GIFT Harry Eas t Miller ROBINSON'S MATHEMATICAL SERIES. THE PROGRESSIVE HIGHER ARITHMETIC, FOB SCHOOLS, ACADEMIES, AND MERCANTILE COLLEGES. FORMING A COMPLETE TREATISE ON ARITHMETICAL SCIENCE, AND ITS COMMERCIAL AND BUSINESS APPLICATIONS. EDITED BY DANIEL W. FISH, A.M., EDITOR or ROBINSON'S PROGRESSIVE SERIES OF ARITHMETICS, AND THE SHORTER COURSE. IVISON, BLAKEMAN, TAYLOR & CO., NEW YORK AND CHICAGO. 1878. . ROBINSON'S Mathematical Series. Graded to the wants of Primary, Intermediate, Grammar, Normal, and High Schools, Academies, and Colleges, Progressive Table Book. Progressive Primary Arithmetic. Progressive Intellectual Arithmetic. Rudiments of Written Arithmetic. JUNIOR-CLASS ARITHMETIC, Oral and Written. NEW. Progressive Practical Arithmetic. Key to Practical Ai*ithrnetic. Progressive Eiglier Arithmetic. Key to Higher Arithmetic, New Elementary Algebra. Key to Neif Elementary Algebra. New University Algebra. Key to New University Algebra. New Geometry and Trigonometry. In one vol. Geometry, Plane and Solid. In separate vol. Trigonometry, Plane and Spherical. In separate vol. New Analytical Geometry and Conic Sections. New Surveying and Navigation. New Differential and Integral Calculus. University Astronomy Descriptive and Physical. Key to Geometry and Trig., Analyt. Geometry and Conic Sect., Surveying and Navigation. Copyright, 1860, 1863, & 1875, DANIEL W. FISH. Electrotyped by SMITH & McDouGAL, 82 Beekman St.,N. Y. GIFT OF >-~*A PREFACE. THIS work is intended to complete a well graded and progressive series of Arithmetics, and to furnish to ad- vanced students a more full and comprehensive text-book on the Science of Numbers than has before been published , a work that shall embrace those subjects necessary to give the pupil a thoroughly practical and scientific arithmetical education, either for the farm, the workshop, or a profes- sion, or for the more difficult operations of the counting- room and of mercantile and commercial life. There are two general methods of presenting the ele- ments of arithmetical science, the Synthetic and the Ana- lytic. Comparison enters into every operation, from the simplest combination of numbers to the most complicated problems in the Higher Mathematics. Analysis first generalizes a subject and then develops the particulars of which it consists; Synthesis first presents particulars, from which, by easy and progressive steps, the pupil is led to a general and comprehensive view of the subject. Analysis separates truths and properties into their ele- ments or first principles; Synthesis constructs general principles from particular cases. Analysis appeals more to the reason, and cultivates the desire to search for first principles, and to understand the reason for every process rather than to know the rule. Hence, the leading method in an elementary course of instruction should be the Synthetic, while in an advanced course it should be the Analytic. The following characteristics of a first class text-book will be obvious to all who examine this work : the typogra- M81911 (iii) iv PREFACE. ' phy and mechanical execution; the philosophical and scientific arrangement of the subjects ; dear and concise definitions ; full and rigid analyses ; exact and compre- hensive rules ; brief and accurate methods of operation : the wide range of subjects and the large number and prac- tical character of the examples in a word, SCIENTIFIC AC- CUEACY combined with PEACTICAL UTILITY, throughout the entire work. Much labor and attention have been devoted to obtain- ing correct and adequate information pertaining to mer- cantile and commercial transactions, and the Government Standard units of measures, weights, and money. The counting-room, the bank, the insurance and broker's office, the navy and ship-yard, the manufactory, the wharves, the custom-house, and the mint, have all been visited, and the most reliable statistics and the latest statutes have been consulted, for the purpose of securing entire accuracy in those parts of this work which relate to these ;ubjects and departments. As the result of this thorough investi- gation, many statements found in most other arithmetics of a similar grade will not agree with the facts presented in this work, and simply because the statements in these other books have been copied from older works, while laws and customs have undergone great changes since the older works were written. New material and new methods will be found in the seve- ral subjects throughout the entire work. Considerable pro- minence has been given to Percentage and its numerous ap- plications, especially to Stocks, Insurance, Interest, Aver- aging Accounts, Domestic and Foreign Exchange, and seve- ral other subjects necessary to quality students to become good accountants or commercial business men. And while this work may embrace many subjects not necessary to the PREFACE. V course usually prescribed in Mercantile and Commercial Colleges, yet those subjects requisite to make good account- ants, and which have been taught orally in that class of institutions from want of a suitable text-book, are fully dis- cussed and practically applied in this work ; and it is there- fore believed to be better adapted to the wants of Mercan- tile Colleges than any similar work yet published. And while it is due, it is also proper here to state that J. C. Porter, A. M. ; an experienced and successful teacher of Mathematics in this State, and formerly professor of Com- mercial Arithmetic, in Iron City Commercial College, Pitts- burgh, Penn., has rendered valuable aid in the preparation of the above-named subjects, and of other portions of the work. He is likewise the author of the Factor Table on pages 72 and 73, and of the new and valuable improvement in the method of Cube Root. Teachers entertain various views relative to having the answers to problems and examples inserted in a text-book. Some desire the answers placed immediately after the ex- amples ; others wish them placed together in the back part of the book; and still others desire them omitted alto- gether. All these methods have their advantages and their disadvantages. If all the answers are given, there is danger that the pupil will become careless, and not depend enough upon the accuracy of his own computations. Hence he is liable to neglect the cultivation of those habits of patient investiga- tion and self-reliance which would result from his being obliged to test the truth and accuracy of his own processes f by proof, the only test he will have to depend upon in all the computations in real business transactions in after life. Besides, the work of proving the correctness of a result is often of quite as much value to the pupil as the work of 1* ^ PREFACE. performing the operation ; as the proof may render simple and clear some part or the whole of an operation that was before complicated and obscure. The improvements in Percentage made necessary by the financial changes of the last few years are especially notice- able. The different kinds of United States' Securities, Bonds, and Treasury Notes are described, and their com- parative value in commercial transactions illustrated by practical examples. The difference between Gold and Currency, and the corresponding difference in prices, ex- hibited in trade, are taught and illustrated, and many other things that every commercial student and business man ought to know and understand. AUGUST, 1860. IMPROVED EDITION. Such changes only have been made in the present edition as were necessary to conform to law and usage, and to meet a demand from many of the best teachers of the country for a full and practical treatise on Mensuration. Hence Foreign Exchange has been so modified in the Tables and Examples as to conform to the Act of March 3, 1873, and to present usage. Thirty -six pages of useful and practical matter on Mensuration and Measurements, have been carefully prepared and substituted at the end of the book for the lengthy treatise of the Metric System heretofore presented, and which is scarcely ever used in this country. To avoid repetition, as well as to put in a more condensed form, the Principles and Applications of the Square and Cube roots that inter- vened between "Evolution" and "Series" in former editions, have been embodied in these thirty-six pages, and also so much of the Metric System as is of any practical value. It is hoped that these improvements will give new life to a book that has already proved its merits by the large and increasing circulation it has obtained. JULY, 1875, CONTENTS. PAGE Definitions 11 Signs 13 Axioms , 14 Notation and Numeration 15 SIMPLE NUMBERS. Addition ! 23 Adding two or more columns at one operation 27 Subtraction ... 30 Two or more subtrahends 33 Multiplication 35 Powers of Numbers 39 Continued Multiplication 40 Contractions in Multiplication... 41 Division , 47 Abbreviated Long Division 50 Successive Division 55 Contractions in Division , 55 General Problems in Simple Numbers 61 PROPERTIES OF NUMBERS. Kxact Divisors 65 Prime Numbers 68 Table of Prime Numbers , 70 Factoring 70 Factor Table . 72 Greatest Common Divisor 76 Least Common Multiple 82 Cancellation : * 86 FRACTIONS. Definitions, Notation and Numeration 89 Reduction , 92 Addition 99 (Vii) Viii CONTENTS. PAQB Subtraction 101 Theory of Multiplication and Division 103 Multiplication 104 Division 107 Greatest Common Divisor Ill Least Common Multiple , 112 DECIMALS. Notation and Numeration 117 Reduction 121 Addition 124 Subtraction 126 Multiplication 127 Contracted Multiplication 128 Division 132 Contracted Division 134 Circulating Decimals 136 Reduction of Circulating Decimals 139 Addition and Subtraction of Circulating Decimals 142 Multiplication and Division of Circulating Decimals 144 UNITED STATES MONEY. Notation and Numeration 145 Reduction 147 Operations 147 Problems ... , 150 Ledger Accounts > 153 Accounts and Bills 153 Continued Fractions 161 COMPOUND NUMBERS. Measures of Extension 164 Measures of Capacity 170 Measures of Weight 171 Measure of Time 175 Measure of Angles 177 Miscellaneous Tables , 178 Government Standards of Measures and Weights 179 English Measures and Weights 182 French Measures and Weights I 84 Money and Currencies 1$*? Reduction 192 Reduction Descending 192 CONTENTS. 'ix PACK Reduction Ascending , 199 Addition 206 Subtraction 209 Multiplication 214 Division 216 Longitude and Time 218 DUODECIMALS. Addition and Subtraction 227 Multiplication 228 Division 230 SHORT METHODS. For Subtraction 232 For Multiplication 233 For Division 241 RATIO. 243 PROPORTION. 247 Cause and Effect ,. , 249 Single Proportion 249 Compound Proportion , 253 PERCENTAGE. Notation 259 General Problems - 260 Applications 268 Commission 268 Stocks , 272 Stock-jobbing 273 Instalments, Assessments, and Dividends 276 Stock Investments 279 Gold Investments 285 Profit and Loss 287 Insurance 291 Life Insurance 293 Life Table 295 Endowment Assurance Table 296 Taxes 298 General Average 301 Custom House Business 303 Simple Interest 307 X CONTENTS. 4 PASS Partial Payments or Indorsements 314 Savings Banks Accounts 319 Compound Interest 321 Discount 328 Banking 330 Exchange 337 Direct Exchange 339 Table of Foreign Coins and Money 342 Arbitrated Exchange 348 Equation of Payments 352 Compound Equations 357 Partnership 364 ALLIGATION 370 INVOLUTION 379 EVOLUTION. Square Root , Contracted Method. Cube Root Contracted Method. SERIES. 398 Arithmetical Progression 400 Geometrical Progression 403 Annuities 408 MISCELLANEOUS EXAMPLES 414 MENSURATION. Lines and Angles 421 Triangles 422 Quadrilaterals 426 Circles 428 Similar Plane Figures 432 Solids 435 Prisms and Cylinders 430 Pyramids and Cones 437 Spheres 438 Similar Solids. 441 Gauging 443 Measurement of Land 445 Boards and Timber 448 Masonry 449 Capacity of Bins. Cisterns, etc 450 METRIC SYSTEM.. .. 453 HIGHER ARITHMETIC. DEFINITIONS. I. Quantity is any thing that can be increased, diminished, or measured ; as distance, space, weight, motion, time. hours, 36 minutes, and 7 bushels 3 pecks. - 15. '"A Power ?.p the product arising from multiplying a number by itself, or repeating it any number of times as a factor. 1G. A Root is a factor repeated to produce a power. 17. A Scale is the order of progression on which any system of notation is founded. Scales are uniform and varying. 18. A Uniform Scale is one in which the order of progression is the same throughout the entire succession of units. 1O. A Varying Scale is one in which the order of progression is not the same throughout the entire succession of units. 20. A Decimal Scale is one in which the order of progression is uniformly ten. 21. Mathematics is the science of quantity. The two fundamental branches of Mathematics are Geometry and Arithmetic. Geometry considers quantity with reference to positions, form, and extension. Arithmetic considers quantity as an assemblage of definite portions, and treats only of those condi- tions and attributes which may be investigated and expressed by numbers. Hence, 22. Arithmetic is the Science of numbers, and the Art of computation. NOTE. When Arithmetic treats of operations on abstract numbers it is a sci- ence, and is then called Pure Arithmetic. When it treats of operations on con- crete numbers it is an art, and is then called Applied Arithmetic. Pure and Applied Arithmetic are also called Theoretical and Practical Arithmetic. 23. A Demonstration is a process of reasoning by which a truth or principle is established. 24. An Operation is a process in which figures are employed to make a computation, or obtain some arithmetical result. 25. A Problem is a question requiring an operation. 26. A Rule is a prescribed method of performing an operation. 27. Analysis, in arithmetic, is the process of investigating principles, and solving problems, independently of set rules. 28. The Five Fundamental Operations of Arithmetic are, Notation and Numeration, Addition, Subtraction, Multiplication, and Division. DEFINITIONS. 13 SIGNS. 29. A Sign is a character indicating the relation of numbers, or an operation to be performed. 30. The Sign of Numeration is the comma (,). It indicates that the figures set off by it express units of the same general name, and are to be read together, as thousands, millions, billions, etc. 31. The Decimal Sign is the period (.). It indicates that the number after it is a decimal. 32. The Sign of Addition is the perpendicular cross, -f, called plus. It indicates that the numbers connected by it are to be added ; as 3 -f 5 + 7, read 3 plus 5 plus 7. 33. The Sign of Subtraction is a short horizontal line, , called minus. It indicates that the number after it is to be sub- tracted from the number before it; as 12 7, read 12 minus 7. 34. The Sign of Multiplication is the oblique cross, x . It indicates that the numbers connected by it are to be multiplied together; as 5 x 3 x 9, read 5 multiplied by 3 multiplied by 9. 35. The Sign of Division is a short horizontal line, with a point above and one below, -5-. It indicates that the number before it is to be divided by the number after it ; as 18 -f- 6, read 18 divided by 6. Division is also expressed by writing the dividend above, and the divisor below, a short' horizontal line. Thus, *g 8 , read 18 divided by 6. 36. The Sign of Equality is two short, parallel, horizontal lines, =. It indicates that the numbers, or combinations of numbers, connected by it are equal; as 4 -f 8 = 15 3, read 4 plus 8 is equal to 15 minus 3. Expressions connected by the sign of equality are called equations. 37. The Sign of Aggregation is a parenthesis, ( ). It indi- cates that the numbers included within it are to be considered together, and subjected to the same operation. Thus, (8 -f 4) X 5 indicates that both 8 and 4, or their sum, is to be multiplied by 5. A vinculum or bar, , has the same signification. Thus, 7x 9-v-3 = 21. 14 SIMPLE NUMBERS. 38. The Sign of Ratio is two points, : . Thus, 7 : 4 is read, the ratio of 7 to 4. 39. The Sign of Proportion is four points, : : . Thus, 3 : 6 : : 4 : 8, is read, 3 is to 6 as 4 is to 8. 40. The Sign of Involution is a number written above, and a little to the right, of another number. It indicates the power to which the latter is to be raised. Thus, 12 s indicates that 12 is to be taken 3 times as a factor; the expression is equivalent to 12 x 12 x 12. The number expressing the sign of involution is called the Index or Exponent. 41. The Sign of Evolution, -y/, is a modification of the letter r. It indicates that some root of the number after it is to be extracted. Thus, v/25 indicates that the square root of 25 is to be extracted; indicates that the cube root of 64 is to be extracted. AXIOMS. An Axiom is a self-evident truth. The principal axioms required in arithmetical investigations are the following : 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 equal quantities be multiplied by the same number, the products will be equal. 4. If equal quantities be divided by the same number, the quo- tients will be equal. 5. If the same number be added to a quantity and subtracted from the sum, the remainder will be that quantity. 6. If a quantity be multiplied by a number and the product divided by the same number, the quotient will be that quantity. 7. Quantities which are respectively equal to any other quantity are equal to each other. 8. Like powers or like roots of equal quantities are equal. 9 The whole of any quantity is greater than any of its parts. 10. The whole of any quantity is equal to the sum of all its parts. NOTATION AND NUMERATION. 15 NOTATION AND NUMERATION. 4:3. Notation is a system of writing or expressing numbers by characters; and, 44. Numeration is a method of reading numbers expressed by characters. 45. Two systems of notation are in general use the Roman and the Arabic. NOTE. The Roman Notation is supposed to have been first used by the Romans : hence its name. The Arabic Notation was first introduced into Europe by the Moors or Arabs, who conquered and held possession of Spain during the llth century. It received the attention of scientific men in Italy at the begin- ning of the 13th century, and was soon afterward adopted in most European countries. Formerly it was supposed to be an invention of the Arabs; but investigations have shown that the Arabs adopted it from the Hindoos, among whom it has been in use more than 2000 years. From this undoubted origin it is sometimes called the Indian Notation. THE ROMAN NOTATION. 4G. Employs seven capital letters to express numbers. Thus, Letters, I V X L C D M Values, one, five, ten. fifty, , e J ,_ fi ! e , one Jj hundred, hundred, thousand. 47. The Roman notation is founded upon five principles, as follows : 1st. Repeating a letter repeats its value. Thus, II represents two, XX twenty, CCC three hundred. 2d. If a letter of any value be placed after one of greater value, its value is to be united to that of the greater. Thus, XI repre- sents eleven, LX sixty, DC six hundred. 3d. If a letter of any value be placed before one of greater value, its value is to be taken from that of the greater. Thus, IX repre- sents nine, XL forty, CD four hundred. 4th. If a letter of any value be placed between two letters, each of greater value, its value is to be taken from the united value of the other two. Thus, XIY represents fourteen, XXIX twenty- nine, XCIY ninety-four. 5th. A bar or dash placed over a letter increases its value one thousand fold. Thus, V signifies five, and V five thousand ; L fifty, and L fifty thousand. 16 SIMPLE NUMBERS. i TABLE OF ROMAN NOTATION. I is One. XX is Twenty. II " Two. XXI " Twenty-one. III " Three. XXX " Thirty. IV " Four. XL " Forty. V " Five. L " Fifty. VI " Six. LX " Sixty. VII'" Seven. LXX " Seventy. VIII " Eight. LXXX " Eighty. IX " Nine. XC " Ninety. X " Ten. C " One hundred. XI " Eleven. CO " Two hundred. XII " Twelve. D " Five hundred. XIII " Thirteen. DC " Six hundred. IV " Fourteen. M " One thousand. [dred. XV " Fifteen. MC " One thousand one hun- XVI " Sixteen. MM " Two thousand. XVII " Seventeen. X " Ten thousand. XVIII " Eighteen. _C " One hundred thousand. XIX " Nineteen. M " One million. NOTES. 1. Though the letters used in the above table have been employed nr the Roman numerals for many centuries, the marks or characters used origi- nally in this notation are as follows : Modern numerals, I V X L C D M Primitive characters, I V X L C N M 2. The system of Roman Notation is not welt adapted to the purposes of nu- merical calculation ; it is principally confined to the numbering of chapters and ections of books, public documents, etc. EXAMPLES FOR PRACTICE. Express the following numbers by the Roman notation: 1. Fourteen. 6. Fifty-one. 2. Nineteen. 7. Eighty-eight. 3. Twenty-four. 8 Seventy-three. 4. Thirty-nine. 9. Ninety-five. 5. Forty-six. 10. One hundred one. 11. Five hundred fifty-five. 12. Seven hundred ninety-eight. 13. One thousand three. 14. Twenty thousand eight hundred forty-five. NOTATION AND NUMERATION. If THE ARABIC NOTATION 48. Employs ten characters or figures to express numbers. Thus, Figures, .0 123456789 Names and ) nau 8 ht one > two three, four, five, six, seven, eight, nine, cipher. values. or 49. The cipher, or first character, is called naught, because it has no value of its own. It is otherwise termed nothing, and zero. The other nine characters are called significant figures, because each has a value of its own. They are also called digits, a word derived from the Latin term diyitus, which signifies finger. 50. The ten Arabic characters are the Alphabet of Arithmetic. Used independently, they can express only the nine numbers that correspond to the names of the nine digits. But when combined according to certain principles, they serve to express all numbers. 01. The notation of all numbers by the ten figures is accom- plished by the formation of a series of units of different values, to which the digits may be successively applied. First ten simple units are considered together, and treated as a single superior unit; then, a collection often of these new units is taken as a still higher unit; and so on, indefinitely. A regular series of units, in ascending orders, is thus formed, as shown in the following TABLE OF UNITS Primary units are called units of the first order. Ten units of the first order make 1 unit " " second " Ten " " " second " " 1 " " " third " Ten " " " third " " 1 " " " fourth " etc., etc. etc., etc. The various orders of units, when expressed by figures, are distinguished from each other by their location, or the place they occupy in a horizontal row of figures. Units of the first order are written at the right hand ; units of the second order occupy the second place ; units of the third order the third place ; and so on, counting from right to left, as shown on the following page : SIMPLE NUMBERS. 000000000 In this notation we observe 1st. That a figure written in the place of any order, expresses as many units of that order as is denoted by the name of the figure used. Thus, 436 expresses 4 units of the 3d order, 3 units of the 2d order, and 6 units of the 1st order. 2d. The cipher, having no value of its own, is used to fill the places of vacant orders, and thus preserve the relative positions of the significant figures. Thus, in 50, the cipher shows the absence of simple units, and at the same time gives to the figure 5 the local value of the second order of units. 54. Since the number expressed by any figure depends upon the place it occupies, it follows that figures have two values, Simple and Local. 55. The Simple Value of a figure is its value when taken alone ; thus, 4, 7, 2. 5G. The Local Value of a figure is its value when used with another figure or figures in the same number. Thus, in 325, the local value of the 3 is 300, of the 2 is 20, and of the 5 is 5 units. NOTE. When a figure occupies units' place, its simple and local values are the same. 57. The leading principles upon which the Arabic notation is founded are embraced in the following GENERAL LAWS. I. AH numbers are expressed by applying the ten figures to dif- ferent orders of units. II. The different orders of units increase from right to left, and decrease from left to right, in a tenfold ratio. III. Every removal of a figure one place to the left, increases its local value tenfold; and every removal of a figure one place to the right, diminishes its local value tenfold. NOTATION AND NUMERATION. 19 08. In numerating, or expressing numbers verbally, the various orders of units have the following names : ORDERS. NAMES. 1st order is called Units. 2d order " " Tens. 3d order " " Hundreds. 4th order " " Thousands. 5th order " " Tens of thousands. 6th order " " Hundreds of thousands. 7th order " " Millions. 8th order " " Tens of millions. 9th order " " Hundreds of millions, etc., etc. etc., etc. 50. This method of numerating, or naming, groups the suc- cessive orders into period* of three figures each, there being three orders of thousands, three orders of millions, and so on in all higher orders. These periods are commonly separated by commas, as in the following table, which gives the names' of the orders and periods to the twenty-seventh place. 3 a no fl 5 g M M .2 . t *9 o o "- 1 - w 3 P-. 8 tw W 3 '-3 M e required to multiply 473 by 138. OPERATION. ANALYSIS. The excess of 473 = 468 -f 5 9 's in 473 is 5, and 473 = 468 138 =135-f3 +5, of which the first part, , AQ QP _ ' no excess of 9's, Partial 5x135= 675 W- The eXC688 f 9 ' S in products. 1 468 X 3 = 1404 138 is 3, and 138 = 135 -f 3, of [ 5x3 = 15 which the first part, 135, con- Entire product, 65274 tains^ no excess of 9's, (I). Multiplying both parts of the multiplicand by each part of the multiplier, we have four partial pro^ ducts, of which the first three have no excess of 9's, because each con- tains a factor having no excess of 9's, (II). Therefore, the excess of 9's in the entire product must be the same as the excess of 9's in the last partial product, 15, which we find to be 1 + 5 = 6. The same may be shown of any two numbers. Hence, to prove multiplication by excess of 9's, Find the excess of 9's in each of the two factors, and multiply them together; if the excess of 9's in this product is equal to the excess of 9's in the product of the factors, the work is supposed to be right. NOTE. If the excess of 9's in either factor is 0, the excess of 9's in the pro- duct will be 0, (II). EXAMPLES FOR PRACTICE. (1.) (2.) (3.) (4.) Multiply 475 3172 9827 7198 By 9 14 _ 84 216 Prod. 4275 44408 825468 1554768 38 SIMPLE NUMBERS. i 5 Multiply 31416 by 175. Ans. 5497800 6. Multiply 40930 by 779. Ans. 31884470. 7. Multiply 46481 by 936. 8. Multiply 15607 by 3094. 9. Multiply 281216 by 978. Ans. 275029248 10. Multiply 30204 by 4267. Ans. 128,880,468. 11. What is the product of 4444 x 2341 ? Ans. 10,403,404. 12. What is the product of 4567 X 9009 ? Ans. 41,144,103. 13. What is the product of 2778588 x 9867? Ans. 27,416,327,796. 14. What is the product of 7060504 x 30204 ? Ans. 213,255,462,816. 15. What will be the cost of building 276 miles of railroad at $61320 per mile ? Ans. $16,924,320. 16. If it require 125 tons of iron rail for one mile of railroad, how many tons will be required for 196 miles ? 17. A merchant tailor bought 36 pieces of broadcloth, each piece containing 47 yards, at 7 dollars a yard ; how much did he pay for the whole ? Ans. $11,844. 18. The railroads in the State of New York, in operation in 1858, amounted to 2590 miles in length, and their average cost was about $52916 per mile; what was the total cost of the rail- roads in New York '( Ans. $137,052,440. 19. The Illinois Central Railroad is 700 miles long, and cost $45210 per mile ; what was its total cost ? 20. The salary of a member of Congress is $3000, and in 1860 there were 303 members ; how much did they all receive ? 21. The United States contain an area of 2988892 square miles, and in 1850 they contained 8 inhabitants to each square mile; what was their entire population ? Ans. 23,911,136. 22 Great Britain and Ireland have an area of 118949 square miles, and in 1850 they contained a population of 232 ^o the square mile; what was their entire population? Ans. 27,596,168. 23. The national debt of France amounts to $32 for each indi- MULTIPLICATION. 39 vidual, and the population in 1850 was 35781628 ; what was the entire debt of France ? Ans. 1,145,012,096. POWERS OF NUMBERS. 8G. We have learned (15) that a power is the product arising from multiplying a number by itself, or repeating it any number of times as a factor; (16), that a root is a factor repeated to pro- duce a power; and (4O) an index or exponent is the number in- dicating the power to which a number is to be raised. 87. The First Power of any number is the number itself, or the root; thus, 2, 3, 5, are first powers or roots. 88. The Second Power, or Square, of a number is the pro- duct arising from using the number two times as a factor; thus, 2 2 ^2x2 = 4; 5 2 =5x5=:25. 89. The Third Power, or Cube, of a number is the product arising from using the number three times as a factor; thus, 4 3 =4x 4x4 = 64. 00. The higher powers are named in the order of their num- bers, as Fourth Power, Fifth Power, Sixth Power, etc. 01. I- What is the third power or cube of 23 ? OPERATION. ANALYSIS. We multiply 23 23 X 23 x 23 = 12167 b J 23 ' and the product by 23; and, since 23 has been taken 3 times as a factor, the last product, 12167, must be the third power or cube of 23. Hence, RULE. Multiply the number by itself as 'many times, less 1, as there are units in the exponent of the required power. NOTE. The process of producing any required power of a number by multi- plication is called Involution. EXAMPLES FOR PRACTICE. 1. What is the square of 72 ? Ans 5184. 2. What is the fifth power of 12 ? Ans. 248832. 3. What is the cube of 25 ? 4. What is the seventh power of 7 ? Ans. 823543. 40 SIMPLE NUMBERS. 5. What is the fourth power of 19 ? Am. 130321. 6. Required the sixth power of 3. Ans. 729. 7. Find the powers indicated in the following expressions; 9 6 , IP, 18 2 , 125 4 , 786 2 , 94 6 , 100 4 , 17 3 , 251.' 8. Multiply 8 3 by 15 2 . Ans. 115200. 9. What is the product of 25 2 X 3 4 ? 10. 7 s ^ 200 = 4* X II 2 , and how many ? Ans. 37,624. GENERAL PRINCIPLES OF MULTIPICATION. 93. There are certain general principles of multiplication, of use in various contractions and applications which occur in sub- sequent portions of this work. These relate, 1st, to changing the factors by addition or subtraction; 2d, to the use of successive factors in continued multiplication. CHANGING THE FACTORS BY ADDITION OR SUBTRACTION. 93. The product is equal to either factor taken as many times as there are units in the other factor. (83, I). Hence, I. Adding 1 to either factor, adds the other factor to the pro- duct. 11. Subtracting 1 from either factor, subtracts the other factor from the product. Hence, III. ADDING any number to either factor, INCREASES the pro* duct by as many times the other factor as there are units in tht number added; and SUBTRACTING any number from either factor, DIMINISHES the product by as many times the other factor as there are units in the number subtracted. CONTINUED MULTIPLICATION. 94. A Continued Multiplication is the process of finding the product of three or more factors, by multiplying the first by the second, this result by the third, and so on. 95. To show the nature of continued multiplication, we observe : 1st. If any number, as 17, be multiplied by any other number, as 3, the result will be 3 times 17; if this result be multiplied by MULTIPLICATION. 41 another number, as 5, the new product will be 5 times 3 times 17, which is evidently 15 times 17. Hence, 17 X 3 X 5 = 17 X 15; the same reasoning would extend to three or more multipliers. 2d. Since 5 times 3 is equal to 3 times 5, (82, 1), it follows that 17 multiplied by 5 times 3 is the same as 17 multiplied by 3 times 5; or 17 X 3 x 5 = 17 X 5 X 3. Hence, the product is not changed by changing the orders of the factors. These principles may be stated as follows : I. If a given number be multiplied by several factors in con- tinued multiplication, the result will be the same as if the given number were multiplied by the product of the several multipliers. II. The product of several factors in continued multiplication will be the same, in whatever order the factors are taken. CONTRACTIONS IN MULTIPLICATION. CASE I. 96. When the multiplier is a composite number. A Composite Number is one that may be produced by multi- plying together two or more numbers. Thus, 18 is a composite number, since (5 x 3 = 18 ; or, 9 x 2 = 18 ; or, 3 x 3 X 2 = 18. 97. The Component Factors of a number are the several numbers which, multiplied together, produce the given number; thus, the component factors of 20 are 10 and 2 (10 X 2 = 20); or, 4 and 5 (4 x 5 = 20) ; or, 2 and 2 and 5 (2 x 2 x 5 = 20). NOTE. The pupil must not confound the factors with the part* of a number. Thus, the factors of which 12 is composed, are 4 and 3 (4 X 3 = 12) ; while the parts of which 12 is composed are 8 and 4 (8 + 4= 12); or 10 and 2 (10 + 2 = 12). The factors are multiplied, while the parts are added, to produce the number. 98. 1. Multiply 327 by 35. ANALYSIS. The factors of 35 are 7 and 5. "We multiply 327 by 7, and this result by 5, and obtain 11445, which must be the same as the product of 327 by 5 times 7, or 35. (95, I). Hence we have the following 42 SIMPLE NUMBERS. RULE. I. Separate the composite number into two or more factors. II. Multiply the multiplicand ~by one of these factors, and that product by another, and so on until all the factors have been used successively ; the last product will be the product required. NOTE. The factors may be used in any order that is most convenient, (95, II). EXAMPLES FOR PRACTICE. 1. Multiply 736 by 24. Ans. 17664. 2. Multiply 538 by 56. Ans. 30128. 3. Multiply 27865 by 84. 4. Multiply 7856 by 144. Ant. 1131264. 5. What will 56 horses cost at 185 each ? 6. If a river discharge 17740872 cubic feet of water in one hour, how much will it discharge in 96 hours ? Ans 1703123712 cubic feet. CASE II. 99. When the multiplier is a unit of any order. If we annex a cipher to the multiplicand, each figure is removed one place toward the left, and consequently the value of the Whole number is increased tenfold, (57, III). If two ciphers are annexed, each figure is removed two places toward the left, and the value of the number is increased one hundred fold ; and every additional cipher increases the value tenfold. Hence, the RULE Annex as many ciphers to the multiplicand as there are ciphers in the multiplier. EXAMPLES FOR PRACTICE. 1. Multiply 364 by 100. Ans. 36400. 2. Multiply 248 by 1000. Ans. 248000. 3. What cost 1000 head of cattle at 50 dollars each? 4. Multiply one million by one hundred thousand ? 5. How many letters will there be on 100 sheets, if each sheet have 100 lines, and each line 100 letters ? Ans. 1000000. MULTIPLICATION. 43 CASE III. 100. When there are ciphers at the right hand of one or both of the factors. 1. Multiply 7200 by 40. OPERATION. ANALYSIS. The multiplicand, factored, is 7200 equal to 72 X 100; the multiplier, factored, is 40 equal to 4 X 10 ; and as these factors taken in 288000 an y or der w iH gi ye tne same product, (95, II), we first multiply 72 by 4, then this product by 100 by annexing two ciphers, and this product by 10 by annexing one cipher. Hence, the following RULE. Multiply the significant figures of the multiplicand by those of the multiplier, and to the product annex as many ciphers as there are ciphers on the right of both factors. EXAMPLES FOR PRACTICE. 1. Multiply 740 by 300. Ans. 222000. 2. Multiply 36000 by 240. Ans. 8640000. 3. Multiply 20700 by 500. 4. Multiply 4007000 by 3002. Ans. 12029014000. 5 Multiply 300200 by 640. CASE IV. 101. "When one part of the multiplier is a factor of another part. 1. Multiply 4739 by 357. OPERATION. ANALYSIS. In this example, 7, one 4700 part of the multiplier, is a factor of 35, the other part. We first find, in the usual manner, the product of the 33173 Prod, by 7 units. mu iti p li ca nd by the 7 units; multi- 165866 Prod, by 35 tens 1691823 Ans. the first figure of the result in tens' place, we obtain the product of the multiplicand by 7 X 5 X 10 = 35 tens ; and the sum of these two par- tial products must be the whole product required. 44 SIMPLE NUMBERS. 2. Multiply 58327 by 21318. OPERATION. ANALYSIS. In this exam 58397 P^ e ' ^ e ^ hundreds is a factor 2131$ of 18, the part on the right of it, and also of 21, the part on Prod, by 3 hundreds. ^ lefl of ^ We fipgt j_ 1049886 Prod, by 18 units. ,. , tiply by 3, writing the first 1224867 Prod, by 21 thousands. . L J * ' * !__ figure in hundreds' place ; 1243414986 Ans. multiplying this product by 6, and writing the first figure in units' place, we obtain the product of the multiplicand by 3 X 6 = 18 units ; multiplying the first partial product by 7, and writing the first figure in thousands' place, we obtain the product of the multipli- cand by 7 X 3 X 1000 = 21 thousands , and the sum of these three partial products must be the entire product required. NOTE. The product obtained by multiplying any partial product is called a derived product. 1O2. From these illustrations we have the following RULE. I. Find the product of the multiplicand by some figure of the multiplier which is a factor of one or more parts of the multiplier. II. Multiply this product ~by that factor which, taken with the figure of the multiplier first used, will produce other parts of the multiplier, and write the .first figure of each result under the first figure of the part of the multiplier thus used. III. In like manner, find the product, either direct or derived, for every figure or part of the multiplier; the sum of all the pro- ducts will be the whole product required. EXAMPLES FOR PRACTICE. 1. Multiply 5784 by 246. Ans. 1422864. 2. Multiply 3785 by 721. Ans. 2728985. 3. Multiply 472856 by 54918. Ans. 25968305808. 4. Multiply 43785 by 7153. Ans 313194105. 5. Multiply 573042 by 24816. Ans. 14220610272. 6. Multiply 78563721 by 127369. 7. Multiply 43 T25652 by 51879 14. MULTIPLICATION. 45 8. Multiply 3578426785 by 64532164. 9. Multiply 2703605 by 4249784. 10. What is the product of 9462108 multiplied by 16824? Ans. 159,190,504,992. EXAMPLES COMBINING THE PRECEDING RULES. 1. A man bought two farms, one containing 175 acres at $28 per acre, and the other containing 320 acres at 837 per acre; what was the cost of both ? Ans. $16,740. 2. If a man receive $1200 salary, and pay $364 for board, $275 for clothing, $150 for books, and $187 for other expenses, how much can he save in 5 years? Ans. $1,120. 3. Two persons start from the same point, and travel in oppo- site directions; one travels 29 miles a day, and the other 32 miles. How far apart will they be in 17 days? Ans. 1,037 miles. 4. A drover bought 127 head of cattle at $34 a head, and 97 head at $47 a head, and sold the whole lot at $40 a head ; what was his entire profit or loss ? Ans. $83 profit. 5. Multiply 675 (77 + 56) by (3 x 156) (214 28). Ans. 152844. 6. Multiply 98 ~\ 6 x (37 + 50) by (64 50; x 5 10. Am. 37200. 7. What is the product of (14 x 25) (9 x 36) + 4324 x (280 112) -f (376~+ 42) x 4 ? Ans 8,00-4,000. 8. In 1850 South Carolina cultivated 29967 farms and planta- tions, containing an average of 541 acres each, at an average value of $2751 for each farm; New Jersey cultivated 23905 farms, con- taining an average of 115 acres each, at an average value of $5030 per farm. How much more were the farming lands of the latter valued at, than those of the former ? 9. There are in the United States 1922890880 acres of land ; of this there were reported under cultivation, in 1850, 1449075 farms, each embracing an average of 203 acres. How many acres were still uncultivated ? 10. Each of the above farms in the United States was valued at an average of $2258, and upon each farm there was an average 46 SIMPLE NUMBERS. of $105 in implements and machinery. What was the aggregate value of the farms and implements ? Ans. $3,424,164,225. Find the values of the following expressions : 11. 2 4 x 5 5 7 3 ? Ans. 49,657. 12. 15 3 (3 2 x 2 5 ) 4- 208 2 9 x 2 4 ? Ans. 46,207. 13. 2 2 + 3 3 4- 4* 4- 5 5 4 6 6 ? 14. In 1852 Great Britain consumed 1200000 bales of American cotton ; allowing each bale to contain 400 pounds, what was its total weight ? 15. If a house is worth $2450, and the farm on which it stands 6 times as much, lacking $500, and the stock on the farm twice as much as the house, what is the value of the whole ? Ans. $21550. 16. A flour merchant bought 1500 barrels of flour at 7 dollars a barrel ; he sold 800 barrels at 10 dollars a barrel, and the re- mainder at 6 dollars a barrel. How much was his gain ? 17. A man invests in trade at one time $450, at another $780, at another $1250, and at another $2275 ; how much must he add to these sums, that the amount invested by him shall be increased fourfold? Ans. $14,265. 18. At the commencement of the year 1858 there were in ope- ration in the United States 35000 miles of telegraph; allowing the average cost to be $115 per mile, what was the total cost? 19. The cost of the Atlantic Telegraph Cable, as originally made, was as follows; 2500 miles at $485 per mile, 10 miles deep- sea cable at $1450 per mile, and 25 miles shore ends at $1250 per mile. What was its total cost ? Ans. $1,258,250. 20. For the year ending June 30, 1859, there were coined in the United States 1401944 double eagles valued at twenty dollars each, 62990 eagles, 154555 half eagles, and 22059 three dollar pieces ; what was the total value of this gold coin? Ans. $29,507,732. DIVISION. 47 DIVISION. 1 OS. Division is the process of finding how many times one number is contained in another. 1O 4. The Dividend is the number to be divided. 105. The Divisor is the number to divide by. 1OO. The Quotient is the result obtained by the process of division. 1 7. The Reciprocal of a number is 1 divided by the number. Thus, the reciprocal of 15 is 1 15, or y 1 ^. NOTES. 1. When the dividend does not contain the divisor an exact number of times, the part of the dividend left is called the Remainder, which must be less than the divisor. 2. As the remainder is always a part of the dividend, it is always of the same name or kind. 3. When there is no remainder the division is said to be exact. 1O8. The method of dividing any number by another depends upon the following principles : I. Division is the reverse of multiplication, the dividend cor- responding to the product, and the divisor and quotient to the factors. II. If all the parts of a number be divided, the entire number will be divided. Since the remainder in dividing any part of the dividend must be less than the divisor, it can be divided only by being expressed in units of a lower order. Hence, III. The operation must commence with the units of the high- est order. 1. Divide 2742 by 6. ANALYSIS. We write the divisor at the left of the dividend, separated from it by a line. As 6 is not contained in 2 thousands, we take 457 Ans. the 2 thousands and 7 hundreds together, and proceed thus ; 6 is contained in 27 hundreds 4 hundred times, and the remainder is 3 hundreds ; we write 4 in hundreds' place in the quotient, and unite the remainder, 3 hundreds, 48 SIMPLE NUMBERS. to the next figure of the dividend, making 34 tens ; then, 6 is con- tained in 34 tens 5 tens times, and the remainder is 4 tens ; writing 5 tens in its place in the quotient, we unite the remainder to the next figure in the dividend, making 42 ; 6 is contained in 42 units 7 times, and there is no remainder ; writing 7 in its place in the quotient, we have the entire quotient, 457. NOTE 1. The different numbers which we divide in obtaining the successive figures of the quotient, are called partial dividends. 2. Divide 18149 by 56. ANALYSIS. As neither 1 nor 18 OPERATION. ... , . ,, ,. . will contain the divisor, we take 56 ) 18149 ( 324/ ff Ans. three figures , igl, for the first par- ^Q tial dividend. 56 is contained in 134 181 3 times, and a remainder ; we write the 3 as the first figure in 229 the quotient, and then multiply 224 the divisor by this quotient figure ; Z 3 times 56 is 168, which subtracted from 181, leaves 13 ; to this re- mainder we annex or bring down 4, the next figure of the dividend, and thus form 134, the next partial dividend; 56 is contained in 134 2 times, and a remainder; 2 times 56 is 112, which subtracted from 134, leaves 22 ; to this remainder we bring down 9, the last figure of the dividend, and we have 229, the last partial dividend ; 56 is contained in 229 4 times, and a remainder ; 4 times 56 is 224, which subtracted from 229, gives 5, the final remainder, which we write in the quotient with the divisor below it, thus completing the division, (35). NOTE 2. When the multiplication and subtraction are performed mentally, ns in the first example, the operation is called Short Division ; but when the work is written out in full, as in the second example, the operation is called Long Division. The principles governing the two methods are the same. 1O9. From these principles and illustrations we derive the following general RULE. I. Beginning at the left hand, take for the first partial dividend the fewest figures of the given dividend that will contain the divisor one or more times ; find how many times the divisor is contained in this partial dividend, and write the result in the quotient; multiply the divisor ly this quotient figure, and subtract the product from the partial dividend used. DIVISION. 49 II. To the remainder bring down the next figure of the dividend, with which proceed as before ; and thus continue till all the figures of the dividend have been divided. III. If the division is not exact, place the final remainder in the quotient, and write the divisor underneath. HO. PROOF. There are two principal methods of proving division. 1st. By multiplication. Multiply the divisor and quotient together, and to the product add the remainder, if any ; if the result be equal to the dividend, the work is correct. (1O8, I.) NOTE. In multiplication, the two factors are given to find the product; in division, the product and one of the factors are given to find the other factor. 2d. By excess of 9's. 111. Subtract the remainder, if any, from the dividend, and find the excess of 9's in the result. Multiply the excess of 9's in the divisor by the excess of 9's in the quotient, and find the excess of 9's in the product; if the latter excess is the same as the former, the work is supposed to be correct. (85.) EXAMPLES FOR PRACTICE. (1.) (2.) (3.) (4.) 6)473832 8)972496 9)1370961 12)73042164 Quotients. 5. Divide 170352 by 36. 4732. 6. Divide 409887 by 47. 8721. 7. Divide 443520 by 84. 5280. 8. Divide 36380250 by 125. 291042. 9. Divide 1554768 by 216. 10. Divide 3931476 by 556. 11. Divide 48288058 by 3094. Re m. 12. Divide 11214887 by 232. 7. 13. Divide 27085946 by 216. 194. 14. Divide 29137062 by 5317. 5219. 15. Divide 4917968967 by 2359. 1255- 5 D 50 SIMPLE NUMBERS. 16. What is the value of 721198 -5- 291 ? Rem. 100. 17. What is the value of 3844449-^-657? 342. 18. What is the value of 536819237 -h 907 ? 403. 19. What is the value of 571943007145 -~ 37149 ? 12214. 20. What is the value of 48659910-^-54001 ? 5009. 21. The annual receipts of a manufacturing company are $147675; how much is that per day, there being 365 days in the jear? Arts. $404|Jf. 22. The New York Central Railroad Company, in 1859, owned 556 miles in length of railroad, which cost, for construction and equipment, $30732518; what was the average cost per mile? Ans. $55,274Jf4. 23. The Memphis and Charleston Railroad is 287 miles in length, and cost $5572470 ; what was the average cost per mile ? An*. $19,416^V 24. The whole number of Post offices in the United States, in 1858, was 27977, and the revenue was $8186793 ; what was the average income to an office? ABBREVIATED LONG DIVISION. llSJo We may avoid writing the products in long division, and obtain the successive remainders by the method of subtraction employed in the case of several subtrahends. (7G.) 1. Divide 261249 by 487. OPERATION. ANALYSIS. Dividing the first partial 487 ) 261249 ( 536 dividend, 2612, we obtain 5 for the first 177 figure of the quotient. We now multi- 313 ply 487 by 5 ; but instead of writing the 217 Kem. product, and subtracting it from the partial dividend, we simply observe what figures must be added to the figures of the product, as we proceed, to give the figures of the partial dividend, and write them for the remainder sought. Thus, 5 times 7 are 35, and 7 (written in the remainder,) are 42, a number whose unit figure is the same as the right hand figure of the partial dividend ; 5 times 8 are 40, and 4, the tens of the 42, are 44, and 7 (written in the remainder,) are 51 ; 5 DIVISION. 51 times 4 are 20, and 5, the tens of the 51, are 25, and 1 (written in the remainder,) are 26. We next consider the whole rejnainder, 177, as joined with 4, the next figure of the dividend, making 1774 for the next partial dividend. * Proceeding as before, we obtain 313 for the second remainder, 217 for the final remainder, and 536 for the entire quotient. Hence, the following RULE. I. Obtain the first figure in the quotient in the usual manner. II. Multiply the first figure of the divisor by this quotient figure, and write such a figure in the remainder as, added to this partial product, will give an amount having for its unit figure the first or right hand figure of the partial dividend used. III. Carry the tens' figure of the amount to the product of the next figure of the divisor, and proceed as before, till the entire remainder is obtained. IV. Conceive this remainder to be joined to the next figure of the dividend, for a new partial dividend, and proceed as with the former, till the work is finished. EXAMPLES FOR PRACTICE. 1. Divide 77112 by 204. Ans. 378. 2. Divide 65664 by 72. Ans. 912. 3. Divide 7913576 by 209. Ans. 37864. 4. Divide 6636584 by 698. 5. Divide 4024156 by 8903. Ans. 452. 6. Divide 760592 by 6791. 7. Divide 101443929 by 25203. Ans. 8. Divide 1246038849 by 269181. Ans. 4629. 9. Divide 2318922 by 56240. 10. Divide 1454900 by 17300. Ans. GENERAL PRINCIPLES OF DIVISION. 11*1. The general principles of division most important in their application, relate; 1st, to changing the terms of division by addition or subtraction ; 2d, to changing the terms of division by multiplication or division ; 3d, to successive division. 52 SIMPLE NUMBERS. 11/4. The quotient in division depends upon the relative values of the dividend and divisor Hence, any change in the value of either dividend or divisor must produce a.change in the value of the quotient; though certain changes may be made in both divi- dend and divisor, at the same time, that will not affect the quotient. CHANGING THE TERMS BY ADDITION OR SUBTRACTION. 115. Since the dividend corresponds to a product, of which the divisor and quotient are factors, we observe, 1st. If the divisor be increased by 1, the dividend must be increased by as many units as there are in the quotient, in order that the quotient may remain the same, (03, I) ; and if the divi- dend be not thus increased, the quotient will be diminished by as many units as the number of times the new divisor is contained in the quotient. Thus, 84 -*- 6 = 14 84 -*- 7 = 14 V = 12 2d. If the divisor be diminished by 1 , the dividend must be diminished by as many units as there are in the quotient, in order that the quotient may remain the same, (93, II) ; and if the dividend be not thus diminished, the quotient will be increased by as many units as the number of times the new divisor is con- tained in the quotient. Thus, 144 _=_ 9 = 16 144 _j_ 8 = 16 + V = 18 These principles may be stated as follows : I. Adding 1 to the divisor taJces as many units from the quotient as the new divisor is contained times in the quotient. II. Subtracting 1 from the divisor adds as many units to the quotient as the new divisor is contained times in the quotient. Hence, III. ADDING any number to the divisor SUBTRACTS as many units from the quotient as the new divisor is contained times in the pro- duct of the quotient by the number added; and SUBTRACTING DIVISION. 53 any number from the divisor ADDS as many units to the quotient as the new divisor is contained times in the product of the quo- tient by the number subtracted. CHANGING THE TERMS BY MULTIPLICATION OR DIVISION. 11O. There are six cases : 1st. If any divisor is contained in a given dividend a certain number of times, the same divisor will be contained in twice the dividend twice as many times; in three times the dividend, three times as many times ; and so on. Hence, Multiplying the dividend l>y any number, multiplies the quotient, by the same number. 2d. If any divisor is contained in a given dividend a certain number of times, the same divisor will be contained in one half the dividend one half as many times ; in one third the dividend, one third as many times ; and so on. Hence, Dividing the dividend by any number, divides the quotient by the same number. 3d. If a given divisor is contained in any dividend a certain number of times, twice the divisor will be contained in the same dividend one half as many times ] three times the divisor, one third as many times ; and so on. Hence, Multiplying the divisor by any number, divides the quotient by the same number. 4th. If a given divisor is contained in any dividend a certain number of times, one half the divisor will be contained in the same dividend twice as many times ; one third of the divisor, three times as many times ', and so on. Hence, Dividing the divisor by any number, multiplies the quotient by the same number. 5th. It a given divisor is contained in a given dividend a cer- tain number of times, twice the divisor will be contained the same number of times in twice the dividend ; three times the divisor will be contained the same number of times in three times the dividend ; and so on. Hence, 5* 54 SIMPLE NUMBERS. Multiplying both dividend and divisor by the same number does not alter the quotient. 6th. If a given divisor is contained in a given dividend a cer- tain number of times, one half the divisor will be contained the same number of times in one half the dividend ; one third of the divisor will be contained the same number of times in one third of the dividend ; and so on. Hence, Dividing both dividend and divisor by the same number does not alter the quotient. NOTE. If a number be multiplied and the product divided by the same num- ber, the quotient will be equal to the number multiplied; hence the 5th case may be regarded as a direct consequence of the 1st and 3d; and the 6th, as the direct consequence of the 2d and 4th. To illustrate these cases, take 24 for a dividend and 6 for a divisor then the quotient will be 4, and the several changes may be represented in theii order as follows : Dividend. Divisor. Quotient. 24 -*- 6 = 4 i 4% n o | Multiplying the dividend by 2 multi- 8 { plies the quotient by 2. n in a of Dividing the dividend by 2 divides 1 \ the quotient by 2. o 9 -19 of Multiplying the divisor by 2 divides - { the quotient by 2. Q f Dividing the divisor by 2 multiplies 8 { the quotient by 2. , -19 . ( Multiplying both dividend and divisor { by 2 does not alter the quotient. fi 1 9 o A f Dividing both dividend and divisor by { 2 does not alter the quotient. 117. These six cases constitute three general principles, which may now be stated as follows : PRIN. I. Multiplying the dividend multiplies the quotient; and dividing the dividend divides the quotient. PRIN. II. Multiplying the divisor divides the quotient; and dividing the divisor multiplies the quotient. DIVISION. 55 PRIN. III. Multiplying or dividing both dividend and divisor by the same number, does not alter the quotient. 118. These three principles may be embraced in one GENERAL LAW. A change in the dividend produce* a LIKE change in the quo- tient; but a change in the divisor produces an OPPOSITE change in the quotient. SUCCESSIVE DIVISION. 11O. Successive Division is the process of dividing one number by another, and the resulting quotient by a second divisor, and so on. Successive division is the reverse of continued multiplication. Hence, I. If a given number be divided by several numbers in succes- sive division, the result will be the same as if the given number were divided by the product of the several divisors, (O5, I). II. The result of successive division is the same, in whatever order the divisors are taken, (95, II). CONTRACTIONS IN DIVISION. CASE I. 12O. When the divisor is a composite number. 1. Divide 1242 by 54. OPERATION ANALYSIS. The component factors of 54 are 6") 1?4 9 6 and 9. We divide 1242 by 6, and the re- sulting quotient by 9, and obtain for the final result, 23, which must be the same as the 23 Ans. quotient of 1242 divided by 6 times 9, or 54, (119, I). We might have obtained the same result by dividing first by 9, and then by 6, (119, II). Hence the following RULE. Divide the dividend by one of the factors, and the quo* 56 SIMPLE NUMBERS. tient thus obtained by another, and so on if there be more than two factors, until every factor has been made a divisor. The last quo- tient will be the quotient required. TO FIND THE TRUE REMAINDER. If remainders occur in successive division, it is evident that the true remainder must be the least number, which, sub- tracted from the given dividend, will render all the divisions exact 1. Divide 5855 by 168, using the factors 3, 7, and 8, and find the true remainder. OPERATION. ANALYSIS. Dividing the 3) 5855 given dividend by 3, we have 7)1951 2 " or a - - remainder of 2. Hence, 2 8) 2 ' 8 ......... 5x3= 15 subtracted from 5855 would 34 ... 6 X 7 X 3 =126 render the first division exact, True remainder ........................... 143 and we therefore write 2 for a part of the true remainder. Dividing 1951 by 7, we have 278 for a quotient, and a remainder of 5. Hence, 5 subtracted from 1951 would render the second division exact. But to dimmish 1951 by 5 would require us to diminish 1951 X 3, the dividend of the first exact division, by 5 X 3 15, (93, III) ; and we therefore write 15 for the second part of the true remainder. Dividing 278 by 8, we have 34 for a quotient, and a remainder of 6. Hence, 6 subtracted from 278 would render the third division exact. But to diminish 278 by 6 would require us to diminish 278 X 7, the dividend of the second exact division, by 6 X 7 ; or 278 X 7 X 3, the dividend of the first exact division, by 6 X 7 X 3 = 126 ; and we therefore write 126 for the third part of the true remainder. Adding the three parts, we have 143 for the entire remainder. Hence the following RULE. I. Multiply each partial remainder by all the preceding divisors. II. Add the several products; the sum will be the true re- mainder. DIVISION. 57 EXAMPLES FOR PRACTICE. 1. Divide 435 by 15 = 3 x 5. Ans. 29. 2. Divide 4256 by 56 = 7 X 8. 3. Divide 17856 by 72 = 9 x 8. 4. Divide 15288 by 42 = 2 x 3 x 7. Am. 364. 5. Divide 972552 by 168 = 8 x 7 x 3. Ans. 5789. 6. Divide 526050 by 126 = 9 x 7 X 2. 7. Divide 612360 by 105 = 7 x 5 x 3. Ans. 5832. 8. Divide 553 by 15 = 3 x 5. R em . 13. 9. Divide 10183 by 105 = 3 x 5 x 7. 103. 10. Divide 10197 by 120 = 2 x 3 x 4 x 5. 117. 11. Divide 29792 by 144 = 3 x 8 x 6. 128. 12. Divide 73522 by 168 = 4 x 6 X 7. 106. 13. Divide 63844 by 135 = 3 x 5 x 9, 124. 14. Divide 386639 by 720 = 2 x 3 x 4 x 5 x 6. 719. 15. Divide 734514 by 168 = 4 x 6 X 7. 18. 16. Divide 636388 by 729 = 9 s . 700. 17. Divide 4619 by 125 = 5 3 . 119. 18. Divide 116423 by 10584 = 3 x 7 2 x 8 x 9. 10583. 19. Divide 79500 by 6125 = 5 3 x 7 2 . 6000. CASE II. 122. When the divisor is a unit of any order. If we cut off or remove the right hand figure of a number, each of the other figures is removed one place toward the right, and, consequently, the value of each is diminished tenfold, or divided by 10, (oT, III). For a similar reason, by cutting off two figures we divide by 100 ; by cutting off three, we divide by 1000, and so on; and the figures cut off will constitute the remainder. Hence the RULE, from the right hand of the dividend cut off as many figures as there are ciphers in the divisor. Under the figures so cut off, place the divisor, and the whole will form the quotient. 58 SIMPLE NUMBERS. EXAMPLES FOR PRACTICE. 1. Divide 79 by 10. Ans. 7 T % 2. Divide 7982 by 100. 3. Divide 4003 by 1000. Ans. 4. Divide 2301050 by 10000. 5. Divide 3600036 by 1000. Ans. CASE III. 123. When there are ciphers on the right hand of the divisor. I. Divide 25548 by 700. OPERATION. ANALYSIS. We resolve 700 into the factors 10 and 7 - Dividing first by 100, the quo- 36 Quotient. 3 2g. 2. Divide 13872 by % 500. 3. Divide 83248 by 2600. Ans. 4. Divide 1548036 by 4300. Ans. 5. Divide 436000 by 300. Ans. 6. Divide 66472000 by 8100. 1. Divide 10818000 by 3600. DIVISION. 59 EXAMPLES COMBINING THE PRECEDING RULES. 1. How many barrels of flour at $8 a barrel, will pay for 25 tons of coal at $4 a ton, and 36 cords of wood at $3 a cord ? Ans. 26. 2. A grocer bought 12 barrels of sugar at $16 per barrel, and 17 barrels at $13 per barrel ; how much would he gain by selling the whole at $18 per barrel ? 3. A farmer sold 300 bushels of wheat at $2 a bushel, corn and oats to the amount of $750 ; with the proceeds he bought 120 head of sheep at $3 a head, one pair of oxen for $90, and 25 acres of land for the remainder How much did the land cost him per acre ? Ans. $36. 4. Divide 450 -f (24 12) x 5 by (90 -i- 6) -f (3 x 11) 18. Ans. 17. 5. Divide 648 x (3 2 x 2 3 ) -*- 9 (2910 -- 15) by 2863 ~ (4375 -TT75) X 4 2 + 3 2 . Ans. 712f . 6. The product of three numbers is 107100; one of the numbers is 42, and another 34. What is the third number ? Ans. 75. 7. What number is that which being divided by 45, the quo- tient increased by 7 2 -f 1> the sum diminished by the difference between 28 and 16, the remainder multiplied by 6, and the pro- duct divided by 24, the quotient will be 12 ? Ans. 450. 8. A mechanic earns $60 a month, but his necessary expenses are $42 a month. How long will it take him to pay for a farm of 50 acres worth $36 an acre ? 9. What number besides 472 will divide 251104 without a re- mainder? Ans. 532. 10. Of what number is 3042 both divisor and quotient ? Ans. 9253764. 11. What must the number be which, divided by 453, will give the quotient 307, and the remainder 109 ? Ans. 139180. 12. A farmer bought a lot of sheep and hogs, of each an equal number, for $1276. He gave $4 a head for the sheep, and $7 a 50 SIMPLE NUMBERS head for the hogs ; what was the whole number purchased, and how much was the difference in the total cost of each ? Ans. 232 purchased ; $348 difference in cost. 13. According to the census of 1850 the total value of the tobacco raised in the United States was $13,982,686. How many school-houses at a cost of $950, and churches at a cost of $7500, of each an equal number, could be built with the proceeds of the tobacco crop of 1850 ? Ans. 1654, and a remainder of $6386. 14. The entire cotton crop in the United States in 1859 was 4,300,000 bales, valued at $54 per bale. If the entire proceeds were exchanged for English iron, at $60 per ton, how many tons would be received ? 15. The population of the United States in 1850 was 23,191,876. It was estimated that 1 person in every 400 died of intemperance. How many deaths may be attributed to this cause in the United States, during that year ? 16. In 1850, there were in the State of New York, 10,593 public schools, which were attended during the winter by 508464 pupils ; what was the average number to each school ? Ans. 48. 17. A drover bought a certain number of cattle for $9800, and sold a certain number of them for $7680, at $64 a head, and gained on those he sold $960. How much did he gain a head, and how many did he buy at first ? Ans. Gained $8 per head; bought 175. 18. A house and lot valued at $1200, and 6 horses at $95 each, were exchanged for 30 acres of land. At how much was the land valued per acre ? 19. If 16 men can perform a job of work in 36 days, in how many days can they perform the same job with the assistance of 8 more men ? Ans. 24. 20. Bought 275 barrels of flour for $1650, and sold 186 bar- rels of it at $9 a barrel, and the remainder for what it cost. How much was gained by the bargain ? Ans. $558. 21 A grocer wishes to put 840 pounds of tea into three kinds of boxes, containing respectively 5, 10, and 15 pounds, using the PROBLEMS. 61 same number of boxes of each kind. How many boxes can he fill? Ans. 84. 22. A coal dealer paid $965 for some coal. He sold 160 tons for $5 a ton, when the remainder stood him in but $3 a ton. How many tons did he buy ? Ans. 215. 23. A dealer in horses gave $7560 for a certain number, and sold a part of them for $3825, at $85 each, and by so doing, lost $5 a head ; for how much a head must he sell the remainder, to gain $945 on the whole ? Ans. $120. 24. Bought a Western farm for $22,360, and after expending $1742 in improvements upon it, I sold one half of it for $15480, at $18 per acre. How many acres of land did I purchase, and at what price per acre ? PKOBLEMS IN SIMPLE INTEGRAL NUMBERS. 124. The four operations that have now been considered, viz., Addition, Subtraction, Multiplication, and Division, are all the operations that can be performed upon numbers, and hence they are called the Fundamental Rules. 12o. I n all cases, the numbers operated upon and the results obtained, sustain to each other the relation of a whole to its parts. Thus, I. In Addition, the numbers added are the parts, and the sum or amount is the whole. II. In Subtraction, the subtrahend and remainder are the parts, and the minuend is the whole. III. In Multiplication , the multiplicand denotes the value of one part, the multiplier the number of parts, and the pro- duct the total value of the whole number of parts. IV. In Division, the dividend denotes the total value of the whole number of parts, the divisor the value of one part, and the quotient the number of parts ; or the divisor the number of parts, and the quotient the value of one part. 126. Every example that can possibly occur in Arithmetic, and every business computation requiring an arithmetical opera- ^ r 62 SIMPLE NUMBERS. tion, can be classed under one or more of the four Fundamental Rules, as follows : I. Cases requiring Addition. There may Le given To find 1. The parts, the whole, or the sum total. 2 The less of two numbers and their difference, or the sub- trahend and remainder, II. Cases requiring Subtraction. There may be y any prime factor ; divide the quotient in the same manner, and so continue the division until the quotient is a prime number. The several divisors and the last quotient will be the prime factors required, PROOF. The product of all the prime factors wil 1 be the given number. EXAMPLES FOR PRACTICE. 1. What are the prime factors of 2150? 2. What are the prime factors of 2445? 3. What are the prime factors of 6300 ? 4. What are the prime factors of 21504? 5. What are the prime factors of 2366 ? 6. What are the prime factors of 1000 ? 7. What are the prime factors of 390625? 8. What are the prime factors of 999999 ? 143. If the prime factors of a number are small, as 2, 3, 5, 7, or 11, they may be easily found by the tests of divisibility, (13G), or by trial. But numbers may be proposed requiring many trials to find their prime factors. This difficulty is obviated, within a certain limit, by the Factor Table given on pages 72, 73. By prefixing each number in bold-face type in the column of Numbers, to the several numbers following it in the same divis- ion of the column, we shall form all the composite numbers less than 10,000, and not divisible by 2, 3, 5, 7, or 11; the numbers in the columns of Factors are the least prime factors of the num- bers thus formed respectively. Thus, in one of the columns of Numbers we find 39, in bold-face type, and below 39, in the same column, is 77, which annexed to 39, forms 3977, a composite num- ber. The least prime factor of this number is 41, which we find at the right of 77, in the column of Factors. 144, Hence, for the use of this table, we have the following RULE. I. Cancel from the given number all factors less than 13, and then find the remaining factors by the table. II. If any number less than 10,000 is not found w the table, and is not divisibk by 2, 3, 5, 7, or 11, it is prime. 72 PROPERTIES OF NUMBERS. FACTOR TABLE. 1 < e .. 2 o 1 5 1 I 1 P j 5 n 5 o 1 t umbers, ictors. |j 1 1 1 umbers, ictors. 1 99 29 11 I 7 43 29 79 37 41 17 83 17 41 23 09 31 17 53 77 31 69 13 9 17 13 57 19 83 13 51 13 97 43 59 17 13 19 27 29 83 71 2 01 17 57 31 61 J 7 89 19 77 13 34 69 53 21 29 47 47 91 29 21 13 23 13 69 13 63 13 91 47 83 19 01 19 87 13 31 61 67 fff 52 47 13 43 23 15 20 25 87 29 03 41 93 17 43 43 69 19 07 41 89 17 49 13 01 19 21 4 j 01 41 93 41 19 ij 39 51 19 71 13 13 13 99 13 61 31 13 17 33 19 07 23 3O 27 zj 01 47 69 17 77 17 19 17 3 89 23 17 37 41 13 09 13 07 31 31 47 37 31 79 29 48 21 23 23 17 10 37 29 47 23 33 17 13 23 39 19 53 59 81 13 11 17 39 13 61 19 03 17 41 23 59 29 37 43 29 13 73 23 59 37 87 41 19 61 49 29 77 13 07 19 77 19 71 19 61 13 43 17 81 59 61 17 93 23 41 47 51 59 91 17 27 13 91 37 77 31 67 17 53 43 97 13 73 29 99 19 ij 51 zj 31 47 97 *9 17 37 07 41 31 37 99 41 03 17 11 17 31 29 27 17 69 17 37 79 85 27 79 13 67 61 43 57 07 Ji 27 61 43 5J 39 41 71 19 43 17 07 47 47 aj 22 19 63 13 07 ij 09 41 33 47 53 17 61 17 81 ji 49 29 0967 57 ij 47 ij 73 29 13 29 19 Z9 39 ij 73 19 63 37 83 43 53 ji 31 19 59 17 53 47 97 97 23 59 37 17 41 ji 89 Z 9 67 53 87 ij 59 41 49 8j 77 47 67 17 99 41 29 17 57 47 57 79 70 73 73 78 77 ij 51 17 83 ij 79 83 98 59 ij 61 61 83 zg 03 47 79 47 01 29 89 19 57 43 89 89 89 41 09 17 67 7J 69 ji 93 19 09 4J 87 8j 07 J7 82 67 ij 93 17 94 27 31 71 29 79 J 7 66 31 79 91 19 11 73 01 59 79 23 90 07 23 41 13 73 2j 87 zj 13 17 33 ij 97 IJ 13 ij 03 ij 87 Ji 17 71 09 97 47 4 3 77 53 91 41 17 ij 37 Ji 74 31 41 07 29 93 ij 19 29 51 13 5359 58 62 23 J7 61 zj 09 ji 37 17 13 43 86 47 8j 69 17 69 71 09 J 7 27 IJ 31 19 67 J 7 21 41 49 47 27 19 11 79 61 ij 81 19 81 41 33 19 33 zj 4129 81 73 23 Ij 59 29 49 73 21 J 7 71 47 87 53 93 13 37 ij 39 17 47 17 87 19 29 17 71 17 51 J7 33 89 73 43 95 99 19 9143 41 79 49 61 93 41 39 43 91 ij 57 zj 39 S j 77 29 03 13 99 93 71 53 ij 67 59 97 47 53 zg 97 53 79 vj 51 41 83 ji 09 37 13 23 99 17 83 61 83 41 99 ji 63 17 79 99 4 j 53 17 896i 17 ji 1747 59 89 19 97 J 7 71 71 ji 13 41 83 71 ij 91 23 89 37 19 09 19 63 67 11 ij 93 59 21 89 03 19 83 19 01 19 29 13 43 61 11 ZJ 13 59 07 19 23 17 75 39 17 21 5 j 87 13 ij 53 41 53 37 17 61 19 71 31 53 41 J7 01 ij 43 ij 33 13 11 ji 31 2J 57 19 59 23 21 ji 31 ij 39 zj 53 zj 19 7 j 57 7 j 39 31 17 2J 39 ij 63 73 71 ij 33 17 41 17 49 17 57 17 31 17 61 19 41 19 49 ij 43 4 i 71 17 79 17 41 ij 71 zj 51 4 j 63 ij 43 19 67 ji 47 17 59 19 67 89 77 61 83 67 47 19 83 ij 57 29 69 67 71 67 69 ij 57 61 73 ji 69 5 j 89 43 91 97 59 59 64 67 67 71 71 97 71 79 79 59 13 77 67 79 67 93 53 97 IJ 63 67 01 J 7 73 ij 81 43 76 81 zj 81 17 91 59 93 29 99 29 69 47 03 19 93 ij 99 zj 13 zj 91 61 83 83 97 19 97 17 96 77 43 07 43 68 72 19 19 99 19 99 37 88 92 07 13 83 ji 09 ij 17 17 01 19 27 zg 80 84 01 ij 11 61 17 59 89 5 j 31 59 21 19 23 31 31 ij 03 5 j 01 ji 09 23 17 13 37 23 93 13 37 41 47 41 41 13 33 17 21 ij 11 41 43 37 23 23 41 31 74 PROPERTIES OF NUMBERS. t 1. Resolve 1961 into its prime factors. OPERATION. ANALYSIS. Cutting off the two 1961 37 = 53 rig^ hand figures of the given 1961 = 37 X 53, Ans. number, and referring to the table, column No., we find the other part, 19, in bold-face type ; and under it, in the same division of the column, we find 61, the figures cut off; at the right of 61, in column Fac., we find 37, the lea'st prime factor of the given number. Dividing by 37, we obtain 53, the other factor. 2. Resolve 188139 into its prime factors. OPERATION. ANALYSIS. "We find by trial 3 7 17 17 188139 *hat the given number is divisible _. by 3 and 7 ; dividing by these fac- tors, we have for a quotient 8959. 8959 By referring to the factor table, 527 we find the least prime factor of this number to be 17 ; dividing by 17, we have 527 for a quotient, 3x7x17x17x31, Ans. Referring again to the tatle, we find 17 to be the least factor of 527, and the other factor, 31, is prime. EXAMPLES FOR PRACTICE. 1. Resolve 18902 into its prime factors. Ans. 2, IS, 727. 2. Resolve 352002 into its prime factors. 3. Resolve 6851 into its prime factors. 4. Resolve 9367 into its prime factors. 5. Resolve 203566 into its prime factors. 6. Resolve 59843 into its prime factors. 7. Resolve 9991 into its prime factors. 8. Resolve 123015 into its prime factors. 9. Resolve 893235 into its prime factors. 10. Resolve 390976 into its prime factors. 11. Resolve 225071 into its prime factors. 12. Resolve 81770 into its prime factors. 13. Resolve 6409 into its prime factors. 14. Resolve 178296 into its prime factors. 15. Resolve 714210 into its prime factors. FACTORING. 75 CASE II. 145. To find all the exact divisors of a number. It is evident that all the prime factors of a number, together with all the possible combinations of those prime factors, will con- stitute all the exact divisors of that number, (142, II). 1 . What are all the exact divisors of 360 ? OPERATION. 360 = 1x2x2x2x3x3x5. 1,2 4 , 8 Combinations of 1 and 2. Ans.< 3 , 6 9 , 18 5 , 10 15 , 30 45 , 90 land 2 and 3. 12 , 24 36 , 72 20 , 40 " " " 1 and 2 and 5 180 360 " "land 2 and 3 and 5- ANALYSIS. By Case I we find the prime factors of 360 to be 1, 2. 2, 2, 3, 3, and 5. As 2 occurs three times as a factor, the different combinations of 1 and 2 by which 360 is divisible will be 1, 1x2 = 2, 1x2x2 = 4, and 1x2x2x2 = 8; these we write in the first line. Multiplying the first line by 3 and writing the products in the second line, and the second line by 3, writing the products in the third line, we have m the first, second and third lines all the different combina- tions of 1, 2, and 3, by which 360 is divisible. Multiplying the first second and third lines by 5, and writing the products in the fourth, fifth and sixth lines, respectively, we have in the six lines together, every combination of the prime factors by which the given number^ 360, is divisible. Hence the following RULE. I. Resolve the given number into its prime factors. II Form a series having 1 for the first term, that prime factor which occurs the greatest number of times in the given number for the second term, the square of this factor for the third term, and so on, till a term is reached containing this factor as many times as it occurs in the given number. III. Multiply the numbers in this line by another factor, jind these results by the same factor, and so on, as many times a* thi* factor occurs in the given number. 76 PROPERTIES OF NUMBERS. IV. Multiply all the combinations now obtained by another factor in continued multiplication, and thus proceed till all the dif- ferent factors have been used. ALL the combinations obtained will be the exact divisors sought. EXAMPLES FOR PRACTICE. 1. What are all the exact divisors of 120 ? Ans. 1,2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. 2. Find all the exact divisors of 84. Ans. 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. 3. Find all the exact divisors of 100. Ans. 1, 2, 4, 5, 10, 20, 25, 50, 100. 4. Find all the exact divisors of 420. 1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, . 30, 35, 42, 60, 70, 84, 105, 140, 210, 420. 5. Find all the exact divisors of 1050. 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 25, 30, 35, 42, 50, 70, 75, 105, 150, 175, 210, 350, 525, 1050. GREATEST COMMON DIVISOR. 94O. A Common Divisor of two or more numbers is a number that will exactly divide each of them. 147. The Greatest Common Divisor of two or more numbers is the greatest number that will exactly divide each of them. 148. Numbers Prime to each other are such as have no com- mon divisor. NOTE. A common divisor is sometimes called a Common Measure ; and the greatest common divisor, the Greatest Common Measure. CASE I. 149. When the numbers can be readily factored. It is evident that if several numbers have a common divisor, they may all be divided by any component factor of this divisor, and the resulting quotients by another component factor, and so on, till all the component factors have been used. GREATEST COMMON DIVISOR. 77 I. What is the greatest common divisor of 28, 140, and 420 ? OPERATION. ANALYSIS. We readily see that 7 7 28 . . 140 . . 420 will exactly divide each of the given * ~~7 M ^ numbers ; and then, 4 will exactly divide each of the resulting quotients. * ' ' ' Hence, each of the given numbers 4x7 = 28, Ans. can he exactly divided by 7 times 4 ; and these numbers must be compo- nent factors of the greatest common divisor. Now, if there were any other component factor of the greatest common divisor, the quotients, 1, 5 and 15, would be divisible by it. But these quotients are prime to each other ; therefore, 7 and 4 are all the component factors of the greatest common divisor sought. From this analysis we derive the following RULE. I. Write the numbers in a line, with a vertical line at the left, and divide by any factor common to all the numbers. II. Divide the quotients in like manner, and continue the divi- sion till a set of quotients is obtained that are prime to each other. II F. Multiply all the divisors together, and the product will be the greatest common divisor sought. EXAMPLES FOR PRACTICE. 1. What is the greatest common divisor of 40, 75, and 100 ? Ans. 5. 2. What is the greatest common divisor of 18, 30, 36, 42, and 54 ? 3. What is the greatest common divisor of 42, 63, 126, and 189? Ans. 21. 4. What is the greatest common divisor of 135, 225, 270, and 315? Ans. 45. 5. What is the greatest common divisor of 84, 126, 210, 252, 294, and 462 ? 6. What is the greatest common divisor of 216, 360, 432, 648, and 936 ? Ans. 72. 7. What is the greatest common divisor of 102 ; 153, and 255 ? Ans. 51. 7* 78 PROPERTIES OF NUMBERS. 8. What is the greatest common divisor of 756, and 1575? 9. What is the greatest common divisor of 182, 864, and 455? 10. What is the greatest common divisor of 2520, and 3240 ? Am. 360. 11. What is the greatest common divisor of 1428, and 1092 ? 12. What is the greatest common divisor of 1008, and 1036? Ans. 28. CASE II. IoO. When the numbers cannot be readily factored. The analysis of the method in this case depends upon the following properties of divisors. I. An exact divisor divides any number of times its dividend. II. A common divisor of two numbers is an exact divisor of their sum. III. A common divisor of two numbers is an exact divisor of their difference. NOTE. The last two properties are essentially the same as 102, II, III. 1. What is the greatest common divisor of 527, and 1207 ? OPERATION. ANALYSIS. We will first describe the pro- 1207 cess, and then examine the reasons for the 527 459 68 68 1054 several steps in the operation. Drawing two vertical lines, we place the greater number 15o on the right, and the less number on the left, 13(3 one line lower down. We then divide 1207, -TI the greater number, by 527, the less, and v. rite the quotient, 2, between the verticals, the product, 1054, opposite the less number and under the greater, and the remainder, 153, below. We next divide 527 by this re- mainder, writing the quotient, 3, between the verticals, the product, 459, on the left, and the remainder, 68, below. We again divide the last divisor, 153, by 68, and obtain 2 for a quotient, 136 for a pro- duct, and 17 for a remainder, all of which we write in the same order as in the former steps. Finally, dividing the last divisor, 68, by the last remainder, 17, we have no remainder, and 17, the last divisor, is the greatest common divisor of the given numbers. Now, observing that the dividend is always the sum of the product and remainder, and that the remainder is always the difference of the GREATEST COMMON DIVISOR. 79 OPERATION. 527 459 68 fW , 1 , 1207 1054 153 136 17 3 2 4 dividend and product, we first trace the work in the reverse order, as indicated by the arrow line in the diagram below. 17 divides 68, as proved by the last division ; it will also divide 2 times 68, or 136, (I). Now, as 17 divides both itself and 136, it will divide 153, their sum, (II). It will also divide 3 times 153, or 459, (I) ; and since it is a com- mon divisor of 459 and 68, it must divide their sum, 527, which is one of the given numbers. It will also divide 2 times 527, or 1054, (I) ; and since it is a common divisor of 1054 and 153, it must divide their sum, 1207, the greater number, (II). Hence, 17 is a com- mon divisor of the given numbers. Again, tracing the work in the direct order, as indicated in the fol- lowing diagram, we know that the greatest common divisor, what- ever it be, must divide 2 times 527, or 1054, (I). And since it will divide both 1054 and 1207, it must divide their difference, 153, (III). It will also divide 3 times 153, or 459, (I) ; and as it 68 I ? , 136 will divide both 459 and 527, it must divide their difference, 68, (HI)- It will also divide 2 times 68, or 136, (I) ; and as it will divide both 136 and 153, it must divide their difference, 17, (III) ; hence, it cannot be greater than 17. Thus we have shown, 1st. That 17 is a common divisor of the given numbers. 2d. That their greatest common divisor, whatever it be, cannot be greater than 17. Hence it must be 17. From this example and analysis, we derive the following RULE. I. Draw two verticals, and write the two numbers, one on each side, the greater number one line above the less. 527 459 1207 1054 153 80 PROPERTIES OF NUMBERS. II. Divide the greater number by the less, writing the quotient between the verticals, the product under the dividend, and the re- mainder below. III. Divide the less number by the remainder ', the last divisor by the last remainder, and so on, till nothing remains. The last divisor will be the greatest common divisor sought. IV. If more than two numbers be given, first find the greatest common divisor of two of them, and then of this divisor and one of the remaining numbers, and so on to the last j the last common divisor found will be the greatest common divisor required. NOTES. 1. When more than two nnmbera are given, it is better to begin with the least two. 2. If at any point in the operation a prime number occur as a remainder, it must be a common divisor, or the given numbers have no common divisor. EXAMPLES FOR PRACTICE. 1. What is the greatest common divisor of 18607 and 417979? OPERATION. 417979 18607 2 37214 , 45839 2 37214 17250 2 8625 1357 6 8142 966 2 483 391 1 391 368 4 ~92 Ans. 23 4 92 2. What is the greatest common divisor of 10661 and 123037 OPERATION. 12303 10661 1 10661 9852 6 1642 Prime 809 A~~ 1 GREATEST COMMON DIVISOR. 81 3. What is the greatest common divisor of 336 and 812 ? Ans. 28. 4. What is the greatest common divisor of 407 and 1067 ? 5. What is the greatest common divisor of 825 and 1372 T 6. What is the greatest common divisor of 2041 and 8476 ? Ans. 13. 7. What is the greatest common divisor of 3281 and 10778 ? 8. Find the greatest common divisor of 22579, and 116939. 9. What is the greatest common divisor of 49373 and 147731 ? Ans. 97. 10. What is the greatest common divisor of 1005973 and 4616175 ? 11. Find the greatest common divisor of 292, 1022, and 1095. Ans. 73. 12. What is the greatest common divisor of 4718, 6951, and 8876? Ans. 7. 13. Find the greatest common divisor of 141, 799, and 940. 14. What is the greatest common divisor of 484391 and 684877 ? Ans. 701. 15. A farmer wishes to put 364 bushels of corn and 455 bushels of oats into the least number of bins possible, that shall contain the same number of bushels without mixing the two kinds of grain ; what number of bushels must each bin hold ? Ans. 91. 16. A gentleman having a triangular piece of land, the sides of which are 165 feet, 231 feet, and 385 feet, wishes to inclose it with a fence having pannels of the greatest possible uniform length; what will be the length of each pannel? 17. B has $620, C $1116, and D $1488, with which they agree to purchase horses, at the highest price per head that will allow each man to invest all his money j how many horses can each man purchase ? Ans. B 5, C 9, and D 12. 18. How many rails will inclose a field 14599 feet long by 10361 feet wide, provided the fence is straight, and 7 rails high, and the rails of equal length, and the longest that can be used ? Ans. 26880. 2 PROPERTIES OP NUMBERS. LEAST COMMON MULTIPLE. 1*51. A Multiple is a number exactly divisible by a given number ; thus, 20 is a multiple of 4. NOTES. 1. A multiple is necessarily composite; a divisor may be either prime or composite. 2. A number is a divisor of all its multiples and a multiple of all its divisors. A Common Multiple is a number exactly divisible by two or more given numbers ; thus, 20 is a common multiple of 2, 4, 5, and 10, 153. The Least Common Multiple of two or more numbers is the least number exactly divisible by those numbers ; thus, 24 is the least common multiple of 3, 4, 6, and 8. 154. From the definition it is evident that the product of two or more numbers, or any number of times their product, must be a common multiple of the numbers. Hence, A common multiple of two or more numbers may be found by multiplying the given numbers together. lo5. To find the least common multiple. FIRST METHOD. From the relations of multiple and divisor we have the following properties : I. A multiple of a number must contain all the prime factors of that number. II. A common multiple of two or more numbers must contain all the prime factors of each of those numbers. III. The least common multiple of two or more numbers must contain all the prime factors of each of those numbers, and no other factors. 1. Find the least common multiple of 63, 66, and 78. OPERATION. ANALYSIS. The 63 = 3 X 3 X 7 number cannot be less 66 = 2x3x11 than 78, because it 78 = 2 X 3 X 13 must contain 78 ; and 2x3x 13 X 11x3x7 = 18018 An*, if it contains 78, it must contain all its prime factors, viz. ; 2 X 3 X 13. LEAST COMMON MULTIPLE. 83 We here have all the prime factors, and also all the factors of 66 ixcept 11. Annexing 11 to the series of factors, 2 X 3 X 13 X 11, and we have all the prime factors of 78 and 66, and also all the fac- tors of 63 except one 3, and 7. Annexing 3 and 7 to the series of factors, 2 x 3 X 13 X 11 X 3 X 7, and we have all the prime factors of each of the given numbers, and no others; hence the product of this series of factors is the least common multiple of the given numbers, (III). From this example and analysis we deduce the following RULE. I. Resolve the given numbers into their prime factors. II. Multiply together all the prime factors of the largest number, and such prime factors of the other numbers as are not found in the largest number, and their product will be the least common multiple NOTR. When a prime factor is repeated in any of the given numbers, it must be taken as many times in the multiple, as the greatest number of times it appears in any of the given numbers. EXAMPLES FOR PRACTICE. 1. Find the least common multiple of 60, 84, and 132. Ans. 4620. 2. Find the least common multiple of 21, 30, 44, and 126. Ans. 13,860. 3. Find the least common multiple of 8, 12, 20, and 30. 4. Find the least common multiple of 16, 60, 140, and 210. Ans. 1,680. 5. Find the least common multiple of 7, 15, 21, 25, and 35. 6. Find the least common multiple of 14, 19, 38, 42, and 57. Ans. 798. 7. Find the least common multiple of 144, 240, 480, 960. SECOND METHOD. 156. 1. What is the least common multiple of 4, 9, 12, 18, and 36 ? 84 PROPERTIES OF NUMBERS. 2 2 3 3 OPERATION 4 . . 9 . . 12 . . 18 .36 2 . . 9 . . 6 . . 9 . 18 9 3 . . 3 . . 9 . 9 3 3 2x2x3x3 = 36 Am. ANALYSIS. Wefirstwrite the given numbers in a se- ries with a vertical line at the left. Since 2 is a fac- tor of some of the given numbers, it must be a factor of the least common mul- tiple sought, (155,111). Di- 2,3 5 2 . . 6 . . 3 . 15 5 viding as many of the numbers as are divisible by 2, we write the quotients, and the undivided number, 9, in a line underneath. Now, since some of the numbers in the second line contain the factor 2, the least common multiple must contain another 2, and we again divide by 2, omitting to write any quotient when it is 1. We next divide by 3 for a like reason, and still again by 3. By this process we have transferred all the factors of each of the numbers to the left of the vertical ; and their product, 36, must be the least common multiple sought, (155, III). 2. What is the least common multiple of 20, 12, 15, and 75? OPERATION. ANALYSIS. We readily 2 , 5 J 20 . . 12 . . 15 . . 75 see that 2 and 5 are among the factors of the given num- bers, and must be factors of the least common multiple ; hence, writing 2 and 5 at the left, we divide every number that is divisible by either of these factors or by their product ; thus, we divide 20 by both 2 and 5 ; 12 by 2 ; 15 by 5 ; and 75 by 5. We next divide the second line in like manner by 2 and 3 ; and afterward the third line by 5. By this process we collect the factors of the given numbers into groups ; and the product of the factors at the left of the vertical is the least common multiple sought. 3. What is the least common multiple of 7, 10, 15, 42, and 70? ANALYSIS. In this operation we omit the 7 and 10, because they are exactly contained in some of the other given numbers ; thus, 7 is contained in 42, and 10 in 70 ; and whatever will contain 42 and 70 must contain 7 and 10. Hence we have only to find th least common multiple of the remaining numbers, 15, 42, and 70. 2x5x2x3x5 = 300, Ans. 3,7 2,5 OPERATION. 15 . . 42 .70 5 . . 2 . 10 3x7x2x5 = 210, Ans. LEAST COMMON MULTIPLE. 85 From these examples we derive the following RULE, I. Write the numbers in a line, omitting such of the smaller numbers as are factors of the larger, and draw a vertical line at the left. II. Divide by any prime factor or factors that may be contained in one or more of the given numbers, and write the quotients and undivided numbers in a line underneath, omitting the Vs. III. In like manner divide the quotients and undivided numbers, and continue the process till all the factors of the given numbers have been transferred to the left of the vertical. Then multiply these factors together, and their product will be the least common multiple required. NOTE. We may use a composite number for a divisor, when it is contained in all the given numbers. EXAMPLES FOR PRACTICE. 1. What is the least common multiple of 15, 18, 21, 24, 35, 36, 42, 50, and 60 ? Ans. 12600. 2. What is the least common multiple of 6, 8, 10, 15, 18, 20, and 24 ? Ans. 360. 3. What is the least common multiple of 9, 15, 25, 35, 45, and 100? Ans. 6300. 4. What is the least common multiple of 18, 27, 36, and 40 ? 5. What is the least common multiple of 12, 26, and 52 ? 6. What is the least common multiple of 32, 34, and 36 ? Ans. 4896. 7. What is the least common multiple of 8, 12, 18, 24, 27, and 36? 8. What is the least common multiple of 22, 33, 44, 55, and 66? 1 9. What is the least common multiple of 64, 84, 96, and 216 ? 10. If A can build 14 rods of fence in a day, B 25 rods, C 8 rods, and D 20 rods, what is the least number of rods that will furnish a number of whole days' work to either one of the four men? Ans. 1400. 86 PROPERTIES OF NUMBERS. 11. What is the smallest sum of money for which I can pur- chase either sheep at $4 per head, or cows at $21, or oxen at $49, or horses at $72 ? Am. $3528 12. A can dig 4 rods of ditch in a day, B can dig 8 rods, and C can dig 6 rods ; what must be the length of the shortest ditch, that will furnish exact days' labor either for each working alone or for all working together ? Am. 72 rods. 13. The forward wheel of a carriage was 11 feet in circumfer- ence, and the hind wheel 15 feet; a rivet in the tire of each was up when the carriage started, and when it stopped the same rivets were up together for the 575th time ; how many miles had the carriage traveled, allowing 5280 feet to the mile ? Am. 17 miles 5115 feet. CANCELLATION. 137. Cancellation is the process of rejecting equal factors from numbers sustaining to each other the relation of dividend and divisor. 138. It is evident that factors common to the dividend and divisor may be rejected without changing the quotient, (117, III). 1. Divide 1365 by 105. OPERATION. ^ ANALYSIS. We first in- 1QA _ , , ,, 1Q dicate the division by wri- _ = * * = 13 ting the dividend above a 105 X X $ horizontal line and the di- visor below. Then factor- ing each term, we find that 3, 5, and 7 are common factors ; and crossing, or canceling these factors, we have 13, the remaining factor of the dividend, for a quotient. 139. If the product of several numbers is to be divided by the product of several other numbers, the common factors should be canceled before the multiplications are performed, for two reasons : 1st. The operations in multiplication and division will thus be abridged. CANCELLATION. 87 2d. The factors of small numbers are generally more readily detected than those of large numbers. 2. Divide 20 times 56 by 7 times 15. OPERATION. ANALYSIS. Having first indi- 4 g cated all the operations required #0 X fc$ 32 by the question, we cancel 7 ~~7 77 = " == ^* from 7 and 56, and 5 from 15 and 20, leaving the factors 3 in the divisor, and 8 and 4 in the dividend. Then 8 X 4 = 32, which divided by 3, gives lOf , the quo- tient required. Hence the following RULE I. Write the numbers composing the dividend above a horizontal line, and the numbers composing the divisor below it. II. Cancel all the factors common to both dividend and divisor. III. Divide the product of the remaining factors of the dividend by the product of the remaining factors of the divisor, and the result will be the quotient. NOTES. 1. When a factor is canceled, the unit, 1, is supposed to take its place. 2. By many it is thought more convenient to write the factors of the dividend on the right of a vertical line, aud the factors of the divisor on the left. EXAMPLES FOR PRACTICE. 1. What is the quotient of!8x6x4x42 divided by 4 x 9 X 3 x 7 x 6? FIRST OPERATION. SECOND OPERATION. *$ x $ x 4 x 4&* _ A = 4 4, Ans. 2. Divide the product of 21 x 8 X 60 x 8 x 6 by 7 X 12 X 3 X 8 x 3. Ans. 80. 3. The product of the numbers 16, 5, 14, 40; 16, 60, and 50, is to be divided by the product of the numbers 40, 24, 50, 20, 7, and 10; what is the quotient? Ans. 32. 88 PROPERTIES OF NUMBERS. 4. Divide the continued product of 12, 5, 183, 18, and 70 by the continued product of 3, 14, 9, 5, 20, and 6. 5. If 213 x 84 x 190 x 264 be divided by 30 X 56 x 36, what will be the quotient ? 6. Multiply 64 by 7 times 31 and divide the product by 8 times 56, multiply this quotient by 15 times 88 and divide the product by 55, multiply this quotient by 13 and divide the pro- duct by 4 times 6. Ans. 403. 7. How many cords of wood at $4 a cord, must be given for 3 tons of hay at $12 a ton ? 8. How many firkins of butter, each containing 56 pounds, at 15 cents a pound, must be given for 8 barrels of sugar, each con- taining 195 pounds, at 7 cents a pound? Ans. 13. 9 A grocer sold 16 boxes of soap, each containing 66 pounds at 9 cents a pound, and received as pay 99 barrels of potatoes, each containing 3 bushels ; how much were the potatoes worth a bushel ? 10. A farmer exchanged 480 bushels of corn worth 70 cents a bushel, for an equal number of bushels of barley worth 84 cents a bushel, and oats worth 56 cents a bushel ; how many bushels of each did he receive ? Ans. 240. 11. A merchant sold to a farmer two kinds of cloth, one kind at 75 cents a yard, and the other at 90 cents, selling him twice as many yards of the first kind as of the second. He received as pay 132 pounds oi butter at 20 cents a pound ; how many yards of each kind of cloth did he sell ? Ans. 22 yards of the first, and 11 yards of the second. 12. A man took six loads of potatoes to market, each load con- taining 20 bags, and each bag 2 bushels. He sold them at 44 cents a bushel, and received in payment 8 chests of tea, each con- taining 22 pounds j how much was the tea worth a pound ? Ans. 60 cents. DEFINITIONS, NOTATION, AND NUMERATION. 89 FRACTIONS. DEFINITIONS, NOTATION, AND NUMERATION. 160. When it is necessary to express a quantity less than a unit, we may regard the unit as divided into some number of equal parts, and use one of these parts as a new unit of less value than the unit divided. Thus, if a yard, considered as an integral unit, be divided into 4 equal parts, then one, two, or three of these parts will constitute a number less than a unit. The parts of a unit thus used are called fractional units ; and the numbers formed from them, fractional numbers. Hence 161. A Fractional unit is one of the equal parts of an inte- gral unit. 162. A Fraction is a fractional unit, or a collection of frac- tional units. 163. Fractional units take their name, and their value, from the number of parts into which the integral unit is divided. Thus, If a unit be divided into 2 equal parts, one of the parts is called one half. If a unit be divided into 3 equal parts, one of the parts is called one third. If a unit be divided into 4 equal parts, one of the parts is called one fourth. And it is evident that one third is less in value than one half, one fourth less than one third, and so on. 164. To express a fraction by figures, two integers are re- quired ; one to denote the number of parts into which the inte- gral unit is divided, the other to denote the number of parts taken, or the number of fractional units in the collection. The former is written below a horizontal line, the latter above. Thus, One half is written A One fifth is written | One third " Two thirds | One fourth " \ Two fourths " Three fourths " | Two fifths " | One seventh " ^ Three eighths " f Five ninths " $ Eight tenths 8* 90 FRACTIONS. 165. The Denominator of a fraction is the number below the line. It denominates or names the fractional unit, and it shows how many fractional units are equal to an integral unit. 166. The Numerator is the numbei above the line. It numerates or numbers the fractional units; and it shows how many are taken. 167. The Terms of a fraction are the numerator and deno- minator, taken together. 168. Since the denominator of a fraction shows how many fractional units in the numerator are equal to 1 integral unit, it follows, I. That the value of a fraction in integral units, is the quo- tient of the numerator divided by the denominator. II. That fractions indicate division, the numerator being a dividend and the denominator a divisor. 169. To analyze a fraction is to designate and describe its numerator and denominator. Thus | is analyzed as follows : 7 is the denominator, and shows that the units expressed by the numerator are sevenths. 5 is the numerator, and shows that 5 sevenths are taken. 5 and 7 are the terms of the fraction considered as an expres- sion of division, 5 being the dividend and 7 the divisor. EXAMPLES FOR PRACTICE. Express the following fractions by figures : 1. Four ninths. Ans. . 2. Seven Jiffy-sixths. Ans. ^. 3. Sixteen forty-eighths. 4. Ninety-five one hundred seventy-ninths. 5. Five hundred thirty-six four hundredths. 6. One thousand eight hundred fifty-seven nine thousand Jive hundred twenty-firsts. 7. Twenty-five thousand eighty-sevenths. 8. Thirty ten thousand eighty-seconds. 9. One hundred one ten millionths. DEFINITIONS, NOTATION, AND NUMERATION. 91 Read and analyze the following fractions : 10- 7 T ft8; *iii; M- n. V; 5 TO; fl; 4%; 'I 5 ; W; % 12. Ill; HI; 'W5 TB%O; 188?- 1 Q 150. 436. 13785. 150072. 100001 id ' 537 > 97^; 47955? 475000 > 2000U2' Fractions are distinguished as Proper and Improper. 170. A Proper Fraction is one whose numerator is less than its denominator; its value is less than the unit 1. i 171. An Improper Fraction is one whose numerator equals or exceeds its denominator; its value is never less than the unit 1. NOTES. 1. The value of a proper fraction, always being less than a unit, can only be expressed in a fractional form , hence, its name. 2. The value of an improper fraction, always being equal to, or greater than a unit, can always be expressed in some other form; hence its name. 172. A Mixed Number is a number expressed by an integer and a fraction. 173. Since fractions indicate division, (1G8, II), all changes in the terms of a fraction will affect the value of that fraction ac- cording to the laws of division; and we have only to modify the language of the General Principles of Division, by substituting the words numerator, denominator, and fraction, or value of the fraction, for the words dividend, divisor, and quotient, respectively, and we shall have the following GENERAL PRINCIPLES OF FRACTIONS. 174. PRIN. I. Multiplying the numerator multiplies the fraction, and dividing the numerator divides the fraction. PRIN. II. Multiplying the denominator divides the fraction, and dividing the denominator multiplies the fraction. PRIN. III. Multiplying or dividing both terms of the fraction by the same number, does not alter the value of the fraction. 1 75, These three principles may be embraced in one GENERAL LAW. A change in the NUMERATOR produces a LIKE change in tht value of the fraction ; but a change in the DENOMINATOR produce* an OPPOSITE change in the value of the fraction. 92 FRACTIONS. REDUCTION. 1 76. The Reduction of a fraction is the process of changing its terms, or its form, without altering its value. CASE I. 177. To reduce fractions to their lowest terms. A fraction is in its lowest terms when its numerator and denomi- nator are prime to each other; that is, when both terms have no common divisor. 1. Reduce the fraction -f^ to its lowest terms. OPERATION. ANALYSIS. Dividing both terms of _6 o __ 1 2 __ 4 the fraction by the same number does ~Q r not alter the value of the fraction, 15) -60 -L 4 (174, HI); hence, we divide both terms of ^ by 5, and both terms of the result, f , by 3, and obtain 4 for the final result. As 4 and 7 are prime to each other, the lowest terms of T 6 ff - are $. Instead of dividing by the factors 5 and 3 successively, we may divide by the greatest common divisor of the given terms, and reduce the fraction to its lowest terms at a single operation. Hence, the RULE. Cancel or reject aU factors common to both numerator and denominator. Or, Divide both terms by their greatest common divisor. EXAMPLES FOR PRACTICE. 1. Reduce T 7 2 2 3 to its lowest terms. Ana. f 2. Reduce f J to its lowest terms. Ans. J. 3. Reduce ||| to its lowest terms. Ans. f . 4. Reduce T 7 g 5 5 to its lowest terms. Ans. . 5. Reduce ^J| to its lowest terms. Ans. |. 6. Reduce ||| to its lowest terms. 7. Reduce ||| to its lowest terms. 8. Reduce 69 to its lowest terms. 9. Reduce f f f to its lowest terms. 10. Reduce f ? to its lowest terms. Ans. NOTE. Consult the factor table. 11. Reduce T 8 2 5 T ^ to its lowest terms. Ans. .REDUCTION. 93 12. Reduce |J|J to its lowest terms. Ans. ff . 13. Reduce f if f to its lowest terms. 14. Reduce f |]| to its lowest terms. Ans. Ji. 15. Reduce |||jj, and |jff to their lowest terms. CASE II. 178. To reduce an improper fraction to a whole or mixed number. 1. Reduce 2 T 9 3 7 to a whole or mixed number. OPERATION. 2 T y = 297 -*- 12 = 24 T % = 24|. ANALYSIS. Since the value of a fraction in integral units is equal to the quotient of the numerator divided by the denominator, (168, 1,) we divide the given numerator, 297, by the given denominator, 12, and obtain for the value of the fraction, the mixed number 24/2 = ^4f . Hence the RULE. Divide the numerator by the denominator. NOTKS. 1. When the denominator is an exact divisor of the numerator, the result will be a whole number. 2. In all answers containing fractions, reduce the fractions to their lowest terms. EXAMPLES FOR PRACTICE. 1. Reduce 9 f to an equivalent integer. Ans. 16, 2. Reduce 6 3 6 to an equivalent integer. 3. Reduce *| 4 to a mixed number. Ans. 17J. 4. Reduce 3 T 4 3 9 to a mixed number. Ans. 26||. 5. Reduce 5 r ^f to a mixed number, Ans. 24|. 6. Reduce 9 ^/ to a mixed number. Ans. 17^|. 7. Reduce 2 || l to a mixed number. 8. Reduce 'f J| 5 to a mixed number. Ans. 156|. 9. Reduce 3 || 2 to a mixed number. 10. Reduce 3 /f/ to a mixed number. Ans. 4|J. 11. Reduce 2 |f | 5 to a mixed number. Ans. 100|. 12. Reduce {{JJgf to a mixed number. 13. In 78 g 59 of a day how many days? Ans. 982| days. 14. In 4 T / of a dollar how many dollars? Ans. $31 A. 15. If 1000 dollars be distributed equally among 36 men, \vhat part of a dollar must each man receive in change ? Ans. J. 94 FRACTIONS. CASE in. 179. To reduce a whole number to a fraction having a given denominator. 1. Reduce 37 to an equivalent fraction whose denominator shall be 5. OPERATION. ANALYSES. Since in each unit there are 37 v 5 = 185 5 fifths, in 37 units there must be 37 times 37 == i|s, Ans. 5 fift h8, or 185 fifths = ] f 5 , The nume- rator, 185, is obtained in the operation by multiplying the whole number, 37, by the given denominator, 5. Hence the RULE. Multiply the whole number by the given denominator ; take the product for a numerator, under which write the given de- nominator. NOTE. A whole number may be reduced to a fractional form by writing 1 9 r under it for a denominator ; thus, = 9 EXAMPLES FOR PRACTICE. 1. Reduce 17 to an equivalent fraction whose denominator shall be 6. Ans. J g 2 . 2. Change 375 to a fraction whose denominator shall be 8. 3. Change 478 to a fraction whose denominator shall be 24. 4. Reduce 36 pounds to ninths of a pound. 5. Reduce 359 days to sevenths of a day. Ann. 25 7 13 . 6. Reduce 763 feet to fourteenths of a foot. Ans. J T 6 ? 82 . 7. Reduce 937 to a fractional form. Ans. 9 p. CASE IV. 18O. To reduce a mixed number to an improper frac- tion. 1. In 12^ how many sevenths? OPERATION. ANALYSIS. In the whole number 12, there are 12f 12 X 7 sevenths = 84 sevenths, (Case III), and 7 84 sevenths -f 5 sevenths =-. 89 sevenths, or 8 , p . 89 Hence the following REDUCTION. 95 RULE. Multiply the whole number by the denominator of the fraction; to the product add the numerator, and under the sum write the denominator. EXAMPLES FOR PRACTICE. 1. Reduce 15| to fifths. Ans. \ 9 . 2. Reduce 24| to an improper fraction. Ans. 9 ^ 9 . 3. Reduce 57f to an improper fraction. 4. Reduce 356 jf to an improper fraction. Ans. 6 ff 4 . 5. Reduce 872 T 5 2 to an improper fraction. 6. Reduce 300^^ to an improper fraction. Ans. 9 ^J 1 . 7. Reduce 434Jf to an improper fraction. Ans. 10 3 3 00 . 8. In 15| how many eighths ? 9. In 135^ how many twentieths? Am. 2 -J$ 3 . 10. In 43| bushels how many fourths of a bushel ? 11. In 760 T 9 8 - The division shows that it is a multiple, and that 5 is the factor which must be employed to produce it. We there- fore multiply both terms of by 5, (174, III), and obtain |, the re- quired result. Hence the RULE Divide the required denominator by the denominator of the given fraction, and multiply both terms of the fraction by the quotient. 9(J FRACTIONS. EXAMPLES FOR PRACTICE. 1. Reduce | to a fraction having 24 for a denominator. Ans. Jf 2. Reduce % to a fraction whose denominator is 96. Ans. f|. 3. Reduce || to a fraction whose denominator is 51. 4. Reduce T s ^ to a fraction whose denominator is 78. 5. Reduce -ff-g to a fraction whose denominator is 3000. Ans. ^V 6. Change 7| to a fraction whose denominator is 8. 7. Change 16 3 7 5 to a fraction whose denominator is 176. 8. Change 5 T 3 T to a fraction whose denominator is 363- 9. Change 36| to a fraction whose denominator is 42. Ans. 'fl 2 . CASE VI. 183. To reduce two or more fractions to a common denominator. A Common Denominator is a denominator common to two or more fractions. 1. Reduce | and ^ to a common denominator. ANALYSIS. We multiply the terms of the OPERATION. first fraction by the denominator of the second, 3 X 9 _ 27 an( j the terms of the second fraction by the 5x9 denominator of the first, (174, III). This 7x5 must reduce each fraction to the same deno- q r === 4 5 minator, for each new denominator will be the product of the given denominators. Hence the RULE. Multiply the terms of each fraction ly the denominators of all the other fractions NOTE. Mixed numbers must first be reduced to improper fractions. EXAMPLES FOR PRACTICE. 1. Reduce | and | to a common denominator. Ans. jf , T ^. 2 Reduce and $ tc a common denominator. REDUCTION. 97 3. Reduce f , T 5 2 and ^ to a common denominator. A- j&, M, Mr 4. Reduce J, 5| and If to equivalent fractions having a com- mon denominator. Ans. ||, L 7 ^ , ||. 5. Reduce T 4 5 and T 3 7 to a common denominator. 6. Reduce ^, ^ and T 1 T to a common denominator. 7. Reduce ^, |, T 7 2 and T 9 ff to a common denominator. CASE VII. 183. To reduce fractions to their least common de- nominator. The Least Common Denominator of two or more fractions is the least denominator to which they can all be reduced. 184. We have seen that all higher terms of a fraction must be multiples of its lowest terms, (181, II). Hence, I. If two or more fractions be reduced to a common denomi- nator, this common denominator will be a common multiple of the several denominators. II. The least common denominator must therefore be the least common multiple of the several denominators. 1. Reduce |, , 7 2 and T \ to their least common denominator. OPERATION. ANALYSIS. We first find the least 3 , 5 12 . . 15 common multiple of the given deno- ^ minators, which is 60. This must be the least common denominator to 3x5x2x2 = 60 which the given fractions can be re- duced, (II). Reducing each frac- tion to the denominator 60, by Case V, we obtain |g, jjj| and B 8 S for the answer. Hence the following RULE. I. Find the hast common multiple of the given denom- j for the least common denominator. 9 G 98 FRACTIONS. II. Divide this common denominator by each of the given de- nominators, and multiply each numerator l>y the corresponding quotient. The products will be the new numerators. NOTES. 1. If the several fractions are not in their lowest terms, they should be reduced to their lowest terms before applying the rule. 2. When two or more fractions are reduced to their least common denominator, their numerators and the common denominator will be prime to each other. EXAMPLES FOR PRACTICE. 1. Eeduce f and r ^ to their least common denominator. Ans. JJ, if. 2. Reduce |, | and | to their least common denominator. 3. Reduce |, -^ and |J to their least common denominator. 4. Reduce |, | and | to their least common denominator. 5. Reduce 3 ^, ^J and || to their least common denominator. 6. Reduce |, T ^> || and ^ to their least common denominator. Atts 52 24 75 8 ^.ns. ?g , ^g, ^ g , ^ 5 . 7. Reduce 2|, T 7 ^, ^ and |J to their least common denomi- nator. Ans. |I, T ^, ^ T %. 8. Reduce |^, 5 9 g and |J to their least common denominator. 9. Reduce |, T \ 5 ^ and | to their least common denominator ^- 88, 88, f I- 10. Reduce 2 ^, -j 7 ^ and ^|| to their least common denomi- nator. Ans. 2 V T , Jjf, 2 6 2 T- 11. Reduce J|J, ||f and j||| to their least common denomi- nator. Ans. ^Vs, &V*, iVA- 12. Reduce 2f , ,\ and 1^ to their least common denominator. 13. Reduce T 9 B V^, f^ and |J|| to their least common denom inator. /n.. ifjfi|, JHf? j, i||H|. 14. Reduce ^, jl, T 2 5 , 2 8 7 , - g 9 3 and |J to their least common denominator. J^. f |gg, f |g, ;jg, |Jg, *f jj, i8- 15. Reduce 4, r ^, rj\, 5 7 3 and ^ to their least common de- nominator. 16. Reduce T \, 7 \, || and 4^ to their least common denomi- nator. Ans. T Vfc, T jj, T %, $$J. ADDITION. 99 ADDITION. 185. The denominator of a fraction determines the value of the fractional unit, (1G5); hence, I. If two or more fractions have the same denominator, their numerators express fractional units of the same value. II. If two or more fractions have diiferent denominators, their numerators express fractional units of different values. And since units of the same value only can be united into one sum, it follows, III. That fractions can be added only when they have a com- mon denominator. 1. What is the sum of \, T 6 3 and T \ ? OPERATION. ' 1J 6 > 12 + 25 + 8 i + A + ts = go = n = f ANALYSIS. "We first reduce the given fractions to a common deno- minator, (III). And as the resulting fractions, -' , f , and ^ have the same fractional unit, (I), we add them by uniting their numerators into one sura, making ^ , the answer. 2. Add 5|, 3J and 4 T 7 3 . ANALYSIS. The sum of the OPERATION. integers, 5, 3, and 4, is 12; the ' 7 4 = 2 s sum of the fractions, -J, }, and 4 + H + 12 ~ ~ J -& T -L, is 2/5 . Hence, the sum of 14 2 \, Ans. both fractions and integers is 12 + 2/ 3 = 14&. 18G. From these principles and illustrations we derive the following general RULE. I. To add fractions. When necessary, reduce the. frac- tions to their least common denominator ; then add the numerators and place the sum over the common denominator. II. To add mixed numbers. Add the integers and fractions separately, and then add their sums. NOTE. All fractional results should be reduced to their lowest terms, and if improper fractions, to whole or mixed numbert. FRACTIONS. EXAMPLES FOR PRACTICE. 1. What is the sum of T 7 2 , T 4 2 , ft and jj ? ^4ws. 2J. 2. What is the sum of if, ft, ft and T 8 3 ? ^Ina. 1 j. 3. What is the sum of / T , 2 8 1? if and f ? 4. What is the sum of 7j J, 8|, 2||, 51 1 and 4|| ? ^Iws. 28|. 5. What is the sum of 37&, 12f J, 13f J and ff ? 6. Add I, | and |. 7. Add |, I, f and ft. J.TIS. 2^. 8. Add |, and ft. 9. Add ft, if and &. ^ 1 j. 10. Add |, I IJ and f J. ^ 7 is. 2|9. 11. Add |, i, ||, If and |f. ^s 4ftV 12. Add 3, 4| and 2ft. 13. Add 16ft and 24ft. Ant. 40^. 14. Add 1^, 2|, 3j, 4| and 5|. 15. Add 4ft, 8 3 6 T and 2&. Ans. 14j|. 16. Add 1, 1, ft and ft. Ans. ff 17. Add J, |, ft and ft. 18. Add J, I, T 3 T , ft and ^|. ^ns. Iff. 19. Add i, ft, ft and | 20. Add 411, 105|, 300{, 241| and 472f ^ w . 1161|J. 21. Add 4|, 21, 1ft, 2&, 5ft, 7|, 4^ and 6|. 22. Four cheeses weighed respectively 36|, 42|, 39 T \ and 51| pounds; what was their entire weight? Ans. 169|| pounds. 23. What number is that from which if 4| be taken, the re- mainder will be 3||? Ans. 8|. 24. What fraction is that which exceeds T 5 5 by 5 5 7 ? 25. A beggar obtained ^ of a dollar from one person, J from another, -^ from another, and ft from another ; how much did he get from all ? 26. A merchant sold 46 j yards of cloth for $127 ft, 64^ yards for $226|, and 76| yards for $31 2| ; how many yards of cloth did he sell, and how much did he receive for the whole ? Ans. l87|j yards, for $666{g. SUBTRACTION. 101 SUBTRACTION. 187. The process of subtracting one fraction from another is based upon the following principles : I. One number can be subtracted from another only when the two numbers have the same unit value. Hence, II. In subtraction of fractions, the minuend and subtrahend must have a common denominator, (185 3 I). 1. From | subtract |. OPERATION. ANALYSIS. Reducing the 4 2 __ 1 2- \ o __ _2 given fractions to a common denominator, the resulting fractions jf and } express fractional units of the same value, (185, I). Then 12 fifteenths less 10 fifteenths equals 2 fifteenths = T 2 ? , the answer. 2. From 238J take 24|. OPERATION. ANALYSIS. We first reduce the frac- 2331 _ 238- 3 . tional parts, \ and , to the common 945 24i denominator, 12. Since we cannot take ]f from ^, we add 1 = jf, to T \, 213 T 5 2 Am. making i| . Then, |j| subtracted from -f 4 leaves fa ; and carrying 1 to 24, the integral part of the subtrahend, (73, II), and subtracting, we have 213/2 f r tne entire remainder. 188. From these principles and illustrations we derive the following general RULE. I. To subtract fractions. When necessary, reduce the fractions to their least common denominator. Subtract the nume- rator of the subtrahend, from the numerator of the minuend, and place the difference of the new numerators over the common denom- inator. II. To subtract mixed numbers. Reduce, the fractional parts to a common denominator, and then subtract the fractional and integral parts separately. NOTK. "We may reduce mixed numbers to improper fractions, and subtract by the rule for fractions. But this method generally imposes the useless labor of reducing integral numbers to fractions, and fractions to integers again. Q* 102 FRACTIONS. EXAMPLES FOR PRACTICE. 1. From T 7 3 take T 3 ^. Am. T 4 ^. 2. From || take f J. Ans. |. 3. From ff take / 3 . 4. From take |. <4ns. ^\. 5 From | take fa 6. From if take fa JLws. f . 7 From T \ take f . ^TW. / F . 8. From |* take f . .4ns. J^. 9. From / T take y 1 ^, 10. From T 7 2 take fe Ans. f . 11. From g \ take y 1 ^. ^ns. ^. 12. From 3% take ^ 13. From 16| take 7J- ^TIS. 9f . 14. From 36J take 8^|. ^Ins. 27|. 15. From 25 T 7 subtract 14{f. 16. From 75 subtract 4|. Ans. 70|. 17. From 18| subtract 5|. 18. From 26 5 \ subtract 25{|. 19. From 28^f subtract 3 T \. ^jw. 24j^. 20. From 78/ 5 subtract 32|. 21. The sum of two numbers is 26^, and the less is 7^ ; what is the greater? Ans. 19^j. 22. What number is that to which if you add 18^, the sum will be 97} ? 23. What number must you add to the sum of 126| and 240 f, to make 560f ? Ans. 193|. 24. What number is that which, added to the sum of , y 1 ^, and T ! H , will make || ? Ans. fa 25. To what fraction must | be added, that the sum may be | ? 26 From a barrel of vinegar containing 31 gallons, 14i gallons were drawn ; how much was then left? AUK. 16f gallons. 27. Bought a quantity of coal for $140f, and of lumber for 1456$ . Sold the coal for $775, and the lumber for $516 T 3 g ; how much was my whole gain ? Ans. $694f|. THEORY OF MULTIPLICATION AND DIVISION. 1Q3 THEORY OF MULTIPLICATION AND DIVISION OF FRACTIONS. 189. In multiplication and division of fractions, the various operations may be considered in two classes : 1st. Multiplying or dividing a fraction. 2d. Multiplying or dividing by a fraction. 190. The methods of multiplying and dividing fractions may be derived directly from the General Principles of Fractions, (174); as follows: I. To multiply a fraction. Multiply its numerator or divide its denominator, (174, I. and IT). II. To divide a fraction. Divide its numerator or multiply its denominator, (174, I. and II). GENERAL LAW. III. Perform the required operation upon the numerator, or the opposite upon the denominator , (174, III). 191. The methods of multiplying and dividing by a fraction may be deduced as follows : 1st. The value of a fraction is the quotient of the numerator divided by the denominator (1O8, I)- Hence, 2d. The numerator alone is as many times the value of the fraction, as there are units in the denominator. 3d. If, therefore, in multiplying by a fraction, we multiply by the numerator, this result will be too great, and must be divided by the denominator. 4th. But if in dividing by a fraction, we divide by the nume- rator, the resulting quotient will be too small, and must be multi- plied by the denominator. Hence, the methods of multiplying and dividing by a fraction may be stated as follows : I. To multiply by a fraction. Multiply by the numerator and divide by the denominator, (3d). II. To divide by a fraction. Divide by the numerator and mul- tiply by the denominator, (4th). 104 FRACTIONS. GENERAL LAW. III. Perform the required operation by the numerator and the opposite by the denominator. MULTIPLICATION. 1. Multiply T % by 4. FIRST OPERATION. ANALYSIS. In the first opera- T 5 2 x 4 = f i = If tion, we multiply the fraction by 4 by multiplying its nume- SECOND OPERATION. ^ by 4. and j n ^ gec()nd T2 X 4 3 = lg operation, we multiply the frac- THIRD OPERATION. tl0n ^ 4 ^ divi ^S it8 d ^ 0m ' 5 v inator by 4, (190, I or III). T X ^ = -| == If In the third operation, we ex- press the multiplier in the form of a fraction, indicate the mul- tiplication, and obtain the result by cancellation. 2. Multiply 21 by f FIRST OPERATION. ANALYSIS. To multiply by 4, 2ix4_ = 84._i2 we must mu ltiply by 4 and di- vide by 7, (191, I or III). SECOND OPERATION. In the first operation, we first 21x4 = 3x4 = 12 multiply 21 by 4, and then di- vide the product, 84, by 7. THIRD OPERATION. In the second operation, we 3 first divide 21 by 7, and then ^A x ^ __ ^2 multiply the quotient, 3, by 4. 1 f In the third operation, we ex- press the whole number, 21, in the form of a fraction, indicate the multiplication, and obtain the result by cancellation. 3. Multiply T \ by |. FIRST OPERATION. ANALYSIS. To multiply by lt step, , x 7 = |f ! we must multiply by 7 and M ,UP, fJ~8=// 3 divide by 8, (Ml, I or III). In the first operation, we muL TT3Z = TB An8 - tiply A by 7 and obtain f | ; MULTIPLICATION. 105 SECOND OPERATION. we then divide ?| by 8 and obtain r $j x | = T 3 T 5 2 = T 5 g fVV which reduced gives T 5 ff , the required product. In the second operation we obtain the same result _ x = J> by multiplying the numerators to- AP 8 gether for the numerator of the pro- duct, and the denominators together for the denominator of the product. In the third operation, we indicate the multiplication, and obtain the result by cancellation. 193* From these principles and illustrations we derive the following general RULE. I. Reduce all integers and mixed numbers to improper fractions. II. Multiply together the numerators for a new numerator, and the denominators for a new denominator. NOTES. 1. Cancel all factors common to numerators and denominators. 2. If a fraction be multiplied by its denominator, the product will be the numerator. EXAMPLES FOR PRACTICE. 1. Multiply I by 8. Am. 2| 2. Multiply * by 27, ^ by 4, and & by 9. 3. Multiply ^ by 15. Ans. . 4. Multiply 8 by |. Ans. 6. 5. Multiply 75 by T 3 g, 7 by 3 8 T , 756 by f, and 572 by 2 \. 6. Multiply f by f . 7. Multiply Ji by ||, and If by f J. 8. Multiply / 2 by |J, and ft by J. 9. Multiply 24 by 3j. Ans. 10. 10. Multiply If by lj|. ^ns. 2J. 11. Multiply A by 2i|. 12. Find the value of f X | X ^ X J. ^*s. 5 \. 13. Find the value of f X J X Jf X T 4 T X 4f -4s. T 1 5 . 14. Find the value of || X 5 7 ^ X jf |. 15. Find the value of 2| x 2^ x r \ X T J H X 1 T 7 5 X 26J. ^s. 2. 16. Find the value of T 7 T x 2 5 T X 4| x 15 x T 3 ff . 17. Find the value of *& x 5 ^ X %. -4n. T ^. 106 FRACTIONS. 18. Find the value of (4 X J) + If X (3} $j). 19. Find the value of 28 + (7| 2|) x X (f + -J). NOTB 3. The word of between fractions is equivalent to the sign of multi- plication; and such an expression is sometimes called a compound fraction. Find the values of the following indicated products : 20. I of < of f. An*. f . 21. of I of 3 \. An*. r V 22. f of A of f 23. }i of A of if of fi. An,. ft. 24. of | of | of I of of * of I of | of T 9j. In the following examples, cancellation may be employed by the aid of the Factor Table. 25. What is the value of f J x |f jf X |||| ? Jr?s. T % 3 T . 26. What is the value of f jj x f f jj | X j jJ ? 4n. 4. 27. What is the value of {}}f x ||f| X |f |f ? 28. What will 7 cords of wood cost at $3f per cord? ^s. $25|. 29. What is the value of (|) 2 X If X (f) 6 ? Ans. 2T | 7 ^. 30. If 1 horse eat | of a bushel of oats in>a day, how many bushels will 10 horses eat in 6 days ? Ans. 25|. 31. What is the cube of 12J ? 32. At $9| per ton, what will be the cost of of | of a ton of hay? Ans. $4. 33. At $ T 9 ff a bushel, what will be the cost of If bushels of corn ? 34. When peaches are worth $J per basket, what is ^ of a basket worth ? 35. A man owning | of 156| acres of land, sold -^ of | of his share; how many acres did he sell? Ans. 47. 36. What is the product of (|) s x (^) 3 X (A) 1 X (3j)*? Ans. fjf. 37. If a family consume 1 barrels of flour a month, how many barrels will 6 families consume in 8 T 9 ^ months? 38. What is the product of 150} (j of 121|+jof 48|) 75^ multiplied by 3 x (| of 1J X 4) 2J ? Ans. 342 T <\^. DIVISION. 107 39. A man at his death left his wife $12,500, which was of | of his estate; she at her death left | of her share to her daughter ; what part of the father's estate did the daughter re- ceive ? Ans. || . 40. A owned | of a cotton factory, and sold | of his share to B, who sold ^ of what he bought to C, who sold | of what he bought to D ; what part of the whole factory did each then own ? Ans. A, fa ; B, T % ; C, / g ; D, -fa. 1 41. What is the value of 2jx~-f of 4^ x (|) 2 +(3) 8 (3f)? Ans. 36Jfj. DIVISION. 194. 1. Divide |J by 3. ANALYSIS. In the first ope- FIRST OPERATION. ration we diyide the f raction by |i -^- 3 = T 7 T 3 by dividing its numerator by SECOND OPERATION. 3 ' *** . in the second P er ation 2 ! . q 2 1 _ 7 we d ^ v ^ e t* 16 fraction by 3 by 25~ ~"75 == 35 multiplying its denominator by 3, (190, II or III). 2. Divide 15 by f ANALYSIS. To divide by 5J-, we FIRST OPERATION. ^ divide by 3 and multiply by 7, (191, II or III). SECOND OPERATION. In the first operation, we first 15 H- | = 105 ~ 3 = 35 divide 15 by 3, and then mul- tiply the quotient by 7. In the second operation we first multiply 15 by 7, and then divide the product by 3. 3. Divide T 4, by f. FIRST OPERATION. ANALYSIS. To divide by 1st step, T we must divide by 3 and 2d step, A X 6 - fl - I Ans. multi P^ b J 5 ' d91, II or III). In the first operation SECOND OPERATION. we first divide T 4 j by 3 by y*j X | == JS = | multiplying the denomina^ tor by 3. We then multi- 108 FRACTIONS. THIRD OPERATION. ply the result, j 4 3 , by 5, by 4 k multiplying the numerator jj x 3" = I by 5, giving | J = for the required quotient. By in- specting this operation, we observe that the result, f , is obtained by multiplying the denomi- nator of the given dividend by the numerator of the divisor, and the numerator of the dividend by the denominator of the divisor. Hence, in the second operation, we invert the terms of the divisor, f, and then multiply the upper terms together for a numerator, and the lower terms together for a denominator, and obtain the same result as in the first operation. In the third operation, we shorten the pro- cess by cancellation. We have learned (1OT) that the reciprocal of a number is 1 divided by the number. If we divide 1 by |, we shall have 1 -7- I = 1 X I - |. Hence 195. The Reciprocal of a Fraction is the fraction inverted. From these principles and illustrations we derive the following general RULE. I. Reduce integers and mixed numbers to improper fractions. II. Multiply the dividend by the reciprocal of the divisor. Noras. 1. If the vertical line be used, the numerators of the dividend and the denominators of the divisor must be written on the right of the vertical. 2. Since a compound fraction is an indicated product of several fractions, its reciprocal may be obtained by inverting each factor of the compound fraction. EXAMPLES FOB PRACTICE. 1. Divide if by 4. if X i - A, An,. 2. Divide j? by 5, and jff by 80. 3. Divide 10 by . Ans. 35. 4. Divide 28 by |. and 3 by , fi 3 . 5. Divide 56 by If. Ans. 36. 6. Divide J by f . 7. Divide 1$ by j, if by fa and 3| by 5f 8. Divide 1J by 1. Ans. If. 9. Divide IJf by jj. Ans. If. DIVISION. 109 10. Divide f of j by J of T V OPERATION. ANALYSIS. The dividend, 3 x 5 __ i reduced to a simple fraction, 7 x _5 __ _5 is ; the divisor, reduced i x i 8 __ 6 _ 1 1 Ans * n **ke manner, is T 5 B ; and * i divided by T % is 14, the ' quotient required. Or, we .3 V 5 V ^ V * ^ _ 1 JL ' & & may apply the general rule directly by inverting both factors of the divisor,. NOTE 3. The second method of solution given above has two advantages. 1st, It gives the answer by a single operation; 2d, It affords greater facility for cancellation. 11. Divide | of T 5 T by T 8 T of T %. Ans. 1. 12. Divide T 7 2 of T 6 3 by J of / T . Ans. l|i. 13. Divide 2 x 7J by 3^ x 3^. 14. Divide 11 by | X 5^ x 7. 15. Divide 3| x 19 by x 7| X If. ^s. 25. 16. Divide T s 3 X if by X j'x T 5 7 X || X f^. Ans. 3 1 5. 17. Divide f f by -%. ^4s. 1 2 V 18. Divide 2 % 3 fV by fj X || X 4. Ans. tf. 19. Divide i X i X | X f by f * f X f X I X j%. 20. What is the value of j| ? OPERATION. ANALYSIS. The fractional form indicates division, the numerator being the dividend and the denominator the divisor, (168, II) ; hence, sve reduce the mixed numbers to improper fractions, and then treat the denominator, 2 -/, as a divisor, and obtain the result, 1^, by the general rule for division of fractions. NOTE 4. Expressions like -^ and -^ are sometimes called complex fractions. 4 7 V 5. In the reduction of complex fractions to simple fractions, if either the numerator or denominator consists of one or more parts connected by + or , the operations indicated by these signs must first be performed, and afterward the division. 21. What is the value of i? Ans. . HO FRACTIONS. 22. What is the value of f X ** ? 4ns. 2. A X 5 i 23. What is the value of ^iM ? 3 _ 24. Reduce j ^ to its simplest form. 3 -r I 5 _ 2 25. Reduce ? - 1 to its simplest form. i X f 26. Reduce J ^7- to its simplest form. Ans. ||. b s TS 27. If 7 pounds of coffee cost $|, how much will 1 pound cost ? 28. If a boy earn $| a day, how many days will it take him to earn $6^? Ans. 17f 29. If | of an acre of land sell for $30, what will an acre sell for at the same rate ? Ans. $67^. 30. At ^ of | of a dollar a pint, how much wine can be bought for $ T % ? Ans. 2| pints. 31. If -j 3 ^ of a barrel of flour be worth $2, how much is 1 barrel worth ? Ans. $7f . 32. Bought of 4^ cords of wood, for | of | of $30; what was 1 cord worth at the same rate? Ans. 33. If 235 acres of land cost $1725|, how much will acres cost? Ans. $918|4f. 34. Of what number is 26| the | part? Ans. 31^. 35. The product of two numbers is 27, and one of them is 2| ; what is the other ? 36. By what number must you multiply 16]4 to produce 148| ? 37. What number is that which, if multiplied by | of | of 2, will produce | ? Ans. l|i. 38. Divide 720 (| X 28^77^) by 40^ -f (^ H- |) X Q)*. 39. What is the value of (3 X (|) 2 + | of ^) 3 -f- (l7j | .GREATEST COMMON DIVISOR. GREATEST COMMON DIVISOR OF FRACTIONS. 196. The Greatest Common Divisor of two or more fractions is the greatest number which will exactly divide each of them, giving a whole number for a quotient. NOTE. The definition of an exact divisor, (128), is general, and applies to fractions as well as to integers. 197. In the division of one fraction by another the quotient will be a whole number, if, when the divisor is inverted, the two lower terms may both be canceled. This will be the case when the numerator of the divisor is exactly contained in the numerator of the dividend, and the denominator of the divisor exactly contains, or is a multiple of, the denominator of the dividend. Hence, I. A fraction is an exact divisor of a given fraction when, its numerator is a divisor of the given numerator, and its denominator is a multiple of the given denominator. And, II. A fraction is a common divisor of two or more given frac- tions when its numerator is a common divisor of the given nume- rators, and its denominator is a common multiple of the given denominators. Therefore, III. The greatest common divisor of two or more given frac- tions is a fraction whose numerator is the greatest common divisor of the given numerators, and whose denominator is the least com- mon multiple of the given denominators. 1. What is the greatest common divisor of |, T 5 2 , and ||? ANALYSIS. The greatest common divisor of 5, 5, and 15, the given numerators, is 5. The least common multiple of 6, 12, and 16, the given denominators, is 48. Therefore the greatest common divisor of the given fractions is ^\, Ans. (HI). PROOF. i -*-.&-) T 5 TT -T- 4 5 g = 4 >- Prime to each other. +*-) 198. From these principles and illustrations, we derive the following 112 FRACTIONS. RULE. Find the greatest common divisor of the given nume- rators for a new numerator, and the least common multiple of the given denominators for a new denominator. This fraction will be the greatest common divisor sought. NOTE. Whole and mixed numbers must first be reduced to improper fractions, and all fractions to their lowest terms. EXAMPLES FOR PRACTICE. 1. What is the greatest common divisor of ^, ^J, and ^f ? Ans. T J 7 . 2. What is the greatest common divisor of 31, 1|, and |J? 3. What is the greatest common divisor of 4, 2|, 2|, and -g 4 ^ ? Ans. %. 4. What is the greatest common divisor of 109^ and 122| ? 5. What is the length of the longest measure that can be exactly contained in each of the two distances, 18| feet and 57^ feet ? Ans. 2-^ feet. 6. A merchant has three kinds of wine, of the first 134f gal- lons, of the second 128| gallons, of the third 115^ gallons; he wishes to ship the same in full casks of equal size; what is the least number he can use without mixing the different kinds of wine ? How many kegs will be required ? Ans. 59. LEAST COMMON MULTIPLE OF FRACTIONS. 199. The Least Common Multiple of two or more fractions is the least number which can be exactly divided by each of them, giving a whole number for a quotient. 200. Since in performing operations in division of fractions the divisor is inverted, it is evident that one fraction will exactly contain another when the numerator of the dividend exactly con- tains the numerator of the divisor, and the denominator oi' the dividend is exactly contained in the denominator of the divisor Hence, I. A fraction is a multiple of a given fraction when its nume- rator is a multiple of the given numerator, and its denominator is a divisor of the given denominator. And LEAST COMMON MULTIPLE. 113 II. A fraction is a common multiple of two or more given frac- tions when its numerator is a common multiple of the given nume- rators, and its denominator is a common divisor of the given denominators. Therefore, III. The least common multiple of two or more given fractions is a fraction whose numerator is the least common multiple of the given numerators, and whose denominator is the greatest common divisor of the given denominators. NOTE. The least whole number that will exactly contain two or more given fractions in their lowest terms, is the least common multiple of their numera- tors, (193, Note 2). 1. What is the least common multiple of |, T 5 2 , and j|? ANALYSIS. The least common multiple of 3, 5, and 15, the given numerators, is 15 ; the greatest common divisor of 4, 12, and 16, the given denominators, is 4. Hence, the least common multiple of the given fractions is l ( 5 = 3, Ans. (III). 2O1. From these principles and illustrations we derive the following RULE. Find the least common multiple of the given numerators for a new numerator, and the greatest common divisor of the given denominators for a new denominator. This fraction will be the least common multiple sought. NOTE. Mixed numbers and integers should be reduced to improper fractions, and all fractions to their lowest terms, before applying the rule. EXAMPLES FOR PRACTICE. 1. What is the least common multiple of f , T 7 ^, j|, and -f-g . Ans. 11 J. 2. What is the least common multiple of -%, ||, and |$ ? 3. What is the least common multiple of 2f J, 1|X, and -f^ ? 4. What is the least common multiple of ^, f , |, |, |, |, J, f , and -fy ? Ans. 2520. 5. The driving wheels of a locomotive are 15 T 5 ff feet in circum- ference, and the trucks 9| feet in circumference. What distance must the train move, in order to bring the wheel and truck in the same relative positions as at starting ? Ans. 459| feet. 10* H FRACTIONS. PROMISCUOUS EXAMPLES. 1. Change J of f to an equivalent fraction having 135 for its denominator. Ans. -ff^. 2. Reduce |, J, |, and || to equivalent fractions, whose denom- inators shall be 48. 3. Find the least common denominator of 1^, f , 2, T 7 ^, | of i, I off of 3 % of 5 5 4. The sum of - p- ^ and 2 5 g t is equal to how many times their difference ? Ans. 2. 5. The less of two numbers is j 7^, and their difference -~ ; s ot s T5 what is the greater number ? -4s. 34 T 4 5 3 3 6. What number multiplied by f of f X 3f , will produce f j ? 7. Find the value of - X + (2 + i) -f- (3 + },) + ^f-- ^. VA- 5 8. What number diminished by the difference between f and ^ of itself, leaves a remainder of 144 ? Ans. 283 J. 9. A person spending -|, |, and ^ of his money, had $119 left; how much had he at first ? 10. What will i of 10| cords of wood cost, at 2 \ of $42 per cord? ~Ans. $31^. 11. There are two numbers whose difference is 25 T 7 -, and one number is f of the other ; what are the numbers ? Ans. 63| and 89 T \. 12. Divide $2000 between two persons, so that one shall have ^ as much as the other. Ans. $1125 and $875. 13. If a man travel 4 miles in | of an hour, how far would he travel in 1^ hours at the same rate ? Ans. 10 miles. 14. At $J a yard, how many yards of silk can be bought for $10|? 15. How many bushels of oats worth $| a bushel, will pay for J of a barrel of flour at $7^ a barrel ? PROMISCUOUS EXAMPLES. 16. If I of a bushel of barley be worth | of a bushel of corn, and corn be worth $f per bushel, how many bushels of barley will 815 buy? Ans. 18. 17. If 48 is | of some number, what is | of the same number? 18. If cloth 1| yards in breadth require 20^ yards in length to make a certain number of garments, how many yards in length will cloth | of a yard wide require to make the same ? 19. A gentleman owning | of an iron foundery, sold -i of his share for $2570| ; how much was the whole foundery worth ? Am. $5141f 20. Suppose the cargo of a vessel to be worth $10,000, and | of f of y 9 ^ of the vessel be worth { of 4- of 1% of the cargo ; what is the whole value of the ship and cargo ? Ans. $22000. 21. A gentleman divided his estate among his three sons as fol- lows : to the first he gave | of it ; to the second | of the remain- der. The difference between the portions of first and second was $500. What was the whole estate, and how much was the third son's share ? f Whole estate, $12000. ^i/15. *s { Third son's share, $2500. 22. If 7^ tons of hay cost $60, how many tons can be bought for $78, at the same rate ? 23. If a person agree to do a job of work in 30 days, what part of it ought he to do in 16^ days ? Ans. ^. 24. A father divided a piece of land among his three sons ; to the first he gave 12^ acres, to the second | of the whole, and to the third as much as to the other two ; how many acres did the third have? Ans. 49 acres. 25. Iff of 6 bushels of wheat cost $4^, how much will f of 1 bushel cost? 26. A man engaging in trade lost f of his money invested, after which he gained $740, when he had $3500 ; how much did he lose? Ans. $1840. 27. A cistern being full of water sprung a leak, and before it could be stopped, | of the water ran out, but | as much ran in at the same time ; what part of the cistern was emptied ? Ans. 116 FRACTIONS. 28. A can do a certain piece of work in 8 days, and B oan do the same in 6 days ; in what time can both together do it ? Ans. 3 1 days. 29. A merchant sold 5 barrels of flour for $32 , which was as much as he received for all he had left, at $4 a barrel ; how many barrels in all did he sell ? Ana. 18. 30. What is the least number of gallons of wine, expressed by a whole number, that will exactly fill, without waste, bottles con- taining either |, J, |, or f gallons ? Ans. 60. 31. A, B, and C start at the same point in the circumference of a circular island, and travel round it in the same direction. A makes | of a revolution in a day, B T 1y, and C - 5 8 T . In how many days will they all be together at the point of starting ? Ans. 178 1 days. 32. Two men are 64 f miles apart, and travel toward each other; when they meet one has traveled 5 miles more than the other; how far has each traveled ? Ans. One 29| miles, the other 35& miles. 33. There are two numbers whose sum is ly 1 ^, and whose dif- ference is |; what are the numbers? Ans. | and 3 7 ^. 34. A, B, and C own a ferry boat; A owns j 1 ^ of the boat, and B owns 7 7 5 of the boat more than C. What shares do B and C own respectively? Ans. B, 7 9 ^ ; C, J!J. 35. A schoolboy being asked how many dollars he had 5 replied, that if his money be multiplied by j|, and of a dollar be added to the product, and f of a dollar taken from the sum, this remainder divided by would be equal to the reciprocal of | of a dollar. How much money had he ? 36. If a certain number be increased by If, this sum diminished by |, this remainder multiplied by 5|, and this product divided by 1*, the quotient will be 7 ; what is the number? Ans. ||. 37. If J of f of 3 times l,be multiplied by |, the product di- vided by , the quotient increased by 4, and the sum diminished by of iteelf, what will the remainder be? Ans. NOTATION AND NUMERATION. - DECIMAL FRACTIONS. NOTATION AND NUMERATION. 202. A Decimal Fraction is one or more of the decimal divisions of a unit. NOTES. 1. The word decimal is derived from the Latin decem, which signi- fies ten. 2. Decimal fractions are commonly called decimals. 203. In the formation of decimals, a simple unit is divided into ten equal parts, forming decimal units of the first order, or tenths, each tenth is divided into ten equal parts, forming decimal units of the second order, or hundredths; and so on ; according to the following TABLE OF DECIMAL UNITS. 1 single unit equals 10 tenths ; 1 tenth " 10 hundredths; 1 hundredth " 10 thousandths ; 1 thousandth " 10 ten thousandths etc. etc. 204. In the notation of decimals it is not necessary to employ denominators as in common fractions; for, since the different orders of units are formed upon the decimal scale, the same law of local value as governs the notation of simple integral numbers, (57)> enables us to indicate the relations of decimals by place or position. 3O5.> The Decimal Sign (.) is always placed before decimal figures to distinguish them from integers. It is commonly called the decimal point. When placed between integers and decimals in the same number, is sometimes called the separatrix. 2O6. The law of local value, extended to decimal units, as- signs the first place at the right of the decimal sign to tenths ; the second, to hundredths; the third, to thousandths; and so on, as shown in the following 118 DECIMALS. DECIMAL NUMERATION TABLE. ii A i S .11- & jf^l! , 3 CQ rS^^^fM i>vii liilitl s-Ssllsl 111 Ilia 043,3 S ^ 0^43 S ^5 3 c_ic^jMn-,h ,^CL_iME_jr~LHri^ SMHHWHP^HKHH WS 5732754. 573256 Integers. Decimals. 2OT. The denominator of a decimal fraction, when expressed, is necessarily 10, 100, 1000, or some power of 10. By examining the table it will be seen, that the number of places in a decimal is equal to the number of ciphers required to express its denomi- nator. Thus, tenths occupy the first place at the right of units, and the denominator of j 1 ^ has one cipher; hundredths in the table extend two places from units, and the denominator of T ^ has two ciphers; and so on. 2O8. A decimal is usually read as expressing a certain number of decimal units of the lowest order contained in the decimal. Thus, 5 tenths and 4 hundredths, or .54, may be read, fifty-four hundredths. For, y 5 ^ -f- T ^ = T 5 _i_ o order to be retained in the pro- duct ; and the other figures of 15.589 , Ans. the multiplier we inverted order, extending to the left. Then, since the product of 3 and 5 is of the third order, or thousandths, the products of the other corresponding figures at the left, 2 and 8, 5 and 7, 7 and 4, etc., will be thousandths ; and we therefore multiply each figure of the multiplier by the figures above and to the left of it in the multipli- cand, carrying from the rejected figures of the multiplicand, as fol- lows : 3 times 6 are 18, and as this is nearer 2 units than one of the next higher order, we must carry 2 to the first contracted product ; 3 times 5 are 15, and 2 to be carried are 17 ; writing the 7 under the 3, and multiplying the other figures at the left in the usual manner, 130 DECIMALS. ; we obtain 14357 for the first partial product. Then, beginning with the next figure of the. multiplier, 2 times 5 are 10, which gives 1 to be carried to the second partial product ; 2 times 8 are 16, and 1 to be carried are 17 ; writing the 7 under the first figure of the former pro- duct, and multiplying the remaining left-hand figures of the mul- tiplicand, we obtain 957 for the second partial product. Then, 5 times 8 are 40, which gives 4 to be carried to the third partial pro- duct ; 5 times 7 are 35 and 4 are 39 ; writing the 9 in the first column of the products, and proceeding as in the former steps, we obtain 239 for the third partial product. Next, multiplying by 7 in the same manner, we obtain 33 for the fourth partial product. Lastly, begin- ning 2 places to the right in the multiplicand, 6 times 7 are 42 ; 6 times 4 are 24, and 4 are 28, which gives 3 to be carried to the fifth partial product; 6 times is 0, and 3 to be carried are 3, which we write for the last partial product. Adding the several partial pro- ducts, and pointing off 3 decimal places, we have 15.589, the required product. 222. From these principles and illustrations we derive the following RULE. I. Write the multiplier with the order of its figures reversed, and with the units' place under that figure of the multi- plicand which is the lowest decimal to be retained in the product. II. Find the product of each figure of the multiplier by the figures above and to the left of it in the multiplicand, increasing each partial product by as many units as would have been carried from the rejected p art of the multiplicand, and one more when the highest figure in the rejected part of any product is 5 or greater than 5 ', and write these partial products with the lowest figure of each in the same column. III. Add the partial products, and from the right hand of the result point off the required number of decimal figures. NOTES. 1. In obtaining the number to be carried to each contracted partial product, it is generally necessary to multiply (mentally) only one figure at the right of the figure above the multiplying figure; but when the figures are large, the multiplication should commence at lenst two places to the right. 2. Observe, that when the number of units in the highest order of the rejected part of the product is between 5 and 15, carry 1; if between 15 and 25 carry 2 ; if between 25 and 35 carry 3 ; and so on. 3. There is always a liability to an error of one or two units in the last place; and as the answer may be either too great or too small by the amount of thii CONTRACTED MULTIPLICATION. error, the uncertainty may be indicated by the double sign, i, read, plus, or minus, and placed after the product. 4. When the number of decimal places in the multiplicand is less than the number to be retained in the product, supply the deficiency by annexing ciphers. EXAMPLES FOR PRACTICE. 1. Multiply 236.45 by 32.46357, retaining 2 decimal places, and 2.563789 by .0347263, retaining 6 decimal places in the product. OPERATION. OPERATION. 236.450 2.563789 75364.23 362 7430. 76914 47290 10255 9458 1795 1419 51 71 15 12 1 2 .089031 7676.02 rb 2. Multiply 36.275 by 4.3678, retaining 1 decimal place in the product. Ans. 158.4 db. 3. Multiply .24367 by 36.75, retaining 2 decimal places in the product. 4. Multiply 4256.785 by .00564, rejecting all beyond the third decimal place in the product. Ans. 24.008 . 5. Multiply 357.84327 by 1.007806, retaining 4 decimal places in the product. 6. Multiply 400.756 by 1.367583, retaining 2 decimal places in the product. Ans. 548.07 . 7. Multiply 432.5672 by 1.0666666, retaining 3 decimal places in the product. 8. Multiply 48.4367 by 2^, extending the product to three decimal places. Ans. 103.418 rfc. 9. Multiply 7-j-fj by 3|||, extending the product to three decimal places. 10. The first satellite of Uranus moves in its orbit 142.8373 + 132 DECIMALS. degrees in 1 day; find how many degrees it will move in 2.52035 days, carrying the answer to two decimal places. Ans. 360.00 degrees. 11. A gallon of distilled water weighs 8.33888 pounds ; how many pounds in 35.8756 gallons ? Ans. 299.16 d= pounds. 12. One French metre is equal to 1.09356959 English yards; how many yards in 478.7862 metres. Ans. 523.58 yards. 13. The polar radius of the earth is 6356078.96 metres, and the equatorial radius, 6377397.6 metres; find the two radii, and their difference, to the nearest hundredth of a mile, 1 metre being equal to 0.000621346 of a mile. DIVISION. 223. In division of decimals the location of the decimal point in the quotient depends upon the following principles : I. If one decimal number in the fractional form be divided by another also in the fractional form, the denominator of the quotient must contain as many ciphers as the number of ciphers in the de- nominator of the dividend exceeds the number in the denominator of the divisor. Therefore, II. The quotient of one number divided by another in the deci- mal form must contain as many decimal places as the number of decimal places in the dividend exceed the number in the divisor. 1. Divide 34.368 by 5.37. ANALYSIS. We first divide as OFiiKATION. c - N ^ or.0 / n A in whole numbers; then, since the O.o7 ) o^doo ( D.4 ,. ., T , 01- -i ! g2 2? dividend has 3 decimal places and the divisor 2, we point off 3 2 = 1 decimal place in the quotient, (II). The correctness of the work is shown in the proof, where the dividend and divisor are written as Jritv/Ul 1 common fractions. For, when we TflUC x 557 = H = 6 - 4 have canceled the denominator of the divisor from the denominator of the dividend, the denominator of the quotient must contain aa DIVISION. 133 many ciphers as the number in the dividend exceeds those in the divisor. S834L Hence the following RULE. Divide as in whole numbers, and from the right hand of the quotient point off as many places for decimals as the decimal places in the dividend exceed those in the divisor. NOTES. 1. If the number of figures in the quotient be less than the excess of the decimal places in the dividend over those in the divisor, the deficiency must be supplied by prefixing ciphers. 2. If there be a remainder after dividing the dividend, annex ciphers, and continue the division : the ciphers annexed are decimals of the dividend. 3. The dividend should always contain at least as many decimal places as the divisor, before commencing the division; the quotient figures will then be inte- gers till all the decimals of the dividend have been used in the partial dividends. 4. To divide a decimal by 10, 100, 1000, etc., remove the point as many places to the left as there are ciphers on the right of the divisor. EXAMPLES FOR PRACTICE. 1. Divide 9.6188 by 3.46. Ans. 2.78. 2. Divide 46.1975 by 54.35. Ant. .85. 3. Divide .014274 by .061. Ans. .234. 4. Divide .952 by 4.76. 5. Divide 345.15 by .075. Ans. 4602. 6. Divide .8 by 476.3. Ans. .001679 + . 7. Divide .0026 by .003. 8. Divide 3.6 by .00006! Ans. 60000. 9. Divide 3 by 450. 10. Divide 75 by 10000. 11. Divide 4.36 by 100000. 12. Divide .1 'by .12. 13. Divide 645.5 by 1000. 14. If 25 men build 154.125 rods offence in a day, how much does each man build ? 15. How many coats can be made from 16.2 yards of cloth, allowing 2.7 yards for each coat ? 16. If a man travel 36.34 miles a day, how long will it take him to travel 674 miles ? Ans. 18.547-|-days. 17. How many revolutions will a wheel 14.25 feet in circum- ference make in going a distance of 1 mile or 5280 feet ? 134 DECIMALS. CONTRACTED DIVISION. To obtain a given number of decimal places in the quotient. In division, the products of the divisor by the several quotient figures maybe contracted, as in multiplication, by rejecting at each step the unnecessary figures of the divisor, (22O). 1. Divide 790.755197 by 32.4687, extending the quotient to two decimal places. FIRST CONTRACTED METHOD. COMMON METHOD. 32.4687)790.755197(24.35 6494 1413 1299 32.4687 ) 790.7 55198 ( 24.35 649 3 74 141 3'811 129 874* 114 97 17 16 115 97 17 16 0639 4061 65787 23435 SECOND CONTRACTED METHOD. 32.4687 ) 790.755197 53.42 141 3 114 17 1 1|42352 ANALYSIS. In the first method of contraction, we first compare the 3 tens of the divisor with the 79 tens of the dividend, and ascertain that there will be 2 integral places in the quotient ; and as 2 decimal places are required, the quotient must contain 4 places in all. Then assuming the four left hand figures of the divisor, we say 3246 is con- tained in 7907, 2 times ; multiplying the assumed part of the divisor by 2. and carrying 2 units from the rejected part, as in Contracted^ Multiplication of Decimals, we have 6494 for the product, which sub- tracted from the dividend, leaves 1413 for a new dividend. Now, since the next quotient figure will be of an order next below the former, we reject one more place in the divisor, and divide by 324, obtaining 4 for a quotient, 1299 for a product, and 114 for a new divi- dend. Continuing this process till all the figures ot the divisor are CONTRACTED DIVISION. 135 rejected, we have, after pointing off 2 decimals as required, 24.35 for a quotient. Comparing the contracted with the common method, we see the extent of the abbreviation, and the agreement of the corres- ponding intermediate results. In the second method of contraction, the quotient is written with its first figure under the lowest order of the assumed divisor, and the other figures at the left in the reverse order. By this arrangement, the several products are conveniently formed, by multiplying each quotient figure by the figures above and to the left of it in the divisor, by the rule for contracted multiplication, (222), and the remainders only are written as in (112). 22G. From these illustrations we derive the following RULE. I. Compare the highest or left hand figure of the divisor with the units of like order in the dividend, and determine how many figures will be required in the quotient. II. For the first contracted divisor, take as many significant figures from the left of the given divisor as there are places re- 'quired in the quotient ; and at each subsequent division reject one place from the right of the last preceding divisor. III. In multiplying by the severed quotient figures, carry from the rejected figures of the divisor as in contracted multiplication. NOTES. 1. Supply ciphers, at the right of either divisor or dividend, when necessary, before commencing the work. 2. If the first figure of the quotient is written under tne lowest assumed figure of the divisor, and the other figures at the left in the inverted order, the several products will be formed with the greatest convenience, by simply multiplying each quotient figure by the figures above and to the left of it in the divisor. EXAMPLES FOR PRACTICE. 1. Divide 27.3782 by 4.3267, extending the quotient to 3 deci- mal places. Ans. 6.328 db. 2. Divide 487.24 by 1.003675, extending the quotient to 2 decimal places. 3. Divide 8.47326 by 75.43, extending the quotient to 5 deci- mal places. 4. Divide .8487564 by .075637, extending the quotient to 3 decimal places. Ans. 11.221 dfc. 13(3 DECIMALS. 5. Divide 478.325 by 1.43f , extending the quotient to 3 deci- mal places. Ans. 332.942 . 6. Divide 8972.436 by 756.3452, extending the quotient to 4 decimal places. 7. Divide 1 by 1.007633, extending the quotient to 6 decimal places. Ans. .992425 . 8. Find the quotient of .95372843 divided by 44.736546, true to 8 decimal places. 9. Reduce f f J-f to a decimal of 4 places. Ans. .7448 . CIRCULATING DECIMALS. 337. Common fractions can not always be exactly expressed in the decimal form ; for in some instances the division will not be exact if continued indefinitely. 338. A Finite Decimal is a decimal which extends a limited number of places from the decimal point. 339. An Infinite Decimal is a decimal which extends an* unlimited number of places from the decimal point. 330. A Circulating Decimal is an infinite decimal in which a figure or set of figures is continually repeated in the same order ; as .3333 + , or .437437437 + . 331. A Repetend is the figure or set of figures continually repeated. When a repetend consists of a single figure, it is in- dicated by a point placed over it; when it consists of more than one figure, a point is placed over the first, and one over the last figure. Thus, the circulating decimals .55555+ and .324324324+, are written, 5 and .324. 333. A repetend is said to be expanded when its figures are continued in their proper order any number of places toward the right; thus, .24, expanded is .2424+, or .242424242+. 333. Similar Repetends are those which begin at the same decimal place or order; as .37 and .5, .24 and .3/5, 1.56 and 24.3. 334. Conterminous Repetends are those which end at the same decimal place or order; as .75 and 1.53, .567, and 3.245. NOTE. Two or more repetcnds are Similar and Conterminous when they begt't* and end at the same decimal places or orders. CIRCULATING DECIMALS. 137 235. A Pure Circulating Decimal is one which contains no figures but the repetend; as .7, or .704. 236. A Mixed Circulating Decimal is one which contains other figures, called finite places, before the repetend; as .54, or .013245, in which .5 and .01 are the finite places. PROPERTIES OF FINITE AND CIRCULATING DECIMALS. 237. The operations in circulating decimals depend upon the following properties. NOTE. 1. The common fractions referred to are understood to be proper frac- tions, in their lowest terms. I. Every fraction whose denominator contains no other prime factor than 2 or 5 will give rise to a finite decimal ; and the num- ber of decimal places will be equal to the greatest number of equal factors, 2 or 5, in the denominator. For, in the reduction, every cipher annexed to the numerator mul- tiplies it by 10, or introduces the two prime factors, 2 and 5, and also gives 1 decimal place in the result. Hence the division will be exact when the number of ciphers annexed, or the number of decimal places obtained, shall be equal to the greatest number of equal factors, 2 or 5, to be canceled from the denominater. II. Every fraction whose denominator contains any other prime factor than 2 or 5, will give rise to an infinite decimal. For, annexing ciphers to the numerator introduces no other prime factors than 2 and 5 ; hence the numerator will never contain all the prime factors of the denominator. III. Every infinite decimal derived from a common fraction is also a circulating decimal; and the number of places in the repetend must be less than the number of units in the denominator of the common fraction. Eor, in every division, the number of possible remainders is limited to the number of units in the divisor, less 1 ; thus, in dividing by 7, the only possible remainders are 1, 2, 3, 4, 5, and 6. Hence, in the reduction of a common fraction to a decimal, some of the remainders must repeat before the number of decimal places obtained equals the number of units in the denominator ; and this will cause the inter- mediate quotient figures to repeat. 12* 138 DECIMALS. NOTES. 2. It will be found that the number of places in the repetend is always equal to the denominator less 1, or to some factor of this number. Thus, the repetend arising from ^ has 7 1 = 6 places ; the repetend arising from f p has i^l = 5 places. 3. A perfect repetend is one which consists of as many places, less 1, as there are units in the denominator of the equivalent fraction. 4. If the denominator of a fraction contains neither of the factors 2 and 5, it will give rise to a pure repetend. But if a circulating decimal is derived from a fraction whose denominator contains either of the factors 2 or 5, it will contain as many finite places as the greatest number of equal factors 2 or 5 in the de- nominator. IV. If to any number we annex as many ciphers as there are places in the number, or more, and divide the result by as many 9's as the number of ciphers annexed, both the quotient and re- mainder will be the same as the given number. For, if we take any number of two places, as 74, and annex two ciphers, the result divided by 100 will be equal to 74 ; thus, 7400 -~ 100 = 74. Now subtracting 1 from the divisor, 100, will add as many units to the quotient, 74, as the new divisor, 99, is contained times in 74, (115, II) ; thus, 7400 -f- 99 = 74 + Jf or 74f ; that is, if two ciphers be annexed to 74, and the result be divided by 99, both quotient and remainder will be 74. In like manner, annex- ing three ciphers to 74, and dividing by 999, we have 74000 -f- 999 = 74^ ; and the same is true of any number whatever. V. Every pure circulating decimal is equal to a common frac- tion whose numerator is the repeating figure or figures, and whose denominator is as many 9's as there are places in the repetend. For, if we take any fraction whose denominator is expressed by some number of 9's, as f $, then according to the last property, annex- ing two ciphers to the numerator, and reducing to a decimal, we have 54 __ 24{$. In like manner, carrying the decimal two places farther, .24ff = .2424^ 5 hence, | = 24. By the same principle, we have =.2 ; T V = -01 ; ^ = .02 ; ifa = .OOi ; %%$ = .324 ; and so on. And it is evident that all possible repetends can thus be derived from frac- tions whose numerators are the repeating figures, and whose denomi- nators are as many 9's as there are repeating figures. CIRCULATING DECIMALS. NOTE 5. It follows from the last property, that any fraction from which a pure repetend can be derived is reducible to a form in which the denominator is some number of 9's ; thus T 8 g- = ^f n| ', & = M- This is true of evei T fraction, whose denominator terminates with 1, 3, 7, or 9. VI. Any repetend may be reduced to another equivalent repe- tend, by expanding it, and moving either the second point, or both points, to the right; provided that in the result they be so placed as to include the same number of places as are contained in the given repetend, or some multiple of this number. For, in every such reduction, the new repetend and the given repe- tend, when expanded indefinitely, will give results which are identical. Thus, .536 = .536536, or .536536536, or .5365, or .53653, or .5365365, or .53653653653 ; because each of these new repetends, when ex- panded, gives .53653653653653653653+. NOTE 6. If in any reduction, the new repetend should not contain the samo number of places, or some multiple of the same number, as the given repetend, we should not have, in the expansions, the same figures repeated in the same order. REDUCTION. CASE I. 238. To reduce a pure circulating decimal to a common fraction. 1. Reduce .675 to a common fraction. OPERATION. ANALYSIS. Since the repetend has 3 h*rk 675 25 places, we take for the denominator of * "n"a7j ~* "5 ? the required fraction the number ex- pressed by three 9's, (237, V). Hence, RULE. Omit the points and the decimal sign, and write the figures of the repetend for the numerator of a common fraction, and as many 9's as there are places in the repetend for the de- nominator. EXAMPLES FOR PRACTICE. 1. Reduce .45 to a common fraction. Ans. T 5 T . 2. Reduce .66 to a common fraction. 3. Reduce .279 to a common fraction. Ans. . 140 DECIMALS. 4. Reduce .423 to a common fraction. Ans. T 4 T 7 T . 5. Reduce .923076 to a common fraction. Ans. jf . 6. Reduce .95121 to a common fraction. 7. Reduce 4.72 to a mixed number. Ans. 4 T 8 T . 8. Reduce 2.297 to an improper fraction. Ans. f f . 9. Reduce 2.97 to an improper fraction. Ans. l ^. NOTE. According to 237, VI, 2.97 = 2.972. 10. Reduce 15.0 to a mixed number. Ans. IS^I^. CASE II. 239. To reduce a mixed circulating decimal to a common fraction. 1. Reduce .0756 to a common fraction. OPERATION. ANALYSIS. Since .756 is equal .0756 = = to , .0756 will be - of , 2. Reduce .647 to a common fraction. OPERATION. ANALYSIS. Reducing the finite 04>r __ _6_4 I 7 part and the repetend separately 640 _ ^ 7 to fractions, we have T V ff + ^. = --- l~rr~7T ^ re duce these fractions to a common denominator, we must _ 640 64 -f 7 multiply the terms of the first by 900 9 ; but the numerator, 64, may 647 _ 64 be multiplied by 9 by annexing 90Q 1 cipher and subtracting 64 from coo ,, I, . . 640 64 = __ Ans. r ' glvmg ( )0 o ? or the first fraction reduced. The ^ r ^ numerator of the sum of the two .647 eiven decimal. fractions will therefore be 640 64 finite figures. - 64 + 7 = 583, and supplying >gq the common denominator, we have ||}g. In the second operation, __ Ans. the intermediate steps are omitted. Hence the following RULE. I. From the given circulating decimal subtract the finite j and the remainder will be the required numerator. CIRCULATING DECIMALS. II. Write as many 9's as there are figures in the repetend, with as many ciphers annexed as there are finite decimal figures, for the required denominator. EXAMPLES FOR PRACTICE. 1. Reduce .57 to a common fraction. Ans. ||. 2. Reduce .048 to a common fraction. Ans. -^. 3. Reduce .6472 to a common fraction. 4. Reduce .6590 to a common fraction. Ans. ||. 5. Reduce .04648 to a common fraction. Ans. -g 4 2 3 ^. 6. Reduce .1004 to a common fraction. 7. Reduce .9285714 to a common fraction. Ans. l|. 8. Reduce 5.27 to a common fraction. Ans. ff . 9. Reduce 7.0126 to a mixed number. Ans. 7^| g . 10. Reduce 1.58231707 to an improper fraction. Ans. |f . 11. Reduce 2.029268 to an improper fraction. CASE III. 24O. To make two or more repetends similar and conterminous. 1. Make .47, .53675, and .37234 similar and conterminous. OPERATION. ANALYSIS. The first of . . . the given repetends begins .47 = .47474747474747 -v at the place of tenths? the .53675 = .53675675675675 V Ans. sec ond at the place of thou- .37234 = .37234723472347 j sandths, and the third at the place of hundredths ; and as the points in any repetend cannot be moved to the left over the finite places, we can make the given repetends similar, only by moving the points of at least two of them to the right. Again, the first repetend has 2 places, the second 3 places, and the third 4 places ; and the number of places in the new repetends must be at least 12, which is the least common multiple of 2, 3, and 4. We therefore expand the given repetends, and place the first point in each new repetend over the third place in the decimal, and the second point over the fourteenth, and thus render them similar and conter- minous. Hence the following 142 DECIMALS. RULE. I. Expand the repetends, and place the first point in each over the same order in the decimal. II. Place the second point so that each new repetend shall con- tain as many places as there are units in the least common mul- tiple of the number of places in the several given repetends. NOTK. Since none of the points can be carried to the left, some of them must be carried to the right, so that each repetend shall have at least as many finite places as the greatest number in any of the given repetends. EXAMPLES FOR PRACTICE. 1. Make .43, .57, .4567, and .5037 similar and conterminous. 2. Make .578, .37, .2485, and 04 similar and conterminous. 3. Make 1.34,4.56, and .341 similar and conterminous. 4. Make .5674, .34, .247, and .67 similar and conterminous. 5. Make 1.24, .0578, .4, and .4732147 similar and conter- minous. 6. Make .7, .4567, .24, and .346789 similar and conterminous. 7. Make .8, .36, .4857, .34567, and .2784678943 similar and conterminous. ADDITION AND SUBTRACTION. 341. The processes of adding and subtracting circulating deci- mals depend upon the following properties of repetends : I. If two or more repetends are similar and conterminous, their denominators will consist of the same number of 9's, with the same number of ciphers annexed. Hence, II. Similar and conterminous repetends have the same denomi- nators and consequently the same fractional unit. 1. Add .54, 3.24 and, 2.785. OPERATION. ANALYSIS. Since fractions can be 54 = 54444 added only when they have the same 3 9d &9A9AV fractional unit, we first make the repe- "^ . "" I 1 tends of the given decimals similar and 2.785 as 2.78527 conterminous. We then add as in finite 6.57214 decimals, observing, however, that the 1 which we carry from the left hand column of the repetends, must also be added to the right hand column ; for this would be required if the repetends were further expanded before adding. CIRCULATING DECIMALS 143 2. From 7.4 take 2. 7852. OPERATION. ANALYSIS. Since one fraction can be subtracted 7 4114. ^ rom anotner on ty when they have the same frac- tional unit, we first make the repetends of the given decimals similar and conterminous. We then sub- 4.6581 tract as in finite decimals ; observing that if both repetends were expanded, the next figure in the subtrahend would be 8, and the next in the minuend 4 ; and the sub- traction in this form would require 1 to be carried to the 2, giving 1 for the right hand figure in the remainder. S5412. From these principles and illustrations we derive the following PwULE, I. When necessary, make the repetends similar and con- terminous. II. To add , Proceed as in finite decimals, observing to increase the sum of the right hand column by as many units as are carried from the left hand column of the repetends. III. To subtract ; Proceed as in finite decimals, observing to diminish the right hand figure of the remainder by 1, when the repetend in the subtrahend is greater than the repetend of the minuend. IV. Place the points in the result directly under the points above. NOTS. When the sum or difference is required in the form of a common frac- tion, proceed according to the rule, and reduce the result. EXAMPLES FOR PRACTICE. 1. What is the sum of 2.4, .32, .567, 7.056, and 4.37 ? Ans. 14.7695877. 2. What is the sum of .478, .321, .78564, .32, .5, and .4326? Ans. 2.8961788070698. 3. From .7854 subtract .59. Ans. .1895258. 4. From 57.0587 subtract 27.31. Ans. 29.7455. 5. What is the sum of .5, .32, and .12 ? Ans. 1. 6. What is the sum of .4387, .863, .21, and .3554 ? 7. What is the sum of 3.6537, 3.135, 2.564, and .53 ? 8. From .432 subtract .25. Ans. .18243. 9. From 7.24574 subtract 2.634. Ans. 4.61. 144 DECIMALS. 10. From .99 subtract .433. Ans. .55656. 11. What is the sum of 4.638, 8.318, .016, .54, and .45? Aiis. 13|f. 12. From .4 subtract .23. Ans. ^. MULTIPLICATION AND DIVISION. S43. 1. Multiply 2.428571 by .063. OPERATION. ANALYSIS. We first re- duce the multiplicand and . . " 7 multiplier to their equiva- .063 = T yu lent fractions, and obtain y X T 1 TJ = T '/o = -154 Am. V 7 and T fr ; then V X T } T = T y o = .154. 2. Divide .475 by .3753. ANALYSIS. The dividend re- OPERATION. . . duced to its equivalent common 47^ = 475 . . ~ fraction is $, and the divisor .61 oO = 7j7j7j(j reduced to its equivalent com- II! X If 38 = L2 ^ -^ s - mon ^action is fjft ; and #$ H- *{f = -H = 1-26. 344. From these illustrations we have the following RULE. Reduce the given numbers to common fractions ; then multiply or divide, and reduce the result to a decimal. EXAMPLES FOR PRACTICE. 1. Multiply 3.4 by .72. An$. 2.472. 2. Multiply .0432 by 18. Ans. .7783. 3. Divide .154 by .2. Ans. .693. 4. Divide 4.5724 by .7. Ans. 5.878873601645. 5. Multiply 4.37 by .27. Ans. 1.182. 6. Divide 56.6 by 137. Ans. .41362530. 7. Divide .428571 by .54. Ans. .7857142. 8. Multiply .714285 by .27. Ans. .194805. 9. Multiply 3.456 by .425. Ans. 1.4710037. 10. Divide 9.17045 by 3.36. Ans. 2.726350506748310. 11. Multiply .24 by .57. Ans. .1395775941230486685032. UNITED STATES MONEY. UNITED STATES MONEY. 245. By Act of Congress of August 8, 1786, the dollar was declared to be the unit of Federal or United States Money ; and the subdivisions and multiples of this unit and their denomina- tions, as then established, are as shown in the TABLE. 10 mills make 1 cent. 10 cents " 1 dime. 10 dimes " 1 dollar. 10 dollars " 1 eagle. 24G. By examining this table we find 1st. That the denominations increase and decrease in a tenfold ratio. 2d. That the dollar being the unit, dimes, cents and mills are respectively tenths, hundredths and thousandths of a dollar. 3d. That the denominations of United States money increase and decrease the same as simple numbers and decimals. Hence we conclude that I. United States money may be expressed according to tJie. deci- mal system of notation. II. United States money may be added, subtracted, multiplied and divided in the same manner as decimals. NOTATION AND NUMERATION. 247. The character $ before any number indicates that it expresses United States money. Thus $75 expresses 75 dollars. 248. Since the dollar is the unit, and dimes, cents and mills are tenths, hundredths and thousandths of a dollar, the decimal point or separatrix must always be placed before dimes. Hence, in any number expressing United States money, the first figure at the right of the decimal point is dimes, the second figure is cents, the third figure is mills, and if there are others, they are ten- thousandths, hundred-thousandths, etc., of a dollar. Thus, $8.3125 13 K 146 DECIMALS. expresses 8 dollars 3 dimes 1 cent 2 mills and 5 tenths of a mil] or 5 ten-thousandths of a dollar. 24:0. The denominations, eagles and dimes, are not regarded in business operations, eagles being called tens of dollars and dimes tens of cents. Thus $24.19 instead of being read 2 eagles 4 dollars 1 dime 9 cents, is read 24 dollars 19 cents. Hence, practically, the table of United States money is as follows : 10 mills make 1 cent. 100 cents " 1 dollar. 25O. Since the cents in an expression of United States money may be any number from 1 to 99, the first two places at the right of the decimal point are always assigned to cents. Hence, when the number of cents to be expressed is less than 10, a cipher must be written in the place of tenths or dimes. Thus, 7 cents is expressed $.07. NOTES. 1. The half cent is frequently written as 5 mills and vice versd. Thus, $.37* = $.375. 2. Business men frequently write cents as common fractions of a dollar. Thus, $5.19 is also written $5^^, read 5 and T ' g 9 ff dollars. 3. In business transactions, when the final result of a computation contains 5 mills or more, they are called one cent, and when lens than 5 they are rejected. Thus, $2.198 would be called $2.20, and $1.623 would be called $1.62. EXAMPLES FOR PRACTICE. 1. Write twenty-eight dollars thirty-six cents. Am. $28.36. 2. Write four dollars seven cents. 3. Write ten dollars four cents. 4. Write sixteen dollars four mills. 5. Write thirty-one and one-half cents. 6. Write 48 dollars If cents. Arts. $48 Olf. 7. Write 1000 dollars 1 cent 1 mill. 8. Write 3 eagles 2 dollars 5 dimes 8 cents 4 mills. 9. Write 6 cents. 10. Head the following numbers : $21.18 $10.01 $ .8125 $164.05 $201.201 $15.08i $7.90 $5.37* $96.005 UNITED STATES MONEY. REDUCTION. Since $1 = 100 cents = 1000 mills, it is evident, 1st That dollars may be changed or reduced to cents by an- nexing two ciphers ; and to mills by annexing three ciphers. 2d. That cents may be reduced to dollars by pointing off two figures from the right; and mills to dollars by pointing off three figures from the right. 3d. That cents may be reduced to mills by annexing one cipher. 4th. That mills may be reduced to cents by pointing off one figure from the right. OPERATIONS IN UNITED STATES MONEY. S55J. Since United States Money may be added, subtracted, multiplied and divided in the same manner as decimals, (24G, II), it is evident that no separate rules for these operations are required. EXAMPLES FOR PRACTICE. 1. Paid 83475.50 for building a house, 6310.20 for painting, $1287. 37 for furniture, and $207. 12 for carpets; how much was the cost of the, house and furniture ? Ans. $5280.20. 2. Bought a pair of boots for $4.62J, an umbrella for SI. 75, a pair of gloves for $.87 1, a cravat for $1, and some collars for $.62J; how much was the cost of all my purchases? 3. Gave $150 for a horse, $175.84 for a carriage, and $62 for a harness, and sold the whole for $390. 37 J; how much did I gain? Ans. $2.035. 4. A man bought a farm for $3800, which was $190. 87 less than he sold it for; how much did he sell it for? 5. A lady bought a dress for $10|, a bonnet for $5, a veil for $2f, a pair of gloves for $.87*, and a fan for $-|. She gave the shopkeeper a fifty dollar bill; how much money should he return to her? ' Ans. $29.875. G. A farmer sold 150 bushels of oats at $.37 ? a bushel, and 4 cords of wood at $3 1 a cord. He received in payment 84 pounds of 148 DECIMALS. , sugar at 61 cents a pound, 25 pounds of tea at $-f a pound, 2 barrels of flour at $5.872, and the remainder in cash; how much cash did he receive? Ana. $39.125. 7. A speculator bought 264.5 acres of land for 86726. He afterward sold 126.25 acres for $311 an acre, and the icmainder for $33.75 an acre; how much did he gain by the transaction ? 8. A merchant going to New York to purchase goods, had $11000. He bought 40 pieces of silk, each piece containing 28 } yards, at $.80 a yard; 300 pieces of calicoes, with an average length of 29 yards, at 11 \ cents a yard ; 20 pieces of broadcloths, each containing 36.25 yards, at $3.875 a yard; 112 pieces of sheeting, each containing 30.5 yards, at $.06i a yard. How much had he left with which to finish purchasing his stock ? Ans. SGiMiU/JA. 9. If 139 barrels of beef cost $2189.25, how much will 1 barrel cost? Ans. $15.75. 10. If 396 pounds of hops cost $44.748, how much are they worth per pound ? Ans. $.113. 11. Bought 10f cords of wood at $44 a cord, and received for it 7.74 barrels of flour; how much was the flour worth per barrel ? 12. If a hogshead of wine cost $287.4, how many hogsheads can be bought for $4885.80 ? Ans. 17. 13. A butcher bought an equal number of calves and sheep for $265; for the calves he paid $3;} a head, and for the sheep $21 a head ; how many did he buy of each kind ? Ans. 40. 14. If 128 tons of iron cost $9632, how many tons can be bought for $1730.75 ? Ans. 23. 15. If 125 bushels of potatoes cost $41.25, how many barrels, each containing 2 bushels, can be bought for $112.20? 16. A grocer on balancing his books at the end of a month, found that his purchases amounted to $2475.36, and his sales to SI 936. 40 ; and that the money he now had was but J of what he had at the beginning of the month; how much money had hu at the beginning of the month? Am*. 8131:7.40. 17. A person has an income of $3200 a year, and his expenses are $138 a month ; how much can he save in 8 years ? UNITED STATES MONEY. 149 18. Sold 120 pieces of cloth at $45 f a piece, and gained thereby $1026 ; how much did it cost by the piece ? Ans. 837.20. 19. A flour merchant paid $3088.25 for some flour. He sold 425 barrels at $0} a barrel, and the remainder stood him in $4.50 a barrel; how many barrels did he purchase ? Ans. 521. 20. If 36 engineers receive $6315.12 for one month's work, how many engineers will $21927.50 pay for one month at the same rate? Ans. 125. 21. A person having $1378.56, wishes to purchase a house worth $2538, and still have $750 left with which to purchase fur- niture; how much more money must he have? Ans. 81900.44. 22. A mechanic earns on an average $1.87? a day, and works 22 days per month. If his necessary expenses are $25:} a month, how many years will it take him to save $1116, there being 12 months in a year? Ans. 6 years 23. Bought 27.5 barrels of sugar for $453.75, and sold it at a profit of $o.62 a barrel; at what price per barrel was it sold ? 24. A man expended $70.15 in the purchase of rye at $.95 a bushel, wheat at $1.37 a bushel, and corn at $.73 a bushel, buying the same quantity of each kind ; how many bushels in all did he . purchase ? Ans 69 bushels. 25. A farmer bought a piece of land containing 375 \ acres, at $22 i per acre, and sold J of it at a profit of $1032! ; at what price per acre was the land sold ? Ans. $27.75. 26 If 3} cords of wood cost $11 37>}, how much will 20i cords cost? Ans. S65.40f. 27. If i of a hundred pounds of sugar cost 86 1, how much can be bought for $46.75, at the same rate ? Ans. 5.5 hundred pounds. 28. A man sold a 'wagon for $62.50, and received in payment 12;y yards of broadcloth at $3J per yard, and the balance in coffee at 12;> cents per pound; how many pounds of coffee did he re- ceive ? Ans. 175 pounds. 29. Bought 320 bushels of barley at the rate of 16 bushels for $10.04, and sold it at the rate of 20 bushels for $17 ; how much was my profit on the transaction ? Ans. $79.20. 13* 150 DECIMALS. PROBLEMS INVOLVING THE RELATION OF PRICE, COST, AND QUANTITY. PROBLEM I. 2o3. Given, the price and the quantity, to find the cost. ANALYSIS. The cost of 3 units must be 3 times the price of 1 unit ; of 8 units, 8 times the price of 1 unit ; of f of a unit, f times the price of 1 unit, etc. Hence, RULE. Multiply the prize of ONE by the quantity. PROBLEM II. 254. Given, the cost and the quantity, to find the price. ANALYSIS. By Problem I, the cost is the product of the price mul- tiplied by the quantity. Now, having the cost, which is a product, and the quantity, which is one of two factors, we have the product and one of two factors given, to find the other factor. Hence, RULE. Divide the cost by the quantity. PROBLEM III. 25.". Given, the price and the cost, to find the quantity. ANALYSIS. Reasoning as in Problem II, we find that the cost is the product of two factors, and the price is one of the factors, Hence, RULE. Divide the cost Ly the price. PROBLEM IV. 2*>6. Given, the quantity, and the price of 100 01 1000, to find the cost. ANALYSIS. If the price of 100 units be multiplied by the number of units in a given quantity, the product will bo. 100 times the required result, because the multiplier used is 100 times the true multiplier. F<>r a similar reason, if the price of 1000 units be multiplied by the number of units in a given quantity, the product will be 1000 times the required result. These errors can be corrected in t\vo ways, l.-t. ]>y dividing the product by 100 or 1000, as the case may bo; or, 2d. By reducing the given quantity to hundreds and decimals of a hundred, or to thousands and decimals of a thousand. Hence, PROBLEMS IN UNITED STATES MONEY. llULE. Multiply the price by the quantify reduced to hundreds and decimals of a hundred, or to thousands and decimals of a thousand. NOTE. In business transactions the Roman numerals C and M are com- monly used to indicate hundreds and thousands, where the price is by the 100 or 1000, PROBLEM V. 257. To find the cost of articles sold by the ton of 2000 pounds. ANALYSIS. If the price of 1 ton or 2000 pounds be divided by 2, the quotient will be the price of J ion or 1000 pounds. We then have the quantity and the price of 1000 to find the cost. Hence, HULE. Divide the price o/ 1 ton by 2, and multiply the quo- tient by the number of pounds expressed as thousandths, EXAMPLES IN THE PRECEDING PROBLEMS. 1. What cost 187 barrels of salt, at $1.32 a barrel? ^l/w. $246.84. 2. What cost 5 firkins of butter, each containing 70^ pounds, at 8 T 3 ff a pound ? Ans. $G6.09|. 3. If the board of a family be $501.87^ for 1 year, how much is it per day? Ans. $1.37|. 4. At $ 10 J a dozen, how many dozen of eggs can be bought for $18.48? Ana. 176^ 5. What is the value of 1 4U sacks of ^uano, each sack contain- ing 162^ pounds, at $17f a ton ? An*. S201.906J. G. What will be the cost O A 3240 peach trees at $16^ per hun- dred? Ans. $534.60. 7. At $66.44 a ton, what will be the cost of 842| tons of rail- road iron? Ans. $55992.31. 8. A gentleman purchased a farm of 325.5 acres for $10660^ ; how much did it cost per acre? Ans. $32.75. 9. What will be the cost of 840 feet of plank at $1.94 per C ; and 1262 pickets at $12^ per M ? Ans. $32.071. 10. At $1^ a bushel, how many bushels of wheat can be bought for 637.68| ? Ans. 25^ bushels. 152 DECIMALS. 11. What will be the cost of 2172 pounds of plaster, at $3.875 a ton? Ans. 84.208f 12. What cost of 456 bushels of potatoes at $.37 J a bushel? 13. If 321 barrels of apples cost 881.25, what is the price per barrel? Ans. $2.50. 14. What must be paid for 24240 feet of timber worth $9.37 per M.? Ans. $227*. 15. At $5| an acre, how many acres of land can be bought for $4234.37i? Ans. 752J. 16. How much must be paid for 972' feet of boards at $20.25 per M, 1575 feet of scantling at $2.87| per C, and 8756 feet of lath at $7 } per M ? Ans. $130.634}. 17. What is the value of 1046 pounds of beef at 84 f per hun- dred pounds? Ans. $48.37f. 18. What is the value of 5840 pounds of anthracite coal at $4.7 a ton, and 4376 pounds of shamokin coal at $5.25 a ton? 19. At $2.50 a yard, how much cloth can be purchased for $2 ? 20. What is the value of 3700 cedar rails at $5| per C ? 21. How much is the freight on 3840 pounds from New York to Baltimore, at $.96 per 100 pounds? Ans. $36.864. 22. What is the value of 9 pieces of broadcloth, each piece containing 271 yards, worth $21 a yard ? Ans. $715.87 J. 23. At $.42 a pound, how many pounds of wool may be bought for $80.745? Ans. 192}. 24. What will be the cost of 327 feet of boards at $15 per M; 672 feet of siding at $1.62J per C, and 1108 bricks at $4} perM? Ans. $20.69|. 25 At $f per yard, how many yards of silk may be bought for $15|? Ans. 18. 26. How much must be paid for the transportation of 18962 pounds of pork from Cincinnati to New York, at $10 a ton? 27. If 15J yards of silk cost $27.9, what is the price per yard ? 28. What cost 27860 railroad tics at $125.38 per thousand ? 29. If .7 of a ton of hay cost $13|, what is the price of 1 ton ? 30. What is the value of 720 pounds of hay at, $12.7:") .-i ton, and 912 pounds of mill feed at 815 J a ton ? Ans. $ll.(irS. ACCOUNTS AND BILLS. 153 LEDGER ACCOUNTS. 258. A Ledger is the principal book of accounts kept by mer- chants and accountants. Into it are brought in summary form the accounts from the journal or day-book. The items often form long columns, and accountants in adding sometimes add more than one column at a single operation, (G8). do 42.17 36.24 18.42 10.71 194.30 347.16 40.00 12.94 86.73 271.19 103.07 500.50 7.59 11.44 81.92 110.10 107.09 207.16 97.20 21.77 150.15 427.26 31642 114.64 81.13 37.50 (20 5 506.76 19432 427.90 173.26 71.32 39.46 152.60 271.78 320.00 709.08 48.50 63.41 56.00 410.10 372.22 137.89 276.44 18.19 27.96 157.16 94.57 177.66 327.40 1132.16 876.57 179.84 (30 52371.67 4571.84 1690.50 2037.09 5094.46 876.54 679.81 155.48 4930.71 3104.13 1987.67 5142.84 27630 522.71 3114 60 1776.82 7152.91 9328.42 472.19 321.42 2423.79 1600.81 5976.27 4318.19 682,45 3174.96 $14763.84 33276.90 47061.39 18242.76 37364.96 8410.31 5724.27 56317.66 81742.73 22431.27 40163.55 32189.60 7063.21 3451.09 9200.00 1807.36 56768.72 63024.27 3618045 90807.08 28763.81 37196.75 4230 61 3719.84 1367.92 8756.47 ACCOUNTS AND BILLS. A Debtor, in business transactions, is a purchaser, c* A person who receives money, goods, or services from another; and 20O. A Creditor is a seller, or a person who parts with money, goods, or services to another. 154 DECIMALS. 961. An Account is a registry of debts and credits. NOTES. 1. An account should always contain the nnmes of both the parties to the transaction, the price or value of each item or article, and the date of the transaction. 2. Accounts may have only one side, which may be either debt or credit; or it may have two sides, debt and credit. 969. The Balance of an Account is the difference between the amount of the debit and credit sides. If the account have only one side, the balance is the amount of that side. 963. An Account Current is a full copy of an account, giving each item of both debit and credit sides to date. 96-4. A Bill, in business transactions, is an account of money paid, of goods sold or delivered, or of services rendered, with the price or value annexed to each item. 96*5, The Footing of a Bill is the total amount or cost of all the items. NOTE. A bill of goods bought or sold, or of services received or rendered at a single transaction, and containing only one date, is often called a BUI of Par- cels ; and an account current having only one side is sometimes called a Bill of Items. 966. In accounts and bills the following abbreviations are in general use : Dr. for debit or debtor ; Cr. for credit or creditor; a | c . or acc't for account; @ for at or by ; when this abbreviation is used it is always followed by the price of a unit. Thus, 3 yd. cloth @ $1.25, sig- nifies 3 yards of cloth at $1.25 per yard; lb. tea @ $.75, signi- fies J pound of tea at $.75 per pound. 967. When an account current or a bill is settled or paid, the fact should be entered on the same and signed by the creditor, or by the person acting for him. The *| e . or bill is then said to be rrrrijttrtf. Accounts and bills may be settled, balanced and receipted by the parties to the same, or by agents, clerks or attor- neys authorized to transact business for the parties. ACCOUNTS AND BILLS. 155 EXAMPLES FOR PRACTICE. Required, the footings and balances of the following bills and accounts. (i.) BiU: receipted by clerk or agent. NEW YORK, July 10, 1860. Mr. JOHN C. SMITH, Bo't of HILL, GROVES & Co., 10 yd. Cassimere, @ $2.85 16 Blk. Silk, 1.12J 72 Ticking, " .14 42 Bid. Shirting, 12 Pressed Flannel, 24 " Scotch Plaid Prints, Rec'd Payment, HILL, GROVES & Co., By J. W. HOPKINS. (20 Bill : receipted by the selling party. CHICAGO, Sept. 20, 1861. CHASE & KENNARD, Bot of McDou0AL, FENTON & Co., 125 pr. Boys' Thick Boots, 275 " Calf 180 " Kip 210 . Brogans, 80 " Women's Fox'd Gaiters, " 95 " " Opera Boots, 175 " " Enameled " 8 cases Men's Calf Boots, 3 " Congress Pump Boots, 1 " Drill, 958 yd., 40 gross Silk Buttons, Rec'd Payment, McDouGAL, FENTON & Co. 156 DECIMALS. (3.) BUI : settled by note. NEW YORK, May 4, 1860. SMITH & PERKINS, Bo't of KENT, LOWBER & Co., 40 chests Green Tea, @ $27.50 25 " Black " 16 " Imperial " 12 sacks Java Coffee, 19.20 48.10 17.75 20 bbl. Coffee Sugar, (A) 26.30 31.85 4.12J " 2.90 15 " Crushed " 36 boxes Lemons, 42 " Oranges, 25 " Eaisins, Rec'd Payment, l>y note at 6 mo. (4-) $3951.00 KENT, LOWBER & Co. Bill : paid by draft, and receipted by Cleric. NEW ORLEANS, April 28, 1861. JAMES CARLTON & Co. Bo't of WILLARD & HALE. 150 bbl. Canada Flour, @ $6.25 275 " Genesee " 170 " Philada. u 326 bu. Wheat, 214 Corn, 300 " Barley, 500 " Rye, 7.16 5.87J 1.621 .82 .91 1.06 $5413.48 Rec'd Payment, by Draft on N. Y. R. S. CLARKE, For WILLARD & HALE. ACCOUNTS AND BILLS. 157 (5.) Account Current ; not balanced or settled. PHILADELPHIA, Nov. 1, 1860. MR. JAMES CORNWALL, To DODGE & SON, Dr. April 15, To 24 tons Swedes Iron, @ $64.30 $ " " 15 cwt. Eng. Blister Steel, " 10.25 June 21, 7 doz. Hoes, (Trowel Steel) " Aug. 10, " 25 " Buckeye Plows, " Oct. 3, " 14 Cross-cut Saws, " " " " 27 cwt. Bar Lead, " " " " 1840 Ibs. Chain, " 7.78 8.45 16.12* 5.90 May 25, July 14, tt tt Sept. 5, it 10 By 20 M. Boards, Or.. @ $17.60 50 M. Shingles, " 3.12* 42 M. Plank, " 9.37 Draft on New York, 75 C. Timber, @ 3.10 36 C. Scantling, " .871 $1000 Dr. Bal. Due DODGE & SON, 35(5.51 (6.) Account Current, another form ; balanced ~by note. WM. RICHMOND & Co. in a|c. current with WOOD & POWELL. Or. I860 18 tO July Aug. 2 17 To 896 pounds butter, $.23 ' 872 " cheese, .09^ Nov. JJ By 61 barrels apple?, $2.25 4i 70 bushels turnips, .22 !4 " 481l< lard, .ll 3 ^ Dec. 1 " 56 < dried apples, .SlU Oct. 4 " 509-% " tallow, J8U J-. " 31 drums figs, -68% '8 31 " 81 dozen esss, -1% " 15 barrels salt, 1.40 1861 Jan. 2 '' Note at 3 mo. to Bal. Dec. 15 " 41 hams, 96S% pounds, .12j^ -= =:S;=: ^^ 1 1 566 2fi ^"" a= ^^ BOSTON, Jan. 1, 1861. WOOD & POWBXL. 158 DECIMALS, ', PROMISCUOUS EXAMPLES. 1. What cost 12| cords of wood S4.87H Ans. $61.54+. 2. At $.371 per bushel, how many barrels of potatoes, each containing 2J bushels, can be purchased for $33.75? Ans. 36. 3. If 36 boxes of raisins, each containing 36 pounds, can be bought for $97.20, what is the price per pound ? Ans. $.075. 4. If .625 of a barrel of flour be worth $5.35, what is a barrel worth ? Ans. $8.56. 5. What is the difference between | of a hundredth, and -i of a tenth ? Ans. .025. 6. What is the product of 814 3 9 9 jj X 26-J-f correct to 2 decimal places ? 7. A drover bought 5 head of cattle @ $75, and 12 head @ $68 ; at what price per head must he sell them to gain $118 on the whole? Ans. $77. 8. If 1 pound of tea be worth $.62, what is .8 of a pound worth? Ans. $.5. 9. A person having $27.96, was desirous of purchasing an equal number of pounds of tea, coffee, and sugar; the tea @ $.872-, the coffee @ $.18f, and the sugar @ $.10J. How many pounds of each could he buy? Ans. 24. 10. If a man travel 13543.47 miles in 365i days, how far does he travel in | of a day ? Ans. 32.445 miles. 11. Bought 100 barrels of flour @ $5.12, and 250 bushels of wheat @ $1.06i. Having sold 75 barrels of the flour @ $6, and all the wheat @ $lf, at what price per barrel must the re- mainder of the flour be sold, to gain $221. 87 on the whole invest- ment? Ans. $6.75. 12. If 114.45 acres of land produce 4580.289 bushels of pota- toes, how many acres will be required to produce 120.06 bushels? Ans. 3. 13. Divide .0172JJ by .03-^, and obtain a quotient true to 4. decimal places. Ans. .5625. 14. Divide 13.5 by 21 hundredths. Ans. 600. 15. A man agreed to build 59.5 rods of wall; having built 8.5 PROMISCUOUS EXAMPLES. 159 rods in 5 days, how many days will be required to finish the wall at the same rate? Ans. 30 days. 16. A farmer exchanged 28 bushels of oats worth $.37* per bushel, and 453 pounds of mill feed worth $.75 per hundred, for 12520 pounds of plaster; how much was the plaster worth per ton ? Ans. $2.25. 17. A farmer sold to a merchant 3 loads of hay weighing re- spectively 1826, 1478, and 1921 pounds, at $8.80 per ton, and 281 pounds of pork at $5.25 per hundred. He received in exchange 31 yards of sheeting @ $.09, 6* yards of cloth @ $4.50, and the balance in money; how much money did he receive? 18. If 35 yards of cloth cost $122.50, what will be the cost of 29 yards? Ana. $101.50. 19. A speculator bought 1200 bushels of corn @ 8.56}. He sold 375 o bushels @ $.60. At what price must he sell the re- mainder, to gain $168.675 on the whole? 20. If a load of plaster weighing 1680 pounds cost $2.856, how much will a ton of 2000 pounds cost? Ans. $3.40. 21. If .125 of an acre of land is worth $15|, how much are 25.42 acres worth ? 22. A farmer had 150 acres of land, which he could have sold at one time for $100 an acre, and thereby have gained $3900; but after keeping it for some time he was obliged to sell it at a loss of $2250. How much an acre did the land cost him, and how much an acre did he sell it for? 23. A lumber dealer bought 212500 feet of lumber at $14.375 per M, and retailed it out at $1.75 per C; how much was his whole gain ? 24. If 10 acres of land can be bought for $545, how many acres can be bought for $17712.50 ? Ans. 325. 25. How much is the half of the fourth part of 7 times 224.56 ? Ans. 196.49. 26. Sold 10450 feet of timber for $169.8125, and gained thereby $39.18| ; how much did it cost per C ? Ans. $1.25. 27. If $6.975 be paid for .93 of a hundred pounds of pork, how much will 1 hundred pounds cost ? 160 DECIMALS. 28. Three hundred seventy-five thousandths of a lot of dry goods, valued at $4000, was destroyed by fire ; how much would a firm lose who owned .12 of the entire lot [ Ans. 8180. 29. Reduce (77 -^-Jf) X | of 1 to a decimal. Ans. .15. 30. If 7.5 tons of hay are worth 375 bushels of potatoes, and 1 bushel of potatoes is worth $.33|, how much is 1 ton of hay worth? Ans. $16.66. 31. A person invested a certain sum of money in trade; at the end of 5 years he had gained a sum equal to 84 hundredths of it, and in 5 years more he had doubled this entire amount. How many times the sum first invested had he at the end of the 10 years? Ans. 3.68 times. 32. A miller paid $54 for grain, T 3 of it being barley at $.62* per bushel, and | of it wheat at $1.87* per bushel; the balance of the money, he expended for oats at $.37 * per bushel. How many bushels of grain did he purchase ? Ans. 40. 33. A merchant tailor bought 27 pieces of broadcloth, each piece containing 19* yards, at $4.31| a yard; and sold it so as to gain $381.87*, after deducting $9.62* for freight. How much was the cloth sold for per yard ? Ans. $5.06?. 34. Bought 1356 bushels of wheat @ Sl.lSf , and 736 bushels of oats @ $.41 ; I had 870 bushels of the wheat floured, and dis- posed of it at a profit of 8235. 87 *, and I sold 528 bushels of tho oats at a loss of $13.62*. I afterward sold the remainder of the wheat at $1.12* per bushel, and the remainder of the oats at $.31 per bushel ; did I gain or lose, and how much ? Ans. I gained $171.07*. 35. The sum of two fractions is |||, and tUeir difference is \ } } 1 ; what are the fractions ? 36. A manufacturer carried on business for 3 years. The first year he gained a sum equal to ^ of his original capital; the second yr-ar he lost jL of what he had at the end of the first year; the third year he gained | of what he had at the end of th> second year, and he then had $28585.70. How much had lu- gained in the 3 years? Ans. $10594.70 CONTINUED FRACTIONS. 161 CONTINUED FRACTIONS. 968. If we take any fraction in its lowest terms, as if, and divide both terms by the numerator, we shall obtain a complex fraction, thus : 13 1 13 Reducing ^, the fractional part of the denominator, in the same manner, we have, 13 1 54 ~~ 4 4- 1 2~ Expressions in this form are called continued fractions. Hence, 269. A Continued Fraction is a fraction whose numerator is 1, and whose denominator is a whole number plus a fraction whose numerator is also 1, and whose denominator is a similar fraction, and so on. 270. The Terms of a continued fraction are the several sim- ple fractions which 'form the parts of the continued fraction. Thus, the terms of the continued fraction given above are, 4, &, and . CASE i. 271. To reduce any fraction to a continued fraction. 1. Reduce ^|| to a continued fraction. OPERATION. ANALYSIS. We divide the denominator, 109 _ 1 339, by the numerator, 109, and obtain 3 339 3 -f 1 for the denominator of the first term of 9 i 1 the continued fraction. Then in the same j-^ manner we divide the last divisor, 109, by the remainder, 12, and obtain 9 for the de- nominator of the second term of the continued fraction. In like man- ner we obtain 12 for the denominator of the final term. Hence the following RULE. I. Divide the greater term by the less, and the last divisor by the last remainder, and so on, till there is no remainder. 14* L 162 CONTINUED FRACTIONS. \ II. Write 1 for the numerator of each term of the continued fraction, and the quotients in their order for the respective denom- inators. . EXAMPLES FOR PRACTICE. 1. Reduce -fiff^ to a continued fraction. Ans. L_ 2. Reduce Jf4? to a continued fraction. 3. Reduce ^?Mnf to a continued fraction. o j o y u i 4. Reduce -f-^ to a continued fraction. CASE II. 272. To find the several approximate values of a continued fraction. An Approximate Value of a continued fraction is the simple fraction obtained by reducing one, two, three, or more terms of the continued fraction. 273. 1. Reduce T 3 g 8 5 to a continued fraction, and find its approximate values. OPERATION. 38 1 = . , the continued fraction. lui 2 + 1 5 1, 1st approx. value. X 2 + 1 _ 3X2 + 1 _ J , 3d 13 X 2 + 4 30' - 4th " ' _ 30X6+13" 103' CONTINUED FRACTIONS. ANALYSIS. We take \, the first term of the continued fraction, for the first approximate value. Reducing the complex fraction formed by the first two terms of the continued fraction, we have T 3 5 for the second approximate value. In like manner, reducing the first three terms, we have ^ for the third approximate value. By exam-- ining this last process, we perceive that the third approximate value, 3 T ff , is obtained by multiplying the terms of the preceding approxima- tion, -pj, by the denominator of the third term of the continued frac- tion, 2, and adding the corresponding terms of the first approximate value. Taking advantage of this principle, we multiply the terms of S 7 iy by the 4th denominator, 5, in the continued fraction, and adding the corresponding terms of T : ^, obtain ^y^, the 4th approximate value, which is the same as the original fraction. Hence the following RULE. I. For the first approximate value, take the first term of the continued fraction. II. for the second approximate value, reduce the complex frac- tion formed ly the first two terms of the continued fraction. III. For each succeeding approximate value, multiply bofh nu- merator and denominator of the last preceding approximation l)ij the next denominator in the continued fraction, and add to the cor- responding products respectively the numerator and denominator of the preceding approximation. NOTES. 1. When the given fraction is improper, invert it, and reduce this result to a continued fraction; then invert the approximate values obtained therefrom. 2. In a series of approximate values, the 1st, 3d, 5th, etc., are greater than the given fraction ; and the 2il, 4th, Oth, etc., are less than the given fraction. EXAMPLES FOR PRACTICE. 1. Find the approximate values of r 6 / F . ^- a. i. if- T%- 2. Find the approximate values of ^\. A,,- 1 4 5 39 83 A.T18. 4, T7 , TTJ, T64 -, 34^. 3. What are the first three approximate values of 4 n i 1 5 All*. TJ, gg 4. What are the first five approximate values of f J ? 5. Reduce | to the form of a continued fraction, and find the value of each approximating fraction. 164 COMPOUND NUMBERS. COMPOUND NUMBEKS. . A Compound Number is a concrete number expressed in two or more denominations, (1O). 27o. A Denominate Fraction is a concrete fraction whose integral unit is one of a denomination of some compound number. Thus, | of a day is a denominate fraction, the integral unit being one day; so are | of a bushel, | of a mile, etc., denominate irac- tions. 27G. In simple numbers and decimals the scale is uniform, and the law of increase and decrease is by 10. But in compound numbers the scale of increase and decrease from one denomination to another is varying, as will be seen in the Tables. MEASURES. 97T. Measure is that by which extent, dimension, capacity or amount is ascertained, determined according to some fixed standard. The process by which the extent, dimension, capacity, or amount is ascertained, is called Measuring; and consists in comparing the thing to bo measured with some conventional standard. Measures are of seven kinds : 1. Length. 4. Weight, or Force of Gravity. 2. Surface or Area. 5. Time. 3. Solidity or Capacity. 6. Angles. 7. Money or Value. The first three kinds may be properly divided into two classes Measures of Extension, and Measures of Capacity. MEASURES OF EXTENSION. 278. Extension has three dimensions length, breadth, and thickness. A Line has only one dimension length. A Surface or Area has two dimensions length and breadth. MEASURES OF EXTENSION. A Solid or Body has three dimensions length, breadth, and thickness. I. LINEAR MEASURE. S79. Linear Measure, also called Long Measure, is used in measuring lines or distances. The unit of linear measure is the yard, and the table is made up of the divisors, (feet and inches,) and the multiples, (rods, furlongs, and miles,) of this unit. TABLE. 12 inches (in.) make 1 foot, ft. 3 feet " 1 yard, yd. 5J yards, or 16 J feet, " 1 rod, rd. 820 rods " 1 statute mile,, .mi. UNIT EQUIVALENTS. ft. in. yd. 1 = 12 rd . 1 = 3 = 36 mi 1 = 5i = 16-i- = 198 1 = 320 = 1760 = 5280 = 63360 SCALE ascending, 12, 3, 5, 40, 8 ; descending, 8, 40, 5, 3, 12. The following denominations are also in use : 3 barleycorns make 1 inch, inches " 1 hand, used by shoemakers in measuring the length of the foot. used in measuring the height of horses directly over the fore feet. 9 " "1 span. 21.888 " " 1 sacred cubit. 3 feet " 1 pace. 6 " " 1 fathom, used in measuring depths at sea. X.152J statute mi. 1 geographic mile, **"** ** 3 geographic " " 1 league. 60 " or | - I f of latitude on a meridian or of 69.16 statute " " j ( longitude on the equator. 360 degrees " the circumference of the earth. NOTES. 1. For the purpose of measuring cloth and other goods sold by the yard, the yard is divided into halves, fourths, eighths, and sixteenths. The old table of cloth measure is practically obsolete. 2. A span is the distance that can be reached, spanned, or measured between the end of. the middle finger and the end of the thumb. Among sailors 8 spans are equal to 1 fathom. 3. The geographic mile is ^ of 3^ or 7, j-'g^Ti of the distance round the center of the earth. It is a small fraction more than 1.15 statute miles. 166 COMPOUND NUMBERS. 4. The length of a degree of latitude varies, being 68.72 miles at the equator, 68.9 to 69.05 miles in middle latitudes, and 69.30 to 69.34 miles in the polar regions. The mean or average length, as stated in the table, is the standard recently adopted by the U. S. Coast Survey. A degree of longitude is greatest at the' equator, where it is 69.16 miles, and it gradually decreases toward the poles, where it is 0. SURVEYORS' LINEAR MEASURE. 38O. A Ghinter's Chain, used by land surveyors, is 4 rods .or 66 feet long, and consists of 100 links. The unit is the chain, and the table is made up of divisors and multiples of this unit. TABLE. 7.92 inches (in.) make 1 link, 1. 25 links " 1 rod, rd. 4 rods, or 66 feet, " 1 chain, . . . ch. 80 chains " 1 mile, ....mi. mi. 1 ch. 1 80 UNIT EQUIVALENTS. rd. 4 = 320 = i. 1 25 100 8000 in. 7.92 198 792 633GO SCALE ascending, 7.92, 25, 4, 80; descending, 80, 4, 25, 7.92. NOTE. The denomination, rods, is seldom used in chain measure, distances being taken in chains and links. II. SQUARE MEASURE. 381. A Square is a figure having four equal sides and four equal corners or right angles. 383. Area or Superficies is the space or surface included within any given lines : as, the area of a square, of a field, of a board, etc. 1 square yard is a figure having four sides of 1 yard or 3 feet each, as shown in the diagram. Its contents are 3x3 = 9 square feet. Hence, The contents or area of a square, or of any other figure hatiiKj d uniform length and a uniform breadth, is found l)ij multiplying the length ly the breadth. 1 yd. = 3 ft. 1 yd. = 3 ft. MEASURES OF EXTENSION. 1(37 Thus, a square foot is 12 inches long and 12 inches wide, and the contents are 12 X 12 144 square inches. A board 20 feet long and 10 feet wide, is a rectangle, containing 20 x 10 = 200 square feet. The measurements for computing area or surface are always taken in the denominations of linear measure, 383. Square Measure is used in computing areas or sur- faces ; as of land, boards, painting, plastering, paving, etc. The unit is the area of a square whose side is the unit of length. Thus, the un,t of square feet is 1 foot square; of square yards, 1 yard square, etc. TABLE. 144 square inches (sq. in.) make 1 square foot,. . . .sq. ft. 9 " feet " 1 " yard,... sq. yd. 30 " yards " 1 " rod, sq. rd. 100 " rods " 1 acre, A. 610 acres " 1 square mile, . . . sq. mi. UNIT EQUIVALENTS. eq. ft. sq. in. eq.yd. 1 = 144 eq.rd. 1 = 9 = 1296 A L 1 = 301 = 2721 = 39204 sq mi. 1 = 160 = 4840 = 43560 = 6272640 1 = 640 = 102400 = 3097GOO = 27878400 = 4014489600 SCALE -ascending, 144, 9, 30, 160, 640; descending, 640, 160, 301 9, 144. Artificers estimate their work as follows : By the square foot: glazing and stone-cutting. By the square yard : painting, plastering, paving, ceiling, and paper-hanging. By the square of 100 square feet : flooring, partitioning, roofing, slating, and tiling. Bricklaying is estimated by the thousand bricks, by the square yard, and by the square of 100 square feet. NOTES. 1. In estimating the painting of moldings cornices, etc., the mea- EuriiiLT-line is carried into all the moldings and cornices. 2. In estimating brick-laying by either the square yard or the square of 100 feet, the work is understood to be 12 inches or H bricks thick. 3. A thousand shingl&s are estimated to cover 1 square, being laid 5 inches tf the weather. 168 COMPOUND NUMBERS. SURVEYORS' SQUARE MEASURE. S84. This measure is used by surveyors in. computing the area or contents of land. TABLE. 625 square links (sq. 1.) make 1 pole, P. 16 poles " 1 square chain, . sq. ch. 10 square chains " 1 acre, A. 640 acres " 1 square mile, .sq. mi. 36 square miles (6 miles square) " 1 township Tp. UNIT EQUIVALENTS. P. sq. ch. sq. 1. = 625 A. 1 = 16 = 10000 sq. mi. 1 = 10 = 160 = 100000 Tp. 1 = 640 = 6400 = 102400 = 64000000 1 = 36 = 23040 == 230400 = 3686400 = 2304000000 SCALE ascending, 625, 16, 10, 640, 36 ; descending, 36, 640, 10, 16, 625. NOTES. 1. A square mile of land is also called a section. 2. Canal and railroad engineers commonly use an engineers' chain, which con- sists of 100 links, each 1 foot long. 3. The contents of land are commonly estimated in square miles, acres, and hundredths; the denomination, rood, is rapidly going into disuse. III. CUBIC MEASURE. 28 5. A Cube is a solid, or body, having six equal square sides or faces. 28G. Solidity is the matter or space contained within the bounding surfaces of a solid. The measurements for computing solidity are always taken in the denominations of linear measure. If each side of a cube be 1 yard, or 3 feet, 1 foot in thickness of this cube will contain 3x3x1 = 9 cubic feet; and the whole cube will contain 3 x 3 X 3 = 27 cubic feet. A solid, or body, may have the three dimensions all alike or all different. A body 4 ft. long, 3 ft. wide, and 2 ft. thick it. = i yd. contains 4 X 3 x 2 = 24 cubic or solid feet. Hence we see that MEASURES OF EXTENSION. TJie cubic or solid contents of a body are found ly multiplying the length, breadth, and thickness together. 287. Cubic Measure, also called Solid Measure, is used in computing the contents of solids, or bodies; as timber, wood stone, etc. The unit is the solidity of a cube whose side is the unit of length. Thus, the unit of cubic feet is a cube which measures 1 foot on each side ; the unit of cubic yards is 1 cubic yard, etc. TABLE. 1728 cubic inches (cu. in.) make 1 cubic foot ....... cu. ft. 27 cubic feet " 1 cubic yard ..... cu. yd. 40 cubic feet of round timber, or } , m 50 hewn j 1 ton or load ......... T. 16 cubic feet " 1 cord foot ....... cd. ft. ..... ca. 24| cubic feet " 1 Pel, SCALE ascending, 1728, 27. The other numbers are not in a regular scale, but are merely so many times 1 foot. The unit equiva- lents, being fractional, are consequently omitted. NOTKS. 1. A cubic yard of earth is called a load. 2. Railroad and transportation companies estimate light freight by the space it occupies in cubic feet, and heavy freight by weight. 3. A pile of wood 8 feet long, 4 feet wide, and 4 feet high, contains 1 cord; and a cord foot is 1 foot in length of such a pile. 4. A perch of stone or of masonry is 16J feet long, 1J feet wide, and 1 foot high. 5. Joiners, bricklayers, and masons, make an allowance for windows, doors, etc., of one half the openings or vacant spaces. Bricklayers and masons, in es- timating their work by cubic measure, make no allowance for the corners of the walls of houses, cellars, etc., but estimate their work by the girt, that is, the entire length of the wall on the outside. 6. Engineers, in making estimates for excavations and embankments, take the dimensions with a line or measure divided into feet and decimals of a foot. The computations are made in feet and decimals, and the results are reduced to cubic yards. In civil engineering, the cubic yard is the unit to which estimates for excavations and embankments are finally reduced. 7. In scaling or measuring timber for shipping or freighting, ^ of the solid contents of round timber is deducted for waste in hewing or sawing. Thus, a log that will make 40 feet of hewn or sawed timber, actually contains 50 cubic feet by measurement; but its market value is only equal to 40 cubic feet of hewn or sawed timber. Hence, the cubic contents of 40 feet of round and 50 feet of hewn timber, as estimated for market, are identical. 15 170 COMPOUND NUMBEKS. MEASURES OF CAPACITY. Capacity signifies extent of room or space. 389. Measures of capacity are all cubic measures, solidity and capacity being referred to different units, as will be seen by com- paring the tables. Measures of capacity may be properly subdivided into two classes, Measures of Liquids and Measures of Dry Substances. I. LIQUID MEASURE. 29O. Liquid Measure, also called Wine Measure, is used in measuring liquids ; as liquors, molasses, water, etc. The unit is the gallon, and the table is made up of its divisors and multiples. TABLE. 4 gtlls (gi.) make 1 pint, pt. 2 pints " 1 quart, qt. 4 quarts " 1 gallon, gal. 31 gallons " 1 barrel, bbl. 2 barrels, or 63 gal. " 1 hogshead, ,. hhd. UNIT EQUIVALENTS. pt. qt. 1 Ri = 4 gal. 1 = 2 Q tibl. 1 = 4 o = 32 1 = 31 = 126 = 252 = 1008 2 = 63 = 252 = 504 = 2016 hhd. 1 SCALE ascending, 4, 2, 4, 31, 2; descending, 2, 31 , 4, 2, 4. The following denominations are also in use : 42 gallons make 1 tierce. 2 hogsheads, or 126 gallons, " 1 pipe or butt. 2 pipes or 4 hogsheads, " 1 tun. NOTES. 1. The denominations, barrel and hogshead, are used in estimating the capacity of cisterns, reservoirs, vats, etc. In Massachusetts the barrel is estimated at 32 gallons. 2. The tierce, hogshead, pipe, butt, and tun are the names of casks, and do not express any fixed or definite measures. They are usually gauged, and have their capacities in gallons marked on them. Several of these denominations aro etill in use in England, (327 330). WEIGHTS BEER MEASURE. 91. Beer Measure is a species of liquid measure used in measuring beer, ale, and milk. The unit is the gallon. TABLE. 2 pints (pt.) make 1 quart, qt. 4 quarts " 1 gallon, gal. 36 gallons " 1 barrel, bbl. 1 barrels, or 54 gallons, " 1 hogshead, . .hhd. UNIT EQUIVALENTS. qt. pt. gal. 1 = 2 bbi. 1 = 4 = 8 hhd . 1 = 36 = 144 = 288 1 = ij = 54 = 216 = 432 SCALE ascending, 2, 4, 36, H ; descending, 1, 36, 4, 2. This measure is not a standard ; it is rapidly falling into disuse. II. DRY MEASURE. 292. Dry Measure is used in measuring articles not liquid ; as grain, fruit, salt, roots, ashes, etc. The unit is the bushel, of which all the other denominations in the table are divisors. TABLE. 2 pints (pt.) make 1 quart, qt. 8 quarts " 1 peck, pk. 4 pecks " 1 bushel, . . bu. or bush. UNIT EQUIVALENTS. qt. pt. P*. 1 . = 2 bu. 1 = 8 = 16 1 = 4 = 32 = 64 SCALE ascending, 2, 8, 4 ; descending, 4, 8, 2. WEIGHTS. T JJ 1 \JT IT. -I O. 393. Weight is the measure of the quantity of matter a body contains, determined by the force of gravity. NOTE. The process by which the quantity of matter or the force of gravity is obtained is called Weighing; and consists in comparing the thing to be weighed with some conventional standard. 172 COMPOUND NUMBERS. Three scales of weight are used in the United States ; namely, Troy, Avoirdupois, and Apothecaries'. I. TROY WEIGHT. 394. Troy Weight is used in weighing gold, silver, and jewels; in philosophical experiments, and generally where great accuracy is required. The unit is the pound, and of this all the other denominations in the table are divisors. TABLE. 24 grains (gr.) make 1 penny weight, .. pwt. or dwt. 20 pennyweights " 1 ounce, ---- ............. oz. 12 ounces " 1 pound, ................ Ib. UNIT EQUIVALENTS. pwt. lb . 1 = 20 = 480 1 = 12 -= 240 == 5760 gr. = 24 SCALE ascending, 24, 20, 12; descending, 12, 20, 24. NOTE. Troy weight is sometimes called Goldsmiths'' Weight. II. AVOIRDUPOIS WEIGHT. 295. Avoirdupois Weight is used for all the ordinary pur- poses of weighing. The unit is the pound, and the table is made up of its divisors and multiples. TABLE. 16 ounces (oz.) make 1 pound, ............ lb. 100 pounds " 1 hundred weight,, .cwt. 20 cwt., or 2000 Ibs., " 1 ton, ............... T. UNIT EQUIVALENTS. lb. OZ. cwt 1 ' as .16 T 1 = 100 = 1600 1 == 20 = 2000 = 32000 SCALE ascending, 16, 100, 20 ; descending, 20, 100, 16. WEIGHTS. 173 NOTE. The long or gross ton, hundred weight, and quarter were formerly in common use; but they are now seldom used except in estimating English goods at the U. S. custom-houses, in freighting and wholesaling coal from the" Penn- sylvania mines, and in the wholesale iron and plaster trade. LONG TON TABLE. 28 Ib. make 1 quarter, marked qr. 4 qr. = 112 Ib. " 1 hundred -weight, " ewt. 20 cwt. = 2240 Ib. " 1 ton, " T. SCALE ascending, 28, 4, 20; descending, 20, 4, 28. 29G. The weight of the bushel of certain grains and roots has been fixed by statute in many of the States ; and these statute weights must govern in buying and selling, unless specific agree- ments to the contrary be made. TABLE OF AVOIRDUPOIS POUNDS IN A BUSHEL, As prescribed by statute in the several States named. COMMODITIES. .5 ^ ~n Connecticut. ij a I ~ .!: ^ 1 | ^ M S 12 3 to I A & e . i 1 r f a s 1 5 5 >H '- { 1 c 2 > | 1 z 8 2 S * 2 K c 50 40 52 32 54 CO 45 5G 28 60 56 56 56 60 48 bo 14 40 46 60 24 33 56 8 44 52 70 4S 80 32 5T 60 54 45 60 2U 48 CO 46 60 25 33 56 44 11 50 70 32 48 60 56 50 45 CO 4S 60 14 5.: 46 % : s 68 35 57 60 56 50 45 CO 20 48 60 14 52 60 56 56 50 33^ 57 60 56 50 45 60 20 32 56 32 32 60 11 50 30 CO 50 4C 46 50 30 52 56 50 CO 48 42 CO 28 2S 56 32 56 60 48 42 CO 28 28 5C 32 56 CO 48 CO 14 52 46 60 24 33 56 44 52 feO 35 57 CO 5C 50 45 CO 20 ' 30 60 48 50 64 55 56 30 60 56 60 48 62 48 00 05 58 32 60 60 56 56 44 60 48 60 56 56 32 56 CO 46 42 CO 28 28 56 34 00 36 60 47 48 56 32 56 60 50 50 CO M 46 46 56 32 60 56 60 45 42 60 28 28 56 36 50 60 56 60 Blue Grass Seed Buckwheat Castor Beans Clover Seed Dried Apples Dried Peaches- Flax Seed Hair Hemp Seed Indian Corn Ind. Corn in ear Ind.Corn Meal. Mineral Coal 1 ... Oats Onions Peas Potatoes Rye Eye Meal Salt* Timothy Seed... Wheat Wheat Bran 1 In Kentucky, 80 Ibs. of bituminous coal or 70 Ibs. of cannel coal make 1 bushel. a In Pennsylvania, 80 Ibs. coarse, 70 Ibs. ground, or 62 Ibs. fine salt make 1 bushel ; and in Illinois, 50 Ibs. common or 55 Ibs. fine salt make 1 bushel. * In Maine, 64 Ibs. of ruta baga turnips or beets make 1 bushel. 15 * [74 COMPOUND NUMBERS. NOTES. 1. The weight of a barrel of flour is 7 quarters of old, or long ton Weight. 2. The weight of a bushel of Indian corn and rye, as adopted by most of the States, and of a bushel of salt is 2 quarters ; and of a barrel of salt 10 quarters, or i of a long ton. The following denominations are also in use : 56 pounds make 1 firkin of butter. 100 100 196 200 280 1 quintal of dried salt fish. 1 cask of raisins. 1 barrel of flour. 1 " " beef, pork, or fish. 1 " " salt at the N. Y. State salt works. III. APOTHECARIES' WEIGHT. Apothecaries' Weight is used by apothecaries and phy- sicians in compounding medicines ; but medicines are bought and sold by avoirdupois weight. The unit is the pound, of which all the other denominations in the table are divisors. TABLE. 20 grains (gr.) make 1 scruple, so. or 9. 3 scruples " 1 dram, dr. or 3. 8 drams " 1 ounce, oz. or ^. 12 ounces " 1 pound, Ib. or ft). UNIT EQUIVALENTS. SC. gr. dr. 1 = 20 . oz . 1 = 3 = 60 lb . 1 = 8 = 24 = 480 1 = 12 = 96 = 288 = 5760 SCALE ascending, 20, 3, 8, 12; descending, 12, 8, 3, 20. APOTHECARIES' FLUID MEASURE. O8. The measures for fluids, as adopted by apothecaries and physicians in the United States, to be used in compounding medi- cines, and putting them up for market, are given in the following TABLE. 60 minims, (^l) make 1 fluidrachm, f%. 8 fluidrachms, " 1 fluidounce, f.^. 16 fluidounces, " 1 pint, 0. 8 pints, " 1 gallon, Cong. TIME. 175 UNIT EQUIVALENTS. f5 1H f3 1 = 60 0. 1 = 8 = 480 Cong. 1 = 16 = 128 = 7680 i 8 = 128 = 2048 = 61440 SCALE ascending, 60, 8, 16, 8 ; descending, 8, 16, 8, 60. MEASURE OF TIME. 299. Time is the measure of duration. The unit is the day, and the table is made up of its divisors and multiples. TABLE. 60 seconds (sec.) make 60 minutes, 24 hours, 7 days, 365 days, 366 days, 12 calendar months, 100 years, 1 minute, min. 1 hour, h. 1 day, da. 1 week, wk. 1 common year, yr. 1 leap year, yr. 1 year, yr. 1 century, C. wk. 1 UNIT EQUIVALENTS. min. h. 1 da. 1 = 60 1 = 24 = 1440 7 = 168 = 10080 8760 = 525600 sec. = 60 = 3600 = 86400 = 604800 = 31536000 = 31622400 yr. mo. f 365 = 1 = 12 : (366 = 8784 = 527040 SCALE ascending, 60, 60, 24, 7 ; descending, 7, 24, 60, 60. The calendar year is divided as follows : No. of month. 1 9 3 4 5 6 7 8 9 10 11 12 Season. Winter, Spring, Summer, Autumn, Winter, Names of months. Abbreviations- No of days. January, Jan. 31 February, Feb. 28 or 29 March, Mar. 31 April, May, Apr. 30 31 June, Jun. 30 July, 31 August, Aug. 31 September, Sept. 30 October, Oct. 31 November, Nov. 30 December, Dec. 31 NOTES. 1. In most business transactions 30 days are called 1 month. 2. The civil day begins and ends at 12 o'clock, midnight. The astrouomi- 176 COMPOUND NUMBERS. i cal day, used by astronomers in dating events, begins and ends at 12 o'clock, noon. The civil year is composed of civil days. BISSEXTILE OR LEAP YEAR. 3OO. The period of time required by the sun to pass from one vernal equinox to another, called the vernal or tropical year, is exactly 365 da. 5 h. 48 min. 49.7 sec. This is the true year, and it exceeds the common year by 5 h. 48 min. 49.7 sec. If 365 days be reckoned as 1 year, the time lost in the calendar will be In 1 yr., 5 h. 48 min. 49.7 sec. " 4 *" 23 " 15 " 18.8 " The time thus lost in 4 years will lack only 44 min. 41.2 sec. of 1 entire day. Hence, If every fourth year be reckoned as leap year, the time gained in the calendar will be, In 4 yr., 44 min. 41.2 sec. "100" ( = 25X4 yr.) 18 h. 37 " 10 " The time thus gained in 100 years will lack only 5 h. 22 min. 50 sec. of 1 day. Hence If every fourth year be reckoned as leap year, the centennial years excepted, the time lost in the calendar will be, In 100 yr., 5 h. 22 min. 50 sec. " 400 " 21 " 31 " 20 " The time thus lost in 400 years lacks only 2 h. 28 min. 40 sec. of 1 day. Hence If every fourth year be reckoned as leap year, 3 of every 4 cen- tennial years excepted, the time gained in the calendar will be, In 400 yr., 2 h. 28 min. 40 sec. " 4000 " 24 h. 46 min. 40 sec. The following rule for leap year will therefore render the calendar correct to within 1 day, for a period of 4000 years. I. Every year that is exactly divisible by 4 is a leap year, the centennial years excepted ; the other years are common years. II. Every centennial year that is exactly divisible by 400 is a leap year ; the other centennial years are common years. NOTES. 1. Julius Caesar, the Roman Emperor, decreed that the year should consist of 365 days 6 hours; that the fi hours should be disregarded for 3 suc- cessive years, and an entire day be added to every fourth year. This day was inserted in the calendar between the 24th and 25th days of February, and is called the intercalary day. As the Romans counted the days backward from the first day of the following month, the 24th of February was called by them sexto CIRCULAR MEASURE. 177 calendas Martii, the sixth before the calends of March. The intercalary day which followed this was called bit-sexto calendas Martii; hence the name bissextile. 2. In 1582 the error in the calendar as established by Julius Caesar had in- creased to 10 days; that is, too much time had been reckoned as a year, until the civil year was 10 days behind the solar year. To correct this error, Pope Gregory decreed that 10 entire days should be stricken from the calendar, and that tlTe day following the 3d day of October, 1582, should be the 14th. Thi brought the vernal equinox at March 21 the date on which it occurred in the year 325, at the time of the Council of Nice. 3. The year as established by Julius Caesar is sometimes called the Julian year ; and the period of time in which it was in force, namely from 46 years B. C. to 1582, is called the Julian Period. 4. The year as established by Pope Gregory is called the Gregorian year, and the calendar now used is the Gregorian Calendar. 5. Most Catholic countries adopted the Gregorian Calendar soon after it was established. Great Britain, however, continued to use the Julian Calendar until 1752. At this time the civil year was 11 days behind the solar year. To cor- rect this error, the British Government decreed that 11 days should be stricken from the calendar, and thrt the day following the 2d day of September, 1752, should be the 14th. 6. Time before the adoption of the Gregorian Calendar is called Old Style (0. S), and since, New Style, (N. S.) In Old Style the year commenced March 25, and in New Style it commences January 1. 7. Russia still reckons time by Old Style, or the Julian Calendar; hence their dates are now 12 days behind ours. 8. The centuries are numbered from the commencement of the Christian era; the months from the commencement of the year; the days from the commence- ment of the month, and the hours from the commencement of the day, (12 o'clock, midnight.) Thus, May 23, 1860, 9 o'clock A.M., is the 9th hour of the 23d day of the 5th month of the 60th year of the 19th century. MEASURE OF ANGLES. 3O1. Circular Measure, or Circular Motion, is used princi- pally in surveying, navigation, astronomy, and geography, for reckoning latitude and longitude, determining locations of places and vessels, and computing difference of time. Every circle, great or small, is divisible into the same number of equal parts : as quarters, called quadrants; twelfths, called signs; 360ths, called degrees, etc. Consequently the parts of different circles, although having the same names, are of different lengths. The unit is the degree, which is -g^ part of the space about a point in any plane. The table is made up of divisors and multiples of this unit. TABLE. 60 seconds ("} make 1 minute, . . . . '. 60 minutes " 1 degree,.... . 30 degrees " 1 sign, S. 12 signs, or 360, " i circle, C. M 178 COMPOUND NUMBERS. ( UNIT EQUIVALENTS. I // 1 = 60 s l = 60 = 3600 c l = 30 = 1800 = 108000 1 = 12 = 360 = 21600 = 1296000 SCALE ascending, 60, 60, 30, 12 ; descending, 12, 30, 60, 60. NOTES. 1. Minutes of the earth's circumference are called geographic or nautical miles. 2. The denomination, signs, is confined exclusively to Astronomy. 3. A degree has no fixed linear extent. When applied to any circle it is always j^ part of the circumference. But, strictly speaking, it is not any part of a circle. 4. 90 make a quadrant or right-angle; 60 " sextant " & of a circle. MISCELLANEOUS TABLES. 3O2. COUNTING. 12 units or things make 1 dozen. 12 dozen " 1 gross. 12 gross " 1 great gross. 20 units " 1 score. 303. PAPER. 24 sheets make 1 quire. 20 quires " 1 ream. 2 reams " 1 bundle. 5 bundles " 1 bale. 304. BOOKS. The terms folio, quarto, octavo, duodecimo, etc., indicate the number of leaves into which a sheet of paper is folded. A sheet folded in 2 leaves is called a folio. A sheet folded in 4 leaves " a quarto, or 4to. A sheet folded in 8 leaves " an octavo, or 8vo. A sheet folded in 12 leaves " a 12mo. A sheet folded in 16 leaves " a 16mo. A sheet folded in 18 leaves " an 18mo. A sheet folded in 24 leaves " a 24mo. A sheet folded in 32 leaves " a 32mo. 3O5. COPYING. 72 words make 1 folio or sheet of common law. 90 " " 1 " *' " " chancery. GOVERNMENT STANDARDS. 179 GOVERNMENT STANDARDS OF MEASURES AND WEIGHTS. 306. In early times, almost every province and chief city had its own measures and weights ; but these were neither definite nor uniform. This variety in the weights and measures of different countries has always proved a serious embarrassment to commerce ; hence the many attempts that have been made in modern times to establish uniformity. The English, American, and French Governments, in establish- ing their standards of measures and weights, founded them upon unalterable principles or laws of nature, as will be seen by ex- amining the several standards. UNITED STATES STANDARDS. 307. In the year 1834 the U. S. Government adopted a uni- form standard of weights and measures, for the use of the custom houses, and the other branches of business connected with the General Government. Most of the States which have adopted any standards have taken those of the General Government. 308. The invariable standard unit from which the standard units of measure and weight are derived is the day. Astronomers have proved that the diurnal revolution of the earth is entirely uniform, always performing equal parts of a revo- lution on its axis in equal periods of duration. Having decided upon the invariable standard unit, a measure of this unit was sought that should in some manner be connected with extension as well as with this unit. A clock pendulum whose rod is of any given length, is found always to vibrate the Bame number of times in the same period of duration. Having now the day and the pendulum, the different standards hereafter have been determined and adopted. STANDARD OF EXTENSION. 3OO. The U. S. standard unit of measures of extension) whether linear, superficial, or solid, is the yard of 3 feet, or 36 inches, 180 COMPOUND NUMBERS. and is the same as the Imperial standard yard of Great Britain. It is determined as follows : The rod of a pendulum vibrating seconds of mean time, in the latitude of London, in a vacuum, at the level of the sea, is divided into 391393 equal parts, and 360000 of these parts are 36 inches, or 1 standard yard. Hence, such a pendulum rod is 39.1393 inches long, and the standard yard is ISfiBi of the length of the pendulum rod. STANDARDS OF CAPACITY. 310. The U. S. standard unit of liquid measure is the old English wine gallon, of 231 cubic inches, which is equal to 8.33888 pounds avoirdupois of distilled water at its maximum density; that is, at the temperature of 39.83 Fahrenheit, the ba- rometer at 30 inches. 311. The U. S. standard unit of dry measure is the British Winchester bushel, which is 18 inches in diameter and 8 inches deep, and contains 2150.42 cubic inches, equal to 77.6274 pounds avoirdupois of distilled water, at its maximum density. A gallon, dry measure, contains 268.8 cubic inches. NOTES. 1. Grain and some other commodities are sold by stricken measure, and in such cases the " measure is to be stricken with a round stick or roller, straight, and of the same diameter from end to end." 2. Coal, ashes, marl, manure, corn in the ear, fruit and roots are sold by heap measure. The bushel, heap measure, is the Winchester bushel heaped in the form of a cone, which cone must be 19J inches in diameter (= to the outside diameter of the standard bushel measure,) and at least 6 inches high. A bushel, heap measure, contains 2747.7167 cubic inches, or 597.2967 cubic inches more than a bushel stricken measure. Since 1 peck contains iis^o.4 = 537.605 cubic inches, the bushel, heap measure, contains 59.6917 cubic inches more than 5 pecks. As this is about 1 bu. 1 pk. If pt., it is sufficiently accurate in practice, to call 5 pecks stricken measure a heap bushel. 3. A standard bushel, stricken measure, is commonly estimated at 2150.4 cubic inches. The old English standard bushel from which the United States standard bushel was derived, was kept at Winchester, England; hence the name. 4. The wine and dry measures of the same denomination are of different capac- ities. The exact and the relative size of each may be readily seen by the fol' lowing 512. COMPARATIVE TABLE OF MEASURES OF CAPACITY. Cubic in. in Cubic in. in Cubic in. in Cubic in. in one gallon, one quart. one pint. one gill. AVine measure, 231 57f 28f 7^ Dry measure (^pk.,).. 268J 67 33 s NOTE. The beer gallon of 282 inches is retained in use onl.v by custom. GOVERNMENT STANDARDS. STANDARDS OF WEIGHT. 3 lt. It has been found that a given volume or quantity of distilled rain water at a given temperature always weighs the same. Hence, a cubic inch of distilled rain water has been adopted as the standard of weight. 314. The U. S. standard unit of weight is the Troy pound of the Mint, which is the same as the Imperial standard pound of Great Britain, and is determined as follows : A cubic inch of dis- tilled water in a vacuum, weighed by brass weights, also in a vacuum, at a temperature of 62 Fahrenheit's thermometer, is equal to 252.458 grains, of which the standard Troy pound con- tains 5760. 31d. The U. S. Avoirdupois pound is determined from the standard Troy pound, and contains 7000 Troy grains. Hence, the Troy pound is f J = j^l of an avoirdupois pound. But the Troy ounce contains 5 yf = 480 grains, and the avoirdupois ounce 7 fig = 437.5 grains; and an ounce Troy is 480 437.5 = 42.5 grains greater than an ounce avoirdupois. The pound, ounce, and grain, Apothecaries' weight, are the same as the like denominations in Troy weight, the only difference in the two tables being in the divisions of the ounce. 316. COMPARATIVE TABLE OF WEIGHTS. Troy. Avoirdupois. Apothecaries'. 1 pound = 5760 grains, = 7000 grains. = 5760 grains, 1 ounce = 480 " = 437.5 " = 480 " 175 pounds, = 144 pounds. = 175 pounds, STANDARD SETS OF WEIGHTS AND MEASURES. 317. A uniform set of weights and measures for all the States was approved by Congress, June 14, 1836, and furnished to the States in 1842. The set furnished consisted of A yard. A set of Troy weights. A set of Avoirdupois weights. 132 COMPOUND NUMBERS. A wine gallon, and its subdivisions. A half bushel, and its subdivisions. 318* State Sealers of Weights and Measures furnish standard sets of weights and measures to counties and towns. A County Standard consists of 1. A large balance, comprising a brass beam and scale dishes, with stand and lever. 2. A small balance, with a drawer stand for small weights. 3. A set of large brass weights, namely, 50, 20, 10, and 5 Ib. 4. A set of small brass weights, avoirdupois, namely, 4, 2, and L Ib., 8, 4, 2, 1, J, and i oz. 5. A brass yard measure, graduated to feet and inches, and the first foot graduated to eighths of an inch, and also decimally ; with a graduation to cloth measure on the opposite side ; in a case. 6. A set of liquid measures, made of copper, namely, 1 gal., gal., 1 qt., 1 pt., pt., 1 gi.; in a case. 7. A set of dry measures, of copper, namely, bu., 1 pk., pk. (or 1 gal.), 2 qt. (or gal.), 1 qt.; in a case. ENGLISH MEASURES AND WEIGHTS. GOVERNMENT STANDARDS. 31Q The English act establishing standard measures and weights, called " The Act of Uniformity," took effect Jan. 1, 1826, and the standards then adopted, form what is called the Imperial System. 32O. The Invariable Standard Unit of this system is the same as that of the United States, and is described in the Act of Uniformity as follows: "Take a pendulum which will vibrate seconds in London, on a level of the sea, in a vacuum; divide all that part thereof which lies between the axis of suspension and the center of oscillation, into 391393 equal parts ; then will 10000 of those parts be an imperial inch, 12 whereof make a foot, and 36 whereof make a yard." ENGLISH MEASURES AND WEIGHTS. 183 STANDARD OF EXTENSION. The English Standard Unit of Measures of Extension, whether linear, superficial, or solid, is identical with that of the United States, (3O9). STANDARDS OF CAPACITY. 322. The imperial Standard Gallon, for liquids and all dry substances, is a measure that will contain 10 pounds avoirdupois weight of distilled water, weighed in air, at 62 Fahrenheit, the barometer at 30 inches. It contains 277.274 cubic inches. 323. The Imperial Standard Bushel is equal to 8 gallons or 80 pounds of distilled water, weighed in the manner above de- scribed. It contains 2218.192 cubic inches. STANDARDS OF WEIGHT. 324. The Imperial Standard Pound is the pound Troy, which is identical with that of the United States Standard Troy pound of the Mint, (314.) 32o. The Imperial Avoirdupois Pound contains 7000 Troy grains, and the Troy pound 5760. It also is identical with the United States avoirdupois pound. TABLES. 326. The denominations in the standard tables of measures of extension, capacity, and weights, are the same in Great Britain and the United States. But some denominations in several of the tables are in use in various parts of Great Britain that are not known in the United States. These denominations are retained in use by common consent, and are recognized by the English common law. They are as fol- lows: 327. MEASURES OF EXTENSION. 18 inches make 1 cubit. 45 inches or 1 -i 11 5 quarters of the standard yard j NOTE. The cubit was originally the length o f a man's forearm and hand; or the distance from the elbow to the e?id of the middle finger. 184 COMPOUND NUMBERS. 328. MEASURES OF CAPACITY. LIQUID MEASURES. 9 old ale gallons make 1 firkin. 4 firkins " 1 barrel of beer. Imperial " "1 firkin. Imperial gallons or 63 wine 70 Imperial gallons or 84 wine " 2 hogsheads, that is 105 Imperial gallons or 126 wine 2 pipes 1 hogshead. 1 puncheon or of a tun. 1 pipe. 1 tun. Pipes of wine are of different capacities, as follows : 110 wine gallons make 1 pipe of Madeira. f Barcelona, 120 " " 1 " \ Vidonia, or I Teneriffe. 130 " " 1 " Sherry. 138 " " 1 " Port, 14.0 tt n i (Bucellas, or | Lisbon. 329. DRY MEASURE. 8 bushels of 70 pounds each make 1 quarter of wheat. 36 " heaped measure, " 1 chaldron of coal. NOTE. The quarter of wheat is 560 pounds, or J of a ton of 2240 pounds. 33O. WEIGHTS. 8 pounds of butchers' meat make 1 stone. 14 " " other commodities " 1 " or of a cwt. 2 stone, or 28 pounds " 1 todd of wool. 70 pounds of salt " 1 bushel. NOTE. The English quarter is 28 pounds, the hundred weight is 112 pounds, and the ton is 20 hundred weight, or 2240 pounds. FRENCH MEASURES AND WEIGHTS. GOVERNMENT STANDARDS. 331. The tables of standard measures and weights adopted by the French Government are all formed upon a decimal scale, and constitute what is called the French Metrical System. FRENCH MEASURES AND WEIGHTS. 185 332. Invariable Standard Unit. The French metrical sys- tem has, for its unit of all measures, whether of length, area, solidity, capacity, or weight, a uniform invariable standard, adopted from nature and called the mitre. It was determined and estab- lished as follows : a very accurate survey of that portion of the terrestrial meridian, or north and south circle, between Dunkirk and Barcelona, France, was made, under the direction of Govern- ment, and from this measurement the exact length of a quadrant of the entire meridian, or the distance from the equator to the north pole, was computed. The ten millionth part of this arc was denominated a metre, and from this all the standard units of measure and weight are derived and determined. STANDARDS OF EXTENSION. 333. ^e French Standard Linear Unit is the me*tre. 334. The French Standard Unit of Area is the Are, which is a unit 10 metres square, and contains 100 square metres. 335. The French Standard Unit of Solidity and Capacity is uhe Litre, which is the cube of the tenth part of the me'tre. STANDARD OF WEIGHT. 33G The French Standard Unit of Weight is the Gramme* which is determined as follows : the weight in a vacuum of a cubic decimetre or litre of distilled water, at its maximum density, was called a kilogramme, and the thousandth part of this was called a Gramme, and was declared to be the unit of weight. NOMENCLATURE OF THE TABLES. 337. It has already been remarked, (331 ), that the tables are all formed upon a decimal scale. The names of the multiples and divisors of the Government standard units in the tables are formed, by combining the names of the standard units with prefixes ; the names of the multiples being formed by employing the prefixes deca, (ten), hecto, (hundred), kilo, (thousand), and myria, (ten thousand), taken from the Greek numerals ; and the names of the divisors by employing the prefixes deci, (tenth), centi, (hundredth), 16* 186 COMPOUND NUMBERS. mili, (thousandth), from the Latin numerals. Hence the name of any denomination indicates whether a unit of that denomination is greater or less than the standard unit of the table. 338. I. FRENCH LINEAR MEASURE. TABLE. 10 millimetres ma 10 centimetres ' 10 decimetres ' 10 metres ' 10 decametres * 10 hectometres ' 10 kilometres ' ke 1 centimetre. 1 decimetre. 1 metre. 1 decametre. 1 hectometre. 1 kilometre. 1 myriametre. NOTES. 1. The metre is equal to 39.3685 inches, the standard rod of brasi on which the former is measured being at the temperature of 32 Fahrenheit, and the English standard brass yard or " Scale of Troughton" at 62. Hence, a metre is equal to 3.2807 feet English measure. 2. The length of a metre being 39.3685 inches, and of a clock pendulum vibrating seconds at the level of the sea in the latitude of London 39.1393 inches, the two standards differ only .2292, or less than of an inch. 339. II. FRENCH SQUARE MEASURE. TABLE. 100 square metres, or centiares (10 metres square) make 1 are. 100 ares (10 ares square) " 1 hectoare. NOTE. A square metre or centiare is equal to 1.19589444 square yards, and an are to 119.589444 square yards. 34O. III. FRENCH LIQUID AND DRY MEASURE. TABLE. 10 decilitres make 1 litre. 10 litres " 1 decalitre. 10 decalitres " 1 hectolitre. 10 hectolitres " 1 kilolitre. NOTES. 1. A litre is equal to 61.53294 cubic inches, or 1.06552 quarts of a U. , S. liquid gallon. 2. A table of Solid or Cubic Measure is also in use in some parts of France, although it is not established or regulated by government enactments or decrees. The unit of this table is a cubic metre, which is equal to 61532.94238 cubic inches, or 35.60934 cubic feet. This unit is called a Stere. TABLE. 10 decisteres make 1 stere. 10 steres " I decastere. MONEY AND CURRENCIES. 187 341. IV. FRENCH WEIGHT. T 10 milligrammes nn 10 centigrammes 10 decigrammes 10 grammes 10 decagrammes 10 hectogrammes 100 kilogrammes 10 quintals ' ABLE. ike 1 centigramme. 1 decigramme. 1 gramme. 1 decagramme. 1 hectogramme. 1 kilogramme. 1 quintal. ( 1 millier, or { 1 ton of sea wa NOTES. 1. A gramme is equal to 15.433159 Troy grains. 2. A kilogramme is equal to 2 Ib. 8 oz. 3 pwt. 1.159 gr. Troy, or 2 Ib. 3 oz. 4.1549 dr. Avoirdupois. 342. COMPARATIVE TABLE OF THE UNITED STATES, ENGLISH, AND FRENCH STANDARD UNITS OF MEASURES AND WEIGHTS. United States. English. French. Extension, Yd. of 3 ft., or 36 in. Same as U. S. Metre, 39.3685 in. n ., ) Wine gal., 231 cu. in. Imp'l gal., 277.274 cu. in. Litre, '61.53294 cu. in. *ciiy, | Wi nc h'r bu., 2150.42 cu. in. Inip'l bu., 2218.192 cu. in. Weight, Troy Ib., 5760 gr. Imperial Ib., 5760 gr. Gramme, 15.433159 T. gr. NOTES. 1. An Imperial gallon is equal to 1.2 wine gallons. 2. An old ale or beer gallon is very nearly 1.221 wine gallons, or 1.017 Im- perial gallons. 3. In ordinary computations 2150.4 cu. in. may be taken as a Winchester bushel, and 2218.2 cu. in. as an Imperial bushel. MONEY AND CURRENCIES. 343. Money is the commodity adopted to serve as the uni- versal equivalent or measure of value of all other commodities, and for which individuals readily exchange their surplus products or their services. 344. Coin is metal struck, stamped, or pressed with a die, to give it a legal, fixed value, for the purpose of circulating as money. NOTE. The coins of civilized nations consist of gold, silver, copper, and' nickel. 345. A Mint is a place in which the coin of a country or government is manufactured. NOTE. In all civilized countries mints and coinage are under the exclusive direction and control of government. 188 COMPOUND NUMBERS. 346. An Alloy is a metal compounded with another of greater value. In coinage, the less valuable or baser metal is not reckoned of any value. NOTE. "Gold and silver, in their pure state, are too soft and flexible for coin- age; hence they are hardened by compounding them with an alloy of baser metal, while their color and other valuable qualities are not materially impaired. 347. An Assayer is a person who determines the composi- tion and consequent value of alloyed gold and silver. The fineness of gold is estimated by carats, as follows : Any mass or quantity of gold, either pure or alloyed, is divided into 24 equal parts, and each part is called a carat. Fine gold is pure, and is 24 carats fine. Alloyed gold is as many carats fine as it contains parts in 24 of fine or pure golfl. Thus, gold 20 carats fine contains 20 parts or carats of fine gold, and 4 parts or carats of alloy. 348. An Ingot is a small mass or bar of gold or silver, in- tended either for coinage or exportation. Ingots for exportation usually have the assayer's or mint value stamped upon them. 349. Bullion is uncoined gold or silver. 3IO. Bank Bills or Bank Notes are bills or notes issued by a banking company, and are payable to the bearer in gold or silver, at the bank, on demand. They are substitutes for coin, but are not legal tender in payment of debts or other obligations. 351. Treasury Notes are notes issued by the General Govern- ment, and are payable to the bearer in gold or silver, at the gene- ral treasury, at a specified time. 3*52. Currency is coin, bank bills, treasury notes, and other substitutes for money, employed in trade and commerce. 353. A Circulating Medium is the currency or money of a country or government. 354. A Decimal Currency is a currency whose denomina- tions increase and decrease according to the decimal scale. I. UNITED STATES MONEY. 355. The currency of the United States is decimal currency, and is sometimes called Federal Money. MONEY AND CURRENCIES. 189 The unit is tlie gold dollar, weighing 25.8 grains; and all the other denominations are either divisors or multiples of this unit. TABLE. 10 mills (m.) make 1 cent ct. 10 cents " 1 dime d. 10 dimes " 1 dollar $. 10 dollars " 1 eagle E. I:NIT EQUIVALENTS. ct. m. d. 1 = 10 *. 1 = 10 = 100 ]: . 1 = 10 = 100 = 1000 1 = 10 = 100 = 1000 = 10000 SCALE uniformly 10. NOTKS. 1. Federal Money was adopted by Congress in 1786. 2. The character $ is supposed to be a contraction of U. 8. (United States), the U being placed upon the S. l>ythe "Coinage Act of 1873," the gold coins are the double eagle, eagle, half eagle, quarter eagle, three dollar, and one dollar pieces. The silver coins are the trade dollar, the half dollar, the quarter dollar, the twenty-cent and the ten-cent pieces. The nickel coins are the five-cent and three-cent pieces. The bronzf coin is the one-cent piece. NOTES. 1. The trade dollar is designed solely for purposes of commerce, and not for currency. Its weight is 420 grains. 2. The mill is a denomination used only in computations : it is not a coin. 3*>G. Government Standard. By Act of Congress, January 18, 1837, all gold and silver coins must consist of 9 parts (.900) pure metal, and 1 part (.100) alloy. The alloy for gold must con- sist of equal parts of silver and copper, and the alloy for silver of pure copper. The nickel coins are 75 parts copper and 25 parts nickel. STATE CURRENCIES. SoT. United States money is reckoned in dollars, dimes, cents, and mills, o'he dollar being uniformly valued in all the States at ICO cents ; but in many of the States money was formerly reckoned in dollars, shillings, and pence. *190 COMPOUND NUHBEKS. NOTE. At the time of the adoption of our decimal currency by Congress, in 1786, the cotonial currency, or bills of credit, issued by the colonies, had depreciated in value, and this depreciation, being unequal in the different colonies, gave rise to the different values of the State currencies. This usage, however, has become nearly, if not quite, obsolete all over the country. Georgia Currency. Georgia, South Carolina, $1 = 4s. 8d, = 56d, Canada Currency. The Dominion of Canada, $1 = 5s. = 60d. New England Currency. New England States, Indiana, Illinois, ] Missouri, Virginia, Kentucky, Tennes- i- $1 = 6s. = 72d. Bee, Mississippi, Texas, J Pennsylvania Currency. New Jersey, Pennsylvania, Delaware, ) A-J - g^ _ QQ Maryland, j " " New York Currency. New York, Ohio, Michigan, ) &i o O/M North Carolina, . . .. J |] II. CANADA MONET. 358. The currency of the Dominion of Canada is decimal, and the table and denominations are the same as those of the United States money. NOTE. The currency of the whole Dominion of Canada was made uniform July 1, 1871. Before the adoption of the decimal system, pounds, shillings and pence were used. The coin of Canada is of silver and of bronze. The silver coins are the 50-cent piece, 25-cent piece, 10-cent piece, and 5-cent piece. The 20-cent piece is no longer coined. The bronze coin is the cent. The gold coin used in Canada is the British sovereign, worth $4.86, and the half sovereign. The intrinsic value of the 50-cent piece in United States money is about 4G cents, of the 25-cent piece 23 ? T cents. In ordinary business transactions they pass the same as United Sta'tes coin of same denomination. 359. Government Standard. The silver coins consist of 925 parts (.925) pure silver and 75 parts (.075) copper. That is, they are .925 fine. MONEY AND CURRENCIES. III. ENGLISH MONET. 36O. English or Sterling Money is the currency of Great Britain. The unit is the pound sterling, and all the other denominations are divisors of this unit. TABLE. 4 farthings (far. or qr.) make 1 penny, .......... ,d. 12 pence " 1 shilling, s. 20 shillings " 1 pound or sovereign . . or sov. UNIT EQUIVALENTS. d. far. 8. 1=4 , or SOT. 1 = 12 = 48 1 = 20 = 240 = 960 SCAEE ascending, 4, 12, 20; descending, 20, 12, 4. NOTES. 1. Farthings are generally expressed as fractions of a penny; thus, 1 far., sometimes called 1 quarter, (qr.)=--|d. j 3 far. = |d. 2. The old/, the original abbreviation for shillings, was formerly written be- tween shillings and pence, and d, the abbreviation for pence, was omitted. Thus 2s. 6d. was written 2/6. A straight line is now used in place of the/ and shil- lings are written on the left of it and pence on the right. Thus, 2/6, 10/3, etc. COINS. The gold coins are the sovereign (= 1) and the half sovereign, (= 10s.) The silver coins are the crown, half crown, florin, the shilling, sixpenny, fourpenny, and threepenny pieces. The copper and bronze coins are the penny, half penny, and farthing. NOTE. The guinea (= 21s.) and the half guinea (= 10a. 6d. sterling) are old gold coins, and are no longer coined. 3G1. Government Standard. The standard fineness of Eng- lish gold coin is 11 parts pure gold and 1 part alloy; that is, it is 22 carats fine. The standard fineness of silver coin is 11 oz. 2 pwt. (= 11.1 oz.) pure silver to 18 pwt. (= .9 oz.) alloy. Hence the silver coins are 11 oz. 2 pwt. fine; that is, 11 oz. 2 pwt. pure silver in 1 Ib. standard silver. This standard is 37 parts (|J = .925) pure silver and 3 parts (^ = .075) copper. NOTE. A pound of English standard gold is equal in value to 14.2878 Ib. 14 Ib. 3 oz. 9 pwt. 1.727 gr. of silver. 192 COMPOUND NUMBERS. IV. FRENCH MONEY. 362. The currency of France is decimal currency. The unit is the franc, of -which the other denominations are di- visors. TABLE. 10 millimes make 1 centime. 100 centimes " 1 franc. SCALK ascending, 10, 100; descending, 100, 10. C oils' s. The gold coins are the 40, 20, 10, and 5 franc pieces. The silver coins are the 5, 2, and 1 franc, the 50 and 20 centime pieces. The bronze coins are the 10, 5, 2, and 1 centime pieces. 363. COMPARATIVE TABLE OF MONEYS. English. U. S. French. US. 1 penny, d. $0.0202 + 1 centime, ct. $0.00193 1 shilling, 8. .2433 + 1 decime, dc. 0.0193 1 florin, fl. .4866 + 1 franc, fr. .193 1 sovereign, sov. 4.8665 REDUCTION. 364:. Reduction is the process of changing a number from one denomination to another without altering its value. Reduction is of two kinds, Descending and Ascending. 365. Eeduction Descending is changing a number of one do- nomination to another denomination of less unit value ; thus, $1 = 10 dimes = 100 cents = 1000 mills. 366. Eeduction Ascending is changing a number cf cne de- nomination to another denomination of greater unit value; thus, 1000 mills = 100 cents = 10 dimes = $1. REDUCTION DESCENDING. CASE I. 367. To reduce a compound number to lower de- nominations. 1. Reduce 3 mi. 57 rd. 2 yd. 1 ft. 8 in. to inches. REDUCTION. 193 OPERATION. ANALYSIS. Since 3 mi. 57 rd. 2 yd. 1 ft. 8 in. in 1 mile there are 320 rd., in 3 miles there are 3 x 320 rd. = 960 rd., and the 57 rd. in the given number add- 508 i ed, makes 1017 rd. 5595$ yd. in 3 mi. 57 rd. 3 Since in 1 rd. there 16787| ft, are5$yd.,in!017rd. 12 there are 1017 x 5$ 2oT458~in. ?<* - 6593$ yd., which plus the 2 yd. in the given number = 5595| yd. in 3 mi. 57 rd. 2 yd. Since in 1 yd. there are 3 ft,, in 5595$ yd. there are 5595| x 3 ft. = 16786$ ft., which plus the 1 ft. in the given number = 16787$ ft, in 3 mi. 57 rd. 2 yd. 1 ft. And since in 1 ft. there are 12 in., in 16787$ ft. there are 16787$ x 12 in. = 201450 in., which plus the 8 in. in the given num- ber = 201458 in. in the given compound number. On examining the operation, we find that we have successively multiplied by the num- bers in the descending scale of linear measure from miles to inches, inclusive. But, as either factor may be used as a multiplicand (82, 1), we may consider the numbers in the descending scale as multipliers. Hence the following RULE. T. Multiply the highest denomination of the given compound number by that number of the scale which will reduce it to the next lower denomination, and add to the product the given number, if any, of that lower denomination. II. Proceed in the same manner with the results obtained in each lower denomination, until the reduction is brought to the denomina- tion required. EXAMPLES FOR PRACTICE. 1. In 16 lb. 10 oz. 18 pwt. 5 gr., how many grains? 17 194 COMPOUND NUMBERS. i 2. In 133 6 s. 8d., how many farthings? Am. 128,000. 3. Change 100 mi. to inches. Ans. G33GOOO in. 4. How many rods of fence will inclose a farm 11 miles square? Ans. 1920 rd. 5. The grey limestone of Central New York weighs 175 Ibs. to the cubic foot; what is the weight of a block 8 ft. long and 1 yd. square ? Ans. 6 T. 6 cwt. 3. What will be the cost of 1 hhd. of molasses at $.28 per gal. ? 7. A man wishes to ship 1548 bu. 1 pk. of potatoes in barrels containing 2 bu. 3 pk. each ; how many barrels must he obtain ? 8. A grocer bought 10 bu. of chestnuts at $3.75 a bushel, and retailed them at $.06 a pint; how much was his whole gain ? 9. Reduce 90 17' 40" to seconds. Ans. 325060". 10. In the 18th century how many days? Ans. 36524 da. 11. At 6^ cts. each, what will be the cost of a great-gross of Writing books ? Ans. $108. 12. How large an edition of an octavo book can be printed from 4 . bales 4 bundles 1 ream 10 quires of paper, allowing 8 sheets to the volume ? Ans. 2970 vol. 13. Suppose your age to be 18 yr. 24 da. ; how many minutes old are you, allowing 4 leap years to have occurred in that time ? 14. How many pence in 481 sovereigns? Ans. 115,440 d. 15. Reduce $7f to mills. Ans. 7375 mills. 16. In 3 P. of Sherry wine, how many qt. ? Ans, 1560 qt. 17. Reduce 37 Eng. ells 1 qr. to yd. Ans. 46 yd. 2 qr. 18. In 6 10s. lOd. how many dollars U. S. currency ? 19. Reduce 6 0. 14fg 3f 45iT[ to minims. 20. Reduce 1 T. 1 P. 1 hhd. to Imperial gallons. Ans. 367 i Imperial gal. 21. How many dollars Canada currency arc equal to 126 12s. 6d.? Ans. $500*. 22. ITow many pint, quart, and two-quart bottles, of each an equal number, may be rilled from a hogshead of wine ? Ans. 72. 23. How many steps of 2 ft. 9 in. each, will a man take, ia walking from Erie to Cleveland, tho distance being 95 mi.? KEDUCTION. 195 24. A grocer bought 12 bbl. of cider at $lf a barrel, and after converting it into vinegar, he retailed it at 6 cents a quart ; how much was his whole gain ? Ans. $69.72. 25. In 75 A. 4 sq. ch. 18 P. 118 sq. 1. how many square links? 26. How many inches high is a horse that measures 16 hands? 27. If a vessel sail 150 leagues in a day, how many statute miles does she sail ? Ans. 517.5. 1 28. If 14 A. be sold from a field containing 50 A., how many square rods will the remainder contain ? Ans. 5,760 sq. rd. 29. A man returning from Pike's Peak has 36 Ib. 8 oz. of pure gold ; what is its value at $1.04 per pwt. ? Ans. 9169.60. 80. A person having 8 hhd. of tobacco, each weighing 9 cwt. 42 Ib., wishes to put it into boxes containing 48 Ib. each ; how many boxes must he obtain ? Ans. 157. 31. A merchant bought 12 bbl. of salt at $1J a barrel, and re- tailed it at f of a cent a pound; how much was his whole gain? 32. A physician bought lib 10^ of quinine at $2.25 an ounce, and dealt it out in doses of 10 gr. at $.12 each; how much more than cost did he receive ? Ans. $82.50. CASE II. 368. To reduce a denominate fraction from a greater to a less unit. 1. Reduce ^ of a gallon to the fraction of a gill. OPERATION. ANALYSIS. To re- A gal. X f x f X f = T 8 T gi. duce gallons to gills, ~ we multiply succes- sively by 4, 2, and 4, 1 the numbers in the de- scending scale. And 2 since the given num- J2 ber is a fraction, we 11 8 = T 8 T gi., Ans. indicate the process, as in multiplication of fractions, after which we perform the indicated operations, and ob- tain -j^j, the answer. Hence, 196 COMPOUND NUMBERS. RULE. Multiply the fraction of the higher denomination l>y the numbers in the descending scale successively, between the given and the required denomination. NOTE. Cancellation may be applied wherever practicable. EXAMPLES FOR PRACTICE. 1. Reduce ^^ of a Ib. Troy to the fraction of a pennyweight. Ans. | pwt. 2. Reduce g | 2 of a hhd. to the fraction of a pint. 3. Reduce 3773 of a mile to the fraction of a yard. Ans. | yd. 4. Reduce ^| 2 of a gallon to the fraction of a gill. 5. What part of an ounce is-g-g^ of f of f of T \ of 3-j[- pounds avoirdupois weight ? Jy/.s-. ^jjj-f-g- oz. 6. Reduce -- of a dollar to the fraction of a cent. 7. Reduce ^ of a rod to the fraction of a link. Ans. | 1. 8. Reduce ^U of a scruple to the fraction of a grain. 9. What fraction of a yard is f of y* T of a rod ? 10. -j 6 5 of a week is | of how many days? Ans. 8| da. 11. What fraction of a square rod is yg 3 ^ of 4| times T 2 ^ of an acre ? Aits. T 3 ^ sq. rd. CASE ITT. 369. To reduce a denominate fraction to integers of lower denominations. 1. What is the value of f of a bushel ? OPERATION. ANALYSIS, f bu. = f of I bu. X 4 = f pk. = If pk. 4 pk., or lg pk. ; 2 pk. = 2 3 p k x 8 = ^ qt. = 44 qt. f 8 qt. = 44 qt. ; and | qt. 4 qt , X2 = f pt. = Hpt. =jof2pt. = lipt. The 1 pk. 4 qt. 1| pt., Ans. Uni * 8 ' I P k -' 4 /" \ ^> with the last denominate fraction, f pt., form the answer. Hence, RULE. I. Multiply the fraction by that number in the scale which will reduce it to the next lower denomination, and if the result be an improper fraction, reduce it to a whole or mixed number. REDUCTION. 197 IT. Proceed with the fractional part, if any, as before, until reduced to the denominations required. III. The units of the several denominations) arranged in their order, will be the required result. EXAMPLES FOR PRACTICE. 1. Reduce T 9 ^ of a yard to integers of lower denominations. Ans. 2 ft. 8 1 in. 2. Reduce j of a month to lower denominations. 3. Reduce {jj J of a short ton to lower denominations. 4. What is the value of J of a long ton ? Ans. 11 cwt. 12 Ib. 7^ oz. 5. What is the value of | of 2^ pounds apothecaries' weight? 6. What is the value of -?% of an acre ? Ans. 86 P. 4 sq. yd. 5 sq. ft. 127 T 5 3 sq. in. 7. Reduce | of a mile to integers of lower denominations. 8. What is the value of ^ of a great gross? Ans. 6 gross 10 doz. 3f . 9. What is the value in geographic miles of T 9 ff of a great circle? Ans. 12150 mi. 10. What is the value of | of 3| cords of wood ? Am. 2 Cd. 5 cd. ft. 9| cu. ft. 11. The distance from Buffalo to Cincinnati is 438 miles; hav- ing traveled | of this distance, how far have I yet to travel ? Ans. 262 mi. 256 rd. 12. What is the value of || f^ ? .4ns. 3 f^ 35 T^. 13. What is the value of f of a sign? Ans. 12 51' 25f". 14. A man having a hogshead of wine, sold T 6 3 of it; how much remained ? Ans. 33 gal. 3 qt. 1 pt. ! T 7 g gi. CASE IV. 3 TO. To reduce a denominate decimal- to integers of lower denominations. 1. Reduce .125 of a barrel to integers of lower denominations. 17* 198 COMPOUND NUMBERS. i OPERATION. ANALYSIS. We first multiply 125 the given decimal, .125 of a barrel, 81.5 by 31.5 (= 31J) to reduce it to o no-- , gallons, and obtain 3.9375 gallons. Omitting the 3 gallons, AVC mul- tiply the decimal, .9375 gal., by 3. / 500 qt. 4 to reduce it to quarts, and obtain 3.75 quarts. We next multiply 1.50 pt. the decimal part of this result by 4 2, to reduce it to pints, and obtain 2~Q { 1.5 pints. And the decimal part 3 gal. 3 qt. lpt. 2 gi., Am. of this r . esult we multi P^ b ? 4 to reduce it to gills, and obtain 2 gills. The integers of the several denominations, arranged in their order, form the answer. Hence, RULE. I, Multiply the given denominate decimal ly that num- ber in the descending scale which will reduce it to the next lower denomination, and point off the result as in multiplication, of decimals. II. Proceed with the decimal part of the, product in the same manner until reduced to the required denominations. The integers at the left will be the answer required. EXAMPLES FOR PRACTICE. 1. What is the value of .645 of a day ? Ans. 15 h. 28 min. 48 sec. 2. What is the value of .765 of a pound Troy ? 3. What is the value of .6625 of a mile? 4. W T hat is the value of .8469 of a degree ? Ans. 50' 48.84". 5. What is the value of .875 of a hhd. ? 6. What is the value of .85251 ? Ans. 17 s. 2.4 -f far. 7. What is the value of .715? Ans. 42' 54". 8. What is the value of 7.88125 acres? Ans. 7 A. 141 P. 9. What is the value of .625 of a fathom ? Ans. 3| ft. REDUCTION. 199 10. What is the value of .375625 of a barrel of pork? 11. What is the value of .1150390625 Cong. ? Ans. 14 f 5 5f^48 m. 12. What is the value of .61 of a tun of wine ? Ans. I P. 27 gal. 2 qt. 1 pt. 3.04 gi. REDUCTION ASCENDING. CASE I. 371. To reduce a denominate number to a com- pound number of higher denominations. 1. Reduce 157540 minutes to weeks. ANALYSIS. Dividing OPERATION. the given number of 60).!57540 min - minutes by CO, because 24 ) 2625 h. -f- 40 min. there are Js as many 7 ^ 1 00 d 4- h hours as minutes, and we obtain 2G45 h. plus a re- 15 wk. -f 4 da. mainder of 40 min. Vie 15 wk. 4 da. 9 h. 40 min., A?is. ncxt divide the 2G45 h. by 24, because there are .Jj as many days as hours, and we find that 2G45 h. = 109 da. plus a remainder of 9 h. Lastly we divide the 109 da. by 7, because there are \- as many weeks as days, and we find that 109 da. = 15 wk. plus a remainder of 4 da. The last quotient and the several remainders annexed in the order of the succeeding denominations, form the answer. 2. Reduce 201458 inches to miles. ANALYSIS. We OPERATION. divide successively 12)201458 in. by the numbers in 3)16788 ft. 2 in. tlie ascending scale of linear measure, 5J or 5.5)5596 yd. in tfae game manner 320)1017 rd. 2 yd. 1 ft. 6 in. as in the last pre- ~3 mi, 57 rd. cedin S operation. But, in dividing the 3 mi. 57 rd. 2 yd. 1 ft, 8 in., Ans. 5596 yd. by 5 or 5.5. we have a re- 200 COMPOUND NUMBERS. ( mainder of 2^ yd., and this reduced to its equivalent compound num- ber, (369) = 2 yd. 1 ft. 6 in. In forming our final result, the 6 in. of this number are added' to the first remainder, 2 in., making the 8 in. as given in the answer. From these examples and analyses we deduce the following RULE. I. Divide the given concrete or denominate number by that number of the ascending scale which will reduce it to the next higher denomination. II. Divide the quotient by the next higher number in the scale / and so proceed to the highest denomination required. The last quotient, with the several remainders annexed in a reversed order, will be the answer. NOTE. The several corresponding cases in reduction descending and reduc- tion ascending, being opposites, mutually prove each other. EXAMPLES FOR PRACTICE. 1. Reduce 1913551 ounces to tons. 2. In 97920 gr. of medicine how many Ib. ? Ans. 17 Ib. 3. Reduce 1000000 in. to mi. 4. How many acres in a field 120 rd. long and 56 rd. wide ? 5. In a pile of wood 60 ft. long, 15 ft. wide, and 10 ft. high, how many cords ? Ans. 70 Cd. 2 cd. ft. 8 cu. ft. 6. How many fathoms deep is a pond that measures 28 ft. 6 in. ? Ans. 4| fath. 7. In 30876 gi. how many hhd. ? 8. How many bushels of corn in 27072 qt. ? Ans. 846 bu. 9. At 2 cts. a gill, how much alcohol may be bought for $'2.54 ? 10. In 1234567 far. how many ? Ans. 1286 If d. 11. Reduce 2468 pence to half crowns. 12. In $88.35 how many francs? Ans. 475. 13. In 622080 cu. in. how many tons of round timber ? 14. In 84621 ty how many Cong. ? 15. If 135 million Gillott steel pens are manufactured yearly^ how many great-gross will they make? Ans. 78125. 16. Reduce 1020300" to S. Ans. 9 S. 13 25' 17. In 411405 sec. how many da. ? REDUCTION. 201 18. During a storm at sea, a ship changed her latitude 412 geographic miles ; how many degrees and minutes did she change ? An*. 6 52'. 19. If a man travel at the rate of a minute of distance in 20 minutes of time, how much time would he require to travel round the earth? Ans. 300 days. 20. In 120 gross how many score ? Ans. 864. 21. How many miles in the semi-circumference of the earth? 22. How much time will a person gain in 36 yr. by rising 45 min. earlier, and retiring 25 ruin, later, every day, allowing for 9 leap years ? Ans. 639 da. 4 h. 30 niin. 23. A grocer bought 20 gal. of milk by beer measure, and sold it by wine measure ; how many quarts did he gain ? Ans. 17|4- 24. How many bushels of oats in Connecticut are equivalent to 1500 bushels in Iowa ? Ans. 1875 bu. 25. Reduce 120 leagues to statute miles. Ans. 414.96 mi. 26. In 1 bbl. 1 gal. 2 qt. wine measure, how many beer gal- lons ? Ans. 27^\. 27. Reduce 150 U. S. bushels to Imperial bushels. Ans. 145.415 + Imp'l. bu. 28. How many squares in a floor 68 ft. 8 in. long, and 33 ft. wide? Ans. 22fjj. 29. How many cubic inches in a solid 4 ft. long 3 ft. wide, and 1 ft. 6 in. thick ? 30. How many acres in a field 120 rd. long and 56 rd. wide ? 31. Change 356 dr. apothecaries weight, to Troy weight. 32. A coal dealer bought 175 tons of coal at $3.75 by the long ton, and sold it at $4.50 by the short ton; how much was his whole gain ? Ans. $225.75. 33. How many acres of land can be purchased in the city of iNew York for $73750, at $1.25 a square foot? Am. 1 A. 56 P. 194 sq. ft. 34. An Ohio farmer sold a load of corn weighing 2492 lb., and a load of wheat weighing 2175 lb. ; for the corn he received $.60 a bushel, and for the wheat $1.20 a bushel; how much did he re- ceive for both loads ? Ans. $70.20. 202 COMPOUND NUMBERS. The following examples are given to illustrate a short and prac- tical method of reducing currencies. 35. What will be the cost of 54 bu. of corn at 5s. a bushel, New England currency ? OPERATION. Or, 54 x 5 270s. -j- 6 270s. $45 9 ANALYSIS. Since 1 bu. costs { 5s.,54bu. cost54x5s. = 270s.; and since Gs. make $1 N. E. :-=. currency, 270 ~- 6 = $45, Ans. OPERATION. 36. What will 270 bu. of wheat cost, @ 8s. 4d. Penn. currency ? ANALYSIS. Multiply the quantity by the price in Penn. currency, and divide the cost by the value of $1 in the same currency; or reduce the shil- lings and pence to a fraction 100 Or, 25 2 POO of a shilling, before multiply- ing and dividing. 37. Bought 5 hhd. of rum at the rate of 2s. 4d. a quart, Geor- gia currency ; how much was the whole cost ? OPERATION. 5 5 ANALYSIS. In this ex- 63 63 ample we first reduce 5 Or, 2 hhd. to quarts by multiply- ^ ing by 63 and 4, and then F proceed as in the preceding examples. $630 $630 38. Sold 120 barrels of apples, each containing 2 bu. 2 pk., at 4s. 7d. a bushel, and received pay in cloth at 10s. 5d. a yard ; how many yards of cloth did I receive ? OPERATION. ANALYSIS. The operation in this example is 12 similar to the preceding examples, except that we 440 divide the cost of the apples by the price of a unit of the article received in payment, reduced to units of the same denomination as the price of a unit of the article sold. The result will be the same in $132 whatever currency. 11 tt REDUCTION. 203 39. What cost 75 yards of flannel at 3s. 6d. per yard. New England currency ? Ans. $43.75. 40. A man in Philadelphia worked 5 weeks at 6s. 4d. a day ; how much did his wages amount to ? As. $25.83. 41. A farmer exchanged 2 bushels of beans worth 10s. Cd. per bushel, for two kinds of sugar, the one at lOd. and the other at lid. per pound, taking the same quantity of each kind; how many pounds of sugar did he receive? Ans. 24 Ib. 42. If corn be rated at 5s. lOd. per bushel in Vermont, at what price in the currency of New Jersey must it be sold, in order to gain $7.50 on 54 bushels? CASE II. 372. To reduce a denominate fraction from a less to a greater unit. 1. Reduce T 8 r of a gill to the fraction of a gallon. OPERATION. ANALYSIS. To re- T 8 T S 1 - * i X X - i = ,V gal. ducc S llls to we divide successively Or > by 4, 2, and 4, the 11 4 i 44 numbers in the as- cending scale. And since the given num- ber is a fraction, we 1 = ^j, Ans. indicate the process, as in division of frac- tions, after which we perform the indicated operations, and obtain i*f , the answer. Hence, RULE. Divide the fraction of the lower denomination by the numbers in the ascending scale successively, between the given and the required denomination. NOTE. The operation mny frequently be shortened by cancellation. EXAMPLES FOR PRACTICE. 1. Keduce f of a shilling to the fraction of a pound. Ans. 204 COMPOUND NUMBERS. 2. Reduce | of a pennyweight to the fraction of a pound Troy. An,.^\\>. 3. What part of a ton is j of a pound avoirdupois weight ? 4. What fraction of an hour is ^ of 20 seconds ? 5. What is the fractional difference between g |^ of a hhd. and | of a pt. ? . Ans. 35'^ hhd. 6. 2 | 4 of l of of a pint is what fraction of 2 pecks ? Ans. | 7. Reduce f of | of T \ of a cord foot to the fraction of a cord. Ans. -$ Cd. 8. What part of an acre is T 3 ^ of T 4 7 of 9^ square rods ? 9. |- of 220 rods is \ of ^ of how many miles? Ans. 12-J mi. 10. A block of granite containing | of ^ of 20 J cubic feet, is what fraction of a perch? Ans. -Ji Pch. 11. What part of a cord of wood is a pile 7^ ft. long, 2 ft. high, and 3| feet wide ? Ans. iff Cd. 12. Reduce f of an inch to the fraction of a yard. CASE III. 373. To reduce a compound number to a fraction of a higher denomination. 1. Reduce 2 oz. 12 pwt. 12 gr. to the fraction of a pound Troy. OPERATION. ANALYSIS. To find 2 oz. 12 pwt. 12 gr. = 1260 gr. what part one compound 1 Ib. Troy 5760 gr. number is of another, if IS lb - = 3 7 2 lb -> Ans - they must be like num * bers and reduced to the same denomination. In 2 oz. 12 pwt. 12 gr. there are 1260 gr., and in 1 Ib. there are 5760 gr. Therefore 1 gr. ia 3 ^ ff Ib., and 1260 gr. are * Ib. = ^ Ib., the answer. Hence, RULE. Reduce the given number to its lowest denomination for the numerator, and a unit of the required denomination to the same denomination for the denominator of the required fraction. NOTE. If the given number contain a fraction, the denominator of this frac- tion must be regarded as the lowest denomination. REDUCTION. 205 EXAMPLES FOR PRACTICE. 1. Reduce 100 P. to the fraction of an acre. Ans. { A. 2. What part of a mile is 266 rd. 3 yd. 2 ft. ? 3. What part of a is 18s. 5/1. 2-^ far. ? Ans. {f . 4. What part of 21 Ib. Apothecaries' weight is 7^ 7 3 23 14 gr. ? Ans. 2 5^ Ib. 5. What part of 3 weeks is 4 da. 16 h. 30 min. ? 6. Reduce 1| pecks to the fraction of a bushel. 7. From a hogshead of molasses 28 gal. 2 qt. were drawn; what part of the whole remained in the hogshead ? Ans. |4- 8. Reduce 4 bundles 6 quires 16 sheets of paper to the frac- tion of a bale. Ans. of a bale. 9. What part of 54 cords of wood is 4800 cubic feet ? 10. What is the value of ^ of a dollar 1 Ant. 36.30. 11. Reduce 3O. 3f 3 If 3 3 6 Iff. to the fraction of a Cong. 12. What part of a ton of hewn timber is 36 cu ft. 864 cu. in. ? CASE IV. 374. To reduce a compound number to a decimal of a higher denomination. 1. Reduce 3 cd. ft. 8 cu. ft. to the decimal of a cord. OPERATION. ANALYSIS. We re- 16 8.0 cu. ft. duce the 8 cu. ft. to the decimal of a cd. ft., by annexing a ci- .4375 Cd., Ans. pher, and dividing Or, by 16, the number of 3 cd. ft. 8 cu. ft. = 56 cu. ft. cu. ft. in 1 cd. ft., an- 1 Cd. a= 128 cu. ft. nexing the decimal jy5 g Cd. = T 7 g Cd. = .4375 Cd., Ans. quotient to the 3 cd. ft. We now reduce the 3. 5 cd. ft. to Cd. or a decimal of a Cd., by dividing by 8, the number of cd. ft. in 1 Cd., and we have .4375 Cd., the answer. Or, we may reduce the 3 cd. ft. 8 cu. ft., to the fraction of a Cd., 18 20C COMPOUND NUMBERS. (as in 373), and we shall have T Vg Cd. = T 7 F Cd., which, reduced to its equivalent decimal, equals .4375 Cd., the same as before. Hence, RULE. Divide the lowest denomination given by that number in the scale which will reduce it to the next higher denomination, and annex the quotient as a decimal to that higher. Proceed in the same manner until the whole is reduced to the denomination required. Or, Reduce the given number to a fraction of the required denomi- nation, and reduce this fraction to a decimal. EXAMPLES FOR PRACTICE. 1. Reduce 5 da. 9 h. 46 min. 48 sec. to the decimal of a week. Ans. .7725 wk. 2. Reduce 3 27' 46.44" to the decimal of a sign. 3. Reduce 51.52 P. to the decimal of an acre. 4. What part of 4 oz. is 2 oz. 1G pwt. 19.2 gr. ? Ans. .71. 5. What part of a mile is 28 rd. 2 yd. 1 ft. 11.04 in. ? 6. Reduce 3J | to the decimal of a pound. 7. Reduce 126 A. 4 sq. ch. 12 P. to the decimal of a town- ship. Ans. .0054893+ Tp. 8. What part of a fathom is 3f ft, ? Ans. .625 fath. 9. What part of 1J bushels is .45 of a peck? Ans. .09, 10. What part of 3 A. 80 P. is 51.52 P. ? Ans. .092. 11. Reduce f of \ of 22| Ib. to the decimal of a short ton. 12. What part of a f% is 5 f^ 36 Tit ? Ans. .1 fg. 13. Reduce 50 gal. 3 qt. 1 pt. to the decimal of a tun. Am. .20188 -f T. ADDITION. 37*5. Compound numbers are added, subtracted, multiplied, and divided by the same general methods as are employed in simple numbers. The corresponding processes are based upon the same principles; and the only modification of the operations and rules is that required for borrowing, carrying, and reducing by a varying, instead of a uniform scale. 876. 1. What is the sum of 50 hhd. 32 gal. 3 qt. 1 pt., 2 hhd. 19 gal 1 pt, 15 hhd. 46J gal., and 9 hhd. 39 gal. 2 qt.? ADDITION. 207 OPERATION. ANALYSIS. Writing the numbers so that hhd. gal. qt. pt. units of the same denomination shall stand 50 32 31 in the same column, we add the numbers of the right ^and or lowest denomination, and find the amount to be 3 pints, which is J? ^ 2 * equal to 1 qt. 1 pt. We write the 1 pt. under 78 11 3 1 the column of pints, and add the 1 qt. to the column of quarts. The amount of the numbers of the next higher denomination is 7 qt., which is equal to i gal. 3 qt. We write the 3 qt. under the column of quarts, and add "he 1 gal. to the column of gallons. Adding the gallons, we find the amount to be 137 gal., equal to 2 hhd. 11 gal. Writing the 11 gal. under the gallons in the given numbers, we add the 2 hhd. to the column of hogsheads. Adding the hogsheads, we find the amount to be 78 hhd., which we write under the left hand denomination, as in simple numbers. 2. What is the sum of ^ wk., f da., and | h. OPERATION. ANALYSIS. We T 7 5 wk. = 4 da. 21 h. 36 min. first find the value | da. = 14 " 24 min. of each fraction in | h. = 22 30 sec. inte ? ers f less de ~ -* nominations, (369), 5 and then add the Or, resulting or equiva- | da, X ^ = 3 3 - wk; lent compound num- | h. X ^ X ^ = -g^ wk ; bers. / 5 wk. -f & wk. -f ^ wk. = { * wk ; Or, we may re- | J wk. = 5 da. 12 h. 22 min. 30 sec. duce the S iven frac ' tions to fractions of the same denomination, (368 or 372), then add them, and find the value of their sum in lower denominations. 377. From these examples and illustrations we derive the following RULE. I. If any of the numbers are denominate fractions, or if any of the denominations are mixed numbers, reduce the frac- tions to inter/en of lower denominations. II. Write the numbers so that those of the same unit value will stand in the same column. III. Beginning at the right hand, add each denomination as in 208 COMPOUND NUMBERS. simple numbers, carrying to each succeeding denomination one for as many units as it takes of the denomination added } to make one of the next higher denomination. NOTE. The pupil cannot fait to see that the principles involved in adding compound numbers are the same as those in addition of simple numbers; and that the only difference consists in the different carrying units. EXAMPLES FOR PRACTICE. (!) (2.) lb. oz. pwt. gr. H>. 5 5 3 gr. 14 6 12 13 10 8 5 1 8 17 5 3 12 7 7 6 2 13 15 9 16 5 11 7 2 7 15 20 21 10 16 13 2 1 19 12 1 2 2 3 4 1 5 21 7 1 19 Sum, 66 11 9 5 58 4 5 2 19 (3.) (4.) mi. rd. ft. in. A. P. sq. yd. sq. ft. 7 26 11 9 140 137 27 6 4 16 7 11 320 70 14 2 36 14 3 111 7 3 1928 214 95 22 7 5 10 1 100 G 1 625 25 76 8 1 15 13 10 104 89 1 4 5. Add 1 T. 17 cwt. 8 lb., 5 cwt. 29 lb. 8 oz., 1 cwt. 42 lb. 6 oz., and 17 lb. 8 oz. Ans. 2 T. 3 cwt. 97 lb. 6 oz. 6. Add 6 yd. 2 ft., 3 yd. 1 ft. 8 in., 1 ft. 10* in., 2 yd. 2 ft. 6* in., 2 ft. 7 in., and 2 yd. 5 in. Ans. 16 yd. 2 ft. 1 in. 7. Add 4 Cd. 7 cd. ft., 2 Cd. 2 cd. ft. 12 cu. ft., 6 cd. ft. 15 cu. ft., 5 Cd. 3 cd. ft. 8 cu. ft., and 2 Cd. 1 cu* ft. 8. What is the sum of If hhd. 42 gal. 3 qt. 1} pt., I gal. 2 qt. f pt., and 1.75 pt. ? Ans. 2 hhd. 23 gal. 2 qt. 3 gi. 9. What is the sum of 145$ A., 7 A., 109J P., 1 A., 136.5 P., and | A.? Ans. 156 A. 39 P. 10. Required the sum of 31 bu. 2 pk., 10-J bu., 5 bu. 6 qt., 14 bu. 2.75 pk., and f pk. Ans. 62 bu. 1 pk. 5 qt If pt. SUBTRACTION. 209 11. Required the value of 42 yr. 7 A mo. -f 10 yr. 3 wk. 5 da. f 9| mo. -J- 1 wk. 16 h. 40 min. -f | mo. + 3| da. Am. 53 yr. 7 mo. 9 da. 23 h. 52 min. 12. Add 3 S. 22 50', 24 36' 25.7", 17' 18.2", 1 S. 3 12' 15.5", 12 36' 17.8", and 57.3". Ans. 6 S. 3 33' 14.5". 13. How many units in 1^ gross 7^ doz., 3 gross 1| doz., | of a great gross, 6| doz., and 4 doz. 7 units ? Ans. 2183. 14. What is the sum of 240 A. 6 sq. ch., 212.1875 sq. ch., and 5 sq. ch. lOf P.? Am. 262 A. 3 sq. ch. 13.8 P. 15. Add 31 Pch. 18 cu. ft., 84.6 cu. ft,, | Pch., and f cu. ft. 16. Add $3f , $25^, $12J, $2f , and $2.54 j. Ans. $47.0725. 17. What is the sum of 3 ft 5 g 4 % 2 9 17 gr., 2 It) 5 5 12 gr.,43 2^19 16 gr.? Ans. 5ft lOg 4^29 5 gr. 18. A N. Y. farmer received $.60 a bushel for 4 loads of corn ; the first contained 42.4 bu., the second 2866 lb., the third 36| bu., and the fourth 39 bu. 29 lb. How much did he receive for the whole? Ans. $100.84-. 19. Bought three loads of hay at $8 per ton. The first weighed 1.125 T., the second If T., and the third 2500 pounds; how much did the whole cost ? Ans. $30.20. 20. A man in digging a cellar removed 140 J cu. yd. of earth, in digging a cistern 24.875 cu. yd., and in digging a drain 46 cu. yd. 20| cu. ft. What was the amount of earth removed, and how much the cost at 18 cts. a cu. yd. ? Ans. 212.425 cu. yd. removed; $38.24 cost. SUBTRACTION. 378. 1. From 18 lb. 5 oz. 4 pwt. 14 gr. take 10 lb. 6 oz. 10 pwt. 8 gr. OPERATION. ANALYSIS. Writing the subtra- ib. oz. pwt. gr. hend under the minuend, placing 5 4 14 units of the same denomination under 10 6 10 8 each other, we subtract 8 gr. from 7 10 14 6 14 gr. and write the remainder, 6 gr., underneath. Since we cannot 18* O 210 COMPOUND NUMBERS. i subtract 10 pwt. from 4 pwt., we add 1 oz. or 20 pwt. to the 4 pwt., subtract 10 pwt. from the sum, and write the remainder, 14 pwt., underneath. Having added 20 pwt. or 1 oz. to the C oz. in the sub- trahend, we find that we cannot subtract the sum, 7 oz., from the 5 oz. in the minuend ; we therefore add 1 Ib. or 12 oz. to the 5 oz., sub- tract 7 oz. from the sum, and write the remainder, 10 oz., underneath. Adding 12 oz. or 1 Ib. to the 10 Ib. in the subtrahend, we subtract the sum, 11 Ib., from the 18 Ib. in the minuend, as in simple numbers, and write the remainder, 7 Ib., underneath. 2. From 12 bar. 15 gal. 3 qt. take 7 bar. 18 gal. 1 qt. OPERATION. ANALYSIS. Proceeding as in the last bar. gal. qt. operation, we obtain a remainder of 4 bar. i;? 28J gal. 2 qt. But, gal. = 2 qt., which added to the 2 qt. in the remainder makes 28 1 gal., and this added to the 28 gal. makes 4 29 29 gal. ; and the answer is 4 bar. 29 gal. 3. From f of a rod subtract f of a yard. OPERATION. ANALYSIS. We first , -, , A . . , . find the value of each | rd. =4 yd. ft. 41 in. 1 yd = 2 " 3 fraction in integers of 4~. ! -- H - lower denominations, 1 lj (369), and then sub- Or, tract the less value from the g reater - Or > 3 Y a x _ a v d v 2 - - 3 rd * J Q < 5 - 4 ) a * TT - *2 m - | rd. -^ rd. = |J rd. ; given fractions to frac- || rc i. = 3 yd. 1 ft. 1J in. tions f tne same de ~ nomination, subtract the less value from the greater, and find the value of the remainder in integers of lower denominations. 379. From these illustrations we deduce the following RULE. I. If any of tJie numbers are denominate fractions, or if any of the denominations are mixed numbers, reduce the frac- tions to integers of lower denominations. II. Write the subtrahend under the minuend, so that units of the same denomination shall stand under each other. III. Beginning at the right hand, subtract each denomination separately, as in simple numbers. SUBTRACTION. 211 IV. // the number of any denomination in the subtrahend ex- ceed that of the same denomination in the minuend, add to the number in the minuend as many units as make one of the next higher denomination, and then subtract / in this case add 1 to the next higher denomination of the subtrahend before subtracting. Proceed in the same manner with each denomination. EXAMPLES FOR PRACTICE. 00 (20 mi. rd. ft. in. A. P. From 175 147 11 4 320 146.4 Take 59 250 12 9 150 111.86 Rem. 115 216 15 1 170 34.54 (3.) (4-j hhd. gal. qt. yr. mo. wk. da. h. 5 36 H 45 1 3 17 2 45 y 10 9 1 22 6.8 5. Subtract 15 rd. 10 ft. 3J in. from 26 rd. 11 ft. 3 in. Ans. 11 rd. llf in. 6. From 1 T. 11 cwt. 30 Ibs. 6 oz. take 18 cwt. 45 Ib. 7. Subtract .659 wk. from 2 wk. 3-jj- da. Ans. 1 wk. 6 da. 5 h. 17 min. 16 sec. 8. From |fj hhd. take .90625 gal. Ans. 32 gal. 9. From | of 3| A. take 142.56 P. 10. Subtract -fa Ib. Troy, from 10 Ib. 8 oz. 8 pwt. 11. From a pile of wood containing 36 Cd. 4 cd. ft, there was old 10 Cd. 6 cd. ft, 12 cu. ft; how much remained? 12. From 5^- barrels take % of a hogshead. Ans. 4 bbl. 11 gal. 1 qt. 1 3. Subtract -J-JJ of a day from f of a week. Ans. 4 da. 49 min. 30 sec. 14. From -| of a gross subtract % of a dozen. Ans. 6|- doz. 15. From % of a mile take fj of a rod. 16. Subtract 2 A. 125.76 P. from 5 A. 64.24 P. Ans. 2 A. 98.48 P. 212 COMPOUND NUMBERS. 17. Subtract .0625 bu. from | pk. 'Ans. 4 qt. 18. From the sum of of 3651 da. and | of 5 wk. take 49} min - Ans. 33 wk. 1 da. 1 h. 10f min. 19. From the sum off of 3| mi. and 17^ rd., take 21 3-J- rd. 20. From 15 bbl. 3.25 gal. take 14 bbl. 24 gal. 3.54 qt. 21. A farmer in Ohio having 200 bu. of barley, sold 3 loads, the first weighing 1457 lb., the second 1578 lb., and the third 1420 lb. ; how many bushels had he left? Ans. 107 bu. 9 lb. 22. Of a farm containing 200 acres two lots were reserved, one containing 50 A. 136.4 P. and the other 48 A. 123.3 P.; the re- mainder was sold at $35 per acre. How much did it brini denominate fraction is the most readily performed by 193, :if'r,<.-r which the product may bo reduced to integers of lower denomina- tions by 389. As the work of multiplying by large prime numbers is somewhat tedious, the following method may often be so modified and adapted as to greatly shorten the operation. MULTIPLICATION. 215 1. How many bushels of grain in 47 bags, each containing 2 bu. 1 pk. 4 qt. ? FIRST OPERATION. 47 = (5 x 9) + 2 2 bu. 1 pk. 4 qt. x 2 o 1 1 bu. o pk. 4 qt. in 5 bags. 9 106 bu. 3 pk. 4 qt. in 45 bags. 4 3 *< "2 " 111 bu. 2 pk.~4 qt. " 47 " SECOND OPERATION. 47 = (6 x 8) _ 1 2 bu. 1 pk. 4 qt, x 1 6 14 bu. 1 pk. in G bags. 114 bu. in 48 bags. 2 " 1 pk. 4 qt. " 1 bag. Ill bu. 2 pk. 4 qt. " 47 bags. ANALYSIS. Multiplying the contents of 1 bag by 5, and the resulting product by 9, we have the contents cf 45 bags, which is the com- posite number next less than the given prime number, 47. Next, multiplying the con- tents of 1 bag by 2, we have the contents of 2 bags, -which, added to the contents of 45 bags, gives us the contents of 45 -f 2 = 47 bags. Or, we may multiply the contents of 1 bag by the fac- tors of the composite num- ber next greater than the given prime number, 47, and from the last product sub- tract the multiplicand. EXAMPLES FOR PRACTICE. (1.) T. cwt. Ib. oz. 12 15 27 9 8 102 2 20 8 (3.) A. P. Bq. yd. Bq. ft. 7 73 21 7 6 mi. 14 (2.) rd. ft. 276 14 9 133 251 Cd. cd. ft. cu. ft. 10 7 13 12 5. Multiply 34 bu. 3 pk. 6 qt. 1 pt. by 14. 6. Multiply 4 Ib. 10 oz. 18.7 pwt, by 27. Ans. 132 Ib. 7 oz. 4.9 pwt 7. Multiply 95 33 2 3 13 gr. by 35. 216 COMPOUND NUMBERS. 8. Multiply 5 gal. 2 qt, 1 pt. 3.25 gi. by 96. 9. Multiply 78 A. 135 P. 15 sq. yd. by 15-f. Ans. 1235 A. 42 P. 23} sq. yd. 10. What is 73 times 9 cu. yd. 10 cu. ft. 1424 cu. in.? 11. Multiply 2 Ib. 8 oz. 13 pwt. 18 gr. by 59. 12. Multiply 4 yd. 1 ft. 4.7 in. by 125. 13. If 1 qt. 2 gi. of wine fill 1 bottle, how much will be re- quired to fill a gross of bottles of the same capacity ? 14. Multiply 7 0. 10 f I 4 f 3 25 1t[ by 24. Ans. 22 Cong. 7 0. 13 f g 2 f 3. 15. Multiply 3 hhd. 43 gal. 2.6 gi. by 17. 16. Multiply 9 T. 13 cwt. 1 qr. 10.5 Ib. by 1.7. NOTE. When the multiplier contains a decimal, the multiplicand may be re- duced to the lowest denomination mentioned, or the lower denominations to a decimal of the higher, before multiplying. The result can be reduced to the compound number required. Ans. 16T. 8 cwt. 2qr. 20.65 Ib. 17. If a pipe discharge 2 hhd. 23 gal. 2 qt. 1 pt. of water in 1 hour, how much will it discharge in 4.8 hours, if the water flow with the same velocity? Ans. 11 hhd. 25 gal. 1 pt. 2.4 gi. 18. What will be the value of 1 dozen gold cups, each cup weighing 9 oz. 13 pwt. 8 gr., at $212.38 a pound ? 19. What cost 5 casks of wine, each containing 27 gal. 3 qt. 1 pt., at $1.371 a gallon? Ans. $191.64 + . 20. A farmer sold 5 loads of oats, averaging 37 bu. 3 pk. 5 qt. each, at $.65 per bushel; how much money did he receive for the grain? Ans. $123.20. DIVISION. 383. 1. Divide 37 A. 60 P. 7 sq. yd. by 8. ANALYSIS. Wri ting the divisor on OPERATION, j the left of the Dividend, divide the 8)37 A._60 P. 7 sq. yd. llighes t denomination, and the quo- 4 107 16 tient is 4 A. and a remainder of 5 A. Write the quotient under the de- nomination divided, and reduce the remainder to rods, making 800 P., which added to the 60 P. of the dividend, equals 860 P. Dividing this, we have a quotient of 107 P. and a remainder of 3 P. Writing the 107 P. under the denomination divided, we reduce the remainder to DIVISION. square yards, making 121 sq. yd., which added to the 7 sq. yd. of the dividend, equals 128 sq. yd. Dividing this, we have a quotient of 16 sq. yd. and no remainder. 2. Divide 111 bu. 2 pk. 4 qt. by 47. OPERATION. 47)lllbu.2pk.4qt.(2bu.lpk.4qt ANALYSIS. The divisor being large, 17 bu. rem. and a prime num- 4 her, we divide by 70 pk. in 17 bu. 2 pk. lon g division, set- 47 ting down the whole work of sub- 2 j>P k - rem - tractingandreduc- ing. 188 qt. in 23 pk. 4 qt 188 From these examples and illustrations we derive the following RULE. I. Divide the highest denomination as in simple num- bers, and each succeeding denomination in the same manner, if there be no remainder. II. If there be a remainder after dividing any denomination, reduce it to the next lower denomination, adding in the given num- ber of that denomination, if any, and divide as before. III. The several partial quotients will be the quotient required. NOTES. 1. When the divisor is large and is a. composite number, we may shorten the work by dividing by the factors. 2. When the divisor and dividend are both compound numbers, they must both be reduced to the same denomination before dividing, and then the process is the smne as in simnle numbers. 3. The division of a denominate fraction is most readily performed by 195, after which the quotient tnoy bo reduced to its equivalent compound number, by 369. EXAMPLES FOR PRACTICE. 5)62 a-) 8. 7 d. 9 far. 3 ib. 9)56 oz (20 pwt. 17 gr. 6 Quotient, 12 19 9 6 8 6Ib. 3 oz. 8 pw. 14 gr. 218 COMPOUND NUMBERS. (3.) (4'.) hhd. gal. qt. pt. T. cwt. qr. Ib. 12)9 28 2 19)373 19 2 4_ "49 2 1 ~19 13 2 16 5. Divide 358 A. 57 P. G sq. yd. 2 sq. ft. by 7. Ans. 51 A. 31 P. 8 sq. ft. 6. Divide 192 bu. 3 pk. 1 qt. 1 pt. by 9. 7. Divide 336 yd. 4 ft. 3^ in. by 21. Ans. 16 yd. 2| in. 8. Divide 77 sq. yd. 5 sq. ft. 82 sq. in. by 13. Ans. 5 sq. yd. 8 sq. ft. 106 sq. in. 9. Divide 678 cu. yd. 1 cu. ft. 1038.05 cu. in. by 67. 10. Divide 1986 mi. 3 fur. 20 rd. 1 yd. by 108. 11. Divide 12 sq. mi. 70 P. by 22. Ans. 341 A. 56f P. NOTE 4. Observe that 22$ = */ ; hence, multiply by 2, and divide the result by 45. 12. Divide 365 da. 6 h. by 240. 13. Divide 3794 cu. yd. 20 cu. ft. 709^ cu. in. by 33f 14. Divide 121 Ib. 3^ 2^ 19 4 gr. by 13|. 15. Divide 28 51' 27.765" by 2.754. Ans. 10 28' 42*". 16. Divide 202 yd. 1 ft. 6| in. by f . Ans. 337 yd. 1 ft, 7 in. 17. Divide 1950 da. 15 h. 15| min. by 100. 18. If a town 4 miles square be divided equally into 124 farms, how much will each farm contain ? Ans. 82 A. 92ff P. 19. A cellar 48 ft. 6 in. long, 24 ft. wide, and 6* ft. deep, was excavated by 6 men in 8 days; how many cubic yards did each man excavate daily? Ans. 5 cu. yd. 22 cu. ft. 1080 cu. in. 20. How many casks, each containing 2 bu. 3 pk. 6 qt., can be filled from 356 bu. 3 pk. 5.qt. of cherries? Ans. 12 LONGITUDE AND TIME. SJ84L Since the earth performs one complete revolution on its axis in a day or 24 hours, the sun appears to pass from east to west round the earth, or through 360 of longitude once in every LONGITUDE AND TIME. 219 24 hours of time. Hence the relation of time to the real motion of the earth or the apparent motion of the sun, is as follows: Time. Longitude. 24 h. = 360 1 h. or 60 min. = 8 ^ = 15 = 900 / 1 min. or 60 sec. = ^ = % (/ = 15' = 900" Isec. = if = V*" = l? x/ Hence, 1 h. of time - = 15 of longitude. 1 min. " = 15 X " " 1 sec. " = 15" " " CASE I. 385. To find the difference of longitude between twc places or meridians, when the difference of time is known. ANALYSIS. A difference of 1 h. of time corresponds to a difference of 15 of longitude, of 1 min. of time to a difference of 15' min. of longitude, and of 1 sec. of time to a difference of 15" of longitude, (384). Hence, the RULE. Multiply the difference of time, expressed in hours, minutes, and seconds, l>y 15, according to the rule for multiplica- tion of compound numbers ; the product will be the difference of longitude in degrees, minutes, and seconds. NOTES. 1. If one place be in east, and the other in west longitude, the dif- ference of longitude is found by adding them, and if the sum be greater than 180, it must be subtracted from 360. 2. Since the sun appears to move from east to west, when it is exactly 12 o'clock sit one place, it will be past 12 o'clock at all places east, and before 12 ;.t nil places west. Hence, if the difference of time between two places, be subtracted from the time at the eagerly place, the result will be the time at the westerly place; and if the difference be added to the time at the westerly place the result will be the time at the easterly place. EXAMPLES FOR PRACTICE. 1. When it is 9 o'clock at Washington, it is 8 h. 7 min. 4 sec. at St. Louis; the longitude of Washington being 77 1', west, what is the longitude of St. Louis? Ans. 90 15' west. 2. The sun rises at Boston 1 h. 11 min. 56 sec. sooner than at New Orleans; the longitude of New Orleans being 89 2' west, what is the longitude of Boston ? Ans. 71 3' west. 220 COMPOUND NUMBERS. i 3. When it is half past 2 o'clock in the morning at Havana, it is 9 h. 13 min. 20 sec. A. M. at the Cape of Good Hope; the longitude of the latter place being 18 28' east, what is the longitude of Havana ? Am. 82 22' west. 4. The difference of time between Valparaiso and Rome is 6 h. 8 min. 28 sec. ; what is the difference in longitude ? 5. A gentleman traveling East from Fort Leavenworth, which is in 94 44' west longitude, found, on arriving at Philadelphia, that his watch, an accurate time keeper, was 1 h. 18 min. 16 sec. slower than the time at Philadelphia; what is the longitude of Philadel- phia ? Ans. 75 10' west. 6. When it is 12 o'clock M. at San Francisco it is 2 h. 58 min. 23 sec. P. M. at Rochester, N, Y; the longitude of the latter place being 77 51' W., what is the longitude of San Francisco? 7. A gentleman traveling West from Quebec, which is in. 71 12' 15" W. longitude, finds, on his arrival at St. Joseph, that his watch is 2 h. 33 min. 53{ J sec. faster than true time at the latter place. If his watch has kept accurate time, what is the longitude of St. Joseph ? Ans. 109 40' 44" V. 8. A ship's chronometer, set at Greenwich, points to 5 h. 40 min. 20 sec. p. M., when the sun is on the meridian ; what is the ship's longitude ? Ans. 85 5' west. NOTK 3. Greenwich, Eng., is on the meridian of 0, and from this meridian longitude is reckoned. 9. The longitude of Stockholm being 18 3' 30" E., when it is midnight there, it is 5 h. 51 min 41 ? sec. p. M. at New York; what is the longitude of New York from Greenwich ? An*. 74 r <>" W 10. A vessel set sail from New York, and proceeded in a south- easterly direction for 24 days. The captain then took an obser- vation on the sun, and found the local time at the ship's meridian to be 10 h. 4 min. 36.8 sec. A. M. ; at the moment of the observa- tion, his chronometer, which had been set for New York time, showed 8 h. 53 min. 47 sec. A. M. Allowing that the chronometer had gained 3.56 sec. per day, how much had the ship changed her longitude since she set sail? Ans. 18 3' 48.6". LONGITUDE AND TIME. 221 CASE II. 386. To find the difference of time between two places or meridians, when the difference of longitude is known. ANALYSIS. A difference of 15 of longitude produces a difference of 1 h. of time, 15' of longitude a difference of 1 min. of time, and 15" of longitude a difference of 1 sec. of time, (384). Hence the RULE. Divide the difference of longitude, expressed in degrees, minutes, and seconds, l>y 15, according to the rule for division of compound numbers; the quotient will be the difference of time in hours, minutes, and seconds. EXAMPLES FOR PRACTICE. 1. Washington is 77 V and Cincinnati is 84 24' west longi- tude; what is the difference of time? Ans. 29 min. 32 sec. 2. Paris is 2 20' and Canton 113 14' east longitude; what is the difference in time ? 3. Buffalo is 78 55' west, and the city of Rome 20 30' east longitude ; what is the difference in time ? Ans. 6 h. 37 min. 40 sec. 4. A steamer arrives at Halifax, 63 36' west, at 4 h. 30 min. p. M. ; the fact is telegraphed to New York, 74 V west, without loss of time ; what is the time of its receipt at New York ? 5. The longitude of Cambridge, Mass., is 71 7' west, and of Cambridge, England, is 5' 2" east; what time is it at the former place when it is 12 M. at the latter ? Ans. 7 h. 15 min. llyf sec. A. M. 6. The longitude of Pekin is 118 east, and of Sacramento City 120 west; what is the difference in time? 7. The longitude of Jerusalem is 35 32' east, and that of Baltimore 76 37' west; when it is 40 minutes past 6 o'clock A. M. at Baltimore, what is the time at Jerusalem? 8. What time is it in Baltimore when it is 6 o'clock p. M. at Jerusalem? Ans. 10 h. 31 min. 24 sec. A. M. 19* 222 COMPOUND NUMBERS. 9. The longitude of Springfield, Mass., is 72 35' 45" W., and of Galveston, Texas, 94 46' 34" W.; when it is 20 inin. past 6 o'clock A. M. at Springfield, what time is it at Galveston ? 10. The longitude of Constantinople is 28 49' east, and of St. Paul 93 5' west; when it is 3 o'clock p. M. at the latter place, what time is it at the former ? 11. What time is it at St. Paul when it is midnight at Constan- tinople? Ans. 3 h. 52 miri. 24 sec. r. M. 12. The longitude of Cambridge, Eng., is 5' 2" E., and of Mobile, Ala., 88 1' 29" W.; when it is 12 o'clock M. at Mobile, what is the time at Cambridge ? PROMISCUOUS EXAMPLES IN COMPOUND NUMBERS. 1. In 9 Ib. 8g 13 29 19 gr. how many grains? 2. How much will 3 cwt. 12 Ib. of hay cost, at $15^ a ton? 3. In 27 yd. 2 qr. how many Eng. ells? Ans. 22. 4. Reduce 818.945 to sterling money Ans. 3 17s. 10-|f|fd. 5. In 4 yr. 48 da. 10 h, 45 sec. how many seconds ? 6. How many printed pages, 2 pages to each leaf, will there be in an octavo book having 24 fully printed sheets ? Ans. 384. 7. At 1/6 sterling per yard, how many yards of cloth may be bought for 5 6s. 6d. ? Ans. 71 yd. 8. In 4 mi. 51 ch. 73 1. how many links? 9. In 22 A. 153 sq. rd. 2j- sq. yd. how many square yards? 10. How many demijohns, each containing 3 gal. 1 qt. 1 pt., can be filled from 3 hhd. of currant wine ? Ans. 56. 11. Paid $375.75 for 2^ tons of cheese, and retailed it at 9 cts. a pound ; how much was my whole gain ? 12. A gentleman sent a silver tray and pitcher, weighing 3 Ib. 9 oz., to a jeweler, and ordered them made into tea spoons, each weighing 1 oz. 5 pwt. j how many spoons ought he to receive ? Ans. 3 doz. 13. What part of 4 gal. 3 qt. is 2 qt, 1 pt. 2 gi. ? Ans. 4 J. 14. Reduce | of T 4 T of a rod to the fraction of yard. 15. How many yards of carpeting 1 yd. wide, will be required to cover a floor 26 ft. long, and 20 ft. wide ? Ans. 58|. PROMISCUOUS EXAMPLES. 223 16. If I purchase 15 T. 3 cwt. 3 qr. 24 Ib. of English iron, by long ton weight, at 6 cents a pound, and sell the same at 8140 per short ton, how much will I gain by the transaction ? 17. What will be the expense of plastering a room 40 ft. long, 30 ft. wide, and 22 J- ft. high, at 18 cents a sq. yd., allowing 1375 sq. ft. for doors, windows, and base board? Ans. $69. 78. 18. How much tea in 23 chests, each weighing 78 Ib. 9 oz. ? 19. Valparaiso is in latitude 33 2' south, and Mobile 30 41 north ; what is their difference of latitude ? Ans. 63 43'. 20. If a druggist sell 1 gross 4 dozen bottles of Congress water a day, how many will he sell during the month of July ? 21. Eighteen buildings are erected on an acre of ground, each occupying, on an average, 4 sq. rd. 120 sq. ft. 84 sq. in. ; how much ground remains unoccupied ? 22. At $13 per ton, how much hay may be bought for $12.02 ? 23. If 1 pk. 4 qt. of wheat cost $.72, how much will a bushel cost? Ans. $1.92. 24. How many bushels, Indiana standard, in 36244 Ibs. of wheat ? 25. At 20 cents a cubic yard, how much will it cost to dig a cellar 32 ft. long, 24 ft. wide, and 6 ft. deep? Ans. $34.13 + . 26. If the wall of the same cellar be laid l feet thick, what will it cost at $1.25 a perch ? Ans. |50.90jf . 27. The forward wheels of a wagon are 10 ft. 4 in. in circum- ference, and the hind wheels 15 1 ft. ; how many more times will the forward wheels revolve than the hind wheels in running from Boston to N. Y., the distance being 248 miles? Ans. 42240. 28. Bought 15 cwt. 22 Ib. of rice at $3.75 a cwt, and 7 cwt. 36 Ib. of pearl barley at $4.25 a cwt. How much would be gained by selling the whole at 4 cents a pound? Ans. $13.255. 29. From f of 3 T. 10 cwt. subtract T 4 ^ of 7 T. 3 cwt. 26 Ib. 30. What is the value in avoirdupois weight of 16 ib. 5 oz. 10 pwt. 13 gr. Troy? Ans. 13 Ib. 8 oz. 11.4-fdr. 31. What decimal of a rod is 1 ft. 7.8 in. ? 32. If a piece of timber be 9 in. wide and 6 in. thick, what length of it will be required to make 3 cu. ft. ? Ans, 8 ft. 224 COMPOUND NUMBERS. 33. If a board be 16 in. broad, what length of it will make 7 sq. ft. ? Ans. 5i ft. 34. If a hogshead contain 10 cubic feet, how many more gal- lons of dry measure will it contain than of beer measure ? 35. How many tons in a stick of hewn timber 60 ft. long, and 1 ft. 9 in. by 1 ft. 1 in. ? Ans. 2.275 tons. 71 36. Subtract ^ bu. -f | of f f of 3i qt. from 5 bu. 3f 1 qt. Ans. 16f pk. 37. What is the difference between -J of 5 sq. mi. 250 A. 120 P., and 3 times 456 A. 134 P. 25 sq. yd.? Ans. 2 sq. mi. 254 A. 106 P. 24-f sq. yd. 38. How many pounds of silver, Troy weight, are equivalent in value to 5.6 Ib. of gold by the English government standard ? Ans. 80 Ib. 2 pwt. 19.2768 gr. 39. If a piece of gold is $ pure, how many carats fine is it ? 40. In gold 16 carats fine what part is pure, and what part is alloy ? 41. A man having a piece of land containing 384J A., divided it between his two sons, giving to the elder 22 A. 1 II. 20 P. more than to the younger ; how many acres did he give to each ? Ans. 203 A. 2 R. 14 P., elder ; 181 A. R. 34 P., younger. 42. 4000 bushels of corn in Illinois is equal to how many bushels in New York ? Ans. 3586/^ bu. 43. The market value being the same in both States, a farmer in New Jersey exchanged 110 bu. of cloverseed, worth $4 a bushel, with a farmer in New York for corn, worth $f a bushel, which he sold in his own State for cash. The exchange being made by weight, in whose favor was the difference, and how much in cash value ? Ans. The N. J. farmer gained 69^ bu. corn, worth $46^-. 44. The great pyramid of Cheops measures 763.4 feet on each side of its base, which is square. How many acres does it cover ? 45. The roof of a house is 42 ft. long, and each side 20 ft. 6 in. wide; what will the roofing cost at ^4.62^ a square ? PROMISCUOUS EXAMPLES. 225 46. If 17 T. 15 cwt. 62 Ib. of iron cost $1333.593, how much will 1 ton cost? 47. How many wine gallons will a tank hold, that is 4 ft. long by of ft. wide., and If ft. deep? Ans. 187^ gal. 48. What will be the cost of 300 bushels of wheat at 9s. 4d. per bushel, Michigan currency ? Ans. $350. 49. What will be the cost in Missouri currency? 50. What will be the cost in Delaware currency ? 61. What will be the cost in Georgia currency? Ans. $600. 52. What will be the cost in Canada currency? Ans. $560. 53. Bought the following bill of goods in Boston : 6 yd. Irish linen @ 5/4 12 flannel " 3/9 8i " calico " 1/7 9 " ribbon " /9 4* Ib. coffee 1/5 6f gal. molasses " 3/8 What was the amount of the bill? Ans. $21.76 -f . 54. How many pipes of Madeira are equal to 22 pipes of sherry ? 55. A cubic foot of distilled water weighs 1000 ounces avoirdu- pois; what is the weight of a wine gallon ? Ans. 8 Ib. 5^-J oz. 56. There is a house 45 feet long, and each of the two sides of the roof is 22 feet wide. Allowing each shingle to be 4 inches wide and 15 inches long, and to lie one third to the weather, how many half-thousand bunches will be required to cover the roof? Ans. 28-^V 57. A cistern measures 4 ft. 6 in. square, and 6 ft. deep; how many hogsheads of water will it hold ? 58. If the driving wheels of a locomotive be 18 ft. 9 in. in cir- cumference, and make 3 revolutions in a second, how long will the locomotive be in running 150 miles? Ans. 3 h. 54 min. 40 sec. 59 In traveling, when I arrived at Louisville my watch, which was exactly right at the beginning of my journey, and a correct P 226 COMPOUND NUMBERS. timekeeper, was 1 h. 6 min. 52 sec. fast; from .-what direction had I come, and how far ? Aits. From the east, 16 43'. 60. How many U. S. bushels will a bin contain that is 8.5 ft. long, 4.25 ft. wide, and 3f ft. deep? 61. Reduce 3 hhd. 9 gal. 3 qt. wine measure to Imperial gal- lons. Am. 165.5807+ Imp' 1 gal. 62. A man owns a piece of land which is 105 ch. 85 1. long, and 40 ch. 15 1. wide; how many acres does it contain ? 63. A and B own a farm together; A owns T 7 2 of it and B the remainder, and the difference between their shares is 15 A. 68 \ P. How much is B's share? Ans. 38 A. 91 J P. 64. At $3.40 per square, what will be the cost of tinning both sides of a roof 40 ft. in length, and whose rafters are 20 ft. 6 in. long? Ans. $55.76. 65. What is the value of a farm 189.5 rd. long and 150 rd. wide, at $3 If per acre? 66. Reduce 9.75 tons of hewn timber to feet, board measure, that is, 1 inch thick. Ans. 5850 ft. 67. How many wine gallons will a tank contain that is 4 ft. long, 34 ft. wide, and 2f ft. deep ? Ans. 299^ gal. 68. If a load of wood be 12 ft. long, and 3 ft. 6 in. wide, how high must it be to make a cord ? 69. In a school room 32 ft. long, 18 ft. wide, and 12 ft. 6 in. high, are 60 pupils, each breathing 10 cu. ft. of air in a minute. In how long a time will they breathe as much air as the room contains ? 70. A man has a piece of land 201| rods long and 4H rods wide, which he wishes to lay out into square lots of the greatest possible size. How many lots will there be ? Ans. 396. 71. A man has 4 pieces of land containing 4 A. 140 P., 6 A. 132 P., 9 A. 120 P., and 11 A. 112 P. respectively. It is re- quired to divide each piece into the largest sized building lots possible, each lot containing the same area, and an exact number of square rods. How much laud will each lot contain ? Ans. 156 P. DUODECIMALS. 227 DUODECIMALS. 387. Duodecimals are the parts of a unit resulting from con- tinually dividing by 12 ; as 1, r J 3 , yl^, iVss* e ^ c - ^ n practice, duodecimals are applied to the measurement of extension, the foot being taken as the unit. In the duodecimal divisions of a foot, the different orders of units are related as follows : V (inch or prime) is ^ of a foot, or 1 in. linear measure. I" ( second ) or -^ of T '- 2 , " T T of a foot, or 1 ' square ' ' V" (third) or -^ of ^/of ^ 2 -, . . " T J- ^ of a foot, or 1 " cubic " TABLE. 12 fourths, (""}, make 1 third I'" 12 thirds " 1 second, 1" 12 seconds " 1 prime, V 12 primes, " 1 foot, ft. SCALE uniformly 12. The marks ', ", '", "", are called indices. NOTES. 1. Duodecimals are really common fractions, and can always be treated as such ; but usually their denominators are not expressed, and they are treated as compound numbers. 2. The word duodecimal is derived from the Latin term duodecim, signifying 12. ADDITION AND SUBTRACTION. 388. Duodecimals are added and subtracted in the same manner as compound numbers. EXAMPLES FOR PRACTICE. 1. Add 12 ft. 7' 8", 15 ft. 3' 5", 17 ft. 9' 7". Am. 45 ft. 8' 8". 2. Add 136 ft. 11' 6" 8'", 145 ft. 10' 8" 5'", 160 ft, 9' 5" 5'", Ans. 443 ft. V 8" 6'". 3. From 36 ft. 7' 11" take 12 ft. 9' 5". Ans. 23 ft. 10' 6". 4. A certain room required 300 sq. yd. 2 sq. ft. 5' of plastering. The walls required 50 sq. yd. 1 sq. ft, 7' 4", 62 sq. yd. 5' 3", 48 sq. yd. 2 sq. ft., and 42 sq. yd. 2 sq. ft. 3' 4", respectively. Re- quired the area of the ceiling. Ans. 97 sq. yd. 5 sq. ft. 1' V. 228 DUODECIMALS. MULTIPLICATION. 38O. In the multiplication of duodecimals, the product of two dimensions is area, and the product of three dimensions is solidity (82, 286). We observe that V X 1ft. = ^ of 1ft. =1'. i x/ x i ft. = T i T of i a = i". V X V = A * T L of 1 ft. = I". \" x V = T h X 3 \ of 1 ft. == 1'". Hence, The product of any two orders is of the order denoted by the sum of their indices. 39O. 1. Multiply 9 ft. 8' by 4 ft. 7'. OPERATION. ANALYSIS. Beginning at the right, 9 ft g/ 8' X 7' = 56" = 8" ; writing 4 ft 7' the S // one place to the right, we re- - serve the 4 X to be added to the next 3 , product. Then, 9 ft. X 7' + 4^ = _ I -- 67 X = 5 ft. 7 / , which we write in the 44 ft. 3' 8", Ans. places of feet and primes. Next mul- tiplying by 4 ft., we have 8 X X 4 ft. = 32 / = 2 ft. 8'; writing the 8 X in the place of primes, we reserve the 2 ft. to be added to the next product. Then, 9 ft. X 4 ft. -f- 2 ft. = 38 ft., which we write in the place of feet. Adding the partial pro- ducts, we have 44 ft. 3 X 8 X/ for the product required. Hence the RULE. I. Write the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier, beginning with the lowest term in each, and call the pro- duct of any two orders, the order denoted by the sum of their in- dices, carrying 1 for every 12. III. Add the partial products ; their sum will be the required answer. EXAMPLES FOR PRACTICE. 1. How many square feet in a floor 16 ft. 8' wide, and 18 ft. 5 f long? 2. How much wood in a pile 4 ft. wide, 3 ft. 8' high, and 23 ft TMong? MULTIPLICATION. 229 3. If a floor be 79 ft. 8' by 38 ft. 11', how many square yards does it contain ? Ans. 344 yd. 4 ft. 4' 4". 4. If a block of marble be 7 ft. 6' long, 3 ft. 3' wide, and 1 ft. 10' thick, what are the solid contents? Ans. 44 ft. 8' 3". 5. How many solid feet in 7 sticks of timber, each 56 ft. long, 11 inches wide, and 10 inches thick? Ans. 299 ft. 5' 4". 6. How many feet of boards will it require to inclose a building 60 ft. 6' long, 40 ft. 3' wide, 22 ft. high, and each side of the roof 24 ft. 2', allowing 523 ft. 3' for the gables, and making no deduction for doors and windows ? Ans. 7880 ft. 5 ; CONTRACTED METHOD. 391. The method of contracting the multiplication of deci- mals may be applied to duodecimals, the only modification being in carrying according to the duodecimal, instead of the decimal, scale. 1. Multiply 7 ft. 3' 5" 8'" by 2 ft, 4' 7" 9"', rejecting all de- nominations below seconds in the product. OPERATION. ANALYSIS. We write 2 ft., the 7 ft. 3' 5" 8'" units of the multiplier, under the 9'" 7" 4' 2 ft lowest order to be reserved in the 14 ft 6' 11" product, and the other terms at the 2ft. 5' 2" left, with their order reversed. Then 4' 3" it is obvious that the product of 5" each term by the one above it is ;jj~ffc J 9"-- ^ nSf seconds. Hence we multiply each term of the multiplier into the terms above and to the left of it in the multiplicand, carrying from tht, rejected terms, thus ; in multiplying by 2 ft., we have 8 /x/ X 2 ft. = 16 //x = 1" 4 //x , which being nearer 1" than 2 X/ , gives 1" to be car- ried to the first contracted product. In multiplying by 4 X , we have 5 // x 4 / = 20'" = 1" 8 //x , which being nearer 2 /x than l /x , gives 2" to be carried to the second contracted product, and so on. EXAMPLES FOR PRACTICE. 1. Multiply 7 ft. 3' 4" 5'" by 5 ft 8 ; 6", extending the pro- duct only to primes. Ans. 41 ft. 7'rfc 230 DUODECIMALS. 2. How many yards of carpeting will cover a floor" 36 ft. 9' 4" long, and 26 ft. 6' 9" wide ? 3. How many cu. ft. in a block of marble measuring 6 ft. 2' V in length, 3 ft. 3' 4" wide, and 2 ft. 8' 6" thick ? 4. Find the product of 7 ft. 6' 8", 3 ft. 2' 11", and 3 ft. 8' 4", correct to within 1'. Ans. 90 ft. 6 f =t. DIVISION. 392. 1. Divide 41 ft. 8' 7" 6'" by 7 ft. 5'. OPERATION. ANALYSIS. Divid- 7 ft. 5')41 ft. 8' 1" 6"'(5 ft. 7' 6" ing the units of the 37 ft. V dividend by the units 4 ft 7' 7" ^ ^ e divisor, we ob- 4 ft. 3' 11" tain 5 ft. for the first ^ 7^7 , term of the quotient, 3' 8" 6'" anc * ^ ^" ^' ^ or a re ~ mainder. Bringing down the next term of the dividend, we have 4 ft. 7 / 7 // for a new dividend. Reducing the first two terms to primes, we have 55 X 7 /x> , whence by trial division we obtain 7 / for the second term of the quo- tient, and 3 X 8 /x for a remainder. Completing the division in like manner, we have 5 ft. 7 / 6 /x for the entire quotient Hence the fol- lowing RULE. I. Write the divisor on the left hand of the dividend, as in simple numbers. II. find the first term of the quotient either by dividing the first term of the dividend by the first term of the divisor, or by dividing the first two terms of the dividend, by the first two terms of the divisor ; multiply the divisor by this term of the quotient, subtract the product from the corresponding terms of the dividend, and to the remainder bring down another term of the dividend. III. Proceed in like manner till there is no remainder, or till a quotient has been obtained sufficiently exact. EXAMPLES FOR PRACTICE. 1. Divide 287 ft. 7' by 17 ft. Ans. 16 ft. 11'. 2. Divide 29 ft. 5' 4" by 6 ft. 8'. Ans. 4 ft. 5'. DIVISION. 231 3. A floor whose length is 48 ft. 6' has an area of 1176 ft. 1' 6" ; what is its width ? Ans. 24 ft. 3'. 4. From a cellar 38 ft. 10' long and 9 ft. 4' deep, were exca- vated 275 cu. yd. 5 cu. ft. V 4" of earth; how wide was the cellar ? Ans. 20 ft. 6'. CONTRACTED METHOD. SOS. Division of Duodecimals may be abbreviated after the manner of contracted division of decimals. 1. Divide 35 ft. 11' 11" by 4 ft. 3' 7" 3'", and find a quotient correct to seconds. OPERATION. 4 ft. 3' 7" 3'" ) 35 ft. 11' 11" ( 8 ft. 4' 5" 34 ft. 4' 10" 1 ft. 1ft. r i" 5' 2" V 11" V 9" 2", rem. ANALYSIS. Having obtained by trial, 8 ft. for the first term of the quotient, we multiply three terms of the divisor, 4 ft. 3' 7 // , carrying from the rejected term, t>'" X 8 = 24 //x = 2 /x , making 34 ft. 4' 10 XX , which subtracted from the dividend leaves 1 ft. 7' l xx for a new divi- dend. In the next division, we reject 2 terms from the right of the divisor, and at the last division, 3 terms, and obtain for the required quotient, 8 ft. 4 X 5 XX . EXAMPLES FOR PRACTICE. 1. Divide 7 ft. 7' 3" by 2 ft. 10' 7", extending the quotient to seconds. Ans. 2 ft. 7' 8"db. 2. Separate 64 ft. 9' 8" into three factors, the first and second of which shall be 7 ft. 2' 4" and 4 ft. 7' 9" 8'" respectively, and obtain the third factor correct to within. 1 second. Ans. 1 ft. 11' 3"db. 3. What is the width of a room whose area is 36 ft. 4' 8" and whose length 7 ft. 2' 11" ? 232 SHORT METHODS < SHORT METHODS. 30 . Under the heads of Contractions in Multiplication and Contractions in Division, are presented only such short methods as are of the most extensive application. The short methods which follow, although limited in their application, are of much value in computations. FOR SUBTRACTION. 195. When the minuend consists of one or more digits of any order higher than the highest order in the subtrahend. The difference between any number and a unit of the next higher order is called an Arithmetical Complement. Thus, 4 is the arithmetical complement of 6, 31 of 69, 2792 of 7208, etc. I. Subtract 29876 from 400000. OPERATION. ANALYSIS. To subtract 29876 from 400000 is the 400000 same as to subtract a number one less than 29876, or 29876 29875, from 399999 (Ax. 2). We therefore diminish 3701^4 *^ e ^ ^ *k e m ^ nuenc ^ by" 1, and then take each figure of the subtrahend from 9, except the last or right- hand digit, which we subtract from 10. Hence the RULE. I. Subtract 1 from the significant part of the minuend and write the remainder, if any-, as a part of the result. II. Proceeding Jo the right, subtract each figure in the subtra- hend from 9, except the last significant figure, which subtract from 10. EXAMPLES FOR PRACTICE. 1. Subtract 756 from 1000. Ans. 244. 2. Subtract 8576 from 4000000. Ans. 3991424. 3. Subtract .5768 from 10. 4. Subtract 13057 from 1700000. 5. Subtract 90.59876 from 64000. 6. Subtract 599948 from 1000000. 7. What is the arithmetical complement of 271 ? Of 18365 ? Of 3401250? FOR MULTIPLICATION. 233 FOR MULTIPLICATION. CASE I. 396. When the multiplier is 9, 99, or any number of 9's. * Annexing 1 cipher to a number multiplies it by 10, two ciphers by 100, three ciphers by 1000, etc. Since 9 is 10 1, any number may be multiplied by 9 by annexing 1 cipher to it and subtracting the number from the result. For similar reasons, 100 times a number = 1 time tne number = 99 times the number, etc. Hence, RULE. Annex to the multiplicand as many ciphers as the multi* plier contains 9's, and subtract the multiplicand from the result. EXAMPLES FOB PRACTICE. 1. Multiply 784 by 99. Ans. 77616. 2. Multiply 5873 by .999. 3. Multiply 4783 by 99999. Am. 478295217. 4. Multiply 75 by 999.999. CASE II. 397. When the multiplier is a number a few units less than the next higher unit. Were we required to multiply by 97, which is 100 3, we could evidently annex 2 ciphers to the multiplicand, and subtract 3 times the multiplicand from the result. Were our multiplier 991, which is 1000 9, we could subtract 9 times the multiplicand from 1000 times the multiplicand. Hence, RULE. I. Multiply by the next higher unit by annexing ciphers. II. From this result subtract as many times the multiplicand as there are units in the difference between the multiplier and th* next higher unit. EXAMPLES FOR PRACTICE. 1. Multiply 786 by 98. Ans. 77028. 2. Multiply 4327 by 96. Ans. 415392. 3. Multiply 7328 by 997. 234 SHORT METHODS 4. Multiply 7873.586 by 9.95. Ans. 78342.18070. 5. Multiply 43789 by 9994. 6. Multiply 7077364 by .999993. CASE ITI. 398. When the left hand figure of the multiplier is the unit, 1, the right hand figure is any digit whatever, and the intervening figures, if any, are ciphers. I. Multiply 3684 by 17. OPERATION. ANALYSIS. If we multiply by the usual 3684 X 17 method, we obtain, separately, 7 times and g9g9g 10 times the multiplicand, and add them. We may therefore multiply by the 7 units, and to the product add the multiplicand regarded as tens, thus : 7 times 4 is 28, and we write the 8 as the unit figure of the product. Then, 7 times 8 is 56, and the 2 reserved being added is 58, and the 4 in the multiplicand, added, is 62, and we write 2 in the product. Next, 7 times 6, plus the 6 reserved, plus the 8 in the multiplicand, is 56, and we write 6 in the product. Next, 7 times 3, plus the 5 reserved, plus the 36 in the multiplicand, is 62, which we write in the product, and the work is done. Had the multiplier been 107, we should have multiplied two figures of the multiplicand by 7, before we commenced adding the digits of the multiplicand to the partial products ; 3 figures had the multiplier been 1007, etc". Hence the RULE. I. Write the multiplier at the right of the multiplicand, with the sign of multiplication between them. II. Multiply the multiplicand by the unit Jiyure of the multi- plier, and to the product add the multiplicand, regarding its local value as a product l>y the left hand figure of the multiplier. EXAMPLES FOR PRACTICE. 1. Multiply 567 by 13. Ans. 7371. 2. Multiply 439603 by 10.5. Ans. 4615831.5. 3. Multiply 7859 by 107. 4. Multiply 18075 by 1008. Ans. 18219600. 5. Multiply 3907 by 10.002. FOR MULTIPLICATION. 235 CASE IV. 399. When the left hand figure of the multiplier is any digit, the right hand figure is the unit, 1, and the intermediate figures, if any, are ciphers. I. Multiply 834267 by 301. OPERATION. ANALYSIS. Regarding the multipli- 834267 X 301 ca,nd as a product by the unit, 1, of the f 7- multiplier, we multiply the multipli-' 14367 cand by 3 hundreds, and add the digits of the multiplicand to the several products as we proceed. Since the 3 is hundreds, the two right hand figures of the multiplicand will be the two right hand figures of the product ; and the product of 3 X 7 will be increased by 2, the hundreds of the multiplicand. Had the multiplier been 31, the tens of the multiplicand would have been added to 3 X 7 ; had the multiplier been 3001 the thousands of the multiplicand would have been added to 3 X 7 ; and so on. Hence the RULE. I. Write the multiplier at the right of the multiplicand, with the sign of multiplication between them. II. Multiply the multiplicand l>y the left hand figure of the mul- tiplier, and to the product acid the multiplicand, regarding its local value as a product by the unit figure of the multiplier. EXAMPLES FOR PEACTICE. 1. Multiply 56783 by 71. 2. Multiply 47. 89 by 60.1. Ans. 2878.189. 3. Multiply 3724.5 by .901 4. Multiply 103078 by 40001. Ans. 4123223078. CASE V. 400. When the digits of the multiplier are all the same figure. 1. Multiply 81362 by 333. OPERATION. ANALYSIS. We first multiply by 999, by 81362000 (396). Then, since 333 is of 999, we take 81362 i of the product, o -\ Qi9er)fQQ Had our multiplier been 444, we would have taken of 999 times the multiplicand. 27093546 Had it been 66, we would have taken = f of 99 times the multiplicand, etc. Hence 236 SHORT METHODS RULE I. Multiply by as many 9's as the multiplier contains digits, by (396). II. Take such a part of the product as 1 digit of the multiplier is part of 9. EXAMPLES FOR PRACTICE. 1. Multiply 432711 by 222. Ans. 96061842. 2. Multiply 578 by 1111. 3. Multiply .6732 by 88.888. Ans. 59.8394016. 4. Multiply 8675 by 77.7. 5. Multiply 44444 by 88888. CASE VI. 4O1. To square a number consisting of only two digits. I. What is the square of 18 ? ANALYSIS. According to (86), we have 18 2 = 18 X 18 Now if one of these factors be diminished by 2, the product will be less than the square of 18 by 2 times the other factor, (93, 1) ; that is, 18 2 =(16 X 18) + (2 x 18). Next, if we increase the other factor, 18, in this result, by 2, the whole result will exceed the square of 18, by 2 times the other factor, 16, (93, III); that is, 18 2 = (16 X 20) + (2 x 18) (2 X 16). But as 2 times 18 minus 2 times 16 is equal to 2 X 2, or 2 2 , we have 18 2 = 16 X 20 + 2 2 . Hence the RULE. I. Take two numbers, one of which is as many units less than the number to be squared as the other is units greater, and one of the numbers taken an exact number of tens. II. Multiply these two numbers together, and to the product add the square of the difference between the given and one of the as- sumed numbers. NOTE. A little practice will enable the pupil to readily square any number less than 100 mentally by this rule. . FOR MULTIPLICATION. 237 EXAMPLES FOR PRACTICE. 1. What is the square of 27 ? Ans. 729. 2. What is the square of 49 ? Ans. 2401. 3. Square 28, 26, 39, 38, 37, 36, and 35. 4. Square 77, 88, 8.6, 99, 98, 69, 68, 6.7, and 62. CASE VII. 402. When the multiplier is an aliquot part of some higher unit. An Aliquot or Even Part of a number is such a part as will exactly divide that number. Thus, 5, 8, and 12 are aliquot parts of 25 and of 100, etc. NOTE. An aliquot port may be either a whole or a mixed number, while n component factor must be a whole number. 403. The aliquot parts of 10 are 5, 3|, 2-J, 2, If, If, 1-j, 1. The aliquot parts of 100, 1000, or of any other number, may be found by dividing the number by 2, 3, 4, etc., until it has been divided by all the integral numbers between 1 and itself. I. Multiply 78 by 3 i, and by 25 separately. OPERATION. ANALYSIS. Since 3 is of 10, 3 ) 780 4 ) 7800 the next higher unit, we multiply ~260 an< * 1950 ^ 8 bv 10 and take ^ of the P r duct. Again, since 25 is \ of 100, we multiply 78 by 100 and take \ of the product. Hence the E-ULE. I. Multiply the given multiplicand ~by the unit next higher than the multiplier, by annexing ciphers. II. Take such a part of this product as the given multiplier is part of the next higher unit. EXAMPLES FOR PRACTICE. 1. Multiply 437 by 25. Ans. 10925. 2. Multiply 6872 by 2. Ans. 17180. 3. Multiply 5734154 by 333f Ans. 1911384666f . 4. Multiply 758642 by 12J. 5. Multiply 78563 by 125. Ans. 9820375. 6. Multiply 57687 by 142f 238 SHORT METHODS CASE VIII. 404. When the right hand figure or figures of the multiplier are aliquot parts of 10, 100, 1000, etc. I. Multiply 2183 by 1233. OPERATION. 218300 12* ANALYSIS. 1233 = 12$ X 100. We there- 797661 f re multi P 1 y ky 100 ' and by 12 ^' in continued ^6196 multiplication. Hence the "26923661 RULE. I. Reject from the right hand of the multiplier such figure or figures as are an aliquot part of some higher unit, and to the remaining figures of the multiplier annex a fraction ivhich expresses the aliquot part thus rejected, for a reserved multiplier. II. Annex to the multiplicand as many ciphers as ore equal to the number of figures rejected from the right hand of the multi- plier, and multiply the result by the reserved multiplier. EXAMPLES FOR PRACTICE. 1. Multiply 43789 by 825. Ans. 36125925. 2. Multiply 58730 by 7125. 3. Multiply 7854 by 34.2*. Ans. 268999.5. 4. Multiply 30724 by 73333$. 5. Multiply 47836 by 712*. Ans. 34083150. 6. Multiply 53727 by 2416f . CASE IX. 405. To find the cost of a quantity when the price is an aliquot part of a dollar. 1. What cost a case of muslins containing 1627 yds., at $.12* per yard ? OPERATION. ANALYSIS. At $1 per yard the case would 8 ) $1627 cost as many dollars as it contained yards ; $203 37* and at ' 12 = per yard ' Jt would cost i as many dollars as it contained yards. We therefore regard the yards as dollars, which we divide by 8. Hence, . FOR MULTIPLICATION. 239 RULE. Take such a part of the given quantity as the price is part of one dollar. NOTK. Since the shilling in most of the different currencies is some aliquot part of the dollar, this rule is of much practical use in making out bills and accounts where the prices of the items are given in State Currency, and the amounts are required in United States Money. EXAMPLES FOR PRACTICE. ] . What cost 568 pounds of butter at 25 cents a pound ? Ans. SI 42. 2. A merchant sold 51 yards of prints at 16f cents per yard, 8 pieces of sheeting, each piece containing 33 yards, at 6i cents per yard, and received in payment 18 bushels of oats at 33 cents per bushel, and the balance in money ; how much money did he re- ceive ? Ans. $19. 3. Required the cost of 28 dozen candles, at 1 shilling pei dozen, New York currency. Ans. $3.50. 4. What cost 576 Ibs. of beef at lOd. per pound, Pennsylvania currency? Ans. $64. 5. If a grocer in New York gain $7.875 on a hogshead of mo- lasses containing 63 gallons, how much will he gain on 576 gallons at the same rate ? Ans. $72. CASE x. 4O6. To find the cost of a quantity, when the quan- tity is a compound number, some part or all of which is an aliquot part of the unit of price. 1. What cost 5 bu. 3 pk. 4 qt. of cloverseed, at $3.50 per bu. ? OPERATION. ANALYSIS. Multiply- 2,, 4,, 8 ) $3.50 price. ing the price 'by 5, we 5 have the cost of 5 bu. $17.50 cost of 5 bu. Dividing the price by 2, 1.75 " " 2 pk we ^ ave t* 10 cost of J bu. .875 " " 1 " =2 pk. Dividing the .4375 " " 4 qt. price by 4, or the cost of $205625, Ans. 2 P k ' ky 2 ' 7 e l v ? * he cost of 1 pk. Dividing the price by 8, or the cost of 2 pk. by 4, or the cost of 1 pk. by 2, we 240 SHORT METHODS have the cost of pk. = 4 qt. And the sum of these,- several values is the entire cost required. 2. At 6 7s. 5d. Sterling per hhd., how much will 4 hhd. 9 gal. 3 qt. of West India Molasses cost ? OPERATION. ANALYSIS. Mul- 7) 6 7s. 5d. price. tiplying the price 4 by 4, we have the " 25 9 10 cost of 4 hhd. cost of 4 hhd - Di - 12) " 18 " 2 " 2 qr. " " 9 gal. viding the price by I Q tt 6 (t (t a 3 qk 7, W e have the cost 26 9 6 " 2|, " 4ws. of 4 hhd - = 9 S al ' Dividing the cost of 9 gal. by 12, we have the cost of -?$ of 36 qt. = 3 qt. And the sum of these several results is the entire cost required. From these illustrations we deduce the following RULE. I. Multiply the price Ly the number of units of the de- nomination corresponding to the price. II. For the lower denominations, take aliquot parts of the price ; the sum of the several results will be the entire cost. NOTE. This method is applicable in certain cases of multiplication, where one compound number is taken as many times as there are units and parts of a unit of a certain kind, in another compound number. This will be seen in the first example below. EXAMPLES FOR PRACTICE. 1. A chemist filtered 18 gal. 3 qt. 1 pt. of rain-water in 1 day; at the same rate how much could he filter in 4 da. 6 h. 30 min. ? OPERATION. ANALYSIS. Multi- 18 gal. 3 qt. 1 pt. in 1 da. P lv ^g the quantity 4 filtered in 1 day by 4, 75 o 4 WG haVG the < l uantit y 4 2 13 ri 6h filtered in 4 day, Di- 1M 1 T V<'30mm. viding the quantity fil- 12 tered m 1 day by 4, 80 2 3/ 3 Ans. we have the quantity filtered in \ da. = 6 h. Dividing the quantity filtered in 6 hours by 12, we have the quan- tity filtered in h. = 30 min. And the sum of these several results is the entire result required. FOR DIVISION. 241 2. What will be the cost of 3 Ib. 10 oz. 8 pwt. 5 gr. of gold at $15.46 per oz. ? Ans. $717.52. 8. A man bought 5 cwt. 90 Ib. of hay at $.56 per cwt. ; what was the cost? Ans. $3.304. 4. AY hat must be given for 3 bu. 1 pk. 3 qt. of cloverseed, at $4.48 per bushel? Ans. $14.98. 5. A gallon of distilled water weighs 8 Ib. 5.42 oz. ; required the weight of 5 gal. 3 qt. 1 pt. 3 gi. Ans. 49 Ib.l2.3506j- oz. 6. At $17.50 an acre, what will 3 A. 1 R. 35.4 P. of land cost? 7. If an ounce of English standard gold be worth 3 17s. 10|d., what will be the value of an ingot weighing 7 oz. 16 pwt. 18 gr. ? Ans. 30 10s. 4.14375d, 8. If a comet move through an arc of 4 36' 40" in 1 day, how far will it move in 5 da. 15 h. 32 min. 55 sec. ? 9. What will be the cost of 7 gal. 1 qt. 1 pt. 3 gi. of burning fluid, at 4s. 8d. per gallon, N. Y. currency? Ans. $4.35-f . 10. What must be paid for 12 J days' labor, at 5s. 3d. per day, New England currency ? FOR DIVISION. CASE I. 407. When the divisor is an aliquot part of some higher unit. 1. Divide 260 by 3J, and 1950 by 25. OPERATION. ANALYSIS. Since 3 is of 10, the next 26 1 19 1 50 higher unit, we divide 2GO by 10 ; and hav- 3 and 4 ing used 3 times our true divisor, we obtain i^ ^ only of our true quotient. Multiplying the result, 26, by 3, we have 78, the true quotient. Again, since 25 is \ of 100, the next higher unit, we divide 1950 by 100 ; and having used 4 times our true divisor, the result, 19.5, is only \ of our true quotient. Multiplying 19.5 by 4, we have 78, the true quotient. Hence the RULE. I. Divide the given dividend by a unit of the order next higher than the divisor, l>y cutting off Iqures from the right. 21 Q 242 SHORT METHODS. II. Take as many times this quotient as the divisor is contained times in the next hiyher unit. EXAMPLES FOB PRACTICE. 1. Divide 63475 by 25. Ans. 2539. 2. Divide 7856 by 1.25. Ans. 6284.S. 3. Divide 516 by 33.3. 4. Divide 16.7324 by 12i. 5. Divide 1748 by .14f. Ans. 12236 6. Divide 576.34 by 1.6. CASE II. 4O8. When the right hand figure or figures of the divisor are an aliquot part of 10, 100, 1000, etc. 1. Divide 26923661 by 1233*. OPERATION. $ ot 100, we multiply both 1233* ) 26923661 dividend and divis P / by 3> (117, HI), and we obtain a 37|00 ) 80771 10'0 ( 2183, Ans. divisor the component fac- 67 tors of which are 100 and 37. 307 We then divide after the manner of contracted divi- sion, (112). 2. Divide 601387 by 1875. OPERATION. ANALYSIS. Multiplying both 1875) 601387 dividend and divisor by 4, we ob- 4 4 tain a new divisor, 7500, having 2 7500 'i 2405548 ciphers on the right of it. Multi- plying again by 4, we obtain a new divisor, 30000, having 4 ciphers on 3|0000)962|2192 the right. Then dividing the new 320i|f I, Ans. dividend by the new divisor, we ob^ tain 320 for a quotient, and 22192 for a remainder. As this remainder is a part of the new dividend, it must be 4 X 4 = 16 times the true remainder; we therefore divide it by 16, and write the result over the given divisor, 1875, and anneJ the fraction thus formed to the integers of the quotient. RATIO. 243 From these illustrations we derive the following RULE. I. Multiply both dividend and divisor by a number or numbers that will produce for a new divisor a number ending in a cipher or ciphers. II. Divide the new dividend by the new divisor. NOTE. If the divisor be a whole number, or a finite decimal, the multiplier will be 2, 4, 5, or 8, or some multiple of one of these numbers. EXAMPLES FOR PRACTICE. 1. Divide 64375 by 2575. 2. Divide 76394 by 3625. Ant. 21, 2 ff %- 8. Divide 7325 by 433i 4. Divide 5736 by 431.25. Ant. 13^f. 5. Divide 42.75 by 566|. 6. Divide 24409375 by .21875. 7. Divide 785 by 3.14f. Ans. 249^. RATIO. Ratio is the relation of two like numbers with respect to comparative value. NOTE. There are two methods of comparing numbers with respect to value; 1st, by subtracting one from the other; 2d, by dividing one by the other. The relation expressed by the difference is sometimes called Arithmetical Ratio, and the relation expressed by the quotient, Geometrical Itatio. 410. When one number is compared with another, as 4 with 12, by means of division, thus, 12-^4 = 3, the quotient, 3, shows the relative value of the dividend when the divisor is considered as a unit or standard. The ratio in this case shows that 12 is 3 times 4; that is, if 4 be regarded as a unit, 12 will be 3 units, or the relation of 4 to 12 is that of 1 to 3. 411. Ratio is indicated in two ways : 1st. By placing two points between the two numbers compared, writing the divisor before and the dividend after the points,. Thus, the ratio of 8 to 24 is written 8 : 24; the ratio of 7 to 5 is written 7 : 5. 244 RATIO. 2d. In the form of a fraction. Thus, the ratio of 8 to 24 is written 2 g 4 ; the ratio of 7 to 5 is f. 4LI$. The Terms of a ratio are the two numbers compared. The Antecedent is the first term; and The Consequent is the second term. The two terms of a ratio taken together arc called a couplet. 418. A Simple Ratio consists of a single couplet; as 5 : 15. 414. A Compound Ratio is the product of two or more sim- ple ratios. Thus, from the two simple ratios, 5 : 1G and 8 : 2, we 5 : 16 8 : 2 may form the compound ratio 5^8 : 10x~2, or ^ X = ^' 2 = 4. 41*5. The Reciprocal of a ratio is 1 divided by the ratio; or, which is the same thing, it is the antecedent divided by the con- sequent. Thus, the ratio of 7 to 9 is 7 : 9 or |, and its reciprocal NOTK. The quotient of the second term divided by the fir?t is sometimes called a Direct Rtu>, and the quotient of the first term divided by the second, fin In Verne or Reciprocal Ratio. 4IO. One quantity is said to vary directly as another, when the two increase or decrease together in the same ratio ; and one quantity is said to vary inversely as another, when one increases in the same ratio as the other decreases. Thus time varies directly as wages ; that is, the greater the time the greater the wages, and the less the time the less the wages. Again, velocity varies in- versely as the time, the distance being fixed; that is, in traveling a given distance, the greater the velocity the less the time, and the less the velocity the greater the time. 41T. Ratio can exist only between like numbers, or between two quantities of the. same kind. But of two unlike numbers or quantities, one may vary either directly or inversely as the other. Thus, cost varies directly as quantity, in the purchase of goods: time varies inversely as velocity, in the descent of falling bodies. In all cases of this kind, the quantities, though unlike in kind, have a mutual dependence, or sustain to each other the relation of cause and effect. RATIO. 245 418. In the comparison of like numbers we observe, I. If the numbers are simple, whether abstract or concrete, their ratio may be found directly by division. II. If the numbers are compound, they must first be reduced to the same unit or denomination. III. If the numbers are fractional, and have a common de- nominator, the fractions will be to each other as their numerators ; if they have not a common denominator, their ratio may be found either directly by division, or by reducing them to a common denominator and comparing their numerators. 419. Since the antecedent is a divisor and the consequent a dividend, any change in either or both terms will be governed by the general principles of division, (117)- "We have only to sub- stitute the terms antecedent, consequent, and ratio, for divitor, dividend, and quotient, and these principles become GENERAL PRINCIPLES OF RATIO. PRIN. I. Multiplying the consequent multiplies the ratio ; divi- ding the consequent divides ihe ratio. PRIN. II. Multiplying the antecedent divides the ratio; dividing the antecedent multiplies the ratio. PRIN. III. Multiplying or dividing Loth antecedent and cnnse-* quent l>y the same number docs not alter the ratio. 4SG. These three principles may be embraced in one GENERAL LAW. A change in the consequent by multiplication or division prodi*- ce* a LIKE change in the ratio ; but a change in the antecedent pr'td iii-ex nn OPPOSITE change in the ratio. 431, Since the ratio of two numbers is equal to the conse- quent divided by the antecedent, it follows, that I. The antecedent is equal to the consequent divided by the ratio; and that, II. The consequent is equal to the antecedent multiplied by the ratio. 21* 246 RATIO. EXAMPLES FOR PRACTICE. 1. What part of 28 is 7 ? 2s = i ; or, 28 : 7 as 1 : ; that is, 28 has the same ratio to 7 that 1 has to \. Ans. . 2. What part of 42 is 6? 3. What is the ratio of 120 to 80 ? Am. f . 4. What is the ratio of 8| to 60 ? Ans. 7. 5. What is the ratio of T \ to 26 ? 6. What is the ratio of 7^ to 21? Ans. f$. 7. What is the ratio of J to y 7 ^ ? Ans. 4i. 8. What is the ratio of 1 mi. to 120 rd.? Ans. -|. 9. What is the ratio of 1 wk. 3 da. 12 h. to 9 wk.? Ans. 6. 10. What is the ratio of 10 A. 60 P. to 6 A. 110 P.? 11. What is the ratio of 25 bu. 2 pk. 6 qt. to 40 bu. 4.5 pk. ? 12. What is the ratio of 18f to 45' 30" ? 12 ?- Of -3 13. What part of -^- is ^- 4 - ? Ans. T f 5 . 14. What is the ratio of to f of T % of ? s 15. Find the reciprocal of the ratio of 42 to 28. Ans. 1^. 16. Find the reciprocal of the ratio of 3 qt. to 43 gal. 17. If the antecedent be 15 and the ratio |, what is the conse- quent? Ans. 12. 18. If the consequent be 3^ and the ratio 7, what is the ante- cedent? Ans. 1|. 19. If the antecedent be i of | and the consequent .75, what is the ratio ? 20.. If the consequent be $6.12 and the ratio 25, what is the antecedent? Ans. $.245. 21. If the ratio be J and the antecedent |, what is the conse- quent ? 22. If the antecedent be 13 A. 145 P. and the ratio |f, v/liat is the consequent? Ans. 6 A. 90 P. PROPERTIES OF PROPORTION. 247 PROPORTION. 422. Proportion is an equality of ratios. Thus, the ratios 5 : 10 and 6 : 12, each being equal to 2, form a proportion. NOTE. When four numbers form a proportion, they are said to be propor- tional. 423. Proportion is indicated in three ways : 1st, By a double colon placed between the two ratios; thus, S : 4 : : 9 : 12 expresses the proportion between the numbers 3, 4, 9, and 12, arid is read, 3 is to 4 as 9 is to 12. 2d. By the sign of equality placed between two ratios; thus, 3 : 4 = 9 : 12 expresses proportion, and may be read as above, or, the ratio of 3 to 4 equals the ratio of 9 to 12. 3d. By employing the second method of indicating ratio; thus, | = *- indicates proportion, and may be read as either of the above forms. 424. Since each ratio consists of two terms, every proportion must consist of at least four terms. Of these The Extremes are the first and fourth terms; and The Means are the second and third, terms. 423. Three numbers are proportional when the first is to the second as the second is to the third. Thus, the numbers 4, 6, and 9 are proportional, since 4:6=6:9, the ratio of each couplet being |, or 1J, 423. When three numbers are proportional, the second term is called the Mean Proportional between the other two. 42*7. If we have any proportion, as 3 : 15 == 4 : 20, Then, indicating this ratio by the second method, we have V = V- Reducing these fractions to a common denominator, 15 X 4 = 20 X 3 ~!2~ 12 And since these two equal fractions have the same denominator, the numerator of the first, which is the product of the means, must be equal to the numerator of the second, which is the product of the extremes ; or, 15 X 4 = 20 X 3. Hence, 248 PROPORTION. I. In every proportion the product of the means equals the product of the extremes. 4.gain, take any three terms in proportion, as 4 : 6^=6 : 9 Then, since the product of the means equals the product of the ex- tremes, 6 2 = 4 x 9. Hence, II. The square of a mean proportional is equal to the product of the other two terms. 4S8. Since in every proportion the product of the means equals the product of the extremes, (4^57, 1), it follows that, any three terms of a proportion being given, the fourth may be found by the following RULE. I. Divide the product of the extremes ~by one of the means, and the quotient will be the other mean. Or, II. Divide the product of the means l>y one of the extremes } and the quotient will be the other extreme. EXAMPLES FOR PRACTICE. The required term in an operation will be denoted by (?), which may be read " how many," or " how much." Find the term not given in each of the following proportions : 1. 4 :26 = 10 : (?). Ans. 65. 2. $8865 : $720 = ( ? ) : 16 A. Ans. 197 A. 3. 4 yd. :(?):: $9.75 : $29.25. Ans. 13 yd. 4. (?) : 21 A. 140 P. : : $1260 : $750. Ans. 36 A. 120 P. 5. 7>50:18 = (?):7 T Voz. 6. 7 oz :(?):: 30 : 407 2s. lOfd. Ans. 7 Ib. 11 oz. 7. (? ) : .15 hhd. : : $2.39 : $.3585. Ans. 1 hhd. 8. 1 T. 7 cwt. 3 qr. 20 Ib. : 13 T. 5 cwt. 2 qr. = $9.50 : ( ?) 9. $175.35 :(?) = : f Ans - $601.20. 10. (?) : 812 = 240| : 149 TT 7 2V Ans - $ 2 g- 11- f yd. :(?):: $i : $59.0625. Ans. 40^ yd. SIMPLE PROPORTION. 249 CAUSE AND EFFECT. 42Q. Every question in proportion may be considered as a comparison of two causes and two effects. Thus, if 3 dollars as o. cause will buy 12 pounds as an effect, 6 dollars as a cause will ouy 24 pounds as an effect. Or, if 5 horses as a cause consume 10 tons as an effect, 15 horses as a cause will consume 30 tons as an effect. Causes and effects in proportion are of two kinds simple and compound. 430. A Simple Cause or Effect contains but one element; as price, quantity, cost, time, distance, or any single factor used as a term in proportion. 431. A Compound Cause or Effect is the product of two or more elements; as the number of workmen taken in connection with the time employed, length taken in connection with breadth and depth, capital considered with reference to the time cm- ployed, etc. 433. Since like causes will always be connected with like effects, every question in proportion must give one of the following statements : 1st Cause : 2d Cause = 1st Effect : 2d Effect. 1st Effect : 2d Effect = 1st Cause : 2d Cause, in which the two causes or the two effects forming one couplet, must be like numbers and of the same denomination. Considering all the terms of a proportion as abstract numbers, we may say that 1st Cause : 1st Effect = 2d Cause : 2d Effect. But as ratio is the result of comparing two numbers or things of the same kind, (417), the first form is regarded as the more natural and philosophical. SIMPLE PROPORTION. 433. Simple Proportion is an equality of two simple ratios, and consists of four terms. Questions in simple proportion involve only simple causes and simple effects. 250 PROPORTION. FIRST METHOD. 1. If $S will buy 36 yards of velvet, how many yards may be bought for $12? $ $ 8 : 12 1st cause. 2d cause. STATEMENT, yds. 36 1st effect. yds. 2d effect 8 X OPERATION. = 12 x 36 ANALYSIS. The re- quired term in this ex- ample is an effect ; and the statement is, $8 is to $12 as 36 yards is to ( ? ), or how many yards. Dividing 12 X 3C, the product of the means, by 8, the given extreme, we have ( ? ) = 54 yards, the re- quired term, (428,11). 2. If 6 horses will draw 10 tons, how many horses will draw 15 tons? STATEMENT. tons. 15 2d effect. 54yd. horses. horses. tons. 6 : (?) = 10 1st cause. 2d cause. 1st effect. OPERATION. ANALYSIS. In this ex- ample a cause is required ; and the statement is, G horses is to ( ? ), or how many horses, as 10 tons ia to 15 tons. Dividing 15 X 6, the product of the ex- tremes, by 10, the given ^ ' ) r mean, we have 9, the re- (?) = 9 horses. quired term, (428, I). 434. Hence the following RULE. I. Arrange the terms in the statement so that the causes shall compose one couplet, and the effects the other, putting (?) in the place, of the required term. II. If the required term be an extreme, divide the product of the means by the given extreme; if the required term be a mean, divide the product of the extremes by the given mean. NOTES. 1. If the terms of nny couplet be of different denominations, they must be reduced to the same unit value. 2. If the odd term be a compound number, it must be reduced either to its lowest unit, or to a fraction or a decimal of its highest unit. 3. If the divisor and dividend contain one or more factors common to both, they should be canceled. If any of the terms of a proportion contain mixed SIMPLE PROPORTION. 251 numbers, they should first be changed to improper fractions, or the fractional part to a decimal. 4. When the vertical line is used, the divisor and (?) are written on the left, and the factors of the dividend on the right. SECOND METHOD. The following method of solving examples in simple proportion without making the statement in form, may be used by those who prefer it. Every question in simple proportion gives three terms to find a fourth. Of the three given terms, two will always be like numbers, forming the complete ratio, and the third will be of the same name or kind as the required term, and may be regarded as the antecedent of the incomplete ratio ; hence the required term may be found by mul- tiplying this third term, or antecedent, by the ratio of the other two, (421, II). From the conditions of the question we can readily determine whether the answer, or required term, will be greater or less than the third term ; if greater, then the ratio will be greater than 1, and the two like numbers must be arranged in the form of an improper frac- tion, as a multiplier ; if less, then the ratio will be less than 1, and the two like numbers must be arranged in the form of a proper frac- tion, as a multiplier. 1. If 4 tons of hay cost $36, what will 5 tons cost? OPERATION. ANALYSIS. In this example, 4 $36 X - = $45, Ans. * ons ^ n d 5 tons are the like terms, and $36 is the third term, and of the same kind as the answer sought. Now if 4 tons cost $36, will 5 tons cost more, or less, than $36 ? Evidently more : and the required term will be greater than the third term, $36, and the ratio greater than 1. We therefore arrange the like terms in the form of an im- proper fraction, {, for a multiplier, and obtain $45, the answer. 2. If 7 men build 21 rods of wall in a day, how many rods will 4 men build in the same time ? OPERATION. ANALYSIS. In this example, 7 21 >< 4 __ 12 rods Ans. men and 4 men are the like terms, and 21 rods is the third term, and of the same kind as the answer sought. Since 4 men will perform less work than 7 men in the same time, the required term will be less than 252 PROPORTION. 21, and the ratio less than 1. We therefore arrange the like terms in the form of a proper fraction, ^, and obtain by multiplication, 12 rods, the answer. 4-SG. Hence the following RULE. I. With the two given numbers, ichich are of the same name or kind, form, a ratio greater or hss than 1, according as the answer is to be greater or less than the third given number. II. Multiply the third number by this ratio j the product will be the required number or answer. NOTES. 1. Mixed numbers should first be reduced to improper fractions, and the ratio of the fractions found according to 418. 2. Reductions and cancellation may be applied as in the first method. The following examples may be solved by either of the fore- going methods. EXAMPLES FOR PRACTICE. 1. If 12 gallons of wine cost $30, what will 63 gallons cost? 2. If 9 bushels of wheat make 2 barrels of flour, how many barrels of flour will 100 bushels make ? A-ns. 22|. 3. If 18 bushels of wheat be bought for $22.25, and sold for $26.75, how much will be gained on 240 bushels, at the same rate of profit? Ans. $60. 4. If 6 bushels of oats cost $3, what will 91 bushels cost? 5. What will 87.5 yards of cloth cost, if If yards cost $.42? 6. If by selling $1500 worth of dry goods I gain $275.40, what amount must I sell to gain $1000 ? 7. If 20 men can perform a piece of work in 15 days, how many men must be added to the number, that the work may be accomplished in 4 of the time ? Ans. 5. 8. If 100 yd. of broadcloth cost $473.07^, how much will 3.25 yd. cost? 9. If 1 Ib. 4 oz. 10 pwt, of gold may be bought for $260.70, how much may be bought for $39.50 ? Ans. 2 oz. 10 pwt. 10. In what time can a man pump 54 barrels of water, if he pump 24 barrels in 1 h. 14 min. ? Ans. 2 h. 46 min. 30 sec. 11. If | of a bushel of peaches cost $^|, what part of a bushel can be bought for $ 3 7 g ? Ans. T 7 3 bu. COMPOUND PROPORTION. 253 12. If the annual rent of 46 A. 134 P. of land be $374.70, how much will be the rent of 35 A. 90 P. ? 13. If a man gain $1870.65 by his business in 1 yr. 3 mo., how much would he gain in 2 yr. 8 mo., at the same rate ? 14. Two numbers are to each other as 5 to 7J, and the less is 164.5, what is the greater? Ans. 246. 7^. 15. If 16 head of cattle require 12 A. 156 P. of pasture during the season, how many acres will 132 head of cattle require? Ans. 107 A. 7 P. 16. If a speculator in grain gain $26.32 by investing $325, how much would he gain by investing $2275? 17. What will be the cost of paving an open court GO. 5 ft. long and 44 ft. wide, if 14.25 sq. yd. cost $34i ? 18. At 6J cents per dozen, what will be the cost of 10 J gross of steel pens ? 19. If when wheat is 7s. 6d. per bushel, the bakers' loaf will weigh 9 oz. ; what ought it to weigh when wheat is 6s. per bushel? Ans. Hi oz. COMPOUND PROPORTION. 437. Compound Proportion is an expression of equality be< tween a compound and a simple ratio, or between two compound ratios. It embraces the class of questions in which the causes, or the effects, or both, are compound. The required term must be either a simple cause or effect, or a single element of a compound cause or effect. FIRST METHOD. 1. If 8 men mow 40 acres of grass in 3 days, how many acres will 9 men mow in 4 days ? STATEMENT. 1st cause. 2d cause. 1st effect. 2d effect = 40 : (?) Or, 8 x 3 : 9 x 4 = 40 : (?) I 8 - J9 (3- J4 254 PROPORTION. OPERATION. ANALYSIS* In this ex- ,,js 9 X 4 X 40 ample the required term "' = 8x3 ' . is the second effect ; and the statement is, 8 men 3 days is to 9 men 4 days, as 40 acres is to ( ? ), or how many acres. Dividing the continued product of all the elements of the means by the ele- ments of the given extreme, we obtain ( ? ) 60 acres. 2. If 6 compositors in 14 hours can set 36 pages of 56 lines each, how many compositors, in 12 hours, can set 48 pages of 54 lines each ? STATEMENT. 1st cadso. 2d cause. 1st effect. 2d effect. ( 6 . {(?).. f 36 . r 48 1 14 { 12 ' ' \ 56 (54 OPERATION. ANALYSIS. In this example, an element of (?) the second cause is required ; and the state- /i?^ ment is, C compositors 14 hours is to ( ?) com- 7^$ positors 12 hours as 36 pages of 56 lines each w is to 48 pages of 54 lines each. Now, since the / \ __ 9^ Ans. required term is an element of one of the means, we divide the continued product of all the ele- ments of the extremes by the continued product of all the given ele- ments of the means. Placing the dividend on the right of the verti- cal line and the divisors on the left, and canceling equal factors wo obtain ( ? ) = 9. 438. From these illustrations we deduce the following RULE. J. Of the given terms, select those which constitute the causes, and those which constitute the effects, and arrange them in couplets, putting (?) in place of the required term. II. Then, if the blank term (?) occur in either of the extremes, divide the product of the means by the product of the extremes; but if the blank term occur in either mean, divide the product of the extremes by the product of the means. NOTKS. 1. The onuses must he exactly alike in the number and "kind of their term? : the same is true of the effects. 2. The snme preparation of the terms by reduction is to be observed as in friuiple proportion. COMPOUND PROPORTION, 255 SECOND METHOD. The second method given in Simple Proportion, is also applicable in Compound Proportion. In every example in compound proportion all the terms appear in couplets, except one, called the odd term, which is always of the same kind as the answer sought. Hence the required term in a compound proportion may be found, by multiplying the cdd term by the com- pound ratio composed of all the simple ratios formed by these couplets, each couplet being arranged in the form of a fraction. The fraction formed by any couplet will be improper when the re- quired term, considered as depending on this couplet alone, should be greater than the odd term ; and proper, when the required term should be less than the odd term. 1. If it cost $4320 to supply a garrison of 32 men with pro- visions for 18 days, when the rations are 15 ounces per day, what will it cost to supply a garrison of 24 men 34 days, wn*en the rations are 12 ounces per day ? OPERATION, men. days. ounces. $4320 x ! X || X || = $4896 ANALYSIS. In this example there are three pairs of terms, or couplets, viz., 32 men and 24 men, 18 days and 34 days, 15 ounces and 12 ounces ; and there is an odd term, $4320, which is of the same kind as = $4896, Ans. the required term. We arrange each coup- let as a multiplier of this term, thus; First, if it cost $4320 to supply 32 men, will it cost more, or less, to supply 24 men ? Less ; we therefore arrange the couplet in the form of a proper fraction as a multiplier, and we have $4320 X j|. Next, if it cost $4320 to supply a garrison 18 days, will it cost more, or less, to supply it 34 days? More ; hence the multiplier is the improper frac- w'on }|, and we have $4320 X $4 X ^. Next, if it cost $4320 to supply a garrison with rations of 15 ounces, will it cost more, or less, when the rations are 12 ounces ? Less ; consequently, the multiplier is the proper fraction {^ and we have $4320 X 4 X f X } =$4896, the required term. Hence the following 256 PROPORTION. RULE. I. Of the terms composing each couplet form a ratio greater or less than 1, in the same manner as if 'the answer de- pended on those two and the third or odd, term. II. Multiply together the third or odd term and these ratios; the product will be the answer souyht. EXAMPLES FOR PRACTICE. 1. If 12 horses plow 11 acres in 5 days, how many horses would plow 33 acres in 18 days? Ans. 10. 2. If 480 bushels of oats will last 24 horses 40 days, how long will 300 bushels last 48 horses, at the same rate ? Ans. 12 J days. 3. If 7 reaping machines can cut 1260 acres in 12 days, in how many days can 16 machines reap 4728 acres ? Ans. 19.7 days. 4. If 144 men in 6 days of 12 hours each, build a wall 200 ft. long, 3 ft. high, and 2 ft. thick, in how many days of 7 hours each can 30 men build a wall 350 ft. long, 6 ft. high, and 3 ft. thick? Ans. 259.2 da. 5. In how many days will 6 persons consume 5 bu. of potatoes, if 3 bu. 3 pk. last 9 persons 22 days ? 6. How many planks lOf ft. long and li in. thick, are equiva- lent to 3000 planks 12 ft. 8 in. long and 2| in. thick? Ans. 6531 J. 7. If 300 bushels of wheat @ $1.25 will discharge a certain debt, how many bushels @ $.90 will discharge a debt 3 times as great? Am. 1250 bu. 8. If 468 bricks, 8 inches long and 4 inches wide, arc required for a walk 26 ft. long and 4 ft. wide, how many bricks will be required for a walk 120 ft. long and 6 ft. wide ? 9. If a cistern 17 J ft. long, 10 J ft. wide, and 13 ft. deep, hold 546 barrels, how many barrels will a cistern hold that is 16 ft. long, 7 ft. wide, and 15 ft. deep? Ans. 384 bbl. 10. If 11 men can cut 147 cords of wood in 7 days, when they work 14 hours per day, how many days will it take 5 men to cut 150 cords, working 10 hours each day? PROMIPCFOUS EXAMPLES 257 PROMISCUOUS EXAMPLES IN PROPORTION. 1. If a staff 4 ft. long cast a shadow 7 ft. in length, what is the liiglit of a tower that casts a shadow of 198 ft. at the same time? Am. 113^ ft. 2. A person failing in business owes $972, and his entire prop- erty is worth but $607.50; how much will a creditor receive on a debt of $11.33 ? Ans. S7.08+. 3. If 3 cwt. can be carried 660 mi. for $4, how many cwt. can be carried 60 mi. ilr $12 ? Ans. 99. 4. A man can perform a certain piece of work in 18 days by working 8 hours a day ; in how many days can he do the same work by working 10 hours a day ? -4ns. 14|. 5. How much land worth $16.50 an acre, should be giv n in exchange for 140 acres, worth $24.75 an acre? 6. If I gain $155.52 on $1728 in 1 yr. 6 mo., how much will I gain on $750 in 4 yr. 6 mo.? Ans. $202.50. 7. If 1 lb. 12 oz. of wool make 2} yd. of cloth 6 qr. wide, how many lb. of wool will it take for 150 yd. of cloth 4 qr. wide ? 8. What number of men must be employed to finish a piece of work in 5 days, which 15 men could do in 20 days? An*. 60. 9. At 12s. 7d. per oz., N. Y. currency, what will be the cost of a service, of silver plate weighing 15 lb. 11 oz. 13 pwt. 17 gr. ? 10. If a cistern 16 ft. long, 7 ft. wide, and 15 ft. deep, cost $36.72, how much, at the same rate per cubic foot, would another cistern cost that is 17 ft. long, 10 ft. wide, and 16 ft. deep? 11. A borrows $1200 and keeps it 2 yr. 5 mo. 5 da.; what sum should he lend for 1 yr. 8 mo. to balance the favor ? 12. A farmer has hay worth $9 a ton, and a merchant has flour worth $5 per barrel. If in trading the former asks $10.50 for his hay, how much should the merchant ask for his flour ? 13. If 12 men, working 9 hours a day for 15| days, were able to execute f of a job, how many men may be withdrawn and the job be finished in 15 days more, if the laborers are employed only 7 hours a day ? -4ns. 4. 14. If the use of $300 for 1 yr. 8 mo. is worth $30, how much is the use of $210.25 for 3 yr. 4 mo. 24 da. worth ? 22* a 258 PROPORTION. 15. What quantity of lining f yd. wide, will it require to line 9 yd. of cloth, H yd. wide? A?is. 15| yd.. 16. If it cost $95.60 to carpet a room 24 ft. by 18 ft., how much will it cost to carpet a room 38 ft. by 22 ft. with the same material? Ans. $185.00+. 17. If 16/ H cords of wood last as long as 11,, tons of coal, how many cords of wood will last as long as 15 T 7 3 tons of coal? 18. A miller has a bin 8 ft. long, 4i ft. wide, and 2^ ft. deep, and its capacity is 75 bu. ; how deep must he make another bin which is to be 18 ft. long and 3| feet wide, that its capacity may be 450 bu. ? Ans. 7^ ft. 19. If 4 men in 2J days, mow Gf acres of grass, by working 8 hours a day, how many acres will 15 men mow in 3 days, by working 9 hours a day ? Ans. 40 jfi acres. 20. If an army of 600 men have provisions for 5 weeks, allowing each man 12 oz. a day, how many men may be maintained 10 weeks with the same provisions, allowing each man 8 oz. a day ? 21. A cistern holding 20 barrels has two pipes, by one of which it receives 120 gallons in an hour, and by the other discharges 80 gallons in the same time; in how many hours will it be filled? 22. A merchant in selling groceries sells 14^ oz. for a pound; how much does he cheat a customer who buys of him to the amount of $38.40 ? Ans. $3.45. 23. If 5 Ib. of sugar costs $.62, and 8 Ib. of sugar are worth 5 Ib. of coffee, how much will 75 Ib. of coffee cost ? 24. B and C have each a farm; B's farm is worth $32.50 an acre, and C's $28.75 ; but in trading B values his at $40 an acre. What value should C put upon his ? 25. If it require $59f reams of paper to print 12000 copies of an 8vo. book containing 550 pages, how many reams will be required, to print 3000 copies of a 12mo. book containing 320 pages? 26. If 248 men, in 5J days of 12 hours each, dig a ditch of 7 degrees of hardness, 23'2i yd. long, S.t yd. wide, and 2 yd. deep; in how many days of 9 hours each, will 24 men dig a ditch of 4 degrees of hardness, 387 yd. long, 5^ yd. wide, and 3* yd deep ? Ans. 155 NOTATION. 259 PERCENTAGE. 440. Per Cent, is a contraction of the Latin phrase per centum, and signifies I?/ the hundred ; that is, a certain part of every hundred, of any denomination whatever. Thus, 4 per cent means 4 of every hundred, and may signify 4 cents of every 100 cents, 4 dollars of every 100 dollars, 4 pounds of every 100 pounds, etc. NOTATION. 441. The character, %, is generally employed in business transactions to represent the words per cent. ; thus C % signifies 6 per cent. 44^. Since any per cent, is some number of hundredths, it is properly expressed by a decimal fraction; thus 5 per cent. s=s 5 % = .05. Per cent, may always be expressed, however, either by a decimal or a common fraction, as shown in the following TABLE. "Words. 1 per cent. 2 per cent. 4 per cent. 5 per cent. 6 per cent. 7 per cent. 8 per cent. 10 per cent. 20 per cent. 25 per cent. Symbols. Decimals. = 1 % = .01 = 2 % = .02 = 4 % = .04 = 5 % - .05 = 6 % = .06 = 7 % = .07 = 8 % = .08 = 10 % = .10 = 20 % = .20 = 25 % = .25 Common tractions. = Toff = TOff Tffff == s'ff j _y 5 jl Tffff 10 Iff* == T3 Tf *" A = S!t 1 100 4 50 per cent. = 50 % = .50 5 = ^ 100 per cent. = 100 % = 1.00 = ins = 1 125 per cent. 125 % = 1.25 ~ T ff 5 ( \ per cent. = 3 2 = .001 = Too == 259 I per cent. I % = .00^ 4 flf = * 12^ per cent = 12J * = .12 '2^ 100 = i 260 PERCENTAGE. EXAMPLES FOR PRACTICE. 1. Express decimally 3 per cent. ; 9 per cent. ; 12 per cent. ; 16 per cent.; 23 per cent. ; 37 per cent.; 75 per cent.; 125 per cent. ; 184 per cent. ; 205 per cent. 2. Express decimally 15 % ; 11 % ; 4* % ; 5^ % ; 8 % ', 20* % ; 25| % ; 35f % ; 241 % ', 130* %. 3. Express decimally \ per cent. ; f per cent. ; per cent. ; | per cent.; f per cent.; 2 7 5 per cent.; -ff^ per cent.; l T 5 g per cent. ; lOi per cent. 4. Express by common fractions, in their lowest terms, 4 % ; 37* % ; 16i % ; 11> % ; 42? % ; 45ft % ; 48j> T #. 5. What per cent, is .0725 ? ANALYSIS. .0725 = .07 = 7J %, Ans. 6. What per cent, is .065? Ans. 6i %. 7. What per cent, is .14375? Ans. 14f %. 8. What per cent, is .0975 ? 9. What per cent, is .014 ? 10. What per cent, is .1025? 11. What per cent, is .004 ? 12. What per cent, is 028 ? 13. What % is .1324? 14. What % is .084f ? 15. What % is .004 Jy ? Ans. T T % 16. What % is .003^ ? GENERAL PROBLEMS IN PERCENTAGE. In the operations of Percentage there are five parts 01 elements, namely : Rate per cent., Percentage, Base, Amount, and Difference. 444. Hate per Cent,, or Rate, is the decimal which denotes how many hundredths of a number are to be taken. NOTKS. 1. Suoh expressions as 6 per cent., nnd 5 %,nre esscntinlly dn-iwah, {he words p^.r ct-ft., or the clmmcter %, indicating the deciinnl deiioiiiiiiiitor. 2. If the di'i-iinal he reluced to !i common fraction in its lowtxt terms, this fraction will still be the equivalent rate, though not the rate per cent. PROBLEMS IN PERCENTAGE. 261 445. Percentage is that part of any number which is indi- cated by the rate. 446. The Base is the number on which the percentage is computed. 447. The Amount is the sum obtained by adding the per- centage to the base. 448. The Difference is the remainder obtained by subtract- ing the percentage from the base. PROBLEM I. 449. Given, the base and rate, to find the per- centage. 1. What is 5 % of 360? ANALYSIS. Since 5 fo of any OPERATION. number is .05 of that number, (442), we multiply the base, 3GO, .'._! by the rate, .05, and obtain the 18.00, Ans. percentage, 18. Or, since the rate Or, is T (T = 2> ff , we have 3GO X J ff == 360 X TT'JJ =18, Ans. 1 8 tne percentage. Hence the fol- lowing RULE. Multiply the base l>y the rate. NOTE 1. Percentage is always a product, of which the base nnd rate are the fuctors. EXAMPLES FOR PRACTICE. 1. What is 4 per cent, of 250 ? Ans. 10. 2 What is 7 per cent, of 3500 ? Ans. 245. 3 What is 16 per cent, of 324 ? Ans. 51.84. 4 What is 12 per cent, of $5600? Ana. -S700. 5. What is 9 % of 785 Ibs.? 6. What is 25 % of 960 mi. ? 7. What is 75 % of 487 bu.? Ans. 365.25 bu. 8. What is 33^ % of 2757 men? 9. What is 125 % of 756? 10. What is | % of $2364 ? Am. $5.91. 262 PERCENTAGE. 11. What is 3 % of $856? Arts. $31,39 -. 12. What is | % O f 5? Amt _i^ 13. What is 14* % of 51 ? 14. If the base is $375, and the rate .05, what is the percent- age ? A tis. $18.75. 15. A man owed $536 to A, $450 to B, and $784 to C; how much money will be required to pay 54 % of his debts ? 16. My salary is $1500 a year; if I pay 15 % for board, 5 % for clothing, 6 % for books, and 8 % for incidentals, what are my yearly expenses ? Ans. $510. NOTK 2. 15 % + 5 % + 6 % + 8 % = 34 %. In all cases where several rates refer to the same base, they may be added or subtracted, according to the conditions of the question. 17. A man having a yearly income of $3500, spends 10 per cent of it the first year, 12 per cent, the second year, and 18 per cent, the third year; how much does he save in the 3 years? 18. A had $6000 in a bank. He drew out 25 % of it, then 30 % of the remainder, and afterward deposited 10 % of what he had drawn ; how much had he then in bank ? Ans. $3435. 19. A merchant commenced business, Jan. 1, with a capital of $5400, and at the end of 1 year his ledger showed the condition of his business as follows : For Jan., 2 % gain ; Feb., 3^ % gain ; March, \ ff loss; Apr., 2 % gain; May, 2 % gain; June, If % loss; July, U % gain; Aug., 1 % loss; Sept., 2| % gain; Oct., 4 % g a i n > Nov., f % loss; Dec., 3 % gain. What were the net profits of his business for the year? Ans. $918. PROBLEM II. 45O. Given, the percentage and base, to find the rate- 1. What per cent of 360 is 18 ? OPERATION. ANALYSIS. Since the percent- 18 -i- 360 = .05 = 5 ffo a S e ^ s always the product of the Q r base and rate, (449), we divide is i 05 = 5 tf the glvcn P ercenta S e > 18 ' hj the given base, 360, and obtain the r.equired rate, .05 = 5 % . Hence the PROBLEMS IN PERCENTAGE. 2G3 RULE. Divide the percentage by the base. EXAMPLES FOR PRACTICE. 1. What per cent, of $720 is $21.00 ? Ans. 3. 2. What per cent, of 1500 Ib. is 234 Ib. ? 3. What per cent, of 980 rd. is 49 rd. ? 4. What per cent, of 320 10s. is 25 12.8s.? Ans. 8. 5. What per cent, of 46 gal. is 5 gal. 3 qt.? Ans. 12 J. G. What per cent, of 7.85 mi. is 5.495 mi.? Ans. 70, 7. What per cent, of T 8 5 is | ? Ans. 75. 8. What per cent, of 4 is ^ ? 9. What per cent, of 560 is 80 ? 10. The base is $578, and the percentage is $26.01 ; what is the rate? Ans. 4 AVG multi - ply the sum obtained for the goods, $6250, which is the base of the commission, (II), by the rate of the commission, .03, and obtain the commission or percent- age, $187.50. 2. A flour merchant remits to his agent in Chicago $3796, for the purchase of grain, after deducting the commission at 4 % ; how much will the agent expend for his employer, and what wil) be his commission ? OPERATION. ANALYSIS. Ac- 1.00 + .04 = 1.04 cording to Prob. $3796 -^ 1.04 = $3650, for grain, IV, (452), we di- $3796 $3650 = $146, commission. vide the remittance, $3796, which is amount, (HI), by 1 plus the rate of commission, or 1.04, and obtain the base of commission, $3650, which is the sum to be expended in the purchase. Subtracting this from the remittance, we have $146, the commission. NOTK 1. It is evident that the whole remittance, $3796, should not be taken as the base of commission ; for that would be computing commission on commis- sion. A person must charge commission only on what he expends or collects, in his capacity as agent. 3. A factor sold real estate on commission of 5 (453), we divide $8075 -4- .95 = $8500, price. the net proceeds, $8075, $8075 = $425, com. which is difference, (IV), by 1 minus the rate of 23* 270 PERCENTAGE. commission, and obtain the base, $8500, which is the price of the property sold ; whence by subtraction, we obtain the commission, $425. 4. An agent sold my house and lot for $8600 ; what was his commission at 21 fy ? Ans. $193.50. 5. A lawyer collects $750.75 ; what is his commission at 3f %? Ans. $28.15 + . 6. My agent in New York has sold 3500 bushels of Indiana wheat @ $1.40, and 3600 bushels of dent corn @ $.74; what is his commission at 2^ % ? 7. A dealer in Philadelphia sells hides on commission of 8J %, as follows: 2000 Ib. Orinoco @ $.23 J, 5650 Ib. Central Ameri- can @ $.22, 450 Ib. Texas @ $.23, and 650 Ib. city slaughter @ $.21; what does he receive for ML services ? Ans. $162.75. 8. A commission merchant sold a consignment of flour and pork for $25372. He charged $132 for storage, and 6i % commis- sion ; what were the net proceeds of the sale ? 9. An agent for a Rochester nurseryman sells 4000 apple trees at $25 per hundred, 2000 pear trees at $50 per hundred, 1600 peach trees at $20 per hundred, 1800 cherry trees at $50 per hundred, and 500 plum trees at $50 per hundred ; what is his commission at 30 %, and how much should he return to his em- ployer as the net proceeds, after deducting $203.50 for expenses? Ans. Commission, $1041 ; Net proceeds, $2225.50. 10. A lawyer having a debt of $785 to collect, compromises for 82 % ', what is his commission, at 5 % ? Ans. $32.185. 11. I purchased in Chicago 4000 bushels of wheat @ $1.25, and shipped the same to my agent in Oswego, N. Y., who sold it @ $1.50; how much did I make, after paying expenses amount- ing to $415, and a commission of 3 % ? Ans. $405. 12. An agent received $63 for collecting a debt of $1260; what was the rate of his commission? Ans. 5 %. 13. My Charleston agent has charged $74.25 for purchasing 26400 Ib. of rice at $4.50 per 100 Ib. ; required the rate of his commission. 14. A house and lot were sold for $7850, and the owner re- COMMISSION. 271 ceived $7732.25 as the net proceeds; what was the rate of com- mission ? 15. A commission merchant in Boston having received 28000 Ib. of Mobile cotton, effects a sale at $.12 per pound. After deducting $35.36 for freight and cartage, $10.50 for storage, and his commission, he remits to his employer $3252.89 as the net proceeds of the sale ; at what rate did he charge commission ? Ans. 5 1 h ow niany shares did he buy ? 9. If 400 shares of the Bank of Commerce sell for $40150, what is the rate of premium? Ans. f %. 10. A broker receives $48447 to be invested in bonds of the Michigan Central railroad, at 94 % ; how much stock can he buy, allowing 1 % brokerage? 11 My agent sells 830 barrels of Genesee flour at $6 per barrel, commission 5 %, and invests the proceeds in stock of the Penn- sylvania Coal Company, at 82f %, charging i % for making the purchase; how many shares do I receive ? Ans. 57. 12. I purchased 18 shares of Ocean Telegraph stock, par value $500 per share, at a premium of 2 % , and sold the same at a dis- count of 28 % what was my loss ? Ans. $2700. NOTE 5. The rate of loss is .02. + .28 = .30, or .30 % . 13. A speculator exchanged $3600 of railroad bonds, at 5 % discount, for 27 shares of stock of the Suffolk Bank, at 3 % premium, receiving the difference in cash ; how much money did he receive ? 14. A merchant owning 525 shares in the American Exchange 276 PERCENTAGE. Bank, worth 104 %, exchanges them for United States bonds worth 105 % ; how much of the latter stock does he receive? 15. I purchased 12 shares of stock at a premium of 5 %, and sold the same at a loss of $96 ; what was the selling price ? 16. Having bought $64000 stock in the Cunard Line, at 2 % premium, at what price must I sell it, to gain $2560 ? . Ans. 106 %. 17. A speculator bought 250 shares in a Carson Valley mining company at 103 %, and 150 shares of the Western Railroad stock at 95 /o ; he exchanged the whole at the same rates, for shares in the N. Y. Central Railroad at 80 %, which he afterward sold at 85 %. How much did he gain? Ans. $2500. 18. I purchased stock at par, and sold the same at 3 % pre- mium, thereby gaining $750 ; how many shares did I purchase ? 19. A broker bought Illinois State bonds at 103 %, and sold at 105 / G . His profits were $240; what was the amount of his purchase? An&. $12000. 20. A man invested in mining stock when it was 4 % above par, and afterward sold his shares at 5 % discount. His loss in trade was $760; how many shares did he purchase? 21. I invested $6864 in Government bonds at 106 %, paying \\ % brokerage, and afterward sold the stock at 112 %, paying 1 /c brokerage ; what was my gain ? Ans. $208. 22. How much money must be invested in stocks at 3 % ad- vance, in order to gain $480 by selling at 7 % advance ? 23. How many shares of stock must be sold at 4 % discount, brokerage i %, to realize $4775? Ans. 50. INSTALLMENTS, ASSESSMENTS, AND DIVIDENDS. 4 81. An Installment is a portion of the capital stock re- quired of the stockholders, as a payment on their subscription. 48S. An Assessment is a sum required of stockholders, to meet the losses or the business expenses of the company. 483. A Dividend is a sum paid to the stockholders from the profits of the business. STOCKS. 277 484. Gross Earnings are all the moneys received from the regular business of the company. 485. Net Earnings are the moneys left after paying expenses, losses, and the interest upon the bonds, if there be any. 48O. In the division of the net earnings, or the apportion- ment of dividends and assessments, the calculations are made by finding the rate per cent, which the sum to be distributed or as- sessed bears to the entire capital stock. Hence, 487'. Dividends and assessments are a percentage computed upon the par value of the stock as the base. EXAMPLES FOR PRACTICE. 1. The Long Island Insurance Company declares a dividend of 6 % ; what does A receive, who owns 14 shares ? ANALYSIS. According to OPERATION. 449, we multiply the base, $1400 X .06 = $84 $1400, by the rate, .06, and obtain the dividend, $84. 2. A canal company whose subscribed funds amount to $84000, requires an installment of $6300 ; what per cent, must the stock- holders pay? OPERATION. ANALYSIS. According to $6300 ~ 84000 = .07J 450, we divide the in- stallment, $6300, which is percentage, by the base, $84000, and obtain the rate, .07 = 7 %. 3. A man owns 56 shares of railroad stock, and the company has declared a dividend of 8 % ; what does he receive ? Am. $448. 4. I own $15000 in a mutual insurance company; how many shares shall I possess after a dividend of 6 % has been declared, payable in stock? Ans. 159 shares. 5. The Pittsburgh Gas Company declares a dividend of 15 % ; what will be received on 65 shares ? 6. A received $600 from a 4 % dividend ; how much stock did he own ? Ans. $15000. 278 PERCENTAGE. 7. The paid-in capital of an insurance company is $536000. Its receipts for one year are $99280, and its losses 'and expenses are $56400; what rate of dividend can it declare ? Am. 8 %. 8. The net earnings of a western turnpike are $3616, and the amount of stock is $56000 ; if the company declare a dividend of 6 O. The calculations in insurance are based upon the fol- lowing relations : I. Premium is percentage (445)- II. The sum insured is the base of premium. III. The sum covered by insurance is difference. EXAMPLES FOR PRACTICE. 1. What premium must be paid for insuring my stock of goods to the amount of $5760 at U % ? 292 PERCENTAGE. OPERATION. ANALYSIS. According to 85760 X .0125 = $72, Ans. Prob. I, (449), we multiply $5700, the base of premium, by .0125, the rate, and obtain $72, the premium. 2. For what sum must a granary be insured at 2 rent expenses. t>O9. Most Companies in this country are mutual, r.nd divMo the profits among the policy holders. The profits result from the Company realizing upon the reserved fun.l more than the assumed rate of interest, four per cent., and from the losses by death being less than was assumed in making the premium, and from the load- ing or margin being more than the expenses. IKSUKA^CE. 295 Dividends are declared at the end of the first, second, third, or fourth year, and may be applied to reduce the annual premium or to increase the policy. NOTES. 1. One-half of the premium is often paid by a note, and the dividends are af.cnvard applied toward canceling the notes. 2. The following table rates have been selected for the different kinds of policies, for the reason that they are based on tin American Table of Mortality. 8. Stock Companies make no dividends to policy holders, but generally charge a rate of premium 20 to SO per cent, less than the 'Mutual Companies, LIFE TABLE. AX.VUAL PREMIUM OX A POLICY OK $1000, Policy parable at Death only. Ago nt : Payments Usue. j during life. Payment for 10 years only. Payment for 5 years only. Single Payment. Age at issue. C3 8I9.S9 $42.56 $73.87 $320.53 25 CO i'o -n 43.07 75.25 832.53 26 27 20 0" 44.22 70.69 833.83 27 23 2143 4,5.19 73.13 845.81 23 CD 2_V,7 40.02 79.74 852.05 29 83 2.' 7;) 4C.97 81.30 859.05 SO 81 20..-:5 47.93 83.05 8C6.8 '. 81 ; 2 24.C5 43.C2 84.83 873.89 82 fiJJ Ci.73 5.). 10 83.G2 881.73 33 84 23.C6 51.22 8S.52 SSXsS 84 83 26.88 52 4;) 00.40 898. 8 1 a 8 ) 30 27.23 53.63 92.54 407.11 86 87 2x17 51.91 94.07 410.21 37 88 20.15 5C..C4 95.SD 425.64 38 ftD 30.19 57.63 9.19 4:35 42 39 4) 81.80 69.09 101.59 415.55 40 41 82.47 60.60 KJ4.0S 456.04 41 4> 88.72 62.19 106.C6 466.89 42 43 85.08 63.84 109.84 478.11 43 44 86.46 65.57 112.13 489.71 44 45 8797 67.37 115.02 501.69 45 43 89.58 6926 118.02 514.04 46 47 41.30 71.25 121.1.5 526.78 47 43 48.13 73.32 124.83 539.88 48 4J 46.09 75.49 127.74 353.83 49 60 47.18 77.77 131.21 567.13 50 51 49.40 80.14 13480 5S1.24 51 5.! 51.73 82.63 133.51 595. G6 r*. 52 M 54.31 8.5.22 142.34 610.36 53 54 57 ()2 S7.f4 143.30 625.88 54 55 59.91 " 90.79 150.88 640.54 55 ;>6 63.00 C3 73 1.54.60 655.09 56 57 66/29 93.91 158.04 671.64 57 K C9.83 100.2 1 163.43 687.48 58 59 7-1 C;) 103. OS 163.07 703.49 59 00 77.G:i 107.3-5 172.^7 719 65 60 01 61.96 1 ! 1 .23 1 77.83 785.02 61 69 86.58 115.82 1S-2.T) 7.52 26 62 c--> OL54 119.06 188.26 768.67 68 64 96 86 124.28 103.77 785.10 64 C5 102.55 129.18 199.4S 601.52 65 296 PEKCEKTAGE. EKDOWMEKT TABLE. ANNUAL PREMIUM TO SECURE flOOO, payable at the end of 35, 30, 25, etc., Years, or at Death if prioj. Prem urns to continue until Policy matures. In In In In In In AliB. 35 years. 30 years. 25 years. years. 15 years. 10 years. AGE. 25 26 $26.33 26.57 $30.61 80.80 $37.17 37.34 $47.68 47.82 $66.02 60.15 $108.91 104.03 25 26 27 26.8=3 81.02 87.52 47.98 66.29 104.16 27 28 27.11 31. '25 37.72 48.15 66.44 104.29 28 29 27.42 31.50 37.92 4833 CO. CO 104.43 29 80 27.76 81.78 33.16 48.53 66.77 1C4.58 1 30 31 28.13 32.09 38.41 48.74 60.06 1(4.75 1 31 32 28.54 32.43 38. 69 48.97 (57.16 1C-) 92 i ;;2 33 28.93 32.79 8.98 49.22 67.36 1(5.11 88 34 29.46 83.19 39.81 49.49 67.CO 105.31 34 35 30. (iO 83. G3 89.63 49.79 67. S5 K '5: 3 35 36 30.58 34.11 40.07 50.11 C8.12 1(5.75 86 37 81.22 34.64 40.50 0.47 C8.41 li. (i ( 87 33 81.93 35.23 40.93 50.86 08.73 K r, -.8 88 39 32.70 85. SS 41.52 51.30 69.09 106.58 89 4u 33.55 3659 42.10 51 .78 6949 106.90 40 41 37.33 42.75 62.81 69.92 107.26 41 42 38.24 43.47 62.89 70.40 107 (55 42 43 39.19 44.26 68JB4 70.92 K 8.08 43 44 40.23 45.12 54.25 71.50 118.55 44 45 41.3T 46.08 65.04 72.14 li 9.07 45 46 47.15 55.91 72.86 1965 4$ 47 48.82 ' 56.S9 78.66 110.30 47 4S 49.61 57.96 74.54 111. (1 48 49 51.04 59.15 75.51 111 61 49 50 52.60 60.45 76.59 112.C8 51 61.90 77.77 118.G4 M 52 63.48 79.07 114.70 i .'2 53 05.22 80.51 li. r 6 ; f 54 67.14 82.09 117.14 M 55 69.24 8:-!. 82 118.C4 56 85.73 120.09 , r 6 57 87.84 121.78 67 58 90.15 1-::i ; 4 59 92.70 126.70 59 60 95.50 127.96 .60 61 13o 45 61 62 188.19 62 63 186.20 68 G4 ]:'.!' 52 C4 G5 143.16 66 EXAMPLES FOR PRACTICE. 1. What sum must a man aged 33 pay annually for life for life policy of $7500? "What sum annually for ten years ? AVhat sum annually for five years? What sum in a single payment ? INSURANCE. 297 OPERATION. ANALYSIS. We multiply $24.78 x 7,500 = $185.85, Ans. the rate per thousand, found 50.10 x 7,500= 375.75, Ans. opposite age 33, Life Table, 86.02 x 7,500 = 649.65, Ans. by the number of thou- 381.73 x 7,500 =2862.985, Ans. sands, expressing the hun- dreds, tens, and units, decimally. 2. A man 30 years of age takes a life policy in a Mutual Com- pany, for $5000, the premiums continuing until death. The divi- dend reduces the annual premium an average of 30%. He dies after making 21 payments; how much more money will his family receive than he has paid to- the Company ? Ans. $3331.55. 3. What annual premium will a man aged 36 years pay to secure an endowment policy for $5000, payable to himself in 20 years, or to his heirs if death occurs prior ? Ans. $250.55. 4. A young man aged 27 takes an endowment policy for $'4000, payable to himself in 25 years. If the dividend increases his policy $2400, how much more will he receive than he has paid the Com- pany ? . Ans. $2648.00. 5. A clergyman aged 45 takes an endowment policy for $3000, payable to himself in 15 years, or to his family at death, and dies after making 13 payments. How much money would he have saved had he taken a policy for the same amount on the continued payment life plan ? Ans. $1332.63. 6. A merchant aged 49 insures for $8000 on the single payment life pl.m, and dies in the fourth year thereafter. How much less would his insurance have cost him had he insured on the 5 pay- ment life plan ? Ans. $338.96. 7. A lawyer aged 31 years insures in the Mutual Life Insurance Company ofN. Y. for $10000, payable to himself in 20 years. If the dividends increase the policy $7000, how much more will he receive than he has paid the Company, reckoning Q% simple inter- est on his payments? Ans. $1110.70. NOTE. 1st payment is at interest 20 years, 2d, 19 years, 3d, 18 years, &c., &c, 8. A has his life insured at the age of 25 ; B insures at the ago of 35, each taking a life policy, premiums payable until death; 298 PERCENTAGE. what will be tho age of each, when the amount of premium paid shall exceed the face of the policy ? Ans. A, 75 yis. ; B, 72 yrs. 0. A person at age of 34 had his life insured for $0000, pay- ments made in 10 years. When he died there was a net gain to his family of $4463.40 ; how many payments had he made? Ans. 5. 10. A gentleman aged 40 insures his life in the Conn. Mutual Life Ins. Co. for $5000, premiums to continue until death. After the* fourth year his premium is reduced one-half by the dividend. What will be the total amount of premiums paid in thirty years? Ans. $2660.50. 11. A clergyman insured for $5000 in 1843, in the Mutual Life Insurance Company of New York, at an annual premium of $175.50, andjdied in 1885. The amount paid to his heirs, includ- ing dividend additions, was $8637.34. How much better WHS this than a compound interest investment at 6 % ? Ans. $565.22. NOTE. The amount oi'$l per annum for 22 years at G% comp. inter, is $45.995. TAXES. 510. A Tax is a sum of money assessed on the person or pro- perty of an individual, for public purposes. 511. A Poll Tax is a certain sum required of each male citi- zen liable to taxation, without regard to his property. Each person so taxed is called a poll. 512. A Property Tax is a sum required of each person own- ing property, and is always a certain per cent, of the estimated value of his property. !$. An Assessment Roll is a list or schedule containing the names of all the persons liable to taxation in the district or com- pany to be assessed, and the valuation of each person's taxable property. 514L Assessors are the persons appointed to prepare the as- sessment roll, and apportion the taxes. 1. In a certain town a tax of $4000 is to be assessed. There are 400 polls to be assessed $.50 each, and the valuation of the taxable property, as shown by the assessment roll, is $950000; what will be the property tax on $1, and how much will be A's tax, whose property is valued at $3500, aud who pays for 3 poih ? TAXES. 299 OPERATION. 8 .50 X 400 = $200, amount assessed on the polls. $4000 $200 = $3800, amount to be assessed on property. $3800 -r- $950000 = .004, rate of taxation ; $3500 x -004 = $14, A's property tax; 8 .50x3 = 1.50, A's poll tax; $15.50, amount of A's tax. Hence the RULE. I. Find the amount of poll tax, if any, and subtract it from the whole tax to be assessed j the remainder will be the prop- erty tax. II. Divide the property tax by the whole amount of taxable property ; the quotient will be the rate of taxation. III. Multiply each man 1 s taxable property by the rate of taxa- tion, and to the product add his poll tax, if any ; the result will be the whole amount of his tax. NOTE. When a tax is to be apportioned nmong a large number of individuals, the operation is greatly facilitated by first finding the tax on $1, $2, $3, etc., to $9; then on $10, $20, $30, etc., to $90, and so on, and arranging the results as in the following TABLE. Prop. Tax. Prop. Tax. Prop. Tax. Prop. Tax. $1 $.00i $10 f.*4 $100 S .40 : $1000 $4.00 2 .003 20 .08 200 .80 I 2000 8. 3 .012 30 12 300 1.20 3000 12. 4 .016 40 .16 400 1.60 4000 16. 5 .020 50 .20 500 2.00 5000 29. 6 .024 60 .21 600 2.40 6000 24. 7 .028 70 .28 700 2.80 7000 28. 8 .032 80 .32 800 3.20 8000 32. 9 .030 90 .36 900 3.60 9000 36. EXAMPLES FOR PRACTICE. 1. According to the conditions of the last example, what would be the tax of a person whose property was valued at 2465, and who pays for 2 polls ? 300 PERCENTAGE. OPERATION. From the table we find that The tax on $2000 is $8.00 " " " 400 " 1.60 n u u 60 (t 24 a a fj a Q2 And " ." " 2 polls " 1.00 Whole tax " $10.86, Ans. 2. What would A's tax be, who is assessed for $8530, and 8 polls? Ans. $35.62. 3. How much will C's tax be, who is assessed for $987, and 1 poll? Ans. $4.448. 4. The estimated expenses of a certain town for one year are $6319, and the balance on hand in the public treasury is $854. There are 2156 polls to be assessed at $.25 each, and taxable pro- perty to the amount of $1864000. Besides the town tax, there is a county tax of 1? mills on a dollar, and a State tax of of a mill on a dollar. Required the whole amount of A's tax, whose property is valued at $32560, and who pays for 3 polls. 5. What does a non-resident pay, who owns property in the same town to the amount of $16840 ? Ans. $79.99. 6. What sum must be assessed in order to raise a net amount of $5561.50, and pay the commission for collecting at 2 %. NOTE. Since the base of the collector's commission is the sum collected, (446), the question is an example under Problem V of Percentage. 7. In a certain district a school house is to be built at an ex- pense of $9120, to be defrayed by a tax upon property valued at $1536000. What shall be the rate of taxation to cover both the cost of the school house, and the collector's commission at 5 % ? 8. The expenses of a school for one term were $1200 for salary of teachers, $57.65 for fuel, and $38.25 for incidentals ; the money received from the school fund was $257.75, and the remain- ing part of the expense was paid by a rate-bill. If the aggregate attendance was 9568 days, what was A's tax, who sent 4 pupils 46 days each? Ans. $19.96+. 9. The expense of building a public bridge was $1260.52, GENERAL AVERAGE. 301 which was defrayed "by a tax upon the property of the town. The rate of taxation was 3} mills on one dollar, and the collector's commission was 3 % ; what was the valuation of the property ? An*. $401920. GENERAL AVERAGE. 5\5. General Average is a method of computing the loss to be sustained by the proprietors of the ship, freight, and cargo, respectively, when, in a case of common peril at sea, any portion of the property has been sacrificed or damaged for the common safety. 51 <5. The Contributory Interests are the three kinds of prop- erty which are taxed to cover the loss. These are, 1st The vessel, at its value before the loss. 2d. The freight, less as an allowance for seamen's wages. 3d. The cargo, including the part sacrificed, at its market value in the port of destination. NOTE. In New York only of the freight is made contributory to the loss. 5\Y. Jettson is the portion of goods thrown overboard. 518. The loss which is subject to general average includes, 1st. Jettson, or property thrown overboard. 2d. Repairs to the vessel, less on account of the superior worth of the new articles furnished. 3d. Expense of detention to which the vessel is subject in port. 1. The ship Nelson, valued at $52000, and having on board a cargo worth 818000, on which the freight was $3600, threw over- board a portion of the goods valued at $5000, to escape wreck in a storm; she then put into port, and underwent repairs amounting to $1200, the expenses of detention being $350. What portion of the loss will be sustained by each of the three contributing interests? What will be .paid or received by the owners of the ship and freight ? What by A, who owned $8000 of the car^o, including $3500 of the portion sacrificed, and by B, who owned $6000 of the cargo, including $1500 of the portion sacrificed, and by C, who owned $4000, or the residue of the cargo ? 26 H02 PERCENTAGE. OPERATION. LOSSES. CONTRIBUTORY , INTERESTS. Jettson, $5000 Vessel, $52000 Repairs, less , 800 Freight, less , 2400 Cost of detention, .... 350 Cargo, 18000 Total, $6150 Total, $72400 $6150 -h $72400 = .0849447+, rate per cent, of loss. $52000 X .0849447 = $4417.13, payable by vessel. 2400 X .0849447 = 203.87, " " freight. 18000 X .0849447 = J.529.00,^ " " cargo. $6150.00, Total contribution. $8000 X .0849447 == $679.56, payable by A. 0000 x .0849447 = 509.67, " " B. 4000 x .0849447 = 339.78, " " C. $4417.13 + $203.87 = $4621.00, payable by owners of vessel and freight. 800.00 4- 350.00= 1150.00, " to 4621.001150.00= 3471.00, balance payable by ship owners. 3500.00679.56=2820.44, " receivable" by A. 1500.00 509.67= 990.33, " " " B. Hence the following RULE. I. Divide the sum of the losses by the sum of the con- tributory interests ; the quotient will be the rate of contribution. II. Multiply each contributory interest by the rate; the products will be the respective contributions to the loss. EXAMPLES FOR PRACTICE. 1. The ship Nevada, in distress at sea, cut away her mainmast, and cast overboard \ of her cargo, and then put into Havana to refit; the repairs cost $1500, and the necessary expenses of deten- tion were 8420. The ship was owned and sent to sea by Georgo Law, and was valued at $25000 ; the cargo was owned by Hayden & Co., and consisted of 2800 barrels of flour, valued at $9 per barrel, upon which the freight was $4200. In the adjustment of the loss by general average, how much was due from Law to Hayden & Co.? An*. $2629.36. 2. A coasting vessel valued at $28000, having been disabled in a storm, entered port, and was refitted at an expense of $270 for repairs, and $120 for board of seamen, pilotage, and dockage, CUSTOM HOUSE BUSINESS. 303 Of the cargo, valued at 35000, $2400 belonged to A, $1850 to B, and 6750 to C ; and the amount sacrificed for the ship's safety was 1400 of A's property, and 8170 of B's; the gross charges for freight were $1500. Required the balance, payable or re- ceivable, by each of the parties, the loss being apportioned by general average. , f 1295 payable by ship owners; $1268 receivable by A; S '| 41.25 " C; 68.25 B. CUSTOM HOUSE BUSINESS. 519. Duties, or Customs, are taxes levied on imported goods, for the support of government and the protection of home industry. 520. A Custom House is an office established by government for the transaction of business relating to duties. It is lawful to introduce merchandise into a country only at points where custom houses are established. A seaport town having a custom house, is called a port of entry. To carry on foreign commerce secretly, without paying the duties imposed by law, is smuggling. XOTK. Customs or duties form the principal source of revenue to the General Government of the United States: by increasing the price of imported goods they operate as an indirect tax upon consumers, instead of a general direct tax. 521. Duties are of two kinds Ad Valorem and Specific. Ad Valorem Duty is a sum computed on the cost of the goods in the country from which they were imported. Specific Duty is a sum computed on the weight or measure of the goods, without regard to their cost. 522. An Invoice is a bill of goods imported, showing the quantity and price of each kind. 523. By the New Tariff Act, approved March 2, 1857, all duties taken at the U. S. custom houses are ad valorem. The principal articles of import arc classified, and a fixed rate is im- posed upon each list or schedule, certain articles being excepted and entered free. In collecting customs it is the design of government to tax only so much of the merchandise as will be available to the im- 304 PERCENTAGE. porter in the market. The goods are weighed, measured, gauged, or inspected, in order to ascertain the actual quantity received in port; and an allowance is made in every case of waste, loss, or damage. 531. Tare is an allowance for the weight of the box or the covering that contains the goods. It is ascertained, if necessary, by actually weighing one or more of the empty boxes, casks, or coverings. In common articles of importation, it is sometimes computed at a certain per cent, previously ascertained by frequent trials by weighing. 525. Leakage is an allowance on liquors imported in casks or barrels, and is ascertained by gauging the cask or barrel in which the liquor is imported. >2G. Breakage is an allowance on liquors imported in bottles. 527. Gross Weight or Value is the weight or value of the goods before any allowance has been made. 52S. Net Weight or Value is the weight or value of the goods after all allowances have been deducted. NOTES. 1. Draft is an allowance for the waste of certain articles, and is made only for statistical purpone ; it does not affect the amount of duty. 2. Long ton me;i.ure is employed in the custom houses of the United States, in estimating goods by the ton or hundred weight. The rates of this allowance are as follows: On 112 lb lib. Above 112 lb. and not exceeding 224 lb., 2 lb. 224 lb. " " " 336 lb., 3 lb. " 336 lb. " " " 1120 lb., 4 lb. " 1120 lb. " " 201(5 lb., 7 lb. 2010 lb 9 lb. 529. In all calculations where ad valorem duties are consid- ered, I. The net value of the merchandise is the worth of the net weight or quantity at the invoice price, allowance being made in cases of damage. II. The duty is computed at a certain legal per cent, on the net value of the merchandise. NOTE. In the following examples the legal rates of duty, according to the New Tariff Act, are given. CUSTOM HOUSE BUSINESS. 305 EXAMPLES FOR PRACTICE. 1. What is the duty, at 24 %, on an invoice of cassimere goods which cost $750 ? ANALYSIS. According to Prob. I, (449), we multiply the invoice, $750, $750 X .24 = $180 w hich is the base of the duty, by the given rate, and obtain the duty, $180. 2. The gross weight of 3 hogsheads of sugar is 1024 lb., 1016 lb., and 1020 lb. respectively; t'ae invoice price of the sugar 7j cents, and the allowance for tare 80 lb. per hogshead; what is the duty, at 24 % ? OPERATION. ANALYSIS. We first find 1024 the gross weight of the 1016 three hhd. from which we subtract the tare, and 3060, gross weight. obtain 2820 lb., the net 80 X 3 = 240, tare. weight. We next find the . , value of the net weight. 2820, net weight. . a Q-M at / cents, the invoice ! L. price, and then compute $211.50, net value. the duty at 24 % on this .24 value, and obtain $50.76, $50.7600, duty. the du ty re q uir ed. 3. Having paid the duty at 8 % on a quantity of Malaga raisins, I find that the whole cost in store, besides freight, is $378 ; what were the raisins invoiced at ? AJ^ALYSIS. According to Prob. IV, (452), we divide the amount, $378 ~ 1.08 = $350 $ 3T8j by i plus the rate> L08> and obtain the base, or invoice, $350. 4. A Boston jeweler orders from Lubec a quantity of watch movements, amounting to $2780 ; what will be the duty, at 4 % ? 5. What will be the duty at 15 % on 1200 lb. of tapioca, in- voiced at 5^ cents per pound ? Am. 9.90. 6. What is the duty at 15 % on 54 boxes of candles, each weighing 1 cwt., invoiced at 8| cents per pound, allowing tare at 3^ per cent. ? 26* 506 PERCENTAGE. 7. A merchant imported 50 casks of port wine, each contain- ing originally 3G gallons, invoiced at $2.50 per gallon. He paid freight at $1.30 per cask, and duty at 30 %, 11 % leakage being allowed at the custom house, and $8.50 for cartage ; what did the wine cost him in store ? Ann. $5903.25. 8. A liquor dealer receives an invoice of 120 dozen bottles of porter, rated at $1.25 per dozen; if 2 % of the bottles are found broken, what will bo the duty at 24 % ? Am. $35.28. 9. The duty at 19 % on an importation of Denmark satin was $319.40, what was the invoice of the goods? Ans. $3200. 10. The duty on GOO drums of figs, each containing 14 lb., invoiced at 5| cents per pound, was $35.28; required, the rate of duty. Am. 8 %. 11. A merchant in New York imports from Havana 200 hhd. of Y\ r . I. molasses, each containing G3 gallons, invoiced at 8.30 per gallon; 150 hhd. of B. coffee sugar, each containing 500 pounds, invoiced at $.05 per pound ; 80 boxes of lemons, invoiced at $2.50 per box; and 75 boxes of sweet oranges, invoiced afe $3.00 per box. What was the whole amount of duty, estimated at 24 % on molasses and sugar, and at 8 % on lemons and oranges ? Ans. $1841.20. 12. A merchant imported 5G casks of wine, each containing 36 gallons net, the duty at 30 % amounting to $907.20 ; at what price per gallon was the wine invoiced ? 13. The duty on an invoice of French lace goods at 24 (f c , was $132, an allowance of 12 % having been made at the custom house for damage received since the goods were shipped ; what was the cost or invoice of the goods. Ans. $625. 14. A quantity of Yalencias, invoiced at $1G54, cost me $1980.50 in store, after paying the duties and $12.24 for freight; what was the rate of duty ? 15. The duty on an importation of Bay rum, after allowing 2 % for breakage, was $823.20, and the invoice price of the rum was $.25 per bottle; how many dozen bottles did the importer receive,duty at 24 % ? Ans. 1143^ doz. SIMPLE INTEREST. SIMPLE INTEREST. 307 53O. Interest u a rar.i paid for the use cf money. >ol. Principal is the sum fl-r the ucc cf -which interest 13 paid. 532. Late per cent, per annum is the sum per cent, paid f jr the use cf any principal for one year. NOTE. The rate per cent, is commonly expressed decimally as hundredth^ (<112). *5O3. Amount is the sum of iho principal and interest. >M. Simple Interest is the sum paid for the use cf Iho princi- pal cnly, during the whole time of the loan or credit. 035. Legal Interest is the rate per cent, established ly law. It varies in different States, as follows : Legal Rate, serceiit. Special by Agreement. Legal Kate, per cent. Special by Agreement. Alabama Arkansas Ciliforaia Connecticut . . Colorado 8 6 10 7 10 c 7 8 G 8 7 10 6 6 6 7 6 5 6 6 6 7 7 Any rate. Any rate. Any rate 18 per ct. 10 per ct. 10 per cf. 10 per ct Any rate. 10 per ct. 10 per ct. 10 per ct. 12 per ct. 10 per ct. 8 per ct Any rate Any rate. 10 per ct. 12 rer ct. Mississippi Any rate. 1 per ct. Any rate. 8 per ct. 15 per ct. Any rate. 10 per ct, 12 per ct. Any rate. Any late. 10 per ct. 12 per ct. Any rate. 12 per ct. 10 per ct. Any rate. Missouri Montana New Hampshire... New JorsL'C3 T Cana 'a New York North Carolina Nebraska Dakota Delaware Dist. Columbia Fioiida 0'iio . . Oregon . .... Idaho Pennsylvania Illinois Rhode Island Indiana South Carolina Tennessee Texas Kansas Kentucky Louisiana Maine Maryland Massachusetts . Miclii"an . ... Utah Vermont . . . Virginia West Virginia W iscon^in Washington Ter.. . England. . Minnesota. . 5SG. Usury is illegal interest, or a greater per cent, than tho Ic^al rate. 17oTK. Tho t:i!;ing of usury is prohibited, under various penalties, in different States. 308 PERCENTAGE. . In the operations of interest there are five parts or ele- ments, namely : I. Rate per cent, per annum ; which is the fraction or decimal denoting how many hundredths of a number or sum of money are to be taken for a period of 1 year. II. Interest; which is the whole sum taken for the whole period of time, whatever it may be. III. Principal ; which is the base or sum on which interest is computed. IV. Amount; which is the sum of principal and interest; and Y. Time. TO COMPUTE INTEREST. CASE I. 588. To find the interest on any sum, at any rate per cent, per annum, for years and months. ANALYSIS. In interest, any rate per cent, is confined to 1 year. Therefore, if the time be more than 1 year, the per cent, will be greater than the rate per cent, per annum, and if the time be less than 1 year, the per cent, will be less than the rate per cent, per annum. From these facts, we deduce the following principles : I. If the rate per cent, per annum be multiplied by the time, expressed in years and fractions or decimals of a year, the product will be the rate for the required time. And II. If the principal be multiplied by the rate for the required time, the product will be the required interest. Hence III. Interest is always the product of three factors, namely, rate per cent, per annum, time, and principal. In computing interest the three factors may be taken in any order ; thus, if the principal be multiplied by the rate per cent, per annum, the product will be the interest for 1 year; and if ihe interest for 1 year be multiplied by the time expressed in years, the result will be the required interest. Hence the following HULE. I. Multiply the principal l>y the rate per cent., and the product will Lfi the interest for 1 year. II. Multiply this product l>y the time in years and fractions of a year; the result will be the required interest. SIMPLE INTEREST. 309 Or, Multiply together the rate per cent, per annum, time, and principal, in such order as is most convenient ; the continued pro- duct will be the required interest. CASE II. 039. To find the interest on any sum, for any time, at any rate per cent. The analysis of our rule is based upon the following Obvious Relations between Time and Interest. I The interest on any sum for 1 year at 1 per cent., is .01 of that sum, and is equal to the principal with the separatrix re- moved two places to the left. II. A month being 7 V of a year, T 1 2 of the interest on any sum for 1 year is the interest for 1 month. III. The interest on any sum for 3 days is ^ = T ^ = ,1 of the interest for 1 month, and any number of days may readily be re- duced to tenths of a month by dividing by 3. IV. The interest on any sum for 1 month, multiplied by any given time expressed in months and tenths of a month, will pro- duce the required interest. These principles are sufficient to establish the following RULE. I. To find the interest for 1 yr. at 1 % : Remove the separatrix in the given principal two places to the left, II. To find the interest for 1 mo. at 1 % : Divide the interest for 1 year by 12. Ill To find the interest for any time at 1 % : Multiply the interest for 1 month by the given time expressed in months and tenths of a month. IV. To find the interest at any rate % : Multiply the interest at 1 fo for the given time by the given rate. CONTRACTIONS. After removing the separatrix in the principal two places to the left, the result may be regarded either as the interest on the given principal for 12 months at 1 per cent., or for 1 month at 12 per cent. If we regard it as for 1 month at 12 per cent., and if the given rate be an aliquot part of 12 per cent., the interest on the 310 PERCENTAGE. given principal for 1 month may readily be found, by taking such an aliquot part of the interest for 1 month as the given rate is part of 12 per cent. Thus, To find the interest for 1 month at G per cent., remove the separa- trix two places to the left, and divide by 2. To find it at 3 per cent., proceed as before, and divide by 4 ; at 4 per cent, divide by 3 ; at 2 per cent., divide by G, etc. SIX PER CENT METHOD.* 340. By referring to 535 it will be seen that the legal rate of interest in 22 States is 6 per cent. This is a sufficient reason for introducing the following brief method into this work : Ax A LYSIS. At G f per annum tho interest on $1 For 12 months is $.00. " 2 months (^=4 of 12 mo.) " .01. " 1 month, or 30 days ( T of 12 mo.) " .OOJ = $.005 ( T V of $.OG) % " G days (i of 30 da.).. ." " .001. " 1 " (i of G da. = 5 \ T of 30 da.) " .000. Hence wo conclude that, 1st. The interest on $1 is $.005 per month, or $.01 for every 2 months; 2d. Tho interest on 81 is $.000 per day, or $.001 for every 6 days. From these principles we deduce the RULE. I. To find the rate: Call every year $.06, cv-ry 2 months $.01, ever?/ G days $.001, and any less number of days sixths of 1 mitt. II. To find the interest: Multiply the principal Ly ike rate. NOTES. 1. To find the interest nt any other rate r any nunu * This method of finding the interest on $1 by inspection was first published in The Scholar's Arithmetic, by Daniel Adams, M. D., in 1801, and froiu it simplicity it has come into very general use. SIMPLE INTEREST. 311 her of days is ns many cents ns CO is contained times in the number of day?. Therefore, if :iny principal be multiplied by the number of days in any given number of months and days, and the product divided by CO. the result will lie the interest, in cents. That is, Multiply the jrinrfjinf. by the number <>f il"if*, divide the prmfnct by 60, find point off ticn ftecfmnl place* in tlif quotient. Tlie retail t will ue the interest in the same dfttotttKOtioH as the principal. EXAMPLES FOR PRACTICE. What is the interest on the following sums for the times given, at 6 per cent. 1 1. $325 ror 3 years, Ans. 858.50. 2. $1600 for 1 yr. 3 mo. Ans. 6120. 3. $36.84 for 5 mo. 4. $35.14 for 2 yr. 9 mo. 15 da. 5. $217.15 for 3 yr. 10 mo. 1 da. Ans. $49.98+. 6. $721.53 for 4 yr. 1 mo. 18 da. 7. $15.125 for 15 mo 17 da. Ans. $1.17+. On the following ac 7 per cent. ? 8. $2000 for 5 yr. 6 mo. 9. $1436.59 for 2 yr. 5 mo. 18 da. Ans. $248.051 + . 10. $224.14 for 8 mo. 13 da. Ans. $11.026. 11. $100/25 for 63 da. Ans. $1.228+. 12. $600 for 24 da. 13. $520 for 5 yr. 11 mo. 29 da. Ans. $218.298. 14. $710.01 for 3 yr. 11 mo. 8 da. On the following at 5 per cent. ? 15. $48 255 for 5 yr. 16. $750 for 1 yr. 3 mo. 17. $347.654 for 4 yr. 10 mo. 20 da. Ans. $158.315 + . 18. $12850 for 90 da. 19. $2500 for ^ mo. 20 da. Ans. $79.86- 20. $850.25 for 8 mo. 21. $48.25 for 1 yr. 2 mo. 17 da. Ans. $2.928 + . On the following at 8 per cent. ? 22. $2964.12 for 11 mo. Ans. $217.368 + 23. $725.50 for 150 da. 24. $360 for 2 yr 6 mo. 12 da. 25. 600 for 3 yr. 2 mo. 17 da. Ana. $154.266. S12 PERCENTAGE. 26. $1700 for 28 da. Ans. $10.58. On the following at 10 per cent. ? 27. 83045.20 for 7 mo. 15 da. Ans. 1 $190.32 + , 28. $1247.375 for 2 yr. 26 da. Ans. $258.48+.' 29. $2450 for 60 da. 30. $375.875 for 3 mo. 22 da. 31. $5000 for 10 da. 32. $127.65 for 1 yr. 11 mo. 3 da. Ans. $24.572. 33. What is the interest of $155.49 for 3 mo., at 6i per cent. ? 34. What is the interest of $070.99 for 6 mo., at 5J per cent. ? 35. What is the amount of $350.50 for 2 yr. 10 mo., at 7 per cent.? Ans. $120.01+. 36. What is the interest of $95.008 for 3 mo. 24 da., at 4-J per cent. ? Ans. $1.353. 37. What is the amount of $145.20 for 1 yr. 9 mo. 27 da., at 12 J per cent. ? Ans. $178.32375. 38. What is the amount of $215.34 for 4 yr. 6 mo., at 3 per cent. ? Ans. $249.256. 39. What is the amount of $5000 for 20 da., at 7 per cent. ? 40. What is the amount of $16941.20 for 1 yr. 7 mo. 28 da., at 4| per cent. ? Ans. $18277.91. 41 If $1756.75 be placed at interest June 29, 1860, what amount will be due Feb. 12, 1863, at 7 % 1 42. If a loan of $3155.49 be made Aug. 15, 1858, at 6 per cent., what amount will be due May 1, 1866, no interest having been paid ? 43. How much is the interest on a note for $257.81, dated March 1, 1859, and payable July 16, 1861, at 7 % ? 44. A person borrows $3754.45, being the property of a minor who is 15 yr. 3 mo. 20 da. old. He retains it until the owner is 21 years old. How much money will then be due at 6 % simple interest? Ans. $5037.22+. 45. If a person borrow $7500 m Boston and lend it in Wis- consin, how much does he gain in a year? 46. A man sold a piece of property for $11320; the terms were $3200 in cash on delivery, $3500 in 6 mo., $2500 in 10 mo., and SIMPLE INTEREST. 313 the remainder in 1 yr. 3 mo. , with 7 % interest ; what was the whole amount paid ? Ans. $11773. 83. 47. May 10, 1859, I borrowed $6840, with which I purchased flour at $5.70 a barrel. June 21, 1860, 1 sold the flour for $6.62 J a barrel, cash. How much did I gain by the transaction, interest being reckoned at 6 % ? 48. If a man borrow $15000 in New York, and lend it in Ohio, how much will he lose in 146 days, reckoning 360 days toi the year in the former transaction, and 365 days in the latter ? 49. Hubbard & Northrop bought bills of dry goods of Bowen, McNamee & Co., New York, as follows, viz.: July 15, 1860, 81250; Oct. 4, 1860, $3540.84; Dec. 1, 1860, $575; and Jan. 24, 1861, $816.90. They bought on time, paying legal interest; how much was the whole amount of their indebtedness, March 1, 1861? 50. A broker allows 6 per cent, per annum on all moneys de- posited with him. If on an average he lend out every $100 re- ceived on deposit 11 times during the year, for 33 days each time at 2 % a month, how much does he gain by interest on $1000? Ans. $182. 51. A man, engaged in business with a capital of $21840, is making 12 } per cent, per annum on his capital; but on account of ill health he quits his business, and loans his money at 7f %. How much does he lose in 2 yr. 5 mo. 10 da. by the change? Ans. $2535.861. 52. A speculator wishing to purchase a tract of land containing 450 acres at $27.50 an acre, borrows th3 money at 5J per cent. At the end of 4 yr. 11 mo. 20 da. he sells f of the land at $34 an acre, and the remainder at $32.55 an acre. How much does he lose by the transaction ? 53. Bought 4500 bushels of wheat at $1.12 J a bushel, payable in 6 months; I immediately realized for it $1.06 a bushel, cash, and put the money at interest at 10 per cent. At the end of the 6 months I paid for the wheat; did I gain or lose by the transac- tion, and how much ? 27 814 PERCENTAGE. PARTIAL PAYMENTS OR INDORSEMENTS. 541. A Partial Payment is payment in part f of a note, bond, or other obligation. 542. An Indorsement is an acknowledgment written on the back of an obligation, stating the time and amount of a partial payment made on the obligation. 543. To secure uniformity in the method of computing in- terest where partial payments have been made, the Supreme Court of the United States has decided that, I. " The rule for casting interest when partial payments have been made, is to apply the payment, in the first place, to the dis- charge of the interest then due. II. " If the payment exceeds the interest the surplus goes to- wards discharging the principal, and the subsequent interest is to be computed on the baiance of the principal remaining due. III. ' If the payment be less than the interest the surplus of interest must not be taken to augment the principal, but the inte-. rest continues on the former principal until the period when the payments, taken together, exceed the interest due, and then the surplus is to be applied towards discharging the principal, and the interest is to be computed on the balance as aforesaid/' Decision of Chancellor Kent. This decision has been adopted by nearly all the States of the Union, the only prominent exceptions being Connecticut, Ver- mont, and New Hampshire. We therefore present the method prescribed by this decision as the UNITED STATES RULE. I. Find the amount of the given principal to the tim: of the first payment, and if this payment exceed the interest then due, subtract it from the amount obtained, and treat the remainder as a new principal. II. But if the interest be greater than any payment, compute the interest on the same principal to a time when the sum of the jjcy- ments shall equal or exceed the interest due, and subtract the sum PARTIAL PAYMENTS. 315 of the payments from the amount of the principal ; the remainder will form a new principal , with which proceed as before. EXAMPLES FOR PRACTICE. BUFFALO, N Y., May 15, 1856. 1. Two years after date I promise to pay to David Hudson, or order, one thousand dollars, with interest, for value received. HENRY BURR. On this note were indorsed the following payments : Sept. 20, 1857, received, $150.60 Oct. 25,1859, " 200.90 July 11, 1861, " 75.20 Sept. 20, 1862, " 112.10 Dec. 5,1863, " 105. What remained due May 20, 1864? OPERATION. Principal on interest from May 15, 1856, $1000 Interest to Sept. 20, 1857, 1 yr. 4 mo 5 da., 94.31 Amount, $1094.31 1st Payment, Sept 20, 1857, 150.60 Remainder for a new principal, $943.71 Interest from 1st paym't to Oct. 25, 1859, 2 yr. 1 mo. 5 da., 138.54 Amount, $1082.25 2d Payment, Oct. 25, 1859, 200.90 Remainder for a new principal, $881.35 Interest from 2d paym't to Dec. 5, 1863, 4 yr. 1 mo. 10 da., 253.63 Amount, $1134.98 3d Payment, less than interest due, $75.20 4th " 112.11 Sum of 3d and 4th payments, less than interest due, $187.31 5th payment, 105.00 Sums of 3d, 4th, and 5th payments, 292.31 Remainder for new principal $842.67 Interest to May 20, 1864, 5 mo. 15 da., 27.04 Balance due May 20, 1864, $869.71 316 PERCENTAGE. $ 12QO - RICHMOND, VA., Oct. 15, 1859. 2. One year after date we promise to pay James Peterson, or order, twelve hundred dollars, for value received, with interest. WILDER & SON. Indorsed as follows: Oct. 15, 1860, $1000; April 15, 1861, $200. How much remained due Oct. 15, 1861 ? Ans. $82.56. $850 T Tjfe. BOSTON, June 10, 1855. 3. Eighteen months after date I promise to pay Crosby, Nich- ols & Co., or order, eight hundred fifty and y 7 ^ dollars, with interest, for value received. 0. L. SANBORN. Indorsed as follows: March 4, 1856, $210.93; July 9, 1857, $140; Feb. 20, 1858, $178; May 5, 1859, $154.30; Jan. 17, 1860, $259 45. How much was due Oct. 24, 1861 ? SAVANNAH, GA., Sept. 4, 1860. 4. Six months after date I promise to pay John Rogers, or order, three hundred eighty-four and T 9 ^j dollars, for value re- ceived, with interest. WAI. JENKINS. This note was settled Jan. 1, 1862, one payment of $126.50 having been made Oct. 20, 1861 ; how much was due at the time of settlement ? $3475. NEW ORLEANS, March 6, 1857. 5 On demand we promise to pay Evans & Hart, or order, three thousand four hundred seventy-five dollars, for value received, with interest. DAVIS & BROTHER. Indorsed as follows- June 1, 1857, $1247.60; Sept 10, 1857, $1400. How much was due Jan. 31, 1858 ? 6. A gentleman gave a mortgage on his estate for $9750, dated April 1, 1860, to be paid in 5 years, with annual interest after 9 months on all unpaid balances, at 10 per cent. Six months from date he paid $846.50; Oct. 20, 1862, $2500; July 3, 1863, $1500; Jan. 1, 1864, $500; how much was due at the expiration of the given time ? PAKTIAL PAYMENTS. 317 $500. PHILADELPHIA, Feb. 1, 1861. 7. For value received, I promise to pay J. B. Lippincott & Co., or order, five hundred dollars three months after date, with interest. JAMES MONROE. Indorsed as follows: May 1, 1861, $40; Nov. 14, 1861,. 88; April 1, 1862, $12 ; May 1, 1862, $30. How much was due Sept. 16, 1862? Ans. $455.57-+. 511. CONNECTICUT RULE.* I. Payments made one year or more from the time the interest commenced, or from another payment, and payments less than the interest due, are treated according to the United States rule. II. Payments exceeding the interest due and made within one year from the time interest commenced, or from a former payment, shall draw interest for the balance of the year, provided the interval does not extend beyond, the settlement, and the amount must be sub- tracted from the amount of the principal for one year; the re- mainder will be the new principal. III. If the year extend beyond the settlement, then find the amount of the payment to the day of settlement, and subtract it from the amount of the principal to that day ; the remainder will be the sum due. *> 15. A note containing a promise to pay interest annually is not considered in law a contract for any thing more than simple interest on the principal. For partial payments on such notes the following is the VERMONT RULE. I. Find the amount of the principal from the time interest com- menced to the time of settlement. II. Find the amount of each payment from the time it was made to the time of settlement. III. Subtract the sum of the amounts of the payments from the amount of the principal; the remainder will be the sum due. * The United States rule is in general use. 27* 318 PERCENTAGE. In New Hampshire interest is allowed on the annual interest if not paid when due, in the nature of .damages for its detention ; and if payments are made before one year's interest has accrued, interest must be allowed on such payments for the balance of the year. Hence the following NEW HAMPSHIRE RULE. I. Find the amount of the principal for one year, and deduct from it the amount of each payment of that year, from the time it was made up to the end of the year ; the remainder will be a new principal, with which proceed as before. II. If the settlement occur less than a year from the last annual term of interest, make the last term~of interest a part of a year, accordingly. EXAMPLES FOR PRACTICEc $ 1QQQ - NEW HAVEN, CONN., Feb. 1, 1856. 1. Two years after date, for value received, I promise to pay to Peck & Bliss, or order, one thousand dollars with interest. JOHN CORNWALL. Indorsed as follows: April 1, 1857, $80; Aug. 1, 1857, $30; Oct. 1, 1858, $10 ; Dec. 1, 1858, $600 ; May 1, 1859, $200. How much was due Oct. 1, 1859 ? Ans. $266.38. BURLINGTON, VT., May 10, 1858. 2. For value received, I promise to pay David Camp, or order, two thousand dollars, on demand, with interest annually. RICHARD THOMAS. On this note were indorsed the following payments : March 10, 1859, $800; May 10, 1860, $400; Sept. 10, 1861, $300. How much was due Jan. 10, 1863 ? 3. How much would be due on the above note, computing by the Connecticut rule ? Am. $831.58. 4. How much, computing by the New Hampshire rule ? By the United States rule ? . ( N. H. rule, $833.21 ; ' I U. S. $831.90. SAVINGS BANK ACCOUNTS. 319 SAVINGS BANK ACCOUNTS. . Savings Banks are institutions intended to receive in trust or on deposit, small sums of money, generally the surplus earnings of laborers, and to return the same with a moderate interest at a future time. 5>48. It is the custom of all savings banks to add to each depositor's account, at the end of a certain fixed term, the interest due on his deposits according to some general regulation for allow- ing interest. The interest term with some savings banks is 6 months, with some 3 months, and with some 1 month. 519. A savings bank furnishes each depositor with a book, in which is recorded from time to time the sums deposited and the sums drawn out. The Dr. side of such an account shows the deposits, and the Cr. side the depositor's checks or drafts. In the settlement, interest is never allowed on any sum which has not been on deposit for a full interest term. Hence, to find the jnount due on any depositor's account, we have the following RULE. At the end of each term, add to the balance of the account one term's interest on the smallest balance on deposit at any one time during that term ; the final balance thus obtained will be the sum due. NOTES. 1. It will be seen that by this rule no interest is allowed for money on deposit during a partial term, whether the period be the first or the last part of the term. 2. An exception to this general rule occurs in the practice of some of the savings banks of New York city. In these, the interest term is 6 months, and the depositor is allowed not only the full term's interest on the smallest balance, but a half terra's interest on any deposit, or portion of a deposit made during the first 3 months of the term, and not drawn out during any subsequent part of the term. EXAMPLES FOR PRACTICE. 1. What will be due April 20, 1860, on the following account, interest being allowed quarterly at 6 per cent, per annum, the terms commencing Jan. 1, April 1, July 1, and Oct. 1 ? Dr. Savings Bank in account with James Taylor. Cr. 1858, Jan. 12, $75 1858, March 5, $30 " May 10, 150 " Aug. 16, 50 " Sept 1, 20 Dec. 1,... 48 1859, Feb. 16, 130 120 PERCENTAGE. OPERATION. Deposit, Jan. 12, 1858, ........................ ...... $75 Draft, March 5, " .......................... .1.. 30 Balance, Apr. 1, I860, .................. $45 Deposit, May 10, 1858, .............................. 150 Int. on $4^, for 3 mo ............. . .................... 68 Balance, July 1, 1860, .................. $195.68 Draft, Aug. 16, 1858, ...... . ......................... 50 Least balance during the current term, ........... $145.68 Deposit, Sept. 1, 1858, .............................. 20.00 Int. on $145.68, for 3 mo ........................... 2.19 Balance, Oct. 1, 1858, ................... $167.87 Draft, Dec. 1, 1858, .................................. 48 Least balance during the current term, ........... 119.87 Int. on $119.87, for 3 mo ................ , ......... 1.80 Balance, Jan. 1, 1860, ................... $121.67 Deposit, Feb. 16,1860, .............................. 130.00 Int. on $121.67, for 3 mo ........................... 1.83 Bal. due after Apr. 1, 1860, ............ $253.50^5. NOTE. In the following examples the terms commence with the year, or on Jan. 1. 2. Allowing interest monthly at 6 % per annum, what sum will be due Sept. 1, 1860, on the book of a savings bank having the following entries ? Bay State Savings Institution, in account with Jane Ladd. Dr. Cr. 1860. 1860. Jan. 3 To cash, 5 75 Jan. 28 By check, 5 00 8 ii 13 45 Feb. 7 Cl tl 8 48 < 20 n 7 60 March 20 (I 10 00 Feb. 20 " check, 16 45 April 11 it il 12 76 27 " cash, 8 40 June 3 tl 3 96 March 6 " check, 14 65 12 (( 10 48 a 29 " cash, 7 98 20 " draft, 17 48 April 25 ti >t 3 49 Aug. 17 check, 5 64 May 7 " draft, 26 50 30 ( 45 79 July Aug. 28 3 " cash, " check, 15 18 68 45 t 26 cash, 4 50 Ans. $116.87. 3. Interest at 7 %, allowed quarterly, how much was due April 4, 1860, on the following savings bank account ? COMPOUND INTEREST. 321 Dr. Detroit Savings Institution, in account with R. L. Selden. Cr. 1859. 1859. Jan. 1 To cash, 47 50 May 12 By check, 50 36 March 12 >< 124 36 Oct. 3 j < 25 78 June 20 <( 130 56 Nov. 16 36 48 Aug. 3 68 75 Dec. 28 1 " " 12 _ 50 1860. Jan. 25 <; 160 80 1 ' 1 Ans. $423.22. 4. How much was due Jan. 1, 1860, on the following account, allowing interest semi-annually, at 6 % per annum ? Irvings Savings Institution, in account with James Taylor. Dr. O. 1858. 1858. June 4 To cash, 175 Sept. 14 By check, 65 Nov. 1 u ^ $55.44, int. of $924 for 1 yr. at 6 %. $151.536 - $55.44 = 2.73 of the game 2.73 yr. = 2 yr. 8 mo. 24 da., Ans. the same rate per cent., for as many years as $55.44 is contained times in $151.536, which is 2.73 times. Reducing the mixed decimal to its equivalent compound number, we have 2 years 8 months 24 days, the required time. Hence the 828 PERCENTAGE. RULE. Divide the given interest by the interest on the principal for 1 year ; the quotient will be the required time in years and decimals. ' EXAMPLES FOR PRACTICE. 1. In what time will $273.51 amount to $312.864, at 7 per cent. ? Ans. 2 yr. 20 da. 2. How long must $650.82 be on interest to amount to $761.44, at 5 per cent. ? Ans. 3 yr 4 mo. 24 da. 3. How long will it take any sum of money to double itself by simple interest at 3, 4J, 6, 7, and 10 per cent. ? How long to quadruple itself? ^ j To double itself at 3 %, 33 yr S ' ( To quadruple itself at 3 %. 100 yr. 4. In what time will $9750 produce $780 interest, at 2 per cent, a month ? 5. In what time will $1000 draw $1171.353 at 6 per cent, com, pound interest ? ANALYSIS. $1171.353-r-1000=$1.171353, the amount of $1 for the required time. From the table, $1, in 2 years, will amount to $1.1236 ; hence $1.171353 $1.1236 $.047753, the interest which must accrue on $1.1236 for the fraction of a year; and $1.1236 X .06 = $.067416 ; $.047753 -7- $.067416 = .7083 yr. = 8 mo. 15 da. Ans. 2 yr. 8 mo. 15 da. 6. In what time will $333 amount to $376.76 at 5 per cent compound interest, payable semi-annually ? 7. In what time will any sum double itself at 6 % compound interest ? At 7 % ? Ans. to last, 10 yr. 2 mo. 26 da. DISCOUNT. 156. Discount is an abatement or allowance made for the payment of a debt before it is due 5,17. The Present Worth of a debt, payable at a future time without interest, is such a sum as, being put at legal interest, will amount to the given debt when it become? due. 1. What is the present worth and what the discount of $642.12 to be paid 4 yr. 9 mo. 27 da. hence, money being worth 7 per cent. ? DISCOUNT. 329 OPERATION. ANALYSIS. Since $1 is the. $1.33775, Amount of 81. present worth of $1.33775 $642.12 -T- 1.33775 = $480 for the given time at the $642. 12, given sum.. given rate of interest, the 480. present worth, present worth of $642.12 $162 \.12, discount. must } as many dollars as $1.33775 is contained times in $642.12. Dividing, and we obtain $480 for the present worth, and subtracting this sum from the given sum, we have $162.12, the dis- count. Hence the following RULE. I Divide the given sum or debt by the amount of $1 for the given rate and time; the quotient will be the present worth of the debt. II. Subtract the present worth from the given sum or debt ; the remainder will be the discount. NOTES. 1. The terms present worth, discount, and. debt, are equivalent to principal, interest, and amount. Hence, when the time, rate per cent., and amount are given, the principal maybe found by Prob. II, (553); and the interest by subtracting the principal from the amount. 2. When payments are to be made at different times without interest, find the present worth of each payment separately. Their sum will be the present worth of the several payments, and this sum subtracted from the sum of the several payments will leave the total discount. EXAMPLES FOR PRACTICE. 1. What is the present worth of a debt of $385. 31|, to be paid in 5 mo. 15 da., at 6 % ? Ans. $375. 2. How much should be discounted for the present payment of a note for $429.986, due in 1 yr. 6 mo. 1 da., money being worth 5i % ? Ans. $32.826. 3. Bought a farm for $2964.12 ready money, and sold it again for $3665.20, payable in 1 yr. 6 mo. How much would be gained in ready money, discounting at the rate of 8 % ? 4. A man bought a flouring mill for $25000 cash, or for $12000 payable in 6 mo. and $15000 payable in 1 yr. 3 mo. He accepted the latter offer ; did he gain or lose, and how much, money being worth to him 10 per cent. ? An*. Gained $238.10. 5. B bought a house and lot April 1, 1860, for which he waa to pay $1470 on the fourth day of the following September, and 28* g3Q PERCENTAGE. $2816.80 Jan 1, 1861. If he could get a discount of 10 per cent, for present payment, how much would he gain by borrowing the sum at 7 per cent., and how much must he borrow? 6. What is the difference between the interest and the discount of $576, due 1 yr. 4 mo. hence, at 6 per cent. ? 7. A merchant holds two notes against a customer, one for $243.16, due May 6, 1861, and the other for $178.64, due Sept. 25, 1861 ; how much ready money would cancel both the notes Oct. 11, I860, discounting at the rate of 7 % ? Am. $401.29. 8. A speculator bought 120 bales of cotton, each bale containing 488 pounds, at 9 cents a pound, on a credit of 9 months for the amount. He immediately sold the cotton for $6441.60 cash, and paid the debt at 8 % discount ; how much did he gain ? 9. Which is the more advantageous, to buy flour at $6.25 a barrel on 6 months, or at $6.50 a barrel on 9 months, money being worth 8 % ? 10. How much may be gained by hiring money at 5 % to pay a debt of $6400, due 8 months hence, allowing the present worth of this debt to be reckoned by deducting 5 % per annum dis- count? Ans. $7.11. BANKING. 558. A Bank is a corporation chartered by law for the pur- pose of receiving and loaning money, and furnishing a paper circulation. 559. A Promissory Note is a written or printed engagement to pay a certain sum either on demand or at a specified time. 5OO. Bank Notes, or Bank Bills, are the notes made and issued by banks to circulate as money. They are payable in specie at the banks. NOTE. A bank which issues notes to circulate as money is called a lank of issue ; one which lends money, a bank of discount ; and one which takes charge of money belonging to other parties, a bank of deposit. Some banks perform two and some all of these duties. 561. The Maker or Drawer of a note is the person by whom the note is signed ; 562. The Payee is the person to whose order the note is made payable; and BANKING. 33] 563. The 'Holder is the owner. 564L A Negotiable Note is one which may be bought and sold, or negotiated. It is made payable to the bearer or to the order of the payee. 565c Indorsing a note by a, payee or holder is the act of writing his name on its back. NOTKS. 1. If a note is payable to the bearer, it may be negotiated without indorsement. 2. An indorsement makes the indorser liable for the payment of a note, if the maker fails to pay it when it is due. 3. A note should contain the words "value received," and the sum for which it is given should be written out in words. 566. The Face of a note is the sum made payable by the note. 567. Days of Grace are the three days usually allowed by law for the payment of a note after the expiration of the time specified in the note. 568 The Maturity of a note is the expiration of 'the days of grace ; a note is due at maturity. NOTE. No grace is allowed on notes payable " on demand," without grace. In some States no grace is allowed on notes, and their maturity is the expira- tion of the time mentioned in them. 569. Notes may contain a promise of interest, which will be reckoned from the date of the note, unless some other time be specified. NOTE. A note is on interest from the day it is'due, even though no mention be made of interest in the note. 570. A Notary, or Notary-Public, is an officer authorized by law to attest documents or writings of any kind to make them authentic. 571. A Protest is a formal declaration in writing, made by a Notary-Public, at the request of the holder of a note, notifying the maker and the indorsers of its non-payment. NOTES. 1. The failure to protest a note on the third day of grace releases the iii- dorsers from all obligation to pay it. 2. If the third day of grace or the maturity of a note occurs on Sunday or a legal holiday, it must be paid on the day previous. 572. Bank Discount is an allowance made to a bank for the payment of a note before it becomes due. 332 PERCENTAGE. 17JI. The Proceeds of a note is the sum received for it discounted, and is equal to the face of the note less the discount. 574, The transaction of borrowing money at banks is con- ducted in accordance with the following custom : The borrower presents a note, either made or indorsed by himself, payable at a specified time, and receives for it a sum equal to the face ; less the interest for the time the note has to run. The amount thus withheld by the bank is in consideration of advancing money on the note prior to its maturity. NOTES. 1. A note for discount at bank must be made payable to the order of some person, by whom it must be indorsed. 2. The business of buying or discounting notes is chiefly carried on by banks and brokers. The law of custom at banks makes the bank discount of a note equal to the simple interest at the legal rate, for the time specified in the note. As the bank always takes the interest at the time of discounting a note, bank discount is equal to simple interest paid in advance. Thus, the true discount of a note for $153, which matures in 4 months at 6 %, is $153 'fsoo = $3.00, and the bank discount is $153 x .02 = $3.06. Since the interest of $3, the true discount, for 4 months is $3 x .02 = $.06, we observe that the bank discount of any sum for a given time is greater than true discount, by the interest on the true discount for the same time. NOTE. Many banks take only true discount. CASE I. 576. Given, the face of a note, to find the discount and the proceeds. RULE. I. Compute the interest on the face of the note for three days more than the specified time; the result will be the discount. II. Subtract the discount from the face of the note; the re- mainder will be the proceeds. NOTES. 1. When a note is on interest, payable at a future specified time, the amount is the face of the note, or the sum made payable, and must be made the basis of discount. 2. To indicate the maturity of a note or draYt, a vertical line ( | ) is used, with the day at which the note is nominally due on the left, and the date of maturity on the right; thus, Jan. 7 | , . BANKING. 333 EXAMPLES FOR PRACTICE. 1. What is the bank discount, and what are the proceeds of a note for $1487 due in 30 days at 6 per cent. ? Ans. Discount, $8.18; Proceeds, $1478.82. 2. What are the proceeds of a note for $384.50 at 90 days, if discounted at the New York Bank? 3. Wishing to borrow $1000 of a Southern bank that is dis- counting paper at 8 per cent., I give my note for $975, payable in 60 days ; how much more will make up the required amount ? 4. A man sold his farm containing 195 A. 2 R. 25 P. for $27.50 an acre, and took a note payable in 4 mo. 15 da. at 7 % interest. Wishing the money for immediate use, he got the note discounted at a bank; how much did he receive ? Ans. $5236.169. 5. Find the day of maturity, the term of discount, and the pro- ceeds of the following notes : $1962-^%. DETROIT, July 26, 1860. Four months after date I promise to pay to the order of James Gillis one thousand nine hundred sixty-two and T 4 ^ dollars at the Exchange Bank, for value received. JOHN DEMAREST. Discounted Aug. 26, at 7%. Ans. Due Nov. 26 | 29 ; term of discount 95 days; proceeds; $1926.20. %. BALTIMORE, April 19, 1859. 6. Ninety days after date we promise to pay to the order of King & Dodge one thousand sixty-six and y 7 ^ dollars at the Citi. zens' Bank, for value received. CASE & SONS. Discounted May 8, at 6 % . Ans. Due July 1 8 | f , ; term of discount, 74 da. ; proceeds, $1053.59. $784/0%. MOBILE, June 20, 1861. 7. Two months after date for value received I promise to pay George Thatcher or order seven hundred eighty-four and T 7 ^ dol- lars at the Traders' Bank. WM. HAMILTON. Discounted July 5, at 8 %. 334 PERCENTAGE. $1845^. CHICAGO, Jan. 31, 1862. 8. One month after date we jointly and severally agree to pay to W. H. Willis, or order, one thousand eight hundred forty 7 five and -f^Q dollars at the Marine Bank. FAYSON & WILLIAMS. Discounted Jan. 31, at 2 % a month. Ans. Due Feb. 28 | March 3; term of discount, 31 da.; pro- ceeds, $1807.36. 9. What is the difference between -the true and the bank dis- count of $950, for 3 months at 7 per cent. ? Ans. $.29. 10. What is the difference between the true and the bank dis- count of $1375.50, for 60 days at 6 per cent. ? CASE II. 577. Given, the proceeds of a note, to find the face. 1. For what sum must I draw my note at 4 months, interest 6 %, that the proceeds when discounted in bank shall be $750 ? OPERATION. ANALYSIS. We $1.0000 first obtain the pro- .0205, disc't on $1 for 4 mo. 3 da. ceeds of $1 by the (T9795, proceeds of $1. last case; then, since $750 nh .9795 = $765.696, Ans. ^ 9795 1S the P r ? - ceeds of $1, $750 is the proceeds of as many dollars as $.9795 is contained times in $750. Dividing, we obtain the required result. Hence the RULE. Divide the proceeds l>y the proceeds of $lybr the time and rate mentioned ; the quotient will be the face of the note. EXAMPLES FOR PRACTICE. 1. What is the face of a note at 60 days, the proceeds of which, when discounted at bank at 6 %, are $1275? Ans. $1288.53. 2. If a merchant wishes to draw $5000 at bank, for what sum must he give his note at 90 days, discounting at 6 per cent. ? Ans. $5078.72. 3. The avails of a note having 3 months to run, discounted at a bank at 7 %, were $276.84; what was the face of the note ? BANKING. 335 4. James T. Fisher buys a bill of merchandise in New York at cash price, to the amount of $1486.90, and gives in payment his note at 4 months at 7^ % ; what must be the face of the note ? 5. Find the face of a 6 mo. note, the proceeds of which, dis- counted at 2 % a month, are $496. AUK. $564.92. 6. For what sum must a note be drawn at 30 days, to net $1200 when discounted at 5 % ? 7. Owing a man $575, I give him a 60 day note; what should be the face of the note, to pay him the exact debt, if discounted at 1^ % a month? Ans. $593.70. 8. What must be the face of a note which, when discounted at a broker's for 110 days at 1 % a month, shall give as its proceeds $187.50? CASE III. 578. Given, the rate of bank discount, to find the corresponding rate of interest. I. A broker discounts 30 day notes at 1^ ^ a month; what rate of interest does his money earn him ? OPERATION. ANALYSIS. If we assume 30 day notes = 33 days' time. $1 as the face of the $100, base. note, the discount for 33 1.65, discount for 33 days. days at 1^ a month will $98.35, proceeds. be $ L65 and the Proceeds $1,65 -f-. 0901541 = 18-fl ? ff 4 , %, Ans. $ 98 - 35 - We then have $98.35 principal, $1.65 in- terest, and 33 days time, to find the rate per cent, per annum, which we do by (554). Hence the KULE. I. Find the discount and the proceeds of $1 or $100 for the time the note has to run. II. Divide the discount by the interest of the proceeds at 1 per cent, for the same time. EXAMPLES FOR PRACTICE. 1. What rate of interest is paid, when a note payable in 30 days is discounted at 6 per cent. ? Ans. 836 PERCENTAGE. 2. A note payable in 2 months is discounted at 2 % a month; what rate of interest is paid? Ans. 25ff-g (a; 5 16> 4.881/2 4.89^ 5 13*^ (ff, 5 12 JX Antwerp . . . ...Francs .. . . 5.17J4 5 IG 1 ^ 5 13^ tjy 5 12 1 /;; Switzerland. . . Francs 5 n% @ 5 ig^- 5 13V 5 12i/ Amsterdam Guilders 41 Vg @. 41 14 4ij^ @, 41 5' Hambur^ Reichsmarks . .. .... 94% 95)s 96 %j^ Frankfort 947^ (g, 951^ c.g (^ 96^ Bremen . . . Reichsmarks 94 7 s @ 95% 96 %X Berlin... ...Reichsmarks ... 94 7 2 (fh 95^ 96 Gh 96^ In the above, u Prime Bankers 1 Bills" are those on the most reliable banking houses; "Good 'Ms applied to those of somewhat inferior credit. : and u Prime Commercial" are merchants' drafts, which usually command a less price in the market. The quota- tions in the first column are those of 60-day bills, and in the second column those of 3 days. 1. What will be the cost in Boston of the following bill of ex- change on Liverpool, the course of exchange being 4.8 7-g- ? 432. BOSTON, June 16, 1875. At sight of this First of Exchange (Second and Third of same tenor and date unpaid), pay to the order of J. Simmons, Boston, Four Hundred Thirty-two Pounds, value received, and charge the same to account of JAMES LOWELL & Co. To RICHARD EVANS & SON, ) Liverpool, England. f OPERATION. ANALYSIS. According $4.875, course of exchange, value of 1. to the course of ex- 432, number of pounds. change, 1 is worth $4.875 x 432 =$2106, Ans. $4.875; hence 432 is worth 432 times $4.875, or $2106, the required cost of the bill. 346 PERCENTAGE. 2. What is the face of a bill on London, that may be purchased in New York for $277.42, exchange being quoted at 4.85? OPERATION. ANALYSIS. Since 1 costs $4.85, as $977 42 $4 85 = 57 2 man y pounds can be bought for $277.42 as $4.85 is contained times in 57.2 = 57 4s., the face. $277.42, or 57.2=57 4s. Hence, etc. 3. What cost in Hamburg a bill on New Orleans for $4500, the course of exchange being 95 ? OPERATION. ANALYSIS. Since $.95 is $4500 -T- $.95 = 4736.84. tlle cost of 4 marks (6O5), 4736.84 x 4 = 18947.36 marks. $ 4500 wil1 cost four times 18947 marks 36 pennies, cost of bill. as man ? marks as $- 95 is contained times in $4500, or 18947.36. Hence the cost of the bill is 18947 marks 36 pennies. GOO. From these illustrations we derive the following RULE. I. To find the cost of a bill, the face being given. Multiply the value of a monetary unit according to the course of exchange, by the number of such units in the face of the bill. II. To find the face of a bill, the cost being given. Divide the cost of the bill by the value of each monetary unit, according to the course of exchange. EXAMPLES FOR PRACTICE. 1. What is the cost in Portland of a bill on Manchester, Eng., for 325 3s. 9d., when sterling exchange is selling at 4.89? Ans. $1591.79 + . 2. What must be paid in Charleston for a bill of exchange on Paris for 6000 francs, the course of exchange being 5.31 ? NOTE. The quotation 5.31 means that number of francs to a dollar. Hence f 1.00-s- 5.31 gives the value of 1 franc, which multiplied by the number of francs (by Rule I) will give the cost of the bill ; or more briefly, the given number of francs divided by the quotation will give the cost of the bill in dollars. 3. What is the cost, m Boston, of a bill on St. Petersburg for 3000 roubles, at $.771; brokerage \% r \ Find also at what per ?cnt. premium the exchange is at that time. (See Table, p. 342.) 4. What will be the cost in Naples of a bill of exchange on New York for $831.12, at $.205 a lira? EXCHANGE. 347 5. A draft on Philadelphia cost 125 in Birmingham, exchange selling for 4.855; required the face of the draft. Ans. $606.875. 6. An agent in Boston having $7536.30 due his employer in England, is directed to remit by a bill on Liverpool ; what is the face of the bill which he can purchase for this money, exchange selling at 4.91 ? Ans. 1534 2s. Id. + 7. A merchant in Cincinnati has 9087-J guilders due him ,in Amsterdam, and requests the remittance by draft. What sum will he receive, exchange on U. S. in Amsterdam selling at 2.41 guilders for $1 ? 8. A trader in London wishes to invest 2500 in merchandise in Lisbon ; if he remits to his correspondent at Lisbon a bill purchased for this sum, at the rate of 4.51 milreis to the pound sterling, what sum in the currency of Portugal will the agent receive ? Ans. 11275 milreis. 9. A draft on Dublin for 360 cost $1736.10; what was the course of exchange ? Ans. 4.82 J. 10. A merchant in Baltimore having received an importation of Madeira wine invoiced at 1 500 milreis, allows his correspondent in Madeira to draw on him for the sum necessary to cover the cost, exchange on the United States being in Madeira 931 reis to the dollar; how much would the merchant have saved by remitting a draft on Madeira, purchased at $1.065 per milreis? Ans. $12.80 + . 11. An importer received a quantity of Leghorn hats, invoiced at 25256.80 lire, which was paid in U. S. gold coin, exported at a cost of 3 % for transportation and insurance, the price of fine gold in Leghorn being 131 lire per ounce Troy. How much more would the goods have cost in store, had payment been made by draft on Leghorn, purchased at the rate of $.198 per lira? Ans. $895.75 -f . NOTE. In U. S. gold coinage, $10 contains 258 x .9=232.2 grains offlne gold (603). 12. When silver is worth in England 61d. per oz. fine, what sum of money in U. S. trade-dollars is equal to l sterling ? Ans. $4.996 + . 13. WTiat is the course of exchange on Berlin when $858.85 ia paid for 890 marks? PERCENTAGE. 6O7. Arbitration of Exchange is the process of computing exchange between two places by means of one or more intermediate exchanges. < NOTES. When there is only one intermediate exchange, the process is called Simple Arbitration ; when there are two or more intermediate exchangee, the process is called Compound Arbitration 2. The arbitrated price is generally either greater or less than the price of direct ex- changes ; and the object of arbitration is to ascertain the best route for making draita or remittances. ,GO8. There are always three methods of receiving money from a place, or of transmitting money to a place, by means of indirect exchange through one intervening place. Thus, If A is to receive money from C through B, 1st. A may draw on B, and B draw on C; 2d. A may draw on B, and C remit to B ; 3d. B may draw on C, and remit to A. If A is to transmit money to C through B, 1st. A may remit to B, and B remit to C ; 2d. A may remit to B, and C draw on B ; 3d. B may draw on A, and remit to C. 1. A man in Albany, N. Y., paid a demand in Paris of 5400 fr., by remitting to Amsterdam at the rate of 1.41-J per guilder, and thence to Paris at the rate of 2.15 francs per guilder. How much Federal money was required ? OPEKATION. $ (?) 5400 francs. 2.15 francs = 1 guilder. 1 guilder = .8.41 J. (?) = $1042.32+, Ans. Or, ANALYSIS. We are to deter- mine how much Federal money is equal to 5400 francs, and the question may be represented thus: $(?)=5400francs. Now, since 2.15 francs = 1 guilder, 5400 divided by 2.15 will give the number of guilders; and that number multiplied by $.415, the value of 1 guilder, will give the required sum. Hence, 5400 and .415 are multipliers and 2.15 is a divisor. The units of currency being canceled, and the work being; abridged also by canceling common factors, we have (?) = $1042.32 + , the sum required. (?) 5400 .43 JU$ 1 1 M$ .083 .43 $448.2 $1042.32 + , Ans. EXCHANGE. 349 Or, since the course of exchange between Amsterdam and Paris gives 1 guilder=2.15 francs^ and the course between Albany and Amsterdam gives $.41^=1 guilder, we multiply the 5400 francs by .415 and divide by 2.15, using the vertical line and cancellation, and obtain $1042.32 + , as before. NOTE. In the first statement, the rates of exchange are so arranged that the same unit of currency shall stand on opposite sides in each two consecutive equations, in order that these factors may all be canceled. 2. A resident of Naples, having a bequest of $8720 made him in Boston, orders the remittance to be made to his agent in London, who remits the proceeds to Naples, reserving his commission of -J-^ on the draft sent. If the course of exchange on London is 8-J-.875 in Boston, and the rate between London and Naples is 25.53 lire to the pound sterling, how much does the man realize from his bequest ? OPERATION. ANALYSIS. We make the statement as (?) lire = $8720 m ^ le ^ rst exam P^ e > according to the <$;4 07^ - - i given rates of exchange. Then, since the '"" * agent is to deduct $ % commission on the 1,5.53 ire. of the draft befofe the urcnase we 05 1 place 1.005 on the left as a divisor (152), (?) = 45438.77 lire, and obtain by cancellation, multiplica- tion, and division, 45438 lire 77 centimes as the proceeds of the exchange. 3. A merchant in Chicago directs his agent in Albany to draw upon Baltimore at 1 % discount, for 81200 due from the sales of produce; he then draws upon the Albany agent, at 2 % premium, for the proceeds, after allowing the agent to reserve -J % for his commission. What sum does the merchant realize from his produce ? OPERATION. ANALYSIS. According to the given / M Q __ i9oo B rates of exchange, 100 dollars in Balti- j QQ g _ QQ A more equal 99 dollars in Albany ; and 100 A - 10^ r* *^ dollars in Albany equal 102 dollars in Chicago ; and since the unit of currency = is the same in each place, being $1, we ( ? ) = $1205.70, Ans. represent its exchange value in each town by the initial letter, and make the state- ment as in the other examples. Then, since the agent is to reserve \ % commission from the avails of his draft, we place 1 . 005 = . 995 on the right as a multiplier, and obtain by cancellation (?) = $1205.70, the answer. 350 PEKCENTACE. From these principles and illustrations we have the following RULE. I. Represent the required sum by ( ? ) f with the proper unit of currency affixed, and place it equal to the given sum on the right. II. Arrange the given rates of exchange so that in any two con- secutive equations the same unit of currency shall stand on opposite sides. III. When there is commission for drawing, place 1 minus the rate on the left if the cost of exchange is required, and on the right if proceeds are required ; and when there is commission for remit- ting, place 1 plus the rate on the right if cost is required, and on the left if proceeds are required. IV. Divide the product of the numbers on the right by the product of the numbers on the left, cancelling equal factors ; the result will be the answer. NOTES. 1. Commission for drawing is commission on the sale of a draft; commis- sion for remitting is commission on the purchase price of a draft. 2. The above method is sometimes called the Chain Rule, or Conjoined Proportion. EXAMPLES FOR PRACTICE. 1. A gentleman in Philadelphia wishes to deposit $5000 in a bank at Stockholm, by remitting to Liverpool and thence to Stockholm ; if the course of exchange on Liverpool is 4.91 in Philadelphia, and the course between Liverpool and Stockholm is 18 crowns to l, how much money will the man have in bank at Stockholm, allow- ing the agent at Liverpool J^ for remitting. Ans. 18792.1 crowns. 2. When exchange at New York on Paris is 5 francs 16 centimes per $1, and at Paris on Hamburg 1.23 fr. per mark, what will be the arbitrated price in New York of 7680 marks of Hamburg? Ans. $1830.69. 3. A gentleman in Cleveland wishes to draw on New Orleans for a bank stock dividend of $750, and exchange direct on New Orleans is \\% discount; how much will he save by drawing on his agent in New York at 1 -J- % premium, allowing his agent to draw on New Orleans at 1 % discount, brokerage at % ? 4. A gentleman in Boston drew on Amsterdam for 6000 guilders at $.415 per guilder; how much more would he have received if he EXCHANGE. 351 had ordered remittance to London, and thence to New York, ex- change at Amsterdam on London being 11.19 guilders per l, and at London on New York 4.88, brokerage at lj$ in London for remitting. Ans. $94.31+. 5. If at Philadelphia exchange on Liverpool is 4.89J-, and at Liverpool on Paris 24 francs 96 J centimes per l, what is the arbi- trated course of exchange between Philadelphia and Paris, through Liverpool? Ans. 5.10. 6. An American resident of Amsterdam wishing to obtain funds from the U. S. to the amount of $6400, directs his agent in Lon- don to draw on the U. S. and remit the proceeds to him in a draft on Amsterdam, exchange on the U. S. being at 4.85 in London, and the course between London and Amsterdam being 18d. per guilder. If the agent charges commission at -% both for drawing and remitting, how much better is this arbitration than to draw directly on the U. S. at 41 cents per guilder? 7. A speculator in Pittsburgh, having purchased 58 shares of railroad stock in New Orleans, at 95 %, remits to his agent in New York a draft purchased at 2 % premium, with orders for the agent to remit the sum due in N. O. Now, if exchange on N. O. is at %fo discount in N. Y., and the agent's commission for remitting is J ( .f. how much does the stock cost in Pittsburgh ? Ans. $5606.08. 8. A merchant in Boston owes 19570 francs in Paris. Which will be the more advantageous to him, to remit directly to Paris at 5.12 or through London at 4.89, buying there exchange on Paris at 25.19 fr. to 1, and paying J^ brokerage? 9. If in London exchange on Paris is 25.71, and in New York on Paris it is 5.1 5 J, what is the arbitrated course of exchange between New York and London ? Ans. 4.987+. 10. A banker in New York remits $3000 to Liverpool, by arbitra- tion, as follows: first to Paris at 5 francs 16 centimes per $1 ; thence to Hamburg at 125 francs per 100 marks; thence to Am- sterdam at 1.71-J marks to the guilder; thence to Liverpool at 11.82 guilders per 1 sterling. How much sterling money will he have in bank at Liverpool, and what will be his gain over direct exchange at 4.91 ? ^ ns j Proceeds in Liverpool, 610 18s. 3d. ( Gain by arbitration, 10s. 9d. 352 PERCENTAGE. EQUATION OF PAYMENTS. 609. Equation of Payments is the process ' of finding the mean or equitable time of payment of several sums, due at dif- ferent times without interest. 610. The Term of Credit is the time to elapse before a debt becomes due. 611. The Average Term of Credit is the time to elapse before several debts, due at different times, may all be paid at once, with- out loss to debtor or creditor. 613. The Equated Time is the date at which the several debts may be canceled by one payment. 613. To Average an Account is to find the mean or equit- able time of payment of the balance. 614. A Focal Date is a date with which all the others are com- pared in averaging an account. NOTE. Each item of a book account draws interest from the time it is due, which may be either at the date of the transaction, or after a specified terui of credit. In averaging, there are two kinds of equations, Simple and Compound. 615. A Simple Equation is the process of finding the aver- age time when the payments or account contains only one side, which may be either a debit or credit. 616. A Compound Equation is the process of averaging when both debts and credits are to be considered. SIMPLE EQUATIONS. CASE I. 617. When all the terms of credit begin at the same date. 1. In settling with a creditor on the first day of April, I find that I owe him $12 due in 5 months, $15 due in 2 months, and $18 due in 10 months ; at what time may I pay the whole amount? EQUATION OF PAYMENTS. 353 OPERATION. ANALYSIS. The $12 X 5 GO "vrhole amount to be 15 X 2= 30 paid, as seen in the ope- lp x 10 = 180 ration, is $45 ; and we oj^ .^YQ are to find how long it 270 ---45 ="6 mo., average credit, sha11 be withheld, or Apr. 1, + 6 mo. = Oc.t. 1, Ans. what term of credit It shall have, as an equiv- alent for the various terms of credit on the different items. Now the value of credit on any sum is measured by the product of the money and time. Therefore, the credit on $12 for 5 mo. = the credit on $00 for 1 mo., because 12 X 5 = 60 X 1. In like manner, we have the credit on $15 for 2 mo. = the credit on $30 for 1 mo. ; and the credit on $18 for 10 mo. = the credit on $180 for 1 mo. Hence, by addition, the value of the several terms of credit on their respective sums equals a credit of 1 month on $270 ; and this equals a credit of 6 months on $45, because 45 X 6 = 270 X 1. Hence the following E,ULE. 1. Multiply each payment l>y its term of credit, and divide the sum of the products by the sum of the payments' the quotient will be the average term of credit. II. Add the average term of credit to the date at which all the credits begin; the result will be the equated time of payment. NOTES. 1. The periods of time used as multipliers must nil be of the same denomination, and the quotient will be of the same denomination as the terms of credit; if these be months, and there be a remainder after the division, con- tinue the division to days by reduction, always taking the nearest unit in the last result. 2 The several rules in equation of payments are based upon the principle of bank discount; for they imply that the discount of a sum paid before it is due equals, the interest of the same amount paid after it is due. EXAMPLES FOR PRACTICE. 1. On the first day of January, 1860, a man gave 3 notes, the first for $500 payable in 30 days; the second for $400 payable in 60 days; the third for $600 payable in 90 days. What was the average term of credit, and what the equated time of payment? An A Term of credit, 62 da.; time of payment, Mar. 3, 1860. 2. A man purchased real estate, and agreed to pay J of the price in 3 mo., t in 8 mo., and the remainder in 1 year. Wishing to cancel the whole obligation at a single payment, how long shall this payment be deferred ? 364 PERCENTAGE. 3. I owe $480 payable in 90 days, and $320 payable in 60 days. My creditor consents to an extension of time to 1 year, and offers to take my note for the whole amount on interest at 6 per cent, from the equated time, or a note for the true present worth of both debts, on interest from date- How much will I gain if I choose the latter condition ? Ans. $1.14. 4. Bought merchandise .April 1, as follows: $280 on 3 mo., $300 on 4 mo., $200 on 5 mo., $560 on 6 mo. ; what is the equated time of payment ? Ans. Aug. 24. CASE II. 618. When the terms of credit begin at different dates. 1. When does the amount of the following bill become due, per average ? CHARLES CROSBY, 1860. To BRONSON & Co., Dr. Jan. 12. To Mdse., $400 " 16. " Mdse. on 2 mo., 600 Apr.20, " Cash, , 375 PIRST OPERATION. SECOND OPERATION. Due Da. Items. Prod. Jan. 12 Mar. 16 Apr. 20 64 99 400 600 375 38400 37125 1375 75525 Due. Da. Items. Prod. Jan. 12 Mar. 16 Apr. 20 99 35 400 600 375 39600 21000 1375 .60600 75525 -:- 1375 = 55 da. 55 da. after Jan. 12, or Mar. 7. 60600 H- 1375 = 44 da. 44 da, before Apr. 20, or Mar. 7. ANALYSIS. The three items of the bill are due Jan. 12, Mar. 16, and Apr. 20, respectively. In the first operation we use the ear/ text maturity, Jan. 12, for a focal date, and find the difference in days between this date and each of the others ; thus, from Jan. 12 to Mar. EQUATION OF PAYMENTS. 355 16 is 64 da. ; from Jan. 12 to Apr. 20 is 99 da. Hence, from Jan. 12 the first item has no" credit, the second nas 64 days credit, and the third 99 days' credit, as appears in the column marked da. We now proceed to find the products as in Case I, whence we obtain the ave- rage credit, 55 da., and the equated time, Mar. 7. In the second operation, the latest maturity. Apr. 20, is taken for a focal date, and the work may be explained thus : Suppose the account to be settled Apr. 20. At that time the first item has been due 99 days, and must therefore draw interest for this time. But interest on $400 for 99 days = the interest on $39600 for 1 day. The second item must draw interest 35 days ; but interest on $600 foi 35 days = interest on $21000 for 1 day. Taking the sum of the products, we find that the whole amount of interest due Apr. 20 equals the interest on $60600 for 1 day ; and this is found, by division, equal to the interest on $1375 for 44 da., which is the average term of interest. Hence the account would be settled Apr 20, by paying $1375, with interest on the same for 44 days. .This shows that the $1375 has been used 44 days, that is, it falls due Mar 7, without interest. Hence we have the following RULE. I. Find the. time at which each item becomes due, by adding to the date of each transaction the term of credit, if any be specified, and write these dates in a column II. Assume either the earliest or the latest date for a focal date, and find the difference in days between the focal date and each of the other dates, and write the results in a second column. III. Write the items of the account in a third column, and mul- tiply each by the corresponding number of days in the preceding column, writing the products in a fourth column. IV. Divide the sum of the products by the sum of the items. The quotient will be the average term of credit or interest, and must be reckoned from the focal date TOWARD the other dates, to find the equated time of payment. NOTES. 1. When dollars and certs are given, it is generally sufficient to take only dollars in the multiplicand, rejecting the cents when less than 50, and carrying 1 to the dollar?, if the cents are more than 50. 2. Months in any terms of credit are understood to he calendar months ; the time must therefore he carried forward to the same day of the month in which the term of credit expires. 356 PERCENTAGE. EXAMPLES FOR PRACTICE. 1. JAMES GORDON, 1860. To HENRY LANCEY, Dr. Mar. 4. To 100 yd. Cassimere, @ $2 50, $250 " 25. " 300C <<' French Prints," .12 360 Apr. 16. 1200 " Sheeting, " .08, 96 " 30. 400 OilCloth, .50, 200 May 17. " Sundries, 350 When is the above bill due, per average ? Am. Apr. 12, 1860. 2. I sell goods to A at different times, arid for different terms of credit, as follows : Sept. 12, 1859, a bill on 30 days' credit, for $180 Oct. 7, " " 30 " 300 Nov. 16, " " 60 " 150 Dec. 20, 90 " 350 Jan. 25, 1860, 30 " 130 Feb. 24, " " 30 " " 140 If I take his note in settlement, at what time shall interest commence ? 3. What is the average of the following account ? 1860, Oct. 1. Mdse., on 60 da............ $240 " Nov. 12. " " 500 " Dec. 25. " " " 436 1861, Jan. 16. " " <" 325 " Feb. 24. " " 436 Mar. 17. " 537 Atis. Mar. 10, 1861. 4. I have 4 notes, as follows: the first for $350, due Aug. 16, 1859 the second for 8250, due Oct. 15, 1859 ; the third for 3 r) >00, due Dec. 14, 1859; the fourth for $248, due Feb. 12, 1860. When shall a note for which I may exchange the four, be made payable ? EQUATION OF PAYMENTS. 35" Dr. COMPOUND EQUATIONS. O1O. 1 Average the following account. John Lyman. Or. 18fiO. 1360. June 12 To Mdse. 530 00 | June 24 | Bv draft at 30 da. 480 00 Sept. 12 U fi 428 00 1 Aug. 20 j cash, 280 00 Oct. 28 " Sundries, 440 00 II Oct. 81 ' 140 00 OPERATION. Dr. Cr. Focal date, Due. June 12 Sept. 12 Oct. 28 Da. Items. Products. Due. Da. Items. Products. 138 46 530 428 440 73140 19688 July 27 Aug. 20 Oct. 8 93 69 20 480 230 140 44640 15S70 2800 Balances, 1398 850 548 9-2828 63310 850 63310 29518 29518 -f- 548 = 54 da., average term of interest. Oct. 28 54 da. = Sept. 4, balance due. ANALYSIS. In this operation we have written the dates of maturity on either side, allowing 3 days' grace to the draft. The latest date, Oct. 28, is assumed as the focal date for both sides, and the two columns marked da. show the difference in days between the focal date and each of the other dates. The products are obtained as in simple equations, and the balance found between the items on the two sides, and also between the products. These balances, being both on the Dr. side, show that there is due on the day of the focal date, $548, with interest on $29518 for 1 day. By division, this interest is found to be equal to the interest on $548 for 54 days. Hence this balance, $548, has been due 54 days ; and reckoning back from the focal date, we obtain the equated time of payment, Sept. 4. Had we taken the earliest maturity, June 12, for the focal date, we should have obtained 84 days for the interval of time ; and since in this case the products would represent the credit to which the several items are entitled after June 12, we should add 84 days to the focal date, which would give Sept. 4, as before. 2. When is the balance of the following account due, per average ? .358 PERCENTAGE. Charles Derby. Dr. Cr. 1859. Jan. 21 Mar. 5 " 22 To Mdse. 82 145 194 00 00 00 1859. Jan. 1 Feb. 4 Mar. 30 By cash, 84 00 " " 40 00 " " 12 f 00 OPERATION. Dr. Cr. Due. Da. Items. Products. Due. Da. Items. Products. Jan. 21 68 32 2176 Jan. 1 88 84 7392 Mar. 5 25 145 3625 Feb. 4 54 40 2160 22 8 194 1552 Mar. 30 12 371 7353 136 9552 136 7353 Balance of account, 235 Balance of products. 2199 2199 -r 235 = 9 da. ; Mar. 30 + 9 da. = Apr. 8, Ans. ANALYSIS. We take the latest maturity, Mar. 30, for the focal date, and consequently the products represent the interest due upon the several items, at that date. We find the balance of the items upon the Dr. side, and the balance of the products upon the Cr. side. The debtor therefore owes, on Mar. 30, $235, but is entitled to such a term of interest on the same as will be equivalent to the interest on $2199 for 1 day, which by division, is found to be 9 da. Hence the balance is due Mar. 30+9 da. = Apr. 8. Thus we see that when the balances are on opposite sides, the interval of time is counted from the other dates. If we take, in this example, the earliest date for the focal date, the balances will both be upon the Dr. side, and the interval of time will be 97 da., which reckoned forward from the focal date, will give the equated time as before. G2O. From these examples we derive the following RULE. I. Find the time when each item of the account is due, and write the dates, in two columns, on the sides of the account to which they respectively belong. II. Use either the earliest or the latest of these dates as the focal date for both sides, ana Jind the products as in the last case. III. Divide the balance of the products by the balance of the account ; the quotient will be the interval of time, which must be reckoned from the focal date TOWARD the other dates when both EQUATION OF PAYMENTS. 359 balances are on the same side of the account, but FROM the oilier dates when the balances are on opposite sides of the account. NOTES. 1. Instead of the products, we may obtain the interest, at any per cent., on the several items for the corresponding intervals of time, and divide the balance of interest by the interest on the balance of the account for 1 day ; the quotient will be the interval of time to be added to, or subtracted from the focal date, according to the rule. The time obtained will be the same, at what- ever rate the interest be computed. 2, There may be such a combination of debits and credits, that the equated time will be earlier or later than any date of the account. EXAMPLES FOR PRACTICE. 1. Required, the average maturity of the following account. Dr. A. Z. Armour. Cr. 1859. 11859. I Feb. 12 To Mdse. 85 March 15 By bal. old acc't. 97 1 36 25 .4 U 36 April 17 " cash, 56 | 00 April 16 (( 174 May 25 li U 25 00 May 20 94 June 8 " sundries. 94 1 75 Dr. OPERATION. Cr. Due. Da. Items. Int Due. Da. Items. Int. Feb. 12 25 April 16 May 20 116 103 53 19 85.75 36.24 17496 94.78 1.66 .62 i 1.55 .30 March 15 April 17 May 25 June 8 85 52 14 97.36 56.00 25.00 94.75 1.33 .49 .06 Balances, 391.73 273.11 4.13 1.93 273.11 1.93 118.62 2.20 Int on $118.62 for 1 da. = $,0198. 2.20-=-.0198=lll da.; June 8111 da.=Feb. 17,1859, Am. ANALYSIS. Taking the latest maturity, June 8, for the focal date, we find the interest of each item, at 6 c /o, from its maturity to the focal date ; then, taking the balance, we find the interest due on the account to be $2.20. Dividing this interest by the interest on the balance of the items for 1 day, we obtain 111 da., the time required for the interest, $2.20, to accrue. The average maturity, therefore, is June 8 111 da. = Feb. 17, 1859. It is evident that when the balances occur on opposite sides, the interval of time will be reckoned as in the method by products. 360 PERCENTAGE. 2. What is the balance of the following account, and when is it due ? Thomas Lardner. Dr. * Cr. 1800. I 1860. March 1 To Sundries, 436 00 March 25 By draft, at 60 da. 400 00 April 12 " Mdse. 548 00 April 6 " 30 " 650 00 July 16 it a 312 00 June 20 " cash, 200 00 Sept. 14 " " 536 00 Aug. 3 84 00 Ans. Balance, $498; due June 22, 1860. 3. When shall a draft for the settlement of the following ac- count be made payable ? David Sanford. Dr Cr. 1859. 1859. Jan. 1 To Mdse. on 3 mo. 54 36 April 1 By cash. 50 00 Feb. 12 ' " " 2 ' 28 45 May 16 draft, at 30 da. 30 00 March 16 " Sundries, 95 75 June 12 H 128 00 June 25 " Mdse. 26 32 " 20 " cash, 150 00 Ans. Aug. 28, 1859. 2>r. Oliver Wainwright. Cr. 1858. 1858. Jan. 1 To Mdse. 36 72 Jan. 10 By cash, 98 72 Feb. 1 a a 48 25 21 tt n, 25 1 84 March 17 ii 72 ' 36 March 23 " sundries, 15 1 17 April 1 (i 98 48 April 6 " " S 1 96 If the above account were settled April 6, 1858, by draft on time, how many days' credit should be given ? Ans. 20 da. 5. I owe $1000 due Apr. 25. If I pay $560 Apr. 1, and $324 Apr. 21, when, in equity, should I pay the balance ? Ans. Aug. 30. NOTE. Make the $1000 the Dr. side of an account, and the payments the Cr. ide, and then average. 6. A man owes $684, payable Aug. 12, and $468, payable Oct. 15. If he pay $839 ; Aug. 1, what will be the equated time for the payment of the balance ? Ans. Dec. 15. 7 . A man holds 3 notes, the first for $500, due March 1, the second for $800, due June 1, and the third for $600, due Aug. 1. He wishes to exchange them for two others, one of which shall be for $1000, payable Apr. 1 ; what shall be the face and when the maturity of the other ? Ans. Face. $900 ; maturity, July 28. EQUATION OF PAYMENTS. 361 8. A owes $500, due Apr 12, and $1000, due Sept. 20, and wishes to discharge the obligation by two equal payments, made at an interval of 60 days ; when must the two payments be made ? Ana, 1st, June 28; 2d, Aug. 27. 9. When shall a note be made payable, to balance the following account ? James Tyler. Dr. CP. 1859. 1859. June 12 To Mdse. on 3 mo. 530 84 Sept. 14 By cash, 436 00 " 20 u u *.t tt 236 48 " 25 *. u 320 00 < ; 30 it t: 739 56 Oct. 3 660 00 July 5 ft it 273 44 " 17 u 370 00 " 16 tt <; it 194 | 78 Nov. 16 840 00 " 29 ti <( tt 536 I 42 1 24 " " 560 00 10. I received goods from a wholesale firm in New York, in parcels, as per bills received, namely : Apr. 1, a bill for $536.78 ; May 16, $2156.94; June 12, $843.75; July 12, $594.37; Sept 18, $856.48. In part payment, I remitted cash as follows : June 3, $500; July 1, $1000; Nov. 1, $1500. When is the balance payable, allowing credit of 2 months for the merchandise ? Ans. July 23. ACCOUNT SALES. G21. An Account Sales is an account rendered by a commis- sion merchant of goods sold on account of a consignor, and con- tains a statement of the sales, the attendant charges, and the net proceeds due the owner. G22* Guaranty is a charge made in addition to commission, for securing the owner against the risk of non-payment, in case of goods sold on credit G23. Storage is a charge made for keeping the goods, and may be reckoned by the week or month, on each article or piece. G24L. Primage is an allowance paid by a shipper or consignor of goods to the master and sailors of a vessel, for loading it G^5o A commission merchant having sold a shipment of goods by parts at different times, and on various terms, makes a final settlement by deducting all charges, and accrediting the owner with the net proceeds. It is evident, therefore, 31 362 PERCENTAGE. I. That commission and guaranty should be accredited to the agent at the average maturity of the sales. II. That the net proceeds should be accredited to the con- signor at the average maturity of the entire account. Hence the following RULE. I. To compute the storage. Multiply each article or parcel by the time it is in store, and multiply the sum of the pro- ducts by the rate per unit; the result will be the storage. II. To find when the net proceeds are due. Average the sales alone, and the result will be the date to be given to the commission and guaranty ; then make the sales the Or. side, and the charges the Dr. side, and average the entire account by a compound equation. NOTE. In averaging, either the product method or the interest method may be used. EXAMPLES FOR PRACTICE. 1. Account sales of 100 pipes of gin, received per ship Hispan- iola, from Havana, on a|c. of Tyler, Jones & Co. I860. April 15 Sold 3 9 Pipe*" 4160 gal @ $105 OD 30 4368 00 May 5 " 40 " 5240 " (3), 1 0'2, cash,. 53-4 80 <' 28 " 3650 " @ 1 OJ " 3650 00 loo 13362 80 April 1 " 1 CHARGES. To Freight and Primage, $136.76 48 54 3207.07 June 28 " StorW from April 1, viz. : On 32 Pipes, 2 wks 64 wks. " 40 " 5 " ... 200 23 " 13 " ... 364 " 100 " cq nil to 628 " @ G c 37 68 *' Commission on $13362 FO at '2% % 33407 " Girinnty on $4368 at 2V % 10920 3873 32 What are the net proceeds of the above account, and when due ? Ans. Net proceeds, $9489.48 ; due, May 20, 1860. NOTE. The time for which storage is charged on each part of the shipment is the internal, reduced to weeks, between Apr. 1, when the pipes were received into store, anO the date of sale. Every fraction of a week is reckoned a full week. 2. A commission merchant in Boston received into his store on May 1, 1859, 1000 bbl. of flour, paying as charges on the same EQUATION OF PAYMENTS. day, freight $175.48, cartage $56.25, and cooperage $8.37. He sold out the shipment as follows: June 3, 200 bbl. @ $6.25; June 30, 350 bbl. @ $6.50; July 29, 400 bbl. @ $6.12 J; Aug. 6, 50 bbl. @ $6.00. Kequired the net proceeds, and the date when they shall be accredited to the owner, allowing commission at 3'i fa) and storage at 2 cents per week per bbl. Ans. Net proceeds, $5614.28 ; due, July 10. SETTLEMENT OF ACCOUNTS CURRENT. To find the cash balance of an account current, at any given date. /. Burns in account current with Tyler & Co. Dr. Or. I860. 1860. Feb. 25 To Mdse. on 3 mo. 360 75 March 1 By cash on ncct. 250 00 March 20 3 240 50 April 20 ' accept, at 30 da. 300 00 April 26 " " 3 875 24 June 12 " Sundries. 375 CO June 24 K u 2 " 235 25 27 " cash on acct. 4UO 00 Required the cash value of the above account, July 1, 1860, interest at 6 . OPERATION. Dr. Cr. Due. Da. Items. Int. Cash ral Due. Da. Items. Int. Cash val. May 25 June 20 July 26 Aug. 24 37 11 25 54 360.75 + 2.22 240.56 -(- .44 87 0.24 3.05 235.252.12 362.97 241.00 871.59 233.13 March 1 May 20 June 12 " 7 122 42 19 4 250.00 + 5 OS 300.00 + 210 o7-"..00 + 1.19 400.00 + .27 255.08 302.10 376.15) 400.27 1708.69 1333.64 $1708.69 $1333.64 = $375.05. Ans. ANALYSIS. For either side of the account we write the dates at which the several items are due, and the days intervening between these dates and the day of settlement, July 1. We then compute the interest on each item for the corresponding interval of time, and add it to the item if the maturity is before July 1, and subtract it from the item if the maturity is after July 1 ; the results must be the cash values of the several items on July 1. Adding the two columns of cash values, and subtracting the less sum from the greater, we have $375.05. the cash balance required. Hence the 364 PARTNERSHIP. RULE. I. Find the number of days intervening between each maturity and the day of settlement. II. Compute the interest on each item for t\ie corresponding interval of time ; add the interest to the item if the maturity is before the day of settlement, and subtract it from the item if the maturity is after the day of settlement ; the results will be the cash Values of the several items. III. Add each column of cash values, and the difference of the amounts will be the cash balance required. EXAMPLES FOR PRACTICE. 1. Find the cash balance of the following account for June 1, 1858, interest at 6 per cent. ? Alvan Parke. Dr. Cr. 1858. 1858. Jan. 12 To check, 500 36 Jan. 1 By bal. from old acct. 536 72 26 (i 250 48 Feb. 3 cash, 486 57 Fob. 13 a u 400 00 March 26 1200 78 March 16 a (t 750 00 April 20 i. 756 36 April 25 it (i 200 00 May 12 c. ti 248 79 Ans. $1106.67. 2. What is the cash balance of the following account on Dec. 31, at 7 per cent. ? James Hanson. Dr. Cr. 1859. ft59. Sept. 3 Oct. 2 To Sundries, " Mdse. on 3. mo. 478 256 36 37 Sept. 17 " 20 By Sundries, " cash on acct. 96 200 54 00 ' 21 .; u 3 375 26 Oct. 3 (( <: t( 325 00 Nov. 12 c. li ti 3 (( 80 00 Nov. 17 i: (l 50 no Dec. 15 " Sundries, 148 76 Dec. 27 ' ' " 84 00 PARTNERSHIP. Partnership is a relation established between two or more persons in trade, by which they agree to share the profits and losses of business according to the amount of capital furnished by each, and the time it is employed. The Partners are the individuals thus associated. NOTE. The terms Capital or Stock, Dividend, and Assessment, have the same lignification in Partnership as in Stocks. PARTNERSHIP. 365 CASE I 629. To find each partner's share of the profit or loss, when their capital is employed for equal periods of time. 1. A and B engage in trade; A furnishes $500, and B $700 as capital ; they gain $96 ; what is each man's share ? OPERATION. ^ ANALYSIS. The whole $ 500 amount of capital em- $ 700 ployed is $500 + $700 $1200, whole stock. =$1200 ; hence, A fur- J LOO = 5 A ' s part of t he stock. nishes TS% =ir f the _7j)0 _ jf B's " " " " capital, and B furnishes $9(Px T we have 4, 1, 2, and 3 for the proportional quantities required. If we compare the prices 7 and 11 fcr the first couplet, and the prices 8 and 14 for the second couplet, as in the second operation, we shall obtain 1 , 4, 3,,, and 2 for the proportional terms. It will be seen that in comparing the simples of any couplet, one of which is greater and the other less than the mean rate, the pro- portional number finally obtained lor either term is the difference between the mean rate and the other term. Thus, in comparing 7 and 14, the proportional number corresponding to the former simple is 4, which is the difference between 14 and the mean rate 10 ; and the proportional number corresponding to the latter simple is 3, which is the difference between 7 and the mean rate. The same is true of every other couplet. Hence, when the simples and the mean rate are integers, the intermediate steps taken to obtain the final pro- portional numbers as in columns 1, 2, 3, and 4. may be omitted, and the same results readily found by taking the difference between each simple and the mean rate, and placing it opposite the one with which it is compared. From these examples and analyses we derive the following RULE. I. Write the prices or qualities of the several ingre- dients in a column, and the mean price or quality at the left. II. Consider any two prices, one of which is less and the other greater than the mern rate, as forming a couplet ; find the differ- ence between each of these prices and the mean rate, and write the reciprocal of each difference opposite the given price in the couplet, as one of the proportional terms. In like manner form the couplets, till all the prices have been employed, writing each pair of propor~ tional terms in a separate column. III. If the proportional terms thus obtained are fractional, mul' tiply each pair by the least common multiple of their denominators, and carry these integral products to a single column, observing to add any two or more that stand in the same horizontal line ; the final results will be the proportional quantities required. NOTES. 1. If the numbers in any couplet or column have a common factor, it may be rejected. 32 374 ALLIGATION. 2. We may also multiply the numbers in any couplet or column by any ran!- tiplier we choose, without affecting the equality of the gains and losses, and thus obtain an indefinite number of results, any one of which being taken will give a correct final result. \ EXAMPLES FOE PRACTICE. 1. What quantities of flour worth $5, $6, and $7f per barrel, must be sold, to realize an average price of $6 per barrel? OPERATION. ANALYSIS. Comparing the 6*4 6 II 4 4| I 4 12 2 2 12 4 first price with the third, we ob- tain the couplet 5 to -f ; and com- paring the second price with the third, we obtain the couplet 4 to . Reducing these proportional terms to integers, we find that we may take 4 barrels of the first kind with 2 of the third, and 12 of the second kind with 2 of the third ; and these two combinations taken together give 4 of the first kind, 12 of the second, and 4 of the third. 2. How much sugar worth 5 cts., 7 cts., 12 cts., and 13 cts. per pound, will form a mixture worth 10 cts. per pound? 3 Ib. of each of the first and third kinds, 2 Ib. of the second, and 5 Ib. of the fourth. 8. How can wine worth $.60 $.90 and $1.15 per gallon be mixed with water so as to form a mixture worth $.75 a gallon ? . ( By taking 3 gal. of each of the first two kinds of 1 wine, 15 gal. of the third, and 8-gal. of water. 4. A farmer has 3 pieces of land worth $40, $60, and $80 an acre respectively How many acres must he sell from the dif- ferent tracts, to realize an average price of $62.50 an acre? 5. How much wine worth $.60, $.50, $.42, $.38, and $.30 pe? pint, will make a mixture worth $.45 a pint ? 6. What relative quantities of silver | pure, | pure, and fy pure, will make a mixture | pure ? Ans. 3 Ib. | pure, 3 Ib. | pure, and 20 Ib. -ft pur3. CASE III. O3*>. When two or more of the quantities are re- quired to be in a certain proportion. 1- A farmer having oats worth $.30, com worth $.00, and wheat ALLIGATION. _ 375 worth $1.10 per bushel, desires to form a mixture worth $.50 pei bushel, which shall contain equal parts of corn and wheat ; in what proportion shall the ingredients be taken ? OPERATION. ANALYSIS. "VVe first obtain 30 50 -j C 60 gillO ill!! A G the proportional terms in col- umns 3 and 4, by Case II. Now, it is evident that the loss and gain will be equal if we take each couplet, or any mul- tiple of each, alone; or both couplets, or any multiples of both, together. Multiplying the terms in column 4 by 2, we obtain the terms in column 5 ; and adding the terms in columns 3 and 5, we obtain the terms in column 6 ; that is, the farmer takes 7 bushels of oats to 2 of corn and 2 of wheat, which is the required proportion. Hence the following RULE. I. Compare the given prices, and obtain the proportional terms by couplets, as in Case II. II. Reduce the couplets to higher or lower terms, as may be re- quired; then select the columns at pleasure, and combine them by adding the terms in the same horizontal line, till a set of pro- portional terms is obtained, answering the required conditions. EXAMPLES FOR PRACTICE. 1. A grocer has four kinds of molasses, worth $.25, $.50, $.62, and $.70 per gallon, respectively ; in what proportions may he mix the four kinds, to obtain a compound worth $.58 per gallon, using equal parts of the first two kinds? Ans. 4, 4, 8 and 11. 2. In what proportions may we take sugars at 7 cts., 8 cts., 13 cts., and 15 cts., to form a compound worth 10 cts. per pound, using equal parts of the first three kinds ? Ans. 5, 5, 5 and 2. 8 A miller has oats at 30 cts., corn at 50 cts., and wheat at 100 cts. per bushel. He desires to form two mixtures, each worth 70 cts. per bushel. In the first he would have equal parts of oats and corn, and in the second, equal parts of corn and wheat; what must be the proportional terms for each mixture ? . ( For the first mixture, 1, 1 and 2. ( For the second mixture, 1, 4 and 4. 376 ALLIGATION. CASE IV. 636. When the quantity of one of the simples is limited. 1 A miller has oats worth $.28, corn worth $.44, and barley- worth $.90 per bushel. He wishes to form a mixture worth $.58 per bushel, and containing 100 bushels of corn. How many bushels of oats and barley may he take ? OPERATION. ANALYSIS. By Case r28 1 58 J 44] | (93JJ 3 1 !, A 7 7 1140 5 6 I 2 100 160 II, we find the pro- portional quantities to be 7 bushels of oats to 5 of corn and 8 of barley. But as 100 bushels of corn, instead of 5, are required, we must take *f = 20 times each of the other ingredients, in order that the gain and loss may be equal ; and we shall therefore have 7 X 20 = 140 bushels 01 oats, and 8 X 20 = 160 bushels of barley. Hence the following RULE. Find the proportional quantities by Case II or Case III. Divide the given quantity by the proportional quantity of this ingredient, and multiply each of the other proportional quan- tities by the quotient thus obtained. EXAMPLES FOR PRACTICE. 1. A dairyman bought 10 cows at $20 a head ; how many must he buy at $16, $18, and $24 a head, so that the whole may cost him an average price of $22 a head ? Ans. 10 at $16, 10 at $18, and 60 at $24. 2. Bought 12 yards of cloth for $15; how many yards must I buy at $lf, and $| a yard, that the average price of the whole may be $li ? Ans. 12 yards at $lf and 16 yards at $|. 3. How much water will dilute 9 gal. 2 qt. 1 pt. of alcohol 96 per cent, strong to 84 per cent. ? Ans. 1 gal. 1 qt. 1 pt. 4. A grocer mixed teas worth $.30, $.55, and $.70 per pound respectively, forming a mixture worth $.45 per pound, having equal parts of the first two kinds, and 12 Ibs. of the third kind; how many pounds of each of the first two kinds did he take ? ALLIGATIOK. 377 CASE V. 637. When the quantities of two or more of the in- gredients are limited. 1. How many bushels of rye at $1.08, and of wheat at $1.44, must be mixed with 18 bushels of oats at $.48, 8 bushels of corn at $.52, and 4 bushels of barley at $.85, that the mixture may be worth $.84 per bushel ? OPERATION. ANALYSIS. Of the given ciOvxio CQ4 quantities there are 18 -f ? 52 x 8 - 4 16 8 + 4 = 30 * " > ' mean or average prce we .80 X 4 == V 30 ) $16.2( We are therefore required to Mean price of the ) * r mix 30 bushels of grain given simples j worth $.54 per bushel, with I 3^ ] 4 ! 2 6 I 30 1 rye at $1.08, and wheat at 5 5 1 25 $1.44, to make a compound 1 1 ' 5 worth $.84 per bushel. Pro- 54 84-! 108 ceeding as in Case IV, we find there will be required 25 bushels of rye, and 5 bushels of wheat. Hence the following RULE. Consider those ingredients whose quantities and prices are given as forming a mixture, and find their mean price ty Case I; then consitler this mixture as a single ingredient ivhose quantity and price are known, and fold the quantities of the other ingredients ly Case IV. EXAMPLES FOR PRACTICE. 1. A gentleman bought 7 yards of cloth @ $2.20, and 7 yards @ $2; how much must he buy @ $1.60, and @ $1.75 that the average price of the whole may be $1.80 ? 2. How much wine, at $1.75 a gallon, must be added to 60 gal- lons at $1.14, and 30 gallons at $1.26 a gallon, so that the mixture may be worth $1.57 a gallon ? Ans. 195 gallons. 3. A farmer has 40 bushels of wheat worth $2 a bushel, and 70 bushels of corn worth $ a bushel. How many oats worth $J a bushel must he mix with the wheat and corn, to make the mix- ture worth $1 a bushel ? Ans. 6| bushels. 32* 378 ALLIGATION. CASE VI. 638. When the quantity of the whole compound is limited. 1. A tradesman has three kinds of tea rated at $.30, 8.45, and 6.60 per pound, respectively; what quantities of each should he take to form a mixture of 72 pounds, worth $.40 per pound? OPERATION. ANALYSIS. By Case II, 2 3 4 5 6 2 1 3 36 2 2 24 1 1 12 6 ~T2 we find the proportional quantities to form the mixture to be 3 Ib. at $.30, 2 Ib. at $.45, and GO ^V 1 1 12 1 lb - at $- 6() - Adding these proportional quanti- ties, we find that they would form a mixture of 6 pounds. And since the required mixture is 7 g 2 = 12 times 6 pounds, we multiply each of the proportional terms by 12, and obtain for the required quantities, 36 Ib. at $.30, 24 Ib. at $.45, and 1 2 Ib. at $.60. Hence the following RULE. Find the proportional numbers as in Case II or Case III. Divide the given quantity by the sum of the proportional quantities, and multiply each of the proportional quantities by the quotient thus obtained. EXAMPLES FOR PRACTICE. 1. A grocer has coffee worth 8 cts., 16 cts., and 24 cts. per pound respectively ; how much of each kind must he use, to fill a cask holding 240 Ib, that shall be worth 20 cts. a pound ? Ans. 40 Ib. at 8 cts , 40 Ib. at 16 cts., and 160 Ib. at 24 cts. 2. A man bought calves, sheep, and lambs, 154 in all, for $154. He paid $3 for each calf, $1* for each sheep, and $ for each lamb ; how many did he buy of each kind ? Ans. 14 calves, 42 sheep, and 98 lambs. 3. A man paid $165 to 55 laborers, consisting of men, women, and boys; to the men he paid $5 a week, to the women $1 a week, nnd to the boys $ a week ; how many were there of each ? Ans. 30 men, 5 women, and 20 boys. INVOLUTION. 379 INVOLUTION. A Power is the product arising from multiplying a number by itself, or repeating it any number of times as a factor," 64O Involution of the process of raising a number to a given power. -U. The Square of a number is its second power. G42; The Cube of a number is its third power. 643. In the process of involution, we observe, L That the exponent of any power is equal to the number of times the root has been taken as a factor in continued multiplica- tion. Hence II. The product of any two or more powers of the same num- ber is the power denoted by the sum of their exponents, and III. If any power of a number be raised to any given power, the result will be that power of the number denoted by the pro- duct of the exponents. 1. What is the 5th power of 6 ? OPERATION. ANALYSIS. W& 6x6x6x6x6 = 7776, Ans. multiply G by ifc- Or, G2 lf an d this pro- 6 x 6 = 6 2 = 36 duct bv G > and G(> 3G x (5 _ 53 __ 216 on until 6 hag 6 3 x 6 ? = 6 5 = 216 x 36 = 7776, Ans. been taken 5 times in continued mul- tiplication ; the final product, 7776, is the power required, (I). Or, we may first form the 2d and 3d powers ; then the product of these two powers will be the 5th power required, (II). 2- What is the 6th power of 12 ? OPERATION. ANALYSIS. We find the cube of the second power, which must bo 144> = 2980984, Ans. 644, Hence for the involution of numbers we have the fol- lowing 380 INVOLUTION. RULE. L Multiply the given number by itself in continued multiplication, till it has been taken as many times as a factor as there are units in the exponent of the required power. Or, II, Multiply together two or more powers of the given number, the sum of whose exponents is equal to the exponent of the required power. Or, III Raise some power of the given number to such a power ^that the product of the two exponents shall be equal to the exponent of the required power. NOTES. 1. A fraction is involved to any power by involving each of its terms separately to the required power. 2 Mixed numbers should be reduced to improper fractions before involution. 3. When the number to be involved is a decimal, contracted multiplication may be applied with great advantage. EXAMPLES FOR PRACTICE. 1. What is the square of 79? Ans 6241. 2. What is the cube of 25.4 ? Ans. 16387.064. 3. What is the square of 1450 ? 4. Raise 16 J to the 4th power. Ans. 79659f 1 5. Raise 2 to the 20th power. Ans. 1048576 6. Raise .4378565 to the 8th power, reserving 5 decimals Ans. .00135 -b 7. Raise 1.052578 to the 6th power, reserving 4 decimals Ans. 1.3600 8. Involve .029 to the 5th power ? Ans. .000000020511149. Find the value of each of the following expressions : 9. 4.367* Ans. 363.691178934721, 10 (J) 3 . Ans 13. 11. (2J) Ans. 7ft. 12 4.6 3 x 25 3 Ans. 1520875. 13. (6f ) 4 7.25 a . 14 (8) 3 x 2.5 2 15. I of (j) 3 of (34)'. Ans. 5f . NOTE. Cancel like powers of the same factor. 16. 7-_3.08. 17 (4" x 5 6 x 12 s ) -f- (4* x 10* x 3 2 ). Ans. 1200 EVOLUTION. 381 EVOLUTION. 645. A Root is a factor repeated to produce a power ; thus, in the expression 7x7x7 = 343, 7 is the root from which the power, 343, is produced. 646. Evolution is the process of extracting the root of a number considered as a power ; it is the reverse of Involution. Any number whatever may be considered a power whose root is to be extracted. 647. A Rational Root is a root that can be exactly obtained. 648. A Surd is an indicated root that can not be exactly ob- tained. 649. The Radical Sign is the character, ^/, which, placed before a number, indicates that its root is to be extracted. 6oO. The Index of the root is the figure placed above the radical sign, to denote what root is to be taken. When no index is written, the index, 2, is always understood. 651* The names of roots are derived from the corresponding powers, and are denoted by the indices of the radical sign. Thus, V'lUO denotes the square root of 100, v/100 denotes the cube root of 100; VlUU denotes the fourth root of 100; etc. 6d2. Evolution is sometimes denoted by a fractional exponent, the name of the root to be extracted being indicated by the deno- minator. Thus, the square root of 10 may be written 10 ; the cube root of 10, 10 , etc. 633. Fractional exponents are also used to denote both invo- lution arid evolution in the same expression, the numerator indi- cating the power to which the given number is to be raised, and the denominator the root of the power which is to be taken ; thus, .2. 7 denotes the cube root of the second power of 7, and is the same as v/7 2 ; so also 7~ = v/7 5 . 634. In extracting any root of a number, any figure cr figures may be regarded as tens of the next inferior order. Thus, in 2546, the 2 may be considered as tens of the 3d order, the 25 as tens of the second order, or the 254 as tens of the first order. 382 EVOLUTION. SQUARE ROOT. 655. The Square Root of a number is one of the two equal factors that produce the number. Thus, the square root of 64 is 8, for 8x8== 64. To derive the method of extracting the square root of a num- ber, it is necessary to determine 1st. The relative number of places in a number and its square root 2d. The relations of the figures of the root to the periods of the number. 3d. The law by which the parts of a number are combined in the formation of its square ; and 4th. The factors of the combinations. 656. The relative number of places in a given number and its square root is shrwn in the following illustrations. Koots Squares. Ecots. Squares. 1 1 1 1 9 81 10 1,00 99 98,01 100 1,00.00 999 99,80,01 1000 1,00,00,00 From these examples we perceive 1st. That a root consisting of 1 place may have 1 or 2 places in the square. 2d. That in all cases the addition of 1 place to the root adds 2 places to the square. Hence, I. If we point of a number into two-figure periods, commencing at the right hand, the number of fall periods and the left hand full or partial period will indicate the number of places in the square root. To ascertain the relations of the several figures of the root to the periods of the number, observe that if any number, as 2345, be de- composed at pleasure, the squares of the left hand parts will be re lated in local value as follow3 : 2000 2 = 4 00 00 00 2300 2 = 5 29 00 00 2340 2 = S 47 56 00 2345 2 = 5 49 90 25: Hence, II. The square of the first figure of the root is contained wholly in the first period of the power ; the square of the first two figures SQUARE ROOT. 383 of the root is contained wholly in the first two periods of the power ; and so on. NOTE. -The periods and figures of the root are counted from the left hand. The combinations in the formation of a square may be shown as follows : If wo take any number consisting of two figures, as 43, and decom- pose it into two parts, 40 + 3, then the square of the number may be formed by multiplying both parts by each cart separately : thus, 40 + 3 40 + 3 120 + 9 160C + 120 43* = 1600 + 240 + 9 = 1849. Of these combinations, we observe that the first, 1600, is the square of 40 the second, 240, is twice 40 multiplied by 3 ; and the third, 9, is the square of 3. Hence, III. The square of a number composed of tens and units is equal to the square of the tens, plus twice the tens multiplied by the units, plus the square of the units. By observing the manner in which the square is formed, we per- ceive that the unit figure must always be contained as a factor in both the second and third parts ; these parts taken together, may therefore be factored, thus, 240 + 9 = (80 + 3) X 3. Hence, IV. If the square of the tens be subtracted from the entire square, the remainder will be equal to twice the tens plus the units multiplied by the units. 1. What is the square root of 5405778576 ? OPERATION. ANALYSIS. Pointing off the 5405778576 ( 73524 given number into periods of 49 two figures each, the 5 periods 143 505 show that there will be 5 fig- 429 ures in the root, (I). Since lQ5~~ 7(377 the square of the first figure 7325 of the root is always contained 14709 35^85 wholly in the first period of 29404 the power, (II), we seek for the 147044" ~ '~588176 reatest sc * ua] ; e in thc first ^ 588176 riod, 54, which we find by trial to be 49, and we place 384 EVOLUTION. its root, 7, as the first figure of the required root, and regard it as tens of the next inferior order, (II). We now subtract 49, the square of the first figure of the root, from the first period, 54, and bringing down the next period, obtain 505 for a remainder. And since the square of the first two figures of the root is contained wholly in the first two periods of the power, (II), the remainder, 505, must contain at least twice the first figure (tens) plus the second figure (units), multiplied by the second figure, (TV). New if we could divide this remainder by twice the first figure plus the second, which is one of the factors, the quotient would be the second figure, or the other factor. But since we have not yet obtained the second figure, the complete divisor can not now be employed , and we therefore write twice the first figure, or 14, at the left of 505 for a trial divisor, re- garding it as tens. Dividing the dividend, exclusive of the right hand figure, by 14, we obtain 3 for the second, or trial figure of the root, which we annex to the trial divisor, 14, making 143, the com- plete divisor. Multiplying the complete divisor by the trial figure 3, and subtracting the product from the dividend, we have 7 6 for a remainder. We have now taken the square of the first two figures of the root from the first two periods ; and since the square of the first three figures ot the root is contained wholly in the first three periods, (II) we bring down the tnird period, 77. to the remainder. 76, and obtain for a new dividend 7677, which must contain at least twiceihe two figures already found plus the third, multiplied by the third, (IV). Therefore to obtain the third figure, we must take for a new trial divisor twice the two figures, 73, considered as tens of the next infe- rior order, which we obtain in the operation by doubling the last fig- ure of the last complete divisor, 143, making 146. Dividing, we ob- tain 5 for the next figure of the root ; then regarding 735 as tens of the next inferior order, we proceed as ; n the former steps, and thus continue till the entire root, 73524, is obtained. G57. From these principles and illustrations we derive the following RULE. 1. Point off the given number into periods of two figures each, counting from units' place toward the left and right. II. Find the greatest square number in the left hand period, and write its root for the first figure in the root ; subtract the square number from the left hand period, and to the remainder briny down the next period for a dividend* SQUARE ROOT. 3S5 III. At the left of the dividend write twice the first figure of the root , for a trial divisor; d/3858.07694409~64. Ans. 62.11342. 8. %/ Ans. .745355 + . 9. x/99225 63504. 10. v/,126736 >/.045369. 11- ^f X V^fg. Ans. jf. 12. %/81 a x 625 x 2*. Ana. 202500. 33 z 386 EVOLUTION. CONTRACTED METHOD. G58. 1. Find the square root of 8, correct to 6 decimal places. OPERATION. ANALYSIS. Extracting the square |2.828427-f-, Ans. root in the usual way until we have 8 000000 obtained the 4 places, 2.828, the 4 corresponding remainder is 2416, and 48 40Q the next trial divisor, with the cipher 384 omitted, is 5656. We now omit to 562 1600 bring down a period of ciphers to 1124 the remainder, thus contracting the 5~648 47600 dividend 2 places ; and we contract 45184 the divisor an equal number of places 5656 2416* ^y omitting to annex the trial figure 2262 f tne root an( * regarding the right "566 hand figure, 6, as a rejected or re- 113 dundant figure We now divide as cZ 77 in contracted division of decimals, 4Q (226), bringing down each divisor in its place, with one redundant figure increased by 1 when the rejected figure is 5 or more, and carrying the tens from the redundant figure in multiplication. We observe that the entire root, 2.828427+5 contains as many places as there are places in the periods used. Hence the following RULE. I If necessary, annex periods of ciphers to the given number, and assume as many figures as there are places required in the root; then proceed in the usual manner until all the assumed figures have been employed ', omitting the remaining figures, if any. II. Form the next trial divisor as usual, but omit to annex to it the trial figure of the root, reject one figure from the right to form each subsequent divisor, and in multiplying regard the right hand figure of each contracted divisor as redundant. NOTES. 1. If the rejected figure is 5 or more, increase the next left hand figure by 1. 2. Always take full periods, both of decimals and integers. EXAMPLES FOR PRACTICE. 1, Find the square root of 32 correct to the seventh decimal place. Ans. 5.6568542+ . CUBE ROOT. 387 x 2. Find the square root of 12 correct to the seventh decimal place- Ana. 3.4641016+ . 3. Find the square root of 3286.9835 correct to the fourth decimal place. Ans. 57.3322 + . 4 Find the square root of .5 correct to the sixth decimal* place. Ans. .745355 + . 5 Find the square root of 6j correct to the sixth decimal place. Ans. 2.563479 + . 6. Find the square root of 1.06 8 correct to the sixth decimal place. Ans. 1.156817 + . 7. Find the value of 1.0125 3 correct to the fourth decimal place. Ans. 1.0188+. 8. Find the value of 1.023375* correct to the sixth decimal place. Ans. 1.011620 . CUBE ROOT. G5O. The Cube Root of a number is one of the three equal factors that produce the number. Thus, the cube root of 343 is 7, since 7x7x7 = 343. To derive the method of extracting the cube root of a number, it is necessary to determine 1st. The relative number of places in a given number and its cube root. 2d. The relations of the figures of the root to the periods of the number. 3d. The law by which the parts of a number are combined in the formation of a cube j and 4th. The factors of these combinations. OOO. The relative number of places in a given -number and its cube, is shown in the following illustrations : Roots. Cubes. Roots. Cubes. 11 11 9 729 10 1,000 99 907,299 100 1,000,000 999 997,002,999 1000 1,000,000,000 From these examples, we perceive, 8S8 EVOLUTION. 1st. That a root consisting of 1 place may have from 1 to 3 places in the cube. 2d. That in all cases the addition of 1 place to the root adds 3 places to the cube. Hence, I. If we point off a number into three-figure periods, com- mencing at the right hand, the number of full periods and the left hand full or partial period will indicate the number of places in the cube root. To ascertain the relations of the several figures of the root to the periods oi the number, observe that if any number, as 5423, be de- composed, the cubes of the parts will be related in local value, as follows : 5000 s = 125 000 000 000 5400 s = 157 464 000 000 5420 s == 159 220 088 000 5423 s = 159 484 621 967. Hence, II The cube of the first figure of the root is contained wholly in the first period of the power ; the cube of the first two figures of the root is contained wholly in the first two periods of the power; and so on To learn the combinations of tens and units in the formation of a cube, take any number consisting of two figures, as 54, and decom- pose it into two parts, 50+4; then having formed the square by 656, III, multiply each part of this square by the units and tens of 54 separately, thus, 54 2 = 50 2 + 2 x 50 x 4 + 4 50+4 50 2 x 4 + 2 x 50 x 4 2 + 4 3 50 3 + 2 X 50* x 4 + 50 x 4* 54 3 = 50 3 + 3 X 50 2 X 4 + 3 x 50 X 4 2 + 4 3 = 156924 Of these combinations, the first is the cube of 50, the second is 3 times the square of 50 multiplied by 4, the third is 3 times 50 multi- plied by the square of 4, and the fourth is the cube of 4. Hence, III. The cube of a number composed of tens and units is equal to the cube of the tens, plus three times the square of the tens multi- plied by the units,, plus three times the tens multiplied by the square of the units, plus the cube of the units. By observing the manner in which the cube is formed, we perceive that each of the last three parts contains the units as a factor; these CUBE ROOT. 389 parts, considered as one number, may therefore be separated into two factors, thus, (3 x 50 2 + 3 X 50 X 4 + 4 2 ) X 4. Hence, IV. //' the cube of the tens be subtracted from the entire cube, the remainder u'ill be composed of two factors, one of which will be three times the square of the tens plus three times the tens multiplied by the units plus the square of the units ; and the other } the units. 1. What is the cube root of 145780726447 ? OPERATION. 145780726447(5263, Ans. I II 125 152 304 7500 20780 7804 15608 1566 9396 811200 5172726 Si.0596 4923576 1 i~ t ~OO > \c\ 83002800 249150- v > i \ ~ i \ i in o mi z.1 \ 15783 47349 cS3050l49 249150447 ANALYSIS. Pointing off the given number into periods of 3 figures each, the four periods show that there will be four figures in the root, (I). Since the cube of the first figure of the root is contained wholly in the first period of the power, (II), we seek the greatest cube in the first period, 145, which we find by trial to be 125, and we place its root, 5, for the first figure of the required root, and regard it as tens of the next inferior order, (654). We now subtract 125, the cube of this figure, from the first period, 145, and bringing down the next period, obtain 20780 for a dividend. And since the cube of the first two figures of the root is contained wholly in the first two periods of the power, (II), the dividend, 20780, must contain at least the product of the two factors, one of which is three times the square of the first figure (tens), plus three times the first figure multiplied by the second (units), plus the square of the second ; and the other, the second fiyure (IV). Now if we could divide this dividend by the first of these factors, the quotient would be the other factor, or the seconcl figure of the root. But as the first factor is composed in part of the second figure, which we have not yet found, we can not now obtain the complete divisor ; and we therefore write three times the square of the first figure, regarded as tens, or 50 2 X 3 7500, at the left of the dividend, for a trial divisor. Dividing the dividend by the trial divisor, we obtain 2 for the second, or trial figure of the root. To 33* 390 EVOLUTION. complete the divisor, we must add to the trial divisor, as a correction, three times the tens of the root already found multiplied by the units, plus the square of the units, (IV). But as 50 X 3 X -f 2 2 == (50 X 3 4- 2) X 2, we annex the second figure, 2, to three times the first figure, 5, and thus obtain 50 X 3 + 2 = 152, the first factor of the correction, which we write in the column marked I. Multiplying this result by the 2, we have 304, the correction, which we write in the column marked II. Adding the correction to the trial divisor, we obtain 7804. the complete divisor. Multiplying the complete divisor by the trial figure of the root, subtracting the product from the dividend, and bringing down the next period, we have 5172726 for a dividend. We have now taken the cube of the first two figures of the root considered as tens of the next inferior order, from the first three periods of the number ; and since the cube of the first three figures of the root is contained wholly in the first three periods of the power, (II), the dividend, 5172726 must contain at least the product of the two factors, one of which is three times the square of the first two figures of the root (regarded as tens of the next order) plus three times the first two figures multiplied by the third, plus the square of the third; and the other, the third figure, (IV). Therefore, to obtain the third figure, we must use for a trial divisor three times the square of the first two figures, 52, considered as tens. And we observe that the significant part of this new trial divisor may be obtained by adding the last complete divisor, the last correction, and the square of the last figure of the root, thus : 7804 = (50 2 x 3) + (50 x 3 X 2) -f 2 2 304 = 50 x 3 X 2 -f 2 2 4= & 8112 = (50 2 -f 00 x 2 + 2*)x3 = 52 2 x 3 This number is obtained in the operation without re-writing the partd, by adding the square of the second root figure mentally, and combining units of like order, thus : 4, 4, and 4 are 12, and we write the unit figure, 2, in the new trial divisor ; then 1 to carry and is 1 ; then 3 and 8 are 11, etc. Annexing two ciphers to the 8112, because 52 is regarded as tens of the next order, and dividing by this new trial divisor, 811200, we obtain 6, the third figure in the root. To complete the second trial divisor, after the manner of completing the first, we should annex the third figure of the root, 6, to three times the former figures, 52, for the first factor of the correction. CUBE ROOT. 391 But as we have in column I three times 5 with the 2 annexed, or 152, we need only multiply the last figure, 2, by 3, and annex the third figure of the root, 6, which gives 1566, the first factor of the correc- tion sought, or the second term in column I. Multiplying this number by the 6, we obtain 9396, the correction sought ; adding the correction to the trial divisor, we have 820596, the complete divisor ; multiplying the complete divisor by the 6, subtracting the product from the divi- dend, and bringing down the next period, we have 249150447 for a new dividend We may now regard the first three figures of the root, 526, as tens of the next inferior order, and proceed as before till the entire root, 5263, is extracted. OO1. From these principles and illustrations we deduce the following RULE. I. Point off the given number into periods of three figures each, counting from units" place toward the left and fight. II. Find the greatest cube that does not exceed the left hand period , and write its root for the first figure in the required root; subtract the cube from the left hand period, and to the remainder briny down the next period for a dividend. III. At the left of the dividend write three times the square of the first figure of the root, and annex two ciphers, for a trial di- visor ; divide the dividend by the trial divisor, and write the quo- tient for a trial figure in the root. IV. Annex the trial figure to three times the former figure, and write the result in a column marked I, one line below the trial divisor, multiply this term by the trial figure, and write the product on the same line in a column marked II; add this term as a correction to the trial divisor, and the result will be the com- plete divisor. V. Multiply the complete divisor by the trial figure ; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. VI. Add the square of the last figure of the root > the last term in column II, and 'he complete divisor together, and annex two ciphers, for a new trial divisor} with which obtain another trial figure in the root. 392 EVOLUTION. VII. Multiply the unit figure of the last term in column I by 3, and annex the trial fijure of the root for the next term of column I; multiply this result by the trial fiyure *of the root for the next term of column II; add this term to the trial divisor for a complete divisor, with which proceed as before. NOTES. 1. If at any time the product be greater than the dividend, diminish the trial figure <.f the root, and correct the erroneous work. 2. If a cipher occur in the root, annex two more ciphers to the trial divisor, and another period to the dividend; then proceed as before with column I, an, nexmg both cipher and trial figure. EXAMPLES FOR PRACTICE 1. "What is the cube root of 389017 ? Am. 73. 2. What is the cube root of 44361864 ? Ans. 354. 3. What is the cube root of 10460353203? Ans. 2187. 4.* What is the cube root of 98867482624 ? Ans. 4624 5. What is the cube root of 30.625 ? Ans. 3 12866 +. 6. What is the cube root of 111 J ? Ans 4.8076 f 7. What is the cube root of .000148877? Ans .053. Find the values of the following expressions. 8. Vl22615327232'f Ans. 4968. 9- ^^7134217728"? Ans. 8. 10. V39304*"? Ans. 1156. 11- ^im X ^mT? Ans. ||. 12 How much does the sum of the cube roots of 50 and 31 exceed the cube root of their sum? Ans. 2.4986 -f . CONTRACTED METHOD. 0G2. In applying contracted decimal division to the extrac- tion of the cube root of numbers, we observe, 1st. For each new figure in the root, the terms in the operation extend to the right 3 places in the column of dividends, 2 places in the column of divisors, and 1 place in column I. Hence, 2d. If at any point in the operation we omit to bring down new periods in the dividend, we must shorten each succeeding divisor 1 place, and each succeeding term in column I, 2 places. 1. What is the cube root of 189, correct to 8 decimal places ? CUBE ROOT. 393 ANALYSIS. We proceed by the usual method to extract the cube root of the given number until we have obtained the three figures, 5.73 : the corres- ponding remainder is 867483, and the next trial divisor with the ciphers omitted is 984987. We now omit to bring down a period of ciphers, thus con- tracting the divid- end 3 places ; and we contract the di- visor an equal num- ber of places by emitting to annex the two ciphers, and regarding the right hand figure, 7, as a redundant figure. Then dividing, we obtain 8 for the next figure of the root. To complete the divisor, we obtain a correc- tion, 1375, contracted 2 places by omitting to anrex the trial figure of the root, 8, to the first factor, 1719, and regarding the right hand figure, 9, as redundant in multiplying. Adding the contraction to the contracted divisor, we have the complete divisor, 986362, the right hand figure being redundant. Multiplying by 8 and subtracting the product from the dividend, we have 78393 for a new dividend. Then to form the new trial divisor, we disregard the square of the root figure, 8. because this square consists of the same orders of units as the two rejected places in the divisor; and we simply add the cor- rection, 1375, and the complete divisor, 986362, and rejecting 1 figure, thus obtain 98774, of which the right hand figure, 4, is redundant. Dividing, we obtain 7 for the next root figure. Rejecting 2 places from the last term in column I, we have 17 for the next contracted term in this column. We then obtain, by the manner shown in the former step, the correction 12, the complete divisor, 98786, the prod- uct, 69150, and the new dividend, 9243. We then obtain the new trial I II OPERATION. |5.73879355db, Ans. 189.000000 125 157 1099 7500 64 000 8599 60 193 1713 5139 974700 3 807000 979839 2939517 1719 1375 984987 867483* 986362 789090 17 12 98774 78393 98786 69150 9880 9243 8892 988 35 i 296 99 55 50 10 5 5 394 EVOLUTION. divisor, 9880; and as column I is terminated by rejecting the two places, 17, we continue the contracted division as in square root, and thus obtain the entire root, 5.73879355 rb, which is correct to the last decimal place, and contains as many places as there are places in the periods used. Hence the following RULE. I. If necessary, annex ciphers to the given number, and assume as many figures as there are places required in the root ; then proceed by the usual method until all the assumed figures have been employed. II. Form the next trial divisor as usual, but omit to annex the two ciphers, and reject one place in forming each subsequent trial divisor. Ill In completing the contracted divisors, omit at first to annex the trial figure of the root to the term in column I, and reject 2 places in forming each succeeding term, in this column. IV. In multiplying, regard the right hand figure of each con- tracted term, in column I and in the column of divisors, as redund- ant. NOTES. 1. After the contraction commences, the square of the last root figure is disregarded in forming the new trial divisors. 2. Employ only full periods in the number. EXAMPLES FOR PRACTICE. 1. Find the cube root of 24, correct to 7 decimal places. Ans. 2.8844992 db. 2. Find the cube root of 12000.812161, correct to 9 decimal places. Ans. 22.89480 1#*4 dr. 3. Find the cube root of .171467, correct to 9 decimal places. Ans. .555554730 rt. 4. Find the cube root of 2. 42999 correct to 5 decimal places. Ans. 1.34442. 5. Find the cube root of 19.44, correct to 4 decimal places. Ans. 2.6888 =b. 6. Find the value of 3/f" to 6 places. Ans. .941035 . 7. Find the value of ^.571428 to 9 places. Ans. .829826686 . HOOT'S OK ANY DKftKXK. 395 8. Find the value of N/1.08674325' 2 to 7 places. Am. 1.057023 db. 9. Find the value of 1.05s to 7 places. Ant. 1.084715 . ROOTS OF ANY DEGREE. 663. Any root whatever may be extracted by means of the square and cube roots, as will be seen in the two cases which follow. CASE I. 664. When the index of the required root contains no other factor than 2 or 3. We have seen that if we raise any power of a given number to any required power, the result will be that power of the given number denoted by the product of the two indices, (643, III). Conversely, if we extract successively two or more roots of a given number, the result must be that root of the given number denoted by the product of the indices. 1. What is the 6th root of 2176782336 ? OPERATION. ANALYSIS. The index of the 6 = 2x3 required root is 6 = 2x3; we v/2176782336 = 46657 therefore extract the square root v/ 4665 6 = 36, Ans. f ^ e given number, and the cube root of this result, and oi> Or ' tain 36, which must be the 6th ^2176782336 = 1296 root required. Or, we first find N/1296 = 36 Ans. the cube root of the given num- ber, and then the square root of the result, as in the operation. Hence the following RULE. Separate the index of the required root into its prime factors, and extract successively the roots indicated by the several factors obtained ; the final result will be the required root. EXAMPLES FOR PRACTICE. 1. What is the 6th root of 6321363049 ? Ans. 43. 2. What is the 4th root of 5636405776? Ans. 274. 396 EVOLUTION. 3. What is the 8th root of 1099511627776? Ans. 32. 4. What is the 6th root of 25632972850442049 ? Ans. " 543. 5. What is the 9th root of 1.577635? Ans. 1 . 051 9.03 + . NOTE. Extract the cube root of the cube root by the contracted method, carrying the root in each operation to 6 decimal places only. 6. What is the 12th root of 16.3939? Ans. 1.2(324 + . 7. What is the 18th root of 104.9617 ? Ans. 1.2950+. CASE II. 665. "When the index of the required root is prime, or contains any other factor than 2 or 3. To extract any root of a number is to separate the number into as many equal factors as there are units in the index of the re- quired root ; and it will be found that if by any means we can separate a number into factors nearly equal to each other, the average of these factors, or their sum divided the number of fac- tors, will be nearly equal to the root indicated by the number of factors. 1. What is the 7th root of 308 ? OPERATION. ANALYSIS. We first /308 _ 2.59+ fincl by Case I, the 6th s/308 = 2 04+ root and a ^ so *ke ^th. 2.59 + 2.04 = 4.63 r t of 308 ; and since 4.63 -r- 2 = 2.31, assumed root. the 7th root must be 2.3 1 6 = 151.93 less than the former 308 ~ 151.93 = 2.0272+ and greater than the ?k 3 LVW'?^ 15 ,- 8 f 2 * latter, we take the ave- 15.88,2 3^= 2.2696, 1st approximation. rage of the tw0j or ono ^l 66 {fS 7 = 8 2.253452+ half of theirsums, 2.31 2.2696 x 6 + 2.253452 = 15.871052 and cal1 li the assumed 15.871052 -r- 7 = 2.267293, 2d approx. the assumed root, 2.31, to the 6th power, and divide the given number, 308, ^y the result, and obtain 2.0272+ for a quotient ; we thus separate 308 into 7 fac- tors, 6 of which are equal to 2.31, and the other is 2.0272. As these 7 factors are nearly equal to each other, the average of them all must be a near approximation to the 7th root. Multiplying the 2.31 by 6, adding the 2.0272 to the product, and dividing this result by 7, we ROOTS OF ANY DEGREE. 337 find the average to be. 2.2696, which is the first approximation to the required root. We next divide 308 by the 6th power of 2.2696, and obtain 2.253452-}- for a quotient ; and we thus separate the given number into 7 factors, 6 of which are each equal to 2.2696, and the other is 2.253452. Finding the average of these factors, as in the former steps, we have 2.267293, which is the 7th root of the given number, correct to 5 decimal places. Hence the following RULE. I. Find by trial some number nearly equal to the re- quired root, and call tins the assumed root. II. Divide the given number by that power of the assumed root denoted by the index of the required root less 1 ; to this quotient add as many times the assumed root as there are units in the index of the required root less 1, and divide the amount by the index of the required root. The result will be the first approxi- mate root required. III. Take the last, approximation for the assumed root, with which proceed as with the former, and thus continue till the re- quired root is obtained to a sufficient degree of exactness. NOTES. 1. The involution and division in all cases will be much abridged by decimal contraction. 2. If the index cf the required root contains the factor?. 2 or 3, we may first extract the square or cube root as many times, successively, as these factors are found in the index, after which we must extract that root of the result which is denoted by the remaining factor of the index. Thus, if the 15th root were re- quired, we should first find the cube root, then the 5th root of this result. EXAMPLES FOR PRACTICE. 1. What is the 20th root of 617 ? OPERATIONS _20 = 2 x 2 x 5. ^617 = 24.839485+. v'24l*39485 = 4.983923 -K v/l9^9T3~ = 1.378L'uti + . Ans. 2. What is the 5th root of 120 ? 3. What is the 7th root of 1.95678 ? 4. What is the 10th root of 743044? 5. What is the 15th root of 15 ? 6 What is the 25th root of 100 ? 7. What is the 5th root of 5 ? 34 398 SERIES. SERIES.* 666. A Series is a succession of num ,ers so related to each other, that each number in the succession may be formed in the same manner, from one or more preceding numbers. Thus, any number in the succession, 2, 5, 8, 11, 14, is formed by adding 3 to the preceding number. Hence, 2, 5, 8, 11, 14, is a series. 667. The Law of a Series is the constant relation existing between two or more terms of the series. Thus, in the series, 3, 7, 11, 15, we observe that each term after the first is greater than the preceding term by 4 ; this constant relation between the terms is the law of this series. The law of a series, and the term or terms on which it depends being given, any number of terms of the series can be formed. Thus, let 64 be a term of a series whose law is, that each term is four times the preceding term. The term following 64 is 64 x 4, the next term 64 x 4 2 , etc. ; the term preceding 64 is 64 4- 4. Hence the series, as far as formed, is 16, 64, 256, 1024. 668* A series is either Ascending, or Descending, according as each term is greater or less than the preceding term. Thus, 2, 6, 10, 14, is an ascending series; 32, 16, 8, 4, is a descending series. 669. An Extreme is either the first or last term of a series. Thus, in the series, 4, 7, 10, 13, the first extreme is 4, the last, 13. 670. A Mean is any term between the two extremes. Thus, in the series, 5, 10, 20, 40, 80, the means are 10, 20, and 40. * The treatise of the " METRIC SYSTEM," as presented at some length at the close of the editions of this book published previous to the year 1875, was of little value, since the system is scarcely any used in this country, and the symbols introduced were different from those authorized or in use. Being frequently requested by teachers to add or substitute an article on Mensuration, as of much more practical value, the editor has carefully prepared and substituted such a treatise at the end of this book, and to avoid repetition and put in more condensed form, has embodied in it the " applications of the square and cube roots " that intervened between " Evolution " and " Series " of former editions. So much of the Metric System as is needful has also been added. The article on " Mensuration " is essentially the same as that presented in the " Complete Arithmetic" of the " Shorter Course," and it is hoped will be entirely satisfactory. PROGRESSIONS. 399 671. An Arithmetical or Equidifferent Progression is a series whose law of formation is a common difference. Thus, in the arithmetical progression, 3, 7, 11, 15, 19, each term is formed from the preceding by adding the common difference, 4. 672. An arithmetical progression is an ascending or descend- ing series, according as each term is formed from the preceding term by adding or subtracting the common difference. Thus, the ascending series, 7, 10, 13, 16, etc., is an arithmetical progression in which the common difference, 3, is constantly added to form each succeeding term; and the descending series, 20, 17, 14, 11, 8, 5, 2, is an arithmetical progression in which the common dif- ference is constantly subtracted-, to form each succeeding term. 673. A Geometrical Progression is a series whose law of formation is a common multiplier. Thus, in the geometrical pro- gression, 3, 6, 12, 24, 48, each term is formed by multiplying the preceding term by the common multiplier, 2. 674. A geometrical progression is an ascending or descending series, according as the common multiplier is a whole number or a fraction. Thus, the ascending series, 1, 2, 4, 8, 16, etc., is a geometrical progression in which the common multiplier is 2; and the descending series, 32, 16, 8, 4, 2, 1, , , etc., is a geo- metrical progression in which the common multiplier is J. 675. The Ratio in a geometrical progression is the common multiplier. 676. In the solution of problems in Arithmetical or Geomet- rical progression, five parts or elements are concerned, viz : In Arithmetical Progression In Geometrical Progression 1. The first term ; 1. The first term ; 2. " last term ; 2. " last term ; 3. " number of terms ; 3. " number of terms ; 4. " common difference ; 4. " ratio ; 5. " sum of the series. 5. " sum of the series. The conditions of a problem in progression may be such as to require any one of the five parts from any three of the four re- maining parts ; hence, in either Arithmetical or Geometrical Pro- gression, there are 5 X 4 = 20 cases, or classes of problems, and no more, requiring each a different solution. 400 SERIES. GENERAL PROBLEMS IN ARITHMETICAL PROGRESSION. PROBLEM I. 677. Given, one of the extremes, the common dif- ference, and the number of terms, to find the other extreme. Let 2 be the first term of an arithmetical progression, and 3 the common difference ; then, 2 =2 =2, 1st term. 2+3 =2+ (3 X 1)= 5, 2d " 2 + 3 + 3 =2+ (3 X 2)= 8, 3d " 2 + 3 + 3 + 3 = 2 + (3 X 3) = 11, 4th " From this illustration we perceive that, in an arithmetical pro- gression, when the series is ascending, the second term is equal to the first term plus the common difference ; the third term is equal to the first term plus 2 times the common difference ; the fourth term is equal to the first term plus 3 times the common difference ; and so on. In a descending series, the second term is equal to the first term minus the common difference ; the third term is equal to the first minus 2 times the common difference ; and so on. In all cases the difference between the two extremes is equal to the product of the common difference by the number of terms less 1. Hence the RULE. Multiply the common difference by the number of terms less 1 add the product to the given term if it be the less extreme, and subtract the product from the given term if it be the greater extreme. EXAMPLES FOR PRACTICE. 1. The first term of an arithmetical progression is 5, the com- mon difference 4, and the number of terms 8 ; what is the last term ? Ans. 33. 2. If the first term of an ascending series be 2, and the com- mon difference 3, what is the 50th term ? 3. The first term of a descending series is 100, the common difference 7, and the number of terms 13 ; what is the last term ? 4. If the first term of an ascending series be f , the common difference $, and the number, of terms 20, what is the last term ? Ans. ?. ARITHMETICAL PROGRESSION. 01 PROBLEM II. 678. Given, the extremes and number of terms, to find the common difference. Since the difference of the extremes is always equal to the common difference multiplied by the number of terms less 1, (677)> we have the following RULE. Divide the difference of the extremes by the number of terms less 1 EXAMPLES FOR PRACTICE. 1. If the extremes of an arithmetical series are 3 and 15, and the number of terms 7, what is the common difference? Ans. 2. 2. The extremes are 1 and 51, and the number of terms is 76; what is the common difference ? 3. The extremes are .05 and .1, and the number of terms is 8; what is the common difference? Ans. .00714285. 4. If the extremes are and 2i, and the number of terms is 18 ; what is the common difference ? PROBLEM III. 679. Given, the extremes and common difference, to find the number of terms. Since the difference of the extremes is equal to the common differ- ence multiplied by the number of terms less 1, (677)> we have the following RULE. Divide the difference of the extremes by the common difference, and add 1 to the quotient. EXAMPLES FOR PRACTICE. 1. The extremes of an arithmetical series are 5 and 75, and the common difference is 5 ; what is the number of terms ? Am. 15. 2. The extremes are | and 20, and the common difference is 6 ; find the number of terms. 402 SERIES. 3. The extremes are 2.5 and .25, and the common difference is .125; what is the number of terms? 4. Insert 5 arithmetical means between 2 and 37. PROBLEM IV. 68O. Given, the extremes and number of terms, to find the sum of the series. Let us take any series, as 2, 5, 8, 11, 14, and writing under it the same series in an inverse order, add each term of the inverted series to the term above it in the direct series, thus : 2 + 5 + 8 + 11 -f 14 = 40, once the sum, 14 + H+ 8+ 5+ 2 = 40, " " " 16 -f 16 -f 16 + 16 + 16 = 80, twice the sum. From this we perceive that 16, the sum of the extremes of the given series, multiplied by 5, the number of terms, equals 80, which is twice the sum of the series ; and 80 -r- 2 = 40, the sum of the series. Hence RULE. Multiply the sum of the extremes by the number of terms, and divide the product by 2. EXAMPLES FOR PRACTICE. 1. Find the sum of the series the first term of which is 4, the common difference 6, and the last term 40. Ans. 154. 2. The extremes are and 250, and the number of terms is 1000 ; what is the sum of the series ? 3. A person wishes to discharge a debt in 11 annual payments such that the last payment shall be $220, and each payment greater than the preceding by $17 ; find the amount of the debt, and the first payment. Ans. First payment, $50. G81. By reversing some one of the four problems now given, or by combining two or more of them, all of the. sixteen remain- ing problems of Arithmetical Progression may be solved or analyzed. GEOMETRICAL PROGRESSION. 403 GENERAL PROBLEMS IN GEOMETRICAL PROGRESSION, PROBLEM I. 682. Given, one of the extremes, the ratio, and the number of terms, to find the other extreme. Let 3 be the first term of a geometrical progression, and 2 the ratio : then, 3 =3 = 3, the 1st term, 3x2 =3 X2* = 6, " 2d " 3x2x2 == 3 X 2 2 = 12, "3d " 3 X 2 X 2 X 2 = 3 X 2 3 = 24, "4th " From this illustration we perceive that, in a geometrical progression, the second term is equal to the first term multiplied by the ratio ; the third term is equal to the first term multiplied by the second power of the ratio; ths fourth term is equal to the first term multiplied by the third power of the ratio ; and so on. The same is true whether the ratio be an integer or fraction. Hence the following RULE. I. If the given extreme be the first term, multiply it by that power of the ratio indicated by the number of terms less 1 ; the result will be the last term. II. If the given extreme be the last term, divide it by that power of the ratio indicated by the number of terms less 1; the result will be the first term. EXAMPLES FOR PRACTICE. 1. The first term of a geometrical series is 6, the ratio 4, and the number of terms 6 ; find the last term. Ans. 6144. 2. The last term of a geometrical series is 192, the ratio 2, and the number of terms 7 ; what is the first term ? 8. If the first term be 6, the ratio -J, and the number of terms 8, what is the last term ? 4. The first term is 25, the ratio ^, and the number of terms 5 ; what is the last term ? Ans. . 404 SERIES. PROBLEM TI. 683. Given, the extremes and number of terms, to find the ratio. Since the last term is always equal to the first term multiplied by that power of the ratio indicated by the number of terms less 1, (682), we have the following RULE. Divide the last term by the first , and extract that root of the quotient indicated by the number of terms less 1 ; the result iv ill be the ratio. EXAMPLES FOR PRACTICE. 1. The extremes are 2 and 512, and the number of terms is 5; what is the ratio ? Ans. 4. 2 The extremes are -$ and 45 T 9 g , and the number of terms is 8; what is the ratio? 3. The extremes are 7 and .0112, and the number of terms is 5 ; what is the ratio ? Ans. 5. 4. Insert 3 geometrical means between 8 and 5000. PROBLEM III. 684. Given, the extremes and ratio, to find the num- ber of terms. Since the quotient of the last term divided by the first term is equal to that power of the ratio indicated by the number of terms less 1, (683), we have the following RULE. Divide the last term by the first, divide this quotient by the ratio, and the quotient thus obtained by the ratio again, and so on in successive division, till the final quotient is I. The number of times the ratio is used as a divisor, plus 1, is the number of terms. EXAMPLES FOR PRACTICE. 1. The extremes are 2 and 1458, and the ratio is 3; what is the number of terms ? Ans. 7. 2. The first term is .1, the last term 100, and the ratio 10; find the number of terms. GEOMETRICAL PROGRESSION. 405 3. The first term is g^, the last term ^, and the ratio 2; what is the number of terms ? 4. The extremes are 196608 and 6, and the ratio is i ; what is the number of terms ? Ans. 6. PROBLEM IV. 685. Given, the extremes and ratio, to find the sum of the series. Let us take the series 5 + 20 + 80 + 320=425, multiply each term by the ratio 4, and from this result subtract the given series term from term, thus : 20 -f 80 + 320 + 1280 = 1700, four times the series, 5 4- 20 + 80 + 320 = 425_,_once the series, 1280 5 = 1275, three times the series, Then 1275 ~ 3 = 425, once the series. Hence the RULE. Multiply the greater extreme, by the ratio, subtract the less extreme from the product, and divide the remainder by the ratio less 1. NOTE. Let every descending series be inverted, and the first term called the last; then the ratio will be greater than a unit. If the series be infinite, the least term is a cipher. EXAMPLES FOR PRACTICE. 1. The extremes are 3 and 384, and the ratio is 2; what is the sum of the series ? Ans. 765. 2. If the extremes are 5 and 1080, and the ratio is 6, what is the sum of the series ? 3. If the first term is 44, the last term ^|^, and the ratio , what is the sum of the series ? Ans. 7^^. 4. What is the sum of the infinite series, 8, 4, 2, 1 7 , i, etc.? PROBLEM V. 686. Given, the first term, the ratio, and the num- ber of terms, to find the sum of the series. If, for example, the first term be 4, the ratio 3, and the number of terms 6, then by Problem I, we have 4 x 3* == the last term. 406 SERIES. Whence by Problem IV, we nave 4 X 36 4 (3 6 1) X 4 - =.- v ' = 1456, the sum of the series, 31 3 1 Hence the following RULE. Raise the ratio to a power indicated by the number of terms, and subtract 1 from the result ; then multiply this remainder by the first term^ and divide the product by th? ratio less 1. EXAMPLES FOR PRACTICE. 1. The first term is 7, the ratio 3, and the number of terms 4 ; what is the sum of the series ? Arts. 280. 2. The first term is 375, the ratio J, and the number of terms 4 ; what is the sum of the series ? 3. The first term is 175, the ratio 1.06, and the number of terms 5; what is the sum of the series? Ans. 986.49-f-- PROBLEM VI. 687. Given, the extremes and the sum of the series, to find the ratio. If we take the geometrical progression, 2, 6, 18, 54, 162, in which the ratio is 3, and remove the first term and the last term, succes- sively, and then compare the results, we have 6 -f 18 4- 54 4- 162 = sum of the series minus the first term. 2 4- 6 ~|- 18 4- 54 = sum of the series minus the last term. Now, since every term in the first line is 3 times the corresponding term in the second line, the sum of the terms in the first line must be 3 times the sum of the terms in the second line. Hence the RULE. Divide the sum of the series minus the first term, by the sum of the series minus the last term. EXAMPLES FOR PRACTICE. 1. The extremes are 2 and 686, and the sum of the series is 800 ; what is the ratio ? Ans. 1. GEOMETRICAL PROGRESSION. 407 2. The extremes are J and 64, and the sum of the series is 1-7|; what is the ratio? 3. If the sum of an infinite series be 4, and the greater ex- treme 3, what is the ratio ? Ans. . O88* Every other problem in Geometrical Progression, that admits of an arithmetical solution, may be solved either by re- versing or combining some of the problems already given. COMPOUND INTEREST BY GEOMETRICAL PROGRESSION. 689. We have seen (55O) that if any sum at compound in- terest be multiplied by the amount of $1 for the given interval, the product will be the amount of the given sum or principal at the end of the first interval j and that this amount constitutes a new principal for the second interval, and so on for a third, fourth, or any other interval. Hence, A question in compound interest constitutes a geometrical pro- gression, whose first term is the principal; the common multiplier or ratio is one plus the rate per cent, for one interval ; the number of terms is equal to the number of intervals -|-1 ; and the last term is the amount of the given principal for the given time. All the usual cases of compound interest and discount computed at compound interest, can therefore be solved by the rules for geo- metrical progression. For example, Find the amount of $250 for 4 years, at 6 % compound interest. OPERATION. $250 x 1.06* = $250 x 1.262477 $316.21925. ANALYSIS. Here we have $250 the first term, 1.06 the ratio, and 5 the number of terms, to find the last term. Then by 682 we find the last term, which is the amount required. EXAMPLES FOR PRACTICE. 1. What is the amount of $350 in 4 years, at 6 % per annum compound interest ? Ans. $441.86. 2. Of what principal is $150 the compound interest for 2 years, at 7 % ? 408 SERIES. 3. What sum at 6 % compound interest, will amount to $1000 in 3 years ? Ans. $839.62. 4. In how many years will $10 amount to $53.24, at 10 % com- pound interest? Ans. 3 years. 5. At what rate per cent . compound interest will any sum double itself in 8 years? Ans. 9.05 -f %. 6. What is the present worth of $322.51, at 5 % compound interest, due 24 years hence ? Ans. $100. ANNUITIES. 690. An Annuity is literally a sum of money which is pay- able annually. The term is, however, applied to a sum which is payable at any equal intervals, as monthly, quarterly, semi-annu- ally, etc. NOTE. The term, interval, will be used to denote the time between payments. Annuities are of three kinds : Certain, Contingent, and Per- petual. 691* A Certain Annuity is one whose period of continu- - ance is definite or fixed. 692. A Contingent Annuity is one whose time of commence- ment, or ending, or both, is uncertain ; and hence the period of its continuance is uncertain. 693. A Perpetual Annuity or Perpetuity is one which con- tinues forever. 694:. Each of these kinds is subject, in reference to its com- mencement, to the three following conditions : 1st. It may be deferred, i. e., it is not to be entered upon until after a certain period of time. 2d. It may be reversionary, i. e., it is not to be entered upon until after the death of a certain person, or the occurrence of some certain event. 3d. It may be in possessionj i. e., it is to be entered upon at tmce. ANNUITIES. 409 695. An Annuity in Arrears or Forborne is one on which the payments were not made when due. Interest is to be reck- oned on each payment of an annuity in arrears, from its maturity, the same as on any other debt. ANNUITIES AT SIMPLE INTEREST. GOG. In reference to an annuity at simple interest, we observe : I. The first payment becomes due at the end of the first inter- val, and hence will bear interest until the annuity is settled. II. The second payment becomes due at the end of the second interval, and hence will bear interest for one interval less than the first payment. III. The third payment will bear interest for one interval less than the second ; and so on to any number of terms. Hence, IV. All the payments being settled at one time, eacl* will be less than the preceding, by the interest on the annuity for one interval. Therefore, they will constitute a descending arithmetical progression, whose first term is the annuity plus its interest for as many intervals less one as intervene between the commencement and settlement of the annuity; the common difference is the in- terest on the annuity for one interval ; the number of terms is the number of intervals between the commencement and settlement of the annuity; and the last term is the annuity itself. 6H7* The rules in Arithmetical Progression will solve all problems in annuities at simple interest. EXAMPLES FOR PRACTICE. 1. A man works for a farmer one year and six months, at $20 per month, payable monthly; and these wages remain unpaid until the expiration of the whole term of service. How much is due to the workman, allowing simple interest at 6 per cent, per annum ? OPERATION. ANALYSIS. Here the $20 + $.10 X 17 = $21.70, first term. last month ' 8 wages, $20 -f $21.70 $ 20 > is the last term ; - 2~ - X 18 = 375.30, sum. t h e number of months, 18, is the number of 410 SERIES. terms ; and the interest on 1 month's wages, $.10, is the common dif- ference ; and since the first month's wages has been on interest 17 months, the progression is a descending series. Then, by 677 we find the first term, which is the amount of the first month's wages for 17 months ; and by 680 we find the sum of the series, which is the sum of all the wages and interest. 2. A father deposits annually for the benefit of his son, com- mencing with his tenth birthday, such a sum that on his 21st birthday the first deposit at simple interest amounts to $210, and the sum due his son to $1860. How much is the deposit, and at what rate per cent, is it deposited ? OPERATION. ANALYSIS. Here the $1860 X 2 $210 X 12 $210, the amount of 12 W' de P slt - - the first deposit, is 21Q _ 100 *^ e nrst term ; 12, jy -- = 10 %, f ate - the number of depo- sits, is the number of terms ; and $1860, the amount of all the deposits and interests, is the sum of the series. By 680 we find the last term to be $100, which is the annual deposit ; and by 678 we find the common difference to be $10, which is the annual rate % . 3. What is the amount of an annuity of $150 for 5 years, pay- able quarterly, at 1J per cent, per quarter? Ans. $3819.75. 4. In what time will an annual pension of $500 amount to $3450, at 6 per cent, simple interest ? Ans. 6 years. 5. Find the rate per cent at which an annuity of $6000 will amount to $59760 in 8 years, at simple interest. Ans. 7 per cent. ANNUITIES AT COMPOUND INTEREST. OO8* Ari Annuity at compound interest constitutes a geomek rical progression whose first term is the annuity itself; the common multiplier is one plus the rate per cent, for one interval expressed decimally; the number of terms is the number of intervals for which the annuity is taken j and the last term is the first term multiplied by one plus the rate per cent, for one interval raised to a power one less than the number of terms. PROMISCUOUS EXAMPLES. 4U 699. The Present Value of an Annuity is such a sum as would produce, at compound interest, at a given rate, the same amount as the sum of all the payments of the annuity at com- pound interest. Hence, to find the present value; First find the amount of the annuity at the given rate and for the given time ly 68G5 then find the present value of this amount by taking out the amount of $1, or divisor, from NOTES. 1. The present value of a reversionary annuity is that principal which will nmount, at the time the reversion expires, to what will then be the present vnlue of the annuity. 2. The present value of a perpetuity is a sum whose interest equals the an- nuity. TOO, Questions in Annuities at compound interest can be solved by the rules of Geometrical Progression. PROMISCUOUS EXAMPLES IN SERIES. 1. Allowing 6 per cent, compound interest on an annuity of $200 which is in arrears 20 years, what is its present amount ? Ans. $7357.11. '2. Find the annuity whose amount for 25 years is $16459.35, allowing compound interest at 6 per cent. Ans. $300. 3. What is the present worth of an annuity of $500 for 7 years, at 6 per cent, compound interest? Ans. $2791.18. 4. What is the present value of a reversionary lease of $100, commencing 14 years hence, and to continue 20 years, compound interest at 5 per cent.? Ans. $629.426. 5. Find the sum of 21 terms of the series, 5, 4|, 4, etc. 6. A man traveled 13 days; his last day's journey was 80 miles, and each day he traveled 5 miles more than on the preceding day. How far did he travel, and what was his first day's journey? Ans. He traveled 650 miles. 7. Find the 12th term of the series, 30, 15, 7, etc. Ans. T if ? . 8. The first term of a geometrical progression is 2, the last 512, and common multiplier 4 ; find the sum of the series. Ans. 682. 412 SERIES. 9. The distance between two places is 360 miles. In how many days can it be traveled, by a man who travels the first day 27 miles, and the last day 45, each day's journey being greater than the preceding by the. same number of miles? Ans. 10. 10. The first term of a geometrical progression is 1, the last term 15625, and the number of terms 7j find the common ratio. Ans. 5. 11. An annual pension of $500 is in arrears 10 years. What is the amount now due, allowing 6 per cent, compound interest ? Ans. $6590.40. 12. Find the first and last terms of an arithmetical progression whose sum is 408, common difference 6, and number of terms 8. Ans. First term, 30 ; last term, 72. 13. A farmer pays $1196, in 13 quarterly payments, in such a way that each payment is greater than the preceding by $12. What are his first and last payments ? Ans. $20, and $164. 14. A man wishes to discharge a debt in yearly payments, mak- ing the first payment $2, the last $512, and each payment four times the preceding payment. How long will it take him to dis* charge the debt, and what is the amount of his indebtedness ? 15. A man dying, left 5 sons, to whom he gave his property as follows : to the youngest he gave $4800, and to each of the otheri 1 J times the next younger son's share. What was the eldest son's fortune, and what the amount of property left ? Ans. Eldest son's share, $24300 ; property, $63300. 16. Find the annuity whose amount for 5 years, at 6 per cent, compound interest, is $2818.546. Am. $500. 17. A merchant pays a debt in yearly payments in such a way that each payment is 3 times the preceding ; his first payment is $10, and his last $7290. What is the amount of the debt, and in how many payments is it discharged ? Ans. Debt, $10930; 7 payments. 18. A man traveling along a road, stopped at a number of stations, but at each station he found it necessary, before proceed- ing to the next, to return to the place from which he first started j PROMISCUOUS EXAMPLES. 413 the distance from the starting place to the first station was D miles, and to the last 25 mile's; he traveled in all 180 miles. How many stations were there on the road, and what was the distance from station to station ? Ans. 6 stations ; 4 miles apart. 19. An annuity of 200 for 12 years is in reversion 6 years What is its present worth, compound interest at 6 % ? Ans. $1182.05 + . 20. A man pays $6 yearly for tobacco, from the age of 16 until he is 60, when he dies, leaving to his heirs $500. What might he have left them, if he had dispensed with this useless habit and loaned the money at the end of each year at 6 % compound interest? Ans. $1698.548+. 21. What is the present worth ot a reversionary perpetuity of $100, commencing 30 years hejipe, allowing 5 per cent, compound interest? An?. $462.75-}-. 22. Two boys, each 12 years old, have certain sums of money left to them ; the sum left to one is put out at V % simple inte- rest, and the sum left the other at 6 % compouiiC interest, paya- ble semi-annually, and the amount cf each, boy's money will be $2000 when he is 21 years old. What is the sum left to each boy? 23. A merchant purchased 8 pieces of cloth, for which he paid $136 ; the difference in the length of any two pieces was 2 yds.^ and the difference in the price $4. He paid $31 for the longest piece, and $1 a yard for the shortest. Find the whole number of yards, and the price per yard of each piece. 24. A farmer has 600 bushels of different kinds of grain, mixed in such a way that the number of bushels of the several kinds con- stitute a geometrical progression, whose common multiplier is 2 ; the greatest number of bushels of one kind is 320. Find the number of kinds of grain in the mixture, and the number of bushels of each kind. Ans. 4 kinds. MISCELLANEOUS EXAMPLES. MISCELLANEOUS EXAMPLES. 1. How many thousand shingles will cover both sides of a roof 36 ft. long, and whose rafters are 18 ft. in length ? 2. From f of 4 of of 70 miles, subtract .73 of 1 mi. 3 fur. 3. What number is that from which if 7 be subtracted, f of the remainder is 91 ? Ans. 144|. 4. What part of 4 is of 6? Ans. f. 5 It is required to mix together brandy at $.80 a gallon, wine at $.70, cider at $.10, and water, in such proportions that the mixture may be worth $.50 a gallon ; what quantity of each must be used ? Ans. 3 gal of water, 2 of cider, 4 of wine, and 5 of brandy. 6. What number increased by J, J, and J of itself equals 125 ? 7. What is the hour, when the time past noon is equal to f of the time to midnight ? Ans. 4 h. 48 min. p. M. 8. A grocer mixed 12 cwt. of sugar @ $10, with 3 cwt. @ $8f, and 8 cwt. @ $7 ; how much was 1 cwt. of the mixture worth ? 9. If $240 gain $5.84 in 4 mo. 26 da., what is the rate ? Ans. At $5 on 6 mo. 22. Sold J of a lot of lumber for what f of it cost ; what fo was gained on the part sold? Ans. 25 ^,. 23. If $500 gain $50 in 1 yr., in what time will $960 gain $60? 24. Received an invoice of crockery, 12 per cent, of which was broken ; at what per cent, above cost must the remainder be sold, to clear 25 per cent, on the invoice? Ans. ^2^. 25. The sum of two numbers is 365, and their difference is .0675 ; what are the numbers ? 26. If the interest of $445.62* be $128.99 for 7 yr., what will be the interest of $650 for 3 yr. 10 mo. 15 da. ? 27. Received from Savannah 150 bales of cotton, each weighing 540 lb., and invoiced at 7d. a pound Georgia currency. Sold it at an advance of 26 ^,, commission lj ^, and remitted the proceeds by draft. What was the face of the draft, exchange being c /c discount? Ans. $12629.28+. 28. A man in Chicago has 5000 francs due him on account in Paris, lie can draw on Paris for this amount, and negotiate the bill at 19 f cents per franc ; or he can advise his correspondent in Paris to remit a draft on the United States, purchased with the sum due him, ex- change on U. S. being at the rate of 5 francs 20 centimes per $1. What sum will the man receive by each method? Ans. By draft on Paris, $970 ; by remittance from Paris, $961.53. 29. What sum must be invested in stocks bearing 6^ fi mi. 67. If stock bought at 5 and B's is $1333^. What sum did A take out at the end of 4 mo. ? Ans. $2400. 83. What sum of money, with its semi-annual dividends of 5 fo invested with it, will amount to $12750 in 2 yr. ? Ans. $10489.450-. 84. If a speculator invests $1500 in flour, and pays 5 fo for freights, 2 fo for commission, and the flour sells at 20 fo advance on cost price, on a credit of 90 days, and he gets this paper discounted at bank at 7 % , and repeats the operation every 15 days, investing all the pro- ceeds each time, how much will be his whole gain in two months ? 420 MISCELLANEOUS EXAMPLES. 85. If a piece of silk cost $.80 per yard, at what price shall it be marked, that the merchant may sell it at 10 % less than the marked price, and still make 20 % proiit? Ans. $1.06f. 86. A merchant bought 20 pieces of cloth, each piece containing 25 yd. at $4f per yard on a credit of 9 mo. ; he sold the goods at $4 per yard on a credit of 4 mo. What was his net cash gain, money being worth 6 ft ? Ans. $173-85. 87. A owes B $1200, to be paid in equal annual payments of $200 each ; but not being able to meet these payments at their maturities, and having an estate 10 years in reversion, he arranges with B to wait until he enters upon his estate, when he is to pay B the whole amount, with 8 % compound interest. What sum will B then re- ceive? Ans. $1996.074+. 88. A gentleman who was entitled to a perpetuity of $3000 a year, provided in his will that, after his decease, his oldest son should receive it for 10 yr., then his second son for the next 10 yr., and a literary institution for ever afterward. What was the value of each bequest at the time of his decease, allowing compound interest at 6 fo ? Ans. To oldest son, $22080.28 ; to second son, $12329.51 ; to insti- tution, $15590.23. 89. B has 3 teams engaged in transportation ; his horse team can perform the trip in 5 days, the mule team in 7 days, and the ox team in 11 days. Provided they start together, and each team rests a day after each trip, how many days will elapse before they all rest the same day? Ans. 23 days. 90. A man bought a farm for $4500, and agreed to pay principal and interest in 4 equal annual installments; how much was the annual payment, interest being 6 fi ? Ans. $1298.67 + . 91. A bought a piece of property of B, and gave him his bond for $6300, dated Jan. 1, 1860, payable in 6 equal annual instalments of $1050, the first to be paid Jan. 1, 1861. A took up his bond Jan. 1, 1864, semi-annual discount at the rate of 6 % per annum on the two payments which fell dne alter Jan. 1, 1864, being deducted ; what sum canceled the bond? Ans. $2972.54 + . 92. A gentleman desires to set out a rectangular orchard of 864 trees, go plfrcad that the number of rows shall be to the number of trees in a STOW, as 3 to 2. If the trees are 7 yards apart, how much ground will the orchard occupy ? Ans. 39445 sq. yd. 93. S. C. Wilder bought 25 shares of bank stock at an advance of 6 % on the par value of $100. From the time of purchase until the end of 3 yr. 3 mo. he received a semi-annual dividend of 4 x30xlO = 600 ft., area. 2. What is the area of an isosceles triangle whose base is 20 ft., each of its equal sides 15 ft. ? Ans. 111.85 sq. ft. RULE, from half the sum of the three sides subtract each side separately ; multiply the half-sum and the three remainders together ; the square root of the product is the area. 3. How many acres in a field in the form of an equilateral tri- angle whose sides measure 70 rods? Ans. 13 A. 41.76 P. 4. The roof of a house 30 ft. wide has the rafters on one side 20 ft. long, and on the other 18 ft. long. How many square feet of boards will be required to board up both gable ends ? TKIANGLES. 425 The following principles relating to right-angled triangles have been established Iby Geometry : 1. The square of the hypothenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. 2. The square of the base, or of the perpendicular, of a right-angled tri- angle is equal to the square of the hypothenuse diminished by the square of the other side. 733. To find the hypothemise, 1. The base of a right-angled triangle is 12, and the perpendicu- lar 16. What is the length of the hypothennse? OPERATION. 12 2 +16 2 = 400 (Prin. 1). /y/400 = 20, hypothenuse. 2. The foot of a ladder is 15 ft. from the base of a building, and the top reaches a window 36 ft. above the base. What is the length of the ladder ? Ans. 39 ft. RULE. Extract the square root of the sum of the squares of the base and the perpendicular ; and the result is the hypothenuse. 3. If the gable end of a house 40 feet wide is 16 ft. high, what is the length of the rafters ? 4. A park 25 ch. long and 23 ch. wide has a walk running through it from opposite corners in a straight line. What is the length of the walk? Ans. 33.97 ch. -f . 5. A room is 20 ft. long, 16 ft. wide, and 12 ft high. What is the distance from one of the lower corners to the opposite upper corner? Ans. 28 ft. 3.36 in. 734L To find the base or perpendicular. 1. The hypothenuse of a right-angled triangle srpendicular 28 ft. What is the base ? OPERATION. 35 2 - 28 2 = 441 (Prin. 2). ^441 = 21, base. 1. The hypothenuse of a right-angled triangle is 35 ft., and the perpendicular 28 ft. What is the base? 426 MENSURATIOX. 2. The hypothenuse of a right-angled triangle is 53 yd. and the base 84 feet. Find the perpendicular. RULE. Extract the square root of the difference between the square of the hypothenuse and the square of the given side y and the result is the required side. 3. A line reaching from the top of a precipice 120 ft. high, on the bank of a river, to the opposite side is 380 ft. long. How wide is the river ? Ans. 360 ft. 6 in. -j- . 4. A ladder 52 feet long stands against the side of a building. How many feet must it be drawn out at the bottom that the top may be lowered 4 feet? Ans. 20 ft. QUADRILATERALS. 735. A Quadrilateral is a plane figure bounded by four straight lines, and having four angles. There are three kinds of quadrilaterals, the Parallelogram, Trapezoid and Trapezium. 736. A Parallelogram is a quadrilateral which has its oppo- site sides parallel. There are four kinds of parallelograms, the Square, Rectangle, Rhom- boid, and Rhombus. 737. A Eectangle is any parallelogram having its angles right angles. 738. A Square is a rectangle whose sides are equal. 73O. A Rhomboid is a parallelogram whose opposite sides only are equal, but whose angles are not right angles. 74O. A Rhombus is a parallelogram whose sides are all equal, but whose angles are not right angles. Square. Rectangle. Rhomboid. Rhombus. QUADRILATERALS. 427 7 4 1 o A Trapezoid is a quadrilateral, two of whose sides are parallel and two oblique. 743. A Trapezium is a quadrilateral having no two sides parallel. 743. The Altitude of a parallelogram or trapezoid is the per- pendicular distance between its parallel sides. The vertical dotted lines in the figures represent the altitude. 744. A Diagonal of a plane figure is a straight line joining the vertices of two angles not adjacent. Parallelogram. Trapezoid. Trapezium. PROBLEMS. 745. To find the area of any parallelogram, 1. Find the area of a parallelogram whose base is 16.25 ft. and altitude 7.5 feet. OPERATION. 16.25 x 7.5 = 121.875 sq. ft., area. 2. The base of a rhombus is 10 ft. 6 in., and its altitude 8 ft. What is its area ? RULE. Multiply the base by the altitude. 3. How many acres in a piece of land in the form of athomboid, the base being 8.75 ch. and altitude 6 ch. ? Ans. 54- A. 746. To find the area of a trapezoid. 1. Find the area of a trapezoid whose parallel sides are 23 and 11 feet, and the altitude 9 feet. OPERATION. 23 ft. + 11 ft.A-2 = 17 ft. ; 17 ft. x 9 153 sq. ft., area. 2. Required the area of a trapezoid whose parallel sides are 178 and 146 feet, and the altitude 69 feet. Ans. 11178 sq. ft. RULE. Multiply one-half the sum of the parallel sides by the altitude. 428 MENSUKATION. 3. How many square feet in a board 16 ft. long, 18 in. wide at one end and 25 in. wide at the other end? 4. One side of a quadrilateral field measures 38 rd. ; the side opposite and parallel to it measures 26 rd., and the distance between the two sides is 10 rd. Find the area. Ans. 2 A. 74:7. To find the area of a trapezium. 1. Find the area of a trapezium whose diagonal is 42 ft. and perpendiculars to this diagonal, as in the diagram, are 16 ft. and 1 8 ft. OPERATION. (18 it. + 16 ft. -r-2) x 42 = 714 sq. ft., area. 2. Find the area of a trapezium whose diagonal is 35 ft. 6 in., and the perpendiculars to this diagonal 9 ft. and 12 ft. 6 in. RULE. Multiply the diagonal by half the sum of the perpendicu- lars drawn to it from the vertices of the opposite angles. 3. How many acres in a quadrilateral field whose diagonal is 80 rd. and the perpendiculars to this diagonal 20.453 and 50.832 rd.? To find the area of any regular polygon, multiply its perimeter, or the sum of its sides, by the perpendicular falling from its centre to one of its sides. To find the area of an irregular polygon, divide the figure into triangles and trapeziums, and find the area of each separately. The sum of these areas will be the area of the whole polygon. THE CIRCLE. 748. A Circle is a plane figure bounded by a curved line, called the circumference, every point of which is equally distant from a point within called the center. 749. The Diameter of a circle is a line pass- ing through its center, and terminated at both ends by the circumference. 75O. The Radius of a circle is a line extending from its center to any point in the circumference. It is one-half the diameter. CISCLE. 429 PROBLEMS. . When either the diameter or the circumference of a circle is given, to find the other dimension of it. 1. Find the circumference of a circle whose diameter is 20 in. OPERATION. 20 in. x 3.1416 = 62.832 in. = 5 ft. 2.832 in., circum. 2. What is the circumference of a wheel 5 ft. 6 in. in diameter. 3. Find the diameter of a circle whose circumference is 62.832 ft. OPERATION. 62.832 ft. -4-3.1416 = 20 ft., diameter. 4. Find the diameter of a wheel whose circumference is 50 ft. RULE 1. Multiply the diameter by 3.1416 ; the product is the cir- cumference. 2. Divide the circumference by 3.1416; the quotient is the diameter. 5. What is the diameter of a tree whose girt is 18 ft. 6 in.? 6. Find the length of tire that will band a wheel 7 ft. 9 in. in diameter. Ans. 24 ft. 4 in. + 7. The diameter of a cylinder is 8 ft. 6 in. ; find its girt. 8. What is the radius of a circle whose circumference is 31.41 6 ft. ? 9. The radius of a circle is 10 ft. ; what is its circumference? 10. Find the circumference of the greatest circle that can be drawn with a string 14 in. long, used as a radius. 7 ft. 3.96 in. . To find the area of a circle, when both its diam- eter and circumference are given, or when either is given. 1. Find the area of a circle whose diameter is 10 ft. and circum- ference 31.416 feet. OPERATION. 31.410 ft. xlO-s-4 = 78.54 sq. ft., area. 2. Find the area of a circle whose diameter is 10 ft. OPERATION. 10 ft. 2 x .7854 = 78.54 sq. ft., area. 3. Find the area of a circle whose circumference is 31.416 ft. OPERATION. 31.416 ft. -5- 3. 1416 = 10 ft., diem.; (10 ft.) 2 x .7854 = 78.54 sq. ft. , area. RULES. To find the area of a circle : 1. Multiply the square of its diameter by .7854. 2. Multiply J of its diameter by the circumference. 430 MENSURATION. 4. What is the area of a circular pond whose circumference is 200 chains? Ans. 318.3 A. 5. The distance around a circular park is 1 J miles. How many acres does it contain ? Ans. 114.59" A. 6. Find the area of the largest circle that can be drawn by using as a radius a striiig 20 in. long. 753. To find the diameter or circumference of a circle, when the area is given. 1. What is the diameter of a circle whose area is 1319.472 ? OPERATION. 1319. 472-J-.7854 = 1680 ; V 1680 = 40.987 + , diameter. 2. What is the circumference of a circle whose area is 19.635 ? OPERATION 19.635 -f- 3.1416 = 6.25 ; ^(^25=2.5, radius; 2.5 x 2 x 8.1416 = 15.708, circumference. RULE 1. Divide the area by .7854 and extract the square root of the quotient ; the result is the diameter. 2. Divide the area by 3.1416 and extract the square root of the quotient ; the result is the radius. The circumference is obtained by Art. (751, 1). Or, 3. Divide the area by .07958, and extract the square root of the quotient. 3. The area of a circular lot is 38.4846 square rods. What is its diameter? Ans. 7 rods. 4. The area of a circle is 286.488 square feet. Required the diameter and the circumference. 754. To find the side of an inscribed square when the diameter of the circle is known. 1. What is the side of a square inscribed in a circle whose diameter is 6 rods? OPERATION. 6 2 -s- 2 = 18 ; \/lS = 4.24 rd., side of square. 2. The diameter of a circle is 200 feet. Find the side of the inscribed square. CIRCULAR EING. 431 RULE. Extract the square root of half the square of the diam- eter. Or, Multiply the diameter by .7071. 3. The circumference of a circle is 104 yards. Find the side of the inscribed square. Ans. 23.4 yd. + . 755. To find the area of a circular ring, formed by two concentric circles. 1. Find the area of a circular ring, when the diameters of the circles are 20 and 30 feet. OPERATION. (30 -f- 20 x 30 20) x .7854 = 392.7 sq. ft., area. 2. Find the area of a circular ring formed by two concentric circles, whose diameters are 7 ft. 9 in. and 4 ft. 3 in. Ans. 32.9868 sq. ft, RULE. Multiply the sum of the two diameters by their difference, and the product by .7854; the result is the area. 3. Two diameters are 35.75 and 16.25 feet ; find the area of the ring. Ans. 796.39 sq. ft. 4. The area of a circle is 1 A. 154.16 P. In the center is a pond of water 10 rods in diameter; find the area of the land and of the water. Ans. Land, 235.62 P. ; water, 78.54 P. 756. To find a mean proportional between two numbers. 1. What is a mean proportional between 3 and 12 ? OPEKATION. -\/12x3 = 6, the mean proportional. When three numbers are proportional, the product of the extremes is equal to the square of the mean. RULE. Extract the square root of the product of the two numbers. Find a mean proportional between 2. 36 and 81. 3. 42 and 168. 4. 64 and 12.25. 5. 8 and 288. 6. ff and 7. f and 8. A tub of butter weighed 36 Ib. by the grocer's scales; but being placed in the other scale of the balance, it weighed only 30 Ib. What was the true weight of the butter? Ans. 32.86+ Ib. 432 MENSURATIOK. SIMILAR PLANE FIGURES. 757. Similar Plane Figures arc such as have the same form; or have angles equal each to each, the same number of sides, and the sides containing the equal angles proportional. All circles, squares, equiangular triangles, and regular polygons of the same number of sides are similar figures. The like dimensions of circles, that is, their radii, diameters, and cir- cumferences, are proportional. PRINCIPLES. 1. The like dimensions of similar plane figures are proportional. 2. The areas of similar plane figures are to each other as the squares of their like dimensions. And conversely, 3. The like dimensions of similar plane figures are to each other as the square root of their areas. The same principles apply also to the surfaces of all similar figures, such as triangles, rectangles, etc. ; the surfaces of similar solids, as cubes, pyramids, etc. ; and to similar curved surfaces, as of cylinders, cones, and spheres. Hence, 4. The surfaces of all similar figures are to each other as the squares of their like dimensions. And conversely, 5. Their dimensions are as the square roots of their surfaces. PROBLEMS. 1. A triangular field whose base is 12 ch. contains 2 A. 80 P. Find the area of a field of similar form whose base is 48 ch. OPERATION. 12 2 : 48* : : 2 A. 80 P. : x = 6400 P = 40 A., area. (PRIN. 2.) 2. The side of a square field containing 18 A. is 60 rd. long. Find the side of a similar field that contains -J- as many acres. OPERATION. 18 A. : 6 A. : : 60 2 : tf = 1200 ; V^OO = 34.64 rd. + , side. (PRIN. 3.) 3. Two circles are to each other as 9 to 16; the diameter of the less being 112 feet, what is the diameter of the greater? OPERATION. 9 : 16 : : 112 2 : x*=3 : 4 : : 112 : 03=149 ft. 4 in., diameter. (PRIN. 2.) 4. A peach orchard contains 720 sq. rd., and its length is to its breadth as 5 to 4 ; what are its dimensions? OPERATION. The area of a rectangle 5 by 4 equals 20 (745). 20 : 720 : : 5 s : tf - 900 ; ^900 = 30 rd., length. 20 : 720 : : 4 9 : z 5 = 576 ; ^576 = 24 rd., width. REVIEW. 433 5. It is required to lay out 283 A. 107 P. of land in the form of a rectangle, so that the length shall be 3 times the width. Find the dimensions. Ans. 369 rd. ; 123rd. 6. A pipe 1.5 in. in diameter fills a cistern in 5 hr. ; find the diam- eter of a pipe that will fill the same cistern in 55 miu. 6 sec. Ans. 3.5 in. 7. The area of a triangle is 24276 sq. ft., and its sides in propor- tion to the numbers 13, 14, and 15. Find the length of its sides ia feet. Ans. 221, 238, and 256 ft. 8. A field containing 6 A. is laid down on a plan to a scale of 1 in. to 20 ft. How much paper will it cover? Ans. 653. 4 sq. in. 9. If it cost $167.70 to enclose a circular pond containing 17 A. 110 P., what will it cost to enclose another ^ as large ? Ans. $75. 10. If a cistern 6 ft. in diameter holds 80 bbl. of water, what is the diameter of a cistern of the same depth, that holds 1 200 bbl. 11. If 63.39 rd. of fence wilt enclose a circular field containing 2 A., what length will enclose 8 A. in circular form ? Ans. 126.78 rd. 758. REVIEW OF PLANE FIGURES. PROBLEMS. 1. The area of a triangle is 270 yd., and the perpendicular 45 ft. Find the base. 2. Find the area of a square whose perimeter is the same as that of a rectangle 48 ft. by 28 feet. 3. A rectangle whose length is 3 times its width contains 1323 P. Find its dimensions. Ans. 21 rd. by 63 rd. 4. Find the area of an equilateral triangle whose sides are 36 ft. 5. The area of a circle is 7569 square feet. Find the length of the side of a square of equal area. Ans. 87 ft. 6. How much less will the fencing of 20 A. cost in the square form than in the form of a rectangle whose breadth is -J- the length, the price being $2.40 per rod? Ans. $185.43. 7. A house that is 50 ft. long and 40 ft. wide has a square or pyramidal roof, whose height is 15 ft. Find the length of a rafter reaching from a corner of the building to the vertex of the roof. 8. Find the length of a rafter reaching from the middle of ono side. Ans. 25 ft. 434 MENSUEATION. 9. Find the length of a rafter reaching from the middle of one end. 10. What is the diameter of a circular island containing 1J sq. miles ? Ans. 403.7 rd. 11. How many rods more of fencing are required to enclose a square field whose area is 5 acres, than to enclose a circular field having the same area? 12. What is the value of a farm, at $75 an acre, its form being a quadrilateral, with two of its opposite sides parallel, one 40 chains and the other 22 chains long, and the perpendicular distance between them 25 chains ? Ans. $5812.50. 13. What is the difference in the area of a grass plat 20 feet square and a circular plat 20 feet in diameter? 14. Find the cost, at 18 cents a square foot, of paving a space in the form of a rhombus, the sides of which are 15 ft., and a per- pendicular drawn from one oblique angle will meet the opposite side 9 feet from the adjacent angle. Ans. $32.40. 1 5. A goat is fastened to the top of a post 4 feet high by a rope 50 ft. long. Find the circumference and area of the greatest circle over which he can graze. 16. How much larger is a square circumscribing a circle 40 rd. in diameter, than a square inscribed in the same circle? 17. What is the value of a piece of land in the form of a tri- angle, whose sides are 40, 48, and 54 rods, respectively, at the rate of $125 an acre? Ans. $724.75. 18. The radius of a circle is 5 feet; find the diameter of anothei circle containing 4 times the area of the first. Ans. 20 ft. 19. Find the difference in the area of a circle 36 feet in diameter, and the inscribed square. 20. How many acres in a semi-circular farm, whose radius i? 100 rods? Ans. 98 A. 28 P. 21. What must be the width of a walk extending around a gar- den 100 feet square, to occupy one-half the ground? 22. An irregular piece of land, containing 540 A. 36 P., is ex- changed for a square piece of the same area ; find the length of one of its sides. If divided into 42 equal squares, what is the length of the side of each? Ans. to last, 45.86 rd. SOLIDS. 435 23. A field containing 15 A. is 30 rd. wide, and is a plane inclining in the direction of its length, one end being 120 ft. higher than the other. Find how many acres of horizontal surface it contains. 24. If a pipe 3 inches in diameter discharge 12 hogsheads of water in a certain time, what must be the diameter of a pipe which will discharge 48 hogsheads in the same time? Ans. 6 in. SOLIDS. 759. A Solid or Body has three dimensions, length, breadth, and thickness. The planes which bound it are called its faces, and their intersections, 76. A Prism is a solid whose ends are equal and parallel polygons, and its sides parallelograms. Prisms take their names from the forms of their bases, as triangular, quadrangular, pentagonal, etc. 761. The Altitude of a prism is the perpendicular distance between its bases. 762. A Parallelopipedon is a prism bounded by six parallelograms, the opposite ones being parallel and cquai. 763. A Cube is a parallelopipedon whose faces are all equal squares. Cube> 764. A Cylinder is a body bounded by a uniformly curved surface, its ends being equal and parallel circles. 1. A cylinder is conceived to be generated by the revolution of a rectan- gle about one of its sides as an axis. 2. The line joining the centres of the bases, or ends, of the cylinder is its altitude, or axis. Triangular Prisin. Quadrangular Pi-ism' Pentagonal Prism. Cylinder. 436 MENSUBATIOtf. PROBLEMS. 765. To find the convex surface of a prism or cylinder. 1. Find the area of the convex surface of a prism whose altitude is 7 feet, and its base a pentagon, each side of which is 4 feet, OPERATION. 4 ft. x 5 = 20 ft., perimeter. 20 ft. x 7 = 140 sq. ft., convex surface. 2. Find the area of the convex surface of a triangular prism, whose altitude is 8-J- feet, and the sides of its base 4, 5, and 6 feet, respectively. OPERATION. 4 ft. + 5 ft. + 6 ft. = 15 ft., perim. 15 ft. x 8 = 127 sq. ft., convex surface. 3. Find the area of the convex surface of a cylinder whose alti- tude is 2 ft. 5 in. and the circumference of its base 4 ft. 9 in. OPERATION. 2 ft. 5 in. =29 in. ; 4 ft. 9 in. = 57 in. 57 in. x 29 = 1653 sq. in. =11 sq. ft. 69 sq. inches, convex surface. RULE. Multiply the perimeter of the base by the altitude. To find the entire surface, add the area of the bases or ends. 4. If a gate 8 ft. high and 6 ft. wide revolve upon a point in its centre, what is the entire surface of the cylinder described by it ? 5. Find the superficial contents, or entire surface of a parallelo- pipedon 8 ft. 9 in. long, 4 ft, 8 in. wide, and 3 ft, 3 in. high. 6. What is the entire surface of a cylinder formed by the revo- lution about one of its sides of a rectangle that is 6 ft. 6 in. long and 4 fee wide? Ans. 263.89 sq. ft. 7. Find the entire surface of a prism whose base is an equilateral triangle, the perimeter being 18 ft., and the altitude 15 feet? PYRAMIDS AKD COXES. 437 766. To find the volume of any prism or cylinder. 1. Find the volume of a triangular prism, whose altitude is 20 ft., and each side of the base 4 feet. OPERATION. The area of the base is 6.928 sq. ft. (729). 6.928 sq. ft. x 20 = 138.58 cu. ft., volume. '2. Find the volume of a cylinder whose altitude is 8 ft. G in., and the diameter of its base 3 feet. OPERATION. 3 2 x .7854 = 7.0686 sq. ft., area of base (752). 7.0686 sq. ft. x 8.5 60.083 cu. ft., volume. RULE. Multiply the area of the base by the altitude. 3. What is the volume of a parallelopipedon, whose base is 9.8 ft. by 7.5 ft., and its height 5 ft. 3 in. 4. What is the volume of a log 18 ft. long and 1J- ft. in diameter? 5. Find the solid contents of a cube whose edges are 6 ft. 6 in.? 6. Find the cost of a piece of timber 18 in. square and 40 ft. long, at $.30 a cubic foot. Ans. $27. 7. Required the solid contents of a cylinder \vhosc altitude is 15 ft. and its radius 1 ft. 3 in. Ans. 73.63 cu. ft. PYRAMIDS AND COXES. 767. A Pyramid is a body, having for its base a polygon, and for its other faces three or more triangles, which terminate in a com- mon point called the vertex. Pyramids, like prisms, take their names from their bases, and are called triangular, square, or quadrangular, pentagonal, etc. Pyramid. Frustum. Cone. Frustum. ^68 > A Cone is a body having a circular base, and whose con- vex surface tapers uniformly to the vertex. A cone is a body conceived to be formed by the revolution of a right- angled triangle about one of its sides containing the right angle. 76O. The Altitude of a pyramid or of a cone is the perpendic- ular distance from its vertex to the plane of its base. 438 MENSURATION. 770. The Slant Height of a pyramid is the perpendicular dis- tance from its vertex to one of the sides of the base; of a cone, is a straight line from the vertex to the circumference of the base. 771. The Frustum of a pyramid or cone is that part -which remains after cutting off the top by a plane parallel to the base. THE SPHERE. 772. A Sphere is a body bounded by a uniformly curved sur- face, all the points of which are equally distant from a point within called the center. 778. The Diameter of a sphere is a straight line passing through the center of the sphere, and terminated at both ends by its surface. 774. The Radius of a sphere is a straight line drawn from the center to any point in the surface. PROBLEMS. 775, To find the convex surface of a pyramid or cone. 1 . Find the convex surface of a triangular pyramid, the slant height being 16 ft., and each side of the base 5 feet. OPERATION. (5 ft. + 5 ft. +5 ft.) x 16^2 =120 sq. ft., convex surface. 2. Find the convex surface of a cone whose diameter is 17 ft. 6 in., and the slant height 30 feet. OPEKATION. 17.5 ft. x 3.1416=54.978 ft., circum. ; 54.978 ft. x 30-=-2= 824.67 sq. ft., convex surface. RULE. Multiply the perimeter or circumference of the base by one-half the slant height. To find the entire surface, add to this product the area of the base. 3. Find the entire surface of a pyramid whose base is 8 ft. 6 in. square, and its slant height 21 feet. 4. Find the entire surface of a cone the diameter of whose base is 6 ft. 9 in. and the slant height 45 feet. Ans. 512.9 sq. ft. 5. Find the cost of painting a church spire, at $.25 a sq. yd., whose base is a hexagon 5 ft. on each side, and the slant height 60 feet. PYRAMIDS AND COKES. 439 77G. To find the volume of a pyramid or of a cone. 1. What is the volume, or solid contents, of a square pyramid whose base is 6 feet on each side, and its altitude 12 feet. OPERATION. 6 x 6 x 12 -f- 3 = 144 cu. ft., volume. 2. Find the volume of a cone, the diameter of whose base is 5 ft. and its altitude 10|- feet. OPERATION. 5 2 ft. x .7854 x 10JT3 = 68.72 cu. ft., volume. RULE. Multiply the area of the base by one-third the altitude. 3. Find the solid contents of a cone whose altitude is 24 ft, and the diameter of its base 30 inches. 4. What is the cost of a triangular pyramid of marble, whose altitude is 9 feet, each side of the base being 3 feet, at $2^- per cubic foot? Ans. $29.25. 5. Find the volume and the entire surface of a pyramid whose base is a rectangle 80 ft. by 60 ft, and the edges which meet at the vertex are 130 feet, Ans. 192000 cu. ft. voL 777. To find the convex surface of a frustum of a pyra- mid or cone. 1. What is the convex surface of a frustum of a square pyramid, whose slant height is 7 ft., each side of the greater base 4 ft., and of the less base 1 8 inches ? OPERATION. The perimeter of the greater base is 16 ft., of the less 6 ft. 16 ft. + 6 ft. x7-r-2 = 77 sq. ft., convex surface. 2. Find the convex surface of a frustum of a cone whose slant height is 15 ft., the circumference of the lower base 30 ft., and of the upper base 16 feet. RULE. Multiply the sum of the perimeters, or circumferences, by one-half the slant height. To find the entire surface, add to this product the areas of both ends, or bases. 3. How many square yards in the convex surface of a frustum of a pyramid, whose bases are heptagons, each side of the lower base being 8 feet, and of the upper base 4 feet, and the slant height 55 feet? Ans. 256-| sq. yd. 440 MEKSUEATIOK". 778. To find the volume of a frustum of a pyramid or cone. 1. Find the volume of a frustum of a square pyramid whose alti- tude is 10 feet, each side of the lower base 12 feet, and of the upper base 9 feet. OPERATION. 12 2 + 9 2 = 225 ; (225 + ^144 x 81) x 10 -T- 3 = 1000 cu. ft., 2. How many cubic feet in the frustum of a cone whose altitude is 6 feet and the diameters of its bases 4 feet and 3 feet? RULE. To the sum of the areas of both bases add the square root of the product, and multiply this sum by one-third of the altitude. 3. How many cubic feet in a piece of timber 30 ft. long, the greater end being 15 in. square, and that of the less 12 in. ? 4. How many cubic feet in the mast of a ship, its height being 50 ft., the circumference at one end 5 ft. and at the other 3 feet. 779. To find the surface of a sphere. 1. Find the surface of a sphere whose diameter is 9 inches. OPERATION. 9 in. x 3.1416 = 28.2744 in., circumference. 28.2744 in. x 9 = 254.4696 sq. in., surface. RULE. Multiply the diameter by the circumference of a great cir- cle of the sphere. 2. What is the surface of a globe 1 6 in. in diameter ? 3. Find the area of the surface of a sphere whose circumference is 31.416 feet. Ana. 314.16 sq. ft. 4. Find the surface of a globe whose radius is 1 foot. 780. To find the volume of a sphere. 1. Find the volume of a sphere whose diameter is 18 inches. OPERATION. 18 in. x 3.1416 = 56.5488 in., circumference. 56.5488 in. x 18 = 1017.8784 sq. in., surface. 1017.8784 sq. in. x 18-f-6 = 3053.6352 cu. in., volume. RULE. Multiply the surface by ^ of the diameter, or ^ of the radius. 2. Find the volume of a globe whose diameter is 30 in. 3. Find the solid contents of a globe whose radius is 5 yards. 4. Find the volume of a globe whose circumference is 31.416 ft. EEVIEW OF SOLIDS. 441 781. To find the three dimensions of a rectangular solid, the volume and the ratio of the dimensions being given. 1. What are the dimensions of a rectangular solid, whose volume is 4480 cu, ft., and its dimensions are to each other as 2, 5, and 7 ? OPERATION. ^/4480-*-(2 x 6 x 7) = 4 ft. ; 4 x 2 = 8 ft., height; 4x5 = 20 ft., width ; 4 x 7 = 28 ft., length. RULE. I. Divide the volume by the product of the terms propor- tional to the three dimensions, and extract the cube root of the quo- tient. II. Multiply the root thus obtained by each proportional term ; the products will be the corresponding sides. 2. What are the dimensions of a rectangular box whose volume is 3000 cu. ft., and its dimensions are to each other as 2, 3, and 4 ? 3. A pile of bricks in the form of a parallelepiped contains 30720 cu. ft., and the length, breadth, and height are to each other as 3, 4, and 5. What are the dimensions of the pile ? 4. Separate 405 into three factors, which shall be to each other as 2, 2J, and 3. Ann. 6, 7, and 9. SIMILAR SOLIDS. 782. Similar Solids are such as have the same/orw, and differ from each other only in volume. PRINCIPLES. 1. The volumes of similar solids are to each other as the cubes of their like dimensions. If the volume of a ball 3 in. in diameter is 27 cii. in., what is the volume of a ball 7 in. in diameter? OPERATION. 3 3 : 7 3 : : 27 cu. in. : x = 343 cu. in., volume. 2. The like dimensions of similar solids are to each other as the tube roots of their volumes. If the diameter of a ball whose volume is 27 cu. in. is 3 in., what is the diameter of a ball whose volume is 343 cu. in.? OPERATION. 27 : 343 : : 3 s : x* = 373 ; ^/373 = 7 in., diameter. 44:2 MENSUBATION. 783. REVIEW OF SOLIDS. PROBLEMS. 1. What is the edge of a cube whose entire surface is 1050 sq. ft., and what is its volume ? 2. What must be the inner edge of a cubical bin to hold 1250 bushels of wheat? Ans. 11 ft. 7 in. 3. How many globes 4 in. in diameter are equal to one whose diameter is 12 inches? 4. How many gallons will a cistern hold, whose depth is 7 ft., the bottom being a circle 7 ft. in diameter and the top 5 feet in diameter? Ans. 1494.25 gal. O 5. What is the value of a stick of timber 24 ft. long, the larger end being 15 in. square, and the less 6 in., at 28 cents a cubic foot? 6. If a cubic foot of iron were formed into a bar -J an inch square, without waste, what would be its length ? Ans. 576 ft. 7. How many barrels of 31-J gal. will a cistern hold that is 8.3 ft. in diameter, and 7 feet deep? 8. If a log 18 ft. long and 3 ft. in diameter is hewn square, how many cubic feet does it contain ? 9. Find the volume of a cube, the area of whose entire surface is 7 sq. ft. 6 sq. inches. Ans. 1 cu. ft. 469 cu. in. 10. If a marble column 10 in. in diameter contains 27 cu. ft., what is the diameter of a column of equal length that contains 81 cubic feet? Ans. 14.42 in. 11. Supposing the earth to be a perfect sphere 7912 miles in diameter, what is its volume in cubic miles ? 12. How many board feet in a post 11 ft. long, 9 in. square at the bottom, and 4in. square at the top? Ans. 40 ft. 7-f. 13. The surface of a sphere is the same as that of a cube, the edge of which is 12 in. Find the volume of each. 14. The contents of a cubical block of marble are 4913 cu. ft. Find the superficial contents or surface. Ans. 1014 sq. ft. 15. Find the dimensions of a bin that holds 450 bu. of grain, if the width and depth are equal, and the length 3 times the width. 16. A ball 4.5 in. in diameter weighs 18 oz. Avoir.; what is the weight of a ball of the same density, that is 9 in. in diameter? GAUGING. 443 17. In what time will a pipe supplying 6 gal. of water a minute, fill a tank in the form of a hemisphere, that is 10 ft. in diameter? 18. If the altitude of a cone that weighs 640 Ib. is 8 ft,, what is the altitude of a similar cone that weighs 270 Ib. ? 19. If a stack of hay 8 ft. high weighs 8 cwt., what is the weight of a similar stack that is 24 ft. high? Ans. 216 cwt. 20. The diameter of a cistern is 8 ft. ; what must be its depth to contain 75 hhd. of water? Ans. 12.56 ft. 21. If a cable 3 inches in circumference supports a weight of 2500 pounds, what must be the circumference of a cable that will support 4960 pounds? Ans. 3.77 in., nearly. 22. How many bushels in a heap of grain in the form of a cone, whose base is 8 ft. in diameter and altitude 4 feet ? GAUGING. 784. Gauging is the process of finding the capacity or volume of casks and other vessels. For ordinary purposes the diagonal rod is used, which gives only approximate results. A cask is equivalent to a cylinder having the same length and a diameter equal to the mean diameter of the cask. 785. To find the mean diameter of a cask (nearly). RULE. Add to the head diameter -|, or, if the staves are but little curved, .6, of the difference between the head and bung diameters. 786. To find the volume of a cask in gallons. RULE. Multiply the square of the mean diameter by the length (both in inches) and this product by .0034. 1. How many gallons in a cask whose head diameter is 24 in., bung diameter 30 in., and its length 34 inches? OPERATION. 24 + (30 24 x f ) = 28 in., mean diameter. 28 2 x 34 x .0034 = 90.63 gal., capacity. 2. What is the volume of a cask whose length is 40 in., the diameters 21 and 30 in., respectively? Ans. 99.14 gal. 3. How many gallons in a cask of slight curvature, 3 ft. 6 in, long, the head diameter being 26 in., the bung diameter 31 in. ? 444 MENSTJKATIOH. 787. CIRCLES. 1. The diameter of any circle Multiplied), j 3.1416, the product ) ,, . tv M a f &y } o-ioo i f == the circumference. Divided ) J ( .3 183, the quotient j Multiplied ) , j .8862, the product ) Divided | ^ 1 1.1284, the quotient f = ** ** fan <*"* Multiplied ) ^ j .8660, the product ) = the side of an inscribed Divided [ y ( .1547, the quotient j equilateral triangle. Multiplied [ ^ j .7070, the product ) = the side of an inscribed Divided f . ( 1.4142, the quotient [ square. 2. The radius of any circle Multiplied ) , ( 6.28318) -n- i j f b 7 ) , rrn-, r r the circumference. Divided ) 3 ( .15915 j 3. The square of the diameter of any circle Multiplied ) , ( .7854, the product ) Divided rn 1.2732, the quotient | = thc "^ 4. The circumference of any circle Multiplied ) ( .3183, tlie product | = Divided j J ( 3.1416, the quotient \ Multiplied), ( .2821, the product ) T^ -j j r by 1 r ,. h = theme of aw equal sof re, Divided ) " ( 3.5450, the quotient j Multiplied) . j .2756, the product ) = the side of the inscribed Divided j ( 3.6276, the quotient j equilateral triangle. Multiplied ) . j .2251, the product ) = the side of an inscribed Divided J ^ ( 4.4428, the quotient j square. 5. The square of the circumference of any circle Multiplied | ( .07958, the product ) ^ Divided j J ( 12.5663, the quotient j 6. The area of any circle Multiplied | ( 1.2732, the product ) = ^ re o/ ^ rf .^ Divided ) ^ ( .7854, the quotient j ( The square of the radius of any circle x 3.1416 ^ 7. < Half the circumference of a circle x \ its diameter > = area. ( Square of the circumference of a circle x .07958 ) LAND. 445 788. SPHERES. r Circumference x its diameter. Radius 2 x 12.5664. 1. The Surface H Diameter* x 3.1410. I Circumference 2 x .3183. f Surface x -J- its diameter. Radius 3 x 4.1 888. 2. The Volume = < _.. I Diameter 3 x.5236. I Circumference 3 x .0169. 3. The Diameter = j A/Of surface x .5642. A/Of volume x 1.2407. ( A/Of surface x 1.77255. 4. The Circumference = 1 3 . ( A/Of volume x 3.8978. ( A/Of surface x .2821. 5. The Radius = -J 3 , ( A/Of volume x .6204. ( Radius x 1.1 547. 6.Thesideofamnscnbedcube= 4 DiametcrXe , LAND. ?"8O. The Unit of land measure is the acre. Measurements of land are commonly recorded in square miles, acres, and hundredths of an acre. PROBLEMS. 1. What is the value of a farm 189.5 rods long and 150 rods wide, at $42J an acre ? 2. A man having a field 70 rd. square appropriated 5 A. of it to corn, 100 sq. rd. to garden vegetables, and the remainder to meadow. What fractional part of the whole field did the meadow comprise ? 3. I bought a piece of land 16 ch. long and 15 ch. wide, at $100 an acre, and dividing it into lots of 6 rods by 5 rods, sold them at $50 each. What was my gain ? Ans. $4000. 4. At $2.75 a rod, how much less will it cost to fence a piece of land 80 rods square, than if the same were in the form of a rectangle twice as long and one-half as wide ? Ans. $220. 446 MENSURATION. T9O. Government Lands are usually surveyed into rectangular tracts, bounded by lines conforming to the cardinal points of the compass. A Base-line on a parallel of latitude, and a Principal Meridian intersecting it, are first established. Other lines are then run six miles apart, each way, as nearly as possible. The tracts thus formed are called Townships, and contain, as near as may be, 23040 acres. A line of townships extending north and south is called a Range* The ranges are designated by their number east or west of the princi- pal meridian. The townships in each range are designated by their number north or south of the base-line. Since the earth's surface is convex, the principal meridians converge as they proceed northward. This tends to throw the townships and sec- tions out of square, and necessitates occasional lines of offset, called " correction lines." Townships are subdivided into Sections, and sections into Half- Sections, Quarter- Sections, Half -Quarter- Sections, Quarter- Quarter- Sections, and Lots. Diagram No. 1. A TOWNSHIP. N Diagram No. 2. 6 5 4 3 2 1 7 8 9 10 11 12 18 17 16 15 14 13 19 20 21 22 23 24 30 29 28 27 26 25 31 32 33 34= 35 36 A SECTION. N N.W. ^ N. W. Y 4O A. E,X of N.W.H 80 A. N, E, H ISO A. S. W. H of N. W. * 8.* 320 A. Diagram No. 1 shows the sub-divisions of a Township into Sections, and how they are numbered, commencing at the N. E. corner. Diagram No. 2 shows the sub-divisions of a Section, on an enlarged icale, and how they are named. LAND. 447 TABLE. 6 mi. x 6 mi. = 36 sq. mi. = 23040 Acres = 1 Township. 1 " x 1 " = 1 " = 640 " =1 Section. 1 " x " = | " = 320 " = 1 Half-Section. $ " x $ " = " = 160 " = 1 Quarter-Section. I " x I " = J " = 80 " =1 Half -Quarter-Section. $ " x -} : " = T V " = 40 " 1 Quarter-Quarter-Section, A Lot is a subdivision of a section, usually of irregular form, on account of bordering upon a navigable river or lake containing as near as may be the area of a Quarter-Quarter-Section, and described as lot No. 1, 2, 3, etc., of a particular section. City and village plats are usually sub-divided into Blocks, and these Into Lots. PROBLEMS. 1. If a township of land is equally divided among 288 families, how many acres does each receive ? What part of a section ? . 2. What number of rails will enclose a quarter-section of land with a fence 6 rails high, and 3 lengths for every 2 rods ; and what will be the cost of the rails, at $40 per thousand? 3. A man bought the S. -|- of a section of land at $2J an acre, and afterward sold the E. -| of what he bought at $4.3 7J an acre. What did he gain on what he sold? Ans. 8340. 4. If I buy the N. E. J and the E. of N. W. J- of a section of land, how many acres do I purchase ? What part of a whole sec- tion ? How are the parts located in respect to each other ? 5. A speculator bought of the 111. Central R. R. Co., the S. of Section 4, township 10 north, range 6 east, at $2 an acre. He after- ward sold the E. -J- of S. E. J at 82.75 an acre ; the N. E. of S. E. J- at $3 an acre ; and the N. of S. W. at $3.84 an acre. How many acres has he left? What was his gain on the purchase price of the whole? Draw diagram. Ans. $27.20. 6. A man having purchased a section of land from the U. S. Government at $1.25 an acre, sold the S. \ of S. W. J at $2.50 an acre; the N.W. J of N.W. J at 81.75 an acre; the W. \ of S.E. } at $2 an acre; and the W. \ of S.W. J of N.E. at $3 an acre. How many acres has he remaining, and what is his gain, provided the remainder is sold at 82-J- an acre ? Draw diagram and explain. 448 MENSURATION. BOARDS AND TIMBER. 791. A Board Foot is I ft. long, 1 ft. wide, and 1 inch thick. Hence 12 board feet make 1 cubic foot. Board feet are changed to cubic feet by dividing by 12, and cubic feet are changed to board feet by multiplying by 12. 1. In Board Measure all boards are assumed to be 1 in. thick. 2. Lumber and sawed timber, as plank, scantling, etc., are usually esti- mated in board measure, hewn and round timber in cubic measure. When lumber is not more than 1 inch thick : RULES. 1. Multiply the length in feet by the width in inches, and divide the product by 12. When more than 1 inch thick : 2. Multiply the length infect by the width and thickness in inches, and divide the product by 12. PROBLEMS. 1. What must be the width of a board 16 ft. long to contain 12 board feet ? OPERATION. 16 ft. = 192 in. ; 144 x 12 -f- 192 = 9 in., the width. 2. What must be the width of a piece of board 5 ft. 3 in. long, to contain 7 square feet? 3. Find the cost of 8 pieces of scantling, 3 in. by 4 in. and 14 ft. long, at $9.50 per thousand board feet. 4. A piece of timber is 10 in. by 16 in. ; what length of it will contain 1 5 cubic feet ? Ans. 1 3 ft. 5. How many board feet in a stick of timber 36 ft. long, 10 in. thick, 1 2 in. wide at one end and 9 in. wide at the other end ? How many cubic feet? Ans. 26J cu. ft. 6. A rectangular field, 16 ch. long and 8 ch. wide, is enclosed by a post and board fence ; the posts are set 8 ft. apart, the boards are 16 ft. long, and the fence is 5 boards high. The bottom board is 12 in. wide, the top board 6 in., and the other three 9 in. wide, The posts cost $25 per C., and the boards $14.80 per M. Required the number of posts, the amount of lumber, and the cost of both. MASONRY. 449 MASONRY. 792. Masonry is estimated by the cubic foot, and by the perch ; also by the square foot and the square yard (287). 1. Brickwork is generally estimated by the thousand bricks; sometimes in cubic feet. 2. When stone is built into a wall, 22 cubic feet make a perch, 2f cubic feet being allowed for mortar and filling. 3. Philadelphia bricks are 8| in. x 4- x 2f ; and Milwaukee bricks 8Hn.x4-x2*. 793. To find the number of bricks in a cu. ft. of masonry. PROBLEMS. 1. How many Milwaukee bricks in a cubic foot of wall 12| in. wide, laid in courses of inortar of an inch thick ? OPEKATION. 8.5 + .25=8.75 in.=length of brick and joint. 2.375 +.25= 2.625 in. = thickness of brick and joint. 8.75 x 2.625 = 22.96875 sq. in. = area of its face. 12.75-^3 (number of bricks in width of wall) = 4.25 m.=icidthoi brick and mortar. 22.96875 x 4.25 = 97.617+ = cubic inches in a brick. 1728-^-97.617+ = 17.7+ = number of bricks in a cubic foot. RULES. I. Add to the face dimensions of the kind of bricks used, one-half the thickness of the mortar or cement in which they are laid, and compute the area. II. Multiply this area by the quotient of the thickness of the wall divided by the number of bricks of which it is composed, the product will be the volume of a brick and its mortar in cubic inches. III. Divide 1728 by this volume, and the quotient will be the num- ber of bricks in a cubic foot. 2. How many bricks, 8 in. x 4 x 2, will be required to build a wall 42 ft. long, 24 ft. high, and 16-J in. thick, laid in courses of mortar J of an inch thick? Ans. 31278^. 3. How many perches of stone, laid dry, will build a wall 60 ft. long, 16 J ft. high, and 18 in. thick? Ans. 60 Pch. 450 MENSURATION. RULES. 1. Multiply the number of cubic feet in the wall, or work to be done, by the number of bricks in a cubic foot ; the product will be the number of bricks required. 2. Divide the number of cubic feet in the work to be done by 24.75 ; the quotient will be the number of perches. 4. How many perches of masonry in a wall 120 ft. long, 6 ft. 9 in. high, and 18 in. thick? 5. At $.56 a cubic yard, what will it cost to remove an embank- ment 240 ft. long, 38 ft. wide, and 8.5 ft. high? 6. Find the cost of digging and walling the cellar of a house, whose length is 41 ft. 3 in., and width 33 ft. ; the cellar to be 8 ft. deep, and the wall 1-|- ft. thick. The excavating will cost $.50 a load, and the stone and mason work $3.75 a perch. Ans. $47 If. 7. What will be the cost of building a wall 60 feet long, 21-J feet high, and 17 inches thick, of Philadelphia bricks, laid in courses of mortar J of an inch thick, at $12 per M. ? Ans. $423.53. 8. How many cubic feet of masonry in the wall of a cellar 37-J-feet long, 26 feet wide, and 9 feet deep, the wall being 2 feet thick, allowing one-half for the corners ; and what will be the cost, at $3.85 a perch? Ans. 2214 cu. ft.; $344.40. CAPACITY OF BINS, CISTERNS, ETC. 794. The Standard Bushel of the United States contains 2150.42 cu. in., and is a cylindrical measure 18J inches in diameter and 8 inches deep (511). 1. Measures of capacity are all cubic, measures, solidity and capacity being measured by different nnits, as seen in the tables. 2. Grain is shipped from New York by the Quarter of 480 Ib. (8 U. S. bu.), or by the ton of 33* U. S. bushels. 3. It is sufficiently accurate in practice to call 5 stricken measures equal to 4 heaped measures. 795. To find the exact capacity of a bin in bushels. RULES. 1. Divide the contents in cubic inches by 2150.42 ; the quotient will represent the number of bushels. Sincn a standard bushel contains 2150.42 cu. in., and a cubic foot con- tains 1728 cu. in., a bushel is to a cubic foot nearly as 4 to 5 ; or a bushel is equal to 1] cu. ft., nearly. Hence for all practical purposes, CAPACITY OF BINS, CISTERNS, ETC. 451 2. Any number of, cubic feet diminished by -J will represent an equivalent number of bushels. Thus, 250 cu. ft. A of 250 cu. ft., or 50 cu. ft. = 200, the number of bushels in 250 cubic feet 3. Any number of bushels increased by will represent an equiva* lent number of cubic feet. Thus, 200 bu. + i of 200 bu., or 50 bu. = 250, the number of cubic feet in 200 bushels. PROBLEMS. 1. How many bushels of wheat can be put in a bin 8 ft. long, 6 ft. 6 in. wide, and 3 ft. 4 in. deep ? 2. What must be the depth of a bin to contain 240 bu., its length being 10 feet and its width 5 feet? OPERATION. 240 bu. + 60 bu. = 300 ; 300-^10 x 5 = 6 ft., the depth. RULE. Divide the contents in cubic feet or inches by the product of the two dimensions, in the same denomination. 3. What must be the length of a bin that is 6 feet wide and 4-| feet deep, to contain 324 bushels? Ans. 15 ft. 4. How many bushels of apples will a bir hold, that is 12 ft. long, 3 ft. wide, and 2 ft. 6 in. deep ? How many of barley ? 5. A bin 20 ft. long, 12 ft. wide, and 5 ft. deep, is full of wheat. What is its value at $2 a bushel ? Ans. $1920. 6. A bin 7 ft. long, 6 ft. wide, and 5 ft. deep, is J full of rye. What is its value at $1.37 a bushel? 7. A crib, the inside dimensions of which are 15 ft. long, 7 ft. 4 in. wide, and 8 ft. high, is full of corn in the ear. If 2 bushels of ears make 1 bushel of shelled corn, what is the value of the whole, when shelled, at -$.92 a bushel ? Ans. $259.07. 8. If 1 bu. or 60 Ib. of wheat make 48 Ib. of flour, how many barrels of flour can be made from the contents of a bin 10 ft. long, 5 ft. wide, and 4 ft, deep, filled with wheat ? Ans. 39^ bbl. 9. Dunkley & Co. bought 12400 bu. of wheat, delivered in New Y"ork, at $1,50 a bushel. They shipped the same to Liverpool, paying 6s. sterling per quarter freight, and sold the entire cargo at 12s. per cental. Making no allowance for exchange or for waste, what was the gross gain in U.S. Money, the being valued at &4.666J-? 452 MENSUKATIOIST. 796. To find the exact capacity of a vessel or space in gallons. PROBLEMS. 1. How many gallons of water will a cistern hold, that is 4 feet square and 6 feet deep ? OPERATION. (4 x 4 x 6 x 172)-^231 - 718|f gal., capacity. RULE. Divide the contents in cubic inches by 231 for liquid gal- lons, or by 2 6 8. 8 for dry gallons. 2. How many cubic feet in a space that holds 48 hhd. ? 3. How many hogsheads will a cistern 11 ft. long, 6 ft. wide, and 7 feet deep contain ? Ans. 54& hhd. 4. How many gallons will a space contain that is 22.5 ft. long, 3.25 ft. wide, and 6.4 ft. deep? 5. A man constructed a cistern to hold 32 hogsheads, the bottom being 6 ft. by 8 ft. What was its depth ? Ans. 5 ft. 7f in. 6. A tank in the attic of a house is 6 ft. 6 in. long, 4 ft. wide, and 3 ft. 6 in. deep. How many gallons of water will it hold, and what will be its weight? Ans. 680-^- gal. ; 5672^ r Ib. 7. If 64 quarts of water be put into a vessel that will exactly hold 64 quarts of wheat, how much will the vessel lack of being full? Ans. 604.8 cu. in. 8. If a man buy 10 bu. of chestnuts at $5 a bushel, dry measure, and sell the same at 25 cents a quart, liquid measure, how much does he gain? Ans. $43.09. 9. A cistern 5 ft. by 4 ft. by 3 ft. is full of water. If it be emptied by a pipe in 1 hr. 30 min., how many gallons are discharged through the pipe in a minute ? Ans. 4-J-f- gal. 10. A vat that will hold 5000 gallons of water, will hold how many bushels of corn ? Ans. 537-^ bu. 11. A cellar 40 ft. long, 20 ft. wide, and 8 ft. deep is half-full of water. What will be the cost of pumping it out, at 6 cents a hogshead? Ans. $22.80. 12. A reservoir 24 ft. 8 in. long by 12 ft. 9 in. wide is full oi water. How many cubic feet must be drawn off to sink the sur* face 1 foot? How many gallons? Ans. 2352^-f gal. THE METEIC SYSTEM OF WEIGHTS AND MEASURES. 797. The Metric System was adopted in France in 1795; its use was authorized in Great Britain in 1864; and in 1866, Con- gress authorized the Metric System to be used in the United States by passing the following bills : AN ACT TO AUTHORIZE THE USE OF THE METRIC SYSTEM OP WEIGHTS AND MEASURES. Be it enacted by the Senate and House of Representatives of the United States of America in Congress assembled, That from and after the passage of this Act, it shall be lawful throughout the United States of America to employ the Weights and Measures of the Metric System ; and no con- tract or dealing, or pleading in any court, shall be deemed iuvalid, or liable to objection, because the weights or measures expressed or referred to therein are weights or measures of the Metric System. SECTION 2. And be it further enacted, That the tables in the schedule hereto annexed shall be recognized in the construction of contracts, and in all legal proceedings, as establishing, in terms of the weights and measures now in use in the United States, the equivalents of the weights and measures expressed therein in terms of the Metric System ; and said tables may bs lawfully used for computing, determing, and expressing in customary weights and measures, the weights and measures of the Metric System. 708. The Metric System of weights and measures is based upon the decimal scale. 700. The Meter is the base of the system, and is the one ten* millionth part of the distance on the earth's surface from the equa- tor to either pole, or 39.37079 inches. 8O. From the Meter are made the Are (air), the Stere (stair), the Liter (leeter), and the Gram ; these constitute the primary or principal units of the system, from which all the^ others are derived. 8O1. The Multiple Units, or higher denominations, are named by prefixing to the name of the primary units the Greek numerals, Deka (10), Hecto (100), Kilo (1000), and Myra (10000). 454 MENSUKATION. The Sub-multiple Units, or lower denominations, are named by prefixing to the names of the primary units the Latin numerals, Deci (fa), Centi (^), Milk ( T ^). Hence, it is apparent from the name of a unit, whether it is greater or less than the standard unit, and also how many times. MEASURES OF EXTENSION. 8o The Meter is the unit of length, and is equal to 39.37 in. nearly. TABLE. Metric Denominations. U. S. Value. 1 Millimeter = .03937079 in. 10 Millimeters, mm. = 1 Centimeter = .3937079 in. 10 Centimeters, cm. = 1 Decimeter = 3.937079 in. 10 Decimeters, dm. = 1 METER - 39.37079 in. 10 METERS M. = 1 Dekameter = 32.808992 ft. 10 Dekameters, Dm. 1 Hectometer = 19.927817 rd. 10 Hectometers, Hm. = 1 Kilometer = .6213824mi. 10 Kilometers, Km. 1 Myrianieter (Mm.) = 6.213824 mi. Tlie meter, like our yard, is used in measuring cloths and short dis- tances. The kilometer is commonly used for measuring long distances, and is about of a common mile. The Are is the unit of land measure, and is a square whoso side is 10 meters, equal to a square dekameter, or 119.6 sq. yards. TABLE. 1 Centiare, ca. = (1 Sq. Meter) = 1.196034 sq. yd. 100 Centiares, " = 1 ARE = 119. 6034 sq. yd. 100 ARES A. = 1 Hectare (Ha.) = 2.47114 acres. The Square Meter is the unit for measuring ordinary surfaces ; as flooring, ceilings, etc. TABLE. 100 Sq. Millimeters, sq.mm. = 1 Sq. Centimeter = .155+ sq.in. 100 Sq. Centimeters, sq.cm. = 1 Sq. Decimeter = 15.5+ sq. in. 100 Sq. Decimeters, sq. dm. = 1 SQ. METEE (8q.M.) = 1.196+ sq. yd METBIC SYSTEM. 455 The Stere is the unit of wood or solid measure, and is equal to a cubic meter, or .2759 cord. TABLE. 1 Decistere = 3.531+ cu. ft. 10 Decisteres, dst. = 1 STERE = 35.316 + cu. ft. 10 STEKES 8t. = 1 Dekastere (DSt.) = 13.079+ cu. yd. 807. The Cubic Meter is the unit for measuring ordinary solids; as excavations, embankments, etc. TABLE. 1000 Cu. Millimeters, cu. mm. = 1 Cu. Centimeter = .001 + cu. in. 1000 Cu. Centimeters, CM. cm. = 1 Cu. Decimeter = 61.026 + " " 1000 Cu. Decimeters, cu. dm. = 1 Cu. METER = 35.316 + cu. ft. MEASURES OF CAPACITY, 808. The Liter is the unit of capacity, both of Liquid and of Dry Measures, and is a vessel whose volume is equal to a cube whose edge is one-tenth of a meter, equal to 1.G5673 qt. Liquid Measure, and .9081 qt. Dry Measure, TABLE. 10 Milliliters, ml. . . . = 1 Centiliter. 10 Centiliters, cl. . . . 1 Deciliter. w 10 Deciliters, ctt. . ... =1 LITER. 10 LITERS L. ... =1 Dekaliter.- 10 Dekaliters, Dl. . . . = 1 Hectoliter. 10 Hectoliters, El. . . . = 1 Kilcliter, or Stere. 10 Kiloliters, El. . . . = I Myrialiter (Ml.). The Hectoliter is the unit in measuring liquids, grain, fruit, and roots in large quantities. 8 SO. EQUIVALENTS IX UNITED STATES MEASURES. Metric Denominations. Cubic Measure. Dry Measure. Wine Measure. 1 Myrialiter = 10 Cubic Meters = 283.72+ bu. = 2641.4+ gal. 1 Kilpliter = 1 Cubic Meter = 28.372+ bu. - 264.17 gal. 1 Hectoliter -^ Cubic Meter = 2.8372+ bu. = 26.417 gal. 1 Dekaliter = 10 Cu. Decimeters = 9.08 quarts = 2.6417 gal. 1 LITER 1 Cu. Decimeter = .908 quart = 1.0567 qt. 1 Deciliter = ^ Cu. Decimeter = 6.1022 cu. in. = .845 gill. 1 Centileter = 10 Cu. Centimeters = .6102 cu. in. = .338 fluid oz. 1 MUliliter = 1 Cu. Centimeter = .061 cu. in. = .27 fluid dr. 456 MENSURATION. MEASURES OF WEIGHT. 8 1 0. The Gram is the unit of loeight, and equal to the weight of a cube of distilled water, the edge of which is one-hundredth of a meter, equal to 15.432 Troy grains. TABLE. 10 Milligrams, 10 Centigrams, 10 Decigrams, 10 GRAMS 10 Dekagrams, 10 Hectograms, mg. 1 Centigram eg. = 1 Decigram dg. = 1 GRAM G. =1 Dekagram Dg. = 1 Hectogram rj- ^ ( Kilogram, Jig. L -i .15432+ gr. Troy. 1.54324+ " " 15.43248+ " " .35273+ oz. Avoir. 3.52739+ " " 10 Kilograms, Kg. 10 Myriagrams, oi,Mg. ) 100 Kilograms ) 10 Quintals, or ) 1000 KILOS ) = 1 1 Myriagram = 22.04621+ " 1 Quintal = 220.46212+ " ( Tonneau, ( or TON j- =2204.62125 The Kilogram, or Kilo, is the unit of common weight in trade and is a trifle less than 2^ Ib. Avoirdupois. The Tonneau is used for weighing very heavy articles, and is abou T 204 Ib. more than a common ton. 811. Units of the Common System maybe readily changed to units of the Metric System by the aid of the following TABLE. 1 Inch 1 Foot : 1 Yard ; 1 Rod : 1 Mile 1 Sq. inch ; 1 Sq. foot 1 Sq. yard 1 Sq. rod 1 Acre 1 Sq. mile 2.54 Centimeters. 30.48 Centimeters. .9144 Meter. 5.029 Meters. 1.6093 Kilometers. 6.4528 Sq. Centimet. 929 Sq. Centimeters. .8361 Sq. Meter. 25.29 Centiares. 40.47 Ares. 259 Hectares. i -t 9* Cu. inch Cu. foot : Cu. yard : Cord : Fl. ounce: Gallon : Bushel ; 1 Troy gr. : 1 Troy Ib. : 1 Av. Ib. : 1 Ton ; : 16.39 Cu. Centimet. 28320 Cu. Centimet. .7646 Cu. Meter. 3.625 Steres. 2.958 Centiliters. 3.786 Liters. .3524 Hectoliter. 64.8 Milligrams. .373 Kilo. .4530 Kilo. .907 Tonneau. RETURN TO the circulation desk of any University of California Library or to the NORTHERN REGIONAL LIBRARY FACILITY Bldg. 400, Richmond Field Station University of California Richmond, CA 94804-4698 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS 2-month loans may be renewed by calling (415) 642-6753 1-year loans may be recharged by bringing books to NRLF Renewals and recharges may be made 4 days prior to due date DUE AS STAMPED BELOW APR 2 5 1992 YB 17385 M81911 Q A ico. THE UNIVERSITY OF CALIFORNIA LIBRARY