd. 2qr. 2na. 2in.; and 22yd. lqr. 3na. Ans. 229yd. 3qr. 3na. l£in. 11. Add together 671E.E. lqr. Ina. lin.; 161E.E. 3qr. 3na. 2in. ; 617E.E. 3qr. Ina. 2in. ; 178E.E. 3qr. 2na. lin. ; 717E.E. 2qr. Ina. 2in. ; and 166yd. 3qr. 2na. lin. 12. Add together 761A. 3R. 37p. 260ft. 125in.; 131A. 2R. 16p. 135ft. 112in.; 613A. 1R. 14p. 116ft. 131in.; 161A. 3R. 13p. 116ft. 123in.; 321A. 2R. 31p. 97ft. 96in.; and 47 A. 3R. 19p. 91ft. 48in. Ans. 2038A. 1R. 13p. 2ft. 95in. 13. Add together 38A. 1R. 39p. 272ft.; 61A. 3R. 38p.- 167ft. ; 35A. 3R. 19p. 198ft.; 47 A. 3R. 16p. 271ft.; 86A. 2R. 13p. 198ft.; and 46A. 1R. 14p. 269ft. 14. Add together 17m. 7fur. 9ch. 3p. 241.; 16m. 3fur. 4ch. 10* 114 ADDITION OP COMPOUND NUMBERS. lp. 151. ; 27m. 4fur. 6ch. 2p. 171.; 18m. 6fur. 3ch. 3p. 211..; 61m. 7fur. 7ch. 2p. 161.; and 17m. lfur. 8ch. 2p. 191. Ans. 160m. Ofur. lch. lp. 121. 15. Required the sum of 27m. 4fur. 3ch. lp. 211.; 29m. 3fur. lch. 3p. 231.; 67m. 3fur. 3ch. lp. 191.; 21m. 7fur. lch. 3p. 161.; 16m. 7fur. 9ch. 3p. 131.; and 31m. 4fur. 8ch. lp. 201. 16. Required the value of 29T. 36ft. 1279in. + 69T. 19ft. 1345in. + 67T. 18ft. 1099in. + 71T. 14ft. 1727in. + 43T. 35ft. 916in. + 53T. 17ft. 1719m. Ans. 335T. 23ft. 1173in. 17. Add together 61C. 127ft. 1161in.; 37C. 89ft. 1711in.; 61C. 98ft. 1336in.; 43C. 56ft. 1678m. 91T. 119ft. 1357in. and 81C. 115ft. 1129in. 18. Required the value of 61 tuns 3hhd. 62gal. 3qt. lpt. -[- 39 tuns 2hhd. 16gal. lqt. lpt. + 68 tuns 3hhd.57gal. 2qt. lpt. -f- 87 tuns 3hhd. 45gal. 3qt. lpt. -f- 47 tuns 2hhd. 59gal. 3qt. lpt. + 47 tuns 3hhd. 39gal. 2qt. lpt. Ans. 354 tuns Ohhd. 30gal. lqt. 19. Required the value of 67hhd. 15gal. 3qt. lpt. -f 16hhd. 16gal. 3qt. + 39hhd. 16gal. 3qt. -f 47hhd. 62gal. lqt. lpt. -f 43hhd. 57gal. 3qt. + 71hhd. 61gal. 3qt. lpt. 20. Add together of beer measure 161hhd. 53gal. 3qt. lpt. ; 371hhd. 52gal. 3qt. lpt.; 98hhd. 19gal. lqt.; 47hhd. 43gal. lqt. Opt.; 61hhd. 43gal. lqt. lpt.; and 42hhd. 27gal. 3qt. lpt. 21. Find the amount of 37bu. 3pk. 5qt. lpt. -|- 61bu. 2pk. 7qt. lpt. -f- 32bu. 3pk. 2qt. + 71bu. lpk. 6qt. lpt. + 61bu. lpk. 3qt. lpt. -j- 32bu. 3pk. 3qt. lpt. Ans. 298bu. Opk. 4qt. lpt. 22. Required the value of 31bu. 3pk. 3qt. + 31bu. 3pk. lqt. + 16bu. 3pk. lqt. + 15bu. 3pk. -f 17bu. 3pk. lqt. + 14bu. 3pk. lqt. 23. Required the value of 57y. llmo. 27d. 23h. 29m. 55s. + 31y. llmo. 18d. 19h. 19m. 39s. + 46y. 9mo. 23d. 17h. 28m. 56s. + 43y. lOmo. 16d. 18h. 17m. 48s. + 32y. 9mo. 19d. 16h. 23m. 28s. + 14y. lmo. 29d. 21h. 28m. 16s. Ans. 227y. 7mo. 16d. 21h. 28m. 2s. 24. Required the value of 23w. 6d. 23h. 59m. 58s. -f 51w. 3d. 18h. 51m. 17s. + 29w. 5d. 21h. 47m. 49s. + 28w. 4d. ADDITION OF COMPOUND NUMBERS. 115 23h. 56m. 18s. + 19w. 6d. lOh. 18m. 53s. -f 86w. Id. 20h. 40m. 51s. 25. Add together 4S. 29° 59' 59"; 6S. 17° 17' 29"; US. 16° W 58"; 9S. 13° 46' 51"; 5S. 27° 16' 42"; and 2S. 25° 11' 17". Ans. 5S. 10° 35' 16". 26. Add together US. 11° 16' 51"; 6S. 6° 6' 16"; 9S. 14° 56' 56"; 3S. 29° 29' 49"; 9S. 17° 18' 58"; and 6S. 13° 13' 52". Note. — We divide the sum of the signs, in the last two questions, by 12, and write down the remainder only. Since the circumference of a circle can- not exceed 12 signs (Art. 143). 27. Bought of a London tailor a vest for 1£. 13s. 4d., a coat for 7£. 12s. 9d., pantaloons for 2£. 3s. 9d., and surtout for 9£. 8s. Od. ; what was the whole amount ? Ans. 20£. 17s. lOd. 28. Bought a silver tankard, weighing lib. 8oz. 17pwt. 14gr., a silver can, weighing lib. 2oz. 12pwt, a porringer, weighing lloz. 19pwt. 20gr. ; and 3 dozen of spoons, weighing lib. 9oz. 15pwt. lOgr. ; what was the whole weight? Ans. 51b. 9oz. 4pwt. 20 gr. 29. What is the weight of a mixture of 31b 4§ 25 23 14gr. of aloe, 2tb 7§ 65 19 13gr. of picra, and lit) 105 15 23 17gr. of saffron? Ans. 71b 10 § 35 19 4 gr. 30. Sold 4 loads of hay ; the first weighed 27cwt. 3qr. 181b. ; the second, 31cwt. lqr. 151b. ; the third, 19cwt. lqr. 151b. ; and the fourth, 38cwt. 2qr. 241b. ; what is the weight of the whole ? 31. Bought 5 pieces of broadcloth; the first contained 17yd. 3qr. 2na. ; second, 13yd. 2qr. lna. ; the third, 87yd. lqr. 3na. ; the fourth, 27yd. lqr. 2na.; and the fifth, 29yd. lqr. 2na. ; what was the whole quantity purchased ? Ans. 175yd. 2qr. 2na. 32. A pedestrian travelled, the first week, 371m. 3fur. 37rd. 5yd. 2ft. 10in.; the second week, 289m. 2fur. 18rd. 3yd. 1ft. 9in. ; and the third week, 399m. 7fur. 3ft. llin. ; how many miles did he travel? Ans. 1060m. 5fur. 16rd. 5yd. 1ft. 33. A man has three farms; the first contains 186A. 3R. 14p. ; the second, 286A. 17p.; and the third, 115 A. 2R. ; how much do they all contain? Ans. 588 A. 1R. 31p. 116 SUBTRACTION OF COMPOUND NUMBERS. 34. The Moon is 5S. 18° 14' 17" east of the Sun; Jupiter is 7S. 10° 29' 28" east of the Moon; Mars is US. 12° ll 7 56" east of Jupiter; and Herschel is 78. 18° 38' 15" east of Mars ; how far is Herschel east from the Sun ? Ans. 7S. 29° 33' 56". OPERATION. £ 8. d. Min. 617 11 8 Sub. 181 15 5 SUBTRACTION OF COMPOUND NUMBERS. 146. Subtraction of Compound Numbers is the process of finding the difference between two compound numbers. Ex. 1. From 617£. lis. 8d. take 181£. 15s. 5d. Ans. 435£. 16s. 3d. Having placed the less number un- der the greater, pence under pence, shillings under shillings, &c, we be- gin with pence, thus: 5d. from 8d. "R A o c T~7 o~ leaves 3d., which we set under the Kem. 4 o o lb o co lumn of pence. As we cannot take 15s. from lis., we add 20s. = l£. to the lis., making 31s., and then subtract the 15s. from it, and set the remainder, 16s., under the column of shillings. Then, having added l£. = 20s. to the 18l£., to compensate for the 20s. added to the lis. in the minuend, we subtract the pounds as in subtraction of simple numbers, and obtain 435£. for the remainder, and as the result com- plete, 435£. 16s. 3d. Rule. — Write the less compound number under the greater, so that units of the same denomination shall stand in the same column. Subtract as in subtraction of simple numbers. If any number in the subtrahend is larger than that above it, add to the upper number as many units as make one of the next higher denomi- nation before subtracting, and carry one to the next lower number before subtracting it. Proof. — The proof is the same as in simple subtraction. Examples. £ s. d. 87 16 3J 19 17 9 i £ s. d. 617 1 1 H 181 15 8* 67 18 5f SUBTRACTION OF COMPOUND NUMBERS. 117 4. 5. T. cwt. qr. lb. oz. dr. cwt. qr. lb. oz. 71 18 1 13 1 13 73 1 15 13 19192168 5 19119 15 51 : L8 2 21 6. 9 8 7. lb. 71 oz. pwt. 3 12 gr. 15 lb. 58 oz. pwt, 5 12 10 - 16 10 17 20 19 9 17 21 54 4 14 8. 19 9. lb 71 § 3 3 1 3 1 gr. 14 lb 15 §39 2 2 gr- 15 18 6 7 2 19 9 9 1 1 18 52 6 3 1 10. 15 11. m. fur. 16 7 rd. ft. 18 3 in. 1 deg. 38 m. 41 fur. rd. 3 29 rd. ft. 2 1 in. 7 9 7 19 16 8 29 36 5 31 3 1 9 6 7 38 2| 5 6 7 38 2 11 Note. — As half a foot is equal to 6 inches, we add them to the 5 inches, which make 11 inches. 12. From 67yd. lqr. lna. lin. take 18yd. 2qr. 2na. 2in. Ans. 48yd. 2qr. 2na. l£in. 13. From 51E.E. 2qr. 3na. take 19E.E. 3qr. lna. 14. From 56A. 1R. 19p. 119ft. HOin. take 17A. 3R. 13p. 127ft. 113in. Ans. 38A. 2R. 5p. 264ft. 33in. 15. Find the value of 13A. 1R. 15p. 19yd. 1ft. 17in. — 9A. 3R. 16p. 30yd. 5ft. 17in. 16. Subtract 19m. 2fur. lch. 3p. 211. from 21m. lfur. 3ch. 2p. 191. Ans. lm. 7fur. lch. 2p. 231. 17. From 28m. 6fur. lch. 2p. 181. take 15m. 7fur. 3ch. lp. 191. 18. Required the value of 49T. 13ft. 1611in. — 18T. 15ft. 1719in. Ans. 30T. 37ft. 1620in. 118 SUBTRACTION OF COMPOUND NUMBERS. 19. Required the value of 361C. 47ft. 1178in. — 197C. 121ft. 1617in. 20. Subtract lltun lhhd. 28gal. 2qt. lpt. from 79tun 3hhd. 19gal. lqt. lpt. Ans. 68tun lhhd. 53 gal. 3qt. 21. Subtract of beer measure 191hhd. 19gal. 3qt. from 769hhd. 18gal. lqt. 22. From 56ch. 2bu. lpk. take 38ch. 3bu. lpk. Ans. 17ch. 35bu. 23. Required the value of 25bu. 3pk. lqt. — 12bu. 3pk. 5qt. Ans. 12bu. 3pk. 4qt. 24. Required the difference between 6mo. 16d. 13h. 27m. 19s. and lmo. 22d. 16h. 41m. 37s. Ans. 4mo. 23d. 20h. 45m. 42s. 25. From 48y. Omo. 15d. 19h. 27m. 31s. take 19y. lOmo. 29d. 21h. 38m. 56s. 26. From 6S. 11° 12' 48" subtract 9S. 8° 15' 56". Ans. 9S. 2° 56' 52". 27. Take IS. 22° 19' 28" from 4S. 19° 41' 22". 28. I have 73A. of land ; if I should sell 5A. 3R. lp. 7ft., how much should I have left ? Ans. 67A. OR. 38p. 265Jft. 29. A owes B 100£. ; what will remain due after he has paid him 3s. 6£d. ? Ans. 99£. 16s. 5£d. 30. It is about 25,000 miles round the globe ; if a man shall have travelled 43m. 17rd. 9in., how much will remain to be travelled? Ans. 24,956m. 7fur. 22rd. 15ft. 9in. 31. Bought 7 cords of wood; 2 cords 78ft. having been stolen, how much remained ? 32. I have 15 yards of cloth ; having sold 3yd. 2qr. lna., what remains? Ans. 11yd. lqr. 3na. 33. If a wagon loaded with hay weighs 43cwt. 2qr. 181b., and the wagon is afterwards found to weigh 9cwt. 3qr. 231b., what is the weight of the hay ? Ans. 33cwt. 2qr. 201b. 34. Bought a hogshead of wine, and by an accident 8gal. 3qt. lpt. leaked out ; what remains ? 35. I had 10A. 3R. lOp. of land ; and I have sold two house- lots, one containing 1A. 2R. 13p., the other 2 A. 2R. op.; how much have I remaining ? Ans. 6 A. 2R. 32p. 36. The Moon moves 13° 10' 35" in a solar day, and the Sun 59' 8" ; now supposing them both to start from the MULTIPLICATION OP COMPOUND NUMBERS. 119 same point in the heavens, how far will the Moon have gained on the Sun in 24 hours ? Ans. 12° 11' 27". 37. A farmer raised 136bu. of wheat; if he sells 49bu. 2pk. 7qt. lpt., how much has he remaining ? Ans. 86bu. lpk. Oqt. lpt. 38. If from a stick of timber containing 2T. 18ft. 1410in. there be taken 38ft. 1720in., how much will be left ? Ans. IT. 19ft. 1418in. MULTIPLICATION OF COMPOUND NUMBERS. 147. Multiplication of Compound Numbers is the pro- cess of taking a compound number any proposed number of times. Ex. 1. "What will 6 bales of cloth cost, at 7£. 12s. 7d. per bale ? Ans. 45£. 15s. 6d. operation. Having written the multiplier under * s ' Jjj the lowest denomination of the multi- Multiplicand 7 12 7 pii can( i, we multiply thus : 7d. X 6 Multiplier 6 = 42d. = 3s. 6d. We write the 6d. Product 4~5 3~5 6 un der the number multiplied, and re- serve the 3s. to be added to the pro- duct of the shillings. Then, 12s. X 6 = 72s., and 3s. (carried) = 75s. = 3£. 15s. We write the 15s. under the column of shillings, and reserve the 3£. to be added to the product of the pounds. Again, 7£. X 6 = 42£., and 3£. (carried) = 4 5£. This, placed under the column of pounds, gives 45£. 15s. 6d. Rule. — Multiply each denomination of the compound number as in multiplication of simple numbers, and carry as in addition of compound numbers. Proof — Write down by themselves the several products obtained by multiplying each denomination of the multiplicand by the multiplier, and these partial products added together will equal the entire product, if the work be right. (Art. 60.) Note. — Going a second time carefully over the work is a good way of testing its accuracy. On learning Division of Compound Numbers, the pupil will find that rule a better method of proving multiplication of compound numbers. 120 MULTIPLICATION OF COMPOUND NUMBERS. Examples. 2. Multiply 1£. 8s. 7d. 2far. by 7. Ans. 10£. 0s. 4d. 2far. PROOF BY ADDITION. 9£d. x 8 = 76d. = 0£. 6s. 4d. 8s. x 8 = 64s. = 3£. 4s. Od. Ans. 19 10 4 2£. x 8 = 16£. = 16£. 0s. Od. 2£. 8s. 9id. x 8 = 19£. 10s. 4d. Note. — The answers to the following examples may be found in corre- sponding numbers of examples in Division of Compound Numbers. 3. 4. T. cwt qr. lb. oz. dr. lb. oz. pwt. gr. Gl 19 3 17 15 15 7 11 14 15 9 5 OPERATION £. 8. d. 2 8 n 8 5. 6. lb. oz. pwt. gr. Jb 5 5 3 gr- 32 8 17 12 38 10 5 2 14 8 11 7. 8. deg. m. fur. rd. ft. in. m. fur. ch. p. 1. 71 38 2 13 14 4 17 7 9 3 23 12 12 9. Multiply 16A. 2R. 4p. 19yd. 7ft. 79in. by 11. 10. Multiply 10yd. 3qr. 3na. by 5. 11. Multiply 17tun 2hhd. 50gal. lqt. by 7. 12. Multiply 29hhd. 61gal. 3qt. lpt. 3gi. by 7. 13. Multiply 19bu. 2pk. 7qt. lpt. by 6. 14. What is the value of 13y. 31 6d. loh. 27m. 39s. X 8 ? 15. Multiply 16deg. 39m. 3fur. 39rd. 5yd. 2ft. by 9. 16. If a man gives each of his 9 sons 23A. 3R. 19^p., what do they all receive ? 17. If 12 men perform a piece of labor in 7h. 24m. 30s., how long would it take 1 man to perform the same task ? 18. If 1 bag contain 3bu. 2pk. 4qt., what quantity do 8 bags contain ? OPERATION. T. cwt. qr. 2 12 lb. 7 weight of 1 load. 14 10 3 17 = 5 weight of 7 loads. MULTIPLICATION OP COMPOUND NUMBEKS. 121 148. When the multiplier is a composite number, and none of its factors exceed 12. Ex. 1. What will 35 loads of coal weigh, if each load weighs 2T. lcwt. 2qr. 61b. ? We find the num- ber 35 equal to the product of 7 and 5 ; we therefore multiply the weight of 1 load by 7, and then that product by 5 ; and the 72 14 2 10 = weight of 35 loads, last product is the an- swer. Hence, when the multiplier is a composite number, Multiply by its factors in succession. Examples. 2. Bought 90 hogsheads of sugar, each weighing 12cwt. 2qr. 111b. ; what was the weight of the whole ? 3. What cost 18 sheep at 5s. 9Jd. apiece ? 4. What cost 21 yards of. cloth at 9s. lid. per yard? 5. What cost 22 hats at lis. 6d. each ? 6. If 1 share in a certain stock be valued at 13£. 8s. 9^-d., what is the value of 96 shares ? 7. If 1 spoon weighs 3oz. 5pwt. 15gr., what is the weight of 120 spoons ? 8. If a man travel 24m. 7fur. 4rd. in 1 day, how far will he go in 1 month ? 9. If the earth revolve 0° 15' per minute, how far does it revolve per hour ? 10. Multiply 39A. 3R. 17p. 30yd. 8ft. lOOin. by 32. 11. If a man be 2d. 5h. 17m. 19sec. in walking 1 degree, how long would it take him to walk round the earth, allowing 3 65 J days to a year ? 149. When the multiplier is not a composite number, and exceeds 12 ; or when a composite number one of whose factors exceeds 12. Ex. 1. What is the value of 453 tons of iron at 18£. 17s. lid. a ton? N 11 122 MULTIPLICATION OP COMPOUND NUMBERS. OPERATION. £. s. d. £. s. d. lton= 18 17 1 1, X 3 = 5613 9 = Value of 3 tons. 10 10 tons = 188 19 2,XO= 9 44 15 1 = Value of 50 tons. H) 100tons=1889 11 8,X4=7558 6 8 — Value of 400 tons. Ans. 8559 16 3= Value of 453 tons. Since 453 is not a composite number, we cannot resolve it into factors ; but we may separate it into parts, and find the value of each part separately: Thus, 453 = 400 -{- 50 -|- 3. In the operation, we first multiply by 10, and obtain the value of 10 tons, and this product we multiply by 10, and obtain the value of 100 tons. Then, to find the value of 400 tons, we multiply the last product by 4 ; and to find the value of 50 tons, we multiply the value of 10 tons by 5 ; and to find the value of 3 tons, we multiply the value of 1 ton by 3. Add- ing the several products, we obtain 8559£. 16s. 3d. for the answer. Hence, Having resolved the multiplier into any convenient parts, as of units, tens, §*c, multiply by these several parts, and add together the products thus obtained for the required result. Examples. 2. Multiply 2hhd. 19gal. Oqt. lpt. by 39. 3. Multiply 3bu. lpk. 4qt. lpt. lgi. by 53. 4. Multiply 16ch. 7bu. 2pk. Oqt. Opt. by 17. 5. What will 57 gallons of wine cost at 8s. 3£d. per gallon ? 6. Bought 29 lots of wild land, each containing 117 A. 3R. 27p. ; what were the contents of the whole ? 7. Bought 89 pieces of cloth, each containing 37yd. 3qr. 2na. 2in. ; what was the whole quantity ? 8. Bought 59 casks of wine, each containing 47gal. 3qt. lpt. ; what was the whole quantity ? 9. If a man travel 17m. 3fur. 13rd. 14ft. in one day, how far will he travel in a year ? 10. If a man drink 3gal. lqt. lpt. of beer in a week, how much will he drink in 52 weeks ? 11. There are 17 sticks of timber, each containing 37ft. 978in. ; what is the whole quantity ? 12. There are 17 piles of x wood, each containing 7 cords 98 cubic feet ; what is the whole quantity ? DIVISION OP COMPOUND NUMBERS. 123 DIVISION OF COMPOUND NUMBERS. 150. Division of Compound Numbers is the process of dividing compound numbers into any proposed number of equal parts. Ex. 1. Divide 139£. 13s. lid. 2far. equally between 5 per- sons. Ans. 27£. 18s. 9d. 2far. operation. Having divided 139£. by 5, we find £. s. d. far. the quotient to be 27£., and 4£. re- 5)139 13 11 2 maining. We place the quotient 27£. — under the 139£., and the remainder 2 7 18 9 2 4£. reduced to shillings = 80s. ; 80s. -f- the 13s. in the dividend = 93s. ; 93s. -5- 5 = 18s. and a remainder of 3s. We write the quotient 18s. under the shillings in the dividend; and the remainder 3s. re- duced to pence = 36d. ; 36d. -f- lid. in the dividend = 4 7d. ; 47d. -5- 5 — 9d. and a remainder of 2d. We write the quotient 9d. under the pence in the dividend ; and the remainder 2d. reduced to farthings = 8far., -\- the 2far. in the dividend = lOfar. ; lOfar. -h 5 = 2far. The quotient 2far. we write under the farthings in the dividend; and thus find the answer to be 27£. 18s. 9d. 2far. Rule. — Divide as in division of simple numbers, each denomination in its order, beginning with the highest. If there be a remainder, reduce it to the next lower denomination, adding in the number already contained in the dividend of this de- nomination, if any, and divide as before. Proof. — The same as in simple numbers. Note. — When the divisor and dividend are both compound numbers, they must be reduced to the same denomination, and the division then is that of simple numbers. Examples. Note. — The answers to the following examples are found in the corre- sponding numbers of examples in Multiplication of Compound Numbers. 5) 2. £. s. d. 8)19 10 4 T. cwt. 9)557 19 3. qr. lb. oz. 11115 dr. 7 4. lb. oz. pwt. gr. ) 39 10 13 3 lb. 8)261 5. oz. pwt. gr. 110 9. A. R. p. yd. ft. in. 181 3 11 6 4 41 124 DIVISION OP COMPOUND NUMBERS. 6. . 7. ft) <5 5 3 gr. deg* m- fur - rd « ft * in - 11)427 10 2 14 12)858 44 4 6 7 8. m. fur. ch. p. 1. 12) 215 7 9 3 1 1 10. Divide 54yd. 2qr. 3na. equally among 5 persons. 11. Divide 123tun 3hhd. 36gal. 3qt. by 7. 12. Divide 209hhd. 55gal. 3qt. Opt. lgi. by 7. 13. What is the value of 118bu. lpk. 5qt. -r- 6 ? 14. What is the value of HOy. 343d. 3h. 41m. 12s. — 8 ? 15. Divide 149deg. 9m. 5fur. 13rd. 3yd. 1ft. by 9. 16. A man divides his farm of 214A. 3R. 12p. equally among his 9 sons ; how much does each receive ? 17. If one man perform a certain piece of labor in 3d. 16h. 54m., how long would it take 12 men to perform the same work ? 18. A farmer has 29 bushels of oats which he wishes to put in 8 sacks ; how much must each sack contain ? 151 , When the divisor is a composite number, and none of its factors exceed 12. Ex. 1. If 35 loads of coal weigh 72T. 14cwt. 2qr. 101b., what will 1 load weigh ? operation. We find the fac- T. cwt. qr. lb. tors f 35 to be 5 5)72 14 2 10 = weight of 35 loads, and 7. We there- 7 ) 14 10 3 17 = weight of 7 loads. ^ ** ^ 2 12 6 = weight of 1 load. by 5, and obtain the weight of 7 loads ; and the weight of 7 loads we divide by 7, and thus find the weight of 1 load. Hence, when the divisor is a composite number, Divide hy its factors in succession. Examples. 2. If 90 hogsheads of sugar weigh 56T. 14cwt. 3qr. 151b., what is the weight of 1 hogshead ? 3. What will be the price of 1 sheep, if 18 cost 5£. 4s. 3d. ? DIVISION OF COMPOUND NUMBERS. 125 4. If 21 yards of cloth cost 10£. 8s. 3d., what is the price of 1 yard ? o. What is the value of 1 hat, when 22 cost 12£. 13s. Od. ? 6. When 96 shares of a certain stock are valued at 1290£. 4s. Od., what would be the cost of 1 share ? 7. If 120 spoons weigh 321b. 9oz. 15pwt., what does 1 weigh ? 8. If a man in 1 month travels 746m. 5fur., how far does he go in 1 day ? 9. If the earth revolves 15° on its axis in 1 hour, how far does it revolve in 1 minute ? 10. Divide 1275A. 2E. 16p. 22yd. 8ft. 32in. equally among 32 men. 11. If a man walk round the earth in 2y. 68d. 19h. 54m., how long would it take him to walk 1 degree, allowing 3 65 J days to a year ? 152» When the divisor is not a composite number, and exceeds 12, or when a composite number one of whose factors exceeds 12, the whole operation can he written, as in the follow- ing example. Ex. 1. Divide 360£. 8s. 4d. by 173. Ans. 2£. Is. 8d. OPERATION. £. S. d. 17 3)360 8 4 ( 2£. We divide the pounds by 173, and 3 4 6 obtain 2£. for the quotient, and 14£. remaining, which we reduce to shil- 1 4 lings, and add the 8s., and again divide 2 by 173, and obtain Is. for the quotient. i 7 o \ 9 o o / i The remainder, 115d., we reduce to U^J^««(is. pence, and add the 4d., and again di- 1 * 3 vide by 173, and obtain 8d. for the \\§ quotient. Thus, the method is the ^ 2 same as by general rule (Art. 150). By writing the several quotients, we 173)1384(8d. obtain 2£. Is. 8d. for the answer. 1384 2. Divide 89hhd. 52gal. 3qt. lpt. by 39. 3. Divide 179bu. 3pk. 5qt. Opt. lgi. by 53. 4. Divide 275ch. 19bu. 2pk. equally among 17 men. 11* 126 PRINCIPLES AND APPLICATIONS. 5. If 57 gallons of wine cost 23£. lis. 5£d., what cost 1 gallon ? 6. Divide 3419A. 2R. 23p. by 29. 7. If 89 pieces of cloth contain 3375yd. 3qr. lna. 0£in., how much does 1 piece contain ? 8. If 59 casks contain 44hhd. 52gal. 2qt. lpt. of wine, what are the contents of 1 cask ? 9. If a man travel in 1 year (365 days) 6357m. 5fur. 14rd. 11^-ft., how far is that per day ? 10. When 175gal. 2qt. of beer are drunk in 52 weeks, how much is consumed in 1 week ? 11. When 17 sticks of timber measure 15T. 38ft. 1074in., how many feet does 1 contain ? 12. Divide 132 cords 2ft. by 17. 13. Divide 697T. 18cwt. 3qr. 141b. by 146. Ans. 4T. 15cwt. 2qr. 10-J-fflb. 14. Divide 916m. 3fur. 30rd. 10ft. 6in. by 47. Ans. 19m. 3fur. 39rd. 13ft. 2|fin. 15. Divide 718A. 3R. 37p. by 29. Ans. 24A. 3R. 6Jfp. 16. Divide 815A. 1R. 17p. 200ft. by 87. Ans. 9A. 1R. 19p. 139|fft. 17. Divide 144A. 3R. 18p. 3yd. 1ft. 36in. by 11. 4 Ans. 13A. OR. 27p. 3yd. Oft. 45 T 9 T in. PRINCIPLES AND APPLICATIONS. DIFFERENCE BETWEEN DATES. 153. To find the time between two different dates. Ex. 1. What is the difference of time between May 16, 1819, and March 4, 1857 ? Ans. 37y. 9mo. 18d. Commencing with January, first operation. t h e fi rst mon th in the year, and y - m0 * d - counting the months and days in Min. 18 5 7 2 4 the later date up to March 4th, Sub. 1819 4 16 we find that 2mo. and 4d. have elapsed. We therefore write the Rem. 3 7 9 18 numbers for subtraction as in the first operation. SECOND OPERATION. Min. 18 5 7 3 Sub. 1819 5 4 16 Rem. 3 7 9 18 PRINCIPLES AND APPLICATIONS. 127 The same result, however, could be obtained, as some pre- fer, by reckoning the number of the given months instead of the number of months that have elapsed since the beginning of the year, which would require the numbers to be written as in the second operation. Written either way, the earlier date, being placed under the later, is subtracted from it. Note. — In finding the difference between two dates, and in computing in- terest for less than a month, 30 days are considered a month. In legal trans- actions, however, a month is reckoned from any day in one month to the same day of the following month. The above process, which is that ordinarily used by business men, does not give always the exact time between two different dates. The result obtained by it may deviate sometimes a day, and, less often, two days, from the exact difference. But for practical purposes it is generally regarded as sufficiently accurate. Examples. 2. What is the time from June 3d, 1854, to April 19th, 1857? 3. A note was given October 26th, 1856, and paid June 12th, 1857 ; how long was it on interest ? 4. The Pilgrims landed at Plymouth December 22d, 1620, N. S., and the Declaration of Independence was made July 4th, 1776 ; what is the difference of time between these events? 5. General Washington was born February 22d, 1732, and died December 14th, 1799 ; how long did he live? Ans. 67y. 9mo. 22d. 154. To find the exact number of days between two differ- ent dates. Ex. 1. How many days from January 28 to July 30, com- mon year? Ans. 183 days. OPERATION. January to July = 6mo. 6 X 31 =186 days. For Feb. 3d., April Id., June Id., 3 + 1 + 1= 5 " Tsi For difference between 30 and 28, 30 — 28 = 2 Ans. 18 3 days. The difference between January and July we find to be 6 mo., which, multiplied by 31, the greatest number of days in any month in the year, gives 186 days. But since in the interval of time in- cluded between the given dates several months end that do not con- 128 PRINCIPLES AND APPLICATIONS. tain 31 days, we make deductions for these, which, in all, amount to 5 days, and have left 181 days ; and as the difference between the given dates is the difference between 30 and 28 more than exactly 6mo., we add 2 to the 281 days, thus obtaining 183 days, the differ- ence of time required. Hence, to find the number of days between two different dates, Find the number of months ending between the given dates, and mul- tiply that number by 31, and from the product make the necessary de- duction for the months counted that do not contain 31 days, if any. Should the later date end later in the month than the earlier, add the difference of days ; but should it end earlier, subtract the same. Note. — The exact difference in days between two different dates can also be obtained by use of the table in Note 2, Art. 142. Examples. 2. How many days has a note to run dated November 15, 1856, and made payable February 13, 1857 ? Ans. 90 days. 3. How many days from June 18, 1855, to May 1, 1856 ? 4. How many days from March 4 to May 3 of the same year? 5. From November 4, 1856, to April 4, 1857, how many days? Ans. 151 days. 6. In a leap year, how many days are there from the 7th of January to the 11th of December ? Ans. 339 tlays. 155. To find the day of the week corresponding to any given day of the month, when the day of the week of some other date is given. Ex. 1. If the 16th day of May be on Saturday, what day of the week will the next 25th of December be ? Ans. Friday. OPERATION. From May 16 to December 25 = 223 days. 223 days -r- 7 = 31 weeks and 6 days. 6 days after Saturday = Friday, Ans. Having found the difference of time in days between the given dates, we bring the days to weeks by dividing by 7, and obtain 31 weeks and 6 days. The 25th of December, therefore, must come 6 days after Saturday, or on Friday. Hence, we Reduce the days between the given dates to weeks. Should there be no remainder, the day given will be the same as that sought, but should there be a remainder, it will indicate the number of days that the day sought is after the day given. PRINCIPLES AND APPLICATIONS. 129 Examples. 2. If the 2d day of April be on Wednesday, what day of the week will the following 4th of July be ? Ans. Friday. 3. If a leap year commence on Tuesday, on what day will the 17th of June, the anniversary of the battle of Bunker Hill, happen ? 4. If in a common year the 25th of December, or Christmas, be on Tuesday, on what day did the year commence ? Ans. Monday. 5. If the 4th day of November be on Tuesday, what day of the next February will be the second Monday of that month? Ans. The 9th. 6. A bill was dated on Thursday, December 20th, 1855, and made payable 90 days after date. In what year and month, and on what day of the month and week, did it become due? Ans. Wednesday, March 19, 1856. DEFFEKENCE OF LATITUDE. 156. Latitude is the distance of any place from the equator, north or south. It is reckoned in degrees, minutes, and sec- onds, from the equator to either pole of the earth ; and cannot exceed 90 degrees, or one fourth of the earth's circumference. Places north of the equator are said to be in north latitude, and those south of the equator, in south latitude. Note. — The shape of the earth not being that of a perfect sphere, but some- what flattened toward the poles, the degrees of latitude differ a little from each other in length toward either pole. Thus the 1st degree is about 68^ statute miles in length; the 40th degree, about 68jq 8 q miles; and the 89th degree about 69^ miles. 157* To find the difference of latitude of any two places. Ex. 1. The latitude of London is 51° SV north, and that of Boston is 42° 23' north. What is their difference of lati- tude ? Ans. 9° 8'. operation. The two places being Lat. of London = 5 1° 3 V N. both in north latitude, we Lat. of Boston =4 2° 2 3' N. ^tract the less latitude from the greater, it the Dif. of Latitude = 9° 8' N. latitude of the one place had been north and that of 130 PRINCIPLES AND APPLICATIONS. the other south, we should have added the two latitudes together for the difference required. Hence When the latitudes of two places are either both north or both south, subtract the less latitude from the greater for their difference; and when the latitude of the one place is north and the other south, add the latitudes together for their difference. Note. — In north latitude, when the sailing is southerly, or in south latitude, when the sailing is northerly, if the difference of latitude be subtracted from the latitude left, the remainder will be the latitude in ; or, in north latitude, when the sailing is northerly, or in south latitude, when the sailing is southerly, if the difference of latitude be added to the latitude left, the sum will give the latitude in. Examples. 2. The latitude of Quebec is 46° 48' north, and that of New Orleans 29° 57' north. What is their difference of lati- tude ? Ans. 16° 51'. 3. The latitude of Washington City is 38° 53' north, and that of Cape Horn 55° 58' south. What is their difference of lati- tude ? Ans. 94° 51'. 4. Valparaiso is in latitude 33° 2' south, and San Francisco 37° 48' north. What is their difference of latitude ? 5. Captain James Francis, sailing southerly from New York City, whose latitude is 40° 42' north, found on reaching Havana that his latitude differed 17° 33' from that he left. What lati- tude was he then in ? Ans. 23° 9' north. 6. Philadelphia is 9° 15' of latitude north of Mobile, whose latitude is 30° 41' north ; what is the latitude of Phila- delphia ? Ans. 39° 56' north. DIFFERENCE OF LONGITUDE. 158. Longitude is the distance of any place from a given meridian, east or west. It is reckoned in degrees, minutes, and seconds, and cannot exceed 180 degrees, or one half of a circle. Note 1. — A degree of longitude on the equator is about 69^ statute miles, and, in general, a degree is 5 J C of any circle of latitude. The meridians all con- verge from the equator to the poles to a point, so that the degrees of longi- tude under different parallels of latitude vary, diminishing with the circles of latitude, till at the poles the longitude becomes nothing. (Art. 133, Note 4.) PRINCIPLES AND APPLICATIONS. 131 Note 2. — The mariners of Great Britain and the United States reckon longitude from the meridian of Greenwich in England, as do the nautical books of both countries for the most part. American maps, however, very generally have longitude reckoned both from the meridian of Greenwich, and from the meridian of Washington. 159. To find the difference of longitude of any two places. Ex. 1. What is the difference of longitude between Boston, which is 71° 4' west, and Detroit, which is 82° 58' west. Ans. 11° 54'. operation. The two places being Long, of Detroit = 8 2° 5 8' both in west longitude, we Long, of Boston =7 1° 4' subtract the less longitude ° from the greater. If, how- Dif. of Longitude =11° $4? ever, one of the places had been in east longitude, and the other in west, we should have added the two longitudes together for the difference required. Hence, When the longitudes of two places are either both east or both west, sub- tract the less longitude from the greater for their difference ; and when the longitude of the one place is east, and that of the other west, add the longitudes together for their difference. Note. — In adding together two longitudes, should their sum exceed 180 degrees, subtract it from 360 degrees, and the remainder will be the correct difference of longitude. Examples. 2. What is the difference of longitude between the city of Washington, whose longitude is 77° 16-' west, and Paris, whose longitude is 2° 20' east? Ans. 79° 36'. 3. The United States extend from the St. Croix River, longitude 67° 2' west, to Cape Flattery, longitude 124° 43' west. Over how many degrees of longitude does the Union extend ? 4. What is the difference of longitude between Raleigh, whose longitude is 78° 48' west, and Sacramento City, whose longitude is 120° west ? Ans. 41° 12'. 5. What is the difference of longitude between Hartford, whose longitude is 72° 40' west, and Fort Leavenworth whose longitude is 94° 44' west ? 6. The longitude of Honolulu is 157° 52' west, and that of Canton 113° 14' east. What is their difference of longi- tude ? Ans. 88° 54'. 132 PRINCIPLES AND APPLICATIONS. LONGITUDE AM) TIME. 160» To find the time corresponding to degrees and min- utes of longitude. Ex. 1. The difference of longitude between Boston and London being 71° 4', what is their difference of time ? Ans. 4h. 44m. 16sec. operation. Si ncc |« f longitude Dif. of longitude = 71° 4' corresponds to 4sec. of 4 time, and 1° of longitude tw. r ± - ^ 7* to 4m. of time (Art. 143, Dif. of time = 284m. 16sec. Note 2)> 4 , co v rre sponds 284m. 16sec. = 4h. 44m. 16sec. Ans. to 16sec, and 71° to 284m.; and 284m. 16sec. = 4h. 44m. 16sec., the answer required. The apparent motion of the sun being west, the time at Boston is as much earlier than that of London as the difference of time between them. Thus, when it is noon at London, it wants 4h. 44m. 16sec. of noon at Boston (Art. 143, Note 2). Therefore, to find the time corresponding to degrees and minutes of longitude, Multiply the minutes of longitude by 4, for seconds of time. Multiply the degrees of longitude by 4, for minutes of time. Note 1. — Should the seconds of time, when found, be 60 or more, they may- be reduced to minutes; and should, also, the minutes be 60 or more, they may be reduced to hours. The difference of time between two places is exactly as much as the true clock time of the one is fast or slow as compared with that of the other. Examples. 2. Galveston in Texas is 14° 43' west of Pittsburg. When it is 12 o'clock at Galveston, what is the time at Pittsburg ? 3. The longitude of Valparaiso is 71° 37' west, and the lorn gitude of Rome is 20° 30' east. When it is llh. 15m. A. M, at Valparaiso, what is the time at Rome ? Ans. 23m. 28sec. past 5 o'clock, P. M. 4. The longitude of Jerusalem is 35° 32' east, and the Ion, gitude of Baltimore 76° 37' west. When it is 9 o'clock, A.M, at Jerusalem, what time is it at Baltimore ? Ans. lh. 31m. 24sec. A. M. 161 • To find the longitude corresponding to hours, minutes^ and seconds of time. PRINCIPLES AND APPLICATIONS. 133 Ex. 1. The difference of time between Boston and London is 4h. 44m. 16sec. ; what is the difference of longitude ? Ans. 71° 4'. operation. Since lh. of time corresponds to 1 5° X 4 = 6 0° 15o of longitude, 4m. of time to 1° 4 4-7-4=11° of longitude, and 4sec. of time to 1 6 -j- 4 == 4/ 1' of longitude (Art. 143, Note 2), 4h. corresponds to 60°, 44m. to 11°, Dif. of longitude =7 1° 4' and 16sec. to 4'; and 60°+ ll°-f 4/ = 71° 4', the answer required. Hence, to find the longitude corresponding to hours, minutes, and seconds of time, Multiply the hours of time by 15, and divide the minutes of time by 4, for degrees of longitude; and divide the seconds of time by ±,for min- utes of longitude. The several results added together will be the differ- ence of longitude. Examples. 2. The difference of time between Washington and Cincin- nati is 29m. 36sec. ; what is the difference of longitude ? 3. A ship-captain sailing from New York to Europe, after being at sea some days, on taking an observation, found that the sun at noon was 2h. 20m. 40sec. earlier than the New York time, as shown by his chronometer. How many degrees east had he sailed ? Ans. 35° 10'. 4. A gentleman travelling west from Philadelphia, whose longitude is 75° 10' west, found, on arriving at St. Louis, that his watch, an accurate timekeeper, which was right when he left Philadelphia, was lh. 20sec. earlier than the time at St. Louis. What, then, is the longitude of St. Louis ? Ans. 90° 15 / west. 5. The difference of time between Baltimore and New Or- leans is 53m. 30sec. ; what is the difference of longitude ? 6. When it is noon at St. Paul's, longitude 93° b' west, it is at Bangor lh. 37m. 12sec. P.M.; what is the longitude of » Bangor ? Ans. 68° 47' west. 7. When it is 11 A. M. at a place 30° east of Greenwich, it is 3h. 44m. 20sec. A. M. at Buffalo ; what is the longitude of Buffalo ? Ans. 78° 55' west. 8. The difference of time between Cambridge Observatory and Greenwich is 4h. 44m. 32sec. ; what is the difference of longitude? Ans. 71° 8'. N 12 134 MISCELLANEOUS EXAMPLES. MISCELLANEOUS EXAMPLES. 1. If the population of the world be as follows: America, 57,650,000; Europe, 263,517,496; Asia, 626,400,000; Africa, 100,000,000 ; Australia, 1,445,000 ; Polynesia, 1,500,000 ; and the average length of life be 33 years, what must be the aver- age number of deaths annually ? Ans. 31,833,712. 2. A farmer has in 3 bins 755 bushels of grain; there being in the first 125 bushels, and in the second 96 bushels, more than in the third ; how many bushels in the second and third ? Ans. 363 in the second, 267 in the third. 3. There is a certain island 30 miles in circumference. If A and B commence travelling round it, A at the rate of 3 miles an hour, and B at the rate of 5 miles an hour, how far apart will they be at the end of 30 hours ? 4. Having money to invest, I purchased two farms at $ 1,750 each, and 19 shares of bank stock at $ 103 per share, and have left $ 113 ; how much money had I ? Ans. $ 5,570. 5. It has been agreed by 12 men to gather 960 bushels of cranberries, and receive for their labor one half of the quantity gathered ; after one half was gathered, one third of the men withdrew, leaving the others to complete the job. How many bushels should each man receive ? Ans. Those who left, 20 bushels each ; those who remained, 50 bushels each. 6. A drover made $ 652.00 by selling a lot of sheep, at a profit of 50 cents each ; how many did he sell ? 7. What will it cost to carpet a floor that is 18 feet wide and 27 feet long, provided the carpeting cost $ 2.25 per yard ? 8. If a young man, by early rising and economy of time, can save for study and improvement of mind two and a half hours a day, how many years' study, of 12 hours per day, can thus be gained in 20 years ? Ans. 4y. 60d. 10^-h. 9. From 4 piles of wood, the first containing 7c. 76ft. 1671in., the second 16c. 28ft. 56in., the third 29c. 127ft. 1000in., the fourth 29c. 10ft. 121 6in., I have sold 45 cords and 6 cord feet; how much remains ? # Ans. 37c. 19ft. 487in. 10. Boston is in north latitude 42° 21' ; Portland is in lati- tude 1° 15' north of Boston ; and Charleston is in latitude MISCELLANEOUS EXAMPLES. 135 10° 40' south of Portland. What is the latitude of Charles- ton? Ans. 32° 5 6' north. 11. If 1 cubic foot of anthracite coal weighs 54 pounds, how many cubic feet of space are required to stow 2 tons of 2000 pounds each ? Ans. 74/ T cubic feet. 12. Two engineers, A and B, surveyed a certain house-lot. A made its contents 3R. 18p. 0yd. 6ft. 64sq. in., but B made its contents 3R. 17p. 30yd. 8ft. lOOsq. in. How much did the one differ from the other ? 13. The products of the industry of 250,000 persons in Mas- sachusetts, during the year 1855, amounted to $ 295,300,000. What was the average amount to each individual, and how much was added to the capital of the State, if one fourth of the whole amount was saved ? Ans. $ 1,181.20 to each in- dividual ; $ 73,825,000 added to the capital. 14. The capacity of a certain cistern is 216 cubic feet ; how many hogsheads of water will it contain ? 15. What day of the month and what day of the year is the second Monday of May, in a common year commencing on Thursday ? Ans. 11th day of May ; 131st day of the year. 16. Purchased 18T. 17cwt. 3qr. 201b. of copperas, at 4 cents per pound. I sold 4T. 6cwt. lqr. 141b. at 5 cents per pound, and 7T. lcwt. 3qr. 101b. at 6 cents per pound. Moses Atwood purchased one fourth of the remainder at 6 cents per pound. One half of what then remained I sold to J. Gale at 10 cents per pound. The remainder I sold to J. Smith at 12 cents per pound ; but he has become a bankrupt, and I lose half my debt. What have I gained by my purchase ? Ans. $ 894.07J-. 17. The distance between Boston and San Francisco is 2691 miles. If Nathan Swift of San Francisco and Oliver Fleet of Boston, on Thursday, the first day of January, 1857, set out to meet each other, Swift travelling 3 miles 7 furlongs 29 rods 15 feet per hour, and Fleet 5 miles 10 rods and 1£ feet per hour, both travelling 6 J hours per day, commencing at 8 o'clock, A. M., provided they rest on the Sabbath, in what year and tonth, and on what day of the month, and at what time of the day, will they meet, and how far will each have travelled ? Ans. On Monday, February 23, 1857, 2h. 30m. P. M. Swift, 1186m. 4fur. 22rd. 13ft. 6in.; Fleet, 1504m. 3fur. 17rd. 3ft. 136 PEINCIPLES AND APPLICATIONS. EXAMPLES BY ANALYSIS. 1. If 7 pairs of shoes cost $ 8.75, what will one pair cost? what will 20 pairs cost? Ans. $ 25.00. 2. If 5 tons of hay cost $ 85, what will 1 ton cost ? what will 17 tons cost? Ans. $289.00. 3. When $ 0.75 are paid for 3gal. of molasses, what is the value of lgal. ? What cost 37 gal. ? 4. Gave $ 1.92 for 41b. of tea; what cost lib.? what cost 371b.? Ans. $17.76. 5. For 121b. of rice I paid $ 1.08 ; what was paid for lib. ; and what must I give for 251b. ? Ans. $ 2.25. 6. Gave S. Smith $ 63.00 for 9 tubs of butter ; what was the cost of 1 tub ? What cost 27 tubs ? Ans. $ 189.00. 7. T. Swan can walk 20 miles in 5 hours ; how far can he walk in 1 hour ? How long would it take him to walk from Bradford to Boston, the distance being in a straight line 28 miles ? 8. If a hungry boy would eat 49 crackers in 1 week, how many would he eat in 1 day ? how many would be sufficient to last him 19 days? Ans. 133 crackers. 9. Gave $ 20 for 5 barrels of flour ; what cost 1 barrel ? what cost 40 barrels ? Ans. $ 160.00. 10. For 31b. of lard there were paid 36 cents ; what was the cost of 371b. ? 11. Paid F. Johnson 72 cents for 9 nutmegs ; how many cents were paid for 1 nutmeg ; and what should be charged for 37 nutmegs? Ans. $2.96. 12. Paid 2£. 17s. 5d. for 521b. of sugar; what cost lib.? what cost 761b. ? 13. Paid 4£. 3s. lid. for 761b. of sugar; what cost 521b.? 14. If a man walk 17m. 4fur. 28rd. in 6 days, how far will he walk in 100 days ? Ans. 293m. lfur. 15. If a farmer feed to his stock in 7 months 41 bu. 3pk. 4qt. lpt. of grain, how much is required for 1 month ? how much for 7 years? Ans. 502bu. 2pk. 6qt. 16. A field containing 39 A. 2R. 5p. 8yd. 6ft. 108in. will pasture 8 cows during the season. How large a field will pas- ture 1 cow ? How large a field 72 cows ? PROPERTIES OF NUMBERS. 137 17. If 4 casks of vinegar contain 63gal. 3qt., what are the contents of one cask? What are the contents of 37 casks ? Ans. 589gal. 2qt. lpt. 2gi. 18. When 5yd. 3qr. lna. of cloth cost $ 4, how much cloth can be bought for $ 1 ? How much for $ 36 ? Ans. 52yd. lqr. lna. 19. If 11T. 3cwt. 2qr. of hay be sufficient to keep 4 horses 7 T 3 T months, how much will keep 1 horse the same time ? How much 23 horses? Ans. 64T. 5cwt. 121b. 8oz. 20. If 12 men can dig a certain ditch in 286 days 4h. 33m., how long will it require 1 man to do the same labor ? How long 72 men? Ans. 47 days 16h. 45m. 30sec. 21. If 27yd. lqr. of cloth be required to make 21 coats, how many yards will be required to make 11 coats ? 22. If a train of cars move at the rate of 174m. 26rd. in 7 hours, how far will it move in 1 hour? How far in 10 hours? Ans. 248m. 5fur. 20rd. 23. If 4 cases of shoes, containing 60 pairs each, cost $ 192, what will 1 pair cost ? What will 25 cases cost ? 24. When 3A. 2R. 20rd. of land will buy 4 hogsheads of molasses, how much land will buy 1 hogshead ? How much 30 hogsheads ? Ans. 27A. OR. 30rd. 25. If a man can travel 20deg. 49m. 5fur. 35rd. 5yd. 3in. in 9 weeks, how far would he travel in 1 week ? How far in 90 weeks ? Ans. 207deg. 13m. lfur. 25rd. 5yd. PROPERTIES OF NUMBERS. DEFINITIONS. 162. An integer is a whole number ; as 1, 7, 16. All whole numbers are either prime or composite. 163. A prime number is a number which can be exactly divided only by itself and 1 ; as 1, 3, 5, 7, 11. A composite number is a number which can be exactly di- vided by some number besides itself and 1 ; as 6, 9, 14, 18. 12* 138 PEOPERTIES OF NUMBERS. 164. A factor of a number is such a number as will, by- being taken an entire number of times, produce it ; as, 3 is a factor of 9, and 4 a factor of 16. 165. A prime factor of a number is a prime number that will exactly divide it ; thus the prime factors of 10 are the prime numbers 1, 2, and 5. Note. — Unity or 1 is not generally regarded as a prime factor, since mul- tiplying or dividing any number by 1 does not alter its value. It therefore will be omitted when speaking of the prime factors of numbers. A composite factor of a number is a composite number that will exactly divide it ; thus, G and 8 are composite factors of 48. 166. Numbers are prime to each other when they have no factor in common ; thus, 4, 9, and 23 are prime to each other. 167. An aliquot part of a number is such a part as will exactly divide it; as, 1, 3, and 5 are aliquot parts of lo. Note. — The aliquot parts of a number include all its factors, prime and composite. An aliquant part of a number is such a part as will not ex- actly divide it ; as 2, 4, 5, 7, and 8 are aliquant parts of 9. 168. The reciprocal of a number is the quotient arising from dividing 1 by the number ; thus, the reciprocal of 2 is 4-. 169. The power of a number is the product obtained by taking the number a certain number of times as a factor ; thus 25 is a power of 5. Note. — When the number is taken once, it is called its first power; when taken twice, as a factor, the product is called its second power; and so on. The second power of a number is sometimes termed its square, and the third power, its cube. 170. The exponent of a power is a figure written at the right of a number, and a little above it, to show how many times it is taken as a factor ; thus, in the expression 4 2 , the exponent is the 2, and the whole is read 4 second power ; and in 7 3 , it is the 3, and the whole read 7 third power. Note. — The first power of a number being always the number itself, its exponent is not expressed. PROPERTIES OP NUMBERS. 139 PROPERTIES OF PRIME NUMBERS. 171. No direct process of detecting prime numbers has been discovered. Note. — A few facts, such as are given below, if kept in mind, will aid somewhat in ascertaining whether a number is prime or not. 172. The only even prime number is 2 ; since all other even numbers, as 4, 6, 8, and 10, it is evident, can be exactly divided by 2, and therefore must be composite. 173. The only prime number having 5 for a unit or right- hand figure is 5 ; since every other whole number thus termi- nating, as 15, 25, 35, and 45, can be exactly divided by 5, and therefore must be composite. 174. Every prime number, except 2 and 5, must have 1, 3, 7, or 9 for the right-hand figure ; since all other numbers are composite. 175. Every prime number above 3, when divided by 6, must leave 1 or 5 for a remainder ; since every prime number above 3 is either 1 greater or 1 less than 6, or some exact number of times 6. 176. In a series of odd numbers written in their proper or natural order, if beginning with 3 every third number, with 5 every fifth, with 7 every seventh, be cancelled, as composite, the remaining numbers, with 2, will be the prime numbers of the natural series. Thus, in the series 1, 3, 5, 7, 0, 11, 13, X$, 17, 19, U, 23, ft$, n, 29, 31, £3, £& 37, #0, 41, 43, £$, 47, ^0, every third number from the 3, every fifth from the 5, every seventh from the 7, every ninth from the 9, and so on, being cancelled, the remaining numbers, with 2, are all the prime numbers under 50. Note 1. — In the series, every third number from the 3 contains that num- ber as a factor; every fifth number from the 5, that number as a factor; and so on. Note 2. — The whole number of prime numbers from 1 to 100,000 is 9,593. Although all of these, except 2 and 5, end in 1, 3, 7, or 9, there are, within the same range, no less than 30,409 composite numbers terminating with some one of the same figures. 140 PROPERTIES OF NUMBERS. 177. All the prime numbers not larger than 4049 are in- cluded in the following TABLE OF PRIME NUMBERS. 1 233 557 883 1249 1607 2003 2389 2791 3229 3631 2 239 563 887 1259 1609 2011 2393 2797 3251 3637 3 241 569 907 1277 1613 2017 2399 2801 3253 3643 5 251 571 911 1279 1619 2027 2411 2803 3257 3659 7 257 577 919 1283 1621 2029 2417 2819 3259 3671 11 263 587 929 1289 1627 2039 2423 2833 3271 3673 13 269 593 937 1291 1637 2053 2437 2837 3299 3677 17 271 599 941 1297 1657 2063 2441 2843 3301 3691 19 277 601 947 1301 1663 2069 2447 2851 3307 3697 23 281 607 953 1303 1667 2081 2459 2857 3313 3701 29 283 613 967 1307 1669 2083 2467 2861 3319 3709 31 293 617 971 1319 1693 2087 2473 2879 3323 3719 37 307 619 977 . 1321 1697 2089 2477 2887 3329 3727 41 311 631 983 1327 1699 2099 2503 2897 3331 3733 43 313 641 991 1361 1709 2111 2521 2903 3343 3739 47 317 643 997 1367 1721 2113 2531 2909 3347 3761 53 331 647 1009 1373 1723 2129 2539 2917 3359 3767 59 337 653 1013 1381 1733 2131 2543 2927 3361 3769 61 347 659 1019 1399 1741 2137 2549 2939 3371 3779 67 349 661 1021 1409 1747 2141 2551 2953 3373 3793 71 353 673 1031 1423 1753 2143 2557 2957 3389 3797 73 359 677 1033 1427 1759 2153 2579 2963 3391 3803 79 367 683 1039 1429 1777 2161 2591 2969 3407 3821 83 373 691 1049 1433 1783 2179 2593 2971 3413 3823 89 379 701 1051 1489 1787 2203 2609 2999 3433 3833 97 383 709 1061 1447 1789 2207 2617 3001 3449 3847 101 389 719 1063 1451 1801 2213 2621 3011 3457 3851 103 397 727 1069 1453 1811 2221 2633 3019 3461 3853 107 401 733 1087 1459 1823 2237 2647 3023 3463 3863 109 409 739 1091 1471 1831 2239 2657 3037 3467 3877 113 419 743 1093 1481 1847 2243 2659 3041 3469 3881 127 421 751 1097 1483 1861 2251 2663 3049 3491 3889 131 431 757 1103 1487 1867 2267 2671 3061 3499 3907 137 433 761 1109 1489 1871 2269 2677 3067 3511 3911 139 439 769 1117 1493 1873 j 227:1 2683 3079 3517 3917 149 443 773 1123 1499 1877' 22 M 2687 3083 3527 3919 151 449 787 1129 1511 1879 2287 2689 3089 3529 3923 157 457 797 1151 1523 1889 2293 2693 3109 3533 3929 163 461 809 1153 1527 1901 2297 2699 3119 3539 3931 167 463 811 1163 1529 1907 2309 2707 3121 3541 3943 173 467 821 1171 1533 1913 2311 2211 3137 3547 3947 179 479 823 1181 1539 1931 2333 2713 3163 3557 3967 181 487 827 1187 1547 1933 2339 2719 3167 3559 3989 191 491 829 1193 1551 1949 2341 2729 3169 3571 4001 193 499 839 1201 1553 1951 2347 2731 3181 3581 4003 197 503 853 1213 1559 1973 2351 2741 3187 3583 4007 199 509 857 1217 1571 1979 2357 2749 3191 3593 4013 211 521 859 1223 1579 1987 2371 2753 3203 3607 4019 223 523 863 1229 1583 1993 2377 2767 3209 3613 4021 227 541 877 1231 1597 1997 2381 2777 3217 3617 4027 229 547 881 1237 1601 1999 2383 2789 3221 3623 4049 PROPERTIES OP NUMBERS. 141 FACTORING. 178. Factoring is the process of resolving a quantity into its factors. 179. Every number that is not prime is composed of prime factors, since all numbers are either prime or composite ; and, if composite, can be separated into factors, which, if themselves composite, can be further separated into those that shall be prime. 180. To resolve a composite number into its prime factors. Ex. 1. It is required to find the prime factors of 42. Ans. 2, 3, 7. operation. *\y" e divide by 2, the least prime number greater 2 4 2 than 1, and obtain the quotient 21 ; and, since 21 is o Try a composite number, we divide this by 3, and obtain - — for a quotient 7, which is a prime number. The 7 several divisors and the last quotient, all being prime, constitute all the prime factors of 42, which, multi- plied together, they equal. Hence Divide the given number by any prime number that will exactly divide it, and the quotient, if a composite number, in the same manner; and so continue dividing, until a prime number is obtained for a quotient. The several divisors and the last quotient will be the prime factors required. Note 1. — The composite factors of any number may be found by multiply- ing together two or more of its prime factors. Note 2. — Such prime factors as two or more numbers may have alike, are termed prime factors common to them ; and these may be readily determined after the numbers are resolved into their prime factors. Examples. 2. What are the prime factors of 105 ? Ans. 3, 5, 7. 3. Resolve 220 into its prime factors. 4. What are the prime factors of 936 ? Ans. 2, 2, 2, 3, 3, 13. 5. What are the prime factors of 1953 ? 6. Resolve 12462 into its prime factors. Ans. 2, 3, 31, 67. 7. Resolve 19987 into its prime factors. Ans. 11, 23, 79. 8. What are the prime factors common to 225, 435, and 540 ? Ans. 3, 5. 142 PROPERTIES OF NUMBERS. 9. What are the prime factors common to 960, 1568, and 5824? 10. What are the prime factors common to 2340, 11934, 12987, and 14859 ? Ans. 3, 3, 13. 11. A man has 105 apples, which he wishes to distribute into small parcels, each of equal numbers ; what are the smallest whole numbers, greater than 1, into which they may be exactly divided ? Ans. 3, 5, and 7. DIVISIBILITY OF NUMBERS. 181. One number is said to be divisible by another, when the latter will divide the former without a remainder. Thus, 9 is divisible by 3. 182. One number is divisible by another, when it contains all the prime factors of that number. Thus, 12, which contains all the factors of 4, is divisible by 4. 183. All even numbers, or such as terminate with 0, 2, 4, 6, or 8, are divisible by 2, since each of them contains 2 as a fac- tor. Thus, 10, 24, 36, 58, are each divisible by 2. 184. AU numbers which terminate with or 5 are divisible by 5, since each of them contains 5 as a factor. Thus, 20, 25, 50, are each divisible by 5. 185. Every number is divisible by 4, or any other number that will exactly divide 100, when its two right-hand figures are divisible by the same* For any figure on the left of the two right-hand figures must express one or more hundreds, and a factor of one hundred is a factor of any number of hundreds ; so, if the sum exactly divides the units and tens of a number, the entire number will be divisible by it. Thus, 116 is divisi- ble by 4; 140, by 20 ; 225, by 25 ; and 450, by 50. 186. Every number is divisible by 8, or any other number that will exactly divide 1000, when its three right-hand figures are divisible by the same. For any figure on the left of the three right-hand figures must express one or more thousands, and a factor of one thousand is a factor of any number of thousands ; so, if the sum exactly divides the units, tens, and PROPERTIES OP NUMBERS. 143 hundreds of a number, the entire number will be divisible by- it. Thus, 1824 is divisible by 8 ; 1840, by 40 ; 3375, by 125 ; 2750, by 250 ; and 4500, by 500. 187. Every number the sum of whose digits 3 or 9 will exactly divide, is divisible by 3 or 9. For 10, or any power of 10, less 1, gives a number, as 9, 99, 999, &c., which is divisible by 3 and by 9. Hence, any number of tens, hundreds, thousands, &c, less as many units, must be divisible by 3 and by 9 ; and if the excess of units denoted by the significant figures, in the aggregate, is likewise divisible by 3 and by 9, it follows that the entire number is thus divisible. For exam- ple, 7542 is a number, the sum of whose digits is divisible by 3 and by 9 ; and separated into tens, hundreds, and thousands, it is equal to 7000 + 500 + 40 + 2. Now, 7000 = 7 X 1000 = 7 X (999 + 1) = 7 X 999 + 7 ; 500 = 5 X 100 = 5 X (99 + 1) = 5 X 99 + 5 ; and 40 = 4 X 10 = 4 X (9 + 1) = 4 X 9 + 4. Therefore, 7542 = 7 X 999 + 5X 99 + 4X9 + 7 + 5 + 4 + 2. The remainders 7 + 5 + 4 + 2, corresponding with the significant figures of the number, added together, equal 18, which sum being divisible by 3 and by 9, it is evident that 7542 is divisible in like manner. Note. — Upon the property of 9 now explained depends the method of proving, by excess of nines, multiplication (Art. 63), and division (Art. 75). The same method of proof may be resorted to in addition and in subtraction. Thus, To prove Addition. Find the excess of nines in each num- ber added, and then the excess of nines in the sum of these results ; which, if the work be right, will equal the excess of nines in the answer. To prove Subtraction. Find the excess of nines in the sub- trahend, and also in the remainder, and then the excess of nines in the sum of these results ; which, if the work be right, will equal the excess of nines in the minuend. 188i Every number occupying four places, in which two like significant figures have two ciphers between them, is divisi- ble by 7, 11, and 13. Thus, 9009, 1001, 300300, 4004, &c, are each divisible by 7, 11, and 13. 144 PROPERTIES OF NUMBERS. 189. Every number is divisible by 11, in which the sum of the digits in the odd places is equal to the sum of the digits in the even places, or in which the difference of their sums can be exactly divided by 11. Thus, 8305, in which 3 + 5 = 8 + 0, and 628001, in which 2 + + 1 and 6 + 8 + differ by 11, are each divisible by 11. 190. Every number divisible by two\or more numbers, which are prime to each other, is divisible by their product For, being prime to each other, dividing" by one of the numbers does not cancel the others as factors. Thus, 770, being divisi- ble by 2, 5, and 7, which are prime to each other, is divisible by 70, their product. 191. Every even number, the sum of whose digits 6 wiU exactly divide, is divisible by 6. For being even, it is divisible by 2 (Art. 183), and its digits being divisible by twice 3, or 6, are evidently divisible by once 3, so that the number is also divisible by 3 ; and as the 2 and the 3 are prime to each other, the number is divisible by their product, or 6 (Art. 190). Thus, 174, 6312, are each divisible by 6. 192. Every number terminating with or 5 that 3 will exactly divide, is divisible by 15, and every number that 9 wiU exactly divide, is divisible by 45. For, terminating with or 5, it is divisible by 5 (Art. 184), and, as 3 and 9 are each prime to 5, if it can be exactly divided by 3 or by 9, it must be divisible by 5 X 3 = 15, or by 5 X 9 = 45. Thus, 75, which 3 will exactly divide, is divisible by 15 ; and 90, which 9 will exactly divide, is divisible by 45. Divisors or Measures. 193. A divisor or measure of a number is any number that will divide it without a remainder. Thus, 3 is a divisor or measure of 6, and 5 a divisor or measure of 10. 194. To find all the divisors or measures of a number. Ex. 1. Required all the divisors of 60. Ans. 1, 2, 3, 4, 5, 6, 10, 15, 20, 30, and 60. OPERATION. 60 rrr 2 X 2 X 3 X 5. 1 2 4=2X2 3 6 12 = 2 X 2 X 5 10 20 = 2 X 2 X 15 30 60 = 2 X 2 X PROPERTIES OF NUMBERS. 145 Resolving the number into its prime factors, we find 60 = 2X2X3X5. o JNow,anynum- 3X5. ^ er * s divisible by 1, and ev- Ans. 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. ery composite number by its prime factors, and every product these factors can form. Now 1, being a divisor of every number, is a divisor of 60, and, since 2 en- ters twice as a factor into 60, it is evident that 2, and 4 = 2 x 2, are also divisors of 60. These divisors we arrange on a horizontal line, and determine the other divisors by multiplying those on this line by the factor 3, for the second line of divisors, by 5 for the third line, and by 3 x 5 for the fourth line, and thus obtain all the pos- sible divisors of the given number. The whole number of divisors is 12, which corresponds to the pro- duct arising from multiplying together the exponents, each increased by 1, of the different prime factors of 60 ; thus, of the different prime factors, since 2 enters twice, its exponent is 2, -\- 1 = 3 ; 3 enters once, its exponent is 1, -j- 1 = 2 ; 5 enters once, its exponent is 1 5 _|_ l = 2 ; and 3 x 2 X 2 = 12, the number of divisors. The same holds true in all cases. Hence, in any composite number, To find the Number of Divisors. — Multiply together the ex- ponents, each increased by 1 , of the different prime factors of the given number, and the product ivill be the number of divisors required. And To find the several Divisors. — Form from the prime factors of the number all the products possible, and these factoids (including 1) and products will be the divisors required. Examples. 2. What are the divisors of 72 ? Ans. 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. 3. Required the divisors of 105 ? 4. How many divisors has 1764 ? Ans. 27. 5. How many divisors has 3528 ? 6. How many divisors has 5880 ? Ans. 48. Common Divisors or Measures. 195. A common divisor or measure of two or more num- bers is any number that will divide them without a remainder ; thus, 2 is a common divisor of 4, 8, and 10. N 13 146 PROPERTIES OP NUMBERS. 196. A common divisor of two numbers is a divisor of their sum, and also of their difference. Thus, 6, a common divisor of 12 and 18, is a divisor of their sum, 40, and of their dif- ference, 6. 197 # A common divisor of the remainder and the divisor is a divisor of the dividend. Thus, in a division having 8 for a remainder, 16 for divisor, and 24 for dividend, 8, a common di- visor of 8 and 1G, is also a divisor of the 24. 198. To find all the divisors common to two or more num- bers. Ex. 1. Required all the common divisors of 45 and 135. Ans. 1, 3, 5, 9, 15, and 45. operation. Resolving the given numbers into their 5 # prime factors^ we find they have of these 3, 3, and 5 in common, and these common" X 5. I )rime factors tcith 1, and all the product* ire are able to form from them (Art. 194), give all the common divisors required. When only the number of com- mon divisors is required, it may readily be found by multiplying to- gether the exponents, each increased by I, of the different common prime factors. (Art. 194.) Examples. 2. What are the common divisors of 51, 153, and 255 ? 3. Required the several common divisors of 180 and 360. Ans. 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, GO, 90, and 180. 4. How many common divisors have 2025, 6075, and 8100? Ans. 15. 5. How many common divisors have 4500 and 9000 ? Ans. 36. The Greatest Common Divisor or Measure. 199. The greatest common divisor or measure of two or more numbers is the greatest number that will divide each of 4 5 1 3 5 = 3 = 3 X 3 X 3 X 5. X 3 X Common f 1 Divisors ( 5 Ans. 3 15 1,3, 9=3X3 45=3X3 5, 9, 15, and 45, PROPERTIES OF NUMBERS. 147 them without a remainder. Thus, 4 is the greatest common divisor of 8, 12, and 16. 200. To find the greatest common divisor of two or more numbers. Ex. 1. Required the greatest common divisor or measure of 24 and 88. Ans. 8. first operation. Resolving the numbers into their 24 = 2X2 X^X 3 prime factors, thus, 24 = 2 X 2X2 XQ—9V9V9V 11 X3,and88 = 2x2X2XH = 9 9 9 _ o a 88 ' We find the fact0rS 2 X 2 x 2 ^ X 2 X 2 — 8. Ans. are com mon to both. Since only these common factors, or the pro- duct of two or more of such factors, will exactly divide both numbers, it follows that the product of all their common prime factors must be the greatest factor that will exactly divide both of them. Therefore, 2x2 X 2 = 8, the greatest common divisor required. The same result may be obtained by a sort of trial process, as by the second operation. It is evident, since 24 cannot be ex- second operation. actly divided by a number greater than 2 4)88(3 itself, if it will also exactly divide 88, 7 2 it will be the greatest common divisor sought. But, on trial, we find 24 will 16)24(1 not exactly divide 88, there being a re- 1 6 mainder, 16. Therefore 24 is not a — o~\ i n / 9 common divisor of the two numbers. / 1 We know that a common divisor of 16 16 and 24 will, also, be a common di- visor of 88 (Art. 197). We next try to find that divisor. It cannot be greater than 16. But 16 will not ex- actly divide 24, there being a remainder, 8; therefore 16 is not the greatest common divisor. As before, the common divisor of 8 and 16 will be the common di- visor of 24 and 88 (Art. 197) ; we make trial to find that divisor, knowing that it cannot be greater than 8, and find 8 will exactly divide 16. Therefore 8 is the greatest common divisor required. The last method may be often contracted, if third operation. there should be observed to be any prime factor 24)88(3 in a remainder which is not common to the pre- 7 2 ceding divisor, by cancelling said factor. Thus, in the third operation, the remainder 16 being <£ observed to contain the prime factor 2, common 8)24(3 to the preceding divisor 24, we cancel it, and, 2 a having left the composite factor 8, we divide 24 by that factor. There being no remainder, 8 is the greatest common divisor, as before obtained. 148 PROPERTIES OF NUMBERS. Rule 1. — Resolve the given numbers into their prime factors. The product of all the factors common to the several numbers will be the greatest common divisor. Or, Rule 2. — Divide the greater number by the less, and if there be a remainder divide the preceding divisor by it, and so continue dividing until nothing remains. The last divisor will be the greatest common di- visor. Note 1. — When the greatest common divisor is required of more than two numbers, find it of two of them, and then of that common divisor and of one of the other numbers, and so on for all the given numbers. The last common di- visor will be the greatest common divisor required. Note 2. — If any remainder be such a prime number as will not exactly divide the preceding divisor, or if the last remainder be 1, the numbers, being prime to each other, will have no common divisor. Examples. 2. What is the greatest common divisor of 56 and 1G8 ? 3. What is the greatest common divisor of OG and 128 ? 4. What is the greatest common measure of 57 and 285 ? Ans. 57. 5. What is the greatest common measure of 1G9 and 175 ? G. What is the greatest common measure of 175 and 455? Ans. 35. 7. What is the greatest common divisor of 1 69 and 8GG? Ans. 1. 8. What is the greatest common measure of 47 and 478? Ans. 1. 9. What is the greatest common measure of 84 and 10G8? Ans. 12. 10. What is the greatest common divisor of 75 and 1G5? Ans. 15. 11. What is the greatest common measure of 78, 234, and 4G8 ? Ans. 78. 12. I have three fields ; one containing 16 acres ; the second, 20 acres ; and the third, 24 acres. Required the largest-sized lots, containing each an exact number of acres, into which the whole can be divided. Ans. 4 acre lots. 13. A farmer has 12 bushels of oats, 18 bushels of rye, 24 bushels of corn, and 30 bushels of wheat. Required the largest bins, of uniform size, and containing an exact number of bushels, into which the whole can be put, each kind by itself, and all the bins be full. PROPERTIES OF NUMBERS. 149 LEAST COMMON MULTIPLE. 201 • A common multiple of two or more numbers is a number that can be divided by each of them without a remain- der ; thus, 14 is a common multiple of 2 and 7. The least common multiple of two or more numbers is the least number that can be divided by each of them without a remainder ; thus, 12 is the least common multiple of 4 and 6. 2®2t A multiple of a number contains all the prime factors of that number; the common multiple of two or more num- bers contains all the prime factors of each of the numbers ; and the least common multiple of two or more numbers con- tains only each prime factor taken the greatest number of times it is found in any of the several numbers. Hence, 1. The least common multiple of two or more numbers must be the least number that will contain all the prime factors of them, and none others. 2. The least common multiple of two or more numbers, which are prime to each other, must equal their product. 3. The least common multiple of two or more numbers must equal the product of their greatest common divisor, by the factors of each number not common to all the numbers. 4. The least common multiple of two or more numbers, di- vided by any one of them, must equal the product of those factors of the others not common to the divisor. 203. To find the least common multiple of two or more numbers. Ex. 1. What is the least common multiple of 8, 16, 24, 32, 44. Ans. 1056. first operation. Resolving the 8 = 2 X 2 X 2 numbers into their 16 = 2X2X 2X2 prime factors, we 24 = 2X2 X 2X3 ^ nci the ^ r differ- 3 2 = 2 X 2 X 2X2X2 ent P rime factors 4 4 = 2X2X11 * b L 2 ' 3 ' ? nc * ^ 11. The greatest 2X2X2X2X2X3X11= 1056 Ans. number of times the 2 occurs as a factor in any of the given numbers is 5 times ; the greatest num- 13* 2)8 SECOND OPERATION. 16 24 32 44 2)4 8 12 16 22 2)2 4 6 8 11 2)1 2 3 4 11 150 PROPERTIES OF NUMBERS. ber of times 3 occurs in any of the numbers is once; and the greatest number of times the 1 1 occurs in any of the numbers is once. Hence, 2, 2, 2, 2, 2, 3, and 11 must be all the prime factors necessary in composing 8, 16, 24, 32, and 44; and consequently, 1056, the product of these factors, is the least common multiple re- quired (Art. 202). Having arranged the numbers on a horizon- tal line, we divide by 2, a prime number that will divide two or more of them without a re- mainder, and write the o « quotients in a line be- ~ ~~ n low ; and we continue 2X2X2X2X2X3XH ==10o6 Ans. to divide by a prime number as before, till the divisor and remainders are all prime to each other. Then, these, since they include all the factors necessary to form the given num- bers and no others, we multiply together for the required least com- mon multiple, and obtain 1056, as before. The least common multiple of two or more numbers may be found generally by a process much shorter than either of the above meth- ods, by cancelling any number that is a factor of any other of the given numbers, and also by dividing the numbers by such a com- posite number as may be observed to be their common or greatest common divisor. third operation. Thus, in the third operation, 4 ) $ X 24 32 44 8 being a factor of several of — o~\~ a o — tTT ^ e num ^ )ers » an( l 16 being a w ' oil factor of one other number, we 3 4 11 cancel them ; and observing that 4 is the greatest common 4X2X^X^X11 = 10ob Ans. divisor of the remaining num- bers, we divide them by it. We next divide by 2, as in the second operation. The numbers in the lower line then being prime to each other, we multiply them and the divisors together, and obtain 1056 as the least common multiple. fourth operation. The fourth operation exhibits $ X 24132 44 a process yet more contracted. — ~ — t-i The 8 and 16 being factors each of one or more of the other num- 24X8X11 = 1056 Ans. bers, we cancel them, as in the third operation. Of the remain- ing numbers we cut off 24 by a short vertical line from the rest as a factor of the least common multiple sought. We then strike out of the two remaining numbers the largest factor each has in common with the 24, by dividing each of them by the greatest common divisor PROPERTIES OF NUMBERS. 151 between it and 24, and write the result beneath. The numbers in the lower line having no factor in common, we carry the process no further. The continued product of the number cut off by the num- bers in the lower line gives 1056, the least common multiple, as by the other methods. In this instance we cut off the 24, but either the 32 could have been separated from the rest, or the 44 cut off, and the needless factors striken out with like result. If, however, we had cut off the 44, the numbers placed in the second line would have contained factors common to each other, so that it would have been necessary in that line to have cut off and stricken out factors as before. The reason for this abridged process is, that by the sepa- rating off, and by the striking out of factors, we get rid, in an expe- ditious way, of the factors not required to form the least common multiple sought. Rule 1. — Resolve the given numbers into their prime factors. The product of these factors, taking each factor only the greatest number of times it occurs in any of the numbers, will be the least common multiple. Or, Rule 2. — Having arranged the numbers on a horizontal line, can- cel such of them as are factors of any of the others, and separate some convenient one from the rest. Reject from each of the numbers remain- ing the greatest factor common to it and that number, and write the result in a line below. Should there be in the second line numbers hav- ing factoids in common, proceed as before ; and so continue until the numbers written below are prime to each other. The continued product of the number or numbers separated from the others with those in the last line will be the least common multiple. Note 1. — Some give a preference to the following rule for finding the least common multiple: Having arranged the numbers on a horizontal line, divide by such a prime number as will exactly divide two or more of them, and write the quotients and undivided numbers in a line beneath. So continue to divide until the quotients shall be prime to each other. Then the product of the divisors and the numbers of the last line will be the least common multiple. Note 2. — The least common multiple of two or more numbers that are prime to each other is found by multiplying them together (Art. 202). Note 3. — When a single number alone is prime to all the rest, it may be separated off, and used only as a factor of the least common multiple sought. Note 4. — When the least common multiple of several numbers, and all the numbers except one, which is prime to the others, are given, to find the un- known number, divide the least common multiple given by that of the known numbers (Art. 202). Examples. 2. What is the least common multiple of 3, 13, 37, and 91. 3. What is the least common multiple of 9, 14, 30, 35, and 47 ? Ans. 29610. 152 PROPERTIES OF NUMBERS. OPERATION. 9 14 3 | 3 5 4 7. 9 ft\ 6 3 47 X 35 X X 3 = 29610 Ans. 4. What is the least common multiple of 6, 8, 10, 18, 20, and 24 ? 5. What is the least common multiple of 14, 19, 38, and 57 ? Ans. 798. 6. What is the least common multiple of 20, 36, 48, and 50 ? Ans. 3600. 7. What is the least common multiple of 15, 25, 35, 45, and 100 ? Ans. 6300. 8. What is the least common multiple of 100, 200, 300, 400, and 575 ? Ans. 27600. 9. The least common multiple of 1, 2, 3, 4, 5, 6, 8, 9, and one other number prime to them, is 2520. What is that other number ? Ans. 7. 10. What is the least common multiple of 18, 24, 36, 126, 20, and 48 ? 11. I have four different measures; the first contains 4 quarts, the second 6 quarts, the third 10 quarts, and the fourth 12 quarts. How large is a vessel, that may be filled by each one of these, taken a certain number of times full ? Ans. 60 quarts. 12. What is the smallest sum of money with which I can purchase a number of oxen at $ 50 each, cows at $ 40 each, or horses at $ 75 each ? Ans. § 600. MISCELLANEOUS EXAMPLES. 1. How many times does 7 occur as a factor of 6174 ? Ans. 3 times. 2. Required the largest prime factor of 5775. 3. Required the largest composite factor of 19929/ Ans. 6643. 4. Required the quotients of 2338 divided by its two prime factors next larger than 1. Ans. 1169 ; 334. 5. Required all the prime numbers that will divide 17385 without a remainder. PROPERTIES OF NUMBERS. 153 6. A farmer has 3000 bushels of grain ; which are the three smallest-sized bags, and the three largest-sized bins, holding an exact number of bushels, that will each measure the same without a remainder ? Ans. Bags of 1, 2, or 3 bushels each; and bins of 1500, 1000, or 750 bushels each. 7. A teacher having a school consisting of 152 ladies and 136 gentlemen, divided it in such a manner that each class of ladies equalled each class of gentlemen, and the classes were the largest the school would admit of, and have them all of the same size. Required the number of classes, and the number in each class. Ans. 19 classes of ladies, 17 classes of gentlemen, and 8 pupils in a class. 8. At noon the second, minute, and hour hand of a clock are together ; how long after will they be again, for the first time, in the same position ? 9. J. Porter has a four-sided garden, the first side of which is 348 feet in length ; the second, 372 feet ; the third, 444 feet ; and the fourth, 492 feet. Required the length of the longest rails that can be used in fencing it, allowing the end of each rail to lap by the other 9 inches, and all the panels to be of equal length ; also, the number of rails, if 5 rails be allowed to each panel. Ans. Length 12ft. 9in. ; and 690 rails. 10. L. Ford has 5 pieces of land, the first containing 3 A. 2R. lp.; the second, 5A. 3R. 15p.; the third, 8A. 29p. ; the fourth, 12A. 3R. 17p.; and the fifth, 15A. 31p. Required the largest sized house-lots, containing each an exact number of square rods, into which the whole can be divided. Ans. 1A. 27p. each. 11. What three numbers between 30 and 140 have 12 for their greatest common divisor, and 2772 for their least common multiple. Ans. 36, 84, and 132. 12. Four men, A, B, C, and D, are engaged in making reg- ular excursions into the country, between which each stays at home just 1 day ; and A is always absent exactly 3 days, B 5 days, and C and D 7 days. Provided they all start off on the same day, how many days must elapse before they can all be at home again on the same day ? Ans. 23 days. 154 COMMON FRACTIONS. COMMON FRACTIONS. 204. A fraction is an expression denoting one or more equal parts of a unit. 205. A fractional unit is one of the equal parts into which the whole thing or integral unit has been divided. Thus halves, thirds, &c, being equal parts of integral units or whole things, are fractional units. 206. The unit of a fraction is the unit or whole thing from which its fractional parts have been derived. 207. A Common Fraction is expressed by two numbers one above the other, with a line between them. 208* The number below the line is called the denominator. It shows into how many parts the whole number has been divided. It gives name to the fraction and value to the frac- tional unit. Thus, in the expression f , the denominator is 7, indicating that the unit of the fraction has been divided into 7 equal parts, and that the value of the fractional unit is one seventh. The number above the line is called the numerator. It shows how many parts have been taken, or numbers the frac- tional units expressed by the fraction. Thus, in the expression f , the numerator is 2, indicating that the fractional unit, which is one seventh, has been taken 2 times. 209* The terms of a fraction are its numerator and denomi- nator. Thus, the terms of the fraction f are the numerator 2 and the denominator 3. 210. A proper fraction is one whose numerator is less than the denominator ; as f , §, J. 211. An improper fraction is one whose numerator is equal to, or greater than, the denominator ; as JJ-, fy 9 -\ 5 -. 212. A mixed number is a whole number with a fraction ; as 3£, 16 J, 90 £. 213. A simple or single fraction has but one numerator and one denominator. It may be either proper or improper; as COMMON FRACTIONS. 155 214* A compound fraction is a fraction of a fraction, or two or more fractions connected by the word of; as J- of J of ^, J off of $. 215 • A complex fraction is a fraction having a fraction or a mixed number for its numerator or denominator, or both ; as f h' 13' h' 216. A fraction is an expression of division ; the numer- ator answering to the dividend, and the denominator to the divisor (Art. 67) ; and the value of a fraction is the quotient arising from the division of the numerator by the denominator (Art. 80). Thus, in the fraction J T 5 -, the numerator 15 is the dividend, the denominator 7 is the divisor, and the value ex- pressed 2 .J, or the quotient arising from the division of the 15 by the 7. 217. Since a fraction is an expression of division, it follows, 1. That, if the numerator be multiplied, or the denominator be divided, by any number, the fraction is multiplied by the same number (Art. 81). 2. TJiat, if the numerator be divided, or the denominator mul- tiplied, by any number, the fraction is divided by the same num- ber (Art. 82). 3. That, if the numerator and denominator be both multiplied, or both divided, by the same number, the fraction will not be changed in value (Art. 83). REDUCTION OF COMMON FRACTIONS. 218. Reduction of fractions is the process of changing their form of expression without altering their value. 219. A fraction is in its lowest terms, when its numerator and denominator are prime to each other (Art. 166). 220. To reduce a fraction to its lowest terms. Ex. 1. Reduce J-| to its lowest terms. Ans. -£. birst operation. By dividing both terms of the fraction by 4 ) ft == T2" 4, a factor common to them both, it is re- 4 ) tV = i ^ns. duced to &. Dividing both terms of j% by 4, a factor common to them both, it is re- 156 COMMON FRACTIONS. duced to £. Now, as 1 and 3 are prime to each other, the fraction £ is in its lowest terms. The same result is often more readily ob- second operation. tained by dividing the terms of the fraction 1^ ) if == "3" Ans. by their greatest common divisor, as by the second operation. Since, by dividing the numerator and denominator by the same number, we cancel equal factors in both, and diminish them in the same ratio, their relation to each other evidently is not changed, and the value of the fraction remains the same (Art. 217). Rule. — Divide the numerator and denominator by any number greater than 1 that tvill divide them both ivithout a remainder, and thus proceed until they are prime to each other. Or, Divide both the numerator and denominator by their greatest common divisor. Examples. 2. Reduce £f to its lowest terms. Ans. 5 4 T . 3. Reduce || to its lowest terms. Ans. £ . 4. Reduce $ £ to its lowest terms. Ans. f . 5. Reduce y 1 ^ to its lowest terms. 6. Reduce -jV^ to its lowest terms. Ans. ^. 7. Reduce ££f to its lowest terms. Ans. ^j. 8. Reduce a^Vg to i ts lowest terms. 9. Reduce fff to its lowest terms. Ans. |||. 10. Reduce t VtV to * ts lowest terms. Ans. t YtV 11. Reduce §-f| to its lowest terms. 12. Reduce T VrV t0 * ts lowest terms. Ans. f g|. 13. Reduce £§ g to its lowest terms. Ans. ffe. 221 # To reduce an improper fraction to an equivalent whole or mixed number. Ex. 1. How many yards in -^V/- of a yard ? Ans. 6 T 3 F . Since 19 nineteenths make one yard, operation. it is evident there will be as many 1 9 ) 1 1 7 ( 6 T 3 g- Ans. yards in 117 nineteenths as 19 is con- 114 tained times in 117, which is 6 A times. o Therefore, 6^ yards is the answer re- ** quired. Rule. — Divide the numerator by the denominator. Note. — Should there a remainder occur, write it over the denominator, and make this fraction a part of the answer. COMMON FRACTIONS. 157 Examples. 2. Reduce iff- to a mixed number. Ans. 11 T 2 5. 3. Reduce VtV" to a m i xea * number. Ans. 14 T f F . 4. Reduce -^j 1 - to a mixed number. 5. Reduce f ^{ to a mixed number. Ans. 3J-ff . 6. Change li V°-- to a mixed number. Ans. 1 1 J J. 7. Change -£§- to a mixed number Ans. 9 Iff. 8. Change -*f * to a whole number. Ans. 125. 9. Change -|f to a whole number. 222. To reduce a whole or mixed number to an improper fraction. Ex. 1. Reduce 19 to a fraction whose denominator shall be 7. operation. Since there are 7 sevenths in 1 19 X 7 = 133. whole one, 19 whole ones = 133 133 sevenths ±= *f* , Ans. s ^enths = if*. 2. Reduce 17f to an improper fraction ? Ans. - 8 5 8 -. OPERATION. 5 Since there are 5 fifths in 1 whole one, «* in 17 whole ones there are 85 fifths, and o £ adding 3 fifths for the fraction, we have 8 | as the equivalent of 1 7|. Hence the o 8 8 fifths == *£, Ans. Rule. — Multiply the whole number by the given denominator, and to the product add the numerator of the fractional part, if any ; and write the result over the denominator. Note. — A whole number may be expressed in its simplest fractional form, by taking it for a numerator with Ifor a denominator. Thus, 4 may be written 4 , and read 4 ones. Examples. 3. Reduce 15 to fourths. Ans. - 6 ¥ -. 4. Reduce 161 T ^ to sixteenths. 5. Reduce 171|| to an improper fraction. Ans. -H-V 4 — • 6. Change 1 1 to a fractional form. Ans. *£■* 7. Change 100 to an improper fraction. N 14 158 COMMON FRACTIONS. 8. Change 5 to a fraction whose denominator shall be 17. Ans. f- f- . 9. Reduce 98§f to an improper fraction. Ans. -^yf--. 10. Reduce 116$f to an improper fraction. Ans. x £? z . 11. 718||- equal how many ninety-sevenths ? Ans. —g- 6 /- 1 . 12. Reduce 100i§J to an improper fraction. Ans. 2£g. uaB- 30 40 45 48 ^ "50-» ^0~> SJJ' An« 594 308 3 15 441 ^• ns » -6"FTJJ "6"93") ^F3"> "6"9TT* Ans. Ans, 798 T"6"8- 70 63 TFF? T6B? 2 4 15 21 < Ans. f J, i| > ¥2' ¥; Ane 224 147 238 24 ^ nS - 252? 252? 2 525 2.52* A n « 570 140 165 12 ^Lns. -g-uQ-, -g-^Q-, -go-^j, -ggxr' And 210 140 105 84 AUS. 2-yg-J 2£3"> 2¥0"> 2¥(T Reduce the following fractions to a common denominator : — tV 16. Reduce f, f, and ' nominator. 17. Reduce f , f , and T %- 18. Reduce T 6 T , f , and 19. Reduce T V, T % and 7f . 20. Reduce -ff, f, and T V 21. Reduce f, T 4 T , and 11 T V 22. Reduce £, §, £, and 8. 23. Reduce f, T 7 T , and f of 7|. Reduce ^, f , £, and 17. Reduce $J, £ of 6, and 21 J. to fractions having a common de- A^q 80 96 105 Allb. J2(T) T2 T2TT' Ans. Ans. 360 560 189 USO"? -BSU) ^UTT- _5 4 6_ _5 7 2 _6 1 6 10"0~1> 10~0T> TO~0~T' Ant- 1485 1020 612 ■ AnS « 2 2 9"5J 22F"5> 22 ?? Ans Ans Anq 528 7 56 6039 AnS * TTF8 - ? TTF8"? TTBB" 204 _5_4 0_ 22F5? 22F5? 21 28 24 ¥2? ¥2? ¥2? 263 16 "2 2 ¥ 5"' 24. 25. 26. Ans. 110 T2U> 5.X6 2- T20? T w- Reduce f, T 4 T , T % f , and J. Anc 12012 5096 5390 8008 7007 Ans. T ¥IFT¥> T4 T¥tfT¥> T¥T5T¥? T¥ITT¥- 27. Reduce T %, $f$, and v ^. An« 28506816 37088064 722813 Ans. TT¥4^T5 5 2? TTF¥U15 5 2? TtW^T552' ADDITION OF COMMON FRACTIONS. 227. Addition of fractions is the process of finding the value of two or more fractions in one sum. Note. — Only units of the same kind, whether integral or fractional, can be collected into one sum ; if, therefore, the fractions to be added do not express the same fractional unit, they require to be brought to the same, by being re- duced to a common or the least common denominator. 228. To add together two or more fractions. 14* 162 COMMON FRACTIONS. Ex. 1. Add T 3 2-, T 5 2-, T 7 2-, and {§ together. Ans. f f = 2£. operation. These fractions A + A + tV + U = f I = ¥ = H- all being te*#fc, that is, having 12 for a common denominator, we add their numerators together, and write their sum, 26, over the common denominator, 12. Thus we obtain |-|, which, being reduced, = 2^, the sum required. 2. What is the sum of }, fo \& and £g ? Ans. 2$j$. $ 12 16 120 3 4 8 12 20 X 4 X 3 = 240 16 20 OPERATION. 2 4 = least common denominator. 30 X 7 = 210 20X 5 = 100 I new numera- 15X11 = 165 f tors. 12x13 = 156 Sum of numerators, 6 3 1 ' 21 51 Ans. Least com. denom., 2 4 2?TT ' ' The given fractions not expressing the same kind of fractional unit, we reduce them to their least common denominator, and thus make the fractional parts all of the same kind. The fractions now all expressing two-hundrcd-fortieths, we add their numerators, and write the result, 631, over the least common denominator, 240, and obtain f|^ = 2^fJ, the answer required. Rule. — Reduce the fractions, if necessary, to a common, or the least common denominator, and write the sum of the numerators over their common di nominator. Note 1. — Mixed numbers must be reduced to improper fractions, and com- pound fractions to simple fractions, and each fraction to its lowest terms, be- fore attempting to obtain the common denominator. Note 2. — In adding mixed numbers, the fractional parts may be added separately, and their sum added to the amount of the whole numbers. Examples. 3. Add T 5 T , T 6 T , T 9 T , \\, -J4, and |f together. Ans. 3-ff , 4. Add ,fe &, H, i%, and £f together. Ans. 2*J. 5. What is the sum of ^ T , ^-f , |f , and £f ? 6. What is the sum of T W, T \% T \% and |ff ? Ans. If. 7. What is the sum of |Jf, |§ j, |f {, and fa ? Ans. mi. 8. Add J, £, 1£, and f together. 9. Add -J T , /g-, -J, and \ together. Ans. lffj. 10. Add T \, y 5 F > f$, and J together. Ans. 2£f£. COMMON FRACTIONS. 163 11. Add T 5 T , &> rtfe an( * i together. Ans. 1. 12. Add -f-f , J-J, £& and -& together. Ans. 3£9. 13. Add }, i, & I, & and j- together. Ans. 1 T \\. 14. Add f , $, and 5f together. Ans. 6f§f. 15. Add -^ T , /^ and 9 T 3 T together. 16. Add f, |, and 4f together. Ans. 6 J. 17. Add |, 7£, and 8f together. Ans. 17^V 18. Add £, 3£, and 5f together. Ans. 9^J. 19. Add 6f, 7f, and 4f together. Ans. 18£^. 20. What is the sum of 17$, 14£, and 13 f ? 21. What is the sum of 16§, 8£, 9f, 3£, and 1 J- ? Ans. 40 T V 22. What is the sum of 371-J-J, 614 Jg-, and 81f ? Ans. 1068fJ. 23. Add f of 18 T 3 T , and ■& of f of 6 T 3 T together. Ans. 12ff£. 24. Add £ of 18, and T 4 T of -fJ- of 7 T \ together. 229. To add any two fractions, whose numerators are alike. Ex. 1. Add I to l> Ans. ^. operation. \Ye first find the Sum of the denominators, 5 -}- 4 = 9 sum of the denom- Product of the denominators, 4x5 = 20 i ^ tor ^ ™ hic }\ . is 9, and then their product, which is 20 ; and the 9 being written as a numerator of a fraction, and the 20 as its denominator, the result, -^, is the answer required. The reason of the operation is, that the process reduces the fractions to a common denominator, and then adds their numerators. Hence, to add two fractions whose numerators are a unit, Write the sum of the given denominators over their product. 2. Add £• to f Ans. lfa Sum of the denominators X OPERATION. by one of the numerators, (4 + 5) X 3 27 " _ ! '_ 1 7 AriS Product of the denominators, 4x5 "20 By multiplying the sum of the denominators by one of the numer- ators for a new numerator, and the denominators together for a new denominator, we reduce the fractions to a common denominator, and add their numerators, and thus obtain -|J = l^g-, the answer re- quired. Hence, to add fractions whose numerators are alike, and greater than a unit, Write the product of the sum of the given denominators by one of the numerators over the product of the denominators. 164 COMMON FRACTIONS. Examples. 3. Add & to i, i to i i to £, i to*, * to i, $ to jfc i to f 4. Add T V to J, T V to i, T \ to J, J T to |, J T to i, T \ to f o. Add T V to i, T V to i T V to |, T V to |, T V to J, T V to f 6. Add J to J, J- to £, i to i, J to i, i to fc i to |, J- to J. 7. Add * to $, | to J, i to i, | to £, i to i, i to i, i to J. 8. Add } to *, | to J, | to i, | to i, | to i, | to 4, -f to £. 9. Add iU>i,i to J, 4 to i, i to *, 4 to & i to |, i to f 10. Add f to T 8 T , f to &, f to A, f to T «V, f to T »^, f to jft-. 11. Add ft to *, | to f, | to T S f to?, f tof, f to T 2 T , f to f 12. Add § to -ft-, f to f, & to f £ to f, £ to £, £ to f> T . 13. Add f to f, f tO -ft-, } tO ^ T 6 T tO T % T 6 T tO f 7 , T 6 T to T %. u. Add f to T « T , t b t to -ft., a to T ^, a to ^ A to jy, a to T v 15. Add A to T 9 T , A to T 9 ¥ , A to Ai A to T % A to A- SUBTRACTION OF COMMON FRACTIONS. 230. Subtraction of Fractions is the process of finding the difference between two fractions. Note. — When the fractions express different fractional units, they require to be brought to those of the same kind before the subtraction can be per- formed. • To subtract one fraction from another. Ex. 1. From \$ take A- Ans - t? — h operation. The fractions both being twelfths, having 12 jj- — A = tV f° r a common denominator, we subtract the less numerator from the greater, and write the difference, 6, over the common denominator, 12. Thus, we have A as the difference required. 2. From tf take f Ans. f | = ££ . operation. The given frac- 7 7 common denominator. tions not express- -i -i [ 7 v -i s\ n r\ \ i n g the same kind 7iiv A-.A At new numerators. of fractional unit, / 1 1 i X * — 44 ) we reduce them 2 6 dif. of numerators. to a common de- — - . nominator, and 4 8 common denominator, thus make the fractional parts all of the same kind. We next find the difference of the new numera- tors, which we write over the common denominator, and obtain ff , the answer required. COMMON FRACTIONS. 165 Rule. — Reduce the fractions, if necessary, to a common, or the least common denominator. Write the difference of the numerators over their common denominator. Note. — If the minuend or subtrahend, or both, are compound fractions, they must be reduced to simple ones. Examples. 3. Subtract T 6 T from ±f. Ans. T 7 T . 4. Subtract T 4 F from -£§ . Ans. T 9 F . 5. From §f take ^ 8 T . 6. From £f take £$. Ans. &. 7. From f f take £$. Ans. ^. Ans. £. 8. From §f take £f 9. Subtract ^-f- from f J. jsois. £. 10. Subtract T \\ from T 3 ¥ 6 ¥ . Ans. -fa. 11. Subtract T ^ from ^-fa. Ans. ^ 12. Subtract T 3 T 2 / F from T 5 T 9 / B . Ans. T 2 T %. 13. Subtract T £§ ^ from T V°xyV Ans. W^. 14. From \\ take ^. 15. From -/^ take x 5 ^. Ans. ^J. 16. From \\ take T V Ans. §f£. 17. From T | take &. Ans. f |f. 18. From ^ take T 4 T . Ans. J^ T . 19. From f£ take T %. 20. From ^z take ^ Ans. y 3 ^. 21. From £$ take T 5 g. Ans. &, 22. From ^ take ^. Ans. -fe. 23. From 7f take f of 9. Ans. l^. 24. What is the value of f of 8f — § of 5 ? 25. What is the value of i of 3 — £ of 2 ? Ans. jfc 281. To subtract a proper fraction or a mixed number from a whole number. Ex. 1. From 7 take 3f. Ans. 3f. operation. Since we have no fraction from which to subtract From 7 * ne 1 » we must add 1, or its equal, -§, to the minuend, Take 3 6 an( ^ sa ^ "§" ^ rom I" ^ eaves f • W e WI "ite the | below * the line, and carry 1 to the 3 in the subtrahend, and Rem. 3 §- subtract as in subtraction of simple whole numbers. The result will be obtained, if we 166 COMMON FRACTIONS. » Subtract the numerator from the denominator of the fraction, and under the remainder write the denominator, and adding one to the ichole number in the subtrahend, subtract the sum from the minuend. Note. — When the subtrahend is a mixed number, we may reduce it to an improper fraction, and change the whole number in the minuend to a fraction having the same denominator, and then proceed as in Art. 230. Examples. 2. 3. 4. 5. 6. 7. From 3 2 16 671 385 1 6 18 Take 5| H o» im om If Eem. 2 6£ ll* 6 7 0& 3 68ff i*m 16* 8. 9. 10. 11. 12. 13. From 1 9 27 169 711 46 81 Take 1 3f ** &1** 8 0ff 15f 4 9W FIRST OPERATION. From 8f = -m Take H- - 4.28 232. To subtract one mixed number from another mixed number. Ex. 1. From 8? take 4|. Ans. 3ff. We first reduce the fractional parts to a com- mon denominator, and obtain as their equivalents 4-| and |4. Now, since we cannot take -f | from -||, we add 1, equal to |4, to the \% in the min- Rem. 3ff uen d, and obtain f£. From f£ taking ff , we have left $-§, which we write below the line, and carry 1 to the 4 in the subtrahend, and subtract from the 8 above as in subtraction of simple whole numbers. second operation. In this operation, we reduce From 8f = -^ = 2$£- the mixed numbers to improper Take 4^ = 2Jt - = -'t^- 8 - fractions, and these fractions to a common denominator. We Rem. *f£- = 3§f then subtract the less fraction from the greater, and, reducing the remainder to a mixed number, obtain 3||-, as before. Hence, in performing like examples, we may Reduce the fractional parts, if necessary, to a common denominator, and subtract the fractional parts and the ichole numbers separately. Increase the fractional part of the minuend, when otherwise it would be less than the subtrahend, before subtracting, by as many parts as it takes to make a unit of the fraction (Art. 208), and carry 1 to the whole number of the subtrahend before subtracting it. Or, Reduce the mixed numbers to improper fractions, then to a common denominator, and subtract the less fraction from the greater. CL COMMON FRACTIONS. 167 Examples. 2. 3. 4. 5. 6. 7. cwt. Tuns. 1 lb. OZ. Miles. From 1 8| 7 3| 6 7* 2 9lf 144£ 171M Take 9f 16 T2" 16| 10. 15| 9 9|x »1* 8. 9. 11. 12. 13. Furlongs. Rods. Inches. Feet. Bushels. Pecks. From 1 1£ 165^ *7« 8 4f| 6 7Hii 17«? Take 9 3| 9 8J l»i» 1«« 18 3|1 «m 14. From a hogshead of wine there leaked out 7 T 9 T gallons ; what quantity remained ? Ans. 55 T ^-gal. 15. A man engaged to labor 30 days, but was absent 5^ days ; how many days did he work ? 16. From 144 pounds of sugar there were taken at one time 17f pounds, and at another 28^ pounds ; what quantity remains ? Ans. 97^-f lb. 17. A man sells 9 J- yards from a piece of cloth containing 34 yards ; how many yards remain ? Ans. 24|yd. 18. The distance from Boston to Providence is 40 miles. A, having set out from Boston, has travelled -f Y of the distance ; and B, having set out at the same time from Providence, has gone T ^- of the distance ; how far is A from B ? Ans. 28 T | T m. 19. From \ of a square yard take ^ of a yard square. 233. To subtract one fraction from another, when their numerators are alike. Ex. 1. From £ take |. Ans. A- OPERATION. 7 — 3 = 4, difference of the denominators. 7 X 3 = 21, product of the denominators. We first find the product of the denominators, which is 21, and then their difference, which is 4, and write the former for the denom- inator of the required fraction, and the latter for the numerator. By this process the fractions are reduced to a common denominator, and their difference found. Hence, to subtract one fraction from another, whose numerators are a unit, we may Write the difference of the denominators over their product. 168 COMMON FEACTIONS. 2. Take f from f . Ans. &. OPERATION. 'Ans. Difference of the denominators mul- tiplied by one of the numerators, (7 — 3) X 2 8 Product of the derominators, 3x7 21 We multiply the difference of the denominators by one of the numerators for a new numerator, and the denominators together for a new denominator, by which process the fractions are reduced to a common denominator, and the difference of their numerators is found. Hence, when the given fractions have their numerators alike and greater than a unit, we may Write the product of the difference of the given denominators, by one of the numerators, over the product of the denominators. 17. Take T 8 T from f- ; T 8 T from f ; T 8 T from | ; 18. Take T 2 T from f ; T 2 T from f ; T 2 T from § ; 19. Take ft from Xp- ; ft from ft ; ft from ft ; ft from 20. Take \ 2 - from ■£ ; -V 2 - from *£ ; - 1 / from ^ ; ft from MISCELLANEOUS EXAMPLES IN ADDITION AND SUB- TRACTION OF ERACTIONS. 1. A benevolent man has given to one poor family J of a cord of wood, to another £ of a cord, and to a third ^ of a cord ; how much has he given to them all ? Ans. 2£ cords. COMMON FRACTIONS. 169 2. I have paid for a knife $ f , for a Common School Arith- metic $ £, for a slate $ -£, and for stationery $ J ; what did I pay for the whole ? 3. R. Howland travelled one day 20 T 7 - miles, another day 19 J- miles, and a third day 22 T 1 6 - miles ; what was the whole distance travelled ? Ans. 62§£ miles. 4. I have bought 6 J tons of anthracite coal, 19|- tons of Cumberland coal, and 3 j- tons of cannel coal ; what is the whole quantity purchased ? Ans. 30^ tons. 5. There is a pole standing £ in the mud, ^ in the water, and the remainder above the water ; what portion of it is above the water ? 6. F. Adams, having a lot of sheep, sold at one time § of them, and at another time £ of the remainder ; what portion of the original number had he then left ? Ans. ^. 7. From a piece of calico containing 31£ yards there have been sold llf yards, 9£ yards, and 3f- yards; how much re- mains ? 8. From a cask of molasses containing 84| gallons, there were drawn at one time 4f- gallons, at another time 11 gal- lons ; at a third time 2 6 J- gallons were drawn, and £ of 7J gallons returned to the cask ; and at a fourth time 13 T 8 T gallons were drawn, and 3 J- gallons of it returned to the cask. How much then remained in the cask? Ans. 35 Jf- Jgal. 9. A merchant had 3 pieces of cloth, containing, respectively, 19 j- yards, 36^- yards, and 33 1 yards. After selling several yards from each piece, he found he had left in the aggregate 71 f yards. How many yards had he sold ? Ans. 18^. MULTIPLICATION OF COMMON FRACTIONS. 234 1 Multiplication of Fractions is the process of multi- plying when the multiplier, or multiplicand, or both, are frac- tional numbers. Note. — If the multiplier is less than 1, only such a part of the multiplicand is taken as the multiplier is of 1. Therefore, the product resulting from multiplying a number by a proper fraction is not larger, but less, than the multiplicand. 235. To multiply when one or both of the factors are frac- tions. N 15 170 COMMON FRACTIONS. Ex. 1. Multiply ^ by 9. Ans. ff = 3*. first operation. ft i s evident that the fraction T3" X 9 == f| = J = H Ans. iV is multiplied by 9 by multi- plying its numerator by 9, since the parts taken, 63, are 9 times as many as before, while the parts into which the unit of the fraction is divided remain the same. second operation. ft is evident, also, that the fraction tV X = % = 3£ Ans. X is multiplied by 9 by dividing its denominator by 9, since the parts into which the unit of the fraction is divided are only \ as many, and con- sequently 9 times as large, as before, while the parts taken remain the same. Therefore, Multiplying the numerator or dividing the denominator of a fraction by any number multiplies the fraction by that number (Art. 217). 2. Multiply 14 by f Ans. G. first operation. By dividing the whole number, 14, by 7)14 7 > the denominator of the fraction, we obtain ^ of 14 = 2, which multiplied by 2X3=6 Ans. 3, the numerator of the fraction, gives $ of 14 = 6. second operation. By multiplying the whole number, 14, 14 by 3, the numerator of the fraction, we 3 obtain 42, a product 7 times as large as ~rr „ . it should be, as the multiplier was not 3, 4 ^ "*" ' == " Ans. a ^ota number, but -| , or 3 -f- 7 ; hence, we divide the 42 by 7 ; and thus obtain ^ of 14 = 6, as before. Therefore, Multiplying by a fraction is taking the part of the multiplicand de- noted by the multiplu r. 3. Multiply i by f . Ans. f J. operation. To multiply -J by f is to take f of the J X -| = f tj Ans. multiplicand, -|. Now, to obtain -| of -|, we multiply the numerators together for a new numerator, and the denominators together for a new denominator (Art. 226). Therefore, Multiplying one fraction by another is the same as reducing com- pound fractions to simple ones. When either of the factors is not a fraction, as in examples first and second, it may be reduced to a fractional form, and then the operation may be like that in the last example. Hence the general Rule. — Reduce whole or mixed numbers, if any, to improper frac- tion*. Multiply the numerators together for a new numerator, and the denominators together for a new denominator. Note. — "When there are common factors in the numerators and denomina-; tors, the operation may be shortened by cancelling those factors. COMMON FRACTIONS. 171 Examples. N 4. Multiply T 7 T by ft Ans. -J-. OPERATION. jr a l s/ — — • 11 %t 3 3 5. Multiply 12 by f x Ans. 8f 6. Multiply f by 12. 7. Multiply \\ by J$. Ans. £. 8. Multiply f by |£ Ans. yW 9. Multiply f by T 6 T . Ans. ||. 10. Multiply A- by if Ans. T 7 9V 11. Multiply T 8 T by f, Ans. f f 12. Multiply f by -if. 13. Multiply if by |J. Ans. |f 14. Multiply T 9 ^ by 14. Ans. 12f. 15. Multiply 13 by f. Ans. 7f 16. Multiply 16 by T V Ans. 2 T §. 17. Multiply 11 by f. Ans. 6f . 18. Multiply ■& by 14. 19. Multiply | by 19. Ans. 16f. 20. Multiply TT by ff. Ans. §. 21. Multiply -| by if. Ans. T V 22. Multiply T \ by § f . Ans. -&. 23. Multiply % by TT . 24. Multiply ^ by T ^. Ans. y^. 25. Multiply $ of T 7 T of H by 100. Ans. 12£. 26. Multiply i of f of £ by 11. Ans. 3ft. 27. What cost ^ of a ton of hay, at $ 17 per ton ? Ans. $ 9yi. 28. What cost ^ of an acre of land, at $ 37 per acre ? 29. At § of a dollar per foot, what cost 7 cords of wood? 30. Multiply leift by 19££. Ans. 3136f 3 -. 31. Multiply -f- by 8f . Ans. 3f . 32. Multiply T % by 17 T 3 T . Ans. 15 TT . 33. Multiply f by 71|, Ans. 63I 7 ;. 34. Multiply J of 9 ¥ by § of 17. Ans. 78 1. 35. Multiply & of 7 by i| of 87 T 3 T . 36. Multiply 8 by J. Ans. 6f . 172 COMMON FRACTIONS. 37. Multiply 12 by f Ans. 8f 38. Multiply 15 by T 6 T . Ans. 8 T 2 T . 39. A merchant owning £ of a ship sells T 4 T of his share to A. What part is that of the whole ship ? 40. Multiply 3£ by lOf Ans. 39J-f. 41. Multiply § of 1\ by \ of llf. Ans. 49|f|. 42. Multiply f of 9 by f of 17. Ans. 26^. 43. Multiply £ of S T \ by \ of 9f Ans. 25 ? VV 236. When one of the factors is a whole number, and the other a mixed number, we may Multiply the fractional -part and the whole number separately, and add together the products. Examples. I. Multiply 7£- by 9. 2. Multiply 12 by 3£. OPERATION. OPERATION. 7* 12 JL -if | X 9 = V-= 5| £ofl2= 9 7x9 =63_ 12X3 = 36 68f Ans. 45 Ans. 3. Multiply 8f by 7. Ans. 60£. 4. Multiply 17 by 3£. 5. Multiply 13 by 8f Ans. 109f . 6. Multiply 37 by 13^. Ans. 507|f. 7. Multiply llf by 8. Ans. 94f 8. What cost 7 T 6 T lb. of beef at 5 cents per pound ? 9. What cost 23 T y)bl. of flour at $ 6 per barrel ? Ans. $141$. 10. What cost 8|yd. of cloth at $ 5 per yard ? Ans. $ 41 J. II. What cost 9 barrels of vinegar at $ 6f per barrel? Ans. $ 57f . 12. What cost 12 cords of wood at % 6.37$ per cord? Ans. $7G.50. 13. What cost llcwt. of sugar at $ 9f per cwt. ? 14. What cost 4| bushels of rye at $ 1.75 per bushel ? Ans. $ 7.G5f . 15. What cost 7 tons of hay at $ 11 -J- per ton ? Ans. $ 83-|-. COMMON FRACTIONS. 173 16. What cost 9 dozen of adzes at $ 10§ per dozen? 17. What cost 5 tons of timber at $3£ per ton? Ans. $15 1. 18. What cost 15cwt. of rice at $7.62£ per cwt. ? Ans. $114,371. 19. What cost 40 tons of coal at $ 8.37 J per ton ? Ans. $ 335. DIVISION OF COMMON FRACTIONS. 237. Division of Fractions is the process of dividing when the divisor or dividend, or both, are fractional numbers. Note. — If the divisor is less than 1, the quotient arising from the division will be as many times the dividend as the divisor is contained times in 1. Therefore, the quotient arising from dividing a whole or mixed number by a proper fraction will always be larger than the dividend. 238. The reciprocal of a fraction is the number resulting from taking its numerator as denominator, and its denominator as numerator, since any two numbers, whose product is 1, are the reciprocals of each other. Thus, the reciprocal of -^ is that fraction inverted, or t 9 q-, since J F °- X tit = *• 239. To divide when the divisor or dividend, or both, are fractions. Ex. 1. Divide -ff by 7. Ans. T 2 T . first operation. ft is evident that the fraction if is di- T f -5- 7 = T 3jr Ans. vided by 7 by dividing its numerator by 7, since the size of the parts, as denoted by the denominator, remains the same, while the number of parts taken is only T as large as before. second operation. ft i s evident the fraction is also di- ^|. -a- 7 =b y 1 ^- = T 2 T Ans. vided by 7 by multiplying its denomi- nator by 7, since the number of parts taken, as denoted by the numerator, remains the same, while the size of the parts is only T as large as before. Therefore, Dividing the numerator or multiplying the denominator of a fraction by any number divides the fraction by that number (Art. 217). 2. Divide |f by T %. Ans. 2. operation. Si nce tne f rac tional units of the two t! •*" A = ^ Ans. fractions are of the same kind, it, is evident that 12 thirteenths contain 6 15* 174 COMMON FRACTIONS. thirteenths as many times as 6 is contained in 12; 12 -r- 6 = 2, Ans, Therefore, When the fractions have a common denominator, the division can be performed as in whole numbers, by dividing the numerator of the divi- dend by the numerator of the divisor, 3. Divide } by f. Ans. ljf first operation. Having reduced the frae- ^-T-f = f|-7-fJ= 1^4" Ans. tions to a common denomi- nator, we divide the nu- merator 32 of the dividend by the numerator 21 of the divisor, as in working the last example, and obtain as the required result 1^\. second operation. In the second oper- ^-f-f = 4-Xf = ff == l^x Ans. ation, we invert the divisor, and then pro- ceed as in multiplication of fractions (Art. 235). The reason of this process, which in effect reduces the fractions to a common denomi- nator, and divides the numerator of the dividend by that of the divisor, will be seen, if we consider that the divisor, -|, is an expres- sion denoting that 3 is to be divided by 8. Now, regarding 3 as a whole number, we divide the fraction ^ by it, by multiplying the denominator; thus, 4 x 8 = -fc. But the divisor 3 is 8 times as large as it ought to be, since it was to be divided by 8, as seen in the origi- nal fraction ; then the quotient, fa, is £ as large as it should be, and must be multiplied by 8 ; thus, fa * 8 = £? = 1%\, the answer, as before. By this operation we have multiplied the dividend by the reciprocal of the divisor, the denominator of the dividend having been multiplied by the numerator of the divisor, and the numerator of the dividend by the denominator of the divisor. Therefore, Dividing by a fraction is the same as multiplying by its reciprocal When either divisor or dividend is not a fraction, it may be changed to a fractional form, and the division performed by the last method. Hence the general Rule. — Invert the divisor, and then proceed as in multiplication of fractions. Note 1. — When either divisor or dividend is a whole or mixed number, or a compound fraction, it must be reduced to the form of a simple fraction be- fore dividing. Note 2. — Factors common to both numerator and denominator should be cancelled. Note 3. — When the given fractions have a common denominator, the answer may be obtained by dividing the numerator of the dividend by that of the divisor. Also, if the fractions have numerators alike, the answer may be ob- tained by dividing the denominator of the divisor by that of the dividend. Note 4. — When the numerator of the divisor will exactly divide the numer- ator of the dividend, and the denominator of the divisor exactly divide the denominator of the dividend, the division can be effected in that way. COMMON FRACTIONS. 175 Examples. 4. Divide T 7 <5- by &. Ans. 6|. OPERATION. 7 7 tf 62 6_2_ H) "*" 62 = l0 X 7" = T~0 = 6i AnS - Or A ■*• *;- ^ -*- 10 - 6i Ans. 5. Divide 2 % by f . Ans. f . OPERATION. 3 28^7~^ X 0~4 AnS * 4 9 3 9-r-3 = 3 A Or — -j — = — — Ans. 28 7 28h-7=4 6. Divide T 7 T by 18. Ans. ^. 7. Divide | by f Ans. f f . 8. Divide 18 by T 7 T . 9. Divide ^ T by f. Ans. £. 10. Divide £$ by f . Ans. f — If 11. Divide ^ by 28. Ans. ^ T V 12. Divide TT by 27. 13. Divide T \ by 128. Ans. ^. 14. Divide \\ by 98. Ans. T ^. 15. Divide £§■ by 19. Ans. £fc. 16. Divide f by 167. Ans. T ^. 17. Divide ££ by 49. Ans. T iJ T . 18. Divide y 1 ^ by 15. Ans. ^is- 19. Divide 27 by TT . 20. Divide 128 by &. Ans. 960. 21. Divide 98 by T f Ans. 151 TT . 22. Divide 19 by Jf Ans. 31 T y. 23. Divide 167 by T |. Ans. 200§. 24. Divide 49 by £§. 25. Divide 15 by T V Ans. 225. 26. Divide f f by ^. Ans. 4. 27. Divide fy by ^ Ans. 3£f 28. Divide £ by f . Ans. -ff . 29. Divide f| by T 7 T . Ans. lyW 176 COMMON FRACTIONS. 30. Divide ^ by |f 31. Divide £f by T V Ans. llf 32. Divide & by 7J. Ans. T 3 T V 33. Divide T 8 T by 16f. Ans. ^V 34. Divide llf by f 35. Divide 21f by 18f Ans. l^J. 36. Divide 17 T 3 T by 28£f Ans. ff^-g. 37. Divide 161-& by 14f Ans. ll T Jf T . 88. Divide -fj of £ by $ of T 8 T . Ans. If 39. Divide £ of 7 T 3 T by T 4 T of 17f Ans. f f J. 40. Divide T Ans. t ^ e di Y i sor (Art. 216), we divide ^ the numerator, f, by the denomi- nator, -|, as in division of fractions (Art. 239). 7 2. Eeduce — to a simple fraction. . Ans. ^ = 4J. If operation. "We reduce the nu- 7 _ i 7 3 __ _ . . merator, 7, and the p — 5 — T A 5 — 5- — *5> ^ nb - denominator, If, to 3 improper fractions, and then proceed as in Ex. 1 . Hence, to reduce complex to simple fractions, Consider the denominator as a divisor, and the numerator as a divi- dend, and proceed as in division of fractions (Art. 239). Note. — Another and often a ready method of reducing a complex fraction is to multiply both its terms by the least common multiple of their denomi- nators. Examples, jl 3. Reduce -^- to a simple fraction. Ans. ^. OPERATION. iXA-A= ttV, Ans. h -y JL3 — T ^ TS — TZ = 2S> 178 COMMON FRACTIONS, Or, multiply by the least common multiple of the denominators, 1 X 4 =2 6£ X 4 = 26 Ans. 4f 4. Reduce -£ to a simple fraction. Ans. 6 T 9 ¥ . 5. Reduce —- to a simple fraction. 52 7 6. Reduce — to a simple fraction. Ans. 1£ 4| 7. Reduce -— to a simple fraction. Ans. f J. 8 6? 8. Reduce -| to a simple fraction. Ans. § §. 3 2 9. Reduce f to a simple fraction. "5 Q 10. Reduce - to a whole number. Ans. 24. is i 11. Reduce - to a simple fraction. Ans. f. 2S 5 1 12. Reduce -^ to a mixed number. Ans. 12#. 3 13. Reduce ^ toa simple fraction. 3 14. Reduce — - to a mixed number. Ans. lh 31 15. Reduce ~ to a simple fraction. Ans. -Jf. 11 ? 16. Reduce --— | to a simple fraction. Ans. f f. 17. If 7 were to be the denominator of the fraction whose 77 numerator is -~> what would be its value? U| 18. If J is the numerator of the fraction whose denominator 3. is £ what is its value ? Ans. 6£f . 5 COMMON FRACTIONS. 179 243. Complex fractions, after being reduced to simple ones, may be added, subtracted, multiplied, and divided, according to the respective rules" for simple fractions. Examples. 3 6 39 A 1. Add i of f- of 28^| to 3—^. Ans. 6- 7 V 2. Add i, 2f, —i and Q together. Ans. 3£t§$|$%|. 495. 343 3. What is the difference between — ~ and 97 145 T 3 T Ans. Mfflfr. 4. What is the continued product of the following numbers 27 87 l * and KfeV 37| 98£ 2i 128 5. Divide § of 7f by f of 11 T 4 T . Ans. |^. 6. Divide f of 91 by T % of 87. Ans. f f J?-. MISCELLANEOUS EXAMPLES IN MULTIPLICATION AND DIVISION OF FRACTIONS. 1. At 2f bushels to an acre, how many bushels of wheat will be required to sow 7 T 4 T acres ? Ans. 17f . 2. Bought 8^- bushels of apples for $ 4.68f-; what did they cost per bushel ? Ans. $ 0.57|. 3. Bought a bale of cloth for $ 96f ; I dispose of it for § of the cost, and by so doing I lose $ 2 on a yard ; required the number of yards in the bale. Ans. 18 T f^-yd. 4. If a dividend be 18 J- times f and a quotient 6 J- times §, what was the divisor ? 5. By what number must If be multiplied, that the product o shall just equal 1 ? Ans. f . 6. Bought a horse and chaise for $ 250, and paid for the harness -fc of what I paid for the horse. The chaise cost |4- the value of the horse. What was the price of each ? Ans. Horse, $130£§; chaise, $119£f; harness, $83^1-. 180 COMMON FRACTIONS. 7. S. Walker has engaged to work at yearly wages of $ 200 and a suit of clothes. At the end of 9 months, falling sick, and being unable to labor longer, he receives the suit of clothes and $ 144, as the amount justly due. What was the cost of the clothes? Ans. $24. 8. What will be the result if £ of f of 3£ be multiplied by £ of itself, and the product divided by £ ? 9. Bought 13| acres of land at $ 25£ per acre, and paid for it in wheat at $ 2-J per bushel. How many bushels did it require ? Ans. 137££g bushels. 10. How long will it take a man to travel 553 miles, pro- vided he travels 3 £ miles per hour, and 9 £ hours per day ? 11. If $-2$ per cord is paid E. Holmes for sawing into three pieces, wood that is 4 feet long, how much more should he receive per cord for sawing into pieces of the same length wood that is 8 feet long ? Ans. $ 0.22£. 12. A steamboat leaves New Orleans, January 1st, bound up the river to a place distant 2317£ miles. Her forward motion is at the rate of 9| miles per hour for 1GJ hours each day, and she lies at anchor in the night for fear of running upon a snag. But having lost her anchor on the fifth day, she each succeeding night drifts backward, at the rate of 2 miles per hour. On what day of January will she reach her point of destination ? Ans. 15th day. A PROPOSED NUMERATOR, OR DENOMINATOR. 244. To reduce one fraction to another of equal value, hav- ing a proposed numerator, or denominator. Ex. 1. Eeduce f to an equivalent fraction having 4 for a numerator. Ans. -^r-. operation. The proposed numerator, 4, is such a 4 part of the given numerator as 5 divided g" X $ by 4, or | . Now, as the numerator pro- _ = _ ^ ns# posed is only | as large as the given 4 5f numerator, in order that the value of the 5 X ' two fractions be the same, the denomi- nator of the proposed fraction should be only £ as large as the denominator of the given fraction. Taking -f of the given denominator, 7, we obtain COMMON FRACTIONS. > 181 4 5§, which, written under the proposed numerator, gives — as the fraction required. 2. Eeduce f to a fraction of equal value having 12 for a denominator. Since the proposed denominator, 12, is 1£ of the given denominator, 9, we find ^ 2 - of the given numerator, 8, for numerator of the proposed fraction ; J 2. f 8 = 101, which, written over 10# the proposed denominator, gives — $- l u as the fraction required. Rule. — Take of both terms of the given fraction such a fractional part as the proposed numerator, or denominator, is of the given numer- ator, or denominator, and the result will be the required fraction. Examples. OPERATION. 12 9 X8 103 3 Ans. 12 12 T X0 3. Change JJ to a fraction whose numerator shall be 34. Ans. f&. 4. Change 3f to a fraction whose numerator shall be 9. Ans -i 5. Reduce 4 to a fraction whose numerator shall be 5. 6. Reduce -ff to a fraction having 12 for its denominator. 7. Change f to fifteenths. Ans. T 9 ^. 8. Reduce J to halves. 33^ 9. Reduce £$ to thirty-fifths. Ans. ^~ DO \10. J. Holton owns -Jf- of a wood-lot, and his brother ^/t °f the same lot; what fraction whose denominator shall be 12 will express the part each owns ? Ans. T 4 ^. A COMMON NUMERATOR. 245. A common numerator of two or more fractions is a common multiple of their numerators. 246. To reduce fractions to a common numerator. Ex. 1. Change f, f, f, and -^ to other fractions of the same value, having a common numerator. Ans. f |, ff, ff, £#. N 16 182 COMMON FRACTIONS. OPERATION. 36, least common multiple of the numerators, = common nu- - 3 tj 6 - of 4 = 48, new denominator, f = f-§~] [merator. ^ of 5 mm 45^ new denominator, t = f f ag. of 7 = 42, new denominator, f — ff * Ans# - 8 F 6 - of 10 = 40, new denominator, y 9 ^ = f^ "We find the least common multiple of all the numerators, which is 36, for the common numerator; and to obtain the several new de- nominators we take such a part of the given denominators, respec- tively, as the. common numerator, 36, is of each given numerator. Thus, both terms of each fraction being proportionably increased, its value is not changed. Rule. — Find the least common multiple of the given numerators for a common numerator. Take, for the new denominator of each fraction, respectively, such a part of its given denominator as the common numer- ator is of its given numerator. Note. — Compound fractions, or whole and mixed numbers, must be re- duced to simple fractions, and all to their lowest terms, before finding the common numerator. Kxami-i.ks. 2. Reduce f , % , £ , and f to other fractions of equal value having a common numerator. Ans. § $■, f£, §£, §§. 3. Change £, 2£, and 1^ to fractions having a common nu- merator. 4. A can travel round a certain island, which is 50 miles in circumference, in 4^ days, B in G§ days, and C in G§ days. If they all set out from the same point, and travel round the island the same way, in how many days will they all meet at the point from which they started, and how many times will each have gone round the island ? Ans. They will meet in 320 days ; A will have gone round the island 75 times ; B, 50 times ; and C, 48 times. GREATEST COMMON DIVISOR OF FRACTIONS. 247 • The greatest common divisor of two or more frac- tions is the greatest number that will divide each of them, and give a whole number for the quotient. 248? To find the greatest common divisor of two or more fractions. COMMON FUACTIONS. 183 Ex. 1. What is the greatest common divisor of ^, 2§, and 5£. Ans. ■£%. OPERATION. 4_ 02 5i JL_ 20 16 = J~2 100 _240 Greatest common divisor of the numerators = 4 ) Greatest com- Least common denominator of the fractions = 45 j required. Having reduced the fractions to equivalent fractions having the least common denominator, -we find the greatest common divisor of the numerators 12, 100, and 240 to be 4. Now, since the 12, 100, and 240 represent forty-fifths, their greatest common divisor is not 4, a whole number, bet 4 forty-fifths ; therefore we write the 4 over the least common denominator, 45, and have -£$ as the answer. Rule. — Reduce the fractions, if necessary, to their least common denominator. The greatest common divisor of the numerators, written over the least common denominator, will give the greatest common divisor required. Examples. 2. What is the greatest common divisor of f , -f-, f , and if? An*. #5, 3. What is the greatest common divisor of 12§, 9£, and 8^? 4. What is the greatest common divisor of ^-, f , f- , and f ? Ans. ^ 5. What is the greatest common divisor of 3f, 5 T 7 ^, and 2^-? 6. A farmer has 33f- bushels of corn, 67^- bushels of rye, 70 J bushels of wheat. He wishes to put this grain, without mixing, into the largest bags, each of which shall contain the same quantity. Required the number of bags and the quantity each will contain. Ans. The capacity of each bag, 3f bushels ; and the number of bags, 51. 7. I have three fields; the first contains 73 T 7 T acres, the second 88-^ T acres, the third 139|f acres. Required the lar- gest-sized house-lots of the same extent into which the three fields can be divided, and also the number of lots. Ans. Size of each lot, 7 T 4 T acres ; number of lots, 41. LEAST COMMON MULTIPLE OF FRACTIONS. 249. The least common multiple of two or more fractions is the least number that can be divided by each of them, and give a whole number for the quotient. 184 COMMON FRACTIONS. 250. To find the least common multiple of two or more fractions. Ex. 1. What is the least common multiple of f , T \, and 2 T \? Ans. 8 J. OPERATION. 3 6 91 3 3 3.3 Least common multiple of the numerators = 33 ( Least com " . . „ . — = 84- i nion multi- Greatest common divisor of denominators = 4 4 V pie required. Having reduced the fractions to their simplest form, we find the least common multiple of the numeratow, 3, 3, and 33, to be 33. Now, since the 3, 3, and 33 are, from the nature of a fraction, divi- dends, of which their respective denominators, 4, 8, and 16, are the divisors (Art. 216), the least common multiple of the fractions is not 33, a whole number, but so many fractional parts of the greatest common divisor of the denominators. This common divisor we find to be 4, which, written as the denominator of the 33, gives ^ = 8 \ as the least number that can be exactly divided by the given fractions. Rule. — Reduce the fractions, if necessary, to their lowest terms. Then find the least common multiple of the numerators, which, written over the greatest common divisor of the denominators, will give the least common multiple required. Note. — The least whole number that will contain two or more fractions an exact whole number of times, is the least common multiple of their numera- tors. Thus, 4, the least common multiple of the numerators of | and |, is the least whole that can be divided by those fractions, and give a whole number for a quotient; 4 -r § = 6 ; 4-7-5 = 5. Examples. 2. What is the least common multiple of f , f , and f ? Ans. 2± = 24. 3. Find the least number that 3^f , 7£, and 5J will divide without a remainder. Ans. 15 J . 4. What is the least common multiple of §, f , and T 9 ^ ? 5. What is the smallest sum of money with which I could purchase a number of sheep at § 2£ each, a number of calves at $ 4£ each, and a number of yearlings at $ 9f each ? and how many of each could I purchase with this money ? Ans. $ 112 \ ; 50 sheep ; 25 calves ; 12 yearlings. 6. There is a certain island 80 miles in circumference. A, B, and C agree to travel round it. A can walk 3 J miles in an hour, B 4f miles, and C h\ miles. They start from the same point and travel round the same way, and continue COMMON FRACTIONS. 185 their travelling 8 hours a day, until they shall all meet at the point from which they started. In how many days will they all meet, and how far will each have travelled ? Ans. In 17-f days ; A 480m., B 640m., and C 720m. 7. How many times the least common multiple of 3J, 4|, and 5£, is the least whole number that 3^, 4§, and 5£ will exactly divide. DENOMINATE FRACTION. 251. A Denominate Fraction is one in which the unit of the fraction is a denomination of a compound number ; as, § of a pound, |- of a mile, and I of a gallon. REDUCTION OF DENOMINATE FRACTIONS. 252. Reduction of denominate fractions is the process of changing fractions from the unit of one denomination to that of another, without altering their value. 253. To reduce a denominate fraction from a higher de- nomination to a lower. Ex. 1. Reduce ^J^ of a pound to a fraction of a penny. Ans. f d. OPERATION. 1 X 20 = 20 20 X 12 = 240 , _ 3 , A 640 640 s ' 5 640 640 ~ 8 3 n„ 1 X^X^ = 3, . a Since 20s. make apound, 0?0 8 ttere wil1 be 20 times as -^ ^ many shillings as pounds, *^ or -g 2 4°o-s. ; and since 12d. make a shilling, there will he 12 time's as many pence as shillings, orff£d. = fd. Rule. — Multiply the given fraction by the same numbers that would be employed in the reduction of whole numbers to the lower denomina- tion required. Examples. 2. Reduce y^W of a pound to the fraction of a farthing. 3. Reduce ^(j of a pound troy to the fraction of a grain. 4. Reduce 2 As" °f a pound, apothecaries' weight, to the fraction of a scruple. 5. Reduce - s $j$ of a cwt. to the fraction of an ounce. 16* 186 COMMON FRACTIONS. 6. Reduce ^y^u" of a ton to the fraction of a pound. 7. What part of an inch is ■%£ ^ of an ell English ? 8. What part of an inch is TT uVsTy of a mile ? 9. Reduce ^(jtf oi " a league to the fraction of an inch. 10. Reduce 3-5 a y ear ? 14. Reduce ^^tt °f a hundred-weight to a fraction of an ounce. 254. To reduce a denominate fraction from a lower de- nomination to a higher. Ex. 1. Reduce § of a penny to a fraction of a pound. Ans. s£ tkere yill be 7 TJ as man ) r shillings as 4 pence, or -As. ; and since 20s. make a pound, there will be -fa as many pounds as shil- lings, or ^ £. Ans. Rule. — Divide the fraction by the same numbers that would be em- ployed in the reduction of whole numbers to a higher denomination. Examples. 2. What part of a pound is f of a farthing ? 3. What part of a pound is f of a grain troy ? 4. Reduce |- of a scruple to the fraction of a pound. 5. Reduce T 6 T of an ounce to the fraction of a hundred- weight. 6. Reduce £ of a pound to the fraction of a ton. 7. Reduce £ of an inch to the fraction of an ell English, 8. Reduce f of an inch to the fraction of a mile. 9. Reduce ^ of an inch to the fraction of a league. 10. Reduce f of an inch to the fraction of an acre. COMMON FRACTIONS. 187 11. Reduce J of a quart to the fraction of a tun, wine measure. 12. Reduce f of a pint to the fraction of a bushel. 13. Reduce ^ of a minute to the fraction of a year (365£ days). 14. "What part of a hundred-weight is f of an ounce ? Ans. sgjjxi' 255 . To find the value of a fraction in whole numbers of lower denominations. Ex. 1. What is the value of T 3 T of a £. Ans. 5 s. 5d. l T 9 T far. OPERATION. 3£. 20 1 1 ) 6 s. (5 s. S i nce l£ . = 2 0s., A- of a £. is ■& 5 5 of 20s. = ff s. = 5Xs. ; and since ~~ 5 S# Is. = 12d., T 5 T of a shilling is T 5 T of I o * 12d. = ffd. = 5 T 5 T d. ; and, since Id. — = 4far . , T 5 T of a penny = T 5 T of 4far. 1 1 ) 6 d. ( 5 d. = |f far. = l T Var. Therefore, T \£. 5 5 = 5s. 5d. l T 9 T far. This is equiva- lent to multiplying the numerator of 5d. the fraction by the numbers required 4 to reduce it to successive lower de 11)20 far ( l- 9 far nominations, beginning with the high- i i ' ll ' est , and dividing each product by the denominator, as in the operation. T 9 T far. Ans. 5s. 5d. l T 9 T far. Rule. — Multiply the numerator of the given fraction by the number required to reduce it to the next lower denomination, and divide the product by the denominator. TJien, if there is a remainder, proceed as before, until it is reduced to the denomination required. Examples. 2. What is the value of fa of a shilling ? Ans. 3£d. 3. What is the value of J of a guinea, at 28 shillings ? Ans. 21s. 9d. l^far. 4. What is the value of T 7 T of a cwt. ? 5. What is the value of f of a lb. avoirdupois ? Ans. 7oz. ljdr. 6. What is the value of § of a lb. troy ? Ans. lOoz. 13pwt. 8gr. 188 COMMON FRACTIONS. 7. What is the value of -fa of a lb. apothecaries' weight ? Ans. 35 55 19 12 T 4 jgr. 8. What is the value of f of an ell English ? Ans. 2qr. 3na. Ojin. 9. What is the value of -f£ of a mile ? Ans. 6fur. 30rd. 12ft. 8 T 4 3-in. 10. What is the value of f of a furlong ? Ans. 35rd. 9ft. 2in. 11. What is the value of fa of an acre ? Ans. 2R. 6rd. 4yd. 5ft. 127 T yn. 12. What is the value of T 9 T of a rod ? 13. What is the value of T Vj of a cord ? Ans. 9ft. 1462 T Vn. 14. What is the value of T ^- of a hhd. of wine ? Ans. 6gal. 2qt. lpt. T %gi. 15. What is the value of £ of a hhd. of beer ? Ans. 42gal. 16. What is the value of ££ of a year (365 J days) ? Ans. 174d. 16h. 26m. 5^sec. 3 3 17. What is the value of 7-£ of a dollar? 256# To find the value of whole numbers in a fraction of a higher denomination. Ex. 1. What part of a £. are 5s. 5d. 1^-far. ? operation. W e reduce the 3s. 5s. 5d. l T 9 T far. = 2880 3 5d. l^tkr. to elevenths 2£ __ 10560 — TT of a farthing, the low- est denomination in the question, for the numerator of the required fraction, and 1 £. to the same denomination for the denominator. We then have the fraction Toi^o^" ==*-$£& as the answer. Rule. — Reduce the given numbers to the lowest denomination men- tioned in either of them. Then write the number which is the fractional part for the numerator, and the other number for the denominator, of the required fraction. Examples. 2. Reduce 3^-d. to the fraction of a shilling. Ans. ^fc. 3. Reduce 21s. 9d. l^far. to the fraction of a guinea. Ans. J. COMMON FRACTIONS. 189 4. Reduce 2qr. 131b. lOoz. 2££dr. to the fraction of a cwt. Ans. T 7 T . 5. What part of a pound avoirdupois are 7oz. ljdr. ? 6. What part of a pound troy are lOoz. 13pwt. 8gr. ? Ans. f . 7. Express 1 gallon liquid measure as a fraction of a gallon dry measure. Ans. ^f f f . 8. What part of a yard are 2qr. Ona. l T 5 3 in. ? Ans. T 7 ^. 9. What part of an ell English are 2qr. 3na. 0£in. ? Ans. f. 10. What part of a mile are 6fur. 30rd. 12ft. 8^11. ? 11. Reduce 35rd. 9ft. 2in. to the fraction of a furlong. Ans. f. 12. What part of an acre are 2R. 6rd. 4yd. oft. 127-^. Ans. T V 13. What part of a square rod are 144ft. 19 T 1 7 in. ? Ans. - 9 -. 14. What part of a cord are 9ft. 1462 T 2 ^in. ? 15. What part of a hogshead of wine are 6gal. 2qt. lpt. T %gi.? Ans. T V 16. What part of a pound avoirdupois is 1 pound troy ? Ans. iff. 17. What part of a year (365J days) are 174d. 16h. 26m. 5^ 5 3 sec. ? Ans. J£. ADDITION OF DENOMINATE FRACTIONS. 257. To add denominate fractions. Ex. 1. Add T 7 ^ of a £. to T 9 T of a £. Ans. 1£. 7s. Id. 2-^far. FIRST OPERATION. , ^ ^ ^ ^ °{^ X of a £. = 10s. 9d. O^ffar. ^ act \ on separately and add the two values together, T 9 T of a £. = 16s. 4d. 1-^fa r. according to the rule for Ans. 1£. 7s. Id. 2 T ^far. ^^^I?^ 111 " 1 numbers * SECOND OPERATION. A £ - + tt £ - = tB £ - — 1 £ - 7s. Id. 2 T %far. Ans. By the second operation, we first add the two fractions together, and then find the value of their sum. (Art. 255.) 190 COMMON FRACTIONS. Examples. 2. Add together -^ T of a ton and £ J of a cwt. Ans. 13cwt. 2qr. 3. Add together f of a yard, \ of an ell English, and f- of a qr. 4. Add together T \- of a mile, -£$ of a furlong, and ^ of a yard. Ans. 5fur. 16rd. Oft. 3-j^in. 5. A has three house-lots ; the first contains £ of an acre, the second § of an acre, and the third -£J of an acre. How many acres do they all contain ? Ans. 2A. 1R. 9p. 142ft. 87fin. 6. A man travelled 18^ miles the first day, 23-j-^ miles the second day, and 192T miles the third day. How far did he travel in the three days ? Ans. 61m. 2fur. 3rd. 13ft. 4f in. 7. Add \± of a gallon of wine to y 1 ^ of a hhd. 8. Add y\ of a week to -J of a day. Ans. 2d. 9h. 18m. 9. Add £ of a square foot to J a foot square. Ans. 1 foot. 10. Add G inches to llrd. 16ft. oin. Ans. 12rd. Oft. oin. SUBTRACTION OF DENOMINATE FRACTIONS. 258 • To subtract one denominate fraction from another. Ex. 1. From -fa of a £. take f of a £. Ans. 8s. 3d. ljffar. fiust operation. We find the value of each fraction -^Y of a £. = 12s. 8 T 8 T d. separately, and subtract one from the 2 of a £. = 4s. 5 } d. other, according to the rule for sub- — — — tracting compound numbers. (Art. Ans. 8s. 3£§d. 146.) SECOND OPERATION. By tllC SCCOnd T 7 T £. — f £. = f ^£. = 8s. 3d. l-J-ffar. Ans. operation, we first subtract the less fraction from the greater, and then find the value of their difference. (Art. 255.) Examples. 2. From % of an ell English take f of a yard. Ans. 3qr. Ona. 2-^in. 3. Take -fy of a furlong from § of a mile. Ans. lfur. 5rd. 10ft. lOin. COMMON FRACTIONS. 191 4. Take -f of a mile from f of a degree. Ans. 48m. 6fur. 17rd. 8ft. 7f-in. 5. From T 4 T of an acre take f of a rod. Ans. 1R. 17p. 22yd. 2ft. 108in. 6. From T 9 ^ of a cord take -f T of a cord. 7. From t 7 tj of a hogshead of wine there leaked out $ of it ; what remained ? Ans. 6gal. 3qt. Opt. 1-g^-gi. 8. From Boston to Concord, N. H., the distance is 72 miles ; f of this distance having been travelled, how much remains ? Ans. 30m. 6fur. 34rd. 4ft. 8f-in. 9. From f of a year take f of a week. Ans. lOlda. 5h. 54m. 17|sec. 10. From T 4 r of an acre take J of a foot. MISCELLANEOUS EXAMPLES IN FRACTIONS. 1. How far will a man walk in 17 T 3 T hours, provided he goes at the rate of 4 J- miles an hour ? Ans. 82m. 4fur. 8rd. 1ft. 4in. 2. How much land is there in a field which is 29 t 7 t T rods square ? Ans. 5A. 1R. 32p. 141ft. 109f|fin. 3. How much wood in a pile which is 17J feet long, 7-jL- feet high, and 4f feet wide ? Ans. 4C. 66|$Jft. 4. What is the value of 19 J barrels of flour, at $ 6f a bar- rel ? o. What is the value of 376|£ acres of land, at $75§ per acre?. Ans. $28387.06£. 6. What cost 17-j2j3j quintals of fish, at $ 4.75 per quin- tal ? Ans. $ 81.55 T 6 T 5 2-. 7. What cost 1670 T % pounds of coffee, at 12f cents per pound? Ans. $212.99|f. 8. What cost 28^ T tons of Lackawana coal, at $llf a ton ? Ans. $ 333.27 T 3 T . 9. Bought 37^ hogsheads of molasses, at $ 17.62J a hogs- head ; w T hat was the whole cost ? Ans. $ 655.202- 5 T . 10. What cost £ of a cord of wood, at $ 5.75 a cord ? 11. What are the contents of a field which is 139f- rods long, and 38$ rods wide ? Ans. 33A. 3R. 15£f p. 12. Bought 15 loads of wood, each containing llf feet, cord 192 COMMON FRACTIONS. measure. I divide it equally between 9 persons ; what does each receive ? Ans. 19-Lft. 13. If the transportation of 18f tons of iron costs $48.15f, what is it per ton ? Ans. $ 2.62^%. 14. If a hogshead of wine costs $ 98 J, what is the price of one gallon ? 15. If 5 bushels of wheat cost $8§, what will a bushel be worth ? Ans. $ 1.64|. 16. What will 11 hogsheads and 17J gallons of wine cost, at 19f cents a gallon ? Ans. $ 140.32J . 17. How many bottles, each containing If pints, are suffi- cient for bottling a hogshead of cider ? Ans. 288. 18. I have a shed which is 18 T 7 j feet long, 10^ feet wide, and 7^4 feet high ; how many cords of wood will it con- tain ? Ans. 1 1 C. 1 24 T %9 F ft. 19. What will G| pounds of tea cost, at 65 J cents per pound? Ans. $4.52^2- 20. How many cubic feet does a box contain, that is 8f feet long, 5 T 7 2- feet wide, and 8 feet high ? Ans. 146 T 9 ^ft. 21. How many feet of boards will it take to cover a side of a house which is 46^- feet long and 17£ feet high ? 22. Required the number of square feet on the surface of 7 boxes, each of which is 5£ feet long, 2 T ^- feet high, and 3 T \j feet wide ; required also the number of cubic feet they would occupy ? Ans. 527f f ft. ; 286f f $ cubic feet. 23. A certain room is 12 feet long, 11£ feet wide, and 7 J feet high ; how much will it cost to plaster it, at 2f cents per square foot ? Ans. $ 13.48 J. 24. A man has a garden that is 14 J rods long, and 10£ rods wide ; he wishes to have a ditch dug around it, that shall be 3 feet wide and 4£ feet deep ; what will be the expense, if he give 2 cents per cubic foot ? Ans. $ 223.7 6£. 25. How many bushels of grain will a box contain which is 14 T 7 2- feet long, 5-}-^- feet deep, and 4 J feet wide, there being 2150! cubic inches in a bushel ? Ans. 294|f f f bu. 26. Which will contain the most, and by how much, a box that is 10 feet long, 8 feet wide, and 6 feet deep, or a cubical one, each of whose sides measures 8 feet ? Ans. The last contains 32 cubic feet the most. COMMON FRACTIONS. 193 27. Which will contain the most gallons, a cistern that is 7£ feet long, 6 feet wide, and 5J feet deep, or one that is 9^- feet long, 4J- feet wide, and 5J- feet deep ? Ans. The first cistern contains 92|- gallons most. 28. My field has four sides. The first side is 31 rods 13^ feet in length ; the second, 41 rods ly 9 ^ feet ; the third, 38 rods 0-i feet; and the fourth, 45 rods 12 T 7 ^- feet. I wish to enclose this field with a rail-fence four rails high, using rails of equal length. Required the length of the longest rails that can be used, allowing that the rails lap by each other y 7 ^ of a foot ; also the number of rails it will take to fence it. Ans. Length, 13-|- feet ; number, 808. 29. Eequired the least number of yards of velvet, expressed by a whole number, that can be cut up without waste, into vest patterns of f , f , or T yards each ? Ans. 30 yards. 30. If a company whose capital stock is divided into 100 equal shares should conclude to divide the same stock into only 30 shares, how much larger would a share of the latter be than one of the former size ? 31. D. Ripley's farm contains 31 A. 3R. 6 p., and J. Ford's farm contains 39A. 2R. 37^-p. What is the fraction, in its sim- plest form, that will express their comparative size ? Ans. f . 32. Bought 68 barrels of flour, at $ 1\\ per barrel ; what was the amount of the whole ? Ans. $ 538 J. 33. What cost 8§ acres of land, at $ 42f per acre ? Ans. $369.20. 34. How shall four 3's be arranged, that their value shall be nothing ? 35. I have a room 20 feet long, 15 feet wide, and 8J feet high. This room contains 4 windows, each of which is 5£ feet in height and 3-J- feet in width. There are two doors 7 feet high and 3 feet wide. The mop-boards are § of a foot wide. A mason has agreed to plaster this room at 6|- cents per square yard ; a painter is to lay on the paper at 9 cents per square yard ; the paper which I wish to have laid on is 2f feet wide, for which I pay 5 cents per yard. What is the amount of my bill for plastering, for papering, and for paper ? Ans. For plastering, $5.1 Iff; for papering, $4.37; for paper, $ 2.80^ N 17 194 COMMON FRACTIONS, EXAMPLES TO BE PERFORMED BY ANALYSIS. 1. If -J- of a bushel of corn cost 63 cents, what cost a bushel ? what cost 15 bushels ? Ans. $ 10.80. Illustration. — If 7 eighths of a bushel cost 63 cents, 1 eighth will cost 1 seventh of 63 cents = 9 cents ; and 8 eighths will cost 8 times 9 cents — 72 cents, and 15 bushels will cost 15 times 72 cents — $ 10.80. 2. If 4 Jib. of pepper cost $ 2.15, what cost 1 pound ? what cost 301b. ? Ans. $ 13.50. Illustration. — In 4£lb. there are ^lb. Then, if 43 ninths lb. cost $2.15, 1 ninth will cost 1 forty-third of $2.15 = $ 0.05, and 9 ninths or lib. will cost 9 times $ 0.05 — $ 0.45, and 301b. will cost 30 times $0.45 = $ 13.50. 3. When $ 1728 are paid for 30 T \ tons of iron, what cost 1 ton ? what cost 7££ tons ? Ans. $ 432. 4. When $ 432 are paid for 7^J tons of iron, what quantity should be received for $ 1728 ? 5. For 7 -J- J tons of iron there were paid $ 432 ; what sum will it require to pay for 30^ tons ? 6. For 30 T ^ tons of iron S 1728 were paid; what quantity should be received for $ 432 ? 7. Gave 7^ bushels of rye for a barrel of flour ; how much rye will it then require to purchase 6| barrels of flour? Ans. 49 J|- bushels. 8. Divide $ 1728 among 17 boys and 15 girls, and give each boy -^j as much as a girl ; what sum will each receive ? Ans. Each girl, $ 66f f ; each boy $ 42f f . 9. If £ of a ton of hay cost $ 14.49, what cost 4f tons ? 10. If 44 tons of hay cost $ 82.50^, what part of a ton will $ 14.49 buy ? 11. If $ 14.49 will buy J of a ton of hay, how much hay can be obtained for $ 82.50f ? 12. When $ 82.50"f are paid for 4f tons of hay, what will be the cost of J of a ton ? 13. When 14 J tons of copperas are sold for $500, what is the value of 1 ton ? what is the value of 9f£ tons ? COMMON FRACTIONS. 195 14. When 9-J£ tons of copperas are sold for $ 333.33£, what is the value of 14|- tons ? 15. Gave $ 333.33^- for 9f i tons of copperas ; what quantity of copperas should be received for $ 500 ? '16. For 14§- tons of copperas $500 were paid ; how much might be purchased for $ 333.33^ ? 17. Purchased 97J gallons of molasses for $31.32; what cost 1 gallon ? what cost 763-g- gallons ? 18. Sold 763f gallons of molasses for $ 244.36 ; what should I receive for 97-J gallons ? 19. If $244.36 will buy 763f gallons of molasses, what quantity can be obtained for $ 31.32 ? 20. Gave 19751b. of flax for 40 barrels of flour; how many pounds were given for 1 barrel ? how many pounds would it require to buy 144 barrels ? Ans. 71101b. 21. If 17 bushels of rye cost $ 15.75, what cost 1 bushel ? what cost 9 1 bushels ? Ans. $ 8.5 6f J. 22. If 9 barrels of flour cost $ 5 Of- , what cost 1 barrel ? what cost 87f barrels ? Ans. $ 492^. . 23. If 13 boarders consume a barrel of pork in 78 days, how long would it last, if 7 more boarders were added to their num- ber ? Ans. 50^ days. 24. If a man by laboring 10 hours a day can, in 9 days, per- form a certain piece of work, how many days would it require to do the same work, were he to labor 15 hours a day ? 25. If a man, by laboring 15 hours a day, in 6 days can per- form a certain piece of work, how many days would it require to do the same work by laboring 10 hours a day ? 26. If a man, by laboring 10 hours a day, can in 9 days per- form a certain piece of work, how many hours must he labor each day to perform the same work in 6 days ? 27. Sold 17 T 3 T bushels of corn for $ 5f- ; what was received for 1 bushel ? what should I have charged for 97f bushels ? Ans. $30fff. 28. Bought 9f tons of hay at $ 19$ per ton; for what must it be sold per cwt. to gain $ 7 on my bargain ? Ans. $ l^ff ^. 29. If I sell hay at $ If per cwt., what should I give for 9f tons, that I may make $ 7 on my bargain ? Ans. $ 329. 30. How many bushels of corn at $0.75 per bushel will 196 COMMON FRACTIONS. it require to purchase 47 T 3 T bushels of wheat at $2§ per bushel? Ans. 168g%bu. 31. If 15 cords of wood cost $ 57 T 9 T , what cost 1 cord? what cost 19 J cords ? 32. If 19 1 cords of wood cost $ 76 T 6 ft, how many cords may- be obtained for $ 57 T 9 T ? 33. At 7 T 3 IJ shillings per yard, what cost 47-J yards ? Ans. 17£. 5s. 6§d. 34. When 172£. 15s. Ofd. are paid for 47£ yards of broad- cloth, what is the value of 1 yard ? Ans. 3£. 12s. llfffd. 35. If lib. of sugar cost ^ of a dollar, what is the value of 43flb.? Ans. $ 23.61ft. 36. If 17glb. of sugar cost $ 2 T 7 T , what cost 501b. Ans. $ 7.58-HhB • 37. Bought 87^ yards of broadcloth for $ 612 ; what was the value of 14ft- yards ? Ans. $ 102.90. 38. If £ of an acre of land cost $ 43.75, what cost 10 acres ? 39. When $ 500 are paid for 10 acres of land, how much might be obtained for § 43.75 ? 40. If 9 hogsheads of sugar cost S 71.87, what cost f of a hogshead ? 41. Paid $ 4.56§ J for f of a hogshead of sugar ; what ought to be given for 9 hogsheads ? 42. If 19 men can grade a certain road in 111 days, how long would it require 47 men to perform the same labor ? 43. When 47 men can grade a certain road in 44|| days, how long would it require 19 men to perform the same labor ? 44. If T 4 T of a ton of hay cost $ 9.20, what cost 17 tons ? 45. When $ 430.10 are paid for 17 tons of hay, what cost T 4 T of a ton ? 46. If ft- of a tub of butter cost $ 7.15, what cost 7 tubs ? 47. When $ 114.40 are paid for 7 tubs of butter, what cost -ft of a tub ? 48. If a horse eat 19 f- bushels of oats in 87f- days, how many will 7 horses eat in 60 days ? Ans. 93£ bushels. 49. Henry Smith can reap a field in 10 days, by laboring 8 hours a day. His son John can reap the same field in 9 days, by laboring 12 hours a day. How long would it take both to reap the field, provided they labored 8 hours a day ? Ans. 5£f days. DECIMAL FRACTIONS. 197 DECIMAL FRACTIONS. 259. A Decimal Fraction is a fraction whose denom- inator is some power often. It, therefore, originates from dividing a unit, first into 10 equal parts, and then each of these parts into 10 other equal parts, and so on indefinitely, so that its fractional units are tenths, hundredths, thousandths, or some like order of parts. 260. Decimal fractions, their denominators being obvious, are commonly expressed by writing the numerator only, with the decimal point (.) before it, care being taken to put a cipher in any decimal place not requiring a digit ; thus, . T 9 (5- may be written .9 and be read 9 tenths. T V^ " .13 " 13 hundredths. T^ny " .005 " 5 thousandths. T#Hfo " - 0105 " 105 ten-thousandths. 261 • By examining the foregoing fractions, it will be seen that, — 1. The denominator of a decimal fraction is 1 with as many ciphers annexed as the numerator has places of figures, 2. In writing a decimal fraction without its denominator, every decimal place not having a significant figure must he filled by a cipher. 3. The first figure or place of a decimal fraction on the right of the decimal point is tenths ; the second, hundredths ; the third, thousandths ; the fourth, ten-thousandths ; fyc. 4. Each figure in the expression of decimal fractions, as in whole numbers, represents value, according to its distance from the place of units. 262. A whole number and a decimal fraction, in a single expression, constitute a mixed number. Thus, 17.63 is a mixed number, and is read seventeen, and decimal sixty-three hundredths ; 150.302, read one hundred and fifty, and decimal three hundred and two thousandths. Note. — For the sake of brevity, especially in reading mixed numbers, as in the instance just given, a decimal fraction is commonly called simply a deci- mal. 17* 198 DECIMAL FRACTIONS. 263. If ciphers are placed on the left of decimal figures, between them and the decimal point, those figures change their places, each cipher removing them one place to the right, and thus diminishing the value represented tenfold. Thus, .9 = t 9 q, but .09 = TJfo, and .009 — TX fa. 264. If ciphers are placed on the right of decimal figures, or are taken away, since their places remain the same, the value represented is not changed. Thus, .7 = ^ and .70 = 265. By regarding the dollar as a unit, we may consider cents and mills of United States money as fractional parts of a decimal character. Thus, 3 dollars and 25 cents, is 3 dollars and 25 hundredths of a dollar, or $3.25; also, 10 dollars 12 cents and 5 mills, is 10 dollars and 125 thousandths of a dol- lar, or $ 10.125. NOTATION AND NUMERATION OF DECIMALS. 266. The relation of decimals to whole numbers and to each other, and also the names of their different orders and places, are shown by the following Table. 8 1 ill o O UJ G U O 3 T3 a •8 1 4 I s 3 -i k i J * I 3 ^ ■? -2 ^ "2 .2 o^s § a s o s Ja S ^ 3 o >2 fl HHKHPPHWHHW§HWpq 654321.234567892 0) O O 03 O Q) Cs(r> -S "S^^^^^rS^^ t~ q? lO rf CO -OOC5 Whole Numbers. Decimals. DECIMAL FRACTIONS. 199 Of the mixed number expressed in the table, the part on the left of the decimal point is the whole number, and that on the right the decimal. The decimal part is numerated from the left to the right, and the value represented is expressed in words thus : Two hundred thirty-four million five hundred sixty-seven thousand eight hundred ninety-two billionths. And the mixed number thus : Seven million six hundred fifty-four thousand three hundred twenty-one, and decimal two hundred thirty-four million five hundred sixty-seven thousand eight hun- dred ninety-two billionths. 267. From the table we deduce the following rules : 1. Read a decimal as though it were a whole number, adding the name of the right-hand order. 2. Write a decimal as though it were a whole number, sup- plying with ciphers such places as have no significant figures. Examples. Express orally, < [>r write in words the followin g numbers : — 1. .056 6. 1.631 11. 1.000007 2. .1003 7. 48.07 12. 5.101016 3. .2786 8. 1.315 13. 1.000327 4. .16302 9. 5.6001 14. 0.000001 o. .97500 10. 87.0006 15. 16.000000007 Express in the c lecimal form by figures : — 16. T V 3 o- 17 6 18 - TirVW 19. 20. 21. 406 To~(T(nro~ i TT5"0"0"T5"UT5" 10 3 1 Tff(T(T 22. 23. 24. 7 17 333^^^ iTotftjxnnr 25. Three hundred twenty-five, and seven tenths. 26. Four hundred sixty-five, and fourteen hundredths. 27. Ninety-three, and seven hundredths. 28. Twenty-four, and nine millionths. 29. Two hundred twenty-one, and nine hundred-thousandths. 30. Forty-nine thousand, and forty-nine thousandths. 31. Seventy-nine million two thousand, and one hundred five thousandths. 32. Sixty-nine thousand fifteen, and fifteen hundred-thou- sandths. 200 DECIMAL FRACTIONS. 33. Eighty thousand, and eighty-three ten-thousandths. 34. Nine billion nineteen thousand nineteen, and nineteen hundredths. 35. Twenty-seven, and nine hundred twenty-seven thou- sandths. 36. Forty-nine trillion, and one trillion th. 37. Twenty-one, and one ten-thousandth. 38. Eighty-seven thousand, and eighty-seven millionths. 39. Ninety-nine thousand ninety-nine, and nine thousand nine billionths. 40. Seventeen, and one hundred seventeen ten-thousandths. 41. Thirty-three, and thirty-three hundredths. 42. Forty-seven thousand, and twenty-nine ten-millionths. 43. Fifteen, and four thousand seven hundred-thousandths. 44. Eleven thousand, and eleven hundredths. 45. Seventeen, and eighty-one quadrillionths. 46. Nine, and fifty-seven trillionths. 47. Sixty-nine thousand, and three hundred forty-nine thou- sandths. 268. Decimals, since they increase from right to left, and decrease from left to right, by the scale of ten, as do simple whole numbers, may be added, subtracted, multiplied, and divided in the same manner. ADDITION OF DECIMALS. 269, Ex. 1. Add together 23.61, 161.5, 2.6789, and 61.111. Ans. 248.8999. OPERATION. .^ . 2 3 6 1 e wnte tne num bers so that figures of the - £ " k same decimal place shall stand in the same col- o^ _ Q a unin, and then, beginning at the right hand, add 2.6 7 8 J them as whole numbers are added, and place 6 1.1 1 1 the decimal point in the result directly under 2 4 8.8 9 9 9 those above * Rule. — Write the numbers so that figures of the same decimal place shall stand in the same column. Add as in whole numbers, and point off in the sum, from the right hand, as many places for decimals as equal the greatest number of deci- mal places in any of the numbers added. .DECIMAL FRACTIONS. 201 Proof. — The proof is the same as in addition of whole numbers. Examples. 2. Add together the following numbers : 81.61356, 6716.31, 413.1678956, 35.14671, 3.1671, 314.6. Ans. 7564.0052656. 3. "What is the sum of the following numbers : 1121.6116, 61.87, 46.67, 165.13, 676.167895? Ans. 2071.449495. 4. Add 7.61, 637.1, 6516.14, 67.1234, 6.1234 together. Ans. 7234.0968. 5. Add 21.611, 6888.32, 3.6167 together. Ans. 6913.5477. 6. Add together $ 15.06, $ 107.09, $ 1.625, and $ 93.765. 7. I have bought a horse for $ 137.50, a wagon for % 55.63, a whip for $ 1.375, and a halter for % 0.87£ ; what did they all cost? Ans. $195.38. 8. What is the sum of twenty-three million ten ; one thou- sand, and five hundred-thousandths ; twenty-seven, and nineteen millionths ; seven, and five tenths ? Ans. 23001044.500069. 9. Add the following numbers: fifty-nine, and fifty-nine thousandths ; twenty-five thousand, and twenty-five ten-thou- sandths ; five, and G.Ye millionths ; two hundred five, and five hundredths. Ans. 25269.111505. 10. What is the sum of the following numbers : twenty-five, and seven millionths ; one hundred forty-five, and six hundred forty-three thousandths ; one hundred seventy-five, and eighty- nine hundredths ; seventeen, and three hundred forty-eight hundred-thousandths? Ans. 363.536487. 11. A farmer has sold at one time 3 tons and 75 hundredths of a ton of hay, at another time 11 tons and 7 tenths of a ton, and at a third time 16 tons and 125 thousandths of a ton. How much has he sold in all ? Ans. 31.575. 12. Add together 73 and 29 hundredths, 87 and 47 thou- sandths, 3005 and 116 ten-thousandths, 28 and 3 hundredths, 29000 and 5 thousandths. 13. Add two hundred nine thousand and forty-six millionths, ninety-eight thousand two hundred seven and fifteen ten-thou- sandths, fifteen and eight hundredths, and forty-nine ten-thou- sandths, together. Ans. 307222.086446. 202 DECIMAL FEACTIONS. SUBTRACTION OF DECIMALS. 270. Ex. 1. From 61.9634 take 9.182. Ans. 52.7814. operation. Having written the less number under the great- 6 1.9 6 3 4 er, so that figures of the same decimal place stand 9.1 8 2 in the same column, we subtract as in whole num- bers, and place the decimal point in the result, as in 5 2.7 8 1 4 addition of decimals. Rule. — Write the less number under the greater, so that figures of the same decimal place shall stand in the same column. Subtract as in whole numbers, and point off the remainder as in addition of decimals. Proof. — The proof is the same as in subtraction of whole numbers. Examples. 2. 3. 4. 5. 3 9.3 5. 6.1 41.7 1.678 9 1.6 7 8 1.99999 21.9767 3 7.62 11 3.3 2 2 4.10001 19.7233 6. From 29.167 take 19.66711. Ans. 9.49989. 7. From 91.61 take 2.6671. Ans. 88.9429. 8. From 96.71 take 96.709. 9. Take twenty-seven and twenty-eight thousandths from ninety-seven and seven tenths. Ans. 70.672. 10. Take one hundred fifteen and seven hundredths from three hundred fifteen and twenty-seven ten-thousandths. Ans. 199.9327. 11. From twenty-nine million four thousand and five take twenty-nine thousand, and three hundred forty-nine thousand two hundred, and twenty-four hundred-thousandths. Ans. 28625804.99976. 12. From one million take one millionth. Ans. 999999.999999. 13. From $ 19 take $ 1.375. Ans. $ 17.625. 14. A merchant bought flour to the amount of $316.87£, and sold it for $ 400 ; how much did he gain by the sale ? 15. From 19 million take 19 billionths. Ans. 18999999.999999981. DECIMAL FRACTIONS. 203 16. Charles Washburne has in one farm 93.45 acres, in another 124 acres, in a third 244.285 acres, and in wood-lots 216.136 acres; how many acres more would he require to have exactly 1000 acres ? MULTIPLICATION OF DECIMALS. 271. Ex. 1. Multiply 76.81 by 3.2. Ans. 245.792. operation. w e multiply as in whole numbers, and point off 7 6.8 1 on the right of the product as many figures for deci- 3.2 mals as there are decimal figures in the multiplicand ^ and multiplier counted together. The reason for 1 6 b L pointing off the decimals in the product, as in the 2 3 4 3 operation, will be seen, if we convert the multipli- 2 4 5 7 9 2 cand and multiplier into common fractions, and mul- tiply them together. Thus, 76.81 = 76^ = ^1 ; PT1 /| Q9 Q2 82 Thpn .7 68 1 y 3 2. 24 5792 94.5 7 9_2__ _ ana 6.L — 6^-§ — t -q. j_nen ^oo * .To — 1000 — Z40 ToFo — 245.792, Ans., the same as m the operation. 2. Multiply .1234 by .0046. Ans. .00056764. operation. Since the number of figures in the product .12 3 4 is not equal to the number of decimals in the .0046 multiplicand and multiplier, we supply the de- ficiency by placing ciphers on the left hand. 7 4 4 ■j'jjg rea son of this process will appear, if we 4 9 3 6 perform the operation thus : .1234 = ^AVo ; 00056764 an * - ?? 6 = to 4 Ao. Then^AX^Aor .v v v o o i o * 56764 qo = .00056764, Ans., the same as in the operation. Rule. — Multiply as in whole numbers, and point off as many figures for decimals, in the product, as there are decimal figures in the multiplicand and multiplier. If there be not so many figures in the product as there are decimal figures in the multiplicand and multiplier, supply the deficiency by pre- fixing ciphers. Proof, — The proof is the same as in multiplication of whole numbers. Examples. 3. Multiply 61.76 by .0071. Ans. .438496. 4. Multiply .0716 by 1.326. Ans. .0949416. 5. Multiply .61001 by .061. 6. Multiply 71.61 by 365. Ans. 26137.65. 7. Multiply .1234 by 1234. Ans. 152.2756. 204 DECIMAL FRACTIONS. 8. Multiply 6.711 by 6543. Ans. 43910.073. 9. Multiply .0009 by .0009. Ans. .00000081. 10. What is the product of one thousand and twenty-five, multiplied by three hundred and twenty-seven ten-thousandths ? Ans. 33.5175. 11. What is the product of seventy-eight million two hun- dred five thousand and two, multiplied by fifty-three hun- dredths ? Ans. 41448651.06. 12. Multiply one hundred and fifty-three thousandths by one hundred twenty -nine millionths. Ans. .000019737. 13. What will 26.7 yards of cloth cost, at $5.75 a yard? Ans. $153,525. 14. What will 14.75 bushels of wheat cost, at $ 1.25 a bushel ? Ans. $ 18.4375. 15. What will 375.6 pounds of sugar cost, at $0,125 per pound ? 16. What will 26.58 cords of wood cost, at $5,625 a cord? Ans. $149.512£. 17. What will 28.75 tons of potash cost, at $ 125.78 per ton? Ans. $3616.175. CONTRACTIONS IN MULTIPLICATION OF DECIMALS. 272. To multiply a decimal by 10, 100, 1000, &c. Re- move the decimal point as many places to the right as there are ciphers in the multiplier, annexing ciphers if required. Thus, 1.25 X 10 = 12.5 ; and 1.6 X 100 = 160. Examples. 1. Multiply 131.634 by 1000. Ans. 131634. 2. Multiply 3478.9 by 100. Ans. 347890. 3. Multiply one thousandth by one thousand. 4. What is the profit on one million yards of cotton cloth, at $ 0.007 per yard. Ans. $ 7000. 273. When it is not necessary that all the decimal places of the product should be retained, tedious multiplications may often be obviated, by contracting the work as. follows : — DECIMAL FRACTIONS. 205 Write the units' place of the multiplier under that figure of the multiplicand whose place it is proposed to retain in the pro- duct, and dispose of all the other figures of the multiplier in an order contrary to the usual one. Then, in multiplying, begin, for each partial product, with that figure of the multipli- cand which stands above the multiplying figure, observing to add to the product the number nearest to that which would have been carried if the places at the right had not been rejected. Write down the several partial products, so that the right-hand figure of each shall be in the same column, and their sum will be the product required. Examples. 1. Multiply 3.141592 by 52.7438, retaining only four places for decimals in the product. Ans. 165.6995 FIRST OPERATION. 3.1 4 1 5 9 2 == Multiplicand 8 3 4 7.2 5 = Multiplier reversed. = Product by 5, -|~ 1 = Product by 2, + 2 = Product by 7, -f- 4 = Product by 4, -- 1 = Product by 3, + 1 = Product by 8, -f- 1 1 6 5.6 9 9 5 = Product sought. 1570796 62832 21991 1257 94 25 SECOND OPERATION. 3.1 4 1 5 9 2 5 2.7 4 3 8 25113 27 3 6 9424776 1256 6368 21991 62831 1570796 16 5.69 95 144 84 001296 By comparison of the two methods of solution, it will be seen that the common one, as shown in the second operation, gives ten places of decimals, or six more than are required by the question, thus ren- dering unnecessary the several figures on the right of the vertical line. By the contrasted way, the multiplier, for convenience, has its figures reversed, or placed contrary to the usual order, so that the product of each figure by the one of the multiplicand above it, must be of the order of ten-thousandths. The first figure, at the right, of each partial product, being of the order of ten-thousandths, is writ- ten in the same column. To the product by 5 we add 1, since, if the 2 in the multiplicand had not been rejected, there would have been 1 to carry to the product of the 9 by the 5 ; to the product by 2 we add 2, since the product of the rejected figures, 92, by 2, approximates to 2 hundred, which would require 2 to be carried ; to the product by 7 we add 4, since the product of the two re- jected figures, 59, by 7, would require 4 to be carried ; to the pro- N 18 206 DECIMAL FRACTIONS. duct by 4 we add 1, since the product of the two rejected figures, 15, by 4, approximates to 1 hundred, which would require 1 to be carried ; and so on, it being sufficient to increase the partial product only by such a number as approximates most nearly to that which would have been carried, provided the two rejected figures next to the figure of the multiplicand had been retained. 2. Multiply 325.701428 by .7218393, retaining only three places of decimals in the product. Ans. 235.104. 3. Multiply 56.7534916 by 5.376928, retaining only five places of decimals in the product. 4. Multiply 843.7527 by 8634.175, retaining only the in- tegers in the product. Ans. 7285109. DIVISION OF DECIMALS. 274. Ex. 1. Divide 1.728 by 1.2. Ans. 1.44. operation. "We divide as in whole numbers, 1.2 ) 1.7 2 8 ( 1.4 4 Ans. and, since the divisor and quotient 1 2 are the two factors, which, being mul ■7. tiplied together, produce the dividend, ^ we point olf two decimal figures in the 4 8 quotient, to make the number in the a o two factors equal to the number in the product or dividend. The reason for pointing off will also be seen by performing the ex- ample with the decimals in the form of common fractions. Thus, 1.728 = 1 T V& = «» ; and 1 .2 = 1 ft = }§ . Then ft ft - ft = tttt X « - \UU = m= *AV = I-**, Ans. as before. 2. Divide 36.6947 by 589. Ans. .0623. operation. . "W e divide as in whole 589 ) 3 6. 6947 (.0 623 Ans. numbers, and since we have 3 5 3 4 . but three figures in the quo- — tient, we place a cipher be- 1 " ° ^ fore them, and thus make the 117 8 decimal places in the divisor ^ rr n rj and quotient equal to those .j _ c „ of the dividend. The reason for prefixing the cipher will appear more obvious by solving the example in the form of common fractions. Thus, 86.6947 = 86^^ = V&W < and 589 — *P- Then 366947 » £8.2. __ 366047 y .1 3 6 6 9 4 7 623 Ofi^S 100 00 7~ 1 — "TWFiT * X8^ 5~8" 9 001 IT ~ TOOOlT — -^-3, Ans. as before. Hence the following DECIMAL FRACTIONS. 207 Rule. — Divide as in ivhole numbers, and point off as many figures in the quotient as the number of decimal places in the dividend exceeds the number in the divisor ; but if there are not as many, supply the deficiency by prefixing ciphers. Note 1. — When the decimal places in the divisor exceed those in the divi- dend, make them equal by annexing ciphers to the dividend, and the quotient will be a whole number. Note 2. — When there is a remainder after dividing the dividend, ciphers may be annexed, and the division continued ; the ciphers thus annexed being regarded as decimals of the dividend ; and to indicate in any case that the division does not terminate, the sign plus ( + ) can be used. Proof. — The proof is the same as in division of whole numbers. Examples. Ans. 13.861+. Ans. 749.084. Ans. 3124.3. 3. Divide 780.516 by 2.43. Ans. 321.2. 4. Divide 7.25406 by 9.57. Ans. .758. 5. Divide .21318 by .38. 6. Divide 7.2091365 by .5201. 7. Divide 56.8554756 by .0759. 8. Divide 119109094.835 by 38123.45. 9. Divide 1191090.94835 by 3812345. 10. Divide 11910909483.5 by 38.12345. 11. Divide 11.9109094835 by 381234.5. 12. Divide 1191.09094835 by 3.812345. 13. Divide 11910909483.5 by .3812345. 14. Divide 1.19109094835 by 3.812345. 15. Divide .119109094835 by .3812345. 16. Divide 30614.4 by .9567. Ans. 32000. 17. Divide .306144 by 9567. Ans. .000032. 18. Divide four thousand three hundred twenty-two and four thousand five hundred seventy-three ten-thousandths by eight thousand and nine thousandths. Ans. .5403—)—. 19. How many yards of calico at $0.0775 per yard can be purchased for $ 10.85 ? 20. What costs 1 acre of woodland when 19.65 acres are sold for $ 982.50 ? Ans. $50. 21. Divide three hundred twenty-three thousand seven hun- dred sixty-five by five millionths. Ans. 64753000000. 208 DECIMAL FRACTIONS. CONTRACTIONS IN DIVISION OF DECIMALS. 275, To divide a decimal by 10, 100, 1000, &c. Remove the decimal point as many places to the left as there are ciphers in the divisor, and if there be not figures enough in the number, prefix ciphers. Thus, 2.15 -f- 10 = .215 ; and 1.9 -T- 100 — .019. Examples. 1. Divide 31.675 by 10. 2. Divide 916.05 by 100. 3. Divide 7.0461 by 100000. 4. Divide 70.461 by 100000. 5. Divide 704.61 by 100000. 6. Divide 7046.1 by 100000. 7. Divide 70460 by 100000. 8. Divide .70460 by 100000. 9. Divide 196.5 by 1000000. Ans. .0001965. 10. If $3500 are paid for 1000 yards of broadcloth, what is it a yard ? Ans. $ 3.50. 11. When $ 1025 are paid for 40 boxes of sugar, each con- taining 250 pounds, what is the cost of 1 pound ? Ans. $0.10£. 276. When the divisor contains many decimal places, and only a certain number of decimals are required to be retained in the quotient, the work may be contracted as follows : — First consider how many figures, in all, it is necessary for the quotient to contain. Then, by using the same number of figures from the left of the divisor, find the first figure of the quotient, and, instead of bringing down a new figure from the dividend, or annexing a cipher to the remainder, reject a figure on the right of the divisor at each successive division, and make the other figures a divisor. In multiplying such a divisor by the quotient figure, observe to add to the product the number nearest to that which would have been carried if no figures had been rejected. DECIMAL FRACTIONS. 209 Examples. 1. Divide 695.57270875 by 52.35775, and retain in the quo- tient three places of decimals. FIRST OPERATION. 5 2.3 5 7 7 5 ) 6 9 5.5 7 2 7 8 7 5 ( 1 3.2 8 5. 52358 = product by 1, + 1. 17 19 9 15707 = product by 3, + 2. 1492 10 4 7:= product by 2, -f- 1. product by 8, -\- 3. ~26 2 6= product by 5, -f- 1- SECOND OPERATION. 5 2.3 5775)695.5 7270875(13.2 8 5. 5 2 35775 445 419 17199 15707 1492 1047 520 325 Ans. 13.285. By inspection, it is evident that the first quotient figure will be of the order of tens, and there- fore the quotient will contain two places of whole numbers ; and as there are to be three places of decimals, it must contain five fig- ures. Hence, we divide at first by five figures of the given di- visor, counting them from the left toward the right, thus using the 52.357 and rejecting the figures, 75, on the right. In multiplying each contracted divi- sor by its quo- tient figure we increase the pro- duct by having regard to rejected figures, as in contracted multiplica- tion of decimals (Art. 273). The nature and extent of the contraction will be seen by compari- son with the common method as shown in the second operation, in which the vertical line cuts off the figures not required. Note. — When the given divisor does not contain as many figures as are re- quired in the quotient, we must begin the division in the usual way, and con- tinue till the deficiency is made up, after which begin the contraction. 2. Divide 4327.56284563 by 873.469, and retain five decimal places in the quotient. Ans. 4.95445. 3. Divide 252070.520751 by 591.57, and terminate the opera- tion with four decimal places in the quotient. Ans. 426.1043. 18* 1958 1550 04087 86200 178875 178875 210 DECIMAL FEACTIONS. 4. Divide 70.23 by 7.9863, and retain in the answer four decimals. 5. Divide 12193263.1112635269 by 123456789, and let the quotient contain as many decimal places, plus one, as there will be integers in it. Ans. 9876.54321. EEDUCTION OF DECIMALS. 277. To reduce a decimal to a common fraction. Ex. 1. Reduce .125 to its equivalent common fraction. Ans. £. OPERATION. •125 = rifift- = jflfr = A = i An8 ' Erasing the decimal point and supplying the denominator, which is understood, we have ^(HF* which reduced to its lowest terms equals J, the answer required. Rule. — Erase the decimal point, and write under the numerator its decimal denominator, and reduce the fraction to its lowest terms. Examples. 2. Reduce .875 to a common fraction. Ans. £. 3. Reduce .9375 to a common fraction. Ans. \% . 4. What common fraction is equivalent to .08125? Ans. t^j. 5. Change .00075 to the form of a common fraction. 6. Express 31.75 by an integer and a common fraction. Ans. 31£. 7. Express 96.024 by an integer and a common fraction. Ans. 96 T ^. 8. Express 163.04 by an integer and a common fraction. 9. Express 1001.4375 by an integer and a common frac- tion. Ans. 1001^. 10. Express 1457.222 by an integer and a common fraction. 11. Express 19678.36 by an integer and a common fraction. 12. Express 9163.8755 by an integer and a common frac- tion. Ans. 9163£$££. 278 • To reduce a common fraction to a decimal. Ex. 1. Reduce & to a decimal. Ans. .375. DECIMAL FRACTIONS. 211 OPERATION. Since we cannot divide the numer- ator, 3, by 8, we reduce it to tenths by annexing a cipher, and then di- viding, we obtain 3 tenths and a re- mainder of 6 tenths. Reducing this remainder to hundredths by annexing a cipher, and dividing, we obtain 7 hundredths and a remainder of 4 hundredths ; which being reduced to thousandths by annexing a cipher, and then divided, gives a quotient of 5 thousandths. The sum of the sev- eral quotients, .375, is the answer. To prove that .375 is equal to f, we change it to the form of a com- mon fraction, by writing its denomi- nator, and reducing it to its lowest terms. Thus, .375 = -f^-fa = f. 8 ) 3.0 ( 3 tenths. 24 8)60(7 hundredths. 5_6 8)40(5 thousandths. 4j0 Ans. .375. Or thus : 8 ) 3.0 .3 7 5 Ans. Rule. — Annex ciphers to the numerator, and divide by the denomi- nator. Point off in the quotient as many decimal places as there have been ciphers annexed. Note. — It is not usually necessary that the decimals should be carried to more than six places. When a decimal does not terminate, the sign plus (+) is generally annexed. Thus, in the expression .333+, the sign annexed indicates that the division could be carried further. Examples. 2. Reduce ■§■ to a decimal. Ans. .625. 3. Reduce J to a decimal. 4. Change A- to a decimal. Ans. .09375. 5. Change T ^ to a decimal. Ans. .076923-f-. 6. Reduce 19|- to an equivalent decimal expression. 7. Reduce $ 31 5 J to an equivalent decimal expression. Ans. $315,875. 8. Reduce $ 1163f to an equivalent decimal expression. Ans. 1163.75. Note. — A decimal with a common fraction annexed constitutes what is called a complex decimal; as, .87 j, .311, and .182. In such expressions, instead of the common fraction, its equivalent decimal, with the decimal point omitted, may be substituted. Thus, Afa = .404. 9. Reduce .62^- to a simple decimal. Ans. .625. 10. Reduce .37-^ to a simple decimal. Ans. .370625. 11. Reduce $ 4.31 J- to a simple decimal expression. Ans. $ 4.3125. 212 DECIMAL FKACTIONS. 12. Reduce $ 60.18J to a simple decimal expression. Ans. $ 60.1875. 13. What decimal expression is equivalent to £ of — of 2.04? o 14. What decimal expression is equivalent to 2^-, -J- 0.37 J, + £ of $ of 4, — 1.05 ? Ans. 2.9875. 279. To reduce a simple or compound number to a deci- mal of a higher denomination. Ex. 1. Reduce 15s. 9d. 3far. to the decimal of a £. Ans. .790625. operation. "VVe commence with the 3far., which 4 3.0 far. we reduce to hundredths by annexing 1 9 9 7 5 1 two °ip ners 5 an d then, to reduce these to the decimal of a penny, we divide by 2 1 5.8 12 5 0s. 4 ? since there will be | as many hun- dredths of a penny as of a farthing, and .790625 £. obtain .75d. Annexing this to the 9d., we divide by 12, since there will be -fa as many shillings as pence; and then, the 15s. and this quotient by 20, since there will be -£$ as many pounds as shillings, and obtain .790625£. for the answer. Hence the following Rule. — Divide the lowest denomination, annexing ciphers if neces- sary, by that number which will reduce it to one of the next higher de- nomination. Then divide as before, and so continue dividing till the decimal is of the denomination required. Note 1. — The given numbers may also be first reduced to a common frac- tion (Art. 256), and then the fraction changed to a decimal. Thus, if it be required to reduce 15s. 6d. to a decimal o£a£.: 15s. 6d. = 186d. ; 1£. = 280d. ; Mo £ ' = lo £ ' = * 775 £ ' Answer. Note 2. — Shillings, pence, and farthings may be readily reduced to a deci- mal of three places, by inspection, thus : Call half of the greatest even number of shillings tenths, and, if there be an odd shilling, call it 5 hundredths; re- duce the pence and farthings to farthings, and increase them by 1, if they amount to 24 or more, for thousandths. Thus, if it be required to reduce, by in- spection, 19s. lOd. 2far. to the decimal of a £.; half of 18s. = 9s., which de- note a value of .9£. ; the Is. denotes a value of .05<£. ; and lOd. 2far. = 42far., -which increased by lfar. = 43far., which denote a value of .043£. ; .9£. + .05£. + .043£. = .993£. Answer. The reason for this process is, that 2s. equal a tenth of a £. ; 1 shilling equals 5 hundredths of a £., and 1 farthing equals §£ 5 £., or so nearly a thousandth of a £. that 24 farthings exactly equal 25 thousandths of a £. ; and therefore farthings require to be increased only by 1 when they amount to 24 or more, to denote with sufficient accuracy their value in thousandths of a £. DECIMAL FRACTIONS. 213 Examples. 2. Eeduce 9s. to the fraction of a pound. Ans. .45. 3. Reduce 15cwt. 3qr. 141b. to the decimal of a ton. 4. Eeduce 2qr. 211b. 8oz. 12dr. to the decimal of a cwt. Ans. .71546875. 5. Eeduce lqr. 3na. to the decimal of a yard. Ans. .4375. 6. Eeduce 5fur. 35rd. 2yd. 2ft. 9in. to the decimal of a mile. Ans. .73603219-f. 7. Eeduce 3gal. 2qt. lpt. of wine to the decimal of a hogs- head. Ans. .0575396+. 8. Eeduce lpt. to the decimal of a bushel. Ans. .015625. 9. Eeduce 2E. 16p. to the decimal of an acre. Ans. .6. 10. Eeduce 175 cubic feet to the decimal of a ton of timber. Ans. 4.375. 11. Eeduce 3.755 pecks to the decimal of a bushel. Ans. .93875. 12. What decimal part of a degree is 25' 34".6? 13. Eeduce 12T. 3cwt. 2qr. 201b. to hundred-weight and the decimal of a hundred- weight. Ans. 243.7. 14. Eeduce 2hhd. 30gal. 2qt. lj-pt. to gallons and the decimal of a gallon. Ans. 156.6875. 15. Eeduce to the decimal of a pound, 19s. llfd., 16s. 9£d., and 17s. 5|-d., and find their sum. Ans. 2.71041 6-f-. 280. To find the value of a decimal in whole numbers of lower denominations. Ex. 1. What is the value of .790625 £. ? Ans. 15s. 9d. 3far. operation. There will be 20 times as many mil- .79062 5£. lionths of a shilling as of a pound ; there- 2 fore, we multiply the decimal, .790625, a by 20, and reduce the improper fraction 1 o.o 1 2 o Os. t a m ixed number by pointing off six 1 2 figures on the right, which is dividing by 9 7 5 Od * ts denominator, 1000000. The figures ' on the left of the point are shillings, and those on the right, the decimal of a shilling. 3.0 Ofar. ^ n * s decimal of a shilling we multiply by A i * o i q** * 2 ' anc *' P omt i n » off as before, obtain 9d., Ans. los. Jd. otar. an( j a decimal of a penny. The decimal 214 MISCELLANEOUS EXAMPLES. of a penny we multiply by 4, and pointing off have 3 farthings, which, taken with the other denominations obtained, gives 15s. 9d. 3far. for the answer. Rule. — Multiply the decimal by that number which icill reduce it to the next lower denomination, and point off as in multiplication of deci- mals. Then, multiply the decimal part of the product, and point off as be- fore. So continue till the decimal is reduced to the denominations required. The several whole numbers of the successive products ivill be the answer. Note. — When there is a decimal in the last product, it may be changed to a common fraction. Examples. 2. What is the value of .625 of a shilling ? Ans. 7Jd. 3. What is the value of .6725 of a cwt. ? Ans. 2qr. 171b. 4oz. 4. What is the value of .9375 of a yard ? 5. What is the value of .7895 of a mile ? Ans. 6fur. 12rd. 10ft. 6Jfin. 6. What is the value of .9378 of an acre ? Ans. 3R. 30p. 13ft. 9 T 9 ^in. 7. Reduce .5615 of a hogshead of wine to its value in gal- lons, &c. Ans. 35gal. lqt. Opt. 3||fgi. 8. Reduce .367 of a year to its value in days, &c. Ans. 134d. lh. 7m. 19£sec. 9. What is the value of .6923828125 of a cwt. ? Ans. 2qr. 191b. 3oz. 13d.. 10. What is the value of .015625 of a bushel ? 11. What is the value of .55 of an ell English ? Ans. 2qr. 3na. 12. What is the value of .6 of an acre ? Ans. 2R. 16p. MISCELLANEOUS EXAMPLES. 1. What is the value of 7cwt. 2qr. 181b. of sugar, at $ 11.75 per cwt. ? Ans. $ 90.24. 2. What cost 19cwt. 3qr. 141b. of iron, at $ 9.25 per cwt. ? MISCELLANEOUS EXAMPLES. 215 3. What cost 39A. 2R. 15p. of land, at $87,375 per acre ? Ans. $ 3459.503|f . 4. What would be the expense of making a turnpike 87m. 3fur. 15rd., at $ 578.75 per mile ? Ans. $ 50595.41^ ¥ . 5. What is the cost of a board 18ft. 9in. long, and 2ft. 3£in. wide, at $ .053 per foot ? Ans. $ 2.277^. 6. Goliath of Gath was 6 \ cubits high ; what was his height in het, the cubit being 1ft. 7.168in. ? Ans. 10ft. 4.592in. 7. If a man travel 4.316 miles in an hour, how long would he be in travelling from Bradford to Boston, the distance being 29J miles ? Ans. 6h. 50m. 6sec.-f 8. What is the cost of 5yd. lqr. 2na. of broadcloth, at $ 5.62 J per yard ? Ans. $ 30.234f . 9. Bought 17 bags of hops, each weighing 4cwt. 3qr. 71b., at $ 5.87£ per cwt. ; what was the cost ? 10. Purchased a farm, containing 176A. 3R. 25rd., at $ 75.37£ per acre ; what did it cost ? Ans. $ 13334.308 J£. 11. What cost 17625 feet of boards, at $12.75 per thou- sand ? Ans. $ 224.718J . 12. How many square feet in a floor 19ft. 3in. long, and 15ft. 9in. wide ? Ans. 303ft. 27in. 13. How many square yards of paper will it take to cover a room 14ft. 6in. long, 12ft. 6in. wide, and 8ft. 9in. high? 14. How many solid feet in a pile of wood 10ft. 7in. long, 4ft. wide, and 5ft. lOin. high ? Ans. 246||ft. 15. How many garments, each containing 4yd. 2qr. 3na., can be made from 112yd. 2qr. of cloth? 16. Bought lgal. 2qt. lpt. of wine for $ 1.82 ; what would be the price of a hogshead ? Ans. $ 70.56. 17. Bought 125jyd. of lace for $ 15.06 ; what was the price of 1 yard? Ans. $0.12. 18. What cost 17cwt. 3qr. of wool, at $35.75 per hundred- weight? Ans. $634.5 62 J. 19. What cost 7hhd. 47gal. of wine, at $87.25 per hogs- head ? Ans. $ 675.84 F %. 20. How many solid feet in a stick of timber 34ft. 9in. long, lft. 3in. wide, and 1ft. 6in. deep ? Ans. 65.15625ft. 21. If 18yd. lqr. of cloth cost $ 36.50, what is the price of 1 yard ? Ans. $ 2.00. 216 MISCELLANEOUS EXAMPLES. 22. If $ 477.72 be equally divided among 9 men, what will be each man's share ? Ans. $ 53.08. 23. A man bought a barrel of flour for $ 5.375, 7gal. of mo- lasses for $ 1.78, 9gal. of vinegar for $ 1.1875, lgal. of wine for $ 1.125, 141b. of sugar for $ 1.275, and 51b. of tea for $ 2.625 ; what did the whole amount to ? Ans. $ 13.367^. 24. A man purchased 3 loads of hay; the first contained 2| tons, the second 3 J tons, and the third l T *g- tons ; what was the value of the whole, at $ 17.625 a ton ? Ans. $ 128.882{£ . 25. How many hogsheads of water will it take to fill a cistern which is 15.25 feet long, 8.4 feet wide, and 10 feet deep ? Ans. 152hhd. 6 T 6 T gal. 26. At $13,625 per cwt., what cost 3cwt. 2qr. 71b. of sugar? Ans. $48,641^. 27. At $ 125.75 per acre, what cost 37 A. 3R. 35rd. ? Ans. $ 4774.570 T V 28. At $11.25 per cwt., what cost 17cwt. 2qr. 211b. of rice? Ans. $199,237^. 29. What cost 7 J bales of cotton, each weighing 3.37cwt., at $ 9.374- per cwt. ? 30. What cost 7hhd. 49gal. of wine, at $97,625 per hogs- head ? Ans. $ 759.305§f . 31. What cost 7yd. 3qr. 3na. of cloth, at $4.75 per yard? Ans. $37.703£. 32. What cost 27T. 15cwt. lqr. 3£lb. of hemp, at $183.62 per ton ? •• Ans. $ 5098.071£££. 33. What is the cost of constructing a railroad 17m. 3fur. 15rd., at $ 1725.875 per mile ? Ans. $ 30067.978f £, 34. When $ 624.53125 are paid for 17A. 3R. 15p. of land, what is the cost of one acre ? 35. Paid $494.53125 for 19T. locwt. 2qr. 141b. of hay; what was the cost per ton ? Ans. $ 24.9993y/ T . 36. How much land, at $ 40 per acre, can be obtained for $ 1004.75 ? Ans. 25A. OR. 19p. 37. How many cords of wood can be put into a space 20.5 feet long, 12.75 feet wide, and 7.6 feet high ? Ans. 15 cords 66£j cubic feet 38. How many bushels of corn at $ 0.62 J per bushel must a farmer exchange for 31 yards of sheeting at $ 0.08£ per yard, and 7 J yards of broadcloth at S 2.75 per yard ? Ans. 37^. CIRCULATING DECIMALS. 217 39. I have expended $ 42.875 for a quantity of grain, -^ of it being corn, at $ 0.75 a bushel ; ^ of it wheat, at $ 2 a bushel ; and the balance oats, at $ 0.40 a bushel, to the amount of $ 3.50. Eequired the number of bushels of each kind pur- chased. , 40. If a mason, in constructing a drain 250.35 feet long, begin with a width of 8 inches, and increase T 5 ^ of an inch in every foot of length, how many times the width of the be- ginning of the drain will its end be ? Ans. 16.646875. 41. A gentleman gave \ of his property to his son James ; £ of it to his son "William ; \ of the remainder to his daughter Mary ; and the balance to his wife. It appeared that Mary received $ 2243.26 less than James. What was the amount divided, and how much did each receive ? Ans. Amount, $13459.56; James, $3364.89; William, $4486.52; Mary, 1121.63 ; wife, $ 4486.52. CIRCULATING DECIMALS. 281. A Circulating Decimal is a decimal in which one or more figures are continually repeated in the same order. Thus, in reducing ^ to an equivalent decimal, on annexing ciphers and dividing by the denominator, the result obtained, .333-)-? is a circulating decimal ; for, however far the division might be carried, the same figure would continue to be repeated without the decimal terminating. Such decimals are sometimes called infinite, or repeating ; and, for sake of distinguishing, those decimals that terminate are sometimes termed finite. 282. A repetend is a figure, or a series of figures, contin- ually repeated. To mark a repetend, a point (.) is placed over a single repeating figure, or over the first and last of a series of repeating figures. Thus, in .3, the point denotes that the 3 is a repetend; and in .72, that the 72 is a repetend. 283. A single repetend is one in which only one figure is N 19 218 CIRCULATING DECIMALS. repeated; as in .1111— |— denoted by .1; and 2222-f-, denoted by 2. 284. A compound repetend is one in which the same set of figures is repeated; as in .135135— 1— ? denoted by .135, and .30363036+, denoted by 3086. 285. A pure repetend is one which contains only the figures of the repetend ; as, .3, .02, and .123. 286. A mixed repetend is one in which a repetend is preceded in the same fraction by one or more figures. The figures preceding the repetend are called the finite part. Thus, .416 is a mixed repetend, of which the figure 6 is the repetend, and the figures 41 the finite part ; also, 1.728 is a mixed repetend, of which the figures 28 are the repetend, and the figures 1.7 the finite part. 287. A perfect repetend is a pure repetend containing the same number of figures as there are units in its denominator less one. Thus, ^ reduced to a decimal gives .142857, which, as it contains as many figures as there are units in the denomi- nator, 7, less one, is a perfect repetend. 288. Similar repetends are those which begin at the same distance from the decimal point; as .3 and .6; or 5.123 and 3.478. 289. Dissimilar repetends are those which begin at differ- ent distances from the decimal point ; as .986 and .4625 ; or .5920 and .0423436. 290. Conterminous repetends are those which terminate at the same distance from the decimal point; as .631 and .465, or .0753 and .4752. 291. Similar and conterminous repetends are those which both begin and end at the same distance from the decimal point ; as .354 and .425 ; or .5757 and 5723. 292. Repetends always arise from common fractions, which, when in their lowest terms, contain in their denominator other factors than 2 and 5. For when a common fraction is in its lowest terms, its numerator and denominator are prime to each CIRCULATING DECIMALS. 219 other (Art. 219), and the annexing of one or more ciphers to the numerator makes the same a multiple of 10, but does not render it divisible by any factor, except 2 and 5, the factors of 10. Therefore, when the denominator of a fraction, in its lowest terms, contains other factors than those of 10, the deci- mal resulting from dividing the numerator with ciphers an- nexed, will not terminate, but will contain one or more figures that constantly repeat. 293. A pure repetend is always equivalent to a common fraction whose numerator is the repeating figure or figures, and whose denominator as many places of nines as there are repeat- ing figures. For, by reducing ^ to a decimal, we obtain as its equivalent the repetend .1; and since .1 is equivalent to ^, .2 will be equivalent to f , .3 to §, and so on, till .9 is equal to f or 1. Again, ^, and F ^, being reduced, give .01, and 001 ; that is, -fa = .01, and F J F = .001 ; therefore, ¥ 2 F = .02, and ^f 3- = ^02, anc * so on 5 tne same principle holding true in all like cases. 294. A mixed repetend is equivalent to a complex decimal (Art. 278), or to a complex fraction. Thus, the mixed repe- tend .2412 is equivalent to the mixed decimal .24^§, which is 2412 equal to the complex fraction -^—p 295 • JRepetends are of the same denomination only when they are similar and conterminous. For then alone, by hav- ing a common denominator, do they express fractional parts of the same unit. REDUCTION OF REPETENDS. 296. To reduce a repetend to an equivalent common frac- tion. Ex. 1. Reduce .123 to an equivalent common fraction. Ans. jfa . . operation. W e write the figures of the given .123 = -J§§ = ££g Ans. repetend with the decimal point omitted for the numerator, and as many nines as places in the repetend for the denominator, of a com- mon fraction (Art. 293), and obtain Iff, which, reduced to its low- est terms, = ^^, the answer required. 220 CIRCULATING DECIMALS. 2. Reduce .138 to an equivalent common fraction. Ans. -sV operation. The mixed repetend .138 \ 3 8 is equivalent to the mixed .13 8 ==— -^ = *§§■ = & Ans. decimal .13$ (Art. 294), 1 U which we readily change to the form of a complex frac- tion by erasing the decimal point and writing the denominator, 100, 1 3 8. which is understood ; and thus obtain ~g , which, reduced to its simplest form, gives -^ , the answer required. Hence, If the given repetend be simple, make the repeating figure or figures the numerator, and take as many nines as the repetend has figures for the denominator. If the given repetend be mixed, change it to an equivalent complex fraction, and that fraction to its simplest fo)~m. Note. — Any circulating decimal may be transformed into another decimal, having a repetend of the same number of figures; as, .78 = .787, and .534 = .5345 = .53453. Thus, when such expressions as 12.5 or 17.56 occur, they may be also transformed; as 12.5 = 12.52, and 17.56 = 17.567 = 17.5675, &c. Examples. 3. Required the common fraction equal to .6. Ans. | = §. 4. Reduce 1.62 to its equivalent mixed number. Ans. Iff. 5. Change .53 to an equivalent common fraction. 6. What common fraction is equivalent to .769230 ? Ans. -if. 7. What common fraction is equivalent to .5925 ? 8. Change 31.62 to an equivalent mixed number. Ans. 31ff 9. Reduce .008497133 to an equivalent common fraction. AnS. -g-y^g-. 297. To determine the kind of decimal to which a given common fraction can be reduced. Ex. 1. Required to find whether the decimal equal to xtVo be finite or circulating ; and if finite, of how many places the decimal will consist. Ans. Finite, of 4 places. CIECULATING DECIMALS. 221 operation. We reduce 3 the given frac- tWu — l\ = 9 v o v o v 9 ^ - 1 8 7 5 - tion to its lowest J X ^ X & X ^ terms, and then resolve the de- nominator, 16, of the fraction obtained, T 8 g, into its prime factors, which we find to be 2 X 2 X 2 X 2. Now, since the denominator contains no prime factor other than 2 or 5, it is evident that, by an- nexing ciphers to the numerator, 3, and dividing by the denomina- tor, 16, the decimal arising will terminate, and thus be finite (Art. 292). Since, in reducing a common fraction to its equivalent decimal, we annex ciphers to the numerator and divide by the denominator (Art. 278), every 10, or 2 and 5, that enter into the denominator as fac- tors must produce one decimal place, and no more, and therefore every other factor 2 or 5 must give one, and only one, decimal place. The denominator, 16, contains only the factor 2 taken 4 times, or 2 4 ; and the exponent of the 2 indicates that the decimal equivalent to A must contain exactly 4 decimal places, which we verify by reduc- ing the ^q to its equivalent decimal, .1875. 2. Find whether the decimal equal to ■&%% be finite or cir- culating ; and if circulating, of how many places the finite part, if any, and the circulating part, will each consist. Ans. Circulating : the finite part, 2 places ; the repetend, 6 places. We reduce OPERATION. 4 7 . the given iWir = ^ = 2v 5 v , v7 =.13428571. fraction to ^^° x ' its lowest terms, and obtain $fc. The denominator, 350, = 2 X 5 x 5X 7, contains a prime factor, 7, other than 2 and 5 ; therefore the decimal equivalent to -g^j. will contain a repetend ; and as, of the factors 2 and 5, the higher exponent of either, that of 5, is 2, the decimal will have 2 finite places before the repetend commences. This we verify by re- ducing $fc to its equivalent decimal, .13428571. Hence, to deter- mine whether the decimal to which a given common fraction can be reduced is finite or circulating, and the number of finite decimal places, if any, Having reduced the given common fraction to its lowest terms, resolve the denominator into its prime factors. If these factors be not other than 2 or 5, the decimal icill be finite; if other prime factors occur with 2 or 5, the decimal will be a mixed repetend ; and if neither 2 nor 5 occurs as factor, the decimal will be a pure repetend. Whichever factor 2 or 5 occurs in the denominator with the higher exponent will by its exponent denote the number of finite decimal places. 19* 222 CIRCULATING DECIMALS. Note. — The number of figures of which a repetend will consist may be dis- covered by dividing 1 with ciphers annexed by the factors other than 2 or 5 of the denominator, until there is a remainder 1. Thus, if it be required to discover the number of figures in the repeating part of the decimal equivalent to g 4 ^j, we divide 1 with ciphers annexed by 7, the only prime factor in the denominator other than 2 or 5, until there is a Remainder of 1, which occurs after the sixth division, thereby indicating that the repeating part will consist of six figures. We have seen that these must be preceded by two places of finite decimals, so that the mixed repetend equal to ^ must consist of eight places in all. Examples. 3. To what kind of a decimal can ^ be reduced ? Ans. A pure repetend, of 2 places. 4. How many places of decimals, finite and repeating, will be required to express T §| ^ ? Ans. 5 places ; 3 finite and 2 repeating. 5. To what kind of a decimal can J-? J be reduced? 6. Keduce 13J-J to a mixed repetend. Ans. 13.37. 7. Change T £f f^ to a mixed repetend. Ans. .008407133. 8. Of how many figures will the repetend consist that corre- sponds to £§ ? Ans. 28 figures. TRANSFORMATION OF REPETENDS. 298. Any finite decimal may be considered as a mixed repetend by making ciphers continually recur ; thus, .42 = .420 = .4200 = .42000, &c. 299. Any circulating decimal may be transformed into an- other having the same number of repeating figures; thus, .127 = .1272 = .12727, &c. 300 • Any circulating decimal having as repetend any num- ber of figures may be transformed to another having twice or thrice that number of figures, or any multiple thereof; thus, .5925, having a repetend of three figures, may be transformed to one having 6, 9, 12, &c places ; therefore .5925 = .5925925 = 5925925925 = 5925925925925925, &c 301 . The value of a decimal is not changed by any of the above transformations, as may be seen by reducing the given ► Ans. CIRCULATING DECIMALS. 223 repetends to their equivalent common fractions (Art. 296) and comparing them together. Hence, they can be used in making dissimilar repetends similar and conterminous. 302. To make any number of dissimilar repetends similar and conterminous. Ex. 1. Make similar and conterminous 9.167, 14.6, 3.165, 12.432, 8.181, and 1.307. OPERATION. Dissimilar. Similar. Similar and Conterminous. 9.16 7 = 9.167 6 = 9.61767676] 14.6 =14.600 =14.60000000 3.1 6 5 ma 3.1 6 5 = 3.1 6 5 5 5 5 5 5 1 2.4 32 =1 2.4 3243 = 1 2.4 3243243 8.1 8 i = 8.18 18 = 8.1 8 i 81818 1.3 7 = 1.3 7 3 6 = 1.3 7 3 7 3 6 , We make the finite mixed decimal, 14.6, a mixed repetend by annexing recurring ciphers, and make it and all the given repetends similar, by extending the figures to the right, so that the circulating part of each may begin at the same distance from th^ decimal point as does that repetend which is preceded by the most finite decimal places. Then, to make conterminous the repetends that have thus been rendered similar, as some of them consist of 1, some of 2, and the others of 3 places, we extend the repeating figures of each repe- tend till those of each occupy as many places as there are units in the least common multiple of 1, 2, and 3, which is 6. Hence, to make dissimilar repetends similar and conterminous, Transform the given repetends so that the circulating parts shall com- mence at the same distance from the decimal point, and shall consist of as many circulating places as there are units in the least common mul- tiple of the number of repeating figures found in the given decimals. Examples. 2. Make 3.671, 1.007i, 8.52, and 7.616325 similar and con- terminous. 3. Make 1.52, 8.7156, 3.567, and 1.378 similar and conter- minous. 4. Make .0007, .141414, and 887. i similar and conterminous. 5. Make .3123, 3.27, and 5.02 similar and conterminous. 6. Make 17.0884, 1563.0929, and 15.i2345 similar and con- terminous. 224 CIRCULATING DECIMALS. ADDITION OF CIRCULATING DECIMALS. 303. Ex. 1. Add 2.765 r 7.16674, 3.671, .7, and .i728 together. Ans. 14.55436. operation. Having made the given Dissimilar. Similar and Conterminous. repe tends similar and COn- 2.7 6 5 = 2.7 6 5 6 5 terminous (Art. 302), we 716674 = 716674 a< ^ ^ m aa *dition of 3.6 7 1 = 3.67 13 6 whole numbers, and obtain 14.55433. The right-hand .7 = .77777 figure of this result we in- -J728 17281 crease by such a number - — . as would have been car- Ans. 1 4.5 5 4 3 6 ried, if the repetends had been continued farther to the right. In that case we should have had to carry 3 after finding the amount of the first left-hand column of the repetends continued. We therefore increase the sum as first found, and thus have the true amount as in the operation, 14.55436. Rule. — Make the given repetends, wlu n dissimilar, similar and con- terminous. A(M as in addition of finite decimals, observing to increase the repetend of the amount by the number, if any, to be carried from the left-hand column of the repetends. Examples. 2. Add 3.5, 7.65i, 1.765, 6.173, 51.7, 3.7, 27.63i,and 1.003 together. Ans. 103.2591227. 3. Reduce -J, -f, and £ to decimals, and find their sum. 4. Find the sum of 27.56, 5.632, 6.7, 16.356, .11, and 6J234. Ans. 63.1690670868888. 5. Add together .165002, 31.64, 1.06, .34634, and 13. 6. Add together .87, .8, and .876. Ans. 2.644553. 7. Required the value of .3 + .45 -|- .45 -f- .35 i + .6468 + .6468 -f .6468, and .6468. Ans. 4.1766345618. 8. Find the value of 1.25 + 3.4 + .637 + 7.885 -f 7.875 + 7.875 + 11.X. Ans. 40.079360724. 9. Add together 131.613, 15.001, 67.134, and 1000.63. 10. Find the value of 5.i6345 + 8.6381 + 3.75. Ans. 17.55919120847374090302. (JIKCULATING DECIMALS. 225 SUBTRACTION OF CIRCULATING DECIMALS. 304. Ex. 1. From 87.1645 take 19.479167. Ans. 67.685377. operation. Having made the repe- Dissimilar.^ Similar and Conterminous. ten( j s s ; im l ar an( J conter- 8 7.1 6 4 5 =8 7.1 6 4 5 4 5 minous, we subtract as in H479H7 - iQ4.7qifi7 whole numbers, regarding, 1 y.4 I J 1 b ( — 1 a. 4= / J 1 b / however, the right-hand Ans. 6 7.6 8 5 3 7 7 % ure of the subtr anend as increased by 1, since 1 would have been carried to it in subtracting, if the repetends had been continued farther to the right, as is evident from the circulating part of the subtrahend being greater than that of the minuend. Rule. — Make the repetends, when dissimilar, similar and conter- minous. Subtract as in subtraction of finite decimals; observing to regard the repetend of the subtrahend as increased by 1, when it exceeds that of the minuend. Examples. 2. From 7.i take 5.02. Ans. 2.08. 3. From 315.87 take 78.0378. Ans. 237.838072095497. 4. Subtract | from f . Ans. .079365. 5. From 16.1347 take 11.0884. Ans. 5.0462. 6. From 18.1678 take 3.27. Ans. 14.8951. 7. From 3.123 take 0.7 i. Ans. 2.405951. 8. From f take T 2 T . Ans. .246753. 9. From § take f . Ans. .i58730. 10. From fr take f r . Ans. .i76470588235294i. 11. From 5.i2345 take 2.3523456. Ans. 2.7711055821666927777988888599994. MULTIPLICATION OF CIRCULATING DECIMALS. 305, Ex. 1. Multiply .36 by 25. Ans. 0929. operation. "We change the ' 6h ~ 99 = U ; lb ~ F6 ~~ 90" their equivalent com- . . mon fractions, and, tt Xff= t& 2 A> a^n fo -N&J&r- 8. What are the first four approximate values of 1.27 ? Ans. i, |, |, tf> M> or 1, 1*, lfc 1 T 3 T , 1&. RATIO. 31(h Ratio is the relation, in respect to magnitude or value, which one quantity or number bears to another of the same kind. 31 1# The comparison by ratio is made by considering how often one number contains, or is contained in, another. Thus, the ratio of 10 to 5 is expressed by 2, the quotient arising from the division of the first number by the second, or it may be expressed by -fij = £, the quotient arising from the division of the second number by the first, as the second or the first number shall be regarded as the unit or standard of comparison. In general, of the two methods, the first is regarded as the more simple and philosophical, and therefore has the preference in this work. Note. — Which of the two methods is to be preferred, is not a question of so much importance as has been by some supposed, since the connection in which ratio is used is usually such as to readily determine its interpretation. 312t The two numbers necessary to form a ratio are called N 20 % 230 RATIO. the terms of the ratio. The first term is called the antecedent, and the last, the consequent. The two terms taken together are called a couplet ; and the quotient of the two terms, the index or exponent of the ratio. 313. The ratio of one number to another may be expressed either by two dots ( : ) between the terms ; or in the form of a fraction, by making the antecedent the numerator and the con- sequent the denominator. Thus, the ratio 6 miles to 2 miles may be expressed as 6 : 2, or as § . 314* The terms of a ratio must be of the same kind, or such as may be reduced to the same denomination. Thus, cents have a ratio to cents, and cents to dollars, &c. ; but cents have not a ratio to yards, nor yards to gallons. 315. A simple ratio is that of two whole numbers ; as, 3 : 4, 8 : 16, 9 : 36, &c. 316. A complex ratio is that of two numbers, of which one or both are fractional ; as, 6 : 4£, f : -^, 4 J : 2£, &c. 317* A compound ratio is the product of two or more ra- tios. Thus, the ratio compounded of 4 : 2 and 6 : 3 is J X f = 2£ = 4, or 4 X 6:2 X 3 = 4. A compound ratio is generally expressed by writing the ra- tios composing it, in a column, with the antecedents in one vertical line, and the consequents in another ; thus, fi \ o [• ex- presses a compound ratio. Note. — If a ratio be compounded of two equal ratios, it is called a dupli- cate ratio; of three ratios, a triplicate ratio, &c. 318* A ratio is either direct or inverse, A direct ratio is the quotient of the antecedent by the consequent ; an inverse ratio, or reciprocal ratio, as it is sometimes called, is the quo- tient of the consequent by the antecedent, or the reciprocal of the direct ratio. Thus the direct ratio of 6 to 2 is §■ or 3 ; and the inverse or reciprocal ratio of 6 to 2 is f or ^-, which is the same as the reciprocal of 3, the direct ratio of 6 to 2. Note 1. — One quantity is said to vary directly as another, when both in- crease or decrease together in the same ratio ; one quantity is said to vary in- RATIO. 231 versely as another, when the one increases in the same ratio as the other de- creases. Note 2. — The word ratio, when used alone, means the direct ratio. 319. When the antecedent and consequent of a ratio are equal, the ratio equals 1, and is called that of equality. Thus, the ratio of 6 : 6 = f = 1, and the ratio of 6 X 4 : 8 X 3 = J4 === 1, are ratios of equality. But if the antecedent is larger than the consequent, the ratio is that of greater inequal- ity, and if the antecedent is smaller than the consequent, the ratio is that of less inequality. Thus, the ratio of 15 : 5 = \ b - = 3, is a ratio of greater inequality ; and the ratio of 7 : 14 = T 7 ^ = £, is a ratio of less inequality. 320. The ratio of two fractions having a common numerator is the same as the inverse ratio of their denominators. Thus, the ratio of f- : § is f- -r- ■§ = 2, which is the inverse ratio of the denominator 4 to the denominator 8. 321 • The ratio of two fractions having a common denomi- nator is the same as the ratio of their numerators. Thus, the ratio of f : f is f- ~ f- = 2, which is the ratio of the numera- tor 6 to the numerator 3. 322. The inverse or reciprocal ratio of two numbers de- notes what part or multiple the antecedent or dividend is of the consequent or divisor. Thus, inquiring what part of 4 is 3, or what part 3 is of 4, is the same as inquiring the inverse or re- ciprocal ratio of 4 : 3. The inverse ratio of 4 : 3 is f , and 3 is f of 4. 323. In order to compare one number with another, by ratio, it is necessary that they should not only be of the same kind, but of the same denomination. Thus, to compare 2 days with 12 hours, it is necessary that the days be reduced to hours, before we can indicate the ratio, which is 48 hours : 12 hours. 324. If the antecedent of a ratio he multiplied, or the conse- quent divided, the ratio is multiplied. Thus, the ratio of 6 : 3 is 2, but 6 X 2 : 3 is 4 ; or 6 : 3 — 2 is 4. 325. If the antecedent of a ratio be divided, or the conse- quent multiplied, the ratio is divided. Thus, the ratio of 18 : 6 is 4, but 18 ~ 3 : 6 is 1 ; or 18 : 6 X 3 = 1. 232 RATIO. 326. If both the antecedent and consequent of a ratio be multiplied or divided by the same number, the ratio is not altered. Thus, the ratio of 8 : 2 is 4 ; of 8 X 2 : 2 X 2 is 4; and of 8 -7- 2 : 2 -r- 2 is 4. EEDUCTION AND COMPARISON OF RATIOS. 327. Ratios, being of the nature of fractions, may be reduced, compared, and otherwise operated upon like them. 328. To reduce a ratio to its lowest terms. Ex. 1. Reduce 18 : 9 to its lowest terms. Ans. 2:1. operation. "VYe cancel in the two terms the 18 : 9 = J F 8 - = f = 2 : 1. common factor 9, and obtain \ = 2:1, the answer. Hence Cancel in the given ratio all factors common to its terms. Examples. 2. Reduce to its lowest terms 63 : 72. Ans. £. 3. Reduce to its lowest terms 66 : 24. 4. Reduce to its lowest terms 4X6X3:8X9X2. Ans. £. 5. What are the lowest terms of 19 X 5 X 2 X 3 : 15 X 12 X 38? 329. To reduce a complex or a compound ratio to a simple one. Ex. 1. Reduce 5£ : £■ to a simple ratio. Ans. 22 : 3. operation. "VV e express the 5A :* = # = *>=*?= 22 :SAns. f iven ra , tio in the 2 * 3 6 TJ f orm f a com _ plex fraction, which, changed to a simple fraction (Art. 242), and reduced to its lowest terms, gives %£■ = 22 : 3, the answer required. 8 * 5 ) 2. Reduce n \ 0yl > to a simple ratio. Ans. 7 : 15, 7:24^ OPERATION. 3 We express the given ratio in the form of a compound fraction, RATIO. 233 which, reduced to a simple one (Art. 329), gives T \ = 7 : 15, the answer required. Hence, to reduce a complex or a compound ratio to a simple one, Proceed as in like operations with fractions. Examples. 3. Eeduce f : f to a simple ratio. Ans. 35 : 24. 4. Eeduce 13^- : 27 to a simple ratio. Ans. 1 : 2. 5. Eeduce 6.25 : 3.125 to a simple ratio. Ans. 2 : 1. 6. Eeduce ok ! in i to a smi P* e rat i°« Ans. 5 : 8. 3: 6] 7. Eeduce 9 : 27 >to a simple ratio. Ans. 3 : 2. 108 : 12 ) 8. Eeduce 76 2 5 *. 25 5 i to a sim P le ratio# Ans. 6 : 1. 330. To find the ratio of one number to another. Ex. 1. Eequired the direct ratio of 108 to 9. Ans. 12. operation. Since 9 is the unit or standard of 108:9=- 1 g-^ = 12 Ans. comparison, we make it the conse- quent (Art, 111) and the 108 the antecedent of the ratio, and obtain i$& =12 Ans. 2. Eequired the inverse ratio of 72 to 8. Ans. £. operation. We divide* the consequent 72 : 8 inverted = ^ == i Ans. 8 Dv tne antecedent 72, or, which is the same thing, find the reciprocal of the direct ratio of 72 : 8 (Art. 318), by inverting its terms, and thus obtain -f% = ^ Ans. Hence, The direct ratio is found by dividing the antecedent by the consequent, and the inverse ratio by dividing the consequent by the antecedent. Note 1. — Eatios expressed by fractions having different denominators must be reduced to a common denominator, in order to be compared; and then they are to each other as their numerators (Art. 323). Note 2. — When a ratio is expressed in terms inconveniently large and prime to each other, we may find the approximate values of the ratio ex- pressed in smaller numbers, as in other fractional expressions (Art. 309). Examples. 3. What is the ratio of 39 to 13 ? Ans. 3. 4. What is the ratio of 2 yards 2 quarters to 9 yards ? 2(1* 234 RATIO. 5. What is the ratio of 21 gallons to J of a hogshead? Ans. 1. 6. What is the ratio of £ of £ of $ 2 : \ of $ 0.50 ? Ans. f . 7. What is the inverse ratio of 24 : 6 ? Ans. £. 8. What part of 36 is 4 ? Ans. £. 9. What part of a farm of 94A. 2R. 16rd. is 11 A. 3R. ? 10. Which is the greater, the ratio of 17 to 9, or of 39 to 19? Ans. 39:19. 11. By how much does the ratio of 36 X 4 X 3 : 12 X 16 X 2 exceed that of 60 -*- (3 X 5) : 20 X 2 -r- 8 ? Ans. £g. 12. What is the inverse ratio of .02 : 2.503 ? 13. Which is the greater, the ratio of \ of \ : -J of £, or that of5:4? 14. The height of Bunker Hill Monument is 220 feet, and that of the great pyramid, Egypt, 500 feet ; what is the ratio of the height of the former to that of the latter ? Ans. ££. 15. A certain farm contains 180 acres, and the township of which it forms a part is 36 square miles in extent. What is the ratio of the latter to the former ? 16. Find approximate values for the ratio of 4900 to 11283. Ans. i, f , ft f f> tW, &c. 17. The ratio of the circumference of a circle to its diameter is 3.141592. Required approximate values for this ratio. Ans. 3, -V-, m> Hh & c., or 3, 3|, 3 T V\, 3 T \% &c. ANALYSIS BY EATIO. 331. Operations by analysis may often be much abridged by ratio. Thus, frequently, it is more convenient to multiply or divide by the ratio a number bears to a unit of the same kind, than to multiply or divide by the number itself. This form of analysis is much used by business men ; and, like that by aliquot parts (Art. 114), is sometimes called Prac- tice. Examples. 1. What cost 14 tons 15cwt. 3qr. 201b. of iron, at $ 60 a ton? Ans. $887.85. KATIO. 235 OPERATION. $ 60.00 = cost of 1 ton. 14 $ 840.00 =t " 14 tons. (lOcwt. : 1 ton = £) ; £ of $ 60 = 30.00 =s « lOcwt. (5cwt. : lOcwt. = £) ; £ of $ 30 = 15.00 * " 5cwt. (2qr. : 5cwt. et T V) ; T V of $ 15 = 1.50 = " 2qr. (lqr. : 2cwt. = £) ; £ of $ 1.50 = 0.75 =* « lqr. (151b. : 3qr. == |) ; $ of $ 2.25 == 0.45 = « 151b. (51b. : 151b. = i) ; £ of $ 0.45 = 0.15 s « 51b. Ans. $887.85= " 14T. 3qr. 201b. Since 1 ton costs $ 60, 14 tons will cost 14 times $ 60, or $ 840. 15c\vt. = lOcwt. -f- 5cwt. Since the ratio of lOcwt. to 1 ton or 20cwt. = -|, lOcwt. will cost ^ as much as 1 ton, or $ 30 ; and as the ratio of 5cwt. to lOcwt. = J, 5cwt. will cost £ as much as lOcwt. or $ 15. 3qr. = 2qr. -\- lqr. Since the ratio of 2qr. to 5cwt. or 20qr. = ^q, 2qr. will cost X as much as 5c wt., or $ 1.50 ; and as the ratio of lqr. to 2qr. = ^, lqr. will cost i as much as 2qr., or $0.75. 20lb. = 15lb. + 5lb. Since the ratio of 15lb. to 3qr. or 75lb. = ^, 15lb. will cost -^ as much as 3qr., or $0.45; and as the ratio of 5lb. to 15lb. = £, 5lb. will cost £ as much as 15lb., or $ 0.15. The cost of the several parts equals the cost of the whole, or $887.85, Ans. 2. What is the value of 17 acres 3 roods 35 rods of land, at $ 80 per acre ? Ans. $ 1437.50. 3. What cost 16cwt. 3qr. 101b. of guano, at $ 2.50 per cwt. ? 4. What cost 27cwt. lqr. 201b. of coffee, at $14 per cwt. ? Ans. $ 384.30. 5. If 1 yard of cloth cost $5.60, what will 7yd. 3qr. 2na. cost? Ans. $44.10. 6. What cost 7 tons 13cwt. 2qr. of hay, at $ 20 per ton ? 7. What cost 99bu. lpk. 4qt. of wheat, at $1.92 per bushel? Ans. $191.80. OPERATION. $1.92 = 100 cost of lbu. (2pk. (4qt: :lbu. :2pk. = 1); $ 192.00 = J of $1.92 = 0.96 > = £of $0.96 = 0.24) = u 100bu. 2pk. 4qt. Ans. $191.80 = a 99bu. lpk. 4qt 236 RATIO. The quantity being nearly 100 bushels, we find the cost of 100 bushels by annexing two ciphers to $ 1.92, the cost of 1 bushel, and obtain $192, from which we subtract the cost of 2pk. 4qt., the differ- ence of quantity between that given and 100 bushels; the cost of 2pk. = $0.96; and that of 4qt. = $0.24; $192 — $0.96 -j- $0.24 = $191.80 Ans. 8. What cost 19yd. 3qr. 2na. of cloth, at $ 4.40 per yard ? Ans. $87.45. 9. How much must be paid for 24A. 3R. 20p. of land, at $ 32 per acre ? 10. How much must be paid for 1991b. 12oz. of butter, at $ 0.30 per lb. ? Ans. $ 59.925. 11. What cost 714 yards of broadcloth, at 15s. 6d. per yard ? Ans. 553£. 7s. 12. How much must be paid for the services of a man 2y. 9mo. 15da-, at $ 450 yer year ? Ans. $ 1256.25. 13. If 1 acre of land cost $ 80.50, what will 25 acres 2 roods 35 rods cost ? Ans. $ 2070.35-f. 14. What cost 4981b. of tea, at 2s. 6d. per lb. ? 15. If lcwt. 2qr. 121b. of alum can be purchased for $ 4.05, how much can be purchased for $ 28.35 ? Ans. lOcwt. 3qr. 91b. OPERATION. $28.35: $4.05 = 7; lcwt. 2qr. 121b. X? = lOcwt. 3qr. 91b. Ans. Since the ratio of $ 28.35 to $ 4.05 = 7, $ 28.35 will purchase 7 times as much as $4.05. By multiplying what the latter will pur- chase by the ratio, we have the answer required. 16. If llgal. 3qt. lpt. of molasses cost $5.83f, what will 35gal. 2qt. lpt. cost? Ans. $17.51|-. 17. If 24yd. 3qr. of cloth cost $ 49.50, what will 12yd. lqr. 2na. cost? 18. If 17bu. 2pk. 4qt. of oats be paid for 14bu. 3pk. of salt, what quantity of oats must be paid for 73bu. 3pk. of salt ? Ans. 88bu. Opk. 4qt. 19. If $ 9.75 will purchase IT. 2cwt. 2qr. 151b. of coal, how much will $ 3.25 purchase ? 20. If a train of cars move at the average velocity of 27m. 3fur. 20rd. per lh. 20m., how far will it move in 4h. ? Ans. 82m. 2fur. 20rd. 7mA I. PROPORTION. 23T PROPORTION. 332. A proportion is an equality of ratios. Any four numbers are in proportion, when the ratio of the first to the second is the same as that of the third to the fourth. Thus, the ratios 9 : 3 and 6 : 2, being equal to each other, when written, 9:3= 6 : 2, or § = § , form a proportion. Proportion is written with the sign of equality (==), or, as is more common, with four dots (: :), between the ratios. Thus, 9:3= 6 : 2, or 9 : 3 : : 6 : 2, expresses a proportion, and is read, The ratio of 9 to 3 is equal to the ratio of 6 to 2, or 9 is to 3 as 6 is to 2. 333. The terms of a proportion are the four numbers which form the proportion. These numbers are also called propor- tionals. The first and third terms, or proportionals, are called antecedents, the second and fourth are called consequents ; the first and last are called the extremes, the second and third the means ; the first and second compose the first couplet, the third and fourth compose the second ; and when the ratio of the first of three terms is to the second as the ratio of the second is to the third, the second term is called a mean 'proportional to the other two terms. 334. A direct proportion is an equality between two direct ratios ; an inverse or reciprocal proportion is an equality be- tween a direct and an inverse or reciprocal ratio. Thus, the numbers 4, 2, 6, 3 are, as they stand, in direct proportion, de- noting 4 : 2 : : 6 : 3 ; but in the order 4, 2, 3, 6, are in inverse proportion, denoting that 4 : 2 : : -£ : £, or the direct ratio of 4 to 2 is equal to the inverse ratio of 3 to 6. Note. — The term proportion, used alone, always means direct proportion. 335. In any proportion, if the antecedents or consequents, or both, are divided, or multiplied, by the same number, they are still proportionals. Thus, dividing the antecedents of the pro- portion 4 : 8 : : 10 : 20 by 2, we have 2 : 8 : : 5 : 20 ; dividing the consequents by 2, we have 4 : 4 : : 10 : 10 ; and dividing both the antecedents and consequents by 2, we have 2:4:: 5:10; each of which results is a proportion, since if we divide 238 I PROPORTION. the second term of each by the first, and the fourth by the third, the two quotients will be equal. The effect is the same when the terms are multiplied by the same number. 336. In every proportion the product of the two extremes is equal to the product of the two means. Thus, the proportion 16 : 8 :: 20 : 10 may be expressed *£- = %%. Now, if we re- duce these fractions to a common denominator, we have *£g- =* V 6 cr 5 Dut m this operation we multiplied together the two ex- tremes of the proportion, 16 and 10, and the two means, 8 and 20; thus, 16 X 10 = 8 X 20. Hence, 1. If the extremes and one of the means are given, the other mean may he found by dividing the product of the extremes by the given mean ; or, 2. If the means and one of the extremes are given, the other extreme may be found by dividing the product of the means by the given extreme. SIMPLE PROPORTION. 337. Simple Proportion is an equality between two simple ratios. Note. — Simple Proportion is sometimes called the Rule of Three, and formerly was termed by arithmeticians the Golden Rule. 338. The object of that part of simple proportion which is usually included in arithmetics, is to find a fourth proportional to three given numbers, or, in other words, to find the fourth term of a proportion, when the other three terms are given. Ex. 1. If a man travel 243 miles in 9 days, how far will he travel in 24 days ? Ans. 648 miles. operation. Since 9 days have the same Extreme. Mean. Mean. Extreme. ratIo tQ 24 d ag ^3 mileg> 9 da. : 2 4 da. : : 2 4 3 m. : — m. t he distance of travel in 9 days, 2 4 have to the distance of travel q r~ £> in 24 days, we have the first jo /. three terms of a proportion given, namely, the two means 9)5832 an d one of the extremes, from which to find the required ex- Ans. 6 4 8m. Extreme, treme. Now, to arrange the given numbers in the order of PROPORTION. 239 a proportion, or state the question, we make the 243 miles the third term, because it is of the same kind as the required fourth term, and as from the nature of the question the latter must be greater than the third term, we make the greater of the other two numbers the second term, and the less the first; and, then, the product of the means di- vided by the given extreme gives the required extreme (Art. 336). By Analysis. — If a man travel 243 miles in 9 days, he will in 1 day travel \ of 243 miles =27 miles ; then, if he travel 27 miles in 1 day, in 24 days he will travel 24 times 27 miles = 648 miles, the answer, as before. By Katio. — 9 : 24 = -fa = f ; 243 miles -*- f = 648 miles, Ans. 2. If 15 yards of cloth cost $ 48.90, what will 5 yards cost ? Ans. $16.30. operation. "We state the question by making 1 5 : 5 : : $ 4 8.9 : $ — $ 48.90 the third term, because it is of the same kind as the required term. $ X 4 8. 9 n q n Then, since the answer must be less _ == 1 6.3 0. than $ 48.90, because 5 yards will cost *■+* less than 15 yards, we make 5 yards, 3 Ans. $ 16. oO. ^he less of the two numbers, the second term, and 15 yards the first ; and pro- ceed as in the first example, except that we abridge the work by can- cellation. By Analysis. — If 15 yards cost $ 48.90, 1 yard will cost -^ of $48.90 == $3.26; then, if 1 yard cost $3.26, 5 yards will cost 5 times $3.26 =$16.30. Rule. — Write the given number that is of the same kind as the re- quired fourth term, or answer, for the third term of the proportion. Of the other two numbers write the larger for the second term, and the less for the first, when the answer should exceed the third term ; hut write the less for the second term, and the larger for the first, when the answer should be less than the third term. Multiply the second and third terms together, and divide their product by the first; or divide the third term by the ratio of the first term to the second. Note 1. — When the first and second terms are of different denominations, they must be reduced to the same denomination ; and when the third term is a compound number, it must be reduced to the lowest denomination men- tioned in it. The answer will be of the same denomination as the third term. Note 2. — To shorten the operations, factors common to the dividend and divisor may be cancelled. Note 3. — The pupil should perform these questions by analysis, as well as by proportion, and introduce cancellation when it will abbreviate the oper- ation. 240 PROPORTION. Examples. 3. If 16 acres of land cost $ 720, what will 197 acres cost ? Ans. $ 8865. 4. If $ 8865 buy 197 acres, how many acres may be bought for $ 720 ? 5. What will 84hhd. of molasses cost, if 15hhd. can be pur- chased for $ 175.95 ? Ans. $ 985.32. 6. If $ 100 gain $ 6 in 12 months, how much would it gain in 40 months ? Ans. $ 20. 7. If a certain vessel has provisions sufficient to last a crew of 10 men 45 days, how long would the provisions last if the vessel were to ship 5 new hands ? Ans. 30 days. 8. If 7 and 9 were 12, what, on the same supposition, would 8 and 4 be ? 9. If 9 men can perform a certain piece of labor in 17 days, how long would it take 3 men to do it ? Ans. 51 days. 10. If 3 men can perform a piece of labor in 51 days, how many must be added to the number to perform the labor in 17 days ? Ans. 6. 11. A rectangular piece of land containing an acre is 5£ rods in breadth. What is its length ? Ans. 29 T J T rods. 12. If $100 gain $ 6 in a year, how much will $850 gain ? 13. If $ 100 gain $ 6 in a year, how much would be suffi- cient to gain $ 32 in a year? Ans. $ 533.33^. 14. If 20 gallons of water weigh 1671b., what will 180 gal- lons weigh? Ans. 15031b. 15. If a staff 3 feet long cast a shadow of 2 feet, how high is that steeple whose shadow is 75 feet? Ans. 112 £ feet. 16. If 5fcwt. be carried 36 miles for $4.75, how far might it be carried for $ 160 ? Ans. 1212|§ miles. 17. If 100 workmen can perform a piece of work in 12 days, how many men are sufficient to perform the work in 8 days ? Ans. 150. 18. If T 7 2- of a yard cost /^ of a dollar, what will f of a yard cost ? Ans. $ 0.48. 19. What must be paid for 21 A. 3R. 20p. of land, if 36A. 3R. cost $ 1260 ? Ans. $ 750. PROPORTION. 241 20. What is the value of 20001b. of standard gold, the eagle, or $ 10 piece, weighing lOpwt. 18gr. ? 21. If 4£ yards of cloth cost $9.75, what will 13£ yards cost ? Ans. $ 29.25. 22. "What is the length of a rectangle whose contents are 1 sq. ft. and whose breadth is 2^ inches ? Ans. 57f inches. 23. If T 7 F of a ship cost 51£., what are -^ of her worth ? Ans. 10£. 18s. 6fd. 24. If the moon moves 13° 10' 35" in one day, in what time does she perform one revolution ? Ans. 27da. 7h. 43m. -\- 25. If 71b. of sugar cost f of a dollar, what are 121b. worth? Ans. $ 1.28f 26. If $ 1.75 will buy 71b. of loaf-sugar, how much will $ 213.50 buy ? Ans. 8cwt. 2qr. 41b. 27. If 7 ounces of gold are worth 30£., what i^;he value of 71b. lloz. Ans. 407^2s. lOfd. 28. A friend borrowed of me $ 500 for 6 months ; how long ought he to lend me $ 600, to requite the favor ? 29. If the penny loaf weighs 7oz. when flour is $ 8 per barrel, how much should it weigh when flour is $7.50 per barrel ? Ans. 7 T \ ounces. 30. If a regiment of soldiers, consisting of 1000 men, are to be clothed, each suit to contain 3 J yards of cloth that is 1| yards wide, and to be lined with flannel 1£ yards wide, how many yards will it take to line the whole ? Ans. 5625yd. 31. If by working 14 hours per day C. Simmons can plant half of a field in 9 days, in what time will he plant the remain- der, working 10 hours per day at the same rate each hour ? Ans. 12f days. 32. If 75 gallons of water fall into a cistern containing 500 gallons, and 40 gallons run out, in an hour, in what time will it be filled ? Ans. 14h. 17m. 8f sec. 33. How many dozen pairs of gloves, at $0.56 per pair, can be bought for $ 120.96 ? Ans. 18 dozen. 34. A certain cistern has three pipes ; the first will empty it in 20 minutes, the second in 40 minutes, and the third in 75 minutes ; in what time would they all «mpty it ? Ans. 11m. 19 if sec. 35. A can mow a certain field in 5 days, and B can mow it N 21 242 PROPORTION. in 6 days ; in what time would both of them together mow it ? Ans. 2 T 8 T days. 36. A wall, which was to be built 32 feet high, was raised 8 feet by 6 men in 12 days ; how many men must be employed to finish the wall in 6 days ? 37. A can build a boat in 20 days, but with the assistance of C he can do it in 12 days ; in what time would C do it alone ? Ans. 30 days. 38. In a fort there are 700 men provided with 1840001b. of provisions, of which each man consumes 51b. a week ; how long can they subsist ? Ans. 52 weeks 4 days. 39. If 25 men have f- of a pound of beef each, three times in a week, how long will 31501b. last them ? Ans. 56 weeks. 40. How many tiles 8 inches square will lay a floor 20 feet long and lGrfeet wide ? Ans. 720. 41. Ilovrmany stones 10 inches long, 9 inches broad, and 4 inches thick, would it require to build a wall 80 feet long, 20 feet high, and 2£ feet thick ? Ans. 17280 stones. 42. If there be paid for 1 ton 7cwt 3qr. 201b. of coal $ 9.50, what will 13 tons 5cwt 2qr. cost? 43. If 61.3 pounds of tea cost $44.9942, what is the price per pound ? Ans. $ 0.734. 44. What is the value of .15 of a hogshead of lime, at $ 2.39 per hogshead ? Ans. $ 0.3585. 45. If .75 of a ton of hay cost $ 15, what is it per ton ? Ans. $ 20. 46. How many yards of carpeting that is half a yard wide will cover a room that is 30 feet long and 1 8 feet wide ? Ans. 120 yards. 47. If a man perform a journey in 15 days, when a day is 12 hours long, in how many days will he do it when a day is but 10 hours long? 48. If 450 men are in a garrison, and their provisions will last them but 5 months, how many must leave the garrison that the same provisions may be sufficient to supply the remaining men 9 months ? Ans. 2(|p men. 49. The hour and minute hands of a watch are together at 12 o'clock ; when will they next be together ? Ans. lh. 5m. 27 T \see. PROPORTION. 243 50. A and B can perform a piece of work in 5 T 5 T days, B and C in 6§ days, and A and C in 6 days ; in what time would each of them perform the work alone, and how long would it take them to do the work together ? Ans. A would do the work in 10 days ; B, in 12 days ; C, in 15 days ; A, B, and C together, in 4 days. 339. To divide a number or quantity into parts, which are proportional to given numbers. Ex. 1. Divide $250 into two parts which shall be one to the other as 2 to 3. Ans. '$ 100 and $ 150. operation. Since the parts 2 -\- 3 = 5 are to be propor- 5 : 3 : : $ 250 : $ 150, the greater part, ) A t[ °™ 1 to 2 and 3 > 5 : 2 : : $ 250 : $ 100, the less part, f Ans * 7[ h ? se g* » 5 > r J it is evident that the sum of the two numbers, 5, will have the same ratio to the greater of them, 3, as the amount to be divided, $ 250, has to the greater of the required parts ; and that the sum, 5, will have the same ratio to the less num- ber, 2, as the $ 250 has to the less of the required parts ; we there- fore make two statements, and then find the required term of each proportion as in Art. 337. Hence, As the sum of the given numbers is to any one of them, so is the whole quantity to be divided to the part corresponding to the number used as the second term. Note. — This application of proportion is sometimes called Distributive or Partitive Proportion. Examples. 2. A farmer divides between his three sons 246A. IK. 32p. of land, sharing it between them as the numbers 3, 4, and 5. What were the shares ? Ans. 61A. 2R. 18p. ; 82A. 0E. 24p. ; 102A. 2R. 30p. 3. Divide 319 into four parts, that shall be to each other as the numbers 4£, 6£, 6-f , and 7. Ans.55ff£; 85 ¥ y T ; 86f f f ; 91fff. 4. Standard gold for coinage consists of 9 parts of pure gold, and 1 part alloy. Allowing the alloy to be silver and copper in equal parts, how much pure gold, silver, and copper are con- tained in a double eagle, its weight being loz. lpwt. 12gr. ? Ans. 19pwt. 8fgr. gold; lpwt. lfgr. silver; lpwt. lfgr. copper. 244 PROPORTION. 5. The half-dollar of the United States coinage weighs 192 grains Troy, and consists of 9 parts pure silver and 1 part of copper. How much pure silver and how much copper in 20 half-dollars? Ans. 7oz. 4pwt. silver; 16pwt. copper. 6. Divide $ 600 between three men, so that the second man shall receive one third more than the first, and the third man shall receive two thirds more than the second. 7. A, B, and C freight a steamer ; A puts on board 98 tons, B 86 tons, and C 64 tons. Owing to danger of being wrecked, there were thrown overboard while at sea 93 tons. "What should be the number of tons lost by each ? Ans. A, 36 j tons ; B, 32£ tons ; and C, 24 tons. 8. A and B start together by railroad from Chicago for Ga- lena; A travels by freight train at the rate of 15 miles per hour, and B by passenger train at the rate of 25 miles per hour. C leaves Galena for Chicago at the same time by ex- press train, whose velocity is at the rate of 32 miles per hour. Allowing the distance between the two places to be 160 miles, how far from Chicago will A and B each be, when C passes them ? Ans. A, 51^ miles ; B 70-^ miles. COMPOUND PROPORTION. 340. A Compound Proportion is an expression of equality between a compound and a simple ratio. Thus, 4*3) k [ „ V : : 60 : 63, is a compound proportion. Compound proportion is employed in the solution of such questions as would require two or more statements in Simple Proportion. Ex. 1. If 8 men spend $ 32 in 13 weeks, what will 24 men spend in 52 weeks ? Ans. $ 384. operation. In stating the Extreme. Mean. question, we make 8 a a \ Mean. Extreme. 1 „ ~ ■ • 1 • t* men : 24 men I . . <& 30 • the ^ quotient ft arising from dividing the percentage by the number denoting the basis must equal the rate per cent. We therefore divide 12 by the 50, and obtain \% = -^, which, expressed decimally, equals .24, or 24 per cent. Since the question, evidently, is the same as to find if of 100 per cent, we multiply the 12 by 100, or annex two ciphers and divide by 50, and obtain the same result as before. Rule. — Annex two ciphers to the number denoting the percentage, and divide by the number on which the percentage is reckoned ; and the quotient will be the rate per cent. Or, As the given basis of percentage is to the given percentage, so is 100 per cent, to the rate per cent, required. Examples. 2. What per cent, of 16 is 2 ? Ans. 12J per cent. 3. What per cent, of 110 is 11 ? Ans. 10 per cent. 4. What per cent, of 2£ is £ ? 5. £ of 18 per cent, is what per cent, of 24 per cent. ? Ans. 25 per cent. 6. What per cent, of $ 150 is 25 per cent, of $ 36 ? N 22 Ans. 6 per cent. 254 PERCENTAGE. 7. 36 bushels is what per cent, of 48 bushels ? Ans. 75 per cent. 8. What per cent, of 4 years is 1 year 6 months ? 9. 31 gallons 2 quarts is what per cent, of 1 hogshead ? Ans. 50 per cent. 10. Of 160 yards of cloth there have been sold 128 yards ; what per cent, of the whole remains unsold ? Ans. 20 per cent. 11. What per cent, of J- of f of § is -J ? Ans. 20 per cent. 12. In a certain school the number of pupils studying ge- ography is 40 per cent, more than the number studying gram- mar. What per cent, less is the number studying grammar than the number studying geography ? Ans. 28f- per cent. 13. If a miller takes out 4 quarts toll from every bushel he grinds, what per cent, does he take for toll ? 14. If a certain coin is made of 22 parts copper and 3 parts nickel, what per cent, of it is copper, and Avhat per cent, nickel? Ans. 88 per cent, copper ; 12 per cent, nickel. 15. In a certain orchard there are 250 trees, of which 40 per cent, are apple-trees, 12 per cent, cherry-trees, 8 per cent, plum-trees, and the remainder, with the exception of 25 pear- trees, consist of peach-trees. What per cent, of the whole are the peach-trees ? Ans. 30 per cent. 349t To find a number when a given number is known to be a certain per cent, of it. Ex. 1. I have bought two house-lots ; for the one I paid $ 300, which was 60 per cent, of what I paid for the other. What did I pay for the latter ? Ans. $ 500. operation. Since $300 $300 -r- 60 = $5 ; $5 X 100 = $500, Ans. is 60 per cent. 5 of the unknown * w^ ^,-v sum > 1 P er 0r W0X1OO _S5OO. cent, of it 00 must equal fo w of $300, or $ 5 ; and 100 per cent., or the whole of it, must equal 100 times $ 5, or $ 500, Ans. And since the question is evidently the same as to find the value of ^°/ of $ 300, we multiply the $ 300 by 100, and divide by 60, and obtain the same result as before. Rule. — Annex two ciphers to the number denoting the percentage, and divide by that denoting the- rate per cent. Or, PERCENTAGE. 255 As 100 per cent, is to the given rate per cent., so is the given per- centage to the basis of percentage required. Examples. 2. 25 is 10 per cent, of what number? Ans. 250. 3. 16£ is 8 per cent, of what number ? Ans. 203|. 4. 72 is 12 per cent, of what number ? 5. J is 40 per cent, of what number ? Ans. 2 T ^-. 6. $1.62£ is 12 i per cent, of how many dollars ? Ans. $ 13. 7. $ 1^ is £ per cent, of what sum ? Ans. $ 140. 8. A flock of sheep has lost 15-J- per cent, of its number, 17 sheep having been killed by the dogs, and 6 more having been drowned. What was the original number ? Ans. 150. 9. If a man owning 45 per cent, of a mill should sell 33^ per cent, of his share for $ 450, what would be the value of the whole mill ? 10. 12^ per cent, of the length of a certain railroad is equal to 3m. lfur. lrd. What is its entire length ? Ans. 25m. Ofur. 8rd. 11. Gave to a benevolent society 19 bushels of corn, which was 17^- per cent, of all I raised. How many bushels had I left ? Ans. 9 If bushels. 12. Dalton says to Turner, $36.89 is 13| per cent, of the sum you borrowed of me ; and Turner replies, It is just 16| per cent, of the amount I have repaid you. How much of the money that was borrowed remains unpaid? Ans. $ 57.66. 350. To find the number on which the percentage is reck- oned, when the amount, or the remainder, and the rate per cent, are given. Ex. 1. Sold a horse for $ 200, which was 25 per cent, more than he cost. What did he cost ? Ans. $ 160. operation. Since the 1 _|_ .25 = 1.25 ; $ 200 — 1.25 = $ 160, Ans. $ 200 is ev- idently the cost and 25 per cent, of the cost, it must equal the cost taken 125 times. Therefore, the given amount, $ 200, divided by 1 increased by .25, or the given per cent, expressed decimally, equals $ 160, the basis of percentage required. 256 PERCENTAGE. 2. Sold 160 cords of wood, which was 20 per cent, less than the whole number owned. How many cords were owned ? Ans. 200 cords. operation. Since the 1 — .20 == .80; 160 -f- .80 = 200 cords, Ans. 160 cords are 20 per cent, less than the whole number owned, it must equal the whole ' number of cords taken .80 times. Therefore the given remainder, 160 cords, divided by 1 decreased by .20, or the given per cent, expressed deci- mally, equals 200 cords, the basis of percentage required. Rule. — Divide the given amount by 1 increased by the given per cent., or the given remainder by 1 decreased by the given per cent, ex- pressed decimally. Or, As 1 increased by the given per cent., or 1 decreased by the given per cent., expressed decimally, is to 1, so is the given amount or the given re- mainder to the basis of percentage required. Examples. 3. 126 is 5 per cent, more than what number ? Ans. 120. 4. 328£ is 10 per cent, less than what number ? Ans. 365. 5. $ 19.50 is 7 per cent, less than how many dollars? 6. $ 34.40 is £ per cent, more than how many dollars ? Ans. $34.27^|. 7. A man expends in a week $ 24, which exceeds by 33 J per cent, his earnings in the same time. What were the earn- ings? Ans. $18. 8. D. Chandler lives distant from the village 2m. 6 fur. 24rd., which is 12 J per cent, nearer than is J. Mitchel's residence. At what distance does Mitchel live from the village ? Ans. 3m. lfur. 33frd. 9. Bought a carriage for $ 123.16, which was 16 per cent, less than I paid for a horse. How much was paid for the horse ? 10. A carpet having been cut from a piece of carpeting, there were left 6yd. l T Vqr., which was 75 per cent, less than the quantity cut off. How many yards were there in the piece at first? Ans. 31yd. lqr. 2na. 11. A steamer with a cargo of flour, having been lightened, during a storm, of 10 per cent, of her freight, at the end of the PERCENTAGE. 257 voyage had but 279 barrels to deliver. How many barrels were taken aboard, and how many were lost ? Ans. Taken 310 barrels ; lost 31 barrels. 12. A and B each received the same sum. A spent 86J- per cent, of his money for land, and B lost of it by gambling as much as would equal 27^- per cent, of what both received. They then together had left just $ 36.85^. What was the sum received by each, and how much had each left ? Ans. Each received $ 63 ; A had left $ 8.50J- ; and B had left $ 28.35. MISCELLANEOUS EXAMPLES. 1. What is 7 per cent, of 1672 ? Ans. 117^. 2. What is 5£ per cent, of $ 3266 ? Ans. 174ff • 3. Find 312 per cent, of £ of 2 J. 4. Find 1 J per cent, of 180. Ans. 2 T V 5. $$ is what per cent, of 3 ? Ans. 31 6§. 6. -|- is what per cent, of T 5 B ? Ans. 40. 7. Find what per cent. 3.50 is of 50. 8. 13 is 16 per cent, of what number? Ans. 81 J. 9. $ 66 is 100 per cent, of what number ? Ans. $ 66. 10. ^ is 61 per cent, of what number ? Ans. 5f -|. 11. $ 21.28£ is 3£ per cent, less than what sum? Ans. $ 22.00. 12. 191b. 12oz. is 16£ per cent, of how many pounds? Ans. 117|-f pounds. 13. A grocer bought 6 boxes of eggs, each containing 30 dozen, and found that 15 per cent, of the whole were bad ; how many eggs did he lose ? Ans. 324 eggs. 14. There is paid for sawing a cord of wood $ 0.69, which is 12 per cent, of the cost of the wood. What did the wood cost? 15. Three men agreed to excavate 40500 cubic feet of earth; by the first week's labor they excavate 200 cubic yards ; by the second, 6000 cubic feet ; and by the third, 25 per cent, of what remained at the end of the second week. They then called the work half done ; but how many cubic feet did the job lack of being half completed ? Ans. 1575 cubic feet. 22* 258 PERCENTAGE. 16. 25 per cent, of £ of a ship is how many per cent, of £ of it ? Ans. 16§ per cent. 17. If molasses cost 20 per cent, less than $ 0.50 per gallon, and it be sold at 25 per cent, more per gallon than it cost, at what price is it sold ? Ans. $ 0.50 per gal. 18. I have 20 yards of yard-wide cloth, which will shrink on sponging 4 per cent, in the length, and 5 per cent, in the width ; how much less than 20 square yards will there be of it after sponging ? Ans. l^f yards. 19. A gentleman having a large farm gave 15 per cent, of it to his oldest daughter, 10 per cent, of what remained and ^ of an acre he gave to his oldest son, and 25 per cent, of the re- mainder he gave to his wife. The residue he divided equally among his other 5 children, who received each 39 acres. How many acres did his farm contain ? Ans. 340 acres. 20. If the population of the United States in 1858 be 30,500,000, what will it be in 1868, allowing the increase should be at the rate of 34£ per cent. ? t ° 21. In a certain battle in which the English, French, and v ' Turks were allied against the Russians, there were 33£ per cent, more French than English, and the Turks were 8£ per cent, more than the French and 1600 more than the English, llcquired the whole number of the allies, and the per cent, the English, the French and the Turks each were of that number. Ans. Whole number 13600 ; English, 26 T 8 7 per cent. ; French, 35 T ^ per cent. ; and Turks, 38^ T per cent. 22. Bought a cargo of flour, consisting of 560 barrels, at $7.25 per barrel, less 10 per cent., and sold the same at 10 per cent, more than $ 7.25 per barrel. At what per cent, above the cost was the flour sold? How much was made by the operation ? 23. The population of a certain city, whose gain of inhab- itants in 5 years has been 25 per cent., is 87500 ; what was it 5 years ago ? Ans. 70000. 24. Bought a horse, buggy, and harness for $ 500. The horse cost 37 J per cent, less than the buggy, and the harness cost 70 per cent, less than the horse. "What was the price of each ? Ans. Buggy $ 275|f ; horse, $ 172Jf ; and harness, $ 51f J. INTEREST. 259 INTEREST. 351. Interest is the compensation which the borrower of money makes to the lender ; and it is generally reckoned as a certain rate per cent, for any given time, but usually for one year. The principal is the sum lent, on which interest is computed. The amount is the interest and principal added together. Simple interest is that reckoned on the principal only ; and is that meant when the term interest is used alone. Legal interest is the rate per cent, established by law. Usury is a higher rate per cent, than is allowed by law. The legal rate per cent, varies in the different States and in different countries. In Maine, New Hampshire, Vermont, Massachusetts, Rhode Island, Connecticut, New Jersey, Pennsylvania, Delaware, Maryland, Virginia, North Carolina, Tennessee, Kentucky, Ohio, Indiana, Illinois, Iowa,. Nebraska, Missouri, Kansas, Ar- kansas, Mississippi, Florida, District of Columbia, and on debts or judgments in favor of the United States, it is 6 per cent. In New York, Michigan, Wisconsin, Minnesota, Georgia, and South Carolina, it is 7 per cent. In Alabama and Texas, it is 8 per cent. In California, it is 10 per cent. In Louisiana, it is 5 per cent. In Canada, Nova Scotia, and Ireland, it is 6 per cent. In England and France, it is 5 per cent. Note. — The legal rate, as above, in some of the States, is only that which the law allows, when no particular rate is mentioned. By special agreement between parties, in Ohio, Indiana, Michigan, Illinois, Iowa, Nebraska, Missouri, Kansas, Arkansas, Louisiana, and Mississippi, interest can be taken as high as 10 per cent. ; in Florida, as high as 8 per cent. ; in Texas and Wisconsin, as high as 12 per cent. ; and in California and Minnesota, any per cent. In New Jersey, by a special law, 7 per cent, may be taken in Jersey City and the township of Hoboken. In Vermont 7 per cent, may be taken on .railroad bonds. Banks in Illinois cannot take above 7 per cent., and in Ohio, not above 6 per cent. In Mississippi, above 6 per cent, can be taken only for money lent. 352 • When the rate of interest is 6 per cent, per annum, the interest for 1 year or 12 months will be 6 cents on every 260 INTEREST. 100 cents on which it is reckoned, or T § 7 of the principal. Hence, for 2 months or £ of a year, it will be 1 cent on every 100 cents, or T ^ of the principal ; and for 1 month or £ of 2 months it will be 5 mills, or ^g- of the principal. Now, since the interest on 100 cents for 1 month, or 30 days, is 5 mills, for 6 days, or -i- of 30 days, it will be 1 mill, or jjfajj of the prin- cipal ; and since the interest on 100 cents for 2 months is 1 cent, or -lis of the principal, for 100 times 2 months, or 200 months, or 16 years 8 months, it will be 100 cents, or equal the whole principal ; and in the same ratio for any other length of time. The interest of $ 1, at 6 per cent, and the ratio of the inter- est to the principal, for 200 months, and other convenient parts of time, is shown in the following Table. Interest of $ 1 For 200 mo. = 16yr. 8mo. is $ 1.00 equal the whole principal. 100 mo. = 8yr. 4mo. " 0.50 " £ of the principal. 66fmo. — 5yr. 6|mo. " 0.333J " | " " 50 mo. — i 4yr. 2mo. " 0.25 " | " " 40 mo. = 3yr. 4mo. " 0.20 « f " " 33jmo. = 2yr. 9£uio. " 0.166f " | " " 25 mo. = 2yr. lmo. " 0.125 " $- " " 20 mo. = lyr. 8mo. " 0.10 " fa " " 16|mo. mm lyr. 4§mo. " 0.0831 " fa " " 12 mo. = lyr. Omo. " 0.06 " ^ " " 10 mo. = fofayr. " 0.05 " fa " « 8 mo. = fofayr. « 0.04 "fa " " 6fmo. = fofayr. « 0.033 J " fa « « 6 mo. = |ofayr. " 0.03 Toir 5 mo. = -fofayr. " 0.025 " fa 4 mo. = Jofayr. u 0.02 " fa 2 mo. = -Jofayr. " 0.01 " ^ 1 mo. = fofayr. « 0.005 " ^ 6 d. = i of a mo. " 0.001 " jfa^ 5 d. — i of a mo. " O.OOOf « ^ 4d. = ftofamo." O.OOOf « ^ 3 d. _ ^ of a ma « O.OOOf « ^ 2d. - ^ of a mo. « 0.000£ « rfto 1 d. = fa of a mo. « 0.0001 « ^ INTEREST. 261 353 # The principal, the rate per cent., the time, and the in- terest have such a relation to each other that, any three of these terms being given, the fourth can be readily found. The com- putations in interest, therefore, admit of the following problems among others : — I. To find the interest ; II. To find the prin- cipal ; III. To find the rate per cent. ; IV. To find the time. 354. To find the interest of any sum for any time at 6 per cent. Ex. 1. What is the interest of $ 2640 for 2 years 7 months and 26 days, at 6 per cent. ? Ans. $ 420.64. FIRST OPERATION. Interest of $ 1 for 2 years Interest of $ 1 for 7 months Interest of $ 1 for 26 days Int. of $ 1 for 2yr. 7mo. 26d. = $0.12 = 0.0 3 5 se 0.0 44- 0.1 5 9 £ The interest of $2640 for 2yr. 7mo. 26d. will be 2640 times as much as the in- terest of $ 1 for the given time. The interest of $1 for 2 years will be twice as much as for 1 year, equal 12cts.; and since the in- terest for 2 months is 1 cent, for 7 months it will be 31 cents, or 3 cents 5 mills. And as the interest for 6 days is 1 mill, for 26 days it will be 41 mills. These several sums added together give the interest of $ 1 at 6 per cent, for the given time, equal $ 0.1 59|- , which taken 2640 times, by multiplying the given principal by it, gives $ 420.64, the interest required. Principal, $ 2 6 4 23760 13200 2640 880 Interest, $4 20,6 40 Ans. SECOND OPERATION. Principal, -J of the prin. = •^ of the prin. = yeVij-ofthe prin. = $2640 3 3 Interest for 2yr. lmo. 8 8 Interest for 6mo. 20d. 2.6 4 Interest for 6d. Ans. $420.6 4 Interest for 2yr. 7mo. 26d. The time, 2yr. 7mo. 26d., is equal to 2yr. lmo. -j- 6mo. 20d. -|- 6d. Now, since the interest on any sum, at 6 per cent, in 200 months equals the principal, for 2yr. lmo., or i of 200 months, it will equal i of the principal. We therefore take 1 of the principal, $ 2640, equal $ 330, as the interest for 2yr. lmo. Of the balance. of 262 INTEREST. time, 6mo. 20d., or 6§mo., being -^ of 200 months, we take -fa of the principal, equal $88, as the interest for the 6mo. 20d.; and the 6d. being j-fa$ of 200 months, we take 10 1 o6 of the principal, equal $ 2.64, as the interest for the 6d. We add together the interest for the parts of the whole time, and obtain, as by the first operation, $ 420.64 as the whole interest. Rule 1 . — Find the interest of $ 1 for the given time, by reckoning 6 cents for every year, 1 cent for every two months, and 1 mill for every 6 days ; then multiply the given principal by the number denoting that interest, and the product will be the interest required. Or, Rule 2. — Take such fractional part or parts of the principal as the number expressing the time is of 200 months. Note 1. — To find the amount, add the principal to the interest. Note 2. — In computing interest for a fractional part of a month, the month is considered as consisting of 30 days. This has the sanction of general usage and the decisions of the courts, though not entirely accurate. Note 3. — Questions in interest, like other exercises in percentage, may be solved by proportion. The foregoing example admits of a statement and solu- tion by the rule of compound proportion (Art. 340). Note 4. — It is customary among merchants to reject the mills in the re- sults of their computations of interest, increasing, however, the number of cents by 1 when the decimal of a cent exceeds 5. Examples. 2. What is the interest of $ 675 for 1 year ? Ans. $ 40.50. 3. What is the interest of $ 3967.87 for 2 years ? Ans. $ 476.144. 4. What is the interest of $ 896.28 for 3 years ? Ans S 161.33. 5. What is the amount of $ 716.57 for 4 years ? 6. What is the amount of $ 76.47 for 7 years ? Ans. $ 108.587. 7. What is the interest of $ 123.45 for 6 years ? Ans. $ 44.442. 8. What is the interest of $ 750 for 12 years ? Ans. $ 540. 9. What is the interest of $ 130 for 2 months ? 10. What is the interest of $ 85 for 3 months ? Ans. $1,275. 11. What is the interest of $ 19.62 for 7 months ? Ans. $0.6867. 12. What is the interest of $ 637 for 10 months ? INTEREST. 263 13. What is the interest of $ 1671.32 for 14 months ? Ans. $116.99. 14. What is the interest of $ 891.24 for 9 months ? Ans. $ 40.10. 15. What is the interest of $ 91 for 5 days ? Ans. $ 0.0758. 16. What is the interest of $ 324.66 for 18 days ? 17. What is the interest of $ 3246 for 27 days ? Ans. $ 14.607. 18. What is the interest $ 1364.24 for 1 day ? 19. What is the interest of $ 6444 for 29 days ? 20. What is the amount of $ 18.60 for 24 days ? 21. What is the interest of $ 386.19 for 100 months ? Ans. $ 193.09. 22. What is the interest of $ 0.75 for 75 years ? Ans. $ 3.37^ 23. What is the interest of $396.15 for 1 year 1 month and 9 days ? Ans. $ 26.343. 24. What is the interest of $36.18 for 3 months and 7 days ? Ans. $ 0.584. 25. What is the interest of $97.15 for 2 years 11 months and 27 days ? 26. What is the interest of $ 76.89J from January 11, 1852, to July 27, 1863 ? Ans. $ 53.262. 27. What is the interest of $98.25 from July 4, 1856, to October 19, 1859 ? Ans. $ 19.404. 28. What is the interest of $22,763 from February 19, 1836, to July 18, 1860 ? 29. What is the interest of $ 175.07 from January 7, 1855, to October 12, 1859 ? Ans. $ 50.04. 30. What is the interest of $ 197.28J- from December 6, 1852, to January 11, 1854 ? Ans. $ 12.987. 31. What is the amount of $ 4377.15 for 3 years ? 32. What is the interest of $444.60 for 5 years and 6 months ? Ans. $ 146.718. 355 • To find the interest of any sum of money at any rate per cent, for any given time. Ex. 1. What is the interest of $ 84.50 at 7 per cent, for 2 years 5 months and 12 days? Ans. $ 14.49. 264 INTEREST. first operation. Having found the Principal, $ 8 4.5 interest for 1 year, Rate per cent. .0 7 and then for 2 years, the interest for 5 Interest for 1 year, 5.9 1 5 months is obtained 2 by first taking J of Int. for 2 years, 11.8300 j ^ j" 461 " 08 *' t j. r a % .* i , Aff1/ , for 4 months, and Int. for 4mo., or i of lyr. 1.9 7 1 6 + thcn i of this last Int. for lmo., or £ of 4mo. .4929+ interestforl month. Int. for 10d., or £ of lmo. .1 6 4 3+ And since 10 days Int. for 2d., or £ of lOd. .0328+ are £ of 1 month, Int.for2yr.5mo.12d. $ 1 4.4 9 1 6+ Ans. ^ontSLelestVr the interest of 10 days ; and since 2 days are £ of 10 days, we take \ of the last interest for 2 days. The interest as found for the several parts of the whole time, added together, gives the interest required. second operation. We first find the in- Principal, $ 8 4.5 terest on the given sum Int. of $ 1 at 6 per cent. .14 7 at 6 per cent., and then ^ add to this interest the J 1 fractional part of itself, 3 3 8 denoted by the excess 8 4 5 of the rate above 6 per T , , n ■% c% a c% i k. t\ ccnt - This excess is 1 Int. at 6 per cent 1 2.4 2 1 5 cent therefore we i of int. at 6 per cen t. 2.0 7 2 5 a(M j f the interest at Int. at 7 per cent. $ 1 4.4 9 1 7 5 Ans. ? P^ ^ ent - to that in " r terest for the answer. If the rate per cent, had been less than 6 per cent, we should have subtracted the fractional part. Rule 1. — First find the interest for one year by multiplying tlie prin- cipal by the rate per cent, expressed decimally ; and for two or more years multiply this product by the number of years. Find the interest for months by taking the most convenient fractional part or parts o/one year's interest. Find the interest for days by talcing the most convenient fractional part or parts o/one month's interest. Or, Rule 2. — Find the interest of the given sum at 6 per cent., and then add to this interest, or subtract from it, such a fractional part of itself as the given rate is greater or less than 6 per cent. Note 1. — I of the interest at 6 per cent, may be taken for that at 1 per cent. ; ^, for that at 1^ per cent. ; £, for that at 2 per cent. ; £, for that at 3 per cent. From the interest at 6 per cent, may be taken £ of itself for that at 4 per cent. ; £ of itself for that at 4£ per cent. ; and £ of itself for that at 5 per cent. To the interest at 6 per cent, may be added I of itself for that at 7 per INTEREST. 265 cent. ; £ of itself for that at 7£ per cent. ; & of itself for that at 8 per cent ; and £ of itself for that at 9 per cent. If the rate is 12 per cent., the interest at 6 per cent, may be taken twice; if 18 per cent., the interest at 6 per cent, may be taken three times, etc. Note 2. — When in this book tho rate of interest is not given, 6 per cent, is to be understood. Examples. 2. What is the interest of $ 16.75 for 7 months and 17 days, at 7 per cent. ? Ans. $ 0.739. 3. What is the interest of $ 11.10$ from April 17, 1852, to December 7, 1852, at 7 per cent. ? Ans. $ 0.496. 4. What is the interest of $ 12.69, from January 2, 1853, to August 30, 1854, at 7 per cent. ? Ans. $ 1.47 J. 5. What is the interest of $5000 for 2 years 5 months 26 days, at 7 per cent. ? 6. What is the amount of $ 416 for 3 years 16 days, at 7 per cent. ? Ans. $ 504.64. 7. What is the interest of $336 for 15 days, at 5 per cent. ? Ans. $ 0.70. 8. What is the interest of $17869.75 from February 7, 1852, to January 11, 1860, at 5 per cent. ? Ans. $7083.3703. 9. What is the interest of $ 300.50 for 1 year 2 months and 15 days ? Ans. $ 21.786. 10. What the interest of $ 37 for 29 days, at 5 per cent. ? 11. What is the interest of $ 35.61 from November 11, 1861, to December 15, 1863 ? Ans. $ 4.474. 12. What is the interest of $ 16.76 from December 17, 1841, to January 17, 1852 ? Ans. $ 10.558. 13. What is the interest of $ 1728.19 from May 7, 1854, to July 17, 1860, at I per cent. ? Ans. $ 26.762. 14. What is the interest of $ 397.16 for 1 year 6 months and 1 day, at 5£ per cent. ? Ans. $ 32.826. 15. What is the amount of $ 100.25 for 2 months and 29 days, at 4 per cent. ? Ans. $ 101.241. 16. What is the interest of $51.17 for 9 months and 29 days, at 4 per cent. ? Ans. $ 1.699. 17. What is the interest of $42.20 for 1 year and 16 days, at 4£ per cent. ? N 23 266 INTEREST. 18. What is the interest of $ 16.25 for 2 years, at 3 per cent. ? Ans. $ 0.975. 19. What is the interest of $ 96.84 from November 27, 1849, to July 3, 1852, at 7£ per cent. ? Ans. $ 18.883. 20. What is the interest of $ 786.97 from October 19, 1857, to August 17, 1861, at 7£ per cent.? 21. What is the interest of $ 71.091 from July 29, 1853, to June 19, 1857, at 12 per cent. ? Ans. $ 33.175. 22. What is the amount of $ 369.29 for 2 years 3 months and 1 day, at 9 per cent. ? Ans. $ 444.163. 23. What is the interest of $ 76.35 for 1 year 8 months and 18 days? Ans. $7,864. 24. What is the interest of $ 47.15 for 1 month and 19 days, at 13£ per cent.? Ans. $ 0.886. 25. What is the interest of $36.72 from May 16, 1829, to February 18, 1857, at 7 per cent. ? Ans. $ 71.342. 26. What is the interest of $ 35.50 for 3 years 5 months and 20 days, at 7 per cent. ? Ans. $ 8.628. 27. What is the amount of $ 496.30 for 6 months and 20 days, at 7 per cent. ? Ans. $ 515.60. 28. What is the interest of $ 691.04 for 1 month 3 days, at 5 per cent. ? Ans. $3,167. 29. What is the interest of $ 9750 for 4 months, at 2 per cent, a month ? Ans. $ 780. 30. What is the interest of $ 9162 for 3 months, at 1£ per cent, a month ? Ans. $ 412.29. 31. What is the interest of $ 1500 for 7 months 20 days, at 10 per cent. ? Ans. $ 95.833. 32. What is the interest of $ 640.50 for 10 months and 26 days, at 10 per cent. ? Ans. $ 58.00. 33. What is the interest of $3178 for 15 months and 15 days ? 34. If a banker borrow $ 10,000 in Boston at 6 per cent., and let the same in Wisconsin at 7 per cent., how much does he make by the operation in that way in 2 years and 6f months ? Ans. $ 255.555. 356. To reckon interest for any number of days, when 12 months of only 30 days each, or 360 days, are considered a year. INTEREST. 267 Ex. 1. What is the interest of $ 460 for 93 days ? Ans. $7.13. FIRST OPERATION. Principal, i of 93d. = I5i&. We multiply the principal $ 4 6 by i of the number of days, .0 1 5 J- considered as thousandths; since the interest of $ 1 at 6 per cent, for 6 days is 1 mill or 1 thousandth of a dollar, (Art. 352,) and for any 2300 460 230 Interest, $ 7.1 3 Ans. other number of days, at the same per cent, must be one sixth as many mills or thousandths of a dollar as there are days SECOND OPERATION. Principal, y^g- of the principal, J of the interest for 60d. of the interest for 30d. tV $460 4.6 2.3 .2 3 Ans. $7.13 Interest for 60d. Interest for 30d. Interest for 3d. Interest for 93d. 93 days =60 days -\- 30 days -f- 3 days. Now, since the interest of any sum at 6 per cent, for 2 months, or 60 days, equals yj-g- of the principal (Art. 352), ^ of $ 460 = $4.60 will be the interest for 60 days ; \ of the interest for 60 days, or $ 2.30, will be that for 30 days; and J* of the interest for 30 days, or $ 0.23, will be that for 3 days. The interest for the several parts of the time added together must give the interest for the whole time, or 93 days. Rule. — Multiply the principal by one sixth of the number of days expressed decimally as thousandths. Or, Divide the principal by 100, and take such a part or parts of the quo- tient as the given number of days is of 60 days. The result will be the interest at 6 per cent., from which it may be found for any other rate, as in Art. 355. Note 1. — One sixth of the number of days in any number of months of 30 days each is equal to 5 times the number of months. Thus, one sixth of the number of days in 7 months equals 7 X 5, or 35 days ; and one sixth of the number of days in 9 months and 13 days equals (9X5) + (13 -r 6), or 45 + 2* = 471 days. Note 2. — It is a common practice among mercantile men, in calculating interest, to consider a year to consist of only 360 days, but the laws of some of the States require 365 days to a year. 360 days are in fact only 2|° = l| f a common year, and therefore are J g less than what perfect accuracy would re- quire. Hence, when the year is considered to be one of 365 days, the interest as found by the rule must be diminished by -% of itself ; Or, we may multiply the principal by the number of days, and, if the rate be 6 per cent., divide by 6083 J; if the rate be 7 per cent., divide by 5214; and if the rate be 5 per cent., di- vide by 7300. 268 INTEREST. Examples. 2. What is the interest of $96 for 33 days, at 7 per cent.? Ans. $.616. 3. What is the interest of $ 320.40 for 63 days, at 5 per cent.? Ans. $2,803. 4. What is the interest of $ 131.20 for 123 days ? Ans. $2,689. 5. What is the interest of $ 26.60 for 78 days? 6. What is the interest of $5780 for 153 days, at 10 per cent? Ans. $245.65. 7. What is the interest of $ 105.10 for 48 days, at 12 per cent.? Ans. $1,681. 8. What is the interest of $ 13.62 for 93 days, at 7£ per cent. ? 9. What is the interest of $ 4580 for 253 days, at 5£ per cent.? Ans. $ 177.029f|. 10. What is the interest of $3140 for 273 days, at 7 per cent.? Ans. $ 166.681§. 11. What is the interest of $ 10550 for 243 days, at 8 per cent.? Ans. $569.70. 12. What is the amount of $ 33.44 for 333 days ? Ans. $ 35.295. 13. What is the amount of $71.60 for 3 months and 18 days ? 14. What is the interest of $ 92.96 for 4 months and 3 days, at 7 per cent. ? Ans. $ 2.223. 15. What is the interest of $ 144.50 for 144 days, allowing 365 days to the year, at 5 per cent. ? Ans. $ 2.85. 16. What is the interest of $ 761.81 for 165 days, allowing 365 days to the year? Ans. $ 20.662. 17. What is the interest of $560 for 183 days, allowing 365 days to a year, at 7 per cent. ? Ans. $ 19.65. 18. What is the interest of $ 1960 for 93 days ? 19. What is the amount of $1000 for 1 month and 3 days? Ans. $1005.50. 20. What is the amount of $7300 for 18 days, at 6 per cent.? Ans. $7321.90. INTEREST. 269 357. To find the interest on sterling money, at any rate per cent., for any time. Ex. 1. What is the interest of 576£. 5s. 6d. for 1 year and 10 months ? Ans. 63£. 7s. 9d. OPERATION. 576£. 5s. 8£d. = 5 7 6.2 5 5 £. Principal. T iy of the principal, 5 7.6 2 5 5 Int. for lyr. 8mo. T \y of the int. for lyr. 8mo. 5.7 6 2 5 5 Int. for 2mo. Ans. 6 3.3 8 8 5 £. = 63£. 7s. 9d.-f- Int. for lyr.lOmo. We reduce the 5s. 6d. to the decimal of a pound (Art. 279), and annex it to the pounds ; we then find the interest as though the sum expressed dollars and cents ; and, reducing the decimal in the answer to shillings and pence, we have 63£. 7s. 9d.-{- as the interest re- quired. Hence, Reduce the shillings, pence, and farthings, if any, in the principal to the decimal of a pound. Then proceed as in United States money; and, if there be a decimal in the result, reduce it to a compound number. Examples. 2. What is the interest of 179£. 12s. lid. for 1 year and 7 months, at 5 per cent.? Ans. 14£. 4s. 5|-d. 3. What is the interest of 25£. for 1 year and 9 months, at 5 per cent. ? 4. What is the interest of 5440£. 10s. for 3 years and 11 months, at 6 per cent. ? . Ans. 1278£. 10s. 4d. 5. What is the interest of 943£. Is. 8d. from May 1, 1857, to October 21, 1857, allowing 365 days to a year, at 5£ per cent. ? Ans. 23£. 9s. 4d. 358. To find the principal, the interest, the time, and the rate per cent, being given. Ex. 1. What principal will gain $ 120 in 4 years, at 6 per cent. ? Ans. $ 500. operation. The interest for the given $12 0.0 0~-.2 4 = $500 Aois. time and rate is 24 cents on a principal of $ 1 ; therefore, the required principal will be as many times $ 1 as the number denot- ing the given interest contains times .24. Rule. — Divide the given interest by the interest of $ 1 for the given time, at the given rate. 23* 270 INTEREST. Examples. 2. What principal will gain $ 10.08 in 1 year, at 7 per cent.? Ans. $144. 3. What principal will yield an income of $ 13.20 in 1 year and 4 months, at 8£ per cent. ? Ans. $ 120. 4. What sum should be paid for a ground rent paying yearly, at 6 per cent., $ 40.50 ? 5. If the interest on a sum borrowed at 2 per cent, a month is $ 24 for 90 days, what is the sum ? Ans. $ 400. 6. What principal at 7£ per cent, is sufficient to produce $ 206.38£ interest in 183 days ? Ans. $ 5600. 359. To find the rate per cent., the principal, the in- terest, and the time being given. Ex. 1. If the interest of $500 for 4 years is $120, what is the rate per cent. ? operation. "We f m rt the interest on a 120 -r 20= G per cent. Ans. given principal for the given term to be $ 20, at 1 per cent. ; then the required rate will be as many per cent, as 20 is contained times in 120, or 6 per cent. Rule. — Divide the given interest by the interest of the given princi- pal at 1 per cent, for the given time. Examples. 2. The interest of $ 144 for 1 year is $ 10.08 ; what is the rate per cent. ? Ans. 7 per cent. 3. At what rate per cent, must $ 120 be on interest to amount to $ 133.20 in 1 year 4 months? 4. At what rate per cent, must $ l,or any other sum, be on interest to double itself in 14f years? Ans. 7 per cent. 5. At what rate per cent, must any sum of money be on in- terest to quadruple itself in 33£ years ? Ans. 9 per cent. 6. I receive yearly $ 232.50 interest on $ 4650 loaned to the State of Massachusetts ; what is the rate per cent. ? Ans. 5 per cent. 7. At what rate per cent, must $7500 be loaned to gain $ 60.00 in 48 days ? INTEREST. 271 8. At what rate per cent, must $ 280 be on interest to amount to $ 411.95 in 6 years and 6 months ? Ans. 7 £ per cent. 9. At what rate per cent, must $ 480 be on interest to amount to $529.60 in 1 year 3 months and 15 days? Ans. 8 per cent. 300. To find the time, the principal, the interest, and the rate per cent, being given. Ex. 1. How long must $ 500 be on interest, at 6 per cent., to gain $ 120 ? Ans. 4 years. operation. We find that the given princi- 120 — 30 = 4 years, Ans. pal at the given rate will produce $ 30 interest in 1 year ; therefore to produce $ 1 20 interest there will be required as many years as 30 is contained times in 120, or 4 years. Rule. — Divide the given interest by the interest of the given princi- pal for 1 year. Examples. 2. How long must $ 120 be on interest, at 8 j- per cent., to gain $ 13.20 ? Ans. lyr. 4mo. 3. In what time will $ 144 produce $ 10.08 interest, at 7 per cent. ? Ans. 1 year. 4. In what time will $ 240 at interest amount to $ 280, at 6 per cent. ? Ans. 2yr. 9mo. lOd. 5. In what time will $ 1, or any other sum, double itself, at 5 per cent, interest ? Ans. 20 years. 6. In what time wdll any sum double itself at 10 per cent, in- terest? Ans. 10 years. 7. In what time will $ 1500 amount to $ 2250, at 5 per cent, interest ? 8. In what time will $480 at 4^- per cent, amount to $ 561.60 ? • Ans. 3yr. 9mo. lOd. 9. In what time, at 12 per cent., will $1728 amount to $3853.44? Ans. lOyr. 3mo. 10. In what time, at 6 per cent., will $ 240 amount to $ 720 ? Ans. 33yr. 4mo. 11. Borrowed, May 16, 1857, the sum of $400, payable as soon as the principal, increased by the interest at 6 per cent., shall equal $ 500. At what date is it payable ? Ans. July 16, 1861. 272 PROMISSORY NOTES. I PROMISSORY NOTES. 361. A Promissory Note, or note of hand, is an engage- ment, in writing, to pay a specified sum, either to a person named in the note, or to his order, or to the bearer. A joint note is one signed by two or more persons, who to- gether are holden for its payment. A joint and several note is one signed by two or more per- sons, who separately and together are holden for its payment. A negotiable note is one so made that it can be sold or trans- ferred from one person to another. 362. The maker or drawer of a note is the person who signs it. The payee, promisee, or holder of a note is the person to whom it is to be paid. The indorser of a note is the person who writes his name upon its back to transfer it, or as a guaranty of its payment. If the indorser, however, wishes only to transfer the note, he may write before his name the words " without recourse," and then, though by his name he guarantees the genuineness of the note, he is not liable for its payment, should the maker not pay it when due. 363. The face of a note is the sum for which it is given. 364. A note should contain the words "value received," and the sum for which it is given should be expressed in written words, or, as is the general custom, the dollars may be written in words and the cents, if any, be expressed as hun- dredths of a dollar in the form of a common fraction. Note. — The laws of Pennsylvania require that the words " without defal- cation " should be inserted in a promissory note; and in Indiana, notes generally contain the words " without any relief whatever from valuation or appraise- ment laws." 365. A note is said to be one on time, when it is made pay- able on or after a certain date, and the day on which it becomes legally due is called the day of maturity. 366. According to the laws of many of the States, when a particular day is specified in the note for its payment, three PROMISSORY NOTES. 273 days additional are allowed, called days of grace, within which the maker may pay the note, unless it states " without grace." Should the third day of grace, however, fall on Sunday, or some public day, the day of maturity will be a day earlier. 367. When a note is written for months, calendar months are understood. Thus, if a note be dated April 21, for one month, it will be nominally due May 21 ; and if dated January 29, 30, or 31, it being for one month, it will be nominally due February 28, should it be a common year, or-February 29, should it be a leap year. 868. When a note is written without interest, it can only draw interest, if on time, after the time specified for its payment, and not then lawfully, in some States, till after a demand has been made ; or if not on time, after payment has been de- manded. Note. — A note attested or witnessed, in Massachusetts and some other States, is taken out of the statute of limitation. PARTIAL PAYMENTS ON NOTES AND BONDS. 369. Partial or part payments on notes, bonds, or other obligations, being receipted for by entry on the back of the obligation, are termed indorsements. United States Rule. 370. In the United States courts, and the courts of Mas- sachusetts, New York, and several other States, interest on notes and bonds, when partial payments have been made, is reckoned according to the following Rule. — Compute the interest on the principal to the time when the first payment was made, which equals or exceeds, either alone or with preceding payments, the interest then due. Add that interest to the principal, and from the amount subtract the payment or payments thus far made. The remainder will form a new principal ; on which compute the in- terest, proceeding as before. Note. — This rule is on the principle that neither interest nor payment should draw interest. 274 PROMISSORY NOTES. Examples. (1.) $ 165.18. Boston, June 17, 1847. For value received, I promise to pay Nathaniel Ford, or order, on demand, one hundred and sixty-five dollars and eighteen cents, with interest. Attest, Joseph Field. James Peterson. On this note are the following indorsements. December 7, 1847, received eighteen dollars and thirteen cents of the within note. October 19, 1848, re- ceived twenty-eight dollars and sixteen cents. September 25, 1849, received thirty-six dollars and twelve cents. July 10, 1850, received three dollars and eighteen cents. June 6, 1851, received thirty-six dollars and twenty-eight cents. December 28, 1852, received thirty-one dollars and seventeen cents. May 5, 1853, received three dollars and eighteen cents. September 1, 1853, received twenty five dollars and eighteen cents. October 18, 1854, received ten dollars. How much remains due September 27, 1855 ? Ans. $15,417. OPERATION. Principal, carrying interest from June 17, 1847, $ 165.180 Interest from June 17, 1847, to Dec. 7, 1847, 5mo. 20d., 4.680 Amount, 169.860 First payment, Decemlfer 7, 1847, .... 18.130 Balance for new principal, 151.730 Interest from Dec. 7, 1847, to Oct. 19, 1848, lOmo. 12d., 7.889 Amount, 159.619 Second payment, October 19, 1848, . . . 28.160 Balance for new principal, 131.459 Interest from Oct. 19, 1848, to Sept. 25, 1849, llmo. 6d., 7.361 Amount, 138.820 Third payment, September 25, 1849, .... 36.120 Balance for new principal, 102.700 Interest from Sept. 25, 1849, to June 6, 1851, 20mo. lid., 10.458 Amount, 113.158 Fourth pay't, July 10, 1850, a sum less than interest, 3.18 Fifth pay't, June 6, 1851, a sum greater than interest, 36.28 39.460 Balance for new principal, . . . . . . 73.698 Interest from June 6, 1851, to Dec. 28, 1852, 18mo. 22d., 6.903 Amount, 80.601 PROMISSORY NOTES. 275 Amount brought forward, $ 80.601 Sixth payment, December 28, 1852, .... 31.170 Balance for new principal, 49.431 Interest from Dec. 28, 1852, to May 5, 1853, 4mo. 7d., 1.046 Amount, Seventh payment, May 5, 1853, Balance for new principal, .... Interest from May 5, 1853, to Sept, 1, 1853, 3mo. 26d., Amount, 48.211 Eighth payment, September 1, 1853, .... 25.180 Balance for new principal, 23.031 Interest from Sept. 1, 1853, to Oct. 18, 1854, 13mo. 17d., 1.562 Amount, 24.593 Ninth payment, 10.000 Balance for new principal, 14.593 Interest from Oct. 18, 1854, to Sept. 27, 1855, llmo. 9d./ .824 Balance due at the time of payment, . . . . $15,417 (2.) $ 769.87. St. Louis, June 17, 1849. For value received, I promise to pay L. Swan, or order, on demand, seven hundred and sixty-nine dollars and eighty-seven cents, with interest. Samuel Q. Peters. Attest, Moses Haynes. Payments: March 1, 1850, seventy-five dollars and fifty cents; June 11, 1851, one hundred and sixty-five dollars; September 15, 1851, one hundred and sixty-one dollars; Jan. 21, 1852, forty-seven dollars and twenty-five cents; March 5, 1853, twelve dollars and seventeen cents ; December 6, 1853, ninety- eight dollars; July 7, 1854, one hundred and sixty-nine dollars. What remains due September 25, 1855 ? Ans. $ 226.297. (3.) $ 300. Chicago, April 30, 1851. For value received, I promise Kimball fy Hammond to pay them, or order, on demand, three hundred dollars, with interest. Simpson W. Leavet. Payments : June 27, 1852, one hundred and fifty dollars ; December 9, 1852, one hundred and fifty dollars. What was due October 9, 1853 ? Ans. $ 26.735. (4.) $ 54.18. San Francisco, Feb. 11, 1852. For value received, I promise to pay John Trow, or order, on demand, fifty-four dollars and eighteen cents, with interest. Luke M. Sampson. 276 PROMISSORY NOTES. Payments: July 11, 1853, twelve dollars and twenty-five cents; August 15, 1854, two dollars and ten cents; July 9, 1855, three dollars and twelve cents; August 21, 1855, thirty-seven dollars and eighteen cents. What was due December 17, 1855 ? Ans. $ 10.222. (5.) $ 1728. Philadelphia, Jan. 7, 1851. For value received, we jointly and severally promise to pay Jones, Oliver, fy Co., or order, one thousand seven hundred and twenty-eight dollars, on demand, with interest, without defalca- tion. John Wickersham. Attest, Timothy True. James Thickstein. (6.) $ 1000. New York, January 1, 1850. For value received, I promise to pay James Johnson, or order, on demand, one thousand dollars, with interest at seven per cent. Samuel T. Fortune. Indorsements: September 28, 1850, one hundred and forty-four dollars; March 1, 1851, twenty dollars; July 17, 1851, three hundred and sixty dol- lars; August 9, 1851, one hundred and ninety dollars ; September 25, 1852, one hundred and seventy dollars ; December 11, 1853, two hundred dollars; July 4, 1855, seventy-five dollars. What was due June 1, 1857 ? Ans. $ 7.61. Connecticut Rule. 371 • The Supreme Court of the State of Connecticut has adopted the following Rule. — Compute the interest to (lie 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 pay- ment, 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. 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. If any payments be made of a less sum than the interest arisen at the time of such payment, no interest is to be computed, but only on the prin- cipal sum for any period. Note. — If a year extends beyond the time of settlement, find the amount of the remaining principal to the time of settlement ; find also the amount of the indorsement or indorsements, if any, from the time they were paid to the time of settlement, and subtract their sum from the amount of the principal. PROMISSORY NOTES. 277 Example. (1.) $ 900. New Haven, June 1, 1858. For value received, I promise to pay J, Downs, or order, nine hundred dollars, on demand, with interest James L. Emerson. Indorsements: June 16, 1859, two hundred dollars; August 1, 1860, one hundred and sixty dollars ; November 16, 1860, seventy-five dollars; February 1, 1862, two hundred and twenty dollars. What was due August 1, 1862 ? Ans. $ 417.822. OPERATION. Principal, $900.00 Interest from June 1, 1858, to June 16, 1859, \2\ months, 56.25 956.25 First payment, 200.00 756.25 Interest from June 16, 1859, to August 1, 1860, 131 months, 51.046 807.296 Second payment, 160.000 647.296 Interest for 1 vear, 38.837 686.133 Am't of 3d pay't, from Nov 16, 1860, to Aug. 1, 1861, 8-^mo., 78.187 607.946 Interest from Aug. 1, 1861, to Aug. 1, 1862, 12 months, 36.476 644.422 Am't of 4th pay't, from Feb. 1, 1862, to Aug. 1, 1862, 6mo., 226.600 Balance due August 1, 1862, .... $417,822 Merchants' Ktjle. 372. It is customary with merchants and others, when par- tial payments are made of notes or other debts, when the note or debt is settled within a year after becoming due, to adopt the following Rule. — Find the amount of the principal from the time it became due until the time of settlement. Then find the amount of each indorse- ment from the time it teas paid until settlement, and subtract their sum from the amount of the principal. N 24 278 PROMISSORY NOTES. Examples. (1.) $ 1728. Baltimore, January 1, 1853. For value received, I promise to pay Riggs, Peabody, S? Co,, or order, on demand, one thousand seven hundred and twenty- eight dollars, with interest. John Paywell, Jr. Indorsements: March 1, 1853, three hundred dollars; May 16, 1853, one hundred and fifty dollars ; September 1, 1853, two hundred and seventy dol- lars; December il, 1853, one hundred and thirty-five dollars. What was due at the time of payment, which was December 16, 1853? Ans. $948.03 OPERATION. Principal, $1728.00 Interest for 11 months and 15 days, . . . . 99.36 $ 1827.36 First payment, $300.00 Interest for 9 months and 15 days, . 14.25 Second payment, .... 150.00 Interest for 7 months, . . . 5.25 Third payment, .... 270.00 Interest for 3 months and 15 days, . 4.72 Fourth payment, .... 135.00 Interest for 5 days, . . . .11 879.33 Balance remaining due, December 16, 1853, $ 948.03 (2.) $ 700. Montpelier, February 4, 1854. For value received, we jointly and severally promise to pay James Thomas, or order, on demand, seven hundred dollars, with interest Sampson Phillips, Kichard Fletcher. Payments: March 18, 1854, one hundred and sixty dollars; June 24, 1854, two hundred dollars; September 11, 1854, one hundred and twenty dollars ; October 5, 1854, sixty dollars. What was due on this note Nov. 28, 1854 ? Ans. $ 180.43. (3.) $ 500. Detroit, January 1, 1857. For value received, three months after date 1 promise to pay to the order of James Francis five hundred dollars. William Amsden. Indorsement: July 1, 1857, two hundred dollars. What was due April 1, 1858, the rate of interest being 7 per cent.? Ans. $324.50. COMPOUND INTEREST. 279 COMPOUND INTEREST. 373. Compound Interest is interest on the original principal with its interest added when remaining unpaid after becoming due. 374. When the interest is added to the principal at the end of every year, and a new principal is thus formed yearly, it is said to compound annually ; when the interest is added to the principal so as to form a new principal half-yearly, it is said to compound semiannually. 375. Compound interest is based upon the principle, that, if the borrower does not pay the interest as it becomes due at stated times, it is no more than just for him to pay interest for the use of it, so long as he shall have it in his possession. Note. — Compound interest is not favored by the laws, though it is not usurious. A contract or promise to pay money with compound interest can- not generally be enforced, being only valid for the principal and legal interest. 376. To find the compound interest of any sum of money at any rate per cent, for any time. Ex. 1. What is the compound interest of $ 300 for 3 years? FIRST OPERATION. Principal for 1st year, Interest of $ 1 for 1 year, Interest for 1st year, Principal for 2d year, Interest for 2d year, Principal for 3d year, Interest for 3d year, $300 .0 6 18.0 3 0.0 3 1 8.0 .0 6 1 9.0 8 318.0 3 3 7.0 8 .06 2 33 0.2 2 4 8 708 Amount for 3 years, 3 5 First principal, 3 Comp. int. for 3 years, $ 5 7.3 4 8, Ans. 7.3 4 8 0.0 We first multiply the given principal by the number denoting the interest of $ 1 for one year, and add the in- terest thus found to the principal for the amount ; on which as a new principal we find the interest for the sec- ond year, and proceed as before ; and so also with the third year. From the amount of the last year we subtract the first principal, and ob- tain the compound in- terest for 3 years. 280 COMPOUND INTEREST. SECOND OPERATION- Principal for 1st year, $ 3 0.0 J% of the principal, 1 5.0 -£ of the interest at 5 per cent, 3.0 Principal for 2d year, 3 1 8.0 Amount for 1st year at 2*y of the principal, 1 5.9 [6 per cent, i of the interest at 5 per cent., 3.1 8 Principal for 3d year, 3 3 7.0 8 Amount for 2d year at 2-\r of the principal, 1 6.8 4 5 [6 per cent. ■J of the interest at 5 per cent., 3.3 7 3 5 7.3 4 Amount for 3d year First principal, 3 0.0 [at 6 per cent. Compound interest for 4 years, $ 5 7.3 4 Ans. In the second operation the work is somewhat abridged, by finding the interest for each year at 6 per cent, by taking -^ of the principal for the interest at 5 per cent., and \ of that for interest at 1 per cent. Rule. — Find the interest of the given stim to the time the interest becomes due, and add it to the principal. Then, find the interest on this amount as a new principal, and add the interest to it, as before. Proceed in the same maimer for each successive period when the in- terest becomes due until the time of settlement. Subtract the principal from the last amount, and the remainder will be the compound interest. Note. — When partial payments have been made on notes at compound in- terest, it is customary to find the amount of the given principal, and from it to subtract the sum of the several amounts of the indorsements. Examples. 2. What is the amount of $ 500 for 3 years at compound in- terest ? Ans. $ 595.508. 3. What is the compound interest of $970 for 2 years 9 months and 24 days ? Ans. $ 173.295. 4. What is the compound interest of $ 300 for 4 years 6 months, at 7 per cent. ? Ans. $ 107.001. 5. What is the compound interest of $ 316 for 3 years 4 months and 18 days? Ans. $ 69.017. 377 a The computation of compound interest is rendered more expeditious by means of the following COMPOUND INTEREST. 281 TABLE, SHOWING THE AMOUNT OF ONE DOLLAR AT COMPOUND INTEREST FOR ANY NUMBER OF YEARS NOT EXCEEDING FIFTY. No. ~2 3 per cent. 3£ per 3ent. 4 per cent. 5 per cent. 6 per cent. ~i.06(f000 7 per cent. ~1.030~ 000 1.035 000 1.040 000 1.050 000 1.070 000 2 1.060 900 1.071 225 1.081 600 1.102 500 1.123 606 1.144 900 3 1.092 727 1.108 718 1.124 864 1.157 625 1.191 016 1.225 043 4 1.125 509 1.147 523 1.169 859 1.215 506 1.262 477 1.310 796 1 5 1.159 274 1.187 686 1.216 653 1.276 282 1.338 226 1.402 552 6 1.194 052 1.229 255 1.265 319 1.340 096 1.418 519 1.500 730 7 1.229 874 1.272 279 1.315 932 1.407 100 1.503 630 1.605 781 8 1.266 770 1.316 809 1.368 569 1.477 455 1.593 848 1.718 186 9 1.304 773 1.362 897 1.423 312 1.551 328 1.689 479 1.838 459 10 1.343 916 1.410 599 1.480 244 1.628 895 1.790 848 1.967 151 11 1.384 234 1.459 970 1.539 454 1.710 339 1.898 299 2.104 852 12 1.425 761 1.511 069 1.601 032 1.795 856 2.012 196 2.252 192 13 1.468 534 1.563 956 1.665 074 1.885 649 2.132 928 2.409 845 14 1.512 590 1.618 694 1.731 676 1.979 932 2.260 904 2.578 534 15 1.557 967 1.675 349 1.800 944 2.078 928 2.396 558 2.759 032 16 1.604 706 1.733 986 1.872 981 2.182 875 2.540 352 2.952 164 17 1.652 848 1.794 675 1.947 901 2.292 018 2.692 773 3.158 815 18 1.702 433 1.857 489 2.025 817 2.406 619 2.854 339 3.379 932 19 1.753 506 1.922 501 2.106 849 2.526 950 3.025 600 3.616 526 20 1.806 111 1.989 789 2.191 123 2.653 298 3.207 135 3.869 684 21 1.860 295 2.059 431 2.278 768 2.785 963 3.399 564 4.140 562 22 1.916 103 2.131 512 2.369 919 2.925 261 3.603 537 4.430 402 23 1.973 587 2.206 114 2.464 716 3.071 524 3.819 750 4.740 530 24 2.032 794 2.283 328 2.563 304 3.225 100 4.048 935 5.072 367 25 2.093 778 2.363 245 2.665 836 3.386 355 4.291 871 5.427 433 26 2.156 591 2.445 959 2.772 470 3.555 673 4.549 383 5.807 353 27 2.221 289 2.531 567 2.883 369 3.733 456 4.822 346 6.213 868 28 2.287 928 2.620 177 2.998 703 3.920 129 5.111 687 6.648 838 29 2.356 566 2.711 878 3.118 651 4.116 136 5.418 388 7.114 257 30 2.427 262 2.806 794 3.243 398 4.321 942 5.743 491 7.612 255 31 2.500 080 2.905 031 3.373 133 4.538 039 6.088 101 8.145 113 32 2.575 083 3.006 708 3.508 059 4.764 941 6.453 387 8.715 271 33 2.652 335 3.111 942 3.648 381 5.003 189 6.840 590 9.325 340 34 2.731 905 3.220 860 3.794 316 5.253 348 7.251 025 9.978 114 35 2.813 862 3.333 590 3.946 089 5.516 015 7.686 087 10.676 581 36 2.890 278 3.450 266 4.103 933 5.791 816 8.147 252 11.423 942 37 2.985 227 3.571 025 4.268 090 6.081 407 8.636 087 12.223 618 38 3.074 783 3.696 011 4.438 813 6.385 477 9.154 252 13.079 271 39 3.167 027 3.825 372 4.616 366 6.704 751 9.703 507 13.994 820 40 3.262 038 3.959 260 4.801 021 7.039 989 10.285 718 14.974 458 41 3.359 899 4.097 834 4.993 061 7.391 988 10.902 861 16.022 670 42 3.460 696 4.241 258 5.192 784 7.761 588 11.557 033 17.144 257 43 3.564 517 4.389 702 5.400 495 8.149 667 12.250 455 18.344 355 44 3.671 452 4.543 342 5.616 515 8.557 150 12.985 482 19.628 460 45 3.781 596 4.702 358 5.841 176 8.985 008 13.764 611 21.002 452 46 3.895 044 4.866 941 6.074 823 9.434 258 14.590 487 22.472 623 47 4.011 895 5.037 284 6.317 816 9.905 971 15.465 917 24.045 707 48 4.132 252 5.213 589 6.570 528 10.401 270 16.393 872 25.728 907 49 4.256 219 5.396 065 6.833 349 10.921 333 17.377 504 27.529 930 50 4.383 906 5.584 927 7.106 683 11.467 400 18.420 154 29.457 025 Note. —If each of the numbers in the table be diminished by 1, the re- mainder will denote the interest of $ 1, instead of its amount. 24* 282 COMPOUND INTEREST. Ex. 1. What is the compound interest of $ 360 for 5 years 6 months and 24 days ? Ans. $ 138.14. OPERATION. Amount of $ 1 for 5 years, $1.3 38226 Principal, 3 6 8 2*93560 40146780 Amount of $ 360 for 5 years, 4 8 1.7 6 1 3 6 Amount of $ 1 for 6mo. 24d., 1.0 3 4 1927045440 1445284080 481761360 Amount of $ 360 for 5y. 6mo. 24d., 49 8.1 41246240 Principal, 3 6 0. Comp. int. of S 360 for 5y. 6mo. 24d., $ 1 3 8.1 4 Ans. We find the amount of S 1 for 5 years in the table, and, multiplying it and the number denoting the given prineipal together, obtain the amount of the $ 360 for 5 years. On this amount as a new principal we find the amount for the remaining 6 months and 24 days, tyy mul- tiplying by the number denoting the amount of $ 1 for the same time. From the last amount subtracting the original prineipal, we have left the compound interest required. Hence, Multiply the am&unt of $ 1 for the given time and rate, as found in the table, by the number ai noting the given principal. The product tvill be the required amount, from which subtract the given principal, and the remainder icill be the compound interest. Note. — When the given time includes not only the regular periods at which interest becomes due, but also a partial period, as a succession of periods of a year each, followed by one containing months or days, or both, after finding the amount for the regular periods, multiply that amount by the amount of $ 1 for the remaining time or partial period, and the product will be the required amount for the given time. In like manner, when the number of successive pe- riods exceeds the limits of the table, make the computations for a convenient length of time by means of the table, and on the amount thus found make another computation by means of the table, and so on. In making computation for a succession of periods shorter or longer than one year each, use the numbers in the table the same as if the periods were those of one year each. Examples. 2. What is the compound interest of $ 1200 for 11 years at 7 per cent. ? Ans. $ 1325.822. COMPOUND INTEREST. 283 3. What is the compound interest of $ 300 for 10 years 7 months and 15 days ? Ans. $ 257.401. 4. What is the compound interest of $ 5 for 50 years at 7 percent.? Ans. $142,285. 5. What is the amount of $ 480 for 40 years, at compound interest? Ans. $4937.144. 6. What is the compound interest of $ 40 for 4 years, at 7 percent.? Ans. $12,431. 7. What is the compound interest of $ 100 for 100 years? Ans. $ 33830.20. 8. What is the difference between the simple and the com- pound interest of $ 1000 for 33 years and 4 months ? 9. To what sum will $ 50, deposited in a savings bank, amount to at compound interest for 21 years, at 3 per cent., payable semiannually? Ans. $173,034. (10.) $ 100. Boston, September 25, 1853. I For value received, I promise to pay J. D. Forster, or order, on demand, one hundred dollars, with interest, after six months. Allen T. Dawes. On this note are the following indorsements : — June 11, 1854, received fifty dollars; September 25, 1854, received fifty dollars. What was due, reckoning at compound interest, August 25, 1855 ? Ans. $ 2.247. (11.) $ 1000. St. Paul, January 1, 1850. For value received, I promise to pay Stephen Howe, or bearer, on demand, one thousand dollars, with interest at 7 per cent. Wilson Goodhue. Indorsements: — June 10, 1850, seventy dollars; September 25, 1851, eighty dollars; July 4, 1852, one hundred dollars; November 11, 1853, thirty dollars; June 5, 1854, fifty dollars. At 7 per cent, compound interest, what remains due April 1, 1855 ? Ans. $ 1022.34. 378. To find the principal, the compound interest, the time, and the rate being given. Ex. 1. What principal at 6 per cent, compound interest will produce $ 2370 in 10 years ? Ans. $ 3000. 284 COMPOUND INTEREST. operation. "VYe fi n d the compound in- $ 2370 -J- .790 = $ 3000 Ans. terest of S 1 for the given time, and at the given rate ; and pro- ceed as in like cases in simple interest (Art. 359). Rule. — Divide the given compound interest by the compound in- terest of $ 1 for the given time at the given rate. Examples. 2. What principal, at 7 per cent, compound interest, will produce $ 205.90 in 6 years and 6 months ? Ans. $ 372.16. 3. What sum of money, at compound interest, will produce $ 1026.54 in 3 years 2 months and 12 days ? Ans. $ 5000. 4. What sum of money must be invested at compound in- terest at a semiannual rate of 3£ per cent, to produce $ 857.25 in 154- years ? Ans. $ 450. 379t To find the rate per cent., the principal, the inter- est, or the amount, and the time being given. Ex. 1. At what rate per cent, must $ 500 be at compound interest to become $ 703.55 in 7 years ? Ans. 5 per cent. OPERATION. $703.55 -T- 500 a= $1.4071, which for 7 years, in the table, denotes a rate of 5 per cent. Since $ 500 becomes $ 703.55 in 7 years at the required rate, $ 1 in the same time at the same rate will amount to -g-J^ as much, or $ 1.4071. Corresponding to this amount of $ 1 for the given time, we find in the table (Art. 377) 5 per cent, the rate required. Rule. — Divide the amount by the principal, and the quotient will be the amount of $ 1 for the given time and the required rate ; and in the table, over this amount, may be found the rate per cent, required. Note. — If the given time contains a part over an exact number of periods, look in the table, against the number denoting the whole periods or years in the given time, for that amount of $ 1 which comes the nearest to the one found by dividing. Then see if the approximate amount, increased by its rate of interest for the fractional period, will equal the other amount; if so, the rate corresponding to the approximate amount will be the rate per cent, required ; if not, the rate of the approximate amount will be as much greater or smaller than the required rate, as the interest added to the approximate amount is greater or smaller than that required to produce the amount found by dividing. The rule can only be well applied when the rate per cent, sought is within the limits of the table. COMPOUND INTEREST. 285 Examples. 2. At what rate per cent, will $ 400 amount to $ 640.405, at compound interest, in 12 years ? Ans. 4 per cent. 3. At what rate per cent, must $ 2500 be loaned, to produce $ 2096.147 compound interest, in 9 years ? Ans. 7 per cent. 4. At what rate per cent, will any sum of money double itself at compound interest in 11 T 8 ^ years ? Ans. 6 per cent. 5. At what rate per cent, will $ 10,000 amount to $ 31479.70 in 19 years and 8f months ? Ans. 6 per cent. 380. To find the time, the principal, the compound interest, and the rate per cent, being given. Ex. 1. In what time will $500, at 7 per cent, compound interest, amount to $ 655.398 ? Ans. 4 years. OPERATION. $655,398 — 500 == $1.310796, which, at the given rate in the table, denotes 4 years' time. Since $500 amounts to $655,398 at 7 percent, in the required time, $ 1 at the same rate, in the same time, will amount to -g^ as much, or $1.310796. Corresponding to this amount of $ 1 at the given rate, we find in the table (Art. 377) 4 years, the time re- quired. Rule. — Divide the amount by the principal, and the quotient will he the amount of $1 at the given rate for the required time ; and in the table, against this amount, may be found the time required. Note. — If the required time cannot be found exactly in the table, the number against that amount of $ 1 which under the given rate is next less than the amount found by dividing, will denote the whole periods or years. Then, find the fractional period or part of a year, by dividing 1 whole period or year by the ratio of the difference between the amount corresponding to the whole periods or years and that found by dividing to the difference between the former of the amounts and that next larger in the table ; and the value of the fraction obtained as the result may be expressed in months or days, or both. Examples. 2. In what time will $ 400 amount to $ 640.405 at 4 per cent, compound interest? Ans. 12 years. 3. In what time will $ 6000 amount to $ 9021.78 at 7 per cent, compound interest ? 286 DISCOUNT AND PRESENT WORTH. 4. In what time will any sum double itself at 5 per cent, compound interest? Ans. 14y. 2mo. 13d. 5. In what time will any sum double itself at 6 per cent, compound interest? Ans. lly. lOmo. 20-)-d. 6. A gentleman has deposited $ 450, for the benefit of his son, in a savings bank, at compound interest at a semiannual rate of 3£ per cent. He is to receive the amount as soon as it becomes $ 1781.66^-. Allowing that the deposit was made when the son was 1 year old, what will be his age when he can come in possession of the money ? Ans. 21 years. DISCOUNT AND PRESENT WORTH. 381. Discount is an allowance or deduction made for paying money before it is due. 382. The present worth is the amount of ready money that will satisfy a debt before it is due. It is equivalent to a principal which, being put at interest, will amount to the debt at the time of its becoming payable. Thus, $ 100 is the present worth of $ 106 due one year hence at 6 per cent. ; for $ 100 at 6 per cent, will amount to $ 10G in that time ; and $ 6 is the discount. 383. In discount, the rate per cent., the time, and the sum on which the discount is made, are given, to find the present ivorth, which corresponds precisely to the rate per cent., the time and the amount being given, in either simple or corn- pound interest, to find the principal. 384. The interest or percentage of any sum cannot prop- erly be taken for the true discount ; for we have seen (Art. 382) that the interest for one year is the fractional part of the sum at interest, denoted by the rate for the numerator, and 100 for the denominator ; and the discount for one year is the frac- tional part of the sum on which discount is to be made, denoted by the rate for the numerator, and 100 plus the rate for the de- DISCOUNT AND PRESENT WORTH. 287 nominator. Thus, if the rate of interest is 6 per cent., the interest for one year is yg^ of the sum at interest; but if the rate per cent, of discount is 6, the discount for one year is T §^ of the sum on which discount is made. 385. Business men, however, often deduct, or "take off," from the face of a bill or note due at some future time, a greater percentage than the interest would be for the given time at the given rate. Therefore, the true present worth and discount are not obtained by that method, but only a nominal present worth and a nominal discount. The true discount is equal to the interest on the true present worth of the debt, while the nominal discount is equal to the interest on the face of the debt. 386. To find the true present worth of any sum, and the discount, for any time, at any rate per cent. Ex. 1. What is the present worth of $ 12.72, due one year hence, discounting at 6 per cent. ? What is the discount ? Ans. $ 12 present worth ; $ 0.72 discount. OPERATION. Amount of $ 1, 1.0 6 ) $ 1 2.7 2 ( $ 1 2, Present worth. 106 2 12 $ 1 2.7 2, Given sum. 2 12 1 2.0 0, Present worth. $ 0.7 2, Discount. Since $ 1 is the present worth of $ 1.06 due one year hence, at 6 per cent., it is evident that the present worth of $12.72 must be as many dollars as $ 1.06 is contained times in $ 12.72. We thus find the present worth to be $ 12, which, subtracted from the given sum, gives $ 0.72 as the discount. Rule. — Divide the given sum by the amount of $ 1 for the given time and rate, and the quotient will be the present worth. From the given sum subtract the present worth, and the remainder will be the discount. Examples. 2. What is the discount on $ 802.50, at 7 per cent., due one year hence ? Ans. $ 52.50. 3. What is the present worth of $117.60, due one year hence, at 12 per cent. ? 288 DISCOUNT AND PRESENT WORTH. 4. What is the present worth of $769.60, due 3 years and 5 months hence ? Ans. $ 638.672. 5. What is the present worth of $ 678.75, due 3 years 7 months hence, at 7 J per cent. ? Ans. $ 534.975. 6. What is the discount on $ 600, due 5 years hence, at 5 per cent. ? 7. A merchant has given two notes ; the first for $ 79.87, to be paid JaAuary 21, 1856; the second for $87.75, to be paid December 17, 1856. How much ready money will discharge both notes February 10, 1855 ? Ans. $ 154.545. 8. C. Gardner owes Samuel Hall as follows: $365.87, to be paid December 19, 1855; $161.15, to be paid July 16, 1856; $ 112.50, to be paid June 23, 1854 ; $ 96.81, to be paid April 19, 1858. What should Hall receive as an equivalent, January 1, 1854 ? Ans. $ 653.40. 9. What is the present worth of $ 67.25 due 3 years hence ? 10. What is the present worth of $80,095, due 3 years hence at compound interest ? Ans. $ 67.25. 11. What is the discount on $ 110.364 due 5 years hence, at 7 per cent, compound interest ? Ans. S 31.677. 387. To find a nominal present worth of any sum due at some future time, and the discount on the same at a given rate, reckon the interest on the face of the debt for the given time and rate, and the same will be the nominal discount ; and this discount subtracted from the face of the note will give the nominal present worth. Examples. 1. I have bought of Paine and Woodard a bill of goods amounting to $960, on six months, but for ready money they take off from the face of the bill, for the time, 5 per cent. What was the amount paid ? ' Ans. $ 912. 2. How much more is the nominal than the true discount on $ 5000 due one year hence, at 7 per cent. ? } 3. When money is worth 6 per cent, a year, how much may be gained by hiring money to pay $ 4440 due 6 months hence, allowing the present worth of this debt to be reckoned by de- ducting the nominal discount? Ans. $3,996. BANKING. 289 BANKING. 388. Banking is the general business transacted at banks. A bank is a joint-stock company, established for the purpose of receiving deposits, loaning money, dealing in exchange, or issuing bank-notes or bills, as a circulating medium, redeemable in specie at its place of business. ., The capital of a bank is the money paid in by its stockholders, as the basis of business. The affairs of a bank are usually managed by a board of di- rectors chosen by the stockholders, and the principal officers are a president, a cashier, and one or more tellers. Note. — The president and cashier sign the bills issued ; the cashier super- intends the bank accounts; and the tellers receive and pay out money. A bank check is an order drawn on the cashier for money. 389. Bank discount is the simple interest of a note, draft, or bill of exchange, deducted from it in advance, or before it becomes due. The interest is computed, not only for the specified time, but for three days additional, called days of grace. The legal rate of discount is usually the same as the legal rate of interest ; and the difference between bank discount and true discount is the same as the difference between interest and true discount (Art. 384). 390. A note is said to be discounted at a bank, when it is reeeived as security for the money that is paid for it, after de- ducting the interest for the time it was given. The money paid for a note is called its avails, proceeds, or present worth. 391. A note, though nominally due at the time specified in it, is not legally due till the days of grace have been counted. The time a note has to run is counted in days from after the day of its being discounted to the day of its becoming legally due. 392. To find the bank discount and the avails or proceeds of a note or bill for any time or rate per cent. N 25 290 BANKING. Ex. 1. What is the discount on a note for $1000 having 63 days to run, discounted at a bank at 6 per cent. ? How much are the proceeds ? Ans. Discount, $ 10.50 ; proceeds, $ 989.50. operation. wr e find the interest Sum discounted, $ 1 0.0 on the sum discounted , r • . n nr\ i -i a a a as in Art. 354, and this T ^ of sum = int. for 60d. 1 0.0 interest b ^ e bank 2*5 of int. for GOd. = int. for 3d. .5 discount (Art. 389), Bank discount, $10.5 ^hich subtracted from _ the sum discounted Proceeds, or present worth, $ 9 8 9.5 gives the proceeds or present worth. Hence, Find the interest on the note, or sum discounted, for the given rate and time, including three days of grace, and this interest will "be the discount. Subtract the discount from the face of the note, or sum discounted, and the remainder will be the proceeds, or present worth. Examples, 2. What is the bank discount on a note for $ 7800 on 90 days' time ? Ans. $ 120.90. 3. What is the bank discount on a note for $ 1200, payable in GO days, at 7 per cent. ? Ans. $ 14.70. 4. What is the bank discount on $ 8000, payable in 60 days ? What are the proceeds ? Ans. Discount, $84; proceeds, $7916. 5. How much money should be received on a note for $ 760, payable in 5 months, discounted at a bank ? ) &. A merchant sold a cargo of hemp for $7860, for which he received a note payable in 6 months. How much money will he receive at a bank for this note ? Ans. $ 7620.27. 7. If the following note was discounted April 3, 1857, how long had it to run, and what were the proceeds ? $ 160 T V I Hi > ■£ gfl * ■* a» £ W 48 a w e « 72 *W £ W 38.72 28.15 M 38.59 32.70 22.80 22.27 8.16 9.14 1 44.68 36.78 25 37.86 32.33 49 21.81 21.72 73 7.72 8.69 2 47.55 38.74 26 37.14 31.93 50 21.11 21.17 74 7.33 8.25 3 49.82 40.01 27 36.41 31.50 51 20.39 20.61 75 7.01 7.83 4 50.76 40.73 28 35.69 31.08 69 19.68 20.05 76 669 7.40 5 61.25 40.88 29 35.00 30.66 53 18.97 19.49 77 6.40 6.99 G 51.17 40.69 30 34.34 30.25 54 18.28 18.92 78 6.12 6.59 7 50.80 40.47 31 33.68 29.83 17.58 18.35 79 5.80 6.21 8 50.24 40.14 32 33.03 29.43 66 16.89 17.78 80 5.51 5.85 9 49.57 39.72 33 32.36 29.02 57 16.21 17.20 81 5.21 5.50 10 48.82 39.23 34 31.68 28.62 58 15.55 16.63 82 4.93 5.16 11 48.04 38.64 35 31.00 ' 28.22 59 14.92 16.04 83 4.65 4.87 12 47.27 38.02 8fl 30.32 27.78 60 14.34 15.45 84 4.39 4.66 13 46.51 37.41 37 29.64 27.34 61 13.82 14.86 85 4.12 4.57 14 45.75 36.79 38 28.96 26.91 62 13.31 14.26 86 3.90 4.21 15 45.00 36.17 39 28.28 26.47 63 12.81 ; 13.66 B7 3.71 3.90 16 44.27 35.76 40 27.61 26.04 M 12.30 13.05 88 3.59 3.67 17 43.57 35.37 41 26.97 25.61 65 11.79 : 12.43 89 3.47 3.56 18 42 87 34.98 42 26.34 25.19. 66 11.27 11.96 3.28 3.73 19 42.17 34.59 43 25.71 24.77' 67 10.75 11.48 M 3.26 3.32 20 41.46 34.22 ! 44 25.09 24.35 68 10.23 ; 11.01 3.37 3.12 21 40.75 33.84 48 24.46 23.92 69 9.70 : 10.50 1 93 3.48 ! 2.40 22 40.04 33.46 46 23.82 23.37 70 9.18 10.06 M 3.53 j 1.98 23 39.31 33.08 1 47 23.17 22.88 71 8.65 I 9.60 1 95 3.53 1.62 Against any age given in the table may be found the expec- tation of age corresponding to it 452. The transactions of life insurance companies extend- ing as they do over a term of years, the value of money and the average rates of interest, no less than the expectation of life, have an important influence in fixing their rates of insurance. 453. The premiums of life insurance are generally reck- oned at a certain sum on $ 100, payable annually in advance. The rates of annual premium for insuring a healthy life for one year, for seven years, or for the whole period of life, in the sum of $ 100, by the Massachusetts Hospital Life Insurance Company, of Boston, and by the Girard Life Insurance An- nuity and Trust Company, of Philadelphia, are given in the following INSURANCE. 339 Table. Mass. Hospital Life Insurance The Girard Life Insurance Age next Company. Company. Age next Birthday. Birthday. 1 year. 7 years. For Life. 1 year. 7 years. For Life. 15 783~ .85 1.44 .88 1.56 15 16 .84 .86 1.47 .84 .90 1.60 16 17 .85 .87 1.51 .86 .91 1.65 17 18 .86 .88 1.54 .89 .92 1.69 18 19 .87 .90 1.58 .90 .94 1.73 19 20 .88 .91 1.62 .91 .95 1.77 20 21 .89 .92 1.66 .92 .97 1.82 21 22 .90 .93 1.70 .94 .99 1.88 22 23 .91 .95 1.74 .97 1.03 1.93 23 24 .92 .96 1.79 .99 1.07 1.98 24 25 .93 .98 1.84 1.00 1.12 2.04 25 26 .95 .99 1.89 1.07 1.17 2.11 26 27 .96 1.01 1.94 1.12 1.23 2.17 27 28 .98 1.03 2.00 1.20 1.28 2.24 28 29 .99 1.05 2.06 1.28 1.35 2.31 29 30 ' 1.01 1.07 2.12 1.31 1.36 2.36 30 31 1.03 1.09 2.18 1.32 1.42 2.43 31 32 1.05 1.11 2.25 1.33 1.46 2.50 32 33 1.07 1.14 2.32 1.34 1.48 2.57 33 34 1.09 1.16 2.40 1.35 1.50 2.64 34 35 1.11 1.19 2.48 1.36 1.53 2.75 35 36 1.14 1.21 2.56 1.39 1.57 2.81 36 37 1.16 1.24 2.65 1.43 1.63 2.90 37 38 1.19 1.28 2.75 1.48 1.70 3.05 38 39 1.22 1.31 2.85 1.57 1.76 3.11 39 40 1.24 1.36 2.95 1.69 1.83 3.20 40 41 1.27 1.41 3.07 1.78 1.88 3.31 41 42 1.31 1.47 3.19 1.85 1.89 3.40 42 43 1.35 1.54 3.32 1.89 1.92 3.51 43 44 1.40 1.62 3.45 1.90 1.94 3.63 44 45 1.47 1.71 3.60 1.91 1.96 3.73 45 46 1.54 1.80 3.75 1.92 1.98 3.87 46 47 1.62 1.90 3.92 1.93 1.99 4.01 47* 48 1.71 2.02 4.09 1.94 2.02 4.17 . 48 49 1.81 2.14 4.27 1.95 2.04 4.49 49 50 1.91 2.28 4.46 1.96 2.09 4.60 50 51 2.03 2.42 4.67 1.97 2.20 4.75 51 52 2.15 2.59 4.89 2.02 2.37 4 90 52 53 2.29 2.76 5.12 2.10 2.59 5.24 53 54 2.44 2.95 5.36 2.18 2.89 5.49 54 55 2.60 3.15 5.62 2.32 3.21 5.78 55 56 2.78 3.38 5.89 2.47 3.56 6.05 56 57 2.96 3.62 6.19 2.70 4.20 6.27 57 58 3.17 3.87 6.50 3.14 4.31 • 6.50 58 59 3.39 4.17 6.83 3.67 4.63 6.75 59 60 3.64 4.50 7.18 J 4.35 4.91 7.00 60 According to the table, a healthy man who is 42 years old next birthday, by paying the Massachusetts Hospital Insurance Company $ 1.31, would secure to his family, heirs, or to whom- 340 INSURANCE. soever he desires, $ 100, should he die in one year, and in the same proportion for a larger sum. And if he would obtain a life insurance of the Girard Life Insurance Annuity and Trust Company, he must pay annually $ 3.40. The New York Life Insurance Company has a schedule of rates like that of the Girard Company, given in the table. 454. To compute the premium of life insurance for any given amount. Ex. 1. What premium will the Massachusetts Hospital Life Insurance Company require for the insurance of a life one year for $ 1728, the person being thirty years of age next birth- day? Ans. $17.45. operation. By the table we find the $1728 X -0 101 =$17.4 5. premium on $100 to be $1.01. Therefore ${'-%-$ = $ .0101 = the premium on Si for one year, and $ 1728 X .0101 = the premium on $ 1728 for the same time. Hence, Compute the premium on the sum to be insured at a rate propor- tionate to the given premium on $ 100. Examples. 2. What amount of premium must S. C. Kendall pay an- nually to the Massachusetts Hospital Life Insurance Company, to effect an insurance on his life for 7 years for $ 8000, his age being 33 years ? Ans. $ 91.20. 3. Robert Vaux, 60 years of age, wishes to engage in a very profitable speculation; and being obliged to borrow the neces- sary funds, he effects an insurance on his life for 7 years, for $ 78000, at the office of the Girard Life Insurance Company. Required the amount of the annual premium. Ans. $ 3829.80. 4. What will be the yearly premium for insuring a person's life, who is 15 years old, for $ 2000 for 7 years, at the New York Life Insurance Company ? 5. A gentleman 45 years of age, being bound on a long and dangerous voyage, and wishing to secure a competence for his family, obtains an insurance for life of the Girard Life Insur- ance Company, for $ 12000. By an act of Providence he dies in the third year. What is the net gain to his family ? Ans. $ 10657.20. CUSTOM-HOUSE BUSINESS. 341 6. Richard Sears, 50 years old, effects an insurance for life for $ 5000, for which he pays an annual premium of $ 4.60 on each $ 100 insured. If he should die at the age of 80 years, how much less will be the amount of insurance than the pay- ments, allowing the latter to be without interest? Ans. $ 1900. 7. A gentleman, 56 years old, gets his life insured for $ 4000, at the office of the Kentucky Mutual Life Insurance Company, by paying an annual premium of $5.20 on each $ 100 insured ; and dies at the age of 60 years. Reckoning interest on his payments at simple interest, what is gained by the insur- ance? Ans. $3043.20.. 8. Alexander Murray, 28 years of age, effects an insurance on his life for $ 10000, at the office of the Massachusetts Hos- pital Life Insurance Company. If the company loan the pre- mium at 6 per cent, compound interest, and he should die at the age of 40 years, who will gain by the insurance ? Ans. The insured gains $ 6626.01. CUSTOM-HOUSE BUSINESS. 455 • Duties or customs are sums of money required by government to be paid on imported goods. Ports of entry are ports into which merchandise may be im- ported, or from which it may be exported. At each port of entry is an establishment, called a custom- house, at which certain officers, appointed by government, at- tend to the collection of the duties. 456. Duties are either specific or ad valorem. A specific duty is a certain sum paid on a ton, hundred- weight, yard, gallon, &c, without regard to the cost of the article. An ad valorem duty is a certain percentage paid on the ac- tual cost of the goods in the country from which they are im- ported. 29* 342 CUSTOM-HOUSE BUSINESS. Note. — As evidence of the cost value of merchandise subject to duties, the importer, owner, or consignee is required to produce an invoice or manifest, if one has been received, made out in the currency of the place or country •whence imported, and containing a true statement of the actual cost of such goods in such foreign currency. When the currency of a place has a depre- ciated falue compared with that of the United States, it is necessary that a consular certificate showing the rate of depreciation should be attached to the invoice. When, however, an invoice has not been received, the fact must be testified to under oath, and then the imported articles will be entered at an appraised value. 457 • In this country, under the present tariff, duties are levied on the ad valorem principle. 458. Allowances are deductions required to be made before estimating the duties, on account of the weight of the cask, box, bag, &c. in which an article is imported, or on ac- count of breakage, leakage, waste, or other damage. Tare is the allowance made for the weight of the cask, box, &c. containing the commodity. Draft is the allowance made for waste in the weighing of goods. Leakage is the allowance made for waste on liquids imported in casks. Gross weight is the weight of the commodity together with the cask, box, bag, &c. containing it. Net weight is what remains after all allowances have been made. 459. No allowances for tare, draft, breakage, &c. are appli- cable to imports subject to ad valorem duties, except actual tare, or weight of a cask or box, and actual drainage, leak- age, or damage. The collector may cause these to be ascer- tained, when he has any doubt as to what they are. Note. — When the tariff laws of the United States required the collection of specific duties, the allowance for draft was on 1121b., lib.; above 1121b. and not exceeding 2241b., 21b. ; above 2241b. and not exceeding 3361b., 31b. ; above 3361b. and not exceeding 11201b., 41b.; above 11201b. and not exceeding 20161b., 71b. ; above 20161b., 91b. The allowance for tare was deducted after the draft had been deducted. The allowance for breakage was 10 per cent, on all beer and porter in bottles ; and 5 per cent, on all other liquors imported in bottles ; and a dozen bottles of common size were estimated to contain 2| gallons. The allowance for leakage was 2 per cent, on liquors imported in casks. In making the allowances for tare or leakage, a fraction, when equal to, or greater than, one half, was reckoned 1 ; when less, it was omitted. CUSTOM-HOUSE BUSINESS. 343 Specific duties were calculated by deducting all allowances to be made from the given quantity of merchandise, and multiplying the remainder by the duty on a unit of the given quantity. 460 • To calculate ad valorem duties. Ex. 1. What is the duty on 25651b. of sugar, invoiced at $ 256.50, at 24 per cent, ad valorem ? Ans. $ 61.56. OPERATION. $ 2 5 6.5 X .2 4 = $ 61.56, duty. Rule. — Find the percentage on the invoiced value of the goods, at the given rate of tariff, and the result will be the ad valorem duty. !^ote 1. — Other questions in duties beside those requiring the finding of the amount of duty, are likewise solved by some one of the rules in per- centage. Note 2. — In custom-house calculations 22401b. are considered a ton, and 1121b. a hundred-weight. Examples. 2. What is the duty at 8 per cent, on an importation of books invoiced at $ 4350 ? Ans. $ 348.00. 3. Required the duty at 19 per cent, on 7890 pounds of cordage, invoiced at 15 cents per pound. Ans. $ 224.865. 4. What is the duty at 24 per cent, on an invoice of woollen goods, which cost in London 98 6£., the pound sterling being valued at $ 4.84? Ans. $ 1145.3376. 5. $ 112.50 duty is paid on an importation of window-glass whose invoice value was $ 750. What was the rate per cent, of duty ? 6. Robinson & Brother of New York have imported wines from Havre, invoiced as follows : 60 baskets Champagne at 70 francs per basket; 36 baskets port at 35 francs per basket; 50 casks of sherry, each 31 gallons, at 4 franca per gallon. The allowance for breakage on the wine in baskets is 5 per cent. ; and the actual waste of that in casks is 1 gallon to a cask. Re- quired the duties at 30 per cent., allowing the value of a franc to be 18-ft cents. Ans. $ 624.2346. 7. Paid $53.76 duties, at the rate of 8 per cent, on 60 casks of raisins, after the deduction of 121b. to a cask for tare. Al- lowing the gross weight of each cask of raisins to have been 1121b., what was their invoice value per pound? Ans. 11£ cents. 344 CUSTOM-HOUSE BUSINESS. 8. A portion of the cargo of the ship Cuba from Havana was invoiced as follows : 40 hogsheads of molasses, 63 gallons each, at 3 reals plate per gallon ; 24 boxes of brown sugar, 4001b. each, at 1 real vellon per pound; 260 boxes of oranges, at $2 per box; and 410 boxes of cigars, at $7 per box. The tare on the sugar was 10 per cent., and the leakage of the molasses 2 per cent. Allowing the value of a real plate to be 10 cents, and a real vellon 5 cents, and the rate of duties on the molasses and the sugar to have been 24 per cent., on the oranges 8 per cent., and on the cigars 30 per cent., what was the whole amount of duties ? 9. 270 tons of railroad iron, invoiced at $ 50 per, ton, cost, when the duties were paid, exclusive of other charges, $ 1 6740. What was the rate per cent, of duty ? Ans. 24 per cent. 10. A merchant of Baltimore makes an importation of goods invoiced at $ 20560. On goods invoiced at $ 3000 the du- ties were at the rate of 4 per cent. ; on goods invoiced at $ 4200 the duties were at the rate of 8 per cent. ; on goods in- voiced at $ 2100 the duties were at the rate of 15 per cent. ; goods invoiced at $6000 were free of duty; and on the re- mainder the duties were at the rate of 30 per cent. What was the whole amount of the duties ? 11. Willard, Fairbanks, & Co., of Boston, import from Liver- pool 10 pieces of Brussels carpeting, 40 yards each, purchased at 5s. per yard, duty 24 per cent. ; 200 yards of hair-cloth, at 4s. per yard, duty 19 per cent.; 100 woollen blankets, at 2s. 6d., duty 15 per cent. ; and shoe-lasting to the cost of 60£., duty 4 per cent. Required the whole amount of duty, allowing the value of the pound sterling to be $ 4.84. Ans. $ 173.635. 12. A merchant imported from Bremen 32 pieces of linen of 32 yards each, on which he paid for the duties, at 24 per cent., $ 122.88, and other charges to the amount of $ 40.96. What was the invoice value per yard, and the cost per yard after du- ties and charges were paid ? Ans. Invoice value pes yard, $ 0.50 ; cost per yard, $ 0.66, COINS AND CURRENCIES. 345 COINS AND CURRENCIES. 461. Coins are pieces of metal legally stamped, and issued for circulation as money. The currency of a state or country is its money or circulating medium of trade. 462. The former currency of this country had the denomi- nations of sterling money, viz. pounds, shillings, and pence. On the adoption by Congress, in 1786, of the present decimal cur- rency, with the dollar as its unit, there were in circulation colonial notes, or bills of credit, which had depreciated in value. This depreciation, being greater in some States than in others, gave rise to the difference in the States as to the number of shillings equivalent to a dollar, as shown in the following Table, ' New Eng. States,"] $ 1 in^ Virginia, Kentucky, Tennessee, fNew York, . J Ohio, in 1 Michigan, L North Carolina, " Pennsylvania, New Jersey, Delaware, ^ Maryland, . J Georgia, \ South Carolina, i ===== 6s. = i^£., called New Eng. currency ; .j of which l£.= $3J-; Is. = 16 jets. == 8s. ==■££., called New York currency; of which l£. == % 1\ ; Is. = 12£cts. 1 in< i * = 7s. 6d. =| £., called Pennsylvania cur- rency ; of which l£.== $ 2| ; Is. == 13 Jets. | =4s. 8d.=^j-£., called Georgia currency ; J of which l£. = $ 4f : Is. == 21f cts. Note 1. — The State currencies are now merely nominal, accounts being no- where kept in them. Articles, however, are sometimes priced in them, but much less often now than formerly. Note 2. — $ 1 equals about ^s- = hi^. of English or sterling money; and consequently 1£. = $4.84; Is. = $0.24£. Also $ 1 = 6s. == £<£. of Nova Scotia, New Brunswick, Newfoundland, and Canada, called Canada currency; of which 1«£. = $ 4; Is. = 20 cents. 463. The legal tender in payment of debts in the United States is gold and silver. Note. — The gold coins of the United States of coinage prior to 1834 are a legal tender for the payment of all debts, at the rate of 94g cents per pemiy- 346 COINS AND CURRENCIES. weight, or about $ 10.66 for each eagle; and all the gold coins of a subsequent coinage are a legal tender of payment in any sums whatever, according to their nominal values. Of the silver coins coined prior to April 1, 1853, the three-cent pieces are a legal tender of payment in any sums of thirty cents and under, and the dollars, half-dollars, quarter-dollars, dimes, and half-dimes are a legar tender of pay- ment in any sums whatever, at their nominal value, and all the silver coin of a subsequent coinage are a legal tender in sums not exceeding five dollars. 464. Besides the coins of the country, foreign coins, whose value has been fixed by law, and bank-notes, redeemable in specie, pass as money. However, by the act of Congress passed February, 1857, all former acts authorizing the currency of for- eign gold or silver coin, and declaring the same legal tender in payment of debts, were repealed. 465. The intrinsic value of foreign coins is the value de- pending upon the weight and purity of the metal of which they are made ; their legal value is that fixed bylaw ; their' commer- cial value is the price they will bring in the market ; and their exchange value is the nominal price assigned to them in reck- oning exchange between one country and another. 466* The United States mint, at present, purchases stand- ard silver at $ 1.22£ per ounce, and fine silver at $ 1.36 per ounce. Note. — At these rates of purchase, five-franc pieces yield about 99 cents each; Mexican and South American dollars, 106| cents each; old Spanish dol- lars, 105 cents each; half-dollars of the United States coined before 1837, 52£ cents each; half-dollars of the United States coined since 1837, and previous to the change of standard in 1853, 52£ cents each; German florins, 414 cents each; Prussian and Hanoverian thalers, 72 cents each; best manu- factured American plate, from 120 to 122 cents per ounce; and genuine British plate, 125 cents per ounce. 467. By the act of 1843 the gold coins of Great Britain, if not less than tttt in fineness, were rated at 94 T 6 - cents per pen- nyweight, and the gold coin of France of not less than T 8 e 9 <$y in fineness, at 92 T 9 J cents per pennyweight. By a previous act, the gold coins of Portugal and Brazil of not less than 22 carats fine were rated at 94 T 8 IT cents per pennyweight, and the gold coins of Spain, Mexico, and Colombia, of a fineness of 20 carats 3y 7 ^ grains, at the rate of 89 T 9 7 cents per pennyweight. Note. — The relative value of gold and silver in the United States, at pres- COINS AND CURRENCIES. 347 ent, is nearly as 14£ to 1 ; in England, as 14f Q 9 5 to 1 ; in France and in Russia, as 15 to 1 ; in Spain, as 16 to 1 ; in China, as 14£ to 1 ; and in Portugal, as 13^ tol. 468. The value of certain foreign coins and currencies, as fixed by law in the collection of duties at the United States custom-houses, is shown in the following Table. Pound Ster. of G. Britain, $ Pound Ster. of Br. Prov.,^ Nova Scotia, N. Bruns., > and Newfoundland, ) Dollar of Mexico, Peru, Chili, and Cen. Amer., Specie Dollar of Sweden and Norway, Specie Dollar of Denmark, Rix Dollar of Bremen, Thaler of Bremen of 72 grotes, Rix Dollar, or Thaler of Prussia and Northern States of Germany, Ruble, silver, of Russia, Florin of the Austrian Em-" pire and city of Augs- burg, Florin or Guilder of Neth- erlands, 4.84 4.00 1.00 1.06 1.05 • 78f .71 .69 .75 .481 .40 Florin of South of Germ., $ 0.40 Ounce of Sicily, Pagoda of India, Star Pagoda of Madras, Tael of China, Millrea of Portugal, Millrea of Azores, Ducat of Naples, Sicca Rupee of Bengal, or 7 of Bombay, £ Rupee of British India, Mark Banco of Hamburg, Franc of France and Belg., Livre Tournois of France, Leghorn Livre, Lira of Lombardo-Vene tian Kingdom, Lira of Tuscany, Lira of Sardinia, Real Plate of Spain, Real Vellon of Spain, 2.40 1.94 1.94 1.48 1.12 .831 .80 .50 .35 .181 .16 ;■• 16 .16 .10 ,05 REDUCTION OF CURRENCIES. 469. Reduction of Currencies is the process of finding the value of the denominations of one currency in the denomina- tions of another. 470* To find the value of the denominations of one currency in the denominations of another, when the values of a unit of each are known. Ex. 1. What is the value of 18£. 4s. 6d. of the New Eng- land currency in United States money ? Ans. $ 60.75. operation. W e reduce the shillings and pence 18£. 4s. 6d. = 18.225£. to the decimal of a pound, and an- 18.225 -7- -\ = $ 60.75. nex '* to tne pounds. We then di- vide the sum by ^ because 6s., or a dollar in this currency, is ^ of a pound. 348 COINS AND CURRENCIES. Ex. 2. What is the value of $ 60.75 in New England cur- rency ? Ans. 18£. 4s. 6d. operation. Since 6s - > or a dollaT of th[ * ? u r 60.75 X A - 18.225£. **■*■ 1S A of a P ou " d ' ? e *? ultl P ^ 1Q 00 ^ IQr >| *A the g 1VCn SUm ^ the fraCt1011 A» 18.225£. =± 18£. 4s. 6d. and find the value of t i ie decimal in shillings and pence. Rule. — Multiply or divide, as the case may require, by the value of the unit of the given currency expressed in United States money. Note. — When one currency is to be changed to another, and neither of them is United States money, the value of one of them may be found in United States money, and the value of that result be found in the other currency. Examples. 3. Change 46£. 16s. 6d. of the currency of New. York to United States money. Ans. $ 117.06£. 4. Change $ 1032 to the currency of Pennsylvania. Ans. 387£. 5. Find the value of $ 515.70 in Canada currency. Ans. 128£. 18s. 6d. 6. Change 160.50 francs to United States money. 7. Change $ 728.41 to English money. Ans. 150£. 9s. ll T 6 2 J T d. 8. Find the value of 12£. 12s. of the currency of Georgia in United States money. Ans. $ 54. 9. Find the value of 128£. 18s. 6d. of Canada in United States money. Ans. $515.70. 10. Find the value of 740.45 rubles, silver, of Russia, in United States money. Ans. $ 555.33J . 11. Find the value of 46£. 16s. 6d. of New York currency in the currency of New England. Ans. 35£. 2s. 4^-d. 12. Change 151 millreas of Portugal to real plate of Spain. 13. Find the value of 1000 specie dollars of Norway in francs. 14. A merchant of Quebec bought in London 30 pieces of broadcloth of 30 yards each, at 1 5 shillings per yard ; what is the amount of his bill in Canada currency? Ans. 81 6£. 15s. 15. A merchant of Prussia bought in Naples silks to the amount of 410 ducats ; what was the amount of his purchase in thalers ? Ans. 475 fj- thalers. EXCHANGE. 349 EXCHANGE. 471 . Exchange, in commerce, is the paying or receiving money in one place for an equivalent sum in another, by means of drafts, or bills of exchange. 472. A Bill of Exchange is a written order, to some person at a distance, to pay a certain sum, at an appointed time, to another person, or to his order. The maker or drawer of a bill is the person who signs it. The buyer, taker, or remitter of a bill is the person in whose favor it is drawn. The drawee of a bill is the person on whom it is drawn, who is also called the acceptor, after he has accepted it. The indorser of a bill is the person who indorses it. The holder or possessor of a bill is the person in whose legal possession it may be at any time. 473 o Most mercantile payments are made in bills of ex- change, since it is generally more convenient to discharge debts by means of them than by cash remittances. For example, suppose A, of Boston, is creditor to B, of Baltimore, $100 ; and C, of Boston, is debtor to D, of Baltimore, $ 100 ; both these debts may be discharged by means of one bill. Thus, A draws for this sum on B, and sells his bill to C, who remits it to D, and the latter receives the amount, when due, from B. Here, by a transfer of claims, the Boston debtor pays the Boston creditor, and the Baltimore debtor the Baltimore creditor ; and no money is sent from one place to the other. The same would take place if D, of Baltimore, drew on C, of Boston, and sold his bill to B, of Baltimore, who should send it to A, of Boston ; the effect, in either case, being merely a transfer of debt and credit. Bills of exchange pass from hand to hand, until due, like any other circulating medium. 474. The terms of a bill vary according to the agreement between parties, or the custom of countries. Some bills are drawn at sight ; others, at a certain number of days, or months, after sight or after date ; and some, at usance, which is the cus- tomary or usual term between different places. N 30 350 EXCHANGE. Days of Grace. Days of grace are a certain number of days granted the ac- ceptor for paying the bill, after the term of a bill has expired. The usual time allowed in this country is three days. Note 1. — In some States three days of grace are allowed on all bills of ex- change payable at sight, or at a future day certain ; but in other States sight drafts are excepted, as in New York, Pennsylvania, New Jersey, Maryland, Virginia, Missouri, Illinois, Michigan, Connecticut, Rhode Island, Delaware, &c. No days of grace are allowed on bills written payable on demand, or which have no time of payment expressed in them. Note 2. — In reckoning when a bill, payable after date, becomes due, the day on which it is dated is not included; and if it be a bill payable after sight, the day of presentment is not included. When the term is expressed in months, calendar months are understood ; and when a month is longer than the preced- ing, it is a rule not to go in the computation into a third month. Thus, if a bill be dated the 28th, 29th, 30th, or 31st of January, and payable one month after date, the term expires on the last day of February, to which the days of grace must, of course, be added ; and therefore the bill becomes due on the 3d of March. Indorsing Bills. 475. An indorsement of a bill is the act by which the holder of it transfers his right to another. It is usually made on the back of the bill, and must be in writing. - 476. Bills payable to order are transferred only by indorse- ment and delivery, but bills payable to bearer are transferred by either mode. On transfer by delivery, the person making it ceases to be a party to the bill ; but on a transfer by indorse- ment, he is to all intents and purposes chargeable as a new drawer. 477. An indorsement may take place any time after the bill is issued, even after the day of payment has elapsed. Note. — An indorsement may be restrictive, giving authority to the in- dorsee to receive the money for the indorser, but not to transfer the bill to another. The indorsement for a part of the money only is not valid, except with regard to him who makes it. The drawer and acceptor are not bound by it. After the payment of a part, however, a bill may be indorsed over for the residue. The indorsement is said to be in blank, when the indorser simply writes his name upon the back of the bill to make it transferable by delivery ; and is said to be special, when the indorser directs the money to be paid to some particular person, or to his order. If the indorser would avoid all liability, he must qualify his indorsement by the words, " without recourse," or by others of the same import. EXCHANGE. 351 Accepting Bills. 478. An acceptance is an engagement to pay a bill accord- ing to the tenor of the acceptance, which may be either absolute or qualified. 479. An absolute acceptance is an engagement to pay a bill according to its request, which is commonly done by the drawee writing his name at the bottom, or across the body of the bill, with the word accepted. 480. A qualified acceptance is when a bill is accepted con- ditionally. But the holder is not obliged to receive a conditional or partial acceptance. He may act as if an acceptance had been entirely refused. Note. — When a bill is drawn for the account of a third person, and is ac- cepted as such, and he fails without making provision for its payment, the acceptor must discharge the bill, and can have no recourse against the drawer. When a holder, at his own risk, takes a conditional or partial acceptance, he must give immediate notice to all other parties to the bill, or he can have no resort to them in default of payment. All bills payable at sight, or at a day certain, or on demand, should be pre- sented within a reasonable time, or the holder may, from his default, be the loser. Protesting Bills. 481 • The holder of a bill, when acceptance or payment has been refused, should give regular and immediate notice to all the parties to whom he intends to resort for payment ; since, if a loss should be incurred, on account of unnecessary delay, by the failure of any of the parties, he would be obliged to bear the loss. Such a notice of non-acceptance or non-payment, when made by a public officer called a notary or notary public, or by any other legal mode, is called a protest. 482. When the parties to a bill, which the drawee has failed to accept or pay, live in different countries, a protest is indis- pensably necessary, as this instrument is admitted in foreign countries as legal proof of the fact of the refusal, and that the holder intends to recover any damages which he may sustain in consequence. 483. In case of non-payment, when the parties to a bill live in the same country, it is not necessary, although frequently 352 . EXCHANGE; practised, that there should be a regular protest by a public notary. But a notice simply of non-payment is sufficient to en- title the holder to claim interest. 484. The damages incurred by non-acceptance and non-pay- ment of a foreign bill, besides interest commencing from the day of demand, consist usually of the exchange or re-exchange, commission, and postage, together with the expenses of protest and interest. The damages of protested bills in general, how- ever, are regulated in a great measure by the local laws and usages of the different States and countries. Liabilities of the Parties. 485. The drawer, acceptor, and each and every indorser of a bill, are liable to the payment of it ; and though the holder can have but one satisfaction, yet, till such satisfaction is ac- tually had, he may sue any of them, or all of them, either at the same time or in succession, and obtain judgment against them all, till satisfaction be made. 486. Nothing will discharge an indorser from his engage- ment, but the absolute payment of the money ; not even a judg- ment recovered against the drawer, or any previous indorser, or any execution against any of them, unless the money be paid in consequence. Note. — When acceptance is refused, and the bill is returned by protest, an action may be commenced immediately against the drawer, though the regu- lar time of payment be not arrived. His debt, in such a case, is considered as contracted the moment the bill is drawn. In order, however, to make the indorsers liable, it is proper that the holder should present the bill for payment on the day it becomes due. Par of Exchange. 487* The intrinsic par of exchange is the value the coins of one country have, when compared with those of another, with respect both to weight and fineness. 488. The commercial par of exchange is the value the coins of one country sell for in the markets of another ; and is therefore not a fixed, but a variable value. EXCHANGE. 353 Course of Exchange. 489. The course of exchange is the variable price of the money of one country, which is paid for a fixed sum of money of another country. 490. The fluctuations of exchange are occasioned by vari- ous circumstances, both political and commercial, but in general bills rise or fall in their prices, like any other salable articles, according to the relation existing for the time being between the demand and the supply. 491. The limits within which the fluctuations of exchange range, correspond with the cost of making remittances in cash. Therefore, in the time of peace, exchange seldom remains long unfavorable to any country. When unfavorable, it has a ten- dency to correct itself, by giving an unusual stimulus to expor- tation, and by throwing obstacles in the way of importation ; and when favorable, it produces the same effect, by restricting exportation and facilitating importation. INLAND BILLS. 492. An Inland Bill of exchange, or draft, is one of which the drawer and drawee are residents of different parts of the same country. Inland bills are seldom bought or sold at the precise sum specified upon their face, but, according to the course of ex- change, are subject to a discount, or command a premium. 493. To compute inland exchange. Ex. 1. What is the value of the following bill of exchange or draft, at ^ per cent, premium ? Ans. $ 2563.20. $ 2560. New York, April 14, 1857. At sight, pay to Dura Wadsworth, or order, two thousand five hundred and sixty dollars, value received, and charge the same to the account of Cameron, Bashford, & Co. To Messrs, Lawrence & Aspinwall, Merchants, Boston. OPERATION. $25 60 X 1.0 4= $25 63.2 0. 30* 354 EXCHANGE. Rule. — Multiply the face of the hill or draft by 1 increased by the rate per cent, of premium, or by 1 decreased by the rate per cent, of dis- count, expressed decimally. The product will be the value of the given bill or draft. Note. — When there is interest to be computed, it must be reckoned on the face of the bill or draft. When other than the value or cost of the bill or draft is to be found, proceed as in percentage. Examples. 2. What is the value of the following bill, or draft, at J of 1 per cent, premium? Ans. $ 1955.37. $ 1950^. Chicago, August 3, 1857. Thirty days after date, please pay to the order of Robert S. Davis § Co., one thousand nine hundred fifty -j^f dollars, value received, and charge the same to our account Keene & Lee. To George Reed, Broker, Boston. 3. What is the cost of a draft on Philadelphia, at £ per cent premium, the face being $ 2000 ? Ans. $ 2010. 4. What must be the face of a draft, at 2 per cent, discount, to cost $ 1744.40 ? 5. What is the cost of a 60 days' bill on Pittsburg to the amount of $ 600, at 1 per cent, discount, and interest off at 6 per cent. ? Ans. $ 587.70. 6. What is the cost of a 30 days' bill on Boston, at f per cent, premium, and interest off at 6 per cent., the face of the bill being $ 9256.40 ? Ans. $ 9240.20. 7. What must be the face of a 30 days' bill which will yield $ 9240.20 when sold at f per cent, premium, and interest off at 6 per cent. ? 8. A 15 days' draft yielded $ 1190.184 when sold at 1£ per cent, discount, and interest off at 6 per cent. What was the face of the draft ? Ans. $ 1212. FOREIGN BILLS. 494. A Foreign Bill of exchange is one of which the drawer and drawee are residents of different countries. 495. Foreign bills are usually drawn in sets; that is, at EXCHANGE. 355 the same time there are drawn two or more bills of the same tenor and date, each containing a condition that it shall con- tinue payable only while the others remain unpaid. Each bill of a set is remitted in a different manner, and when one of the set has been accepted and paid, the others become worthless. In reckoning the value of foreign bills of exchange, an ac- quaintance with the moneys of account of foreign countries is required. 498. The moneys of account of the principal places of com- mercial importance, with the par value of the unit, expressed in United States money, are shown in the following Table. Cities and Countries. London, Liverpool, &c, Paris, Havre, &c, Amsterdam, Hague, &c., Bremen, Hamburg, Lubec, &c, Berlin, Dantzic, < Belgium, St. Petersburg, Stockholm, Copenhagen, Vienna, Trieste, &c, Naples, Venice, Milan, &c, Florence, Leghorn, &c, Genoa, Turin, &c, Sicily, Portugal, Spain, Constantinople, British India, Canton, Denominations of Money. 12 pence = 1 shilling; 20s. = 1 pound = 100 centimes = 1 franc m 100 cents = 1 guilder or florin = 5 swares = 1 grote ; 72gr. = 1 rix dollar = 12 pfennings = 1 schilling; 16s. = 1 mark banco = 12 pfennings = 1 groschen; 30gr.=l thaler= 100 centimes = 1 franc = 100 kopecks = 1 ruble = 12 rundstycks = 16 skillings ; 48s. = 1 rix dollar specie = 16 skillings = 1 mark ; 6m. = 1 rix dollar = 60 kreutzers = 1 florin = 10 grani = 1 carlino ; lOcar. = 1 ducat = 100 centesimi = 1 lira = 100 centesimi = 1 lira = 100 centesimi = 1 lira = 20 grani = 1 taro ; 30 tari = 1 ounce = 1000 reas = 1 millrea = J 34 maravedis = 1 real vellon = ( 68 maravedis = 1 real plate = 100 aspers = 1 piaster = 12 pice = 1 anna; 16 annas = 1 rupee = 100 candarines = 1 mace ; 10m. = 1 tael = Value. 1.06 1.05 0.48£ 0.80 0.16 0.16 0.186 2.40 1.12 0.05 0.10 0.05 0.44£ 1.48 Note 1. — Grani is the plural of grano, carlini of carlino, centesimi of cente- simo, lire of lira, tari of taro. Note 2. — The moneys of account in Brazil are of the same denominations as in Portugal ; in Mexico, Central America, New Granada, Chili, Venezuela, Bolivia, Peru, and Buenos Ayres, accounts are kept in pesos, or dollars, of 8 reals each ; in Cuba accounts are kept in dollars equal 8 reals plate, or 20 reals vellon, equal $ 1 ; in British American Possessions accounts are kept in the de- nominations of sterling money, but of a depreciated value compared with the currency of the same denominations in England, except in Canada, where, by an act of the Colonial Parliament, passed in 1857, the decimal currency of the United States has been adopted. 356 EXCHANGE. 497 1 The exchange value, in United States money, of the pound sterling of Great Britain is that of its former legal value, or $ 4§- = $ 4.44f , which is considerably below either its in- trinsic or commercial value. The commercial value is gen- erally about 9 per cent, more than this exchange, or nominal par value. Thus, nominal par value being = $ 4.44| Add premium at 9 per cent. = .40 The commercial par value will be = $ 4.84|. Therefore, when the nominal exchange between the United States and Great Britain exceeds 9 per cent, premium, it is above true par ; when less, it is below true par. 498. The quotations of the rates of exchange of the United States on England have always reference to the old par value of the pound sterling in United States money. The course of exchange of the United States on France is so many francs and centimes payable in France for a dollar paid here; on Holland it is so many cents a guilder (of Netherlands) ; on Hamburg, so many cents a mark banco ; on Bremen, so many cents a rix dollar ; and so on. Note 1. — The course of exchange of London on France is so many francs and centimes payable in France for 1£. ; on Amsterdam, it is so many florins for 1£. ; on Hamburg, so many schillings for 1£. ; on Vienna and Trieste, so many florins and creutzers ; on Spain, so many pence paid in England for 1 dollar of plate ( = 8| reals plate) payable in Spain ; on Lisbon, so many pence for 1 millrea ; on Naples, so many pence for 1 ducat ; and so on. Note 2. — The value of 1£. sterling from 7 to lo| premium on the old par value of $ 4.44|, is shown in the following Table. 7 per cent. 7 l u premium, a 7| " a 7| 8 " u u 84- " C( 4 " a S3 U u 9 per cent, premium, $4,844 $4,756 4.767 4.778 4.789 4.80 4.811 4.822 4.833 Note 3. — In this country the quotations of foreign exchanges are usually for bills payable 60 days after sight, and of inland or domestic exchanges, for bills payable at sight. 91 u u 4.856 9 I a u 4.867 9f u u 4.878 10 it a 4.889 ioi u u 4.90 10 i n a 4.911 101 a u 4.922 EXCHANGE. 357 To compute foreign exchange. Ex. 1. What should be paid for the following bill at 9£ per cent, premium ? Ans. $ 486.666. Exchange for £ 100. Philadelphia, May 21, 1857. Sixty days after sight of this, my first Bill of Exchange, {second and third of the same date and tenor unpaid,) pay to Lang don Shannon, or order, one hundred pounds sterling, value received, with or without further advice. William P. Brown. To Messrs. Peabody & Co., Bankers, London. OPERATION. 100 X -V- = $444.44; $444,444 X 1.095 = $486,666. 2. What is the cost at Amsterdam of a bill on New York for $ 340.67, exchange being at $ 0.38 to the guilder ? Ans. 896.50 guilders. OPERATION. 340.67 ~ 38 = 896.5 == 896 guilders 50 cents. 3. What must be paid in Boston for a bill on Paris for 3676 francs, exchange being 5 francs 20 centimes to the dollar ? Ans. $706.92 T \. 4. What is the value of a bill on Hamburg for 3000 marks 10 schillings, exchange being at $ 0.35 to a mark banco ? Ans. $ 1050.213-. 5. How large a bill can be purchased on Liverpool for $81727.75, exchange being at 9 J per cent, premium? 6. Paid $14400.12 for a bill on Havre for 79000 francs; how much was exchange below par ? Ans. 2 per cent. 7. What is the cost of a draft on St. Petersburg for 5763 rubles 75 kopecks, exchange being at 74 cents a ruble ? Ans. $ 4265.175. 8. What must the face of a bill on Lisbon be which costs $ 550.66, exchange being at $ 1.10 a millrea? Ans. 500 millreas 600 reas. 9. What is the cost at Berlin of a draft on Philadelphia for $ 10000, exchange being at $ 0.68 a thaler ? Ans. 14705 thalers 26 groschen 5-J-|-f~ pfennings. 10. How much must be paid in Chicago for a bill on Stock- holm amounting to 400 specie rix dollars 12 skillings ? Ans. $ 424.265. 358 ARBITRATION OP EXCHANGES. 11. What is the value in St. Louis of a draft on Dantzic for 300 thalers 20 groschen 10 pfennings, exchange being at $0.69 a thaler ? Ans. $ 207.47^. 12. What must be the face of a draft on Calcutta which costs in Boston $5694, when exchange is at $0.40 a company rupee ? Ans. 14235 rupees. 13. How much must be paid in Naples for a draft on Balti- more amounting to $ 615.60, the value of a ducat in exchange being $0.80? Ans. 769 ducats 5 carlini. 14. There was paid in New Orleans $7300 for 1500£. draft on Liverpool ; at what per cent, of premium was it purchased ? Ans. 9 J per cent. 15. What must be the face of a draft on Paris that can be bought in London for 868£. 17s. 6d., exchange at 23 francs 60 centimes a pound sterling ? Ans. 20505 francs 45 centimes. 16. How much must be paid in Genoa for a bill on New York whose face is $2640, when exchange is at $0.18 a lire ? Ans. 14666 lire (j$% centesimi. 17. When a bill on Paris for 88128 francs costs $ 17280, at what per cent is the rate of exchange below par ? 18. Robert Anderson, of Cincinnati, has consigned a cargo of pork, valued at 17000£., to Richard Arkwright & Son, Liv- erpool. Robert S. Davis & Co., being about to import an in- voice of books, have purchased of Anderson a bill of exchange, at 8£ per cent, premium, for the value of the said cargo. What should they pay for the bill ? Ans. $ 81977.777. ARBITRATION OF EXCHANGES. 500. Arbitration of Exchanges is the process of finding the proportional exchange of two places by means of one or more given intermediate exchanges. 501* Exchange effected thus, through one or more inter- mediate exchanges, is called circular exchange. The exchange when made through a single intervening ex- ARBITRATION OF EXCHANGES. 359 change is called simple arbitration, and when through two or more intervening exchanges is called compound arbitration. 502. Since the actual course or rate of exchange between any two places is almost always, from various circumstances, different from the arbitrated course, the object of arbitration is to enable an individual in one place to ascertain whether he can most advantageously draw and remit directly between his own place and another, or circuitously through other places. 503. Exchange of merchandise, and the different weights and measures of different countries, may be arbitrated in the same manner as bills of exchange and currencies. 504. The value, in the standards of the United States, of the principal weights and measures of the most important com- mercial places, is shown in the following Table. England. Value. Avoirdupois pound, Old wine gallon, lib. lgal. Imperial gallon, Old ale gallon, 1.20gal. 1.22gal. Old Winchester bushel, lbu. Imp. corn bu. (= 8 imp. gal.), 1.03bu. Quarter of grain (=8 imp. bu.), 8.25bu. Imperial yard, 36in. France. Kilogramme, 2.201b. Quintal = lOOkil., 220.541b. Millier = lOOOkil., 2205.481b. Litre, 2.11pt. Velt, 2gal. Decalitre = 10 litre, 2.64gal. Hectolitre = 100 litre, 26.42gal. Hectolitre = 100 litre, 2.84bu. Metre, 39.36in. Holland and Belgium. Pond, 1.081b. Fr. kilogramme, 2.201b. Last, marine, = 2000 p. 44101b. Vat = lOOkan = 1 hectol. Fr., 26.42s Ahm of wine, 41sral. Mudde=100 kop=l hectolitre, 2.84bu. Last of grain, 85.25bu. Amsterdam ell, 27in. Hamburg. Pound, 1.061b. 100 pounds, 106.81b. Ahm of wine, 38.25gal. Value. Fuder = 6 ahms, 229.5gal. Last of grain, 89.64bu. Stock = l£ last, 134.4bu. Brabant ell, 27.58in. Denmark and Norway. Pound, 1.101b. Centner = 100 pounds, 110.281b. Viertel of wine, 2.04gal. Anker of wine, lOgal. Ahm = 4 ankers, 40gal. Fuder of wine, 237.16gal. Toende or barrel of grain, 3.95bu. Last = 12 toende 47.50bu. Danish ell, 24.66in. Sweden. Pound, .931b. Pound of iron, .751b. Anker of wine, 10.35gal. Eimer of wine, 20.75gal. Ahm = 2 eimers, 41.50gal. Pipe = 3 ahms, 124.25gal. Tun or barrel of grain, 4.16bu. Ell, 23.36in. Bremen. Pound, 1.091b. Centner, 1161b. Viertel of wine, 1.93gal. Anker = 5 viertels, 9.65gal. Oxhoft = 6 ankers, 58gal. Scheffel of grain, 2bu. Last = 40 scheffels, 80.70bu. Ell, 22.75in. 360 ARBITRATION OF EXCHANGES. Trieste. Pound, 100 pounds, Eimer of wine, Staro of grain, Ell for silks, Ell for woollens, Constantinople. Quintal, Alma for liquids, Kisloz of grain, Pik, commercial, Value. 1.231b. 123.601b. 15gal. 234bu. 25.20in. 26.60in. 124.451b. 1.37gal. .94bu. 27in. Cantaro or arroba of wine, Moyo of wine =16 arrojm, Botta = 38ar. of wine = 3&& of oil, Fanega of grain, Cahiz = 12 fanegas, Vara or yard, Talue. 4.25gal. 68gal. ir. 127.5gal. 1.57bu. 18.91bu. 33.37in. Cuba. Quintal, Arroba of wine, Fanega of grain, Vara, 101.751b. 4.1gal. 3bu. 33.34in. Naples. Rottolo, 1.961b. Cantaro grosso=100 rottolo, 196.501b. Cantaro piccolo, 1061b. Calcutta. Maund, 74.661b. Bazaar maund, 82.131b. Guz, 36in. Russia. Pound, .901b. Pood = 40 pounds, 361b. 100 pounds, 90.261b. Wedro of wine, 3.25gal. Sorokovy = 40 wedros, 130gal. Chetwert of grain, 5.95bu. Arsheen, 28in. Sashen, 7 ft. Prussia. Pound, 1.031b. 100 pounds, Dantzic, 103.31b. Quintal = 110 pounds, 113.421b. Eimer of wine, 18.14gal. Ahm, 39.66gal. Scheffel of grain, 1.52bu. Last of grain, 91bu. Berlin Ell, 25.5in. Prussian Ell, 26.28in. Portugal. Pound or arratel, 1.011b. Arroba = 22 arratels, 22.261b. Quintal = 4 arrobas, 89.051b. 100 pounds or arratels, 101.191b. Almude of wine, 4.37gal. Tonelado = 52 almude, 227.25gal. Movo, 23.03bu. Vara, 43.20in. Spain. Pound, 1.011b. Arroba = 25 pounds, 25.381b. Quintal = 4 arrobas, 101.521b. Cantaro or arroba of oil, 3.75gal. Note. — The weights and measures of Mexico, Central America, and of the republics of South America are the same generally as those of Spain ; of Brazil, the same as those of Portugal; of the British North American Provinces, the same, in general, as in England ; and of Hayti, the weights are the same as in the United States, except about 8 per cent, heavier, and the measures the same as in France. Salmaof oil, 42.7".<:al. Cairo of wine 264gal. Carro of grain, 52.20bu. Canna, 83in. Sicily. Cantaro grosso, 192.501b. Cantaro sottile, 1751b. 100 Sicilian pounds, 701b. Tonna, 9.38gal. Salma prr< 9.48bu. Salma generale, 7.62bu. Couna or yard, 38.40in. Genoa. Cantaro grosso, 76.871b. Cantaro sottile, 69.891b. Mt-z/.arola, 39.25gal. Mina of grain, 3.50bu. Canna piCCOla, 87.50in. Canna grossa, 116.70in. Venice. 100 pounds, pesso grosso, 105.181b. 100 pounds, pesso sottile, 66.421b. Miro of oil, 4.02gal. Anfora of wine, 1.37gal. Staja of grain, 2.27bu. Moggio = 4 staji, 9.08bu. Braccio for silks, 24.84in. Braccio for woollens, 26.64in. China. Catty, 1.331b. Pecul, 133.331b. Covid, 14.62in. ARBITRATION OF EXCHANGES. 861 505 # To compute arbitration of exchanges. Ex. 1. When the exchange between New York and Lon- don is at a premium of 9 per cent., and that between London and Paris 25 francs to a pound sterling, how much must be paid in New York for a bill on Paris for 1000 francs ? operation. Since $ 44 = $40 ±= 9£. l£.ofthenomi- 109£. = 100£. nal par of ex- 1 £. = 2 5 fr. change (Article 1 Ofr. = $ — 497), $40=9£.; IQ j0 and 109£.of the yw ^ A \, „ vw^w same value = 40 X 109 X 1 XJ0M = $193J7+ Ang . 100£. at 9 per 9 X *00 X ^ :\ cent * Premium. We wnte $e terms of equivalent value as antecedent and consequent, and proceed as in conjoined proportion (Art. 341). Note. — When it is required to find which of several routes of exchange is the most advantageous, the rate of exchange by each route may be determined first, and the results then compared. Examples. 2. When exchange at Lisbon on Paris is at the rate of 5 francs 95 centimes per millrea, and at Paris on the United States at 5 francs 20 centimes per dollar, how much must be paid in Lisbon to cancel a demand in New York for $ 3500 ? Ans. 3058 millreas 823 T 9 T reas. 3. A merchant in Boston wishes to pay 2000£. in Liverpool. Exchange on Liverpool he finds is at 10 per cent, premium, on Paris 5 francs 20 centimes to a dollar, and on Hamburg 35 cents to a mark banco ; and the exchange between France and Eng- land at the same time is 24 francs to a pound sterling, and that of Hamburg on England 13 J- marks banco to a pound sterling. Which is the most advantageous course of remittance, that di- rect to Liverpool, or that through Paris or Hamburg ? 4. A merchant of St. Louis wishes to pay a debt of $ 5000 in New York. The direct exchange is 1^- per cent, in favor of New York, but on New Orleans it is £ per cent, premium ; and between New Orleans and New York at \ per cent, discount. How much would he save by the circular exchange compared with the direct ? j Ans."$ 87.56£. N 31 362 ARBITRATION OF EXCHANGES. 5. A merchant in Boston owes a debt of 97 GO thalers in Bremen, to pay which he purchases a bill on London, at a pre- mium of 9 per cent., and remits the same to his agent in Eng- land, on whom his creditor is requested to draw. If the ex- change between London and Bremen be at the rate of 3 1<1. sterling per thaler, and the charges for brokerage J per cent., how much must have been the cost of the bill in New York? Ans. $6731.74+. 6. When exchange between New Orleans and Hamburg is at 34 cents per mark banco, and between Hamburg and St. Petersburg is 2 marks 8 schillings per ruble, how much must be paid in St. Petersburg for a bill on New Orleans for $ 650 ? Ans. 764 rubles 70jf kopecks. 7. "When exchange in Philadelphia on Boston is at J per cent, premium, and on Chicago at 2 per cent, discount, if the exchange between Chicago and Boston is at par, how much better is the circuitous route of exchange between Philadelphia and Boston than the direct? 8. A merchant, about to import broadcloth, finds he can ob- tain the quality desired in Amsterdam at 8 guilders per Am- sterdam ell ; in Berlin, at 3 thalers 15 groschen per Berlin ell ; and in England, at 15 shillings per yard. Exchange being on Amsterdam at 40 cents per guilder, on Berlin at 66 cents per thaler, and on England at 9 J per cent, premium, and the freight being the same in each case, from which place can he make the importation to the best advantage ? Ans. Berlin. 9. When exchange between Washington and London is at 8 per cent, premium, and between London and Paris 25.25 francs per pound sterling, what sum in Washington is equal to 7000 francs in Paris ? 10. A merchant in London remits to Amsterdam 1000£. at the rate of 18d. per guilder, directing his correspondent at Amsterdam to remit the same to Paris at 2 francs 10 centimes per guilder, less J per cent, for his commission ; but the ex- change between Amsterdam and Paris happened to be, at the time^-the order was received, at 2 francs 20 centimes per guilder. The merchant at London, not apprised of this, drew upon Paris at 25 francs per pound sterling. Did he gain or lose, and how much per cent. ? Ans. Gain, 16f f per cent. ALLIGATION. 363 ALLIGATION. 505. Alligation is a process employed in the solution of questions relating to the compounding or mixing of articles of different qualities or values. It is of two kinds : Alligation Medial, and Alligation Alter- nate, ALLIGATION MEDIAL. 506. Alligation Medial is the process of finding the mean or average rate of a mixture composed of articles of different qualities or values, the quantity and rate of each being given. 507. To find the average value of several articles mixed, the quantity and rate of, each being given. Ex. 1. A grocer mixed 2cwt. of sugar worth $ 9 per cwt. with lcwt. worth $ 7 per cwt. and 2cwt. worth $ 10 per cwt. ; what is lcwt. of the mixture worth ? Ans. $ 9. $9 X 2=$18 Since 2cwt. at $ 9 per cwt. is worth 7X1— 7 $18, lcwt. at $ 7 per cwt. is worth $ 7, JO v/ 2 = 2 an( * 2cwt - at $ 10 per cwt. is worth $ 20 ; 2cwt. -)- lcwt. -f- 2cwt. = 5cwt. is worth 5 ) $45 $18-j-$7-f$20 = $45; and lcwt. is & n » worth as many dollars as 45 contains times $ 9 Ans - 5,or$9. Rule. — Find the value of each of the articles, and divide the sum of their values by the number denoting the sum of the articles. The quotient will be the average value of the mixture. Examples. 2. If 19 bushels of wheat at $ 1.00 per bushel should be mixed with 40 bushels of rye at $0.66 per bushel, and 11 bushels of barley at $ 0.50 per bushel, what would a bushel of the mixture be worth ? Ans. $ 0.727|. 3. If 3 pounds of gold of 22 carats fine be mixed with 3 pounds of 20 carats fine, what is the fineness of the mix- ture? Ans. 21 carats. 4. If I mix 20 pounds of tea at 70 cents per pound with 15 pounds at 60 cents per pound, and 80 pounds at 40 cents per pound, what is the value of 1 pound of this mixture ? Ans. $ 0.47£§. 364 ALLIGATION. ALLIGATION ALTEENATE. 508. Alligation Alternate is a process of finding in what ratio, one to another, articles of different rates of quality or value must be taken, to compose a mixture of a given mean or average rate of quality or value. 509. To find the proportional quantities of the articles of different rates of value that must be taken to compose a mix- ture of a given mean rate of value. Ex. 1. A merchant has spices, some at 18 cents a pound, some at 24 cents, some at 48 cents, and some at GO cents. How much of each sort must be taken that the mixture may be worth 40 cents a pound ? Ans. lib. at 18c; lib. at 24c; lib. at 8c; Ulb, at 60c. FIRST OPERATION. Given mean, , 40 cents. * lib. at 18c, gain 22c \ = 38c., lft>. at 18c. = 18c lib. at 24c, gain 16c j gain- lib. at 21c = 24c 111), at 48c, loss 8c lib. at 60c, loss 20c. f loss. -i 1 1 > . at 60c, loss 10c. J 4 4 Ib lib. at 48c = 48c 38c., l£lb. at 60c. = 90c 180c 180c -S- 4£ = 40c per lb. Compared with the given mean value, by taking lib. at 18cts. there is a gain of 22cts., by taking lib. at 24cts. a gain of 16ets., by taking lib. at 48cts. a loss of 8cts., and by taking lib. at 60cts. a loss of 20cts. Now it is evident that the mixture, to be of the mean or average value given, should have the several items of gain and loss in the aggregate exactly offset one another. This balance we effect by taking £lb. more of the spice at 60cts. ; and thus have a mixture of the required average value, by having taken, in all, lib. at 18cts., lib. at 24cts., lib. at 48cts., and 1-Jlb. at 60cts. We prove the cor- rectness of this result by dividing the value of the whole mixture by the number of pounds taken. second operation. Having arranged in a column 20 10 the rates of the articles, with the 8 ^^ 4 « given mean on the left, we con- 3 ccc. nect t g e ther terms denoting the j ^ ' rate of the articles, so that a rate less than the given mean is united with one that is greater. We then proceed to find what quantity of each of the two kinds whose rates have been connected can be taken, in making a mixture, so that what shall be gained on the one kind shall be balanced by the loss on the other. By taking lib. at 18cts. the gain will be 22cts. ; hence it will require ALLIGATION. 365 ? yb. to gain let. ; and by taking lib. at 60cts. the loss will be 20cts. ; hence it will require -^-lb. to lose let. Therefore, the gain on -^lb. at 18cts. balances the loss on -^lb. at 60cts. The proportions at these rates are, then, -£% and -^ or (by reducing to a common denom- inator) ¥ 2 4°o and ¥ 2 ¥ 2 q, or (by omitting the denominators, which do not affect the ratio) 20 and 22, which is obviously the same result as would be obtained by placing against each rate the difference be- tween the rate with which it is connected and the mean rate. In like maimer we determine the quantity that may be taken of the other two articles, whose rates are connected together. We thus find that there may be taken -^-lb. at 18cts., -^lb. at 24cts., |lb. at 48cts., and -^lb. at 60cts. ; or, 20lb. at 18cts., 8lb. at 24cts., 161b. at 48cts., and 22lb. at 60cts. By dividing the last set by 2, we obtain another set of results, and by multiplying or dividing any of these results others may be found, all of which can be proved to satisfy the conditions of the question. Hence, examples of this kind admit of an indefinite number of answers. Rule 1. — Take a unit of each article of the proposed mixture, and note the gain or loss ; and then take such additional quantity or quan- tities of the articles as shall equalize the gain and loss. Or, Rule 2. — Write the rates of the articles in a column, with the mean rate on the left, and connect the rate of each article which is less than the given mean ivith one that is greater ; the difference between the mean rate and that of each of the articles, written opposite to the rate with which it is connected, will denote the quantity to be taken of the article corresponding to that rate. Note. — When a rate has more than one rate connected with it, the sum of the differences written against it will denote the quantity to be taken. There will be as many different answers as there are different ways of connecting the rates ; and, by multiplying and dividing, these answers may be varied indefi- nitely. Examples. 2. How much barley at 45 cents a bushel, rye at 75 cents, and w r heat at $1.00, must be mixed, that the composition may be worth 80 cents a bushel ? Ans. 1 bushel of rye, 1 of barley, and 2 of wheat. 3. A goldsmith would mix gold of 19 carats fine with some of 15, 23, and 24 carats fine, that the compound may be 20 carats fine. What quantity of each must he take ? Ans. loz. of 15 carats, 2oz. of 19, loz. of 23, and loz. of 24. 4. It is required to mix several sorts of wine at 60 cents, 80 cents, and $1.20, with water, that the mixture may be worth 75 cents per gallon ; how much of each sort must be taken ? Ans. lgal. of water, lgal. of 60 cents, 9gal. of 80 cents, and lgal. of $1.20. 31* 366 ALLIGATION. 510. When the quantity of one or more of the articles com- posing a mixture of a given mean value is given, to find the quantity of each of the others. Ex. 1. How much gold of 15, 17, and 22 carats fine must be mixed with 5 ounces of 18 carats fine, so that the compo- sition may be 20 carats fine ? Ans. loz. at 15 carats, loz. at 17 carats, 9oz. at 22 carats. operation. By taking loz. at 15 carats {loz. at 15, gain 5 ^ fine there is a gain of 5 carats, loz. at 17, gain 3 >- = 18 by taking loz. at 17 carats a 5oz. at 18, gain 10 ) £*in of 3 carats, by taking 5oz., loz. at 22, loss 2 ) the ? iven quantity, at 18 carats, ( __ ^g a gain of 10 carats, and by tak- 8oz. at 22, loss 16 ) in S loz - at 22 carats a loss of . 2 carats; and to balance the gain and loss we take 8oz. additional, at 22 carats, a loss of 16. We have then for the result loz. at 15 carats, loz. at 17 carats, and 9oz. at 22 carats. Rule. — Take of the limited artiele or articles the quantify or quan- tities given, with a unit of each of the other articles of the proposed mixture, and note the gain or loss ; and tfu n take, if required, such ad- ditional quantity or quantities of the articles not limited, as shall equalize the gain and loss. Examples. 2. How much wine at $1.75 and at $ 1.25 per gallon must be mixed with 20 gallons of water, that the whole may be sold at $ 1 .00 per gallon ? Ans. 20 gallons of each. 3. How much wheat at $ 2.00 per bushel and at $ 1.80 per bushel must be mixed with 4 bushels at $ 2.20 per bushel and 10 bushels at $ 1.70 per bushel to make a mixture worth § 1.90 per bushel ? Ans. 9 bushels at $ 2.00 ; 1 bushel at $ 1.80. 4. How many pounds of sugar, at 8, 14, and 13 cents a pound, must be mixed with three pounds at 9^- cents, 4 pounds at 10J- cents, and 6' pounds at 13^- cents a pound, so that the mixture may be worth 12£ cents a pound ? Ans. lib. at 8cts.; 5£lb. at 13cts.; and 91b. at 14cts. 5. How much barley at 45 cents a bushel must be mixed with 10 bushels of oats at 58 cents a bushel, to make a mixture worth 50 cents a bushel ? ALLIGATION. 367 511. When the quantity and rate of a mixture, with the rates of the articles composing it, are given, to find the quantity of each article which is not limited. Ex. 1. How many gallons of water must be mixed with wine at $1.50 a gallon so as to make a mixture of 100 gallons worth $ 1.20 a gallon ? Ans. 20 gallons. operation. Representing the 1 90 $ *S a l' a * 0-00? g am 1-20 1«20 rate of the water by * ( lgal. at 1.50, loss .30 ) 0.00, we then find, V 1 20 as m Art. 5 ^9, the 3gal. at 1.50, loss .90) quantity required of ° J each article, in com- 1 -|- 4 = 5gal. ; \ of lOOgal. = 20gal. Ans. posing a mixture of the given mean, to be 1 gallon of water and 4 gallons of the wine. Therefore, the quan- tity of water is to the whole quantity of the mixture as 1 to 5. Hence, in a mixture of 100 gallons at the mean rate given, the water must be \ of 100 gallons, or 20 gallons. Rule. — Find the proportional quantities of the several articles, as in Art. 509, or 510, as though the quantity of the mixture were not limited. Then take such a part of the given quantity of the mixture, as each of these proportional quantities is of their sum. Examples. 2. A merchant has sugar at 8 cents, 10 cents, 12 cents, and 20 cents a pound ; with these he would fill a hogshead that would contain 200 pounds. How much of each kind must he take, so that the mixture may be worth 15 cents a pound ? Ans. 33p>. of 8, 10, and 12cts., and 1001b. of 20cts. 3. How much wheat at $ 2.00 and $ 1.80 a bushel must be mixed with 4 bushels at $ 2.20, and 10 bushels at $ 1.70, so as to make a mixture of 48 bushels, worth $ 1.90 per bushel? Ans. 21 bushels $ 2.00; 13 bushels at $ 1.80. 4. How much gold of 15, 17, and 22 carats fine must be mixed with five ounces of 18 carats fine, to make a compo- sition of 5 pounds, that shall be 20 carats fine ? 5. A gentleman's servant having been ordered to purchase 20 animals for $ 20, brought home sheep at $ 4.00, lambs at $ 0.50, and kids at $ 0.25 each. Required the number of each kind. Ans. 3 sheep ; 15 lambs ; and 2 kids. 368 MISCELLANEOUS EXAMPLES. MISCELLANEOUS EXAMPLES. 1. A manufacturer employs a number of men at $ 1.20, and a number of boys at $ 0.80, per day ; and the amount of the wages of the whole is the same as if each had $0.9 7 J- per day. Required the number of men, that of the boys being 9. Ans. 7 men. 2. What is the value of 5000 specie rix dollars 12 skillings of Sweden in United States money ? Ans. $ 5300.265. 3. Exchange between New Orleans and England being in New Orleans at 8 per cent, premium, and in Liverpool at 10 per cent, premium, if L. Sandford of Liverpool owes M. Las- sale of New Orleans for cotton to the amount of 1500£. 15s. sterling, what will be the difference between Lassale drawing or Sandford remitting the amount ? 4. If 17 gallons of spirits at $1.26 per gallon be mixed with 7 gallons at a different price, and 20 per cent, be made by selling the mixture at $ 1.56, what was the price of the latter kind per gallon? Ans. $ 1.39f per gallon. 5. What is the value of 100 ounces 20 tari 10 grani of Naples in lire and centesimi of Leghorn ? Ans. 1510 lire 25 centesimi. 6. If 20 United States gallons equal 1 eimer of Sweden, 3 eimers of Sweden equal 4 eimers of Trieste, 24 eimers of Trieste equal 9 ahms Danish, and 33 ahms Danish equal 5 carri of Naples, which will cost the most in United States money, 170 eimers of Trieste of wine at 1 florin 45 kreutzers per gallon, or 12 carri of wine at 1 ducat of Naples a gallon? 7. When exchange on England is at 8 per cent, premium, and freight at 12d. per United States bushel, how much can be paid per bushel for wheat in Baltimore, in answering an order from Liverpool limited to 60s. per imperial quarter ? Ans. $ 1.50 T 6 T per bushel. 8. A merchant mixes 11 pounds of tea with 5 pounds of an inferior quality, and gains 16 per cent, by selling the mixture at 87 cents per pound. Allowing that a pound of the one cost 12 cents more than a pound of the other, what was the cost of each kind per pound ? Ans. The one 78£cts. ; the other 66Jets. per lb. INVOLUTION. 369 EVOLUTION. 512. Involution is the process of finding the powers of quantities. A power of a number or quantity is the product of that num- ber multiplied into itself one or more times. 513. The number from which a power is derived is called the root of that power. The first power is the root, or the number involved. The second power is the product of the root multiplied by it- self once, or used twice as a factor. The third power is the root used three times as a factor; &c. 514. The index or exponent of a power is a small figure written at the right, above the root, indicating the number of times it is employed as a factor. Thus, the second power of 4 is written 4 2 , the third power of 9 is written 9 3 , and the fourth power of f- is written (J) 4 . Note. — In denoting the power of a fraction, the fraction is included in a parenthesis, in order that the exponent may be regarded as applying to the whole expression, and not to the numerator alone. When no index is written, the number itself is to be considered the first power. The second power is sometimes called the square of a number, the third power the cube, and the fourth power the biquadrate. 515. To raise a number to any required power. 2=2, the first power of 2, is written 2 l or 2. 2X2= 4, the second power of 2, is written 2 2 . 2X2X2= 8, the third power of 2, " " 2 3 . 2 X 2 X 2 X 2 = 1 6, the fourth power of 2, " " 2 4 . 2X2X2X2X2 = 32, the fifth power of 2, " " 2 5 . By examining the several powers of 2 in the examples, it is seen that each has been produced by taking the 2 as a factor as many times as there are units in the exponent of each power raised. Hence the Ktjle. — Multiply the given number into itself, till it has been used as a factor as many times as there are units in the exponent of the power to which the number is to be raised. Note 1. — The number of multiplications will always be one less than the number of units in the exponent of the power to be raised, since in the first 370 INVOLUTION. multiplication the root is used twice; once by being taken as the multiplicand, and once more as the multiplier. Note 2. — A fraction is involved by involving both its numerator and its de- nominator. Examples. 1. What is the 3d power of 8 ? Ans. 512. 2. What is the 5th power of 4 ? Ans. 1024. 3. What is the 3d power of J ? Ans. § J. 4. What is the 4th power of 2J ? Ans. 50$f . 5. What is the 5th power of -J ? Ans. ^j. 6. What is the 6th power of 5 ? Ans. 15625. 7. What is the 6th power of If ? Ans. 16jf w j ; 8. What is the value of 7 10 ? Ans. 282475249. 9. What is the value of .045 4 ? Ans. .000004100625. 516. To raise a number to any required higher power, with- out producing all the intermediate powers. Ex. 1. What is the 7th power of 5 ? Ans. 78125. OPERATION. 12 3 3+2 + 2= 7 5, 2 5, 12 5; 125x25x25 = 7812 5. We raise the 5 to the 2d and to the 3d power, and write above each power its exponent. Then, by adding the exponent 2 to itself, and increasing the sum by the exponent 3, we obtain 7, a number equal to the exponent of the required power; and by multiplying 25, the power belonging to the exponent 2, into itself, and the product thence arising by 125, the power belonging to the exponent 3, we ob- tain 78125, the required 7th power. Therefore, The product of two or more powers of the same number is that power which is denoted by the sum of their exponents. Hence, the Rule. — Multiply together two or more powers of the given number, the sum ofichose exponents is equal to the exponent of the power re- quired, and the product will be that power. Note. — When the number to be involved contains a decimal, it is generally sufficient to retain in the result not more than six places of decimals; and the work may be accordingly contracted as in the multiplication of decimals (Art. 273). Examples. 2. What is the 7th power of 8 ? Ans. 2097152. 3. What is the 9th power of 7 ? Ans. 40353607. 4. What is the 10th power of 6 ? EVOLUTION. 371 5. What is the 5th power of 195 ? Ans. 281950621875. 6. What is the 6th power of § ? Ans. T 6 ^ 4 F . 7. Required the 2d power of 4698. 8. Required the 2d power of 6031. Ans. 36372961. 9. What is the 13th power of 7 ? Ans. 96889010407. 10. What is the 12th power of 6 ? 11. What is the 15th power of 9 ? Ans. 205891132094649. 12. What is the 4th power of 4.367 ? Ans. 363.691179+. 13. Involve the following numbers to the powers denoted by their respective exponents : (2£ ) 5 , 1.04' 5 , and (3f) 4 . Ans. 157 -ft 8 A; 1.800943+ ; 116i|-g-f EVOLUTION. 517 • Evolution, or the extraction of roots, is the process of finding the roots of quantities. It is the reverse of involu- tion. 518. The root of a quantity or number is such a factor as, being multiplied into itself a certain number of times, will pro- duce that quantity or number. The root takes the name of the power of which it is the cor- relative term. Thus, if the number is a second power, the root is called the second or square root; if it is the third power, the root is called the third or cube root ; if it is the fourth power, its root is called the fourth or biquadrate root ; and so on. Rational roots are such as can be exactly obtained. Surd roots are such as cannot be exactly obtained. 519» Roots are usually denoted by writing the radical sign, \/, before the power, with the index of the root over it ; in case, however, of the second or square root, the index 2 is omitted. Thus, the third root of 27 is denoted by ^/27, the second root of 16 is denoted by \/16, and the fourth root of % is denoted by /Z/% % 372 EVOLUTION. Roots are sometimes denoted by a fractional index or expo- nent, of which the numerator indicates the power, or the number of times the number is to be taken as a factor, and the denomina- tor indicates the root, or the number of equal factors into which that product is to be divided. Thus the square or second root of 12 is denoted by 12*, the fourth root of f by (g)±, and the square of the cube root of 27, or the cube root of the square of 27, is denoted by 275. 520» All the rational roots of whole numbers are also whole numbers, since every power of a fractional number is also a fractional number. 521. Prime numbers have no rational roots. A composite number, to have a given rational root, must have the exponent of the power of each of its prime factors exactly divisible by the exponent of that root. Note. — The number of composite numbers that have rational roots is com- paratively small. The number of rational square roots of whole numbers from 1 to 250000 inclusive is only 500, and the number of rational cube roots of whole numbers from 1 to 8000000 inclusive is only 200. 522 . The roots represented by the first ten numbers and their first six corresponding powers are shown in the following Table. 1st Power, 1 2 3 4 5 6 7 8 9, 10 2d Power, 1 4 9 16 25 36 49 64 81| 100 3d Power, 1 8 27 64 125 216 343 512 729: 1000 4th Power, 1 1G 81 256 625 1296 2401 4096 6561! 10000 5th Power, 1 82 243 1024 3125 7776 16807 32768 59049' 100000 6th Power, 1 64 729 4096 15625 46656 117649 262144 531441 1000000 Note. — It will be observed by the table, that a rational square root can only be obtained from numbers ending in 1, 4, 5, 6, or 9; or in an even number of ciphers, preceded by one of these figures. It is true, also, that, if the square number ends in 1, its square root ends in 1 or 9 ; if in 4, its square root ends in 2 or 8 ; if in 9, its square root ends in 3 or 7 ; if in 6, its square root ends in 4 or 6 ; and if in 5, its square root ends in 5. A perfect cube, however, may end in either of the nine digits, and in ciphers if the number of them is three or any multiple of three; also if the cube num- ber ends in 1, its cube root will end in 1 ; if in 8, its cube root ends in 2 ; if in 3, its cube root ends in 7 ; if in 4, its cube root ends in 4 ; if in 5, its cube root ends in 5 ; if in 6, its cube root ends in 6 ; if in 7, its cube root ends in 3 ; if in 8, its cube root ends in 2 ; and if in 9, its cube root ends in 9, EVOLUTION. 373 EXTRACTION OF THE SQUARE ROOT. 523. The extraction of the square root of a number is the process of finding one of its two equal factors ; or of finding such a factor as, when multiplied by itself, will produce the given number. 524 • The method generally adopted for extracting the square root depends upon the following principles : — 1. The square of any number has, at most, only twice as many figures as its root, and, at least, only one less than twice as many. For the square of any number of a single figure consists of either one or two places of figures, as l 2 == 1, and 9 2 = 81 ; the square of any number of two figures consists of either three or four places, as 10 2 = 100, and 99 2 == 9801 ; and the same law holds in regard to numbers of three or more figures. Therefore, when the square number consists of one or two figures, its root will consist of one figure ; when of three or four figures, its root will consist of two figures ; when of five or six figures, its root will consist of three figures; and so on. Hence, if a number be separated into as many periods as pos- sible of two figures each, commencing at the right, to these periods respectively will correspond the units, tens, hundreds, &c. of the square root of the number. 2. The square of a number consisting of tens and units is equal to the square of the tens, plus twice the product of the tens into the units, plus the square of the units. Thus, if the tens of a number be denoted by a and the units by b, the square of the number will be denoted by (a + b) 2 = a 2 + 2 a b + b 2 . Then, by this formula, if a = 3, and 5—6, we have 3 tens + 6 units = 30 + 6 = 36; and 36 2 = (30 + 6) 2 = 30 2 + 2 X (30 X 6) X 6 2 = 1296. Or, analytically, a + b = a + b = (a + b) X « = (a + b)Xb = 30+6 30+6 30 2 +30X6 30X6+6 2 = 30+ 6 = = SO-]- 6 h = 900+180 = = 180+36 = = 900+360+36 = 36 36 1080 216 (a + b) 2 m. 30 2 +2X(30X6)+6 2 1296 374 EVOLUTION. It is evident, as evolution is the reverse of involution (Art. 517), that from the process now given of obtaining a square may be deduced a method of extracting its root. Since the square of (a-\-b) is a 2 -|- 2 a b -\- b 2 , the square root of a 2 -f- 2 a b -j- b 2 must be a -J- b. Now it will be observed that a, the first term of the root, is the square root of a 2 , the first term of the square ; and if a 2 be subtracted, there will remain 2 a b -f- b 2 , from which b, the second term of the root, is to be obtained. But 2 a b -\- b 2 is the same as (2 a -\- b) \ b, therefore the remainder equals (2 a -f- b) X &• But as b, the units, is al- ways much less than 2 a, twice the tens, we consider that 2 a X b is about equal to the whole remainder, and taking 2 a (which we know) as the trial divisor, we obtain b y the units. But as the true divisor is 2 a -\- b, we add the units to twice the tens and multiply the sum by the units, which gives a pro- duct equal to the whole remainder, or 2 a b -f- b 2 . Since every number of more than one figure may be considered as composed of tens and tin its, we may have tens and units of units, tens and units of tens, tens and units of hundreds, &c. Hence, the principle just explained applies equally whether the root contains two or more than two figures. 525. To extract the square or second root of numbers. Ex. 1. What is the square root of 1296 ? Ans. 36. operation. Beginning at the right, we separate the 1 9 Q c qt number into periods of two figures each, by 1 z J o o b placing a point (•) over the right-hand figure " of each period. Since the number of periods (56396 is two, the root will consist of two figures, 3 o, g tens and units. Then 1296 = the square of the tens plus twice the product of the tens into the units, plus the square of the units The square of tens is hundreds, and must therefore be found in the hundreds of the number. The greatest number of tens whose square does not exceed 12 hundreds is 3, which we write as the tens figure of the root. "We subtract the 9 hundreds, the square of the 3 tens, from the 12 hundreds, and there remain 3 hundreds ; after which we write the figures of the next period, and the remainder is 396 = twice the product of the tens into the units plus the square of the units. We have then next to find a number which, added to twice the 3 tens of the root, and mul- tiplied into their sum, shall equal 396. By dividing this remainder EVOLUTION. 375 by twice the three tens of the root, we may obtain the units, a num- ber somewhat too large. But though it may be too large, it cannot be too small, since the remainder 396 contains twice the product of the tens into the units, and also the square of the units. We there- fore make twice the three tens of the root = 6 tens, a trial divisor, with which we divide the 39 tens, exclusive of the 6 units, which cannot form any part of the product of the tens by the units. The quotient figure obtained, 6, must be the units figure of the root, or a number somewhat larger. To determine whether it expresses the real number of the units in the root, we annex it to the 6 tens, and multiply the number 66, thus formed, by it. The product is 396, which being subtracted, there is no remainder. Therefore 1296 is a perfect square, and 36 the root sought. 2. What is the square root of 278784? Ans. 528. operation. Since there are three pe- 278784 25 102 287 204 104* $ 8384 8384 I 5 2 8 riods, the root will contain three figures; the first two may be considered as tens and units of tens. As the square of tens cannot give less than hundreds, we must find that square in the two left-hand periods ; and as we have tens and units of tens, proof. their square = the square of 528 X 528 == 278784 tne tens P ms tw ^ ce tne tens into the units, plus the square of the units =2787 (nearly). We proceed then with the first two periods exactly the same as when the root consists of but two figures, and thus take from the given number the square of the 52 tens, which leaves a remainder of 8384. We now consider the given number 278784, as the square of a number consisting of 52 tens and a certain number of units, which square will of course equal the square of the tens plus twice the tens into the units, plus the square of the units. But the square of the tens, or (52) 2 has already been taken from the given number, leaving a remainder, 8384, which must equal twice the tens into the units plus the square of the units. From this we readily obtain the units, just as when we had but two figures in the root. Rule. — Separate the given number into as many periods as possible of two figures each, by placing a point over the place of units, another over the place of hundreds, and so on. Find the greatest square in the left-hand period ; write the root of it at the right of the given number after the manner of a quotient in di- vision, and subtract the second power from the left-hand period. Bring down the next period to the right of the remainder for a divi- dend, and double the root already found for a trial divisor. Find hoio often this divisor is contained in the dividend, exclusive of the right- hand figure, and write the quotient as the next figure of the root. 376 EVOLUTION. Annex the last root figure to the trial divisor for the true divisor, which multiply by the last root figure and subtract the product from the dividend. To the remainder bring down the next period for a new div- idend. Double the root already found for a new trial divisor, and continue the operation as before, till all the periods have been brought down. Note 1. — When the product of any trial divisor exceeds its corresponding dividend, the last root figure must be made less. If a dividend does not contain its corresponding divisor, a cipher must be placed in the root, and also at the right of the divisor; then, after bringing down the next period, this last divisor must be used as the divisor of the new dividend. Note 2. — When there is a remainder after extracting the root of a number, periods of ciphers may be annexed, and the figures of the root thus obtained will be decimals. Note 3. — If the given number is a decimal, or a whole number and a de- cimal, the root is extracted in the same manner as in whole numbers, except, in pointing off the decimals, either alone or in connection with the whole num- ber, we begin at the separatrix and place a point over every second figure to- ward the right, filling the last period, if incomplete, with a cipher. The number of decimal places in the root will always equal the number of periods of decimals in the power. Note 4. — If the given number is a common fraction, reduce it to its sim- plest form, if it is not so already, and extract the root of both terms, if they are perfect powers ; otherwise, either find their product, extract its root, and di- vide the result by the denominator, or reduce the fraction to a decimal, and extract the root of the decimal. Note 5. — When the given number is a mixed number, it may be changed to the form of a common fraction, or the fractional part may be reduced to a decimal, before attempting to extract the root. Examples. 3. What is the square root of T |-f^? Ans. ¥ 4 T . 4. What is the square root of ¥ 3 «y? Ans. .1936+. 3 X 30 = 240; ^ 2A0 = .1936+. 80 ' Or & = .0375 ; \/.0375 = .1936+. 5. What is the square root of 3444736 ? Ans. 1856. 6. What is the square root of 998001 ? Ans. 999. 7. What is the square root of t 3 oVf ? 8. Extract the square root of 234.09 ? Ans. 15.3. 9. What is the square root of 42£? Ans. 6^. EVOLUTION. 377 10. What is the square root of .000729 ? 11. What is the square root of 17.3056? 12. What is the square root of 52 T 9 g ? 13. What is the square root of 95 T *g-? 14. What is the square root of 363^ T ? 15. How much is \/1.96 ? 16. How much is 6561* ? 17. How much is \/9 3 ? 18. How much is 8*? 19. How much is 9645192360241 ? one of the two Ans. .027. Ans. 4.16. Ans. 7 J. Ans. 9|. Ans. 19 T 1 g-. Ans. 1.4. Ans. 81. Ans. 27. Ans. 64. equal factors of Ans. 3105671. 526. When the square root is to be extracted to many places of figures, the work may be contracted thus : — Having found in the usual ivay one more than half of the root figures required, the rest may be found by dividing the last remainder, with a single figure annexed instead of two, by the last divisor, proceeding as in contracted division of decimals. (Art. 276.) Examples. 1. What is the square root of 785 to five places of decimals ? Ans. 28.01785. CONTRACTED METHOD. 785 4 2 8.0 1 7 8 5+ 48 385 384 5601 10000 5601 5602 43990 39219 4771 4482 289 280 9 COMMON METHOD. 785 4 2 8.0 1 7 8 5 -f 48 385 384 5601 10000 5601 5602' 7 439900 392189 5602348 4771100 4482784 560 35 6 5 28831600 28017825 813775 The nature and extent of the contraction will be seen by compar- ing the contracted method with the common method. 32* 378 EVOLUTION. 2. Extract the square root of 6§ to four places of deci- mals. Ans. 2.5298+. 3. Required the square root of 2 to five places of deci- mals. Ans. 1.41421+. 4. Required the square root of 3.15 to eight places of deci- mals. Ans. 1.77482393+. 5. Required the square root of 373 to seven places of deci- mals. Ans. 19.3132079+. 6. Extract the square root of 8.93 to eight places of deci- mals. Ans. 2.98831055+. EXTRACTION OF THE CUBE ROOT. 527. The extraction of the cube root of a number is the process of finding one of its three equal factors ; or, of finding a factor which, being multiplied into itself twice, will produce the given number. 528. The common method of extracting the cube root de- pends upon the following principles : — 1. The cube of any number has, at most, only three times as many figures as its root, and, at least, only two less than three times as many. For the cube of a number of a single figure consists of, at most, three figures, and, at least, two less than that number, as l 3 = 1, and 9 3 ess 729 ; the cube of a number of two figures consists of, at most, six figures, and, at least, two figures less than that number, as 10 3 = 1000, and 99 3 = 970299 ; and so on. Therefore, when a cube number consists of one, two, or three figures, its root will consist of one figure ; when of four, five, or six figures, its root will consist of two figures, and so on ; and if a number be separated into as many periods as possible of three figures, each commencing at the right, to these periods respectively will correspond the units, tens, hundreds, &c. of the cube root of that number. 2. Tlie cube of a number consisting of tens and units is equal to the cube of the tens, plus three times the square of the tens into the units, plus three times the tens into the square of the miits, plus the cube of the units. 'Jhus, if the tens of a number be denoted by a, and the units by b, the cube of the number EVOLUTION. 379 will be denoted by (a + b) 3 ■= a 3 + 3 a 2 b + 3 a b 2 + & 3 . Then, by this formula, if a = 3, and 6 equal 6, we have 3 fens + 6 units = 30 + 6 = 36, and 3 6 3 == (30 + 6) 3 = 30 3 + 3 (30 2 X 6) + 3 (30 X 6 2 ) + 6 3 = 46656. Or, analytically, « + £ = 30 + 6 =36 a _|- 6 = 30 + 6 =36 (a-\-b) Xa = 30 2 + 30 X 6 = To80 (a+b) Xb= 30 X 6-f6 2 = 216 (a + 6) 2 =30 3 + 2 X30 X 6 + 6 2 = 1296 a_j_6 ,= 30 -j-6 . = 36 (a-J i -b) 2 Xa = 30 3 + 2 X 30 2 X 6 + 30 X 6 2 = 38880 ( a + 6) 2 X b = 3Q 2 X 6 + 2X 30X 6 8 +6 8 == 7776 (a + b) 3 = 30 3 + 3 (30 2 X6)+3(30 X 6 2 )+6 3 = 46656 It is evident, as evolution is the reverse of involution, that from this process of obtaining a cube may be deduced a method of extracting the cube root. Since the cube of a + b is a 3 + 3 a 2 b + 3 a b 2 + b 3 , the cube root of a 3 + 3 a 2 b + 3 a b 2 + b 3 must be a + J. Now a, the first term of the cube root, is the cube root of a 3 , the first term of the cube ; and if a 3 be subtracted, there will remain 3 a 2 b + 3 a b 2 + b 3 , from which b, the second term of the root, is to be obtained. But 3 a 2 b + 3 a b 2 + b 3 is the same as (3 a 2 + 3 a b + b 2 ) X b ; therefore the remainder equals (3 a 2 + 3 a b + b 2 ) X &• But as 3 a b, three times the tens into the units, plus b 2 , the square of the units, is generally much less than 3 a 2 , three times the square of the tens, we consider that 3 a 2 X b is about equal to the whole remainder, and taking 3 a 2 (which we know) as the trial divisor, we obtain b, the units. But as the true divisor is 3a 2 + S ab -\~ b 2 we add three times the tens by the units plus the square of the units, and multiply the sum by the units, which gives a product equal the whole remainder, or 3a a ft + 3ay+ b 3 . Since every number of more than one figure may be con- sidered as composed of tens and units, we may have tens and units of units, tens and units of tens, tens and units of hundreds, &C. Hence, the principle just explained applies equally wheth- er the root contains two or more than two figures. 380 EVOLUTION. 529t To extract the cube or third root of numbers. Ex. 1. What is the cube root of 46656 ? Ans. 36. operation. ^ Beginning at 46656 ! 36 the right, we sep- 3 3 = 27 arate the given Trial div., 3 X 30 2 = 2700 3 x 30 X 6 = 540 6 2 = 36 True divisor, 3276 X 6 number into pe- 196o6 riods, by placing a point over the units figure and 19656 over tne third figure to the left. Since the number PR0 ° F * ■ of periods is two, 36 X 36 X 36 = 46656 the root will con- sist of two figures, tens and units. Then 46656 = the cube of tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. The cube of tens is thousands, and must therefore be found in the thousands of the num- ber. The greatest number of tens whose cube does not exceed 46 thousands is 3, which we write as the tens figure of the root. We then subtract the 27 thousands, the cube of the 3 tens, from the 46 thousands, and there remain 19 thousands; and, annexing the next period, Ave have as the entire remainder, 19656, equal three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units, or the product of three times the square of the tens, plus three times the tens into the units, plus the square of the units, multiplied by the units. By dividing this remainder by three times the square of the tens of the root, we obtain the units, or a number somewhat too large. Although it may be too large, it cannot be too small, since the remainder 19656 con- tains not only three times the square of the tens into the units, but three times the tens into the square of the units, plus the cube of the units. We therefore make three times the square of the tens of the root, = 27 hundreds, a trial divisor, with which we divide the 196 hundreds of the remainder, disregarding the 56 units, since they cannot form any part of the product of the square of the tens by the units. The quotient figure obtained, 7, must be the units figure of the root, or a number somewhat larger. But on undertaking to complete the divisor on the supposition that 7 is the true units figure of the root, we find a divisor too large for the remainder. We therefore take 6, a number one less, and to de- termine whether it expresses the real number of units in the root, we add to the 2 7 hundreds of the trial divisor three times the square of the 3 tens of the root into the 6 units, plus the square of the 6 units ; and multiplying the true divisor, 3276, thus formed by the units, and subtracting the product, 19656, from the remainder, there is nothing left. Hence, 46656 is a perfect cube, and 36 its cube root. EVOLUTION. 381 12326391 8 4326 2. What is the cube root of 12326391 ? FIRST OPERATION. 2 3 = Trial divisor, 3 X 20 2 = 1200 3 X 20 X 3 == 180 3 2 = 9 True divisor, 1389 X 3 = JL 1 Trial divisor, 3 X 230 2 & 158700 3 X 230 X 1 = G90 1 2 = 1 True divisor, 159391 Xl = Ans. 231. 231 15 9 3 9 1 159391 Since there are three periods the root will contain three figures, the first two of which may be considered as tens and units of tens. As the cube of tens cannot give less than thousands, we must find that cube in the two left-hand periods ; and as we have tens and units of tens, their cube will equal the cube of the tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. This we apply to the first two periods exactly the same as when the root consists of but two figures; and thus take from the given number the cube of the 23 tens, which leaves a remainder of 159391. We now consider the given number, 159391, as the cube of a number consisting of 23 tens and a certain number of units, which cube will of course equal the cube of the tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. But the cube of the tens, or (23) 3 , has already been taken from the given number, leaving a remainder, 159391, which must equal three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units ; or equal the product of three times the square of 23, plus three times 23 into the units, plus the square of the units, multiplied by the units. From this we readily obtain the units, just as when we had but two figures in the required root. 231 SECOND OPERATION. 2 3 — Trial divisor, 63X3 = 1200 = 189 True divisor, 3 2 sfcs 1389 x 3 = = 9 Trial divisor, 691 X 1 = 158700 =t= 691 True divisor, 1593 91 x 3r 1-8 8 326291 4326 4167 15 9 3 9 1 159391 382 EVOLUTION. In the second operation the work is somewhat abridged by render- ing shorter the method of finding each divisor after the first, of both kinds. Thus, the second true divisor is obtained by prefixing to the second root figure, 3, three times the part of the root preceding it, or 3 X 2 == 6, and adding 189, the product of the number 63 thus formed by the last root figure, 3, to the preceding trial divisor, or 1200 -j- 189 = 1389, second true divisor. This is equivalent to adding *to the trial divisor, in forming the true divisor, three times the tens into the units, plus the square of the units, or 3 X 20 X 3 + 3 2 . For the second trial divisor we annex two ciphers to the sum found by adding together the square of the last root figure, the last true di- visor, and the number standing over it, or 9 -f- 1389 -f- 189 = 1587, with two ciphers annexed = 158700, the second trial divisor. This is equivalent to taking for the trial divisor three times the square of the tens, or the part of the root already found, or 3 X 230*. The next true divisor is found in like manner as was the second true di- visor. Rule 1. — Separate the given number into as many periods as pos- sible of three figures each, b< ginning at the units' place. Find the greatest cube in the left-hand period, and write its root as the first figure of the required root. From that period subtract the cube, and to the remainder bring down the next period for a dividend. Multiply the square of the root figure by 3, and to the product annex two ciphers for a trial divisor, and see how often it is contained in the dividend, and ivrite the result as the next figure of the root. Add to the trial divisor three ti?nes the product of the tens figure of the root by the units figure with a cipher annexed, and the square of the last figure y for a true divisor. Multiply the true divisor by the last figure of the root ; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. Multiply the square of the root figures already found by 3, and to the product annex two ciphers for a new trial divisor ; and proceed as be- fore until all the periods are brought down. Or, Rule 2. — Having found the first trial divisor and determined the second root figure as by the preceding rule, — Tale three times the part of the root already found, except the last figure, to it annex the last figure of the root, multiply the result by the figure annexed, and write the product below the trial divisor, and add it to the same for a true divisor. Multiply the true divisor by the last figure of the root : subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. To the last true divisor and the number immediately over it, add the square of the last root figure, and to the sum annex two ciphers for a new trial divisor ; then proceed as before. Note 1. — The observations made in Notes 1, 2, 3, and 5, under the rule for the extraction of the square root (Art. f>25), are equally applicable to the ex- EVOLUTION. 383 traction of the cube root, except that two ciphers must be placed at the right of a true divisor when it is not contained in its corresponding dividend, and in pointing off decimals each period must contain three figures. Note 2. — If the given number is a common fraction, reduce it to its sim- plest form, if it is not so already, and extract the root of both terms, if they be perfect powers; otherwise, either find the product of the numerator by the square of the denominator, extract its root, and divide the result by the de- nominator; or reduce the fraction to a decimal, and extract the root of the decimal. Examples. 3. What is the cube root of 77308776 ? Ans. 426. 4. What is the cube root of Jf§? Ans. f. 5. What is the cube root of 84.604519 ? Ans. 4.39. 6. What is the cube root of 54439939 ? Ans. 379. 7. What is the cube root of 60236288 ? Ans. 392. 8. Required the cube root of .726572699. Ans. .899. 9. Extract the third root of 109215352. Ans. 478. 10. What is the third root of £§ $$ ? Ans. J-f . 11. What is the value of ^ffjjif ? Ans. f f. 12. What is the value expressed by ^34965783 ? Ans. 327. 13. What is the value of 122615327232" ? Ans. 4968. 14. What is the value of 436036824287" ? Ans. 7583. 530 • When the cube root is to be extracted to many places of decimals, the work may be contracted thus : — Having found in the usual way one more than half of the root figures required, the rest may he found by dividing the last remainder by its corresponding true divisor, as in contracted division of decimals (Art. 276), observing however at each step to reject two figures from the right of the divisor and one from the right of the remainder. Example. 1. Required the cube root of 2 to four places of decimals. Ans. 1.2599-f. 2. Find the third root of 11 to four places of decimals. Ans. 2.2239. 3. Extract the cube root of 3 to six places of decimals. Ans. 1.442249+. 4. Extract the cube root of 9 to fifteen places of decimals. Ans. 2.08008382301904. 884 EVOLUTION. EXTRACTION OF ANY ROOT. 531. Tlie root corresponding to any perfect power may be obtained by resolving that power into its prime factors, and mul- tiplying together one of each number of equal factors denoted by the exponent of the required root. Thus one of each two equal factors of the power will give the second or square root ; one of each three equal factors w r ill give the third or cube root ; one of each four equal factors will give the fourth root ; and so on. 532. When the index or exponent of the root to be extracted is a composite number, the root may be obtained by successive extractions of the simpler roots denoted by the several factors of that exponent. Thus the fourth root may be obtained by ex- tracting the square root twice in succession ; the sixth root by extracting the square root and then the cube root ; and so on. K\AMI>LES. 1. Required the fourth root of 50625 ? Ans. 15. BY FACTORS. BY SUCCESSIVE EXTRACTIONS. 50625 10125 50025,225 4 52025 X 5~|405 4210G 84 445 2225 2225 381 327 3j9 X~3 5 X 3 = 15, Ans. 2. What is the square root of 998001 ? 3. What is the cube root of 262144 ? 4. What is the fourth root of 43046721 ? 5. What is the fifth root of 14348907 ? 6. What is the sixth root of 11390625 ? 225 1 5125 125 15, Ans. Ans. 999. Ans. 64. Ans. 27. Ans. 15. 533* When the given number is an imperfect power, or otherwise, or the exponent denoting the root is prime, or other- wise, the required root may be found by an elegant process, perfect in principle, called from its inventor, OPERATION. 4 16 4 16 4 32 8 4800 4 625 120 5425 5 650 125 607500 5 4059 130 611559 5 EVOLUTION. 385 Horner's Method. Ex. 1. Kequired the cube root of 9295677. Ans. 453. The greatest cube 92959677 453 contained in the left- £4 hand period we find to be 64, whose root, 4, 28959 we write as the first 27125 figure of the required —— — ■ root. This figure, 4, ±00^0/ / we wn te under the ci- 1834677 pher of the first column ; a and adding it to the cipher obtain 4, which sum multiplied by the 4 gives 16; and the result, 16, we write under the cipher of the second column, and by addition obtain 16, which sum, multiplied by the 4, gives 64 ; and the result, 64, we write in the last column under the left-hand period, 92, of the given number. The 64 sub- 1350 tracted from the 92 above it gives for a remainder 28. 3 We next add the 4 to the last term, 4, of the first _ column, obtaining 8 ; and the result, 8, multiplied by the looo 4j gives 32, which we write under the last term, 16, of the second column, and, adding the same together, obtain 48. We next add the 4 to the last term, 8, of the first column, ob- taining 12 ; and, annexing one cipher to the last term, 12, of the first column, obtaining 120, two ciphers to the last term of the second column, obtaining 4800, and to the remainder in the last column bringing down the next period, 959, obtaining 28959, we complete the work preparatory to the finding of the second root figure. To determine that root figure, we take the last term, 4800, of the second column, for a trial divisor, and the last term, 28959, of the last column, for a dividend; and, dividing, 6 would appear to be the second figure of the root. This, on trial, however, is found to be too large ; we therefore take 5, which answers. This 5 we add to the last term, 120, of the first column, obtaining 125 ; which sum, 125, multi- plied by the 5, gives 625, and that product, added to the last term, 4800, of the second column, gives 5425 ; and this result, 5425, multi- plied by the 5, gives 27125, which, written in the last column and sub- tracted from the figures above it, gives a remainder 1834. Then we add the 5 to the last term, 125, of the first column, obtaining 130 ; and the result, 130, multiplied by the 5, gives 650, which we write under the last term, 5425, of the second column, and by addition ob- tain 6075. We next add the 5 to the last term, 130, of the first column, obtaining 135 ; and, annexing one cipher to the last term, 135, of the first column, obtaining 1350, two ciphers to the last term, 6075, of the second column, obtaining 607500, and to the remainder N 33 386 EVOLUTION. in the last column bringing down the next period, 677, obtaining 1834677, the work is completed preparatory to finding the third root figure. To determine that figure, as before, we divide the last term of the third column by the last term of the second. We thus obtain 3, which, added to the last term of the first column, gives 1353, which sum, multiplied by the 3, gives 4059 ; and that product, being added to the last term of the second column, gives 611559 ; and that sum, multiplied by the 3, gives 1834677, which being exactly as large as the last term of the third column, on being written under it and sub- tracted there is no remainder. The given number is therefore a per- fect power, and the cube root sought is 453. In practice, the work may be performed with less figures, by writ- ing down in the several columns only the results. Rule. — Commence as many columns as there are units in the ex- ponent of the root to be extracted, by writing the given number as the head of the right-hand column, and a cipher as the head of each of the others. Separate the given number into as many periods as possible of as many figures each as the exponent of the root requires ; and having found the nearest root of the left-hand period, write it as the first figure of the required root. Write this figure in the first column, and, having added it to what stands above it, multiply the sum by the same figure, and write the pro- duct in the second column ; add, in like manner, in the second column, and multiply the sum by the same figure, writing the product in the third column; and so proceed, writing the last product in the last column, and subtracting it from what stands above it. Then add the same figure to the last term of the first column, multi- ply the sum by the same figure, and add the product to the last term of the second column; and so on, writing the last product in the last column but one. Repeat the process, stopping each time with one col- umn farther to the left, till the last product shall fall in the second column-. Add the figure found for the root to the last term of the first column ; annex one cipher to the last number in the first column, two ciphers to the last number in the second column, and so on ; and to the last num- ber in the last column bring down the next period for a dividend. Take the last term of the column next to the last for a trial divisor, and see how often it is contained in the dividend, and write the result as the next figure of the root. Add 'this figure to the last term of the first column, multiplying the sum by the same figure, add the product to the second column, and so on ; proceed as before, till all the periods have been brought down, or an answer sufficiently exact has been obtained. Note 1. — When any dividend will not contain its corresponding trial divi- sor, write a cipher in the root, bring down to the dividend another period, an- nex an additional cipher to the last term of the first column, two additional EVOLUTION. 387 ciphers to the last term of the second column, and so on; and use the same trial divisor as before, increased however by the additional ciphers. Note 2. — When the given number does not have an exact root, periods of ciphers may be annexed. Note 3. — When the root is required to many places of decimals, the work may be contracted by rejecting one figure at the right from the number in the column next to the last, two from the number in the column next farther to the left, and so on, and otherwise proceeding as directed in the rule, except that the new figure of the root is not added to the first column. As soon as all the figures are rejected in the number of the first column, the remainder of the work may be performed as in contracted division of decimals (Art. 276). Examples. 2. Required the fourth root of 1.016397 to eight places of decimals. Ans. 1.00407427. OPERATION. 1 1 1 2 3 3 1 1 3 1.0163970011.00407427 1 1 1 1639700 1609632 2 1 4000000 2408- 30068 28338 3 1 60000 2- 402408 2416 1730 1619 400 602 2 "604 2 606 404824 111 81 30 28 ~2 3. What is the cube root of 41673648563 ? Ans. 3467. 4. What is the cube root of 43614208? 5. What is the cube root of 1.05 to six places of deci- mals? Ans. 1.016397. 6. What is the fifth root of 184528125 ? Ans. 45. 7. Required the fourth root of 100 to six places of deci- mals ? Ans. 3.162278. 8. Required the fifth root of the fourth power of 9 to seven places of decimals? Ans. 5.7995466. 388 EVOLUTION. APPLICATIONS OF POWERS AND ROOTS. 534. A Triangle is a figure having three sides and three angles. When one of the sides of a triangle is perpendicular to another side, the opening between them is called a right angle, and the triangle is called a right-angled triangle. The lowest side, A B, is called the base of the triangle ABC, the side B C the perpen- dicular, the longest side, A C, the hypothenuse, and the angle at B is a right angle. Also, the line B C, being perpendicular to the base, is the altitude. Base. 535. The square described upon the hypothenuse of a right angled triangle is equivalent to the sum of the squares de- scribed upon the other two sides. Tims, if the hypothenuse A C be 5 feet, the base AB 4 feet, and the perpendicular B C 3 feet, then 5 a = 4* -f- 3' 2 , or 25 = 16 -J- 9. B 536. To find the hypothenuse, the base and perpendicular being given. Add together the square of the base and the square of the perpendicu- lar, and extract the square root of the sum. Thus, if the base be 4 and the perpendicular 3, the hypothenuse will equal V4* -J- 3 2 = \/lb = 5. 537 # To find the perpendicular, the base and hypothenuse being given. Subtract the square of the base from the square of the hypothenuse, and extract the square root of the remainder. Thus, if the base be 4 and the hypothenuse 5, the perpen- dicular will equal */ 5* — 4 2 =s /^/l) - — 3. 538. To find the base, the hypothenuse and perpendicular being given. EVOLUTION. 389 ; Subtract the square of the perpendicular from the square of the hy- pothenuse, arid extract the square root of the remainder. Thus, if the perpendicular be 3 and the hypothenuse 5, the base will equal „/ 5" — & = VT6 — 4. 539 • All triangles having the same base are to each other as their altitudes. All similar triangles, and other similar rectilineal figures, are to each other as the squares of their homolo- gous or corresponding sides. Thus, the triangles ACE and A C D, having the same base, A C, are to each other as the altitude E C of the one is to the alti- tude D C of the other. Also, the triangles ACE and BCD, having their corresponding angles the same, and their sides in direct proportion, are said to be similar, and are to each other as the squares of their corresponding sides, or as (A E) 2 is to (B D) 2 , (A C) 2 is to (B C) 2 , and (C E) 2 is to (C D) 2 . Like- wise the larger square, of which A C is one of the equal sides, is to the smaller square, of which B C is one of the equal sides, as (A C) 2 is to (B C) 2 . 540. All circles (Art. 143) are to each other as the squares of their diameters, semidiameters, or circumferences. The circumference of a circle is the line which bounds it ; and the diameter is a line drawn through the centre, and terminated by the circumference ; as A B and C D. Then, the larger circle, of which A B is the diameter, is to the smaller, of which C D is the diameter, as (A B) 2 is to (C D) 2 , &c. 541. To find the side, diameter, or circumference of any surface, which is similar to a given surface. State the question as in Proportion, and square the given sides, diam- eters, or circumferences, and the square root of the fourth term of the proportion will be the answer required. Thus, if 12 feet be the length of a side of a triangle whose area is 72 square feet, the length of the corresponding side of a 33* 890 EVOLUTION. similar triangle whose area is 32 square feet would be found as follows : 72 : 32 : : 12 2 =* 144 : 64 ; y^T £s 8 feet, length required. 542. To find the area of any surface which is similar to a given surface. State the question as in Proportion, and square the given .sides, diam- eters, or circumferences, and the fourth term of (lie proportion will be the answer required. Thus, if 72 square feet be the area of a triangle of which 12 feet is one of the sides, the area of a similar triangle of which the corresponding side is 8 feet would be found as follows : 12 2 = 144 : 8 2 = 64 : : 72 sq. ft. : 32 sq. ft., area required. 543. To find the side of a square equal in area to any given surface. Find the square root of tlie given area, and that root ivill be the side of the area required. 544. A sphere is a solid bounded by a continued convex surface, every part of which is equally distant from the point within called the centre. A The diameter of a sphere is a straight line pass- ing through the centre, and terminated by the sur- face ; as A B. o 545. A cone is a solid having a circle for its base, and tapering uniformly to a point, called the vertex. The altitude of a cone is its perpendicular height, or a line drawn from the vertex perpendicular to the plane of the base, as B C. The diameter of its base is a straight line drawn through the centre of the plane of the base from one side of the circle to the other ; as A D. 546. Spheres are to each other as the cubes of their diam- eters, or of their circumferences. Similar cones are to each other as the cubes of their altitudes, or of the diameters of their bases. All similar solids are to each other as the cubes of their homologous or corresponding sides, or of their diameters. EYOLUTION. 391 Note. — Cones and other solids are said to be similar when their corre- sponding parts are in direct proportion to each other. 547. To find the contents of any solid which is similar to a given solid. State the question as in Proportion, and cube the given sides, diam- eters, altitudes, or circumferences, and the fourth term of the -proportion is the answer required. 548. To find the side, diameter, altitude, or circumference of any solid, which is similar to a given solid. State the question as in Proportion, and cube the given sides, diam- eters, altitudes, or circumferences, and the cube root of the fourth term of the proportion is the answer required. 549. To find the side of a cube that shall be equal in solidity to any given solid. Find the cube root of the contents of the given solid, and that root will be the side of the cube required. 550. To find a mean proportional (Art. 333) between any two numbers. Find the square root of the product of the two numbers, and that root will be the mean proportional required. 551. To find two mean proportionals between two given numbers. Find the cube root of the quotient of the greater of the two numbers divided by the less. The product of the less number by that root icill be the least mean proportional , and, the quotient of the greater number by the same root will be the other mean proportional. 552. To find any two numbers, whose sum and product are given. From the square of half the sum of the two numbers subtract their product, and the square root of the remainder will equal half the differ- ence of the two numbers, which added to half their sum will give the larger, and subtracted from half their sum will give the smaller, of the numbers required, 558. To find any two numbers, when their sum and the difference of their squares are given. The difference of their squares divided by the sum of the numbers will give their difference; and half of their difference added to half of their sum will give the larger, and half of their difference subtract half of their sum icill give the smaller, of the numbers required. I 392 EVOLUTION. Examples. 1. A certain general has an army of 141376 men. How many must he place in rank and file to form them into a square? Ans. 376. 2. If the area of a circle be 1760 yards, how many feet must the side of a square measure to contain that quantity ? Ans. 125.857+ feet 3. If a line 144 feet long will reach from the top of a fort to the opposite side of a river 64 feet wide, on whose brink it stands, what is the height of the "fort ? Ans. 128.99-)-. 4. A certain room is 20 feet long, 16 feet wide, and 12 feet high ; how long must a line be to extend from one of the lower corners to the upper corner farthest from it ? Ans. 28.28ft\ 5. A certain field is 40 rods square ; what must be the length of one of the equal sides of another field that shall contain only one fourth as much area ? Ans. 20 rods. 6. The areas of two similar triangular-shaped fields are 60 and 90 acres, and a side of the former is 66 rods. Required the corresponding side of the latter ? 7. If a lead pipe £ of an inch in diameter will fill a cistern in 3 hours, what should be its diameter to fill it in 2 hours ? Ans. .918-f- inches. 8. If a pipe 1J- inches in diameter will fill a cistern in 50 minutes, how long would it require a pipe that is 2 inches in diameter to fill the same cistern ? Ans. 28m. 7^s. 9. If a pipe 6 inches in diameter will draw off a certain quantity of water in 4 hours, in what time would it take 3 pipes of four inches in diameter to draw off twice the quantity ? Ans. 6 hours. 10. The first term of a proportion is 40, and the fourth term 90. Required a mean proportional between them. Ans. 60. 11. In a pair of scales a body weighed 31£ pounds in one scale, and only 20 pounds in the other scale. Required its true weight. Ans. 25 pounds. 12. I wish to set out an orchard of 2400 mulberry-trees, so that the length shall be to the breadth as 3 to 2, and the dis- tance between any two adjacent trees 7 yards. How many trees must there be in the length, and how many in the EVOLUTION. 393 breadth ; and on how many square yards of ground will they stand ? Ans. GO in length ; 40 in breadth ; 112749 sq. yd. 13. The sum of two persons' ages is 50 years, and their pro- duct is 600 years. What are their ages ? Ans. Of the one, 20 years ; of the other, 30 years. 14. Two ships sail from the same port ; one goes due north 128 miles, the other due east 72 miles; how far are the ships from each other? Ans. 146.86-]— 15. There are two columns in the ruins of Persepolis left standing upright ; one is 70 feet above the plane, and the other 50 ; in a straight line between these stands a small statue, 5 feet in height, the head of which is 100 feet from the summit of the higher, and 80 feet from the top of the lower column. Re- quired the distance between the tops of the two columns. 16. The sum of two numbers is 44, and the square of their difference is 16. Required the numbers. Ans. 24 the larger number ; 20 the smaller. 17. A tree 80 feet in height stands on a horizontal plane ; at what height from the ground must it be broken off, so that the top of it may fall on a point 40 feet from the bottom of the tree, the end where it was broken off resting on the stump ? Ans. 30 feet. 18. The height of a tree, growing in the centre of a circular island, 100 feet in diameter, is 160 feet; and a line extending from the top of it to the farther shore is 400 feet. What is the breadth of the stream, provided the land on each side of the water be level? Ans. 316.6-)- feet. 19. A ladder 70 feet long is so planted as to reach a win- dow 40 feet from the ground, on one side of the street, and without moving it at the foot it will reach a window 30 feet high on the other side ; what is the breadth of the street ? 20. If an iron wire -j-V of an inch in diameter will sustain a weight of 450 pounds, what weight might be sustained by a wire an inch in diameter ? Ans. 450001b. 21. A gentleman proposes to plant a vineyard of 10 acres. If he places the vines 6 feet apart, how many more can he plant by setting them in the quincunx order than in the square order, allowing the plat to lie in the form of a square, and no vine to be set nearer its edge than 1 foot in either case ? Ans. 1870 more in the quincunx order. ■ 394 EVOLUTION. 22. Four men, A, B, C, and D, bought a grindstone, the dia- meter of which was 40 inches and the place for the shaft 4 inches in diameter. It was agreed that A should grind off his share first, then in turn B, C, and D. Required how many inches each man will grind off from the semidiameter, provid- ing they each paid the same sum. Ans. A, 2.651in. ; B,3.137in. ; C, 4.064in. ; and D, 8.148in. 23. I have a board whose surface contains 49 §■ square feet; the board is 1£ inches thick, and I wish to make a cubical box of it. Required the length- of one of its equal sides. Ans. 36 inches. 24. A carpenter has a plank 1 foot wide, 22^J feet long, and 2J- inches thick ; and he wishes to make a box whose width shall be twice its height, and whose length shall be twice its width. Required the contents of the box. Ans. 5719 cubic inches. 25. If a ball, 3 inches in diameter, weigh 4 pounds, what will be the weight of a ball that is G inches in diameter ? Ans. 321bs. 26. If a globe of gold, one inch in diameter, be worth $ 120, what is the value of a globe 3£ inches in diameter? 27. If the weight of a well-proportioned man, 5 feet 10 inches in height, be 180 pounds, what must have been the weight of Goliath of Gath, who was 10 feet 4§ inches in height? Ans. 1015.1-f-lb. 28. If a bell, 4 inches in height, 3 inches in width, and \ of an inch in thickness, weigh 2 pounds, what should be the dimensions of a similar bell that would weigh 2000 pounds ? Ans. 3ft. 4in. high, 2ft. 6in. wide, and 2^in. thick. 29. What are the two mean proportionals between 56 and 12096? Ans. 336 and 2016. 30. Having a small stack of hay, 5 feet in height, weighing lcwt, I wish to know the weight of a similar stack that is 20 feet in height. Ans. 64cwt. 31. If a man dig a small square cellar, which will measure 6 feet each way, in one day, how long would it take him to dig a similar one that measured 10 feet each way ? Ans. 4.629+ days. 32. If an ox, whose girth is 6 feet, weighs 6001b., what is the weight of an ox whose girth is 8 feet? Ans. 1422.2-{-lb. PROGRESSION, OR SERIES. 395 33. Four women own a ball of yarn, 5 inches in diameter. It is agreed that each shall wind off her share from the ball. How many inches of its diameter shall each wind off? Ans. First, .45-f- inches ; second, .57-f- inches ; third, .82— j— inches ; fourth, 3.149-(- inches. 34. John Jones has a stack of hay in the form of a quad- rangular pyramid. It is 16 feet in height, and 12 feet wide at its base. It contains 5 tons of hay, worth $ 17.50 per ton. Mr. Jones has sold this hay to Messrs. Pierce, Row, Wells, and Northend. As the upper part of the stack has been injured, it is agreed that Mr. Pierce, who takes the upper part, shall have 10 per cent, more of the hay than Mr. Rowe ; and Mr. Rowe, who takes his share next, shall have 8 per cent, more than Mr. Wells ; and Mr. Northend, who has the bottom of the stack, that has been much injured, shall have 10 per cent, more than Mr. Wells. Required the quantity of hay, and how many feet of the height of the stack, beginning at the top, each receives. Ans. Pierce receives 27^/TjCwt. and 10.3 6 6-|- feet in height; Rowe, 24j-f|cwt. and 2.493 feet; Wells, 22§4Jcwt. and 1.666 feet; Northend, 2o 2 \\cwt. and 1.474 feet. PROGRESSION, OR SERIES. 554. A Series is a succession of numbers that depend on one another by some fixed law. The numbers constituting a series are called its terms; of which the first and last are called extremes, and the other terms the means. ARITHMETICAL PROGRESSION. 555. Arithmetical Progression, or Progression by Difference, is a series that increases or decreases by a con- stant number, called the common difference. The series is said to be an ascending one when each terra 396 PROGRESSION, OR SERIES. after the first exceeds the one before it ; and a descending one when each term after the first is less than the one before it. Thus, 1, 5, 9, 13, 17, 21, 25, 29, 33, is an ascending series, in which each term after the first is derived from the one pre- ceding it by the addition of the common difference 4 ; and 25, 22, 19, 16, 13, 10, 7, 4, 1, is a descending series, in which each term after the first is derived from the one preceding it by the subtraction of the common difference 3. '556. In arithmetical progression, the first term, the last term, the number of terms, the common difference, and the sum of the terms, are so related to each other, that, three of these be- ing given, the two others may be readily determined. 557 • To find the common difference, the extremes and num- ber of terms being given. Ex. 1. The extremes of an arithmetical series are 3 and 45, and the number of terms is 22. Required the common dif- ference. Ans. 2. operation. It is evident that the number of common 45 g differences in any series must be 1 less than . = 2. the number of terms. Therefore, since the 22 — 1 number of terms in this series is 22, the num- ber of common differences will be 22 — 1 = 21, and their sum will be equal to the difference of the extremes; hence the difference of the extremes, 45 — 3 = 42, divided by the number of common differences, 21, gives 2 as the common difference required. Rule. — Divide the difference of the extremes by the number of terms less one, and the quotient ivill be the common difference. Examples. 2. A certain school consists of 19 teachers and scholars, whose ages form an arithmetical series ; *the youngest is 3 years old,* and the oldest 39. What is the common difference of their ages ? Ans. 2 years. 3. A man is to travel from Albany toV certain place in 11 days, and to go but 5 miles the first day, increasing the distance equally each day, so that the last day's journey may be 45 miles. Required the daily increase. Ans. 4 miles. PROGRESSION, OR SEBIB8. 897 558. To find the number of terms, the extremes and com- mon difference being given. Ex. 1. If the extremes of an arithmetical series are 3 and 19, and the common difference is 2, what is the number of terms ? Ans. 9. operation. It is evident, that, if the difference of the 19 — 3 extremes be divided by the common dif- ~ [-1=9. ference, the result will be the number of common differences; thus 19 — 3 = 16 ; 16 -J- 2 = 8. Then, as the number of terms must be 1 more than the number of common differences, 8 -|- 1 = 9 is the number of terms in the series. Rule. — Divide the difference of the extremes by the common differ- ence, and the quotient increased by 1 will be the required number of terms. Examples. 2. A man going a journey travelled the first day 7 miles, the last day 51 miles, and each day increased his journey by 4 miles. How many days did he travel ? Ans. 12. 3. In what time can a debt be discharged, supposing the first week's payment to be S 1, and the payment of every succeed- ing week to increase by S 2, till the last payment shall be S 103 ? Ans. 52 weeks. 559. To find the sum of all the terms, the extremes and number of terms being given. Ex. 1. The extremes of an arithmetical series are 3 and 19, and the number of terms 9. Required the sum of the series. Ans. 99. operation. In an arithmetical 3-J-19 ooa series the sum of the o X 9 = 99j Ans. extremes is equal to the sum of two terms Or, 3 + 19 = 22 ; 22 X H = 09, Ans. that are equally dis- tant from them, or to double the middle term, if the number of terms be odd. Thus, in the series, 3, 5, 7, 9, 11, 13, 15, 17, 19, the sum of 3 and 19 is equal to the sum of 5 and 17, or of 7 and 15, and is double the middle term, 11. The reason of this is evident, since 5 and 7 exceed the less ex- treme by the same quantities by which 17 and 15 are respectively less than the other extreme. N 34 398 PROGRESSION, OR SERIES. Hence, in this latter series, it is evident that, if each terra were made 11, half the sum of the extremes, the sum of the whole would remain the same ; therefore the sum of the series must equal half the sum of the extremes multiplied by the number of terms, or the sum of the extremes multiplied by half the number of terms. Kule. — Multiply half the sum of the extremes by the number of terms. Or, Multiply the sum of the extremes by half the number oftemxs. Examples. 2. If the least term of a series of numbers in arithmetical progression be 4, the greatest 100, and the number of terms 17, what is the the sum of the terms ? Ans. 884. 3. Suppose a number of stones were laid a rod distant from each other, for thirty miles, the first stone being a rod from a basket. What distance will that man travel who gathers them up singly, returning with them one by one to the basket ? Ans. 288090 miles 2 rods. 560» To find the sum of the terms, the extremes and com- mon difference being given. Ex. 1. If the two extremes are 3 and 19, and the common difference is 2, what is the sum of the series ? operation. ^ nas been shown (Art 558) that, 2Q 3 if the difference of the extremes be [-1 = 9; divided by the common difference, 2 the quotient will be the number 19 -|- 3 Q ooa °^ terms * ess * one * Therefore the 2 ^ ' Ans. number of terms less one will be ■ I" == 8 ; and 8 -J- 1 will equal the number of terms. It has also been shown (Art. 559) that, if the number of terms be multiplied by the sum of the extremes, and the product divided by 2, the quotient will be the sum of the series; therefore — j~ X 9 = 99, the answer required. BULB, — Divide the difference of the extremes by the common differ- ence, and to the quotient add 1 ; by this sum multiply half the sum of the extremes, and the product will be the sum required. Examples. 2. If the extremes are 3 and 45, and the common difference 2, what is the sum of the series ? Ans. 528. 3. A owes B a certain sum, to be discharged in a year, by PROGRESSION, OK SERIES. 399 paying 6 cents the first week, 18 cents the second week, and thus to increase every week by 12 cents, till the last payment should be $ 6.18. What is the debt ? Ans. $ 162.24. 561 . To find one of the extremes, when the other extreme and the number and sum of the terms are given. Ex. 1. If 3 be the first term of a series, 9 the number of terms, and 99 the sum of the series, what is the last term ? operation. It has been shown (Art. 559) that, if the 99 X 2 . sum of the extremes be multiplied by the =22, Ans. number of terms, the product will be twice the sum of the series ; therefore, if twice the sum of the series be divided by the number of terms, the quotient will be the sum of the extremes. If from this we subtract the given extreme, the remainder must be the other extreme. Rule. — Divide twice the sum of tlie series by the number of terms ; from the quotient take the given term, and the remainder will be the term required. Examples. 2. The sum of a series of ten thousand even numbers i3 100010000, and the last term of the series is 20000. Required the first term. Ans. 2. 3. A merchant, being indebted to 22 creditors $528, ordered his clerk to pay the first % 3, and the rest sums increasing in arithmetical progression. What is the difference of the pay- ments, and the last payment ? Ans. Difference 2 ; last payment $ 45. 562. To find any number of arithmetical mea?is, the ex- tremes and the number of terms being given. Ex. 1. If the first term of an arithmetical series is 1, the last term 99, and the number of terms 8, what are the second and seventh terms of the series ? Ans. The second term, 15 ; the seventh, 85. We find the common differ- opekation. ence, 14, as in Art. 557, the qq __ -j first term, 1, plus the common — 14 ; difference, 14, gives 15 for 8 — 1 the second term, and the last iiia i^.qo i/i - qz term, 99, minus the common 1 ~r i4: — I0 > JJ — i4 - b0 - difference, 14, gives the sev- enth term. 400 PROGRESSION, OR SERIES. Rule. — Find the common difference, which, added to the less ex- treme, or subtracted from the greater, will give one mean. From that mean derive others in the same v:ay, till those required are found. Examples. 2. The extremes of a series are 4 and 49, and the number of terms 6. Required the middle two terms. Ans. 22 and 31. 3. Insert five arithmetical means between 20 and 30. Ans. 21§, 23£, 25, 26§, and 28£. GEOMETRICAL PROGRESSION. 563 • Geometrical Progression, or progression by quotients, is a series of numbers that increase or decrease by a constant multiplier or divisor, called the common ratio. The series is an ascending one, when each term after the first increases by a constant ratio ; and a descending one, when each term after the first decreases by a constant ratio. Thus 2, 6, 18, 54, 162, 486 is an ascending geometrical series; and 64, 32, 16, 8, 4, 2 is a descending geometrical series. Of the former, 3 is the common ratio, and of the lat- ter, 2. 564 1 In geometrical progression theirs* term, the last term, the number of terms, the common ratio, and the sum of the terms are so related to each other, that, any three of these being given, the other two may be readily determined. 565 1 To find any proposed term, one of the extremes, the ratio, and the number of terms being given. Ex. 1. If the first term of a geometrical series be 3, the ratio 2, and the number of terms 8, what is the last term ? Ans. 384. operation. It i s evident that the successive 3 X 2 7 = 3 X 128 = 384. terms are the result of repeated multiplications by the ratio ; thus the second term must be the product of the first term by the ratio, the third term the product of the second term by the ratio, and so on. The eighth, or last term, therefore, must be the result of seven PROGRESSION, OR SERIES. 401 such multiplications, or the product of the first term, 3, by 2 7 , or 3 X 128 = 384. If the last term had been given and the first required, the process would evidently have been by division, since every less term is the result of a division of the term next larger by the ratio. Rule. — Raise the ratio to the power ivhose index is one less than the number of terms ; by ichich multiply the least term to find the greatest, or divide the greatest to find the least. Note 1. — When the ratio requires to be raised to a high power, the process may be abridged, as in Art 516. Note 2. — The rule may be applied in computing compound interest, the principal being the first term, the amount of one dollar for one year the ratio, the time, in years, one less than the number of terms, and the amount the last term. Examples. 2. If the first term be 5, and the ratio 3, what is the seventh term ? Ans. 3645. 3. If the series be 72, 24, 8, &c, and the number of terms 6, what is the last term ? Ans. / T . 4. If the larger extreme be 885735, the ratio 3, and the number of terms 12, what are the tenth and the eleventh terms ? Ans. 15 and 45. 5. If the seventh term is 5, and the ratio -J, what is the first term ? Ans. 3645. 6. If the first term is 50, the ratio 1.06, and the number of terms 5, what is the last term ? Ans. 63.123848. 7. If I were to buy 30 oxen, giving 2 cents for the first ox, 4 cents for the second, 8 cents for the third, &c, what would be the price of the last ox ? Ans. $ 10737418.24. 8. What is the amount of $ 160.00 at compound interest for 6 years ? Ans. $ 226.96305796096. 9. What is the amount of $ 300.00 at compound interest at 5 per cent, for 8 years ? Ans. % 443.236+. 10. What is the amount of $ 100.00 at compound interest at 6 per cent, for 30 years ? Ans. $ 574.349117291325011626410633231080264584635- 7252196069357387776. 566. To find the sum of a series, the first term, the ratio, and the number of terms being given. 34* 402 PROGRESSION, OR SERIES Ex. 1. If the first term be 1, the ratio 3, and the number of terms 5, required the sum of the terms. Ans. 121. operation. If we multiply the series 1, 3, 9, 27, (81 X 3) — 1 __ - 91 81 by the ratio 3, we shall obtain as a 3 1 ~~ second series 3, 9, 27, 81, 243, whose sum is three times the sum of the first se- ries, and the difference between whose sum and the sum of the first series is evidently twice the sum of the first series. Now it will be observed that the two series have their terms alike, with the excep- tion of the first term in the first series, and the last in the second series. We have then only to subtract the first term in the first series from the last term in the second, and the remainder is twice the sum of the first series ; and half this being taken gives the re- quired sum of the series. Therefore the sum of the first series must be 242 -f- 2 = 121. Rule. — Find the last term as in Art. 565, multiply it by the ratio, and the product less the Jirst term divide by the ratio less 1 ; the result will be the sum of the series. Note 1. — If the ratio is less than a unit, the product of the last term mul- plied by the ratio must be subtracted from the first term, and the remainder divided by unity or 1 decreased by the ratio. Note 2. — When a descending series is continued to infinity, it becomes what is called an infinite series, whose last term must always be regarded as 0, and its ratio as a fraction. To find the sum of an infinite series, — Divide the first term by 1 decreased by the fraction denoting the ratio, and the quotient will be the sum required. This process furnishes an expeditious way of finding the value of circulating decimals, since they are composed of numbers in geometrical progression, whose common ratios are Jj, ^, iq CC , &c. according to the number of fac- tors contained in the repetend. Thus, .3333, &c. represents the geometrical series j , jgg, i^q, &c. whose first term is fa and common ratio jq. Examples. 2. The first term of a series is 5, the ratio §, and the number of terms 6 ; required the sum of the series. (i) 5 x 5 = n%; m x § - m ; s - m = Wj 3. Find the value of the circulating decimal .232323, &c. t 2 o% -*- (1 — To-o-) = ih Ans. 4. What is the sum of the series 4, 1, £, T \, &c., continued to an infinite number of terms ? 1-4- (1 _ £) = 5i Ans. 5. If the first term is 50, the ratio 1.06, and the number of terms 4, what is the sum of the series? Ans. 218.7308. PROGRESSION, OR SERIES. 403 6. A gentleman offered a house for sale on the following terms ; that for the first door he should charge 10 cents, for the second 20 cents, for the third 40 cents, and so on in a geometri- cal ratio. If there were 40 doors, what was the price of the house ? Ans. $ 109951162777.50. 7. If the series f , ^, -|, y 1 ^, ^, &c. were carried to infinity, what would be its sum ? Ans. 1£. 8. A gentleman deposited annually $ 10 in a bank, from the time his son was born until he was 20 years old. Required the amount of the 21 deposits at 6 per cent., compound interest, when his son was 21 years old? Ans. $423.92. 9. Find the value of .008497133, &c., continued to in- finity. Ans. sf h- 10. If a body be put in motion by a force which moves it 10 miles in the first portion of time, 9 miles in the second equal portion, and so, in the ratio of T 9 ^-, for ever, how many miles will it pass over ? Ans. 100 miles. 567 1 To find the ratio, the extremes and number of terms being given. Ex. 1. If the extremes of a series are 3 and 192, and the number of terms 7, what is the ratio ? operation^ It has been shown (Art. 192 -i- 3 = 64 ; / 1.000 ooc 1.000 000 1.000 000 1.000 000 2 2.030 000 ! 2.035 000 2.040 000 2.050 000 2.060 000 2.070 000 3 3.090 900 3.106 225 3.121 600 3.152 500 3.183 600 3.214 900 4 4.183 627 4.214 943 4.246 464 4.310 125 4.374 616 4.439 943 5 5.309 136 5.362 466 5.416 323 5.525 631 5.637 093 5.750 739 6 6.468 410 6.550 152 6.632 975 6.801 913 6.975 319 7.153 291 7 7.662 462 7.779 408 7.898 294 8.142 008 8.393 838 8.654 021 8 8.892 336 9.051 687 9.214 226 9.549 109 9.897 468 10.259 803 9 10.159 106 10.368 496 10.582 795 11.026 564 11.491 316 11.977 989 10 11.463 879 11.731 393 12.006 107 12.577 893 13.180 795 13.816 448 11 12.807 796 13.141 992 13.486 351 14.206 787 14.971 643 15.783 599 12 14.192 030 14.601 962 15.025 806 15.917 127 16.869 941 17.888 451 13 15.617 790 16.113 030 16.626 838 17.712 983 18.882 138 20.140 643 14 17.086 324 17.676 986 18.291 911 19.598 632 21.015 066 22.550 488 15 18.598 914 19.295 681 20.023 588 21.578 564 23.275 970 25.129 022 16 20.156 881 20.971 030 21.824 531 23.657 492 25.670 528 27.888 054 17 21.761 588 22.705 016 23.697 512 25.840 366 28.212 880 30.840 21? 18 23.414 435 24.499 691 25.645 413 28.132 385 30.905 653 33.999 033 19 25.116 868 26.357 180 27.671 229 30.539 004 33.759 992 37.378 965 20 26.870 374 28.279 682 29.778 079 33.065 954 36.785 591 40.995 492 21 28.676 486 30.269 471 31.969 202 35.719 252 39.992 727 44.865 177 22 30.536 780 32.328 902 34.247 970 38.505 214 43.392 290 49.005 739 23 32.452 884 34.460 414 36.617 889 41.430 475 46.995 828 53.436 141 24 34.426 470 36.666 528 39.082 604 44.501 999 50.815 577 58.176 671 25 36.459 264 38.949 857 41.645 908 47.727 099 54.864 512 63.249 030 26 38.553 042 41.313 102 44.311 745 61.113 454 59.156 383 68.676 470 27 40.709 634 42.759 060 47.084 214 54.669 126 63.705 766 74.483 823 28 42.930 923 46.290 627 49.967 583 58.402 583 68.528 112 80.697 691 29 45.218 850 48.910 799 52.966 286 62.322 712 73.639 798 87.346 529 30 47.575 416 51.622 677 56.084 938 66.438 848 79.058 186 94.460 786 31 50.002 678 54.429 471 59.328 335 70.760 790 84.801 677 102.073 041 32 52.502 759 57.334 502 62.701 469 75.298 829 90.889 778, 110.218 154 33 55.077 841 60.341 210: 66.209 527 80.063 771 97.343 165! 118.933 425 34 57.730 177 63.453 152 69.857 909 85.066 959: 104.183 755' 128.258 765 35 60.462 082 66.674 013 73.652 225 90.320 307i 111.434 780; 138.236 878 36 63.271 944 70.007 603| 77.598 314 95.836 323 119.120 867i 148.913 460 37 66.174 223 73.457 869 81.702 246 101.628 139 127.268 119 160.337 400 38 69.159 449 77.028 895 85.970 336! 107.709 546; 135.904 206: 172.561 020 39 72.234 233 80.724 906 90.409 150| 114.095 023! 145.058 458; 185.640 292 40 75.401 260 84.550 278| 95.025 516 120.799 774 ! 154.761 966; 199.635 112 41 78.663 298 88.509 537| 99.826 536' 127.839 763: 165.047 684 214.609 570 42 82.023 196 92.607 371 104.819 598, 135.231 751! 175.950 645 230.632 240 43 85.483 892 96.848 629 110.012 382; 142.993 339 187.507 5771 247.776 496 44 89.048 409 101.238 331 115.412 877) 151.143 006,'l99.758 032 266.120 851 45 92.719 861 105.781 673 121.029 392 159.700 156 212.743 514, 285.749 311 46 96.501 4571 110.484 031 126.870 568 168.685 164226.508 125 306.751 763 47 100.396 601 115.350 973 132.945 390 178.119 422 241.098 612] 329.224 386 48 104.408 396 120.388 297 139.263 206 188.025 393i256.564 529 353.270 093 49 108.540 648 125.601 846 145.833 734 198.426 663j272.958 401 378.999 000 50 112.796 867 130.999 910 152.667 084 209.347 9761290.335 905 406.528 929 ANNUITIES, 407 TABLE, SHOWING THE PRESENT WORTH OF AN ANNUITY OF ONE DOLLAR PER ANNUM, TO CONTINUE FOR ANY NUMBER OF YEARS NOT EXCEEDING FIFTY. g H 1 3 per cent. 3£ per cent. 4 per cent. 5 per cent. 6 per cent. 7 per cent. I l 0.970 874 0.966 184 0.961 538 0.952 381 0.943 396 0.934 579 2' 1.913 470 1.899 694 1.886 095 1.859 410 1.833 393 1.808 017 2 3 2.828 611 2.801 637 2.775 091 2.723 248 2.673 012 2.624 314 3 4 3.717 098 3.673 079 3.629 895 3.545 951 3.465 106 3.387 209 4 5 4.579 707 4.515 052 4.451 822 4.329 477 4.212 364 4.100 195 5 6 5.417 191 5.328 553 5.242 137 5.075 692 4.917 324 4.766 537 6 7 6.230 283 6.114 544 6.002 055 5.786 373 5.582 381 5.389 286 7 8 7.019 692 6.873 956 6.732 745 6.463 213 6.209 744 5.971 295 8 9 7.786 109 7.607 687 7.435 332 7.107 822 6.801 692 6.515 228 9 10 8.530 203 8.316 605 8.110 896 7.721 735 7.360 087 7.023 577 10 11 9.252 624 9.001 551 8.760 477 8.306 414 7.886 875 7.498 669 11 12 9.954 004 9.663 334 9.385 074 8.863 252 8.383 844 7.942 671 12 13 10.634 955 10.302 738 9.985 648 9.393 573 8.852 683 8.357 635 13 14 11.296 073 10.920 520 10.563 123 9.898 641 9.294 984 8.745 452 14 15 11.937 935 11.517 411 11.118 387 10.379 658 9.712 249 9.107 898 15 16 12.561 102 12.094 117 11.652 296 10.837 770 10.105 895 9.446 632 16 17 13.106 118 12.651 321 12.165 669 11.274 066 10.477 260 9.763 206 17 18 13.753 513 13.189 682 12.659 297 11.689 587 10.827 603 10.059 070 18 19 14.323 799 13.709 837 13.133 939 12.085 321 11.158 116 10.335 578 19 20 14.877 475 14.212 403 13.590 326 12.462 210 11.469 421 10.593 997 20 21 15.415 024 14.697 974 14.029 160 12.821 153 11.764 077 10.835 527 21 22 15.936 917 15.167 125 14.451 115 13.163 003 12.041 582 11.061 241 22 23 16.443 608 15.620 410 14.856 842 13.488 574 12.303 379 11.272 187 23 24 16.935 542 16.058 368 15.246 963 13.798 642 12.550 358 11.469 334 24 25 17.413 148 16.481 515 15.622 080 14.093 945 12.783 356 11.653 583 25 26 17.876 842 16.890 352 15.982 769 14.275 185 13.003 166 11.825 779 26 27 18.327 031 17.285 365 16.329 586 14.643 034 13.210 534 11.986 709 27 28 18.764 108 17.667 019 16.663 063 14.898 127 13.406 164 12.137 111 28 29 19.188 455 18.035 767 16.983 715 15.141 074 13.590 721 12.277 674 29 30 19.600 441 18.392 045 17-292 033 15.372 451 13.764 831 12.409 041 30 31 20.000 428 18.736 276 17.588 494 15.592 811 13.929 086 12.531 814 31 32 20.338 766 19.068 865 17.873 552 15.802 677 14.084 043 12.646 555 32 33 20.765 792 19.390 208 18.147 646 16.002 549 14.230 230 12.753 790 33 34 21.131 837 19.700 684 18.411 198 16.192 204 14.368 141 12.854 009 34 35 21.487 220 20.000 661 18.664 613 16.374 194 14.498 246 12.947 672 35 36 21.832 252 20.290 494 18.908 282 16.546 852 14.620 987 13.035 208 36 37 22.167 235 20.570 525 19.142 579 16.711 287 14.736 780 13.117 0171 37 38 22.492 462 20.841 087 19.367 864 16.867 893 14.846 019 13.193 473' 38 39 22.808 215 21.102 500 19.584 485 17.017 041 14.949 075 13.264 928| 39 40 23.114 772 21.355 072 19.792 774 17.159 086 15.046 297 13.331 709| 40 41 23.412 400 21.599 104 19.993 052 17.294 36S 15.138 016 13.394 120 41 42 23.701 359 21.834 883 20.185 627 17.423 208 15.224 543 13.452 449 42 43 23.981 902 22.062 689 20.370 795 17.545 912 15.306 173 13.506 962 43 44 24.254 274 22.282 791 20.548 841 17.662 773 15.383 182 13.557 908 44 45 24.518 713 22.495 450 20.720 040 17.774 070 15.455 832 13.605 522 45 40 24.775 449 22.700 918 20.884 654 17.880 067 15.524 370 13.650 020 46 47 25.024 708 22.899 438 21.042 936 17.981 016 15.589 028 13.691 608 47 48 25.266 707 23.091 244 21.195 131 18.077 158 15.650 027 13.730 474 48 49 25.501 657 23.276 564 21.341 472 18.168 722 15.707 572 13.766 799 49 50 25.729 764 23.455 618 21.482 185 18.255 925 15.761 861113.800 746 50 408 ANNUITIES. 574. To find the amount of an annuity, at compound in- terest, forborne, or in arrears, for any number of years. Ex. 1. What will an annuity of $ 60, unpaid, or in arrears, 4 years, amount to, at 6 per cent, compound interest ? Ans. $262,476. The amounts of the several operation. payments form a geometrical 1.06 4 — 1 series, of which the annuity is J^q 6 2 X 60 = 262.4/6. t i ie fi rst term, the amount of ' $ 1 for one year the ratio, the Or, 4.374616 X GO = 262.476. years the number of the terms, and the amount required is the sum of the series. Hence, Rule. — Find the sum of the series as in geometrical progression. 0r ' Multiply the amount of $ 1 for the given time, found in the table, by the annuity, and the product icdl be the required amount. Note. — The amount of an annuity at simple interest corresponds to the sum of an arithmetical series, of which the annuity is the first term, the interest on the annuity for one term the common difference, and the time in years the number of terms. 2. What will an annuity of $ 500 amount to for 5 years, at 6 per cent, compound interest? Ans. $2818.546. 3. What is the amount of an annuity of $ 80, unpaid, or in arrears, for 9 years, at 5 per cent, compound interest ? Ans. $882,125. 4. What is the amount of an annuity of $ 1000, forborne for - 15 years, at 3£ per cent, compound interest ? Ans. $19295.68. 5. What will an annuity of $ 30, payable semiannually, amount to, in arrears for 3 years, at 7 per cent, compound in- terest ? 6. Suppose a salary of $ 600 per year, payable quarterly, to remain unpaid 5J- years; to what sum will it amount, at 6 per cent, compound interest ? . Ans. $ 5404.295. 575. To find the present worth of an annuity, at compound interest. Ex. 1. What is the present worth of an annuity of $ 60, to be continued 4 years, at 6 per cent, compound interest ? Ans. $ 207.906. ANNUITIES. 409 operation. The present worth re- $ 3.465106 X 60 = $ 207.906-j-. quired evidently may be ob- tained by finding the amount of the given annuity, by the last articles, and then finding in the usual way the present worth of that amount. A more expeditious method, however, is to find, in the table, the present worth of an an- nuity of $ 1 for the given time and rate, and take that sum as many times as there are dollars in the given annuity, as in the operation. Rule. — Multiply the present worth of an annuity of $1 for the given time and rate by the number denoting the given annuity. 2. What is the present worth of an annuity of $100, for 9 years, at 6 per cent. ? Ans. $ 680.169. 3. What is the present worth of an annuity of $ 200, for 7 years, at 5 per cent. ? Ans. $ 1157.27. 4. Eequired the present worth of an annuity of $ 500, to continue 40 years, at 7 per cent. 5. A gentleman wishes to purchase an annuity, which shall afford him, at 6 per cent, compound interest, $ 500 a year, for ten years. What sum must he deposit in the annuity office to produce it ? Ans. $ 3680.04. 6. If a widow be entitled to $ 160 a year, payable semi- annually, from a fund, for 8 years, what is its value at present, at 6 per cent, compound interest? Ans. $ 1004.88. 576. To find the present worth of an annuity in perpetuity. Ex. 1. What is the present worth of a perpetual lease, which yields an income of $ 600, the rate of interest being that of 6 per cent. ? Ans. $ 10000. The question is evidently the same operation. as one requiring what principal in one $ 600 -f- .06 = $ 10000. year, at 6 per cent, interest, will yield $600. Rule. — Divide the given annuity by the number denoting the inter- est of%\ for one year. Note. — When the annuity is payable quarterly, semiannually, or in any other periods less than a whole year, the annuity must be increased by the in- terest which may thus accrue on the parts of the annuity payable before the end of the year, before dividing by the interest of $ 1 for one year. 2. A ground rent in the city of Philadelphia yields an an- nual income of $963, at 6 per cent, interest. What is the value of the estate ? Ans. $ 16050. N 35 410 ANNUITIES. 3. "What is the present value of a perpetual lease, yielding an income of $ 6335, interest being at 7 per cent. Ans. $ 90500. 4. What sum should be paid for a perpetual annuity of $ 1200, payable semiannually, interest being at 5 per cent. ? Ans. $ 24000. 577. To find the 'present worth of an annuity in rever- sion. Ex. 1. What is the present worth of an annuity of $ 300, to commence in 3 years, and to continue 5 years, allowing com- pound interest at 6 per cent.? Ans. $ 1061.03. operation. The present wort li $6.209794 — $2.673012 = $3.536772; of an annuity of Si, $ 3.53677 X 300 = $ 1061.03. . at G per cent., com- mencing at once, and continuing till the termination of the annuity, or for 3 -f- 5 = 8 years, is $ 6.209794 ; and the present worth of the same annuity up to the time of the commencement of the reversion is S 2.673012. The difference of these present worths multiplied by the number of dollars in the given annuity is the present worth of the reversion. Rule. — Find the present worth of an annuity of $ 1, commencing immediately and continuing till the reversion commences, and also till the reversion terminates ; and multiply the difference of these present worths by the number of dollars in the given annuity. The result will be the present worth required. 2. The reversion of a lease of $ 350 per annum, to continue 1 1 years, which commences 9 years hence, is to be sold. What is its worth, allowing the purchaser 6 per cent, per annum for his ready money ? Ans. $ 1633.70. 3. A father presents to his daughter, for 8 years, a rental of $ 70 per annum, payable yearly, and its reversion for the 12 years succeeding to his son. What is the present value of the gift to his son, allowing 4 per cent, compound interest ? Ans. $ 480.03. 4. What is the present worth of the reversion of a perpetuity of $ 240 per annum, payable yearly, but not to come into pos- session till the expiration of 100 years, compound interest being allowed at 6 per cent. ? Ans. $ 11.78. 578. To find the annuity, the present worth, time, and rate being given. PERMUTATIONS AND COMBINATIONS. 411 Ex. 1. What annuity, continued for 4 years, at 6 per cent, compound interest, will amount to a debt of $ 207.90 ? Ans. $ GO. The present value represented by operation. the debt, divided by the present $ 207.90 -r- 3.465 = $ 60. worth of $ 1 for the given time and rate, gives the annuity required. Rule. — Divide the given present worth by the present worth of an annuity of$ 1 for the given time and rate, and the result will be the an- nuity required. Note. — When the amount of an annuity, the time and rate, are given, the annuity may be found by dividing the given amount by the amount of $ 1 for the given time and rate. 2. The present value of an annuity, to be continued 10 years, at 6 per cent, compound interest, payable annually, is $ 3680.04 ; required the annuity. Ans. $ 500. 3. An annuity, remaining unpaid for 9 years, at 5 per cent, compound interest, amounted to $ 882.125 ; what was the an- nuity ? 4. A yearly pension which has been forborne for 6 years, at 6 per cent, amounts to $ 279 ; what was the pension ? Ans. $ 40. PERMUTATIONS AND COMBINATIONS. 579. Permutation is the process of finding the number of changes that can be made in the arrangement of any given number of things. 580. Combination shows how often a less number of things combined can be taken out of a greater, without respect to their order. 581 . To find the number of changes that can be made with any given number of things, taken all at once. Ex. 1. How many changes of order do the first three letters of the alphabet admit of? Ans. 6. 412 x PERMUTATIONS AND COMBINATIONS. opeiiation. By trial we shall find that two are all the 1 X 2 X ^ = 6. possible permutations that can be made of the first two letters of the alphabet ; as, a b and b a. If we take an additional letter, 6 are all the possible permuta- tions; as, a be, acb, bca,bac, cab, cba. Now, the same result may be obtained, in the case of the two letters, by multiplying together the first two digits, and in case of the three letters by multiplying to- gether the first three digits, as in the operation. Rule. — Multiply together all the terms of the natural series of num- bers, from 1 up to the given number, inclusive, and the product will be the number required. 2. How many changes may be rung on 6 bells ? Ans. 720 changes. 3. For how many days can 10 persons be placed in a differ- ent position at dinner ? 4. How many changes may be rung on 12 bells, and how long would they be in ringing, supposing 10 changes to be rung in one minute, and the year to consist of 365 days 5 hours and 49 minutes? Ans. 479001600, and 91y. 2Gd. 22h. 41m. 5. How many changes do the letters of the alphabet admit of? Ans. 403291461126605635584000000. 582. To find how many changes may be made by taking each time any number of different things less than all. Ex. 1. How many sets of 4 letters each may be formed out of 8 different letters ? Ans. 1680. operation. Ifc is evident that 8 X (8-1) X (8-2) X (8-3) S^bJSiS = 8X7X6X5 = 1680. before each of the oth- ers ; therefore 2 out of the 8 letters admit of 8 X 7 permutations. By taking 3 out of the 8 letters, the third letter can be arranged as the first, second, and third, in each of the same permutations, giving 8X7x6 permuta- tions. In like manner for 4 out of 8, we obtain 8X7x6x5 per- mutations. Rule. — Take a series of numbers, beginning with the number of things given, and decreasing by 1, until the number of terms equals the number of things to be taken at a time, and the product of all the terms icill be the answer required. 2. How many changes can be rung with 4 bells out of 6 ? Ans. 360. ANALYSIS BY POSITION. 413 3. How many words can be made out of the 26 letters of the alphabet, 6 being taken at once? Ans. 165765600. 583. To find the number of combinations that can be formed from a given number of different things, taken a given number at a time. Ex. 1. How many combinations can be made of 3 letters out of 4, the letters all being different ? Ans. 4. operation. We find the number of permuta- 4 X & X $ tions which may be made by taking 3 — - = y = 4. out of 4 letters, as in Art. 582, and the 1X^X3 number of permutations by the three letters all at once, as by Art. 581, and, dividing the first by the latter, obtain the number of combina- tions required. Rule. — Take tJie series, 1, 2, 3, 4, 5, fyc, up to ilie less number of things, and find the product of the terms, Take also a series of num- bers beginning with the greater number of things, and decreasing by 1 until the number of terms equals the less number of things, and find the product of the terms. The latter result divided by the former will give the number required. 2. How many combinations can be made of 7 letters out of 10, the letters all being different ? Ans. 120. 3. A successful general, being asked what reward would sat- isfy him for his services, demanded only a cent for every file of 10 men which he could make with a body of 100 men. What would his demand amount to? Ans. $ 173103094564.40. ANALYSIS BY POSITION. 584. Analysis by Position is the process of solving analytical questions, by assuming or supposing one or more numbers, and reasoning from them, operated upon as if they were the number or numbers required to be found. 585 • Questions in which the required number is in any way increased or diminished in any given ratio, or in which it is multiplied or divided by any number, may be solved by means of a single assumption. 35* 414 ANALYSIS BY POSITION. 586. Questions in which the required numbers, or their parts, or their multiples, are increased or diminished by some given number which is no known part or multiple of the re- quired number, or when a power or root of the required num- ber is either directly or indirectly contained in the result given in the question, may be solved by two assumptions. Note. — When the answer is obtained by means of a single assumption, the process is called Position; and when obtained by means of two assumptions, it is called Double Position. Analysis by Position affords often a very compendious method of working questions, whose solution otherwise, except by algebra, would be lengthy and difficult. It is even found very useful in shortening some of the processes of algebra. 587. "When the answer may be obtained by means of a single assumption or supposition. Ex. 1. A schoolmaster, being asked how many scholars he had, replied, that if he had as many more as he now has, and half as many more, he should have 200. Of how many schol- ars did his school consist ? Ans. 80 scholars. operation. By having as many more, and Assumed number, 6 half as many more, he would have As many more, 6 had 2 i times tlie <> rI g. m al number; tt u? o ^ therefore, the required number Half as many more, 3 „, ust bc ag many £, 200 containg \ 5 Q times 2£, or 80. The same result may be obtained, as in the opera- 150:200:: 60:8 0, Ans. tion, thus : We assume the number of scholars to be 60 ; if to 60 as many more, and half as many more, are added, the sum is 150. As this result has the same ratio to the result in the question as the sup* posed number has to the number required, we find the answer by a proportion. Rule. — Assume any convenient number, and proceed with it ac- cording to the nature of the question. Then, if the result be either too much or too little, as the result found is to the result given, so will be the number assumed to the number required. 2. A person, after spending -J and £ of his money, had $ 60 left ; what had he at first? Ans. $ 144. 3. What number is that, which, being increased by £, £, and ^ of itself, equals 125? 4. A's age is double that of B, and B's is triple that of C, and the sum of all their ages is 140. What is each person's age ? Ans. A's 84, B's 42, C's 14 years. ANALYSIS BY POSITION. 415 5. A person lent a sum of money at 6 per cent., and at the end of 10 years received the amount, $ 560. What was the sum lent? ' Ans. $350. 588 1 When two or more assumptions or suppositions are required in finding the answer. Ex. 1. A lady purchased a piece of silk at 80 cents per yard, and lining for it at 30 cents per yard ; the silk and lining contained 15 yards, and the price of the whole was $ 7. How many yards were there of each ? Ans. 5 yards of silk ; 10 yards of lining. OPERATION. Assume 6 yards of silk, $ 4.80 Assume 4 yards of silk, $ 3.20 Lining would be 9yd., 2.70 Lining would be 11yd., 3.30 Their sum, $7.50 Their sum, $6.50 Sum in the question, 7.00 Sum in the question, 7.00 First error, + $ °- 50 Second error, — $ 0.50 .50 -j- .50 : 6 — 4 : : .50 : 1 ; 6 yards — 1 yard = 5 yards ; 15 yards — 5 yards = 10 yards. Since the silk and lining contain 15 yards, cost $ 7.00, the average price per yard is 46§ cents; and 80 cents — 4G§ cents = 33i cents; 46§ cents — 30 cents — 16§ cents; and the quantity of lining will therefore be to that of the silk as 33^- is to 16|, or as to 2 is to 1. Hence, if the given number of yards, 15, be divided into 3 parts, two of those parts, or 10 yards, will be for the lining, and the other part, or 5 yards, will be for the silk. The same result is obtained as in the operation, thus : We assume the quantity of silk to be 6 yards ; then the lining would be 9 yards. We find the cost of each of these at the prices given, and, adding, have as the sum of their costs $ 7.50. This is a result too large by $ 0.50, when compared with the sum in question. By assuming the quantity of the silk to be 4 yards, and proceeding in a like manner, we obtain for the cost of the silk and the lining $6.50, a sum too small by $0.50. The first error, arising from a result too large, is marked with the sign plus (-)-), and the second error, arising from a result too small, is marked with the sign minus ( — ). Then, since one of the results is too large, and the other too small, we then say, as the sum of the errors, or .50 -)- .50, is to the difference of the as? sumptions, or 6 — 4, so is .50 the less error, to 1, the correction; which, being added to 4 (yards), the sum, 5, expresses the number of yards of silk; and consequently that of the lining must be 10 yards. Rule. — Assume two different numbers, perform on them separately the operations indicated in the question, and note the errors of the re- 416 ANALYSIS BY POSITION. suits. Then, as the difference of the errors, if the results be both too great or both too small, or as the sum of the errors, if one result be too great and the other too small, is to the difference of the assumed num- bers, so is either error to the correction to be applied to the number which produced that error. Note 1. — The rule usually given fails in an "important class of questions; but the rule here given, if not the simplest in the resolution of some questions, has the advantage of being applicable in every case. Note 2. — In relation to all questions which in algebra would be resolved by equations of the first degree, the differences between the true and the as- sumed numbers are proportional to the differences between the result given in the question and the results arising from the assumed numbers. But the prin- ciple does not hold exactly in relation to other questions ; hence, when applied to them, the above rule, or any other of the kind that can be given, will only produce approximations to the true results. In which case the assumed num- bers should be taken as nearly true as possible. Then, to approximate more nearly to the required number, assume for a second operation the number found by the first, and that one of the first two assumptions which was nearer the true answer, or any other number that may appear to be still nearer to it. In this way, by repeating the operation as often as may be necessary, the true results may be approximated to any assigned degree of accuracy. This pro- cess is sometimes applied with advantage in extracting the higher roots, when approximate results, differing but slightly from entire correctness, will answer. 2. A and B invested equal sums in trade ; A gained a sum equal to £ ot his stock, and B lost $ 225 ; then A's money was double that of B's. What did each invest ? Ans. $ GOO. 3. A person, being asked the age of each of his sons, replied, that his eldest son was 4 years older than the second, his sec- ond 4 years older than the third, his third 4 years older than the fourth, or youngest, and his youngest half the age of the oldest. What was the age of each of his sons ? Ans. 12, 16, 20, and 24 years. 4. A gentleman has two horses, and a saddle worth $ 50. Now if the saddle be put on the first horse, it will make his value double that of the second horse ; but if it be put on the second, it will make his value triple that of the first. What was the value of each horse ? Ans. The first, $ 30 ; second, $ 40. 5. A gentleman was asked the time of day, and replied, that § of the time past from noon was equal to ■£§ of the time to midnight. What was the time ? Ans. 12 minutes past 3. 6. A and B have the same income. A saves y 1 ^ of his, but B, by spending $100 per annum more than A, at the end of 10 years finds himself $ GOO in debt. What was their income ? Ans. $ 480. SCALES OP NOTATION. 417 7. A gentleman hired a laborer for 90 days on these condi- tions : that for every day he wrought he should receive 60 cents, and for every day he was absent he should forfeit 80 cents. At the expiration of the term he received $ 33. How many days did he work, and how many days was he idle ? Ans. He labored 75 days, and was idle 15 days. 8. There is a fish whose head weighs 15 pounds, his tail weighs as much as his head and £ of his body, and his body weighs as much as his head and tail. What is the weight of the fish? ' Ans. 721b. 9. If 12 oxen eat 3£ acres of grass in 4 weeks, and 21 oxen eat 10 acres in 9 weeks, how many oxen would it require to eat 24 acres in 18 weeks, the grass growing uniformly ? Ans. 36 oxen. 10. What number exceeds three times its square root by 11 ? (Note 2.) Ans. 26.4201648. SCALES OF NOTATION. 589. The scale of any system of notation is the law of relation existing between its units of different orders. 590. The radix of any scale is the number of units it takes of one order to make a unit of the next higher. Thus, 10 is the radix of the decimal or denary system, 2 of the binary, 3 of the ternary, 4 of the quaternary, 5 of the quinary, 6 of the senary, 7 of the septenary, 8 of the octary, 9 of the nonary, 11 of the undenary or undecimal, 12 of the duodenary or duo- decimal, 20 of the vigesimal, 30 of the trigesimal, 60 of the sex- agesimal, and 100 of the centesimal, 591 . In writing any number in a uniform scale, as many distinct characters or symbols are required as there are units in the radix of the given system. Thus, in the decimal or denary scale 10 characters are required, in the binary scale 2 charac- ters, in the duodenary or duodecimal 12 characters, and so on. In the binary scale use is made of the characters 1 and 0, in 418 SCALES OP NOTATION. 6)75432 OPERATION. 12)75432 6)12572 12)6286 6)2095 2 12)523 11, or e 6)349 1 12)43 7 6)58 1 3 7 6)9 4 1 3 the ternary, 1, 2, and 0, &c, the cipher being one of the char- acters in each scale. In the duodenary scale, eleven charac- ters being required beside the cipher, the first nine may be supplied by the nine digits, the tenth by t, the eleventh by e, and the twelfth by 0. 592. To change any number expressed in the decimal scale to any other required scale of notation. Ex. 1. Express the common number 75432, in the senary and duodenary scales. Ans. 1341120 and 377e0. By dividing the given number by 6, it is distribut- ed into 12572 classes, each containing 6, with remain- der. By the second division by 6, these classes are dis- tributed into 2095 classes, each containing 6 times 6, or the second power of 6, with a remainder of 2 of the former class, each containing 6. By the third division the classes last found are distributed into 349 classes, each containing 6 of the latter, which were each the second power of 6, and therefore these are the third power of 6, with a remainder 1 time the second power of 6 , In like manner, the next quotient expresses 58 times the fourth power of 6, with a remainder 1 time the third power of 6 ; the next quotient expresses 9 times the fifth power of 6, with a re- mainder 4 times the fourth power of 6 ; and the last quotient ex- presses 1 time the sixth power of 6, with a remainder 3 times the fifth power of 6. Hence, the given number is found to be equal to 1 X 6 6 + 3 X 6 5 + 4 X 6 4 + 1 X 6 3 + 1 X 6 2 + 2 X 6 + 0, or according to the senary system of notation 1341120. By proceeding in like manner, we find the given number to be equal to 3 X 12 4 + 7 X 12 3 + 7 X 12 2 + 11 X 12 + 0, or, ac- cording to the duodenary scale, 377e0. Rule. — Divide the given number by the radix of the required scale repeatedly ', till the quotient is less than the radix ; then the last quotient, icith the several remainders in the retrograde order annexed, placing ciphers where there is no remainder, will be the the given number ex- pressed in the required scale, 2. Change 37 from the decimal to the binary scale. Ans. 100101. 3. Reduce 1000000 in the decimal scale to the ternary and also to the nonary. Ans. 1212210202001, and 1783661. SCALES OP NOTATION. 419 4. How will 476897 in the decimal scale be expressed in the duodecimal scale ? Ans. Itee95. 593 • To change any number into the decimal scale, when expressed in any other scale of notation. Ex. 1. Change 377#) from the duodecimal to the decimal scale. Ans. 75432. OPERATION. We multiply the left-hand figure by the ra- dix, and add to the product the next figure ; then we multiply this sum by the radix, and add to the product the next figure, and so proceed till all the figures have been em- ployed ; and we thus have, as the values of the several figures collected into one sum, 75432, obtained m a manner similar to the reduction of compound numbers. 75432 Rule. — Multiply the left-Jiand figure of the given number by the given radix, and to the product add the next figure ; then multiply this sum by the radix, and add to this product the next figure; and so proceed till all the figures of the given number have been added. The result will be the given number in the decimal scale. Note. — When it is required to change a number from a scale other than decimal to another scale also other than decimal, first change the number as given into the decimal scale, and then the result into the required scale. 2. Reduce 234 from the quinary to the decimal scale. Ans. 69. 3. Change 21122 from the ternary to the decimal scale. Ans. 206. 4. Change 100101 in the binary scale to a number in the decimal scale. Ans. 37. 5. Reduce 13579 in the duodecimal scale to the undecimal scale. Ans. 190^3. 6. How will 123454321 in the senary scale be expressed in the duodenary scale? Ans. 9873d. 594 # To perform addition, subtraction, multiplication, di- vision, &c. in a scale of notation whose radix is other than 10, we may Proceed as in the common scale of notation, except that the radix of the given scale must be used in the cases wherein the number 10 would be applied in the decimal system. 420 DUODECIMALS. Ex. 1. Required the sum and difference of 45324502 and 25405534 in the senary scale, or scale whose radix is 6. Ans. Sum, 115134440; difference, 15514524. 2. Multiply 2483 by 589 in the undenary scale, or scale whose radix is 11. Ans. 13122^5. 3. Divide 1184323 by 589 in the duodenary scale, whose radix is 12. Ans. 2483. 4. Extract the square root of 11122441 in the senary scale. Ans. 2405. DUODECIMALS. 595. Duodecimals are numbers expressed in a scale whose radix is 12, so that 12 units of each lower order make a unit of the next higher. 596. In finding the contents of surfaces and solids, how- ever, it is customary to apply the term duodecimal to a mixture of the decimal and duodecimal scales. Thus, in admeasure- ments in which the foot is the leading unit, though the differ- ent orders of units are expressed according to the duodecimal scale, the number of units in each order is usually expressed according to the decimal scale. 597. According to this mixed scale, the foot is divided into 12 equal parts, and each of these parts into 12 other equal parts, and so on indefinitely, giving y 1 ^, T¥f , &c. In writing these fractions without their denominations, to distinguish their orders, or denominations, accents, called indices, are written on the right of the numerators. Thus, inches are called primes, and are marked '; the next subdivision is called seconds, marked " ; the next is thirds, marked '" ; and so on. Note. — Numbers expressed by the mixed scale of feet, primes, seconds, &c. may be changed to the pure duodecimal scale, and the operations of addi- tion, subtraction, multiplication, division, and so on, then be performed with them, as in Art. 594, observing to place a point between the unit and its lower duodecimal orders, and in the result changing the figures on the left of the point into the decimal scale, and marking those on the right as primes, sec- onds, &c, according to their places from the order of units. But the operations of adding, subtracting, &c. are usually performed by other methods, such as are given in the articles that follow. DUODECIMALS. 421 ADDITION AND SUBTRACTION OF DUODECIMALS. 598. Duodecimals may be added and subtracted in the same manner as compound numbers. Ex. 1. Add together 121ft. 3' 9", 105ft 11' 8", 80ft. 0' 6", and 15ft. 10' 0" 4"'. Ans. 323ft. V 11" 4'". 2. From 462ft. 4' 9" take 307ft. 9' 1". 3. What is the value of 92ft. 0' 6" — 21ft. 9' 10" -f- 19ft. 10' 3" 6"'? Ans. 90ft. 0' 11" 6'". MULTIPLICATION OF DUODECIMALS. 599. The index of the unit of a product of any two duode- cimal orders is equal to the sum of the indices of those factors. That is, feet multiplied by a number denoting feet produces feet ; feet by a number denoting primes ^produces primes ; primes by a number denoting primes produces seconds, &c. Note. — In multiplication of duodecimals, or in other multiplication, the multiplier is always regarded as an abstract number, though the notation of feet, primes, &c. is usually retained, in order the better to note the different orders of units. For the same reason, in division of duodecimals, the divisor usually retains the notation of feet, primes, &c. 600. To multiply one duodecimal by another. Ex. 1. Required the number of square feet in a platform 6 feet 8 inches long, and 4 feet 5 inches wide. Ans. 29 sq. ft. 5' 4". We first multiply each of the terms in the multi- plicand by the 5' in the multiplier ; thus, 8'X 5'=40" = 3' and 4". Writing the 4" under the multiplier, we reserve the 3' to add to the next product. Then 6ft. X 5' = 30' ; and 30' -|- 3' = 33' = 2ft. and 9', which we write in their order beneath the multiplier. We next multiply by the 4ft., thus: 29sq.ft. 5' 4" 8' X 4 feet = 32' = 2ft. and 8'. We write the 8' under the primes in the other partial product, and reserve the 2ft. to add to the next product ; and 6ft. X 4$. = 24ft. ; 24ft. -\- 2ft. = 26ft., which we write under the feet in the other par- tial product. The two being added together, we have 29 sq. ft. 5' 4"; or (the primes and seconds being changed to a fraction of a foot), 29/^sq. ft. N 36 FIRST OPERATION. 6ft. 8' 4ft. 5' 2 9/4// 26 8' SECOND OPERATION, 6-8 4-5 294 228 422 DUODECIMALS. In the second operation the work is performed as in pure duodecimals (Art. 594). The point (•) separates lower duodecimal orders from those be- ginning with feet. As to the number of feet in the multiplicand and multi- plier, no change is required in that 2 5*5 4 = 29sq. ft. o' &.", p ar t f either of the given numbers in expressing them according to the duo- decimal scale. In performing the multiplication, we make the several reductions required according to the radix 1 2 ; and have, after point- ing off, 25*54 square feet, expressed according to the duodecimal scale. The units, or feet, at the left of the point, are readily- changed to the decimal scale by multiplying the left-hand figure, 2, by 12, the number of units in the radix, and adding the right-hand figure, 5, and giving the figures to the right of the point their proper notation, we have then 29sq. ft. 5' 4" for the answer, as before. Rule. — Write the multiplier under the multiplicand, so that units of the same orders shall stand in the same column. Beginning at the right, multiply each term in the midtiplicand by each term of the multiplier, and write the first term of each partial product under its multiplier, observing to carry a unit for every twelve from each lower order to the next higher. The sum of the partial product will be the product required. 2. How many square feet in a floor 48 feet 6 inches long, and 24 feet 3 inches broad ? Ans. 1176 sq. ft. 1' G". 3. The length of a room being 20 feet, its breadth 14 feet 6 inches, and height 10 feet 4 inches, how many square yards of painting are in it, deducting a fireplace of 4 feet by 4 feet 4 inches, and 2 windows, each 6 feet by 3 feet 2 inches ? Ans. 73/ T square yards. 4. Required the solid contents of a wall 53 feet 6 inches long, 10 feet 3 inches high, and 2 feet thick. 5. There is a house with four tiers of windows, and four windows in a tier ; the height of the first is 6 feet 8 inches ; the second, 5 feet 9 inches ; the third, 4 feet 6 inches ; the fourth, 3 feet 10 inches ; and the breadth is 3 feet 5 inches ; how many square feet do they contain in the whole ? Ans. 283sq. ft. Tin. 6. How many cords in a pile of wood 97 feet 9 inches long, 4 feet wide, and 3 feet 6 inches high ? Ans. 10^£ cords. 7. Required the number of cords of wood in a pile 100 feet long, 4 feet wide, and 6 feet 11 inches high. Ans. 21 1|. DUODECIMALS. 423 DIVISION OF DUODECIMALS. 601 . To divide one duodecimal by another. Ex. 1. A board in the form of a rectangle, whose area is 27 sq. ft. 8' 6", is 1ft. 7in. wide ; what is its length ? Ans. 17ft. Gin. first operation. We find how many times 27 lft. 7') 27sq.ft. 8' 6" ( 17ft. 6' square feet contains the divisor, 26 11 and obtain 17 feet for the quo- tient, we multiply the entire di- 9 6 visor by the 17ft., and subtract 9 6 the product, 26ft. 11', from the corresponding portion of the div- idend, and obtain 9', to which remainder we bring down the 6", and, dividing, obtain 6' for the quotient. Multiplying the entire divisor by 6", we obtain Oft. 9' 6", which, subtracted as before, leaves no remainder. Therefore 1 7 feet 6 inches is the length required. second operation. I n the second operation we re- 1-7 ) 23-86 ( 15-6 as 17ft. 6' duce the feet of the given multi- yj plicand and multiplier to the duo- ■ decimal scale, and thus obtain 88 23-86 and 1-7. We then conduct 7 e the division with reference to the ^TT radix 12, as is ordinarily done with ^ respect to 10 (Art. 594 ). The re- ^k suit obtained is 15-6 in the duo- decimal, which, on changing the figures to the left of the point to the decimal scale, and giving the proper notation to the figure on the right of the point, becomes trans- formed to 17ft. 6' = 17ft. 6in., the answer, as before. Rule. — Find how many times the highest term of the dividend will contain the divisor. By this quotient multiply the entire divisor, and subtract the product from the corresponding terms of the dividend. To the remainder annex the next denomination of the dividend, and divide as before, and so continue till the division is complete. 2. It required 834 sq. ft. 3' of board to cover the side of a certain building. The height was 17ft. 9in. ; what was the length of the side ? Ans. 47 feet. 3. How many feet wide is a plank of uniform width, whose length is 18ft 9in., thickness 3 inches, and solid contents 84ft. 4' 6"? 4. An alley has an area of 792ft. 6' 9" 2 //; . Its width is 12ft. 7' 8". Eequired its length. Ans. 62ft. 8' 6". '424 MISCELLANEOUS EXAMPLES, MISCELLANEOUS EXAMPLES. 1. A merchant engages a clerk at the rate of $ 20 for the first month, $ 25 for the second, $ 30 for the third, &c, thus in- creasing his salary by $ 5 per month. How long must the clerk retain his situation, so as to receive on the whole as much as he would have received had his salary been fixed at $ 52.50 per month ? Ans. 14 months. 2. A mason has plastered 3 rooms ; the ceiling of each is 20 feet by 16 feet 6 inches, the walls of each are 9 feet 6 inches high, and 90 yards are to be deducted for doors, windows, &c. For how many yards must he be paid ? Ans. 251yd. 1ft. Gin. 3. A man of wealth, dying, left his property to his ten sons, and the executor of his will, as follows : to his executor, $ 1024 ; to his youngest son, as much and half as much more ; and increasing the share of each next elder in the ratio of 1J-. What was the share of the eldest? 4. A butcher, wishing to buy some sheep, asked the owner how much he must give him for 20 ; on hearing his price, he said it was too much ; the owner replied, that he should have 10, provided he would give him a cent for each different choice of 10 in 20, to which he agreed. How much did he pay for the 10 sheep, according to the bargain ? Ans. $ 1847.56. 5. If 340 square feet of carpeting are required to cover the floor of a room, how many yards will be required, provided the width of the carpeting is 3 feet 9 inches ? Ans. 30yd. 8in. 6. If a clergyman's salary of $ 700 per annum is 6 years in arrears, how much is due him, allowing compound interest at 6 per cent. ? Ans. $ 4882.72. 7. Suppose a clock to have an hour-hand, a minute-hand, and a second-hand, all turning on the same centre. At 12 o'clock all the hands are together and point at 12. (1.) How long will it be before the second-hand will be be- tween the other two hands, and at equal distances from each? Ajis. 60 t 7 ^° 7 seconds. (2.) Also before the minute-hand will be equally distant be- tween the other two hands? Ans. 61fff seconds. (3.) Also before the hour-hand will be equally distant be- tween the other two hands ? Ans. 59^-f seconds. MENSURATION. 425 MENSURATION. DEFINITIONS. 602. ' A point is that which has neither length, breadth, nor thick- ness, but position only. » 603. A line is length, without breadth or thickness. A straight line is one which has the same direction in its whole extent ; as the line A B. A curved line is one which continually changes its direction ; as the line C D. C^ \d 604. An angle is the inclination or opening of two lines, which meet in a point. A A right angle is an angle formed by a straight line and a perpendicular to it ; as the angle ABC. An acute angle is one less than a right angle ; as the angle E B C. B An obtuse angle is one greater than a right angle ; ^ as the angle F B C. B 605. A surface is that which has length and breadth, without thickness. A plane surface, or simply a plane, is that in which, if any two points whatever be taken, the straight line that joins them will lie wholly in it. Every surface, which is not a plane, or composed of planes, is a curved surface. 606. The area of a figure is its quantity of surface ; and is es- timated in the square of some unit of measure, as a square inch, a square foot, &c. 607. A solid, or body, is that which has length, breadth, and thickness. 608. The solidity, or volume of a solid, is estimated in the cube of some unit of measure ; as a cubic inch, a cubic foot, &c. 609. Mensuration is the process of determining the areas of surfaces, and the solidity or volume of solids. MENSURATION OF SURFACES. 610. A plane figure is an enclosed plane surface ; if bounded by straight lines only, it is called a rectilineal figure, or polygon. The perimeter of a figure is its boundary, or contour. 36* 426 MENSURATION. 61 1 • Three-sided polygons are called triangles ; those of four sides, quadrilaterals ; those of five sides, pentagons, and so on. Triangles. A 612. An equilateral triangle is one whose sides are all equal ; as CAD. Notk. — The line AB, drawn from the angle A perpendic- ular to the base C I), is the altitude of the triangle GAD. q An isosceles triangle is one which has two of its sides equal ; as E F G. A scalene triangle is one which has its three sides unequal ; as II I J. ii A riglit-angled triangle is one which has a right angle ; as KLM. 613. To find the area of a triangle. j\hdtiphj the base by half the altitude, and the product will be the area. Or, Add the three sides together, take half that sum, and from this sub- tract each side separately ; then multiply the half of the sum and these remainders together, and the square root of this product will be the area. Ex. 1. What are the contents of a triangle whose perpendicular height is 12 feet, and whose base is 18 feet ? Ans. 108 feet. 2. There is a triangle, the longest side of which is 15.6 feet, the shortest side 9.2 feet, and the other side 10.4 feet. What are the contents ? Ans. 46.139-f- feet 3. The triangular gable of a certain building has a base of 40 feet and an altitude of 15 feet ; how many square feet of boards will cover it ? Ans. 300 sq. ft. 4. The perimeter of a certain field in the form of an equilateral triangle is 336 rods ; what is the area of the field ? Ans. 33 acres 152 sq. rd. Quadrilaterals. 61 4# A parallelogram is any quadrilateral whose opposite sides d , , c are parallel. 615. A rectangle is any right-angled parallel- ogram ; as ABCD. MENSURATION. 427 H a E 616. A square is a parallelogram whose sides are equal, and whose angles are right angles ; asEFGH. 61 7# A rhombus is a parallelogram whose sides are equal, and whose angles are not right angles ; as IJKL. 618. A rhomboid is a parallelogram whose an- J\ / gles are not right angles ; as M N O P. / \ / M Q N Note. — The altitude of a parallelogram is the perpendicular distance be- tween any two of its parallel sides taken as bases, as the line P Q, drawn between two sides of the rhomboid M N P, and perpendicular to the sides MN and OP. U T K Z 619. A trapezoid is a quadrilateral which has only two of its sides parallel ; as KSTU. 620. A trapezium is a quadrilateral which has no two sides parallel ; asWXYZ. w Note. — A diagonal of a quadrilateral, or of any polygon of more than four sides, is a straight line which joins the vertices of two opposite angles, or of two angles not adjacent; as the line XZ joining vertices of opposite angles of the trapezium \Y X Y Z. 621. To find the area of a parallelogram. Multiply the base by the altitude, and the product will be the area, Ex. 1. What are the contents of a board 15 feet long and 2 feet wide ? Ans. 30 feet. 2. A rectangular state is 128 miles long and 48 miles wide. ^ How many square miles does it contain ? Ans. 6144 miles. 3. The base of a rhomboid being 12 feet, and its height 8 feet, re- quired the area. Ans. 96 feet. 4. Required the area of a rhombus of which one of the equal sides is 358 feet, and the perpendicular distance between it and the oppo- site side is 194 feet. Ans. 69452 sq. ft. 5. The largest of the Egyptian pyramids is square at its base, and measures 693 feet on a side. How much ground does it cover ? Ans. 1 1 acres 4 poles. 6. What is the difference between the area of a floor 40 feet square, and that of two others, each 20 feet square ? Ans. 800 feet. 7. There is a square whose area is 3600 yards ; what are the sides 428 MENSURATION. of the square, and the breadth of a walk along each side and each end of the square, which shall take up just one half of the square ? A ( 42.42-4- yards, side of the square. Ans * j 8.78+ yards, breadth of the walk. 622 • To find the area of a trapezoid. Multiply half of the sum of the parallel sides by the altitude, and the product is the area. Ex. 1. If the parallel sides of a trapezoid are 75 and 33 feet, and the perpendicular breadth 20 feet, what is the area ? Ans. 1080 sq. ft. 2. Required the area of a meadow in the form of a trapezoid, whose parallel sides are 786 and 473 links, and whose altitude is 986 links. Ans. 6 acres 33 rods 3 yards. 623* To find the area of a trapezium. Divide the trapezium into two triangles hy a diagonal, and then find the areas of these triangles ; their sum will be the area of the trape- zium. Ex. 1. Required the area of a garden in the form of a trapezium, of which the four sides are 328, 456, 572, and 298 feet, and the diag- onal, drawn from the angle between the first and second sides, 598 feet. Ans. 3 acres 1 rood 31 rods 29 yards 3.85 feet. 2. Given one of the diagonals of a field, in the form of a trapezium, equal 1 7 chains 56 links, to compute the area, the perpendiculars to that diagonal from the opposite angles being 8 chains 82 links, and 7 chains 73 links. Ans. 14 acres 2 roods 5 rods. Pentagons, Hexagons, &c. 624. A pentagon is a polygon of five sides ; a hexagon, one of six sides ; a heptagon, one of seven sides ; an octagon, one of eight sides ; a nonagon, one of nine sides; and so on for a decagon, undecagon, dodecagon, &c. D 625. A regular polygon is one whose sides and angles are equal ; as the pentagon ABODE. 626. To find the area of a regular polygon. Multiply the perimeter by half the perpendicular let fall from the centre upon one of the sides. Or, Multiply the square of one of the sides by the number against the poly- yon in the following Table. Pentagon, 1.720477 Nonagon, 6.181824 Hexagon, 2.598076 Decagon, 7.694209 Heptagon, 3.633913 Undecagon, 9.365641 Octagon, 4.828427 Dodecagon, 11.196152 MENSURATION. 429 Ex. 1. What is the area of a regular pentagon, of which the side is 250 feet, and the perpendicular from the centre to one side 172.05 feet ? Ans. 107531.25 sq. ft. 2. What is the area of a regular hexagon whose side is 356 yards, and whose perpendicular is 308.305 yards ? Ans. 329269.74yd. 3. The side of a regular octagonal enclosure is 60 yards ; how many acres are included ? Ans. 3 acres 2 roods 14 rods 19 yards. 4. The side of a field, whose shape is that of a regular decagon, is 243 feet ; what is its area ? Ans. 10 acres 1 rood 28 rods 24 yards 6.347 feet. Circles. 627 • A circle is a plane figure bounded by a line, every part of which is equally distant from a point within called the centre ; as AEFGBD. The circumference or periphery of a circle is the line that bounds it. A radius is a line drawn from the centre to the circumference ; as C A, or C D. A diameter is a line which passes through the centre, and is termi- nated by the circumference ; as A B. An arc is any portion of the circumference ; as A D, A E, or E G F. The chord of an arc is the straight line joining its extremities ; as E F, which is the chord of the arc E G F. 628 1 The segment of a circle is the portion included by an arc and its chord; as the space included by the arc EGF and the chord E F. 629. The sector of a circle is the portion included by two radii and the intercepted arc ; as the space ACDA. 630 • A zone is the space between two parallel chords of a circle ; as the space A E F B A. O 631. A lune, or crescent, is the space included between the intersecting arcs of two eccentric circles ; as A C B E A. 632. A circular ring is the space included between the circumference of two concentric A circles ; as the space between the rings A B and CD. 430 MENSURATION. 633 • To find the circumference of a circle, the diameter being given. Multiply the diameter by 3.141592. Ex. 1. If the diameter of a circle is 144 feet, what is the circum- ference ? Ans. 452.389248 feet. 2. If the diameter of the earth is 7964 miles, what is its circum- ference? Ans. 25019.638688-j- miles. 3. Required the circumference of a circle, whose radius is 512 feet. Ans. 4 furlongs 34 rods 5 yards 1 foot. 634. To find the diameter of a circle, the circumference being given. Multiply the circumference by .318309. Ex. 1. Required the diameter of a circle, whose circumference is 1043 feet. Ans. 331.997-J- feet. 2. If the circumference of a circle is 25000 miles, what is its diam- eter? Ans. 7957.74-}- miles. 3. If the circumference of a round stick of timber is 50 inches, what is its diameter ? Ans. 15.91549-f- inches. 635* To find the area of a circle, the diameter, or the circum- ference, or both, being given. Multiply the square of the diameter by .785398. Or, Multiply the square of the circumference by .079577. Or, Multiply half the diameter by half the circumference. Ex. 1. If the diameter of a circle is 761 feet, what is the area ? Ans. 454840.475158 feet 2. There is a circular island, three miles in diameter ; how many- acres does it contain ? Ans. 4523. 89-f- acres. 3. Required the area of a circle, of which the circumference is 1284 yards. Ans. 27 acres 17 rods 0.8-j- yards. 4. Required the area of a circle, of which the diameter is 169, and the circumference 532 inches. Ans. 17 yards 3 feet 13 inches. 636. To find the area of a sector of a circle. Multiply the length of the arc by half the radius of the circle. Or, As 360° are to the degrees in the arc of the sector, so is the area of the circle to the area of the sector. Ex. 1 . Required the area of a sector, of which the arc is 79 and the radius of the circle 47 inches. Ans. 1856.5 inches. 2. Required the area of a sector, of which the arc is 26°, and the radius of the circle 25 feet. Ans. 141.8 square feet. 637 • To find the area of the segment of a circle. Find the area of the sector which has the same arc with the segment ; and also the area of the triangle formed by the chord and the radii drawn to its extremities. The difference of these areas, ichen the seg- ment is less, and their sum, when the segment is greater, than the semi- circle, will be the area of the segment. Or, I MENSURATION. 431 To two thirds of the 'product of the height of the segment by the chord add the cube of the height, divided by twice the chord, Ex. 1. Required the area of the segment ABC A, of which the arc ABC is 49.25°, the chord AC 10 feet, and the radii E A, E B, and E C, each 12 feet. Ans. 7.35 sq. ft. 2. Required the area of a segment whose height is 15 rods and whose chord is 24 rods. Ans. 1 acre 3 roods 30 rods 9.4 yards. 638. To find the area of a zone of a circle. From the area of the whole circle subtract the areas of the segments on the sides of the zone, Ex. 1. Required the area of a zone whose parallel sides are 23.25 and 20.8 feet, in a circle whose radius is 12 feet. Ans. 206 sq. ft. 2. Required the area of a zone included between two chords of 1 6 feet each, the diameter of the circle being 20 feet. Ans. 224.7 sq. ft. 639 # To find the area of a lune or crescent. Find the difference of the areas of the two segments formed by the arcs of the lune and its chord. Ex. 1. If the chord of two intersecting arcs is 72 feet, and the height of one of the segments is 30, and of the other 20 feet, what is the area of the crescent ? Ans. 612 sq. ft. 640 1 To find the area of a circular ring. Multiply the sum of the diameters of the two circles by the difference of the diameters, and that product by .78.54. Ex. 1. What is the area of the ring formed by two circles whose diameters are 10 and 20 yards? Ans. 235.62 sq. yd. 2. In the centre of a circular pond there is an island 128 yards in diameter ; what is the area of the pond, provided the exact distance from any part of the outer side of the pond to the centre of the island is 78^ yards ? Ans. 1 acre 1 rood 14 rods 17 yards 7.4 feet. 641. To find the side of a square that shall equal the area of a circle of a given diameter or circumference. Multiply the diameter of the circle by .886227. Or, Multiply the circumference of the circle by .282094. Ex. 1. I have a round field, 50 rods in diameter ; what is the side of a square field that shall contain the same area ? Ans. 44.31135-f-rods. 2. I have a circular field 360 rods in circumference ; what must be the side of a square field that shall contain the same area V Ans. 101.55-j- rods. 3. John Smith had a farm which was 10,000 roils in circumference, which he sold at $ 71.75 per acre. He purchased another farm con- 432 MENSURATION. taining the same quantity of land in the form of a square ; required the length of one of its sides. Ans. 2820.94-j- rods. 642 1 To find the diameter of a circle that shall contain the area of a given square. Multiply the side of the given square by 1.12838. Ex. 1. The side of a square is 44.31135 rods ; required the diam- eter of a circular field containing the same area. 643. To find the side of the. largest equilateral triangle that can be inscribed in a circle of a given diameter or circumference. Multiply the given diameter by .866025. Or, Multiply the given circumference by .275664. Ex. 1. There is a certain piece of round timber 30 inches in di- ameter ; required the side of an equilateral triangular beam that may be hewn from it. Ans. 25.98-)- inches. 2. How large an equilateral triangle may be inscribed in a circle whose circumference is 5000 feet ? Ans. 1378.323 feet 3. Required the side of an equilateral triangular beam, that may be hewn from a round piece of timber 80 inches in circumference. Ans. 22.05-f- inches. 644 • To find the side of the largest square that can be inscribed in a circle of a given diameter or circumference. Multiply the given diameter by .707106. Or, Multiply the given circumference by .225079. Note. — To find the circumference of a circle required to exactly admit a square of a given side, divide the given side by .225079. Ex. 1. I have a piece of timber 30 inches in diameter ; how large a square stick can be hewn from it? Ans. 21.214- in. square. 2. Required the side of a square that may be inscribed in a circle 80 feet in diameter. Ans. 56.56848-f- fret. 3. I have a circular field whose circumference is 5000 rods ; what is the side of the largest square field that can be made in it ? Ans. 1125.395+ rods. 4. How large a square stick may be hewn from a piece of round timber 100 inches in circumference ? Ans. 22.5-f- inches square. 5. What must be the circumference of a tree that, when hewn, shall be 18 inches square? Ans. 79.9 7— [— inches. 6. I have a garden which is 20 rods square ; required, in feet, the circumference of a circle that will enclose this garden. Ans. 1466.15-[- feet. 645. To find the diameter of the three largest equal circles that can be inscribed in a circle of a given diameter. Divide the given diameter by 2.155. Ex. 1. Required the diameter of each of the largest three circles that can be inscribed in a circle 86.2 inches in diameter. Ans. 40 inches. MENSURATION. 433 Ellipse. 646. An ellipse is a plane figure bounded by a curve, from any point of which the sum of the distances to two fixed points is equal k [ G to a given distance. The two fixed points are called the foci, as GH in the ellipse AECBDFA. The major or transverse axis of an ellipse is its longest diameter, as A B. The minor or conjugate axis of an ellipse is its shortest diameter, as C D. The segment of an ellipse is a portion cut off from the ellipse, as F A E F. 647« To find the area of an ellipse, the two diameters being given. Multiply the two diameters together, and that product by .785398. Ex. 1. What is the area of an ellipse, whose two diameters are 24 and 18 inches ? Ans. 339.2928 inches. 2. What is the area of an elliptical pond, whose longest diameter is 33 feet 5 inches, and whose shortest diameter 20 feet 3 inches ? Ans. 59 sq. yd. 67 sq. in. MENSURATION OF SOLIDS. Prisms and Cylinders. 648* A prism is a figure whose ends or bases are any plane figures which are equal and similar, and parallel to each other, and whose sides are parallelograms. / A A triangular prism is one whose base is a triangle ; as the figure A B. A square prism is one whose base is a square ; a pentagonal prism, one whose base is a pentagon ; and so on, according to the figure of the ends or bases. A parallelopiped is a prism whose ends or bases, as well as its sides, are parallelograms. 649. A cylinder is a round body of uniform diameter, and which has circular bases parallel to each other ; as the figure C D. The perimeter of a prism or cylinder is the line that bounds its end or base ; and the altitude or height is the distance between the ends or bases. The convex surface of a prism or cylinder is the entire surface, exclusive of the two ends or bases. 650. To find the surface of a prism, or of a cylinder. Multiply the perimeter of the given prism or cylinder by the height, and to the product add the area of the two ends. Ex. 1. Required the surface of a triangular prism, of which the dis- n 37 434 MENSURATION. tance between the ends is 13 feet, and the sides of the base 23.34 and 19 inches. Ans. 85.22-}- square feet. 2. Required the surface of a pentagonal prism, whose length is 14 feet, and each side of whose base is 33 inches. Ans. 218.52 sq. ft. 3. Required the surface of a cylinder 13 feet long, the circumfer- ence of whose base is 57 inches. Ans. 65.34 square feet. 4. How often must a cylinder, 5 feet 3 inches long, whose diameter is 21 inches, revolve, to roll an acre ? Ans. 1509.18 times. 5. Required the wall-surface of a square room, whose sides are each 16 feet long and 10 feet high. Ans. 71^- square yards. 651 • To find the contents or volume of a prism or cylinder. Multiply the area of the base of the given prism or cylinder by the height. Ex. 1. What are the contents of a triangular prism, whose length is 12 feet, and each side of whose base is 2^ feet ? Ans. 32.47-j- cu. ft. 2. Required the volume of a triangular prism, whose length is 10 feet, and the three sides of whose triangular end or base are 5, 4, and 3 feet. Ans. 60 cu. ft. *3. How many cubic feet in a block of marble, whose length is 3 feet 2 inches, breadth 2 feet 8 inches, and depth 2 feet 6 inches ? Ans. 21^ cu. ft. 4. What is the volume of a cylinder, whose length is 9 feet, and the circumference of whose base is 6 feet ? Ans. 25.78-(- cu. ft. Pyramids and Cones. 652 • A 'pyramid is a solid having for its base some rectilineal figure, and for its sides triangles meeting in a common point called the vertex ; as the figure A B. The slant height of a pyramid is a line drawn from the vertex to the middle of one of the sides of the base. 653 • A cone is a solid having a circle for its base, and tapering uniformly to a point called the vertex. The slant height of a cone is a line drawn from the vertex to the circumference of the base. 654. The altitude or height of a pyramid or of a cone is a line drawn from the vertex perpendicular to the plane of the base. D 655. The frustum of a solid is the part that remains after cutting off the top by a plane parallel to the base ; as the frustum of a cone C D. 656 • To find the surface of a pyramid or of a cone. Multiply the perimeter or the circumference of the base by half of the slant height, and to the product add the area of the base. MENSURATION. 435 Ex. 1. Required the area of the surface of a square pyramid, whose base is 2 feet 8 inches square, and whose slant height is 3 feet 9 inches. 2. What is the convex surface of a cone, whose slant height is 20 feet, and the circumference of whose base is 9 feet ? Ans. 90 feet. 65 7 • To find the volume of a pyramid, or of a cone. Multiply the area of the base by one third of the altitude. Ex. 1. What is the solidity of a cone, whose height is 12-| feet, and the diameter of whose base is 21 feet ? Ans. 20.45-f- feet. 2. What are the contents of a triangular pyramid, whose height is 14 feet 6 inches, and the sides of whose base are 5, 6, and 7 feet ? Ans. 71.035-f feet 65 8 • To find the surface of a frustum of a pyramid, or of a cone. Multiply the sum of the perimeters or of the circumferences of the two ends by half of the slant height; and to the product add the areas of the two ends. Ex. 1. Required the surface of a frustum of a pentagonal pyramid, whose slant height is 10 inches, and the sides of whose base are 3 and 5 inches. 2. What is the surface of the frustum of a cone, the diameters of the bases being 43 and 23 inches, and the slant height 9 feet ? Ans. 90.72-f- sq. ft. 659. To find the volume of a frustum of a pyramid, or of a cone. Multiply the areas of the two ends together, and extract the square root of the product. To this root add the two areas, and multiply their sum by one third of the altitude. Ex. 1. If the length of a frustum of a square pyramid be 18 feet 8 inches, the side of its greater base 27 inches, and that of its less 16 inches, what is the volume ? Ans. 61.228-f- cu. ft. 2. What are the contents of a stick of timber, whose length is 40 feet, the diameter of the larger end being 24 inches, and of the smaller end 12 inches ? Ans. 73 J cu. ft., nearly. Spheres, Spheroids, &c. 660 • A sphere is a solid, bounded by a curved surface, every part of which is equally distant from a point within, called the centre. A The axis or diameter of a sphere is a line passing through the centre, and terminated by the surface ; as the line A B. The radius of a sphere is a line drawn from the centre to any part of the surface. 436 MENSURATION. 661 • A segment of a sphere is a part of it cut off by any plane ; as the figure ACBD. The plane is the base of the segment ; the per- pendicular distance from the centre of the base to the convex surface is the height of the seg- ment ; as CD. 662. A spherical zone is a part of the surface of a sphere included between two parallel planes, which form its bases ; and the height of a spherical zone is the perpendicular distance between the planes form- ing its bases. 663. A cylindrical ring is a figure formed by bending a cylinder uniformly till the two ends meet ; as A C D B. 664. A spheroid is a figure resembling a sphere, and which may be formed by the revolution of an ellipse about one of its axes ; as AECBDP A. If the ellipse revolves about its longer or transverse axis or diameter, the spheroid is prolate, or oblong ; if about its shorter or con- jugate diameter, the spheroid is oblate, or flat- tened. 665. A segment of a spheroid is a part cut off by any plane, as F AEF. 666. To find the surface of a sphere. Multiply the diameter by the circumference. Ex. 1. Required the convex surface of a globe, whose diameter is 24 inches. Ans. 1809.55-f- inches. 2. Required the surface of the earth, its diameter being 7957J miles, and its circumference 25,000 miles. Ans. 198943750 square miles. 667. To find the solidity of a sphere. Multiply the cube of the diameter by .523598. Ex. 1. What is the solidity of a sphere, whose diameter is 12 inches ? Ans. 904.78+ inches. 2. Required the solidity of the earth, supposing its circumference to be 25,000 miles. Ans. 263858149120.06886875 miles. 668. To find the convex surface of a segment or of a zone of a sphere. Multiply the height of the segment or zone by the circumference of the sphere of which it is a part MENSURATION. 437 Ex. 1. If the diameter of a sphere is 12 \ feet, what is the convex surface of a segment cut off from it, whose height is 2 feet ? Ans. 78.54 sq. ft. 2. If the diameter of the earth, considered as a perfect sphere, is 7970 miles, and the height of each temperate zone is taken at 2143.623553 miles, what is the surface of each temperate zone? Ans. 53673229.81-f- sq. m. 669 • To find the solidity of a segment of a sphere. Multiply the square of the height plus three times the square of the radius of the base, by the height, and this product by .5236. Ex. 1. Required the solidity of a spherical segment, whose height is 3 feet, and the radius of whose base is 4-| feet. Ans. 109.56 cu. ft. . 2. Required the solidity of the segment of a sphere, whose height is & feet, and the diameter of whose base is 20 feet. Ans. 1795.42 cu. ft. 670* To find the surface of a cylindrical ring. Multiply the sum of the thickness and the inner diameter by the thick- ness, and that product by 9.8696. Ex. 1. Required the surface of a cylindrical ring, whose inner diameter is 21 inches, and whose thickness is 4 inches. Ans. 986.96 sq. in. 671 • To find the solidity of a cylindrical ring. Multiply the sum of the thickness and the inner diameter by the square of the thickness, and that product by 2.4674. Ex. 1. Required the solidity of a cylindrical ring, whose inner diameter is 25 inches, and whose thickness is 5 inches. Ans. 1850.55 cu. in. 672 • To find the solidity of a spheroid. Multiply the square of the revolving axis by the fixed axis, and that product by .523598. Ex. 1. If the fixed axis of a spheroid is 32 inches, and the revolv- ing axis 20 inches, what is the solidity ? Ans. 6702.08 cu. in. 2. Required the contents of a balloon in the form of a prolate spheroid, having its longer diameter 48 feet, and its shorter 38 feet. Ans. 36291.76 cu. ft. MENSURATION OF LUMBER. 673* Boards are usually measured by the square foot. Planks, joists, beams, &c. are usually measured by board measure, the board being considered to be 1 inch in thickness. Round timber is sometimes measured by the ton, and sometimes by board measure. 37* 438 MENSURATION. 674. To find the number of square feet in a board. Multiply tlie length of the board, taken in feet, by its width, taken in inches , and the product divided by 12 will give the contents in square feet. Or, Take both the length and width in feet, and their product will be the contents in feet Note. — If the board is tapering, take half the sum of the width of its ends for the width. Ex. 1. What are the contents of a board 2-4 feet long, and 8 inches wide ? Ans. 10 feet 2. What are the contents of a board 30 feet long, and 16 inches wide ? Ans. 40 feet. 3. What are the contents of a tapering board, 30 feet long, whose ends are, the one 26 inches, and the other 14 inches wide ? 675. To find the number of feet, board measure, in a plank, joist, beam, &c. Multiply the ividth taken in inches by the thickness in inches, and this product by the length, in feet ; and the last product divided by 12 icill give the contents in feet, board measure. Note. — If the plank, joist, &c. is tapering in width, take half the sum of the width of the ends for the width ; and if the taper be both of the width and the thickness, the common rule of obtaining the contents in cubic feet is, to multiply half the sum of the areas of the two ends by the length, and divide the product by 144. Ex. 1. How many feet are there in 3 joists, which are 15 feet long, 5 inches wide, and 3 inches thick ? Ans. 56 \ feet 2. How many feet in 20 joists, 10 feet long, 6 inches wide, and 2 inches thick ? Ans. 200 feet. 3. How many feet in a beam 20 feet long, 10 inches thick, whose width tapers from 18 to 16 inches ? Ans. 283 J feet. 676 • To find the contents of round timber. Multiply the length, taken in feet, by the square of one fourth of the mean girth, taken in inches , and this product divided by 144 will give the contents in cubic feet. Note. — The girth of tapering timber is usually taken about £ the distance from the larger to the smaller end. The rule is that in common use, though very far from giving the actual number of cubic feet in round lumber measured by it. 40 cubic feet, as given by the rule, are in fact equal to 50^ true cubic feet. The following rule gives results more nearly accurate, requiring to be diminished by only one foot in 190, to give exact contents. Multiply the square of one fifth of the mean girth, taken in inches, by twice the length, in feet ; and divide by 144. Ex. 1. How many cubic feet in a stick of timber which is 30 feet long, and whose girth is 40 inches ? Ans. 20f feet. 2. If a stick of timber is 50 feet long, and its girth is 56 inches, what number of cubic feet does it contain ? Ans. 68^-g feet. 3. What are the contents of a log 90 feet long, and whose circum- ference is 120 inches ? Ans. 562^ feet. MENSURATION. 439 GAUGING OF CASKS. 677 • Gauging is the process of finding the capacities of casks or other vessels. Casks are generally considered to be of four varieties : 1. Having the staves nearly straight ; 2 , Having the staves very little curved ; 3, Having the staves of a medium curve ; 4. Having the staves con- siderably curved. Note. — Casks of the first variety approach very nearly the form of a cylinder ; those of the third variety are of the shape of a molasses hogshead ; those of the second variety have a curvature of stave between that of the first and third ; and the fourth have a greater curvature than that of the third. 678. In gauging casks, it is necessary first to find the mean di- ameter. This is found by taking the end and middle diameters, and the length in inches ; and then adding to the end diameter the pro- duct of the difference between the end and middle diameters by .55, .60, .65, or .70, as the cask may be of the first, second, third, or fourth, variety. 679* To find the capacity of a cask in gallons, Multiply the square of the mean diameter, in inches, by the length, in inches ; and the product multiplied by .0034 will give the capacity in liquid or wine gallons. Note 1. — If the capacity is required in ale or beer gallons, use for a multi- plier .0028 instead of .0034. If imperial gallons are required; multiply the liquid or wine gallon, as found by the rule, by .833. Note 2. — The contents of any vessel being known in cubic inches, its capa- city in liquid gallons may be found by dividing by 231 ; in ale or beer gallons, by dividing by 282 ; and in bushels, by dividing by 2150.42. Ex. 1. Required the capacity in gallons of a cask of the fourth variety, whose middle diameter is 35 inches, head diameter 27 inches, and length 45 inches. Ans. 162.6. 2. What is the capacity in gallons of a cask of the third variety, whose middle diameter is 38 inches, head diameter 30 inches, and length 42 inches ? 3. What are the contents in liquid measure of a tub 40 inches in diameter at the top, 30 inches at the bottom, and whose height is 50 inches? Ans. 209.66gal. 4. How many wine gallons will a cubical box contain, that is 10 feet long, 5 feet wide, and 4 feet high ? Ans. 1496 T 8 T gal. 5. How many ale gallons will a trough contain, that is 1 2 feet long, 6 feet wide, and 2Jeet high ? Ans. 88 2 Jf gal. 6. How many bushels of grain will a box contain that is 15 feet long, 5 feet wide, and 7 feet high ? Ans. 421. 8bu. TONNAGE OF VESSELS. 680 • The tonnage of a ship is the number of tons burden it will carry, with safety, under the ordinary circumstances of navigation. 440 MENSURATION. The liglit-loaded water-line of a vessel is the line made by the water upon the outside of the hull as it floats without load ; and the deep- loaded water-line is that made in like manner when it is fully laden. The number of cubic feet of the hull between these two water- lines, divided by 35, the number of cubic feet of sea-water which must be taken to weigh a ton, represents the weight of water dis- placed in sinking the vessel from the light to the deep-loaded water- line, and therelbre its true tonnage. 681. Government has by law established a rule by which the custom-house officers are to be guided in collecting tonnage duties. But as it does not always give the actual tonnage, builders, and oth- ers, usually make their estimates by some other rule. Government Rule. For Single-decked Vessels. Tale the length on deck from the forward side of the main stem to the after side of the stern post, and the breadth at the broadest part above the main wales ; take the depth from the under side of the deck plank to the ceiling of the hold ; and deduct from the length three fifths of the breadth ; 7iiultij)ly the re- mainder by the breadth, and the product by the depth ; and divide the last product by 95. For Double-decked Vessels. Proceed as with single-decked vessels, except for the depth take half the breadth. Note. — The government rule is differently construed. The length is usually taken in a line with the deck; the depth at the main hatch. But with regard to the breadth, there is a great want of uniformity among measurers; most take the breadth about 45 inches below the plank-sheer at the broadest part; some consider the upper wale, and others the lower, as the main wale, thus making a considerable difference in their results. The government rule for single-decked vessels operates very well, but the rule for double-decked vessels, which is also intended to include all vessels of more than one deck, often fails to give the true tonnage. A more accurate method would, for double-decked vessels, take the breadth 5 feet below the upper deck, at the broadest part, and for three-decked vessels 7 feet below the upper deck; and in each case for depth of hold three ffths of the breadth. Ex. 1. A. & G. T Sampson, of East Boston, have contracted to build a clipper ship 191 T 6 ^ feet lon< what is the government tonnage 2. What is the government tonnage of the ship length is 184 T C takes 2.06+ft., and pays $ 23.48^f if f ; D takes 1.58-fft, and pays $ 21.74i|-0ff . ^ 24. A, B, and C bought a grindstone" for which they paid $ 10.60. B paid 20 per cent, more than A, and 10 per cent, less than C. The diameter of the stone was 65 inches, and the diameter of the place for the shaft 3 inches. What sum did each pay, and how much must each grind off from the semidiameter to obtain his proper share of the stone ? Ans. A paid $ 3, B $ 3.60, and C$4. A grinds off 5 inches ; B 1\ inches, and C 18-1- inches. 25. A servant draws off a gallon on each day, for 20 days, from a cask containing 10 gallons of wine, each time supplying the deficiency by the addition of a gallon of water ; and then, to escape detection, he again draws off 20 gallons, supplying the deficiency each time by a gallon of wine. How much water still remains in the cask ? Ans. 1.0679577 gallons, or more than a gallon and half a pint. 26. The dimensions of a bushel measure are 18-1- inches wide, and 8 inches deep ; what should be the dimensions of a similar measure that would contain 8 bushels? Ans. 37in. wide, 16in. deep. 27. What is the weight of a hollow spherical iron shell 5 inches in diameter, the thickness of the metal being 1 inch, and a cubic inch of iron weighing iff of a pound ? Ans. 13.2387lb. 28. At a certain time between 2 and 3 o'clock, the minute-hand was between 3 and 4. Within an hour after, the hour-hand and minute-hand had exactly changed places with each other. What was the precise time when the hands were in the first position? Ans. 2h. 15m. 56 T 9 T \s. 29. Required the contents of the largest cube that can be inscribed in a sphere 20 inches in diameter. Ans. 1539.58-]- cu. in. 30. If in a pair of scales a body weigh 90 pounds in one scale, and only 40 pounds in the other, required the true weight, and the propor- tion of the lengths of the two arms of the balance-beam on each side of the point of suspension. Ans. Weight 60lb., and the proportions 3 to 2. 31. In turning a one-horse chaise within a ring of a certain diame- ter, it was observed that the outer wheel made two turns, while the inner wheel made but one ; the wheels were each 4 feet high ; and supposing them fixed at the distance of 5 feet asunder on the axletree, what was the circumference of the track described by the outer wheel ? Ans. 62.83-)- feet. 32. The ball on the top of St. Paul's Church is 6 feet in diameter. What did the gilding of it cost, at 8-Jd. per square inch V 444 MISCELLANEOUS EXAMPLES. 33. There is a conical glass, 6 inches high, 5 inches wide at the top, and which is ^ part filled with water. What must be the diameter of a ball, let fall into the water, that shall be immersed by it ? Ans. 2.44 5-|- inches. 34. A certain lady, the mother of three daughters, had a farm of 500 acres, in a circular form, with her dwelling-house in the centre. Being desirous of having her daughters settled near her, she gave to them three equal parcels, as large as could be made in three equal circles included within the periphery of her farm, one to each, with a dwelling-house in the centre of each; that is, there were to be three equal circles, as large as could be made within a circle that contained 500 acres. How many acres did the fann of each daughter contain, how many acres did the mother retain, how far apart were the dwell- ing-houses of the daughters, and how far was the dwelling-house of each daughter from that of the mother? Ans. Each daughter's fann contained 107 acres 2 roods 31.22-j— rods. The mother retained 176 acres 3 roods 26.34-}- rods. The distance from one daughter's house to the other was 148.1 1981 7-{- rods. The mother's dwelling-house was distant from her daugh- ters' 85.5 1-\- rods. 35. James Page has a circular garden, 10 rods in diameter ; how many frees can be set in it, so that no two shall be within ten feet of each other, and no tree within two and a halt' feet of the fence en- closing the garden ? Ans. 241. 36. A and B engaged to reap a field for 90 shillings ; and as A could reap it in 9 days, they promised to complete it in 5 days. They found, however, that they were obliged to call in C, an inferior work- man, to assist them for the last two days, in consequence of which B received 3s. 9d. less than lie otherwise would have done. In what time could B and G each reap the iield ? Ans. B could reap it in 15 days, and C in 18 days. 37. A merchant tailor bought 40 yards of broadcloth, 2-J- yards wide ; but on sponging it, it shrunk in length upon every 4 yards half a quarter, and in width, one nail and a half upon every 1^ yards. To line this cloth, he bought flannel 5 quarters wide, which, being wet, shrunk the whole width on every 20 yards in length, and in width it shrunk half a nail. Required the number of yards of flannel used in lining the cloth. Ans. 71 T 7 T yards. 38. 1 have a garden in the form of an equilateral triangle, whose sides are 200 feet. At each corner stands a tower ; the height of the fust is 30 feet, the second 40, and the third 50. At what distance from the base of each tower must a ladder be placed, that it may just reach the top of each ? And what is the length of the ladder, the garden being a horizontal plane ? ' Ans. The "foot of the ladder from the base of the first tower 118.811-}- feet; second tower, 1 1 5.82 7-f- feet ; third tower, 111.875-)- feet. Length of the ladder, 1 22.535-f- feet. THE END. -r 'tirr- 1 RETURN EDUO TO—* 2600" VNON- "olman PSYCHOLOGY LIBRARY Hall 642-4209 LOAN PERIOD 1 ~ 1 MONTH 2 3 4 5 6 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS 2- hour books must be renewed in person Return to desk from which borrowed DUE AS STAMPED BELOW nr>r o ft 10 5rt U b I q o la -. • . fejj> '■ - y . . 1 t?T^ m m a "•»/* f^ rmt O CT- 3 I - MAY DC. Kfc. AFTER SEP 15 138 2 ' FORM NO. DD10, 5m, UNIVERSITY OF CALIFORNIA, BERKELEY 3/80 BERKELEY, CA 94720 ©s \ YB 35818 11577021 QA102 G67 1857 Educ. Lib. YJ