X i^^'- I .m A: 'n ./^:: 5)^ K "'-) J a-''' / >^-■ o' -0 / / /) ( ROBINSON'S MATHEMATICAL SERIES. THE PROGRESSIVE HIGHER ARITHMETIC, FOR SCHOOLS, ACADEMIES, AND MERCANTILE COLLEGES. COMBINING THB ANALYTIC AND SYNTHETIC METHODS; AND rOBMING A COMPLETE TREATISE OK ARITHMETICAL SCIENCE, AND ITS COMMERCIAL AND BUSINESS APPLICATIONS. BY HORATIO N. EOBINSON, LL. D., AUTHOR OF WORKS ON ALGEBRA, GEOMETRY AND TRIGONOMETRY, BURYEYINQ AIT* NAVIGATION, CONIO SfiCTIONS, CALCULUS, ASTRONOMY, ETC. NEW YORK: IVISON, PUINNEY, BLA'KEMAN & CO^ CHICAGO : a C. GRIGGS ft 00. 186 7. RO B I N S O N'S TJie most Complete, most Practical, and most Scientific Series of Mathematical Text-Books ever issued in this country, ►» ♦ ■» — ■ Robinson's Progressive Table Book, - - - . . Kobinson's Progressive Primary Arithmetic, - Robinson's Progressive Intellectual AritLmetic, - Robinson's Rudiments of Written Arithmetic, - Robinson's Progressive Practical Arithmetic, - - - Robinson's Key to Practical Arithmetic, Robinson's Progressive Higher Arithmetic, - - - Robinson's Key to Higher Arithmetic, Robinson's Arithmetical Examples, Robinson's New JElementary Algebra, - - - . ^ Robinson's Key to Elementary Algebra, Robinson's University Algebra, Robinson's Key to University Algebra, Robinson's Wew University Algebra, Robinson's JS.ey to New University Algebra, • . - - Robinson's New Geometry and Trigonometry, - Robinson's Surveying and Navigation, Robinson's Analyt. Geometry and Conic Sections, Robinson's DifFeren. and Int. Calculus, (in preparations- Robinson's Elementary Astronomy, - Robinson's University Astronomy, Robinson's Mathematical Operations, Robinson's Key to Geometry and Trigonometry, Conic Sections and Analytical Geometry, Entered, according to Act of Congress, in the year 1S60, by DANIEL W. riSII & J. 11. FRENCH, and again in the year 1S63, by DANIEL W. FI8II, A.M., In the Clerk's Office of the District Court of the United States for the Nortlicrn District of the New York. 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 difiicult 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 Sijnthetio 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 i;vhich 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 understarid 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 ]t should be the Analytic. The following characteristics of a first class text-book will be obvious to all who examine this work: the typogra- (iii) lY PKEFACE. ' pliy and mechanical execxttion ; the philosophical and scieyitific arrangement of the subjects ; clear and concise definitions ; full and rigid analyses ; exact and compre- hensive rules; brief and accurate methods of operation: the zuide range of subjects and the large number and prac- ticcd character of the examples — in a word, scientific ac- CUEACY combined with peactical utility, throughout the entire ivork. 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 subjects 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 qualify students to become good accountants or commercial business men. And while this work may embrace many subjects not necessary to the PREFACE. y 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 w^ork 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 Eoot. 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 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* yi 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. If answers ara placed in the back part of the book, the pupil will at once refer to them whenever he is in any doubt or difficulty in performing an operation. Hence the object aimed at is not accomplished by placing the answers to- gether in this manner. Again, if all the answers are omitted, the pupil may be- come involved in doubt and uncertainty, and acquire a distaste for the study ; and from this discouragement, sub- sequently make but limited advancement in Mathematical Science. In order, therefore, that pupils may receive the advan- tages of both methods, the answers to nearly one half of the examples in this book are omitted. They will be found, together with full and clear solutions of all the examples, in a Key to this work, which has been prepared for the use of teachers and private learners. Many valuable hints and suggestions which have been received from teachers and friends of education, have been incorporated into this work. The author desires to make especial acknowledgment of the valuable services rendered in the preparation of this work by D. W. Fish, A.M., of Rochester, N. Y., a gentleman who has had long and successful experience as a teacher, and an intimate ac- quaintance with the plans and operations of some of the best schools in the country. Augii-st 1, 1860. CONTENTS, 1^- • PAG> Definitions 11 Signs 13 Axioms , 14 Notation and Numeration ,. , 15 SIMPLE NUMBERS. Addition 23 Adding two or more celumns at one operation 27 Subtraction 30 Two or more subtrahends , 33 Multiplication 33 Powers of Numbers 89 Continued Multiplication.. , 40 Contractions in Multiplication 41 Division 47 Abbreviated Long Division 60 Successive Division 55 Contractions in Division 55 General Problems in Simple Numbers , 61 PROPERTIES OF NUMBERS. Exact Divisors 65 Prime Numbers 6S 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 (yii) 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 184 Money and Currencies 187 Reduction 192 Reduction Descending 192 CONTENTS, Jx PAGE Reduction Ascending , 199 Addition 206 Subtraction 209 Multiplication 214 Division ,. 216 Longitude and Time 218 DUODECIMALS. Addition and Subtraction 22'?' Multiplication 223 Division 230 SHORT METHODS. For Subtraction 232 For Multiplication 233 For Division 241 RATIO. ' 243 PROPORTION. 24r Cause and Effect 249 Simple Proportion 249 Compound Proportion 253 PERCENTAGE. Notation 259 General Problems 260 Applications 268 Commission 2G8 Stocks 272 Stock-jobbing 273 Instalments, Assessments, and Dividends 276 Stock Investments 279 Gold Investments 2S5 Profit and Loss 287 Insurance 291 Life Insurance 293 Life Table 295 Endowment Assurance Table 206 Taxes 29S General Average 801 Custom House Business 803 Simple Interest 807 X CONTENTS. PAGa Partial Payments or Indorsements 314 Savings Banks Accounts 319 Compound Interest 321 Compound Interest Table «... 823 Problems in Interest 324 Discount 328 Banking 330 ExchuKge « -•.. S37 Direct Exchange. 339 Table of Foreign Corns and Money ,..^ 342 Arbitrated Exchange , 348 Equation of Payments : 352 Simple Equations 352 Compound Equations 857 Account Sales 861 Settlement of Accounts Current 363 Partnership. c , 864 ALLIGATION. 870 INVOLUTION. 879 EVOLUTION. 881 Square Root 882 Contracted Method 386 Cube Root 887 Contracted Method , 892 Applications of the Square and Cube Roots 898 SERIES. 406 Arithmetical Progression 408 Geometrical Progression 411 Compound Interest by Geometrical Progression 415 Annuities 416 Annuities at Simple Interest. 417 Annuities at Compound Interest 421 Miscellaneous Examples 422 Metric System 429 HIGHER ARITHMETIC. DEFINITIONS. I, Quantity is any thing that can be increased, diminished, or measured ; as distance, space, weight, motion, time. 3, A Unit is one, a single thing, or a definite quantity. 3. A Number is a unit, or a collection of units. -4. The Unit of a Number is one of the collection constituting the number. Thus, the unit of 34 is 1 ; of 34 days is 1 day. 5. An Abstract Number is a number used without reference to any particular thing or quantity; as 3, 24, 756. ©• A Concrete Number is a number used with reference to some particular thing or quantity; as 21 hours, 4 cents, 230 miles. 7. Unity is the unit of an abstract number. 8. The Denomination is the name of the unit of a concrete number. O. A Simple Number is either an abstract number, or a con- crete number of but one denomination; as 48, 52 pounds, 36 days. 10. A Compound Number is a concrete number expressed in two or more denominations ; as, 4 bushels 3 pecks, 8 rods 4 yards 2 feet 3 inches. II. An Integral Number, or Integer, is a number which ex- presses whole things; as 5, 12 dollars, 17 men. 12. A Fractional Number, or Fraction, is a number which expresses equal parts of a whole thing or quantity; as J, f of a pound, ~^^ of a bushel. 13. Like Numbers have the same kind of unit, or express the same kind of quantity. Thus, 74 and 16 are like numbers; so are 74 pounds, 16 pounds, and 12 pounds; also, 4 weeks 3 days, and 16 minutes 20 seconds, both being used to express units of time. 14. Unlike Numbers have different kinds of units, or are used (11) 12 SIMPLE NUMBERS. to express different kinds of quantity. Thus, 36 miles, and 15 days ; 5 hours 36 minutes, and 7 bushels 3 pecks. lo. A Power is the product arising from multiplying a number by itself, or repeating it any number of times as a factor. 10. A Boot 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 varyino-. 18. A Uniform Scale is one in which the order of progression is the same throughout the entire succession of units. 19. 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 ia 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. KoTE yi, 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 ,s an art, «nd ig then cnWed Applied ArithLtic. 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. 20. 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. SIGNS. 13 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^ millionsy hillions, etc. 31. The Decimal Sign is the period (.). It indicates that the number after it is a decimal. 33. The Sign of Addition is the perpendicular cross, +, called plus. It indicates that the numbers connected by it are to be added ) as 3 + 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. 33. The Sign of Division is a short horizontal line, with a point above and one below, -^-, 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 ahove, and the divisor helow, a short horizontal line. Thus, ^g®, read 18 divided by 6. 36. The Sign of Eq[uality is two short, parallel, horizontal lines, =. It indicates that the numbers, or combinations of numbers, connected by it are equal; as 4 + 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 + 4) x 5 indicates that both 8 and 4, or their sum, is to be multiplied by 5. A vinc ulum or bar, , has the same signification. Thus, 7x9-T-3 = 21. 2 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 : G : : 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^ 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, v/, 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; -^64 indicates that the cube root of 64 is to be extracted. AXIOMS. 42. 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. 8. 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. ^5 NOTATION AND NUMERATION. 43. Notation is p. system of writing or expressing numbers by characters ; and, 4:4, 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 Romnn Notntion is supposed to have been first used hy the Roirifins ; hence its name. The Arabic Notation was first introduced into Europe by the Moors or Arabs, who conquered and held possession of Spain during tlio llih century. It received the attention of scientific men in Italy at the bepin- nit)g of the l.'^th century, and was soon afterward adopted in most European countries. Formerly it was supposed to be an invention of the Arabs; but investicrations 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 eometimes called the Indian Notation. THE ROMAN NOTATION. 4:6. Employs seven capital letters to express numbers. Thus, Letters, I Y X L C D M Values, one, five, ten, fifty, ,^Xd, hunrred, thornd. 417. The Roman notation is founded upon five principles, as follows : 1st. Eepeatino^ a letter repeats its value. Thus, II represents two, XX twenty, CCC three hundred. 2d. If a letter of any value be placed a/fer 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 he/ore one of greater value, its value is to be taJcen 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, XI Y represents fourteen, XXIX twenty- nine, XCIY ninety-four. 5th. A bar or dash placed over a letter increases its value one thousand fold. Thus, Y signifies five, and Y five thousand; L fifty, and L fifty thousand. 16 SIMPLE NUMBEES. TABLE or ROMAN NOTATION. I is One. XX is i 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. cc " Two hundred. XII " Twelve. D " Five hundred. XIII - Thirteen. DC " Six hundred. XIV " Fourteen. M " One thousand. [dred. XV '' Fifteen. MC " One thousand one hun- XVI '' • Sixteen. MM ** Two thousand. XVII - • Seventeen. X " Ten thousand. s:viii ' ' Eighteen. C " One hundred thousand. XIX '' Nineteen. M " One million. KoTRS. — 1. Though the letters used in the above table have been employed as the Roman numerals for man^' 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, | V X L C N M 2. The system of Roman Notation is not well adapted to the purposes of nu- merical calculation; it is principally confined to the numbering of chapters and sections 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. 8. Twenty-four. 8. Seventy-three. 4. Thirty-nine. 9. Ninety-five. 6. 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. j'j THE ARABIC NOTATION 48. Employs ten characters or figures to express numbers. Thus, Figures, 123456789 , ^ naught one, two, three, four, five, six, seven, eight, nine, values. \ ^^^^^^ 49. The cipher, or fir^t 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 digitus, 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. 51. 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. 59. 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 : J8 SIMPLE NUMBERS. t •i^ -o ^ "B S S ^ 000000000 53. 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. 56, 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. All numbers are expressed hy 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 2)lace 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. 1} 59. This method of numerating, or naming, groups the suc- cessive orders into^eriWs 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. c3 d "^ '^ .2 ;5 cr" icr" -J5 pO a :3 98, 7 65, 4 32, 109, 876, 556, 789, 012, 3 45 ninth eighth seventh sixth fifth fourth third pecond first period, period, period, period, period, period, period, period. period. NoTR. — This is the French method of numerating, and is the one in pfeneral use in this country. The English numerate by periods of six figures each. 60. The names of the periods are derived from the Latin numerals. The twenty-two given on the following page extend tlie numeration table to the sixty-sixth place or order, inclusive. 20 SIMPLE NUMBERS. PERIODS. NAMES. PERIODS. NAMES. 1st Units. 12th Decillions. 2d Thousands. 13th Undecillions. 3d Millions. 14th Duodecillions. 4th Billions. 15th Tredecillions. 5th Trillions. 16th Quatuordecillions 6th Quadrillions. 17th Quindecillions. 7th Quintillions. 18th Sexdecillions. 8th Sextillions. 19th Septendecillions. 9th Septillions. 20th Octodecillions. 10th Octillions. 21st Novendecillions. 11th Nonillions. 22d Vigintillions. 61. From this analysis of the principles of Notation and Nume- ration, we derive the following rules : RULE FOR NOTATION. I. Beginning at the left hand, write the figures belonging to the highest period. II. Write the hundreds^ tenSy and units of each successive period in their order, placing a cipher wherever an order of units is omitted. RULE FOR NUMERATION. I. Separate the number into periods of three figures each, com^ mencing at the right hand. II. Beginning at the left hand, read each period sej)aratelj/, and give the name to each period, except the last, or period of units. Note. — Omit and in reading the orders of units and periods of a number. EXAMPLES FOR PRACTICE. Write and read the following numbers : — 1 One unit of the 3d order, two of the 2d, five of the 1st. Ans. 125; read, one hundred twenty five. 2. Two units of the 5th order, four of the 4th, five of the 2d, six of the 1st. Ans. 24056 ; read, twenty-four thousand fiffy-six. 3. Seven units of the 4th order, five of the 3d, three of the 2d, eight of the 1st. NOTATION AND NUMERATION. 21 4. Nine units of the 4th order, two of the 3d, four of the 1st. 6. Five units of the 4th order, eight of the 2d. 6. Five units of the 5th order, one of the 3d, eight of the 1st. 7. Three units of the 5th order, six of the 4th, four of the 3d, seven of the 1st. 8. Two units of the 6th order, four of the 5th, nine of the 4th, three of the 3d, five of the 1st. 9. Three units of the 8th order, five of the 7th, four of the 6th, three of the 5th, eight of the 4th, five of the 3d, eight of the 2d, seven of the 1st. 10. Three units of the 9th order, eight of the 7th, four of the 6th, six of the 5th, nine of the 1st. 11. Five units of the 12th order, three of the 11th, six of the 10th. 12. Four units of the 12th order, five of the 10th, eight of the 5th, nine of the 4th, four of the 3d. 13. Three units of the 15th order, six of the 14th, five of the 13th, three of the 9th, six of the 8th, five of the 7th, three of the 3d, six of the 2d, five of the 1st. 14. Five units of the 18th order, three of the 17th, six of the 16th, four of the 15th, seven of the 14th, eight of the 13th, four of the 12th, five of the 11th, six of the 10th, seven of the 9th, eight of the 8th, nine of the 7th, five ot the 6th, six of the 5th, three of the 4th, two of the 3d, four of the 2d, eight of the 1st. 15. Two units of the 20th order, seven of the 19th, four of the 18th, eight of the 13th, five of the 6th, five of the 5th, five of the 4th, nine of the 1st. Write the following numbers in figures: 16. Forty-eight. 17. One hundred sixty-four. 18. Forty-eight thousand seven hundred eighty-nine. 19. Five hundred thirty-six million three hundred forty-seven thousand nine hundred seventy-two. 20. Ninety-nine billion thirty-seven thousand four. 21. Eight hundred sixty-four billion five hundred thirty-eight million two hundred seventeen thousand nine hundred fifty-three. 22 SIMPLE NUMBERS. 22. One hundred seventeen quadrillion two hundred thirty-fivo trillion one hundred four billion seven hundred fifty million sixty- six thousand ten. 23. Ninety-nine quintillion seven hundred forty-one trillion fifty-four billion one hundred eleven million one hundred one. 24. One hundred octillion one hundred septillion one hundred quintillion one hundred quadrillion one hundred trillion one hundred billion one hundred million one hundred thousand one hundred. 25. Four decillion seventy-five nonillion three octillion fifty- two septillion one sextillion four hundred seventeen quintillion ten quadrillion twelve trillion fourteen billion three hundred sixty million tw^enty-two thousand five hundred nineteen. Write the following numbers in figures, and read them : 26. Twenty-five units in the 2d period, four hundred ninety-six in the 1st. Ans. 25,496. 27. Three hundred sixty-four units in the 8d period, seven hundred fifteen in the 2d, eight hundred thirty-two^in the 1st. 28. Four hundred thirty-six units in the 4th period, twelve in the 3d, one hundred in the 2d, three hundred one in the 1st. 29. Eighty-one units in the 5th period, two hundred nineteen in the 4th, fifty-six in the 2d. 30. Nine hundred forty-five units in the 7th period, eighteen in the 5th, one hundred three in the 3d. 31. One unit in the 10th period, five hundred thirty-six in the 9th, two hundred forty-seven in the 8th, nine hundred twenty-four in the 7th. Point off and read the followino- numbers : 82. 564. 37. 2005. 33. 24835. 38. 100103. 34. 2474783. 39. 53000008. 35. 247843112. 40. 1001005003. 86. 23678542789. 41. 750000040003. 42. 247364582327896438542721. 43. 379403270506038 009503070. 44. 20005700032004673000430512500000567304705030040. ADDITION. 23 ADDITION. 69« Addition is the process of uniting several numbers of the same kind into one equivalent number. - 63. The Sum or Amount is the result obtained by the process of addition. 64. When the given numbers contain several orders of units, the method of addition is based upon the following principles : I. If the like orders of units be added separately, the sum of all the results must be equal to the entire sum of the given num- bers. (Ax. 10). II. If the sum of the units of any order contain units of a higher order, these higher units must be combined with units of like order. Hence, III. The work must commence with the lowest unit, in ordei to combine the partial sums in a single expression, at one ope- ration. I. Find the sum of 897, 476, and 873. OPERATION Analysis. We arrange the numbers so that 897 units of like order shall stand in the same column. 476 We then add the first, or right hand column, and o73 find the sum to be 16 units, or 1 ten and 6 units ; 1746 writing "fhe 6 units under the column of units, we add ^e 1 ten to the column of tens, and find the 6um to be 24 tens, or 2 hundreds and 4 tens ; writing the 4 tens under the column of tens, we add the 2 hundreds to the column of hundreds, and find the sum to be 17 hundreds, or 1 thousand and 7 hundreds ; writing the 7 hundreds under the column of hundreds, and the 1 in thousands* place, we have the entire sum, 1746. 65. From these principles we deduce the following HuLE. I. Write the numbers to he added so that all the units of the same order shall stand in the same column) that is, units under units , tens under tenSy etc, II. Commencing at units ^ add each column separately y and write the sum underneath^ if it he less than ten. 24 SIMPLE NUMBERS. III. If the sum of any column he ten or more than ten, write the unit figure only, and add the ten or tens to the next column. IV. Write the entire sum of the last column. Notes. — 1. In adding, learn to pronounce the partial results without naming the Jifjitres separately. Thus, in the operation given for illustration, say 3, 9, 16; 8, 15, 24,- 10, 14, 17. 2. When the sum of any column is greater than 9, the process Of adding the tens to the next column is called currying. OG. Proof. There are two principal methods of proving Addition. 1st. By varying the combinations. Begin with the right hand or unit column, and add the figures in each column in an opposite direction from that in which they were first added ; if the two results agree, the work is supposed to be right. 2d. By excess of 9's. 07. This method depends upon the following properties of the number 9 : * I. If a number be divided by 9, the remainder will be the same as when the sum of its digits is divided by 9. Therefore, II. If several numbers be added, the excess of 9's in the sum must be equal to the excess of 9's in the sum of all the digits in the numbers. 1. Add 34852, 24784, and 72456, and prove the work by the excess of 9's. OPERATION. 34852 24784 72456 ... 8, excess of 9's in all the digits of the numbers. 132092 ... 8, " " " " sum «^ " Analysis. Commencing with the first number, at the left hand, we say 3 and 4 are 7, and 8 are 15 ; dropping 9, the excess is 6, which added to 5, the next digit, makes 11; dropping 9, the excess is 2; then 2 and 2 are 4, and 2 (the left hand digit of the second number) * For a demonstration of these properties, see 186, IX. ADDITION. 25 are 6, and 4 are 10 ; dropping 9, the excess is 1. Proceeding in like manner through all the digits, the final excess is 8 ; and as 8 is also the excess of 9^s in the sum, the work of addition is correct. It is evident that the same result will be obtained by adding the digits in columns as in rows. Hence, to prove Addition by excess of 9*s: — CommenciDg at any figure, add the digits of the given numbers in any order, dropping 9 as often as the amount exceeds 9. If the final excess be equal to the excess of 9^s in the sum, the work is right. Note. — This method of proving addition by the excess of 9's, fails in tKtf- fol- lowing cases : Ist, when the figures of the answer are misplaced; 2d, when the value of one figure is as much too great as that of another is too small. EXAMPLES FOR PRACTICE (1.) (2.) (3.) (4.) 8635 1234567 67 24603 2194 723456 123 298765 7421 34565 4567 47321 5063 45666 89093 • 68653 2196 333 654321 6376 1245 90 1234567 340 26754 2038677 5. 123+456+785+12+345+901 + 567=how many? 6. r2345+67890+8763+347 + 1037 + 198760=how many? 7. 172+4005 + 37Gl + 20472+367012 + 19762=how many? 8. What is ^he sum of thirty-seven thousand six, four hundred twenty-nine thousand nine, and two millions thirty-six ? Ans. 2,466,051. 9. Add eight hundred fifty-six thousand nine hundred thirty- three, one million nine hundred seventy-six thousand eight hun- dred fifty-nine, two hundred three millions eight hundred ninety- five thousand seven hundred fifty-two. Ans. 206,729,544. 10. What is the sum of one hundred sixty-seven thousand, three hundred sixty-seven thoasand, nine hundred six thousand, two hundred forty-seven thousand, seventeen thousand, one hun- dred six thousand three hundred, forty thousand forty-nine, ten thousand four hundred one ? Ans. 1,860,750. 11. What number of square miles in New England, there' 3 26 SIMPLE NUMBERS. being in Maine 31766, in New Hampshire 9280, in Vermont 10212, in Massachusetts 7800, in Ehode Island 1306, and in Connecticut 4674? Ans. 65,038. 12. The estimated population of the above States, in 1855, was as follows; Maine 653000, New Hampshire 338000, Vermont 327000, Massachusetts 1133123, Ehode Island 166500, and Con^ necticut 384000. What was the entire population ? 13 At the commencement of the year 1858 there were in ope- ration in the New England States, 3751 miles of railroad; in New York, 2590 miles; in Pennsylvania, 2546; in Ohio, 2946; in Virginia, 1233 ; in Illinois, 2678 ; and in Georgia, 1233. V>^hat was the aggregate number of miles in operation in all these States? 14. The Grand Trunk Railway is 962 miles long, and cost $60000000 ; the Great Western Eailway is 229 miles long, and cost S14000000; the Ontario, Simcoe and Huron, is 95 miles long, and cost §3300000 ; the Toronto and Hamilton is 38 miles long, and cost §2000000. What is the aggregate length, and what the cost, of these four roads ? Ans. Length, 1,324 miles; cost, $79,300,000. 15. A man bequeathed his estate as follows; to each of his two sons, $12450; to each of his three daughters, $6500; to his wife, $650 more than to both the sons, and the remainder, which was $1000 more than he had left to all his family, he gave to benevolent institutions. What was the whole amount of hig property? A71S, $140,900. 16. How many miles from the southern extremity of Lake Michigan to the Gulf of St. Lawrence, passing through Lake Michigan, 330 miles ; Lake Huron, 260 miles ; Eiver St. Clair, 24 miles; Lake St. Clair, 20 miles; Detroit River, 23 miles; Lake Erie, 260 miles; Niagara River, 34 miles; Lake Ontario, 180 miles; and the River St. Lawrence, 750 miles? 17. The United States exported molasses, in the year 1856, to the value of $154630; in 1857, $108003; in 1858, $115893; and tobacco, during the same years respectively, to the value of $1829207, $1458553, and $2410224. What was the entire valua of the molasses and tobacco exported in these three years ? ADDITION. 27 18. The population of Boston, in 1855, was 162629; Provi- dence, 50000; New York, 629810; Philadelphia, 408815; Brook- lyn, 127618 ; Cleveland, 43740 ; and New Haven, 25000. What •was the entire population of these cities ? Ans. 1,447,612. 19. Iron was discovered in Greece by the burning of Mount Ida, B. C. 1406; and the electro-magnetic telegraph w^as invented by Morse, A. D. 1832. What period of time elapsed between the two events ? Ans. 3,238 years. 20. The number of pieces of silver coin made at the United States Mint at Philadelphia, in the year 1858, were as follows : 4628000 half dollars, 10600000 quarter dollars, 690000 dimes, 4000000 half dimes, and 1266000 three-cent pieces. What was the total number of pieces coined ? 21. The cigars imported by the United States, in the year 1856, were valued at 63741460; in 1857, at $4221096; and in 1858, at $4123208. What was the total value of the importations for the three years ? Ans. $12,085,764. 22. In the appropriations made by Congress for the year ending June 30, 1860, were the following; for salary and mileage of members of Congress, $1557861; to officers and clerks of both Houses, $157639 ; for paper and printing of both Houses, $170000 ; to the President of the United States, $31450 ; and to the Yice President, $8000. What is the total of these items ? ADDING TWO OR MORE COLUMNS AT ONE OPERATION. 68, 1. What is the sum of 4632, 2553, 4735, and 2863 ? OPERATION. Analysis. Beginning with the units and tens of 4632 the number last written, we add first the tens above, 2553 then the units, thus ; 63 and 30 are 93, and 5 are 98, 4735 and 50 are 148, and 3 are 151, and 30 are 181, and __.__ ^ ^^^ ^^^' ^^ *^^^ sum, we write the 83 under the 14783 columns added, and carry the 1 to the next columns, thus ; 28 and 1 are 29, and 40 are G9, and 7 are 76, and 20 are 96, and 5 are 101, and 40 are 141, and 6 are 147, which we write in its place, and we have the whole amount, 14783. 28 SIMPLE NUMBERS. EXAMPLES FOR PRACTICE. -0 (1.) (2.) (3.) (4.) 8450 75634 123456 7349042 5425 86213 47021 2821986 8595 92045 82176 1621873 6731 73461 570914 236719 7963 34719 379623 401963 5143 26054 7542 67254 4561 19732 25320 45067 6783 84160 57644 910732 4746 97013 908176 6328419 2373 34567 73409 1437651 8021 43651 3147 9716420 7273 52170 67039 8191232 71064 719419 2345467 34128358 5. What is the total number of churches, the number of persons accommodated, and the value of church property in the United States, as shown by the following statistics ? No. of No. of persons Value of churches. accommodated. church property Methodist 12484 4220293 $14636671 Baptist 8798 3134438 10931382 Presbyterian 4591 2045516 14469889 Congregational 1675 795677 7973962 Episcopal 1430 631613 11261970 Koman Catholic 1269 705983 8973838 Lutheran 1205 532100 2867886 Christians 812 296050 845810 Friends 715 283023 1709867 Union 619 213552 690065 Universalist 494 205462 1767015 Free Church 361 108605 252255 Moravian :... 331 112185 443347 German Reformed 327 156932 965880 Dutch Reformed 324 181986 4096730 Unitarian 244 137867 8268122 Mennonite 110 29900 94245 Tunkers 52 35075 46025 Jewish... 31 16575 371600 Swedenborgian 15 5070 108100 ADDITION. 29 I^H 6. Give the amounts of the productions of the United States and Territories for the year 1850, as expressed in the following columns : Pounds of Butter. Alabama 4,008,811 Arkansas 1,854,239 California 705 Columb.,Dist. 14,872 Connecticut... 6,498,119 Delaware 1,055,308 Florida 371,498 Georcria 4,640,559 Illinois 12,526,543 Indiana 12,881,535 Iowa 2,171,188 Kentucky.... 9,947,523 Louisiana...'.. 683,069 Maine 9,243,811 Maryland 3,086,160 Massachusetts 8,071,370 Michigan 7,065,878 Mississippi... 4,346,234 Missouri 7,834,359 N.Hampshire 6,977,056 New Jersey... 9,487,210 New York.... 79,766,094 N.Carolina... 4,146,290 Ohio 34,449,379 Pennsylvania 39,878,418 llhode Island 995,670 S. Carolina... 2,981,850 Tennessee.... 8,139,585 Texas 2,344,900 Vermont 12,137,980 Virginia 11,089,359 Wisconsin.... 3,633,750 Territories... 295,984 gi* Pounds of Cheese. Pounds of Wool. Bushels of Wheat. 31,412 657,118 294,044 30,088 182,595 199,639 150 5,520 17,228 1,500 525 17,370 5,363,277 497,454 41,762 3,187 57,768 482,511 18,015 23,247 1,027 46,976 990,019 1,088,534 1,278,225 2,150,113 9,414,575 624,564 2,610,287 6,214,458 209,840 373,898 1,530,581 213,954 2,297,433 2,142,822 1,957 109,897 417 2,434,454 1,364,034 296,259 3,975 477,438 4,494,680 7,088,142 585,138 31,211 1,011,492 2,043,283 4,925,889 21,191 559,619 137,990 203,572 1,627,164 2,981,652 3,196,563 1,108,476 185,658 365,756 375,396 1,601,190 49,741,413 10,071,301 13,121,498 95,921 970,738 2,130,102 20,819,542 10,196,371 14,487,351 2,505,034 4,481,570 15,367,691 316,508 129,692 49 4,970 487,233 1,066,277 177,681 1,364,378 1,619,386 95,299 131,917 41,729 8,720,834 3,400,717 535,955 436,292 2,860,765 11,212,616 400,283 253,963 4,286,131 73,826 71,894 517,562 80 SIMPLE NUMBERS. SUBTRACTION. 60. Subtraction is the process of determining the difference, between two numbers of the same unit value. •J'O, The Minuend is the number to be subtracted from. yi. The Subtrahend is the number to be subtracted. 7^. The Difference or Eemainder is the result obtained by the process of subtraction. 73. When the given numbers contain more than one figure each, the method of subtraction depends upon the following prin- ciples : I. If the units of each order in the subtrahend be taken sepa- rately from the units of like order in the minuend, the sum of the differences must be equal to the entire difference of the given numbers. (Ax. 10 .) II. If both minuend and subtrahend be equally increased^ the remainder will not be changed. 1. From 928 take 275. OPERATION. Analysis. We first subtract 5 units from Minuend, 928 8 units, and obtain 3 units for a partial re- Subtraheud, 275 mainder. As we cannot take 7 tens from 2 Remainder 653 tens, we add 10 tens to the 2 tens, making 12 tens ; then 7 tens from 12 tens leave 5 tens, the second partial remainder. Now, since we added 10 tens, or 1 hundred, to the minuend, we will add 1 hundred to the subtra- hend, and the true remainder will not be changed (II) ; thus, 1 hundred added to 2 hundreds makes 3 hundreds, and this sum sub- tracted from 9 hundreds leaves 6 hundreds ; and we have for the total remainder, 653. NoTTi;. — The process of adding 10 to the minuend is sometimes called hor- rowwrj 10, and that of adding 1 to the next figure of the subtrahend, carrying 1. I' J:. From these principles and illustrations we deduce the following HuLE. I. Write the less number under the greater.^ 'placing units of the same order under each other. SUBTRACTION. gj II. Begin at the right Tiandj and taJce each figure of the suhtra- hend from the figure above it, and ■write the remit underneath. III. If any figure in the subtrahend, he greater than the corres- pond ing figure above it^ add 10 to that upper figure before sub- tractingj and then add 1 to the next left hand figure of the subtra- (lend. 7^* Proof. It is evident that the subtrahend and remainder must together contain as many units as the minuend ; hence, to prove subtraction, we have three methods : 1st. Add the remainder to the subtrahend; the sum will be equal to the minuend. Or, 2d. Subtract the remainder from the minuend ; the difference will be equal to the subtrahend. Or, 3d. Find the excess of 9's in the remainder and subtrahend together, and it will be equal to the excess of 9's in the minuend. EXAMPLES FOR PRACTICE. (1.) (2.) (3.) (4.) From 47965 103767 57610218 '89764321 Take 26714 98731 8306429 83720595 Eem. 21251 5036 49303789 6043726 5. From 180037561 take 5703746. 6. From 2460371219 take 98720342. 7. 89037426175 — 2435036749 = how many? 8. 10000033421 — 999044110 = how many? 9. A certain city contains 146758 inhabitants, which is 3976 more than it contained last year; how many did it contain last year? Ans. 142,782. 10. The first newspaper published in America was issued at Boston in 1704; how long was that before the death of Benjamin Franklin, which occurred in 1790 ? 11. A merchant sold a quantity of goods for $42017, which was $1675 more than they cost him; how much aid they cost him? Ans. $40,342. 12. In 1858 the exports of the United States amounted to 82 SIMPLE NUMBERS. $324644421, and the imports to $282613150; how much did the exports exceed the imports ? Ans. $42,031,271. 13. In 1858 the gold coinage of the United States amounted to $52889800, and the silver to $8233287; how much did the gold exceed the silver coinage ? 14. The South in 1850 produced 978311690 pounds of cotton, valued at $101834616, and 237133000 pounds of sugar valued at $16599310; how much did the cotton exceed the sugar in quan- tity and in value ? Ans. 741,178,690 pounds; $85,235,306. 15. The area of the Chinese Empire is 5110000 square miles, and that of the United States 2988892 square miles ; the esti- mated population of the former is 340000000, and that of the latter in 1850 was 23363714. What is the difference in area and in population ? 16. The population of London in 1850 was 2362000, and that of New York city 515547; how many more inhabitants had London than New York ? A7is. 1,846,453. 17. The total length of railroads in operation in the United States, January 1, 1859, was 27857 miles, and the total length of the canals 5131 miles; how many miles more of railroad than of canal? Ans. 22,726. 18. The entire deposit of domestic gold at the United States Mint and its branches, to June, 1859, was $470341478, of which $451310840 was from California; how much was received from other sources ? Ans. $19,030,638. 19. During the year ending September 30, 1858, the number of letters exchanged between the United States and Great Britain were 1765015 received, and 1603609 sent; between the United States and France, 624795 received, and 639906 sent. How many letters did the exchange with Great Britain exceed those with France? Ans. 2,103,923. 20. The Southern States in 1850 had a population of 6696061, the Middle States 6624988, and the Eastern States 2728116; how many more inhabitants had the Middle and Eastern States than the Southern States ? 21. Having $20000, I wish to know how much more I must SUBTRACTION. 33 accumulate to be able to purchase a piece of property worth $23470, and have $5400 left? Ans. ^8,870. 22. A has §3540 more than B, and §1200 less than C, who has S20600 ; D has as much as A and B together. How much has D ? Ans. §35,260. TWO OR MORE SUBTRAHENDS. •yO. Two or more numbers may be taken from another at a single operation, as shown by the following example : I. A man having 1278 barrels of flour, sold 236 barrels to A, 362 to B, and 387 to C; how many had he left? OPERATION. Analysis. Since the remainder sought, 1278 added to the subtrahends, must be equal to the minuend, we add the columns of the subtrahends, and supply such figures in the remainder as, combined with these sums, will produce the minuend. Thus, 7 and 2 are 9, and 6 are 15, and 3 (supplied in the remainder sought) are 18 ; then, carrying the tens^ figure of the 18, 1 and 8 are 9, and 6 are 15, and 3 are 18, and 9 (supplied in the remainder) are 27; lastly, 2 to carry to 3 are 5, and 3 are 8, and 2 are 10, and 2 (supplied in the remainder) are 12 ; and the whole remainder is 293. Hence, the following BuLE. T. Hamng written the several subtrahends under tJi^- min- uend j add the first coIu7nn of the subtrahends, and supply such a figure in the remainder sought^ as, added, to this partial sum, will give an amount having for its unit figure the figure above in the minuend. II, Carry the tens^ figure of this amount to the next column of the subtrahends J and proceed as before till the entire re.main 19 _i_ Q j^ ( Dividing both dividend and divisor by 1 2 does not alter the quotient. 11^. 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 midtiplies the quotient. Dividend. 24 Divisor -- 6 . Quotient. = 4 . 48 -- 6 = 'r . 12 -H 6 = .{: :. 24 -v-12 = .{: . 24 -- 3 = s{: b DIVISION. 55 Prin. III. Multiplying or dividing both dividend and divisor hy the same numher, does not alter the quotient. 118» These three principles may be embraced in one GENERAL LAW. A change in the dividend produces a like change in the quo- tient ; hut a change in the divisor produces an OPPOSITE change in the quotient. SUCCESSIVE DIVISION. 119. 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, (9S, I). II. The result of successive division is the same, in whatever order the divisors are taken, (9^, II). • CONTRACTIONS IN DIVISION. CASE I. 120. When the divisor is a composite number. 1. Divide 1242 by 54. Analysis. The component factors of 54 are OPERATION. , ^ 6 and 9. We divide 1242 by 6, and the rc- ^ ^ suiting quotient by 9, and obtain for the final 9)207 result, 23, which must be the same as the 23 Ans. quotient of 1242 divided by times 9, or 54, (119, I). We might have obtained the same result by dividing first by 9, and then by G, (119, II). Hence the following KuLE. Divide the dividend hy one of the factors^ and the quos 66 SIMPLE NUMBERS. tient thus obtained hy another , and so on if there he 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. 131. 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 I. Divide 5855 by 168, using the factors 3, 7, and 8, and find the true remainder, OPERATION. Analysis. Dividing the S) 5855 given dividend by 3, we have 7 "i 1951 2 "^^^ ^^^ ^ quotient, and a remainder of 2. Hence, 2 ^)^ 5x3= 15 subtracted from 5855 would 34 ...6x 7x 3=1 26 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 diminish 1951*by 5 would require us to diminish 1951 X 3, the divide*id 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 6x7x3 = 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. L Multiph/ each j)artial remainder hy all the preceding divisors. II. Add the several products; the sum will he the true re^ mainder. DIVISION. 57 1. D 2. D 3. D 4. D 5. D 6. D 7. D 8. D 9. D 10. D 11. D 12. D 13. D 14. D 15. D 16. D 17. B 18. Di 19. D EXAMPLES FOR PRACTICE. ide 435 by 15 = 3 X 5. Ans. 29. de 425G by 56 = 7 X 8. de 17856 by 72 = 9 x 8. [de 15288 by 42 :== 2 x 3 x 7. Ans. 364. [de 972552 by 168 = 8 x 7 X 3. Ans, 5789. [de 526050 by 126 = 9 X 7 X 2. [de 612360 by 105 = 7 x 5 x 3. Ans. 5832. ide 553 by 15 = 3 X 5. Rem. 13. de 10183 by 105 = 3 X 5 X 7. 103. de 10197 by 120 = 2 X 3 X 4 X 5. 117. de 29792 by 144 = 3 x 8 x 6. 128. de 73522 by 168 = 4 x 6 x 7. 106. de 63814 by 135 = 3 X 5 X 9. ^ 124. [de 386639 by 720 = 2 x 3 x 4 x 5 x 6.' 719. de 734514 by 168 = 4 x 6 x 7. 18. [de 636388 by 729 = 9^. 700. de 4619 by 125 = 5^ 119. de 116423 by 10584 = 3 x 7^ x 8 x 9. 10583. de 79500 by 6125 = 5^ x T. 6000. CASE II. 1$53. "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, (37^ III). For a similar reason, by cutting off two figures we divide by 100 ; by cutting off threcj we divide by 1000, and BO on ; and the figures cut off will constitute the remainder. Hence the ^mj^ Rule. Ftotyi 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 quotienL 58 SIMPLE NUMBERS. EXAMPLES FOR PRACTICE. 1. Divide 79 by 10. Ans. 7-i%. 2. Divide 7982 by 100. 3. Divide 4003 by 1000. Ans. 4y^3^^. 4. Divide 2301050 by 10000. 6. Divide 3600036 by 1000. Ans. 3600tMtj- CASE III. 133. When there are ciphers on the right hand of the divisor. I. Divide 25548 by 700. Analysis. "We resolve 700 OPERATION. . , ^ -.^^ -, ^ into the factors 100 and 7. 7|00) 255148 Dividing first by 100, the quo- 36 Quotient. 3 2d rem. tient is 255, and the remainder 3 X 100 + 48 = 348 true rem. 48. Dividing 255 by 7, the final quotient is 36, and the second remainder 3. Multiplying the last remainder, 3, by the preceding divisor, 100, and adding the preceding remainder, we have 300 -f 48 = 348, the true remainder, (121). In practice, the true remainder may be obtained by prefixing the second remainder to the first. Hence the HuLE. I. Cut off the ciphers from the right of the divisor , and as many figures from the right of the dividend. II. Divide the remaining figures of the dividend hy the remain- ing figures of the divisor j for the final quotient. III. Prefix the remainder to the figures cut off^ and the result will he the true remainder. EXAMPLES FOR PRACTICE. 1. Divide 7856 by 900. Ans. 8f §f 2. Divide 13872 by 500. 3. Divide 8||W:8 by 2600. Ans. ^2^^^^, 4. Divide 1548036 by 4300. Ans. ^^^^lU- 5. Divide 436000 by 300. Ans. 1453|g§. 6. Divide 66472000 by 8100. r. Divide 10S18000 by 3600. DIVISION. EXAMPLES COMBINING THE TRECEDING RULES. 59 1. How many barrels of flour at S8 a barrel, will pay for 25 tons of coal at $4 a ton, and 3G 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. x\ 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+ (24 — 12) x 5 by (90 -f- 6) + (3 x llJ^irU.. Ans. 17. 5. D ivide 648 x (3^ x '2') — 9 — (2910 ~- 15) by 2863 -^ (4375 -nj5) X 4^+ 3^ 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. 7d. 7. What number is that which being divided by 45, the quo- tient increased by 7^ + 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 go 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 ot 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. 2U 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 be 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 ? PROBLEMS IN SIMPLE INTEGRAL NUMBERS. 134« 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 Eules. l^O. In 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 Sifhtraction, the subtrahend and remainder are the parts, and the minuend is the whole. III. In Multiplicationy 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. lY. 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. 1S6. Every example that can possibly occur in Arithmetic, and every business computation requiring an arithmetical opera- 6 62 SIMPLE NUMBERS. tion, can be classed under one or more of the four Fundamental Rules, as follows : I. Cases requiring Addition. There may he given To find 1. The parts, the whole, or the sum total. 2 The less of two numbers and ^ ^, . ,.«. ^1 1 I the g-reater number or the their dmerence, or the sub- > P ^ , 1 T . T minuend. trahend and remainder, J II. Cases requiring Subtraction. There may he (jiven To find 1. The sum of two numbers and ) one of them, j *^^ °t^e'^- 2. The greater and the less of ^ two numbers, or the minuend I the difference or remainder and subtrahend, J III. Cases requiring Multiplication. There may he given To find 1. Two numbers, their product. "2. Any number of factors, their continued product. 3. The divisor and quotient, the dividend. IV. Cases requiring Division. There may he given To find 1. The dividend and divisor, the quotient. 2. The dividend and quotient, the divisor. 3. The product and one of two ' factors, 4. The continued product of ^ several factors, and the pro- > that one factor. duct of all but one factor, J I2T. Let the pupil be required to illustrate the following pro- blems by orio^inal examples. Problem 1. Given, several numbers, to find their sum. Prob. 2. Given, the sum of several numbers and all of them but one, to find that one. V the other factor. ^KsmF PROBLEMS. g3 Prob. 3. Given, the parts, to find the whole. Prob. 4. Given, the whole and all the parts but one, to find that one. Prob. 5. Given, two numbers, to find their difierence. Prob. 6. Given, the greater of two numbers and their difierence, to find the less number. Prob. 7. Given, the less of two numbers and their difference, to find the greater number. Prob. 8. Given, the minuend and subtrahend, to find the remainder. Prob. 9. Given, the minuend and remainder, to find the sub- trahend. Prob. 10. Given, the subtrahend and remainder, to find the minuend. Prob. 11. Given, two or more numbers, to find their product. Prob. 12. Given, the product and one of two factors, to find the other factor. Prob. 13. Given, the continued product of several factors and all the factors but one, to find that factor. Prob. 14. Given, the factors, to find their product. Prob. 15 Given, the multiplicand and multiplier, to find the product. Prob. 16. Given, the product and multiplicand, to find the multiplier. Prob. 17. Given, the product and multiplier, to find the mul- tiplicand. Prob. 18. Given, two numbers, to find their quotients. Prob. 19. Given, the divisor and dividend, to find the quotient. Prob. 20. Given, the divisor and quotient, to find the dividend. Prob. 21. Given, the dividend and quotient, to find the divisor. Prob. 22. Given, the divisor, quotient, and remainder, to find the dividend. Prob. 23. Given, the dividend, quotient, and remainder, to find the divisor. Prob. 24. Given, the final quotient of a continued division and the several divisors, to find the dividend. 64 SIMPLE NUMBEKS. Prob 25. Given, the final quotient of a continued division, the first dividend, a,nd all the divisors but one, to find that divisor. Prob. 26. Given, the dividend and several divisors of a con- tinued division, to find the quotient. Prob. 27. Given, two or more sets of numbers, to find the difierence of their sums. Prob. 28. Given, two or more sets of factors, to find the sum of their products. Prob. 29. Given, one or more sets of factors and one or more numbers, to find the sum of the products and the given numbers. Prob. 30. Given, two or more sets of factors, to find the ditler- ence of thoir products. Prob. 31. Given^ one or more sets of factors and one or more numbers, to find the sum of the products and the given number or numbers. Prob. 32. Given, two or more sets of factors and two or more other sets of factors, to find the difference of the sums of the products of the former and latter. Prob. 33. Given, the sum and the difference of two numbers, to find the numbers. Analysis. If the difference of two unequal numbers be added to the less number, the sum wdll be equal to the greater ; and if this sum be added to the greater number, the result will be twice the greater number. But this result is the sum of the two numbers plus their difference. Again, if the difference of two numbers be subtracted from the greater number, the remainder will be equal to the less number ; and if this remainder be added to the less number, the result will be twice the less number. But this result is the sum of the two numbers minus their difference. Hence, I. The sum of two numbers plus their difference is equal to twice the greater number. II. The sum of two numbers minus their difference is equal to twice the less number. EXACT DIVISORS. 65 PROPERTIES OF NUMBERS. EXACT DIVISORS. 1SS8. An Exact Divisor of a number is one that gives an integral number for a quotient. And since division is the reverse of multiplication, it follows that all the exact divisors of a number are factors of that number, and that all its factors are exact divisors. Notes. — 1. Every number is divisible by itself and unity ; but the number itself and unity are not generally considered as factors, or exact divisors of the number. '^ 2. An exact divisor of a number is sometimes called the measure of the number. ISO , An Even Number is a number of which 2 is an exact divisor; as 2, 4, 6, or 8. 130* An Odd Number is a number of which 2 is not an exact divisor; as 1, 3, 5, 7, or 9. 131. A Perfect Number is one that is equal to the sum of all its factors plus 1; as 6 = 3 + 2 + 1, or 28 = 14 + 7 + 4 + 2 + 1. Note —The only perfect numbers known are 6, 28, 496,8128, 33550336, 8589869056, 137438691328, 2305843008139952128, 2417851639228158837784576, 990352031428297183044881 6128. 133. An Imperfect Number is one that is not equal to the sum of yi its factors plus 1 , as 12, which is not equal to 6 + 4 + 3 + 2 + 1. 133. An Abundant Number is one which is less than the sum of all its factors phis 1 ; as 18, which is less than 9 + 6 + 3 + 2 + 1. 134. A Defective Number is one which is greater than the sum of all its factors plus 1 ; as 27^ which is greater than 9 + 3 + 1. 13o. To show che nature of exact division, and furnish tests of divisibility, observe that if we begin with any number, as 4, and take once 4, two times 4, three times 4, four times 4, and so on indefinitely, forming the series 4, 8, 12, 16, etc., we shall have 6* B 66 PROPERTIES OF NUMBERS. all the numbers that are divisible by 4 ; and from the manner of forming this series, it is evident, 1st. That the product of any one number of the series by any integral number whatever, will contain 4 an exact number of times ; 2d. The sum of any two numbers of the series will contain 4 an exact number of times ; and 8d. The difference of any two will contain 4 an exact number of times. Hence, I Any number which will exactly divide one of two numbers will divide their product. II. Any number which will exactly divide each of two numbers will divide their sum. III. Any number which will exactly divide each of two num- bers will divide their difference. 136. From these principles we derive the following properties : I. Any number terminating with 0, 00, 000, etc., is divisible by 10, 100, 1000, etc., or by any factor of 10, 100, or 1000. For by cutting off the cipher or ciphers, the number will be divided by 10, 100. or 1000, etc., without a remainder, (122) ; and a number of which 10, 100, or 1000, etc., is a factor, will contain any factor of 10, 100, or 1000, etc., (I). II. A number is divisible by 2 if its right hand figure is even or divisible by 2. For, the part at the left of the units' placed taken alone, with its local value, is a number which terminates with a cipher, and is divi- sible by 2, because 2 is a factor of 10, (I) ; and if both parts, taken separately, with their local values, are divisible by 2, their sum, which is the entire number, is divisible by 2, (135, II). Note. — Hence, all numbers terminating with 0, 2, 4, 6, or 8, are even, and all numbers terminating with 1, 3, 5, 7, or 9, are odd. III. A number is divisible by 4 if the number expressed by its two right hand figures is divisible by 4. For, the part at the left of the tens^ place, taken alone, with its local value, is a number which terminates with two ciphers, and is divisible by 4, because 4 is a factor of 100, (I) ; and if both parts, EXACT DIVISORS. gf taken separately, with their local values, are divisible by 4, their sum which is the entire number, is divisible by 4, (135, II) TV. A number is divisible by 8 if the number expressed by its three right hand figures is divisible by 8. For, the part at the left of the hundreds' place, taken alone, with its local value, is a number which terminates with three ciphers, and IS divisible by 8, because 8 is a factor of 1000, (I) ; and if both parts, taken separately, with theiv local values, are divisible by 8, their sum, or the entire number, .s divisible by 8, (135, II). Y. A number is divisible by any power of 2, if as many right hand figures of the number as are equal to the index of the given power^ are divisible by the given power. For, as 2 is a factor of 10, any power of 2 is a factor of the corres- ponding power of 10, or of a unit of an order one higher than is indicated by the index of the given power of 2 ; and if both parts of a number, taken separately, with their local values, are divisible by a power of 2, their sum, or the entire number, is divisible by the same power of 2, (135, II). YI. A number is divisible by 5 if its right hand figure is 0, or 5. For, if a number terminates with a cipher, it is divisible by 5, because 5 is a factor of 10, (I) ; and if it terminates with 5, both parts, the units and the figures at the left of units, taken separately, with their local values, are divisible by 5, and consequently their sum, or the entire number, is divisible by 5, (135, II). YII. A number is divisible by 25 if the number expressed by its two right hand figures is divisible by 25. For, the part at the left of the tens' figure, taken with its local value, is a number terminating with two ciphers, and is divisible by 25, because 25 Is a factor of 100, (I) ; and if both parts, taken separately, with their local values, are divisible by 25, their sum, or the entire number, is divisible by 25, (135, II). YIII. A number is divisible by any power of 5, if as many right hand figures of the number as are equal to the index of the given power are divisible by the given power. For, as 5 is a factor of 10, any power of 5 is a factor of the corres- ponding power of 10, or of a unit of an order one higher than is indi- J58 PROPERTIES OF NUMBERS. cated by the index of the given power of 5 ; and if both parts of a number, taken separately, with their local values, are divisible by a power of 5, their sum, or the entire number, is divisible^by the same power of 5, (135, II). IX. A number is divisible by 9 if the sum of its digits is divis- ible by 9. For, if any number, as 7245, be separated into its parts, 7000 -f 200 +40+5, and each part be divided by 9, the several remainders will be the digits 7, 2, 4, and 5, respectively ; hence, if the sum of these digits, or remainders, be 9 or; an exact number of 9^s, the entire number must contain an exact number of 9's, and will therefore be divisible by 9. Note. — Whence it follows that if a number be divided by 9, the remainder will be the same as the excess of 9's in the sum of the digits of the number. Upon this property depends one of the methods of proving the operations in the four Fundamental Rules. X. A number is divisible by a composite number, when it is divisible, successively, by all the component factors of the com- posite number. For, dividing any number successively by several factors, is the same as dividing by the product of these factors, (119, I). XI. An odd number is not divisible by an even number. For, the product of any even number by any odd number is even , and, consequently, any composite odd number can contain only odd factors. XII. An even number that is divisible by an odd number, is also divisible by twice that odd number. For, if any even number be divided by an odd number, the quo- tient must be even, and divisible by 2 ; hence, the given even num- ber, being divisible successively by the odd number and 2, will be divisible by their product, or twice the odd number, (119, I). PRIME NUMBERS. 1S7. A Prime Number is one that can not be resolved or separated into two or more integral factors. Note. — Every number must be either prime or composite. PRIME NUMBERS. C9 138* To find all the prime numbers within any given limit, we observe that all even numbers except 2 are composite ; hence, the prime numbers must be sought among the odd numbers. 130.. If the odd numbers be written in their order, thus; 1, 3, 5, 1, 9, 11, 13, 15 17, etc., we observe, 1st. Taking every third number after 3, we have 3 times 3, 5 times 3, 7 times 3, and so on ; which are the only odd numbers divisible by 3. 2d. Taking every fifth number after 5, we have 3 times 6, 5 times 5, 7 cimes 5, and so on; which are the only odd numbers divisible by 5. And the same will be true of every other number in the series. Hence, 3d. If we cancel every third number, counting from 3, no number divisible by 3 will be left; and since 3 times 5 will be canceled, 5 times 5. or 25, will be the least composite number left in the series. Hence^ 4th. If we cancel every fifth number, counting from 25, no number divisible by 5 will be left; and since 3 times 7, and 5 times 7, will be canceled, 7 times 7, or 49, will be the least com- posite number left in the series. And thus with all the prime numbers. Hence, 140. To find all the prime numbers within any given limit, we have the following Rule. I. Write all the odd riumhers in their natural order. II. Cancel, or cross out, 3 times 3^ or 9, and every third number after it; 5 times 5, or 25, and every fifth number after it; 7 times 7, or 49, and every seventh number after it ; and so on, beginning with the second power of each prime number in succession, till the given limit is reached. The numbers remaining, together tvith the number 2, will be the prime numbers required. Notes. — 1. It is unnecessary to count for every ninth number after 9 times 9, for being divisible by 3, they will be found already canceled; the same may be said of any other canceled, or composite number. 2. This method of obtaining a list of the prime numbers was employed by Eratosthenes (born b. c, 275), and is called Eratoathenea' Sieve, 70 PROPERTIES OF NUMBERS. TABLE OF PRIME NUMBERS LESS THAN 1000. 1 59 139 233 337 439 557 653 769 883 2 61 149 239 347 443 563 659 773 887 3 67 151 241 349 449 569 661 787 907 6 71 157 251 353 457 571 673 797 911 7 73 163 257 359 461 577 677 S09 919 11 79 167 263 367 463 587 683 811 929 13 83 173 269 37d 467 593 691 821 937 17 89 179 271 379 479 599 701 823 941 19 97 181 277 383 487 601 709 827 947 23 101 191 281 389 491 607 719 829 953 29 103 193 283 397 499 613 727 839 967 31 107 197 293 401 503 617 733 853 971 37 109 ]99 307 409 509 619 739 857 977 41 113 211 311 419 521 631 743 859 983 43 127 223 313 421 523 641 751 863 991 47 131 227 317 431 541 643 757 877 997 53 137 229 331 433 547 647 761 881 FACTORING. CASE I. number into its 141, To resolve any composite prime factors. The Prime Factors of a number are those prime numbers which multiplied together will produce the given number. 14:2, The process of factoring numbers depends upon the fol- lowing principles : I. Every prime factor of a number is an exact divisor of that number. II. The only exact divisors of a number are its prime factors, or some combination of its prime factors. 1. What are the prime factors of 798 ? OPERATION. 2 3 7 19 798 399 133 "19 Analysis. Since the given number is even, we divide by 2, and obtain an odd number, 399, for a quotient We then divide by the prime numbers 3, 7, and 19, successively, and the last quotient is 1. The divisors, 2, 3, 7, and 19, are the prime factors required, (II). Hence, the FACTORING. YJ Rule. Divide the given number hy 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 wiP be the given number. EXAMPLES FOR PRACTICE. 1. What arc the prime factors of 2150 'Z 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, (136), 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 BULE. 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, in the tabU, and is not divisible by 2, 3, 5, 7, or 11, it is prime. 72 PROPERTIES OF NUMBERS. FACTOR TABLE. 1 1 i ^ .a Numbers. Factori. 1 i ll 1 1 |1 1 99 29 11 17 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 a 01 17 57 31 61 37 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 67 5a 47 13 43 23 15 ao as 87 29 03 41 93 17 4343 69 19 07 41 89 17 49 13 01 19 21 43 01 41 93 41 19 13 39 51 19 71 13 13 13 99 13 61 31 13 17 33 19 07 23 30 27 23 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 53 43 29 63 19 4 37 17 16 %\ 73 31 71 37 35 77 41 4:4: 47 37 67 23 03 13 73 29 33 23 17 29 81 29 77 17 03 31 79 23 27 19 49 13 87 17 37 19 79 13 43 31 19 13 87 13 97 19 23 13 91 13 2943 53 23 93 67 81 13 81 23 49 17 47 19 99 23 31 51 53 40 39 23 5943 53 93 17 11 51 13 59 17 ae 03 29 69 43 09 19 53 61 67 31 11 47 5 21 19 79 23 71 13 03 19 07 13 87 17 31 29 69 41 83 19 17 13 27 17 39 17 81 41 7341 2343 27 53 89 37 33 37 71 17 91 67 21 17 29 23 47 31 91 19 83 37 27 37 31 31 99 59 43 13 89 67 97 59 2973 33 13 57 13 IT 97 13 41 19 33 13 36 61 31 45 49 39 19 51 19 59 19 03 13 aa 69 17 39 43 01 13 63 17 11 13 01 13 53 53 59 13 89 29 11 29 01 31 ai 4947 11 23 69 13 31 23 13 17 59 23 89 19 la 17 17 09 47 01 37 51 23 29 19 87 61 37 13 27 13 63 31 6 07 17 39 37 27 17 4313 61 29 4941 97 17 41 19 79 13 71 41 11 13 19 23 51 17 31 23 47 41 73 19 53 13 4:1 53 29 81 17 77 19 29 17 41 17 63 41 49 13 59 31 93 31 67 19 17 23 5947 97 19 89 17 67 13 47 29 69 29 57 37 71 17 97 23 79 13 21 13 73 17 50 54 89 13 61 13 81 i; 63 31 7347 3a 83 29 41 41 77 23 17 29 29 61 97 17 71 31 18 7943 as 11 13 37 63 23 79 19 2947 47 13 7 73 19 07 13 91 29 09 53 33 53 1347 71 43 89 13 41 71 59 53 03 19 13 17 23 33 13 29 3941 21 61 81 37 46 53 31 6143^ 13 23 13 13 19 17 23 23 31 19 47 17 87 37 83 47 01 43 57 13 73 13 31 17 33 31 29 31 27 13 39 17 63 13 43 19 87 53 07 17 63 61 91 17 67 13 39 13 43 19 29 17 67 47 77 29 49 23 89 59 19 31 69 37 97 23 79 19 43 17 4943 53 13 69 19 81 17 57 13 99 13 33 41 83 13 55 93 13 49 19 53 17 63 17 73 13 87 19 63 53 4:a 61 59 51 13 37 99 17 57 23 91 31 69 23 81 43 93 37 81 19 2341 67 13 11 19 39 29 8 63 29 19 %\ 99 13 33 91 17 37 19 81 31 2347 43 23 17 19 69 37 09 23 07 29 a9 17 31 99 29 47 31 87 43 29 23 49 31 41 29 87 19 19 19 13 19 11 41 37 47 38 • 67 17 93 13 41 53 61 67 61 23 91 13 21 17 19 41 21 23 41 13 09 13 4:3 99 37 43 37 67 19 71 ij 14: 27 41 49 31 23 37 49 17 11 37 03 13 47 49 19 87 37 93 19 03 23 37 13 61 23 29 29 79 31 27 43 07 59 09 17 61 13 »7 29 FACTORING. 78 FACTOR TABLE — Continued. 1 s a y ii » fx 2 „• 1 (2 2 „• 1 1 1 3 c< 1 1 li li ;2; C 55 ;S ll 1 d 56 60 3947 51 13 61 53 57 13 23 71 1347 51 53 53 19 59 13 03 13 01 17 43 17 59 19 67 13 61 47 27 23 17 19 57 17 59 47 71 19 OD 71 19 13 63 23 77 13 77 19 63 79 33 29 41 23 73 19 63 59 73 17 11 31 23 19 67 29 87 71 79 29 97 43 47' 13 53 79 79 13 69 13 83 23 17 41 31 37 87 13 89 83 89 37 77 51 83 71 43 81 83 71 73 97 27 17 49 23 93 43 93 61 91 23 09 13 77 41 73 37 91 17 87 37 01 89 29 13 59 7J 97 73 69 73 29 59 83 59 79 61 89 99 17 03 31 3343 71 13 99 67 01 67 03 67 39 71 81 83 17 03 29 93 07 17 71 53 77 59 65 13 31 13 71 47 61 19 23 80 13 09 59 01 71 27 71 81 13 61 09 23 29 13 19 13 51 23 31 47 97 29 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 5T 07 31 27 61 43 53 39 41 71 19 43 17 07 47 47 23 22 19 63 13 07 13 09 41 3347 53 17 61 17 81 31 49 29 09 67 57 13 47 13 73 29 13 29 19 29 39 13 73 19 63 37 83 43 58 31 81 19 59 17 53 ^47 97 97 23 59 37 17 41 31 89 29 67 53 87 13 59 41 49 83 77 47 07 17 99 41 29 17 57 47 57 79 TO 73 73 78 77 13 51 17 83 13 79 83 98 59 13 61 61 83 29 03 47 79 47 01 29 89 19 57 43 89 89 89 41 09 17 67 7J 69 31 93 19 0943 87 83 07 37 83 67 13 93 17 94: 27 31 71 29 79 37 66 31 79 91 19 11 73 01 59 79 23 90 07 23 41 13 73 23 87 23 13 17 33 13 97 13 13 13 03 13 87 ii 17 71 09 97 4743 77 5J 91 41 17 13 37 31 74: 31 41 07 29 93 13 19 29 51 13 53 59 58 63 23 37 61 23 09 31 37 17 1343 86 47 83 69 17 69 71 09 37 27 13 31 19 67 37 21 41 49 47 27 19 11 79 61 13 81 19 81 41 33 19 33 23 41 29 81 73 23 13 59 29 49 73 21 37 71 47 87 53 93 13 37 13 39 17 47 17 87 19 29 17 71 17 51 37 33 89 7343 95 99 19 91 4J 41 79 49 61 93 41 39 43 91 13 57 23 39 53 77 29 03 13 99 93 71 53 13 67 59 97 47 53 29 97 53 79 17 51 41 83 31 09 37 13 23 99 17 83 61 83 41 99 31 63 17 79 99 43 53 17 89 61 17 31 1747 59 80 19 97 37 71 71 31 13 41 83 71 13 91 23 89 37 19 09 19 63 6r 11 13 93 59 21 89 03 19 83 19 01 19 29 13 43 61 11 23 13 59 07 19 23 17 75 8J 17 21 53 87 13 13 53 41 53 37 17 61 19 71 31 53 41 37 01 13 43 13 33 13 11 31 31 23 57 19 59 23 21 31 31 13 39 23 ■ 53 23 19 73 57 73 39 31 17 23 39 13 63 73 71 13 33 17 41 17 49 17 57 17 31 17 61 19 41 19 49 13 4341 71 17 79 17 41 13 71 23 51 43 63 13 43 19 67 31 47 17 59 19 67 89 77 61 83 67 47 19 S3 13 57 29 69 67 71 67 69 13 57 61 73 31 69 53 S9 43 91 97 59 59 64: 67 67 71 71 97 71 79 79 59 13 77 67 79 67 93 53 97 13 63 67 01 37 73 13 81 43 76 81 23 81 17 91 59 93 29 99 29 69 47 03 19 99 13 99 23 13 23 91 61 83 83 97 19 97 17 96 7743 07 4? 68 73 19 19 99 19 99 37 88 93 07 13 83 31 09 13 17 17 01 19 27 29 80 84: 01 13 11 61 17 59 89 53 31 59 21 19 23 31 31 13 03 53 01 31 09 23 17 13 37 23 93 13 37 41 47 41 41 13 33 17 21 13 11 41 43 37 23 23 41 31 3 188139 7 62713 17 8959 17 527 31 74 PROPERTIES OF NUMBERS. 1. Eesolve 1961 into its prime factors. OPERATION. Analysis. Cutting off the two 1961 -— 37 = 53 right 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 least 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 that the given number is divisible by 3 and 7 ; dividing by these fac- tors, we have for a quotient 8959. By referring to the factor table, we find the least prime factor of this number to be 17 ; dividing by 17, we have 527 for a quotient, 3x7x17x17x31^ Ans. Preferring again to the table, 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, 13, 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 ? OPERATIOX. 360 = 1x2x2x2x3x3x5. Ans.< 1 , 2 , 4 , 8 Combinations of 1 and 2. 3 , 9 , 6 , 18 , 12 , 24 1 36 , 72 1 (( " 1 and 2 and 3. 5 , 10 , 20 , 40 (( " land 2 and 5. 15 , 45 , 30 , 90 , 60 , 120 1 180 , 360 f ti « land 2 and 3 id 5. Analysis. Bj 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, 1 X 2 X 2 = 4, and 1 X 2 X 2 X 2 = 8; these we write in the first line. Multiplying the first line by 3 and writmg the products in the second line, and the second line by 3, writing the products in the third line, we have in 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 mimher into its prime /actors, II Form a series having \ for the firsts term, that prime factor which occurs the greatest number of times in the given numher 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, and these residts by the same factor, and so on, as many times as this factor occurs in the given number. 76 PROPERTIES OF NUMBERS. TV. Multiply all the comhinations now obtained hy 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 scni72 ? . 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 ? A71S. 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; 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. F 82 PROPERTIES OF NUMBERS. LEAST COMMON MULTIPLE. t^l. 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. 152. 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. lo3. 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 numhers may he found hy multiplying the given numbers together. 155. 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 = 2 X 3 X 11 than 78, because it / 8 = ^ X ^ X lo must contain 78 ; and 2x3x13x11x3x7 = 18018 Ans. if it contains 78, it must contain all its prime factors, viz. ; 2 X 3 X 13. LEAST COMMON MULTIPLE. 33 We here have all the prime factors, and also all the factors of 66 except 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 prinie factors. II. Multiple/ together all the prime factors of the largest numher^ and such prime factors of the other numbers as are not found in the largest number j and their product will be the least common multiple. NoTR. — When a prime factor is repented 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. Alls. 4620. 2. Find the least common multiple of 21, 80, 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. 1^6. 1. What is the least common multiple of 4, 9, 12, 18, and 36 ? 84 PROPERTIES OF NUMBERS. 2 4. OPERATION. . 9 . . 12 . . 18 . . 36 2 2 . . 9 . . 6 . . 9 . 18 3 9. . 3 . . 9 . 9 3 3 3 3 2 X 2 X 3 X 3 = Analysis. We first write 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- 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 tho vertical ; and their product, 3G, 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 see that 2 and 5 are among the factors of the given num- bers, and must be factors of the least common multiple ; 2x5x2x3x5 = 300, Ans. 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? OPERATION. Analysis. In this operation 2,5 20. .12. .15. .75 2,8 2 . . 6. . 3 . . 15 5 5 3,7 2,5 10 15 4*^ 70 ^'® omit the 7 and 10, because they are exactly contained in some of the other given numbers ; 3x7x2x5 = 210, Arts. 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 the least common multiple of the remaining numbers, 15, 42, and 70. LEAST COMMON MULTIPLE. 85 From these examples we derive the following Rule. I. Write the numbers in a Itnej omitting such of the smaller numbers as ai^e 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 j and write the quotients and undivided numbers in a line underneath, omitting the Vs. III. In like manner divide the quotients and undivided numberSy 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 l^ast 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? 9. What is the least common multiple of <34, 84, 96, and 210 ? 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? Aiis. 1400. 8 go 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 872? Ans. $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 ? Ans. 72 rods. 13. The foi-ward 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 ? Ans. 17 miles 5115 feet. CANCELLATION. 157* Cancellation is the process of rejecting equal factors from numbers sustaining to each other the relation of dividend and divisor. 158. 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. ^T,^r>.rT,T^>. Analysis. We first in- OPERATION. -lo/-.- ^ L ^ -ir. dicate the division by wri- = - . = 13 tii^g the dividend above a 105 ?i X fi Xt 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. 151^. 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 abridf^ed. CANCELLATION. g^ 2d. The factors of small numbers are generally more readily detected than those of large numbers. 2. Divide 20 times 66 by 7 times 15. OPERATION. Analysis. Having first indi- 4 g cated all the operations required ^0 X ^i 32 hy the question, we cancel 7 ~T 77 ^^ o~ ^^^^ from 7 and 5G, and 5 from 15 '^q 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 10|, 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 xvill be the quotient. NoTKS. — 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, and the factors of the divisor on the left. EXAMPLES FOR PRACTICE. 1. What is the quotient of 18 X 6 x 4 X 42 divided by 4 x 9 X 3 X 7x 6? FIRST OPERATION. SECOND OPERATION. ' ^^ X X 4 X M' =4 i ^x^x^x^x^ ^ t 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, fs to be divided by the product of the numbers 40, 24, 50, 20, 7, and 10; what is the quotient? ^^s- ^-- 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 3G, 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 of 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 ; how much was the tea worth a pound ? Ans. 60 cents. DEFINITIONS, NOTATION, AND NUMERATION. 89 FKACTIONS. 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 t\iQm^ fractional numbers. Hence 161, A Fractional unit is one of the equal parts of an inte- gral unit. 16 ^« A Fraction is a fractional unit, or a collection of frac- tional units. 1@3. 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. IG^i:* To express a fraction by figures, two integers are re quired j 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 farmer is written below a horizontal line, the latter above. Thus, One fifth is written | One half is written ^ One third " \ Two thirds " | One fourth ^^ | Two fourths " f Three fourths " | 8* Two fifths " I One seventh " ^ Three eighths " | Five ninths ^^ f Eight tenths " i% 90 , FRACTIONS. 16^. The Denominator of a fraction is tlie mimber below tlie line. It denominates or names the fractional unit, and it shows how many fractional units are equal to an integral unit. 'I@@. The Numerator is the number 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. mo. 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 seventlis. 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 EOR PRACTICE. Express the following fractions by figures : — 1. Four ninths. Ans. 2. Seven Ji/fiy-sixths. Aris. -^^, 3 . Sixteen forty-eigliths. 4. Ninety-five one hundred seventy-ninths. 5. Five hundred thirty-six /oitr hundredths. . 6. One thousand eight hundred fifty-seven 7iine thousand Jive hundred twenty-firsts. 7. Twenty-five thousand eighty-sevenths. 8. Thirty ten thousand eighty-seconds. 9. One hundred one ten millionths. I DEFINITIONS, NOTATION, AND NUMERATION. 91 Read and analyze the following fractions: — 10 4 . 7.17. 45_ . J72_ . _48_ . _ 8 1_ . 456 ■*-^' 9^ lH ^ 38^ J 00^ 37 5^ 1009^ 7HG3^ 537* 11 2 0. 87. 95 . 48. 75 . 175. 4_3 6 . _7 6 6 J--*-' 4 ^ 30 > J 00 ^ '12 ^ 4 37 ^ ^ > 60 ^ 4 57"5* 19 4 6 7. 5 3^. 10000. 7 5__. 5007 ^•^' ^3(1^ ;^4§^ 75 ; l^OOO; 30 01" iq 150. 4 36. 137 85. 15 0072. I_0000i ^*^' 53 7^ 972^ 4 7 95(J^ 4 7500 0^ 2000(52* Fractions are distinguished as Proj)er and Improper, 170. A Proper Fraction is one whose numerator is less than its denominator; its value is less than the unit 1. 1*^1. 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 bo 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. ly^. A Mixed E'umber is a number expressed by an integer and a fraction. ITS. Since fractions indicate division, (ISS, II), all changes in the iermH 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 rmmerator^ denominator^ and fraction^ or value of the fraction, for the words dividend, divisor^ and quotient, respectively, and we shall have the followins; GENERAL PRINCIPLES OF FRACTIONS. 174. Prin. I. MidAiphjing tlie numerator multiplies the fraction^ and dioiding the numerator divides the fraction. Prin. II. Midtiplijing the denominator divides the fraction, and dividing the denominator TRultiplies the fraction. Prin. III. Multiplying or dividing Loth terms of the fraction hy the same number, does not alter the value of the fraction. 175. These three principles may be embraced in one general LAW. A change in the numerator produces a like change in the value of the fraction ; hut a change in the denominator j)ro(i?^cej an OPPOSITE change in the value of the fraction. OPERATION. l%% 12 __ 2 1 Or, 4 1 5)tVo = 4 92 FRACTIONS. REDUCTION. "B 70. 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 loivest terms when its numerator and denomi- nator are prime to each other ; that is, when both terms have no common divisor. I. Eeduce the fraction y^^^ to its lowest terms. Analysis. Dividing both terms of the fraction by the same number does not alter the value of the fraction, (174, III); hence, we divide both terms of yg^ 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 j^^ 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 all factors common to hofh numerator and denominator. Or, Divide both terms hy their greatest common divisor, EXAMPLES FOR PRACTICE. 1. Reduce j'^^^^j to its lowest terms. Ans. | 2. Reduce f| to its lowest terms. Ans. |. 8. Reduce H| to its lowest terms. Ans. |. 4. Reduce -^^^ to its lowest terms. Ans. |. 5. Reduce ^if ^^ i^s lowest terms. Ans. |. 6. Reduce ||f to its lowest terms. 7. Reduce ||| to its lowest terms. 8. Reduce j^^.^^ to its lowest terms. 9. Reduce f §|| to its lowest terms. 10. Reduce |^| to its lowest terms. Ans. -||. Note. — Consult the factor table. II. Reduce j^Y? *^ ^^^ lowest terms. Ans. ||. REDUCTION. 93 12. Reduce f^fl to its lowest terms. Ans. |f. 13. Reduce |||f to its lowest terms. 14. Reduce || J^ to its lowest terms. Ans. 15. Reduce |i|||, and ^||f g to their lowest terms. 4 1 CASE II. 17§. To reduce an improper fraction to a whole or mixed number. 1. Reduce ^-f^ to a whole or mixed number. OPERATION. 2_9J7 = 297 -^ 12 = 24-^-c = 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 24y'2 = 24f . Hence the Rule. Divide the numerator hy the denominator. Notes. 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 tbo fractions to their lowest terms. EXAMPLES FOR PRACTICE. 1. Reduce ^^ to an equivalent integer. Ans. 16. 2. Reduce ^-^ to an equivalent integer. 3. Reduce ^g"* to a mixed number. Ans. 17^. 4. Reduce ^^^^ to a mixed number. Ans. 26||, 5. Reduce ^^^ to a mixed number, Ans. 24|. 6. Reduce ^g® to a mixed number. Ans. 17^|. 7. Reduce ^||* to a mixed number. 8. Reduce ^ j^|^ to a mixed number, Ans. 156|. 9. Reduce 3|02 ^q ^ mixed number. 10. Reduce g^/g® to a mixed number. Ans. 4|J. 11. Reduce ^fff^ to a mixed number. Ans. 100|. 12. Reduce j^l^f to a mixed number. 13. In '^^^^ of a day how many (Jays? Ans. 982| days. 14. In ^f^ of a dollar how many dollars? Ans. %^\^^. 15. If 1000 dollars be distributed equally among 36 men, what part of a dollar must each man receive in change ? Ans. J. 94 FRACTIONS. CASE III. 179. To reduce a whole number to a fraction having a given denominator. 1. lleduce 37 to an equivalent fraction whose denominator shall be 5. OPERATION. Analysis. Since in each unit there are S7 X ^ = 1S5 ^ fifths, in 37 units there must be 37 times 37 ^ 185^ j^^. 5 fifths, or 185 fifths = n^ . The nume- rator, 185, is obtained in the operation by multiplying the whole number, 37, by the given denominator, 5. Hence the HuLE. 3fultipli/ the whole number hy the given denominator ; talce the product for a number atorj under which write the given de- nominator. Note. — A whole number may be rerluced to a fractional form by writing 1 under it for a denominator; thus, 9 = ^-. EXAMPLES FOR PRACTICE. 1. Reduce 17 to an equivalent fraction whose denominator shall be 6. Ans, ^g^. 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. Ans^. ^y ^ 6. Reduce 763 feet to fourteenths of a foot. Ans. ^^y\®^. 7. Reduce 937 to a fractional form. Ans. ^^^ . CASE IV. 180. 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 1-7 12 X 7 sevenths = 84 sevenths, (Case III), and ' 84 sevenths -f 5 sevenths =: 89 sevenths, or \^. 89 Hence the following I REDUCTION. 95 Rule. Multiply the whole number hy the denominator of the fraction ; to the product add the numerator, and under the sum write the denominator. EXAMPLES FOR PRACTICE. 1. Reduce 154 to fifths. Ans. ^f. 2. Reduce 24| to an improper fraction. Ans. ^■^. 3. Reduce 57 f to an improper fraction. 4. Reduce 356i| to an improper fraction. Ans. ^^|^. 5. Reduce 872^^3^ to an improper fraction. G. Reduce 800 ^^^ to an improper fraction. ^ Ans. ^|^§^. 7. Reduce 434^| to an improper fraction. Ans. ^%^3^^. 8. In 15| how many eighths? 9. In 135^% how many twentieths? Ans. ^-J^^. 10. In 43| bushels how many fourths of a bushel ? 11. In 760 j^^ days how many tenths of a day? CASE V. 181. To reduce a fraction to a given denominator. We have seen that fractions may be reduced to lower terms by division. Conversely, I. Fractions may be reduced to higher terms by multiplication. II. All higher terms of a fraction must be multiples of its lowest terms. 1. Reduce -| to a fraction whose denominator is 40. OPERATION. Analysis. We first divide 40, the re- 40 -f- 8 = 5 quired denominator, by 8, the denomi- 3^5 nator of the given fraction, to ascertain 8 X 5 "^ ^^' ^^^ ^^ ^* ^^ ^ multiple of this term, 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 hy the denominator of the given fraction^ and multiply both terms of the fraction hy the quotient. 96 rrtACTioNS. EXAMPLES FOR PRACTICE. 1. Eeduce | to a fraction having 24 for a denominator. Alls. ^|. 2. Reduce -^^ to a fraction whose denominator is 96. Ans. f|. 3. Reduce |f to a fraction whose denominator is 51. 4. Reduce y^g to a fraction whose denominator is 78. 6. Reduce gW to a fraction whose denominator is 3000. A'ijv 4 9 6 6. Change 7| to a fraction whose denominator is 8. 7. Change IOt/^ to a fraction whose denominator is 176. 8. Change bj\ to a fraction whose denominator is 363. 9. Change 36f to a fraction whose denominator is 42. Ans. If 42. CASE VI. 18S8. 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. £j,g^ fraction by the denominator of the second, ^ ^ "—=27 and the terms of the second fraction by the 5x9 ^^ denominator of the first, (174, III). This Y v> 5 must reduce each fraction to the same deno- o" w 5 ^^ 4 o rainator, for each new denominator will be the product of the given denominators. Hence the Rule. Multiply/ the terms of each fraction hy 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. H, -j-^. 2. Reduce ^ and | to a common denominator. REDUCTION. 97 3. Reduce |, -f^ and i to a common denominator. JtJS "72 50 60 ji.ns. -j5^, j^^y ^2^. 4. Reduce ^, 5| and 1| to equivalent fractions having a com- mon denominator. Ans. -1|, y^j^, ||. 5. Reduce y^^ and j^^ to a common denominator. Ay-,^ 6 8 3 9 ^^^5- 22T, -z^j- 6. Reduce ^, ^ and Jj to a common denominator. 7. Reduce |, |, j'^2 ^"^ /s ^^ ^ common denominator. /j„jf 768 115 2 896 864 •^"^- To 3 5? 153 5? 75 3 5? T5^S* 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 |, ^^ and f^ to their least common denominator. OPERATION. Analysis. We first find the least 12 . . 15 common multiple of the given deno- . minators, which is 60. This must 4 be the least common denominator to 3 , 5 2 , 2 3x5x2x2 = 60 which the given fractions can be re- duced, (II). Reducing each frac- tion to the denominator 60, by Case V, we obtain f J, || and /„ ^^r the answer. Hence the following Rule. I. Find the least common multiple of the given denoTti" inators, for the least common denominator. 9 G 98 FRACTIONS. II. Divide this common denominator hy each of the given d&- nominators^ and multipli/ each numerator hi/ the corresponding quotient. The products loill be the neic 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 cummon denominator will be prime to each other. EXAMPLES FOR PRACTICE. 1 1. Reduce | and f^ to their least common denominator. /< ^ Q 2 5 12 2. Reduce |, | and | to their least common denominator. 8. Reduce* |, -^^ and |^ to their least common denominator. 4. Reduce |, | and | to their least common denominator. 5. Reduce j^^, i| and || to their least common denominator. Jjjd 3 6 3 5 2 6 6. Reduce |, 7*3? || and -^-^ to their least common denominator. J.JO 5 2 2 4 7 5 8 7. Reduce 2|, ^-^^ ^^ and |g to their least common denomi- v.qfnr Aii^i 312 5 6 25 74 l^^^OX. imS. y2^, J33^, y2^, JTTQ. 8. Reduce |^, g^g and |J to their least common denominator. 9. Reduce |g, ^-^j^ and ^| to their least common denominator- A...|6•i^r 14. Reduce -|, {A, f., ^s^, /^ and |g to their least common flpnmninntnr An400 6 930 1008 2240 1944 32i_3 aenominaior. jins. ^^qq, ^ggo' 7ogo' 7 5G0' 7 56T5? 7560 15. Reduce ^, A, rp^j ^^ and j'g^^ to their least common de- nominator. 16. Reduce j\, t^\, || and 4^ to their least common denomi- r^o+r»T. Aiif: 2 8 7 6 45 5 IiatOr. .^?IS. y^^, y^5, JXJ-, y^jj. ADDITION. 99 ADDITION. 185. The denominator of a fraction determines the value of the fractional unit, (16o) ; 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 different 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 4, j% and ^ ? O^ 1 ^ JO OPERATION. « + « + 2 _ 12 + 2 5 + 8 _ ,,, _ 3 . 5 ' T3Z ' 1 5 — 0Q — gU — ?• Analysis. We first reduce the given fractions to a common deno- minator, (III). And as the resulting fractions, J§, |^, and /j have the same fractional unit, (I), we add them by uniting their numerators into one sum, making || — J, the answer. 2. Add 5|, 3| and 4/3. Analysis. The sum of the OPERATION. integers, 5, 3, and 4, is 12; the ^ + 5 + i^ = 1^ sum of the fractions, J, J, and 34_7_| 7 95 '4'K' ^ ^ H J^ ]2 — ^2i -7- is 24. Hence, the sum of 14^''^, Ans. both fractions and integers is 12 + 2^^ = 14^^. 180, From these principles and illustrations we derive the following general Rule. I. To add fractions. — When necessary, reduce tlie.frac' tioiis to their least common cJenominator ; 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 numbers. 100 FRACTIONS. EXAMPLES FOR PRACTICE. 1. What is the sum of Z^, y^^, /^ and ^ ? Ans. 2\, 2. What is the sum of ||, j\y f^ and f^ ? Ans. 1|. 3. What is the sum of /j, r^-^y ^f and ^f ? 4. What is the sum of 1^, Sfl, 2||, 5^§ and 4|| ? ^rw. 28|. 5. What is the sum of S7^%, 12|5, 13|| and f | ? 6. Add I, I and |. 7. Add I, I, If and j-\. J.n«. 2 ^^• 8. Add 1, I and J3. 9. Add ,% II and /^. ^n^. l|g. 10. Add 1, f 1^ and f |. jiws. 2||. 11. Add I, 1^,11,11 and if ^n.4/,V 12. Add 3^, 4| and 2/3. 13. Add 16^2 and 24 Jg. Ans. 40/^. 14. Add 1^, 2f , 3|, 4| and 5|. 15. Add 4/3, 82^ and 2/^. Ans. 14if 16. Add I, |, jKj ^°^ T7- -^^- It- 17. Add ^, f , j2^ and -\. 18. Add 1, I, y\, 3\ and ^|. Ans. Iff. 19. Add-i, j^V/gandf. 20. Add 41i, 105|, 300|, 241| and 472f Ans. 1161|g. 21. Add 4^, 2|, 1 J^, 2/^, 5/5, 7|, 4» and 6|. 22. Four cheeses weighed respectively 36|, 42|, 30/g and 51| pounds; what was their entire weight? Anfs. 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 /g by ^-f ? 25. A beggar obtained | of a dollar from one person, J from another, ^ from another, and f^ from another; how much did he get from all ? 26. A merchant sold 46| yards of cloth for ^127/5, 64-^4 yards for $226|, and 76| yards for 831 2§ ; how many yards of cloth did he sell, and how much did he receive for the whole ? Ans. 187|J yards, for ?666j|. SUBTRACTION. 201 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 hare the same unit value. Hence, II, In subtraction of fractions, the minuend and subtrahend must have a common denominator^ (ISSj I). 1. From I subtract §. OPERATION'. Analysis. Reducing the 4 2 __ 1 2_r^i __ _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 = j%, the answer. 2. From 238^ take 24|. OPERATION. Analysis. We first reduce the frac- 23 §1 3:; 238-^. tional parts, j and |, to the common 9_j^5 --_ ^410 denominator, 12. Since we cannot -1^/2 -'^^•-^- making ;|. Then, || subtracted from f f leaves f\ ; and carrying 1 to 24, the integral part of the subtrahend, (73, II), and subtracting, we have 213y\ for the entire remainder. ISS. From these principles and illustrations we derive the following general EuLE. I. To subtract fractions. — WJien necessar^y reduce the frai'tions to tJietr kast cojiunon deno7nwafor. Subtract (he nume- rator of the subtrahend from the numerator of the minuendy and phice the difference of the new numerators over the cominon de^wm- I'nafor. II. To subtract mixed numbers. — Reduce the /raetional parts to a co7n7non denoininatory and then subtract the fractional and vite>e numerator. EXAMPLES FOR PRACTICE. 1. Multiply I by 8. Ans. 2| 2. Multiply I by 27, f^ by 4, and 3^^ by 9. 3. Multiply ^\ by 15. Ans. |. 4. Multiply 8 by |. Ans. 6. 5. Multiply 75 by f-., 7 by ^fj, 756 by |, and 572 by ^\, 6. Multiply I by |. 7. Multiply 11 by II, and jf by f ^. 8. Multiply /3,- by 11, and/, by i^. 9. Multiply 24 by 3f. 10. Multiply If by 1]|. 11. Multiply 3% by 21 1. 12. Find the value of | X | X y\ X |. 13. Find the value of | X | X -Jf X ^\ X 4|. 14. Find the value of || X ^'^3 X fff. 15. Find the value of 2| x 24 X fj X y-gg X 1/^ X 26}. Ans. 2. 16. Find the value of y\ x -ij X 4| x 15 X /g. 17. Find the value of ^^^^ x ^izj X V/- ^'^^' lU- Ans, 10 Ans. 91 Ans. 7 34 Ans. 1 X06 FRACTIONS. 18. Find the value of (4-^^ X |) + If X (H — j%)- 19. Find the value of 28^+ (7| — 2|) X § X (f + 4). KoTE 3. — The word of between fractions is equivalent to the sign of multi- plication; and such an expression is sometimes culled ncotnpouud fraction. Find the values of the following indicated products : — 20. 4 of I of |. Ans. f . 21. f of I of 3,3_. Ans. ^\. 22. f of j\ of f . 23. 1^ of ^% of Jf of f|. Alls. Z^. 24. i of I of I of I of I of 4 of J of I of j%. In the following examples, cancellation may be employed by the aid of the Factor Table. 25. What is the value of |f | x i||| X |||| ? Ans. j%\. 26. AVhat is the value of ||o i x ||f i X iUi ? Ans. ^. 27. What is the value of |||f x ||f| X f f |f ? J ^ a 13 3 1 28. AYhat will 7 cords of wood cost at S3| per cord ? ^ns. $25|. 29. What is the value of (|y X ^f X (if? Ans. ^tjI^^. 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 12| ? 32. At S9| per ton, what will be the cost of -| of | of a ton of hay? Ans. $4. 33. At $'^Q a bushel, what will be the cost of 1| bushels of corn ? 34. When peaches are worth $| 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 (f)' x Q)' X (/^)' X (3|y ? Ans. aT4. 37. If a family consume 1| barrels of flour a month, how many barrels will 6 families consume in 8j^^ months? 38. What is the product of 150^— (y QfT21|4-j of 48|j-— 75, multiplied by 3 x C| of li x 4) — 2?? Ans. 342/^^. DIVISION. 207 89. A man at liis death left liis wife §12,500, wliicli was i of 'I 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 : Jin., g^. 40. A owned | of a cotton factory, and sold | of his share to B, who sold i of v/hat he bought to C, who sold | of what he bought to D ; what part of the whole factory did each then own ? ^ns. A, i, ; B, J^, ; C, ^ _^ A^ 41. What is the value of 2i x Hf ^^ H X (|)'+(^3?— ('^3)' • A71S. ^.469 -'54 0' DIVISION. 194. 1. Divide f^ by 3. FIRST OPERATION. SECOND OPERATION. 21 ^5 8 = 21 *^^ — 75 7 23 2. Dinde 15 by |. Analysis. In the first ope- ration we divide the fraction by 3 by dividing its numerator by 3, and in the second operation we divide the fraction by 3 by multiplying its denominator by 3, (190, II or III). Analysis. To divide by ?, we must divide by 3 and multiply by 7, (191, II or III). In the first operation, we first divide 15 by 3, and then mul- tiply the quotient bj 7. In the second operation we first multiply 15 by 7, and then divide the product by 3. 3. Divide X by h 1 o J o FIRST OPERATION. 15 -^ f = 5 X 7=^35 SECOND OPERATION. 15 -f- 1 = 105 -f- 3 = 35 FIRST OPERATION. 1st step, 2d step, -^ A -^ 3 = -4, 4 y 5 — 2 4 SECOND OPERATION. _4 v5 20 4 ^ — 4b — y Analysis. To divide by f , we must divide by 3 and multiply by 5, (191, ' II or III). In the first operation we first divide j- by 3 by multiplying the denomina- tor by 3. We then multi- 108 FRACTIONS. THIRD OPERATION. plj t^G result, $^3, bj 5, by 4 k multiplying the numerator £^ X 3 = I by 5, giving |« = | for the Q required quotient. By in- specting this operation, we observe that the result, |§, 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, |, 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 (107) that the reciprocal of a number is 1 divided by the number. If wg divide 1 by |, we shall have 1 — 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 KuLE. I. Reduce integers and mixed numbers to improper fractions. II. Midtiply the dividend hy the reciprocal of the divisor. Notes. — 1. If the vertical line be U8ed, 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 FOR PRACTICE. 1. Divide if by 4. f I X ^ = 3^^, Ans. 2. Divide jf by 5, and iff by 80. 3. Divide 10 by f . Ans, 35. 4. Divide 28 by |, and 3 by /^. 5. Divide 56 by If. Ans. 36. 6. Divide ^| by |. 7. Divide if by f , 4| by ^^,, and 3| by 5f 8. Divide 1| by 1|. Ans. If. 9. Divide l^f by |f . Ans. If. DIVISION. 109 10. Divide f of f by ? of /^. OPERATION. Analysis. The dividend, 3^5-3! reduced to a simple fraction, 7 V 5 5 is ^ ; the divisor, reduced -g X 1^ — 75 . f . ' 1 X '^ = ^=11 Ans. ^^ ^^^ manner, is j\ ; and Or, ^ divided by j% is 14, the quotient required. Or, we ^ X - X ^ X ^^ = 11 5 y "^ ^ ^ may apply the general rule directly by inverting both factors of the divisor. Note 3. — The second method of solution given above has two advantilgea. 1st, It gives the answer by a single operation; 2d, It afifords greater facility for cancellation. 11. Divide 4 of /^ by j\ of /g. Ans. 1. 12. Divide /^ of f^ by | of .\. Ans. Igi 13. Divide 2^ x 7i by 8i x 8-^%. 14. Divide 11 by | X 5| x 7. 15. Divide 31 x 19 by 1 X 7| x If. -Ans. 25. 16. Divide y'V X If by i X I X 3% X |f X f 1- An,^. 3|f. 17. Divide 11^ by f^^^. Ans. 1^. 18. Divide 3,\Wr by U X || X g|. Ans, ^|. 19. Divide i X I X I X -I by I X -? X i X I X T%. 51 20. What is the value of -f ? OPERATION. ^ _!. U V JL 5 11 Analysis. The fractional form indicates division, the numerator being the dividend and the denominator the divisor, (168, II) ; hence, we reduce the mixed numbers to improper fractions, and then treat the denominator, 2/, as a divisor, and obtain the result, IJ, by the general rule for division of fractions. 51 y Note 4. — Expressions like — and — are sometimes cnWed complex fractiona. 5. In the reduction of complex fractions to simple fractions, if either the numerator or denominator consists of one or more parts connected by -f or — , the operations indicated by these signs must first be performed, and afterward the division. 21. What is the value of f ? Ans. ^. 10 110 FRACTIONS. 22. What is the value of -f ^ l\? Ans. 2. 28. What is the value of -^-^t_M ? Ans. 7^. 3 . 1 5 24. Eeduce ^ ^ to its simplest form. '- + 25. Reduce ^ ^| to its simplest form. 3 >^ 7 ,5 Qf 3 26. Eeduce 7^,^ — p-- to its simplest form. Ans. ||. 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 $(jl ? Ans. 17f 29. If I of an acre of land sell for $30, what will an acre sell for at the same rate ? Ans. $67^. 30. At ^ of I of a dollar a pint, how much wine can be bought for $^% ? ^ Ans. 2| pints. 31. If -^2^ of a barrel of flour be worth $21, how much is 1 barrel worth ? Ans. $7|. 32. Bought I of q cords of wood, for | of ^ of $30; what was 1 cord worth at the same rate? Ans. $4y^. 33. If 235^ acres of land cost $1725|, how much will 125^ acres cost? Ans. $918|lf. 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}^ to produce 148| ? 37. What number is that which, if multiplied by | of g of 2, will produce I ? Ans. l|i. 88. Divide 720 - (§ X 28^^71) by 40| + (/, - f ) x Q)*. 39. What is the value of (b^ X (ff + | of jV ^ (l7^ — | + FKf)'x5)? 40 Divide i°^( t)'x3j |of5|^ ,,,„, GREATEST COMMON DIVISOR. m 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. NoTK. — The definition of .in exact divisor, (128), is general, and applies to fractions as well as to integers. lOT. 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 w^hen 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 ofy the denominator of the dividend. Hence, I. A fraction is an exact divisor of a given fraction w^hen its numerator is a divisor of the given numerator , and its denominator is a viultiple 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 midtij^le 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 |, -^r^, and j|? 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 5=^, Ans. (III). Proof. ■jAj -^- ^^^ = 4 > Prime to each other. T? • 45 — "' -^ 1^8. 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 midtiple of the given denominators for a neiv denominator. This fraction will he 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 ^, i|, and ^f ? Ans. -^1^. 2. What is the greatest common divisor of 31, 1^, and |^? 3. What is the greatest common divisor of 4^ 2|, 2|, and -^j^ ? Ans. ,2_. 4. What is the greatest common divisor of 1091 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 1151 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. 900. 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 of 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- ratorSy 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 j and whose denominator is the greatest common divisor of the given denominators. NoTK. — 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 |, -f^j and ||? 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 '/ = 3 J, Ans. (III). 301. From these principles and illustrations we derive the following KuLE. Find the least common multiple of the given numerators for a new numerator y and the greatest common divisor of the given denominators for a new denominator. This fraction will he the least commo7i midtiple 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 |, y^^, ||, and 3^^^ . Ans. 11^. 2. What is the least common multiple of ^^, ||, and || ? 3. What is the least common multiple of 2||, 1|X, and y^^^^ ? % 4. What is the least common multiple of 1, |, |, i, |, |, |, |, and -^% ? ^ Ans. 2520. 5. The driving wheels of a locomotive are ISy^^j 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 114 FRACTIONS. TROMISCUOUS EXAMPLES. 1. Change J of | to an equivalent fraction having 135 for its denominator. Ans. ■^^■^. 2. Ptcduue |, 1, 1^ and j-l to equivalent fractions^ whose denom- inators shall be 48. 3. Find the least common denominator of 1^, |, 2, ^-q, | of |, 4 nf 1 g 01 4. p of ^ §- of ^ . 4. The sum of ^— -^ and .;' ,. ,^. is equal to how many times their difference ? Ans. 2. 543 |5 5. The less of two numbers is - — tPk-^j and their difference -~ ; 5 o^ H Tg what is the greater number ? Ans. 34y^g3^ 6. "What number multiplied by | of | X 3|j w^il produce || ? ^/iS. |. 7. Find the value of ^-^ x ^' + (l^ + A) -f- (3 + 4) 8. AVhat number diminished by the diiiercnce between ^ and ^ of itself, leaves a remainder of 144 ? ^7^5. 283 1. 9. A person spending -J, |, and ^ of his money, had §119 left; how much had he at first? 10. What will 1 of 10| cords of wood cost, at ^^^ of $42 per cord? ^ ^Ans. §31^. 11. There are two numbers whose difference is 25j'C, and one 1 Ans. 63| and 89f^. 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 §5 a yard, how many yards of silk can be bought for §1G|? 15. How many bushels of oats worth $| a bushel, will pay for I of a barrel of flour at §7| a barrel ? PROMISCUOUS EXAMPLES. 115 16. If f of a bushel of barley be worth | of a bushel of corn, and corn be worth $'j per bushel, how many bushels of barley will $15 buy? ' Ans. 18. 17. If 48 is I of some number, what is | of the same number? 18. If cloth 1| yards in breadth require 20i yards in length to make a certain number of garments, how many yards in length will cloth I 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 ? A?u. $51411. 20. Suppose the cargo of a vessel to be worth $10,000, and | of I of -j^0 of the vessel be worth i of | of If 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 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 J; 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. If I of 6 bushels of wheat cost $4^, how much will f of 1 bushel cost ? ^ 26. A man engaging in trade lost | of his money invested, after which he gained $740, when he had $3500; hovf 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 PEACTIONS. 28. A can do a certain piece of work in 8 days, and B can do the same in 6 days ; in what time can both together do it ? Ans. 8| days. 29. A merchant sold 5 barrels of flour for $32 J, which was | as much as he received for all he had left, at §4 a barrel • how many barrels in all did he sell ? ji^s. 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 |, |, |, or | 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 j\, and C /j. In how many days will they all be together at the point of starting ? Ans. 178} days. 82. Two men are 64| miles apart, and travel toward each other; when they meet one has traveled 5} miles more than the other; how far has each traveled ? A72S. One 29 f miles, the other 35 J miles. 33. There are two numbers whose sum is ly^^, and whose dif- ference is I; what are the numbers? Ajis. | and r^-^, 34. A, B, and C own a ferry boat; A owns ^y^ of the boat, and B owns -^^^ of the boat more than C. What shares do B and C own respectively? Ans. B, -f^; C, ^^. 35. A schoolboy being asked how many dollars he had, replied, that if his money be multiplied by ||, and ^ of a dollar be added to the product, and | of a dollar taken from the sum, this remainder divided by 3^^^ 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 b|k 1 1, the quotient will be 7} ; what is the number? Ans. ^^. 37. If I of 4 of 3} times any number be multiplied by |, the product divided by |, the quotient increased by 4 J, and the sum diminished by | of itself, the remainder will be how many times the number ? Ans, 6 j\^^ times. r NOTATION AND NUMERATION. jj^ DECIMAL FRACTIONS. NOTATION AND NUMERATION. 303. A Decimal Fraction is one or more of the decimal divisions of a unit. NoTRS. — 1. The word decimal is derived from the Latin decern, which signi- fies fen. 2. Decimal fractions are commonly called decimals, ^03. 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. 304. In the notation of decimals it is not necessary to employ denominators as in common fractions; for, since the dificrent orders of units are formed upon the decimal scale, the same law of local value as governs the notation of simple integral numbers, (o"^), enables us to indicate the relations of decimals by place or position. S05j The Decimal Sign (.) is always placed before decimal ^gures to distinguish them frcfm integers. It is commonly called ^1^ decimal point. When placed between integers and decimals in the same number, is sometimes called the separatrix. SOO. 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 followinsc 118 DECIMALS. DECIMAL NUMERATION TABLE. OQ -M ^ ^ ^ g g OQ O DQ 13 1 03 no 1 o €4-1 02 J3 c § 2 i § ^ . ". . '~, q^pJ-HiJ". "-i^,^,^ O^,-^ C^ -^ 13 5732754.57325 ooooggS Sggooo 2©7« 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'^ has one cipher; hundredths in the table extend two places from units, and the denominator of j^-q has two ciphers ; and so on. S08. 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, j\ + j^^ = -f-^%. 20^. From the foregoing explanations and illustrations we derive the following ^^ PRINCIPLES OF DECIMAL NOTATION AND NUMERATION, I. Decimals are governed by the same law of local value that governs the notation of integers. II. The different orders of decimal units decrease from left to right, and increase from right to left, in a tenfold ratio. NOTATION AND NUMERATION. 119 III. The value of any decimal figure depends upon the place it occupies at the right of the decimal sign. , TV. Prefixing a cipher to a decimal diminishes its value ten- fold, since it removes every decimal figure one place to the right. y. Annexing a cipher to a decimal does not alter its value, since it does not change the place of any figure in the decimal. YI. The denominator of a decimal, when expressed, is the unit, 1, with as many ciphers annexed as there are places in the decimal. YII. To read a decimal requires two numerations ; first, from units, to find the name of the denominator; second, towards units, to find the value of the numerator. SIO. Having analyzed all the principles upon which the writing and reading of decimals depend, we will now present these principles in the form of rules. RULE FOR DECIMAL NOTATION. I. Write the decimal the same as a whole number ^ placing cipher's in the place of vacant orders^ to give each significant figure its true local value. II. Place the decimal point he/ore the first figure. RULE FOR DECIMAL NUMERATION. T. Numerate from the decimal pointy to determine the denomi- nator. II. Numerate towards the decimal point, to determine the nu- merator. III. Read the decimal as a whole number, giving it the name of its lowest decimal unit, or right hand figure. wL EXAMPLES FOR PRACTICE. Express the following decimals by figuiHes, according to the decimal notation. 1. Five tenths. Ans. .5. 2. Thirty-six hundredths. Ans. .36, 3. Seventy-five ten-thousandths. Ans. .0075. 120 DECIMALS. 4. Four hundred ninety-six thousandths. 5. Three hundred twenty-five ten-thousandths. 6. One millionth. 7. Seventy-four ten-million ths. 8. Four hundred thirty-seven thousand five hundred forty- nine millionths. 9. Three million forty thousand ten ten-million ths. 10. Twenty-four hundred-millionths. 11. Eight thousand six hundred forty-five hundred-thousandths. 12. Four hundred ninety-five million seven hundred five thou- sand forty-eight billionths. 13. Ninety-nine thousand nine ten-billionths. 14. Four million seven hundred thirty-five thousand nine hun- dred one hundred-millionths. 15. One trillionth. 16. One trillion one billion one million one thousand one ten- trillionths. 17. Eight hundred forty-one million five hundred sixty-three thousand four hundred thirty-six trillion ths. 18. Nine quintillionths. Express the following fractions and mixed numbers decimally : 46.4. 19. j%. Ans. .3. 25. 46/^. Ans. 4 20. I'AV 99 85 ^-'- TCJOOUI)- 9Q 100004 ^^- TTJOO^UIJ- 26. 27. 28. 29. 30. 205,-. '^^TUO(T^%(jnTJO- ^"t^55T5T505TJTJTJT5TJ Read the following numbers : 31. .24. 38. 8.25. 32. .075. 39. 75.368. 33. .503. ^ 40. 42.0637. 34. .00725. 41. 8.0074. 35. .40000004. 42. 30.4075. 36. .0000256. 43. 26.00005. 37. .0010075, 44. 100.00000001. REDUCTION. 121 REDUCTION. CASE I. 311. To reduce decimals to a common denominator. 1. Reduce .5, .24, .7836 and .375 to a common denominator. OPER\Tiox Analysis. A common denominator must contain F.()r\f) as many decimal places as are equal to the' greatest oiQQ number of decimal figures in any of the given deci- .7836 mals. We find that the third number contains four .3750 decimal places, and hence 10000 must be a common denominator. As annexing ciphers to decimals does not alter their value, we give to each number four decimal places, by annexing ciphers, and thus reduce the given decimals to a common denominator. Hence, Rule. Give to each number the same number of decimal placeSj by annexing ciphers. Notes. — 1. If the numbers be reduced to the denominator of that one of the given numbers having the greatest number of decimal places, they will have their least common decimal denominator. 2. An integer m;iy readily be reduced to decimals by placing the decimal point after units, and annexing ciphers ; one cipher reducing it to tenths, two ciphers to hundredths, three ciphers to thousandths, and so on. EXAMPLES FOR PRACTICE. 1. Reduce .18,' .456, .0075, .000001, .05, .3789, .5943786, and .001 to their least common denominator. 2. Reduce 12 thousandths, 185 millionths, 936 billionths, and 7 trillionths to their least common denominator. 3. Reduce 57.3, 900, 4.7555, and 100.000001 to their least common denominator. \ CASE II. 313. To reduce a decimal to a common fraction. 1. Reduce .375 to an equivalent common^^ction. ^ „,^^, Analysis. Writinsr the decimal OPERATION. ° r>^^- .^^ « fiojures, .375, over the common de- •^ < ^ — 1 OUU — #• nominator, 1000, we have yVA = f . Hence, 11 122 DECIMALS. HuLE. Omit the decimal point, and supply the proper denomi- nator. EXAMPLES FOR PRACTICE. 1. Reduce .75 to a common fraction. Ans. |. 2. Reduce .625 to a common fraction. Ans. |. 3. Reduce .12 to a common fraction. 4. Reduce .68 to a common fraction. 5. Reduce .5625 to a common fraction. 6. Reduce .024 to a common fraction. Ans. y|^. 7. Reduce .00032 to a common fraction. Ans. ^j^-^* 8. Reduce .002624 to a common fraction. Ans- y/glj^- 9. Reduce .13| to a common fraction. OPERATION. 131 — ll^ — 4 ^ 2 .xc»3 ^^^ -g^^ — jj^. Note. — The decimal .13^ may properly be called a complex decimal. 10. Reduce .57^ to a common fraction. Ans. ^. 11. Reduce .66| to a common fraction. Ans. |. 12. Reduce .444^ to a common fraction. 13. Reduce .024| to a common fraction. Ans. yfj^. 14. Reduce .984| to a common fraction. 15. Express 7.4 by an integer and a common fraction. Ans. 7|. 16. Express 24.74 by an integer and a cojnmon fraction. 17. Reduce 2.1875 to an improper fraction. Ans. ||. 18. Reduce 1.64 to an improper fraction. 19. Reduce 7.496 to an improper fraction. Ans. f| 7 CASE III. 213. To reduce a common fraction to a decimal, ip 1. Reduce | to tIs equivalent decimal. FIRST OPERATION. ANALYSIS. "VYc first annex 5 5 000 62 5 C9^ ^^^ same number of ciphers to H — H 0^^ — TTJ^U = -^-^ i^oth terms of the fraction ; this does not alter its value, (174, REDUCTION. 123 SECOND OPERATION. Ill) ; we then divide both re- 8 ^ 5.000 suiting terms by 8, the siji;nifi- —~~ cant figure of the denominator, * "^ and obtain the decimal denom- inator, 1000. Omitting the denominator, and prefixing the sign, we have the equivalent decimal, .025. In the second operation, we omit the intermediate steps, and obtain the result, practically, by annexing the three ciphers to the nume- ,rator, 5, and dividing the result by the denominator, 8. 2. Reduce j|- to a decimal. OPERATION. Analysis. Dividing as in the former ex- 125 ) 8.000 ample, we obtain a quotient of 2 figures, 24. 024 But since 3 ciphers have been annexed to the numerator, 3, there must be three places in the required decimal ; hence we prefix 1 cipher to the quotient figures, 24. The reason of this is shown also in the following operation. 3 3000 24 __ 02-t 8141. From these illustrations we derive the following E-ULE. I. Aimex ciphers to the numerator^ and divide hy the denominator. II. Point off as many decimal places in the result as arc equal to the number of ciphers annexed. Note.- If the division is not exact when a sufficient number of decimal figures have been obtained, the sign, +, may bo annexed to the decimal to indi- cate that there is still a remainder. When this remainder is such that the next decimal figure would be 5 or greater than 5, the last figure of the terminated decimal may be increased by 1, and the sign, — , annexed. And in general, + denotes that the written decimal is too small, and — denotes that the written decimal is too large ; the error always being less than one half of a unit in the last place of the decimal. EXAMPLES FOR PRACTICE. 1. Reduce | to a decimal. Ans. .75. 2. Reduce /^ to a decimal. ^ Ans. .3125. 3. Reduce Z to a decimal. 5 4. Reduce ^i to a decimal. 5. Reduce || to a decimal. 6. Reduce ^^ to a decimal. Ans. .04. 7. Reduce ^Jj^ to a decimal. Ans, .068. 124 DECIMALS. 8. Reduce ^^ to a decimal. Ans, .59375 9. Reduce y^g^^^ ^^ ^ decimal. i 10. Reduce ^^ to a decimal. Arts, .29167 — . 11. Reduce -^^j^ to a decimal. 12. Reduce || to a decimal. -4ns. .767857+. 13. Reduce 7^ to the decimal form. Ans. 7.125. 14. Reduce 56/5 to the decimal form. Ans, 56.078125. 15. Reduce 32| to the decimal form. 16. Reduce .24^ to a simple decimal. 17. Reduce 5.78 1§ to a simple decimal. 18. Reduce .3y-^4^ to a simple decimal. Ans. .30088. 19. Reduce ^-^^^ to a simple decimal. Ans. 4.008. 20. Reduce •30y||§^^ to a simple decimal. ' ' AlDDITION. 313. Since the same law of local value extends both to the rio'ht and left of units' place ; that is, since decimals and simple integers increase and decrease uniformly by the scale of ten, it is evident that decimals may be added, subtracted, multiplied and divided in the same manner as integers. 216. 1. What is'the sum of 4.75, .246, 37.56 and 12.248 ? OPERATION. Analysis. We write the numbers so that units of 4.75 like orders, whether integral or decimal, shall stand .246 in the same columns ; that is, units under units, tenths 37.56 under tenths, etc. This brings the decimal points 12.248 directly under each other. Commencing at the right 54 804 hand, we add each column separately, carrying 1 for every ten, according to the decimal scale ; and in the result we place the decimal point between units and tenths, or directly under the decimal points in the numbers added. Hence the fol- lowing 0^ Rule. I. Write the numbers so that the decimal points shall stand directly under each other, II. Add as in whole numhersj and place the decimal pointy in the result J directly under the points in the numbers added. ADDITION". 125 EXAMPLES FOR PRACTICE. 1. Add .375, .24, .536, .78567, .4637, and .57439. Ans. 2.97476. 2. Add 5.3756, 85.473, 9.2, 46.37859, and 45.248377. Ans. 191.675567. 3. Add .5, .37, .489, .6372, .47856, and .02524. 4. Add .46|, .325|, .16^%, and .275/^. Ans. 1.2296625. 5. Add 4.6^, 7.32 3L, 5.3784^, and 2.64878|. 6. Add 4.3785, 2|/5f, and 12.4872. Ans. 24.9609 + . 7. What is the sum of 137 thousandths, 435 thousandths, 836 thousandths, 937 thousandths, and 496 thousandths ? Ans. 2.841. 8. What is the sum of one hundred two ten-thousandths, thir- teen thousand four hundred twenty-six hundred thousandths, five hundred sixty-seven millionths, three millionths, and twenty-four thousand seven hundred-thousandths ? 9. A farm has five corners; from the first to the second is 34.72 rods; from the second to the third, 48.44 rods; from the third to the fourth, 152.17 rods; from the fourth to the fifth, 95.36 rods; and from the fifth to the first, 56.18 rods. What is the whole distance around the farm ? 10. Find the sum of ||, -^.fjr, 3^/^, and yiy^H i^ decimals, correct to the fourth place. Ans. .6GC9 + . Note. — In the reduction of each fraction, carry the decimal to at least the fifth place^ in order to insure accuracy in the fourth place. 11. A man owns 4 city lots, containing 16-,^^ rods, 15^^ rods, 18^1 rods, and ll^''^ rods of land, respectively; how many rods in all ? 12. What is the sum of 4^^ decimal units of the first order, 2| of the second order, 9^ of the third order, and 3^V of the fourth order? aIs. .486929. 13. What is the approximate sum of 1 decimal unit of the first order, J- of a unit of the second order, | of a unit of the third order, I of a unit of the fourth order, i of a unit of the fifth order, J of a unit of the sixth order, and 7^ of a unit of the seventh order ? Ans. .1053605143—. 11* 126 DECIMALS. 917. SUBTRACTION. 1. From 4.156 take .5783. OPERATIOX. 4.1560 .5783 Analysis. We write the given numbers as in addi- tion, reduce the decimals to a common denominator, and subtract as in integers. Or, we may, in practice, omit the ciphers necessary to reduce the decimals to a common denominator, and merely conceive them to be annexed, subtracting as otherwise. Hence the fol- lowing 3.5777 lluLE. I. Write (he 'numbers so that the decimal points shall stand directly/ iinder each other, II. Subtract as in tchole numbers, and place, the decimal point in the result directly under the points in the given numbers. S.0717 Or, 4.156 .5783 EXAMPLES FOR PRACTICE. 9. 10. 11. 12. .9876 .3598 (2.) 48.3676 23.98 Minuend, Subtrahend, Remainder, .6278 24.3876 From 37.456 take 24.367. From 1.0066 take .15. From 1000 take .001. From 36| take 22^1. 4 2 o From From 7 56| take .55j||. take 5/5. (3.) 36.5 35.875632 .624368 .4ns. 13.089. Ans. 999.999. Ans. 14.27. Ans. 1.7708 + . From |§4 take ^J f. From one take one trillionth. Ans. .999999999999. A speculator having 57436 acres of land, sold at different times 536.74 acres, 1756.19 acres, 3678.47 acres, 9572.15 acres, 7536.59 acres, and 4785.94 acres; how much land has he remaining ? 13. Find the difference between f ||f i and ;Jf||f, correct to the fifth decimal place. , Ans. 4.17298+. MULTIPLICATION. MULTIPLICATION. 127 918. In multiplication of decimals, the location of the decimal point in the product depends upon the following principles : I. The number of ciphers in the denominator of a decimal is equal to the number of decimal places, (200, YI). II. If two decimals, in the fractional form, be multiplied to- gether, the denominator of the product must contain as many ciphers as there are decimal places in both factors. Therefore, III. The product of two decimals, expressed in the decimal form, must contain as many decimal places as there are decimals in both factors. 1. Multiply .45 by .7. OPERATION. Analysis. We first multiply ^c as in w^hole numbers ; then, ^T" since the multiplicand has 2 0-, r decimal places and the multi- plier 1, we point ofi* 2 + 1 = 3 PROOF. decimal places in the product, ^4^5^ X 7^5 = -rV/(J = -315 (HI). The reason of this is further illustrated in the proof, a method applicable to all similar cases. 910. Hence the following Rule. Mxdtiply as in whole nvmhers, and from the right hand of the product point off as many figures for decimals as there are decimal places in both factors. Notes. — 1. If there be not as many figures in the product as there are deci- mals in both factors, supply the deficiency by prefixing ciphers. 2. To multiply a decimal by 10, 100, 1000, etc., remove the point as many places to the right as there are ciphers on the right of the multiplier. EXAMPLES FOR PRACTICE. 1. Multiply .75 by .41. Ans. .3075. 2. Multiply .436 by .24. 3. Multiply 5.75 by .35. Ans. 2.0125. , 4. Multiply .756 by .025. Ans. .0189. 5. Multiply 3. 784 by 2.475. 128 DECIMALS. 6. Multiply 7.23 by .0156. An^. .112788. 7. Multiply .0075 by .005. Ans. .0000375. 8. Multiply 324 by .324. 9. Multiply 75.64 by .225. 10. Multiply 5.728 by 100. Ans, 572.8. 11. Multiply .36 by 1000. 12. Multiply .000001 by 1000000. 13. Multiply .576 by 100000. 14. Multiply 7| by 5^. Ans. 42.625. 15. Multiply .63^ by 24. 16. Multiply 4/^ by 7^%. Ans. 31.74. 17. Find the value of 3.425 x 1.265 x 64. Ans. 277.288. 18. Find the value of 32 x .57825 x .25. 19. Find the value of 18.375 x 5.7 X 1.001. Ans. 104.8422375. 20. If a cubic foot of granite weigh 168.48 pounds, what will be the weight of a granite block that contains 271 cubic feet? 21. When a bushel of corn is worth 2.8 bushels of oats, how many bushels of oats must be given in exchange for 36 bushels of corn and 48 bushels of oats ? Ans. 148.8. CONTRACTED MULTIPLICATION. SSO. To obtain a given number of decimal places in the product. It is frequently the case in multiplication, that a greater number of decimal figures is obtained in the product, than is necessary for practical accuracy. This may be avoided by contracting each partial product to the required number of decimal places. To investigate the principles of this method, let us take the two decimals .12345 and .54321, and having reversed the order of tho digits in the latter, and written it under the former, multiply each figure of the direct number by the figure below in the reversed num- ber, placing the products with like orders of units in the same column, thus : CONTRACTED MULTIPLICATION. 129 .12345 direct = .12345 ,54321 reversed = : 12345. .00UU25 = .00005 X .5 .000016 = .0004 X .04 .000009 = .003 X .003 .000004 = .02 X .0002 .000001 = .1 X .00001. In this operation we perceive that all the products are of the same order ; and this must always be, whether the numbers used be frac- tional, integral, or mixed. For, as we proceed from right to left in the multiplication, we pass regularly from lower to higher orders in the direct number, and from higher to lower in the reversed number. Hence 2SI. If one number be written under another with the order of its digits reversed, and each figure of the reversed number be multiplied by the figure above it in the direct number, the prod- ucts will all be of the same order of units. 1. Multiply 4.78567 by 3.25765, retaining only 3 decimal places in the product. OPERATION. Analysis. Since the product . ^c-n'T of any figure by units is of the 4./8o67 "^ ^ .•! ^ -. ^fiy'r^ ^ same order as the figure multi- plied, (82, II,) we write 3, the units of the multiplier, under 5, the third decimal figure of the multiplicand, and the lowest order to be retained in the pro- duct ; and the other figures of the multiplier we write in the 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 G 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, I 14357 = 4785 X 3 + 2 957 = 478 X 2 + 1 239 = 47 X 5 + 4 33 = 4 X 7+A 3 = X 6+f 15.589=1=, Ans. 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 J 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 ofi" 3 decimal places, we have 15.589, the required product. 22S. From these principles and illustrations we derive the following Rule. I. Write the mnltij)lier with the order of its figures reversed, and xcifh the units' place under that figure of the midti- plicand which is the lowest decimal to he retained in the product. II. Find the product of each figure of the multiplier hy the figures ahove and to the left of it in the midtiplicandj increasing each partial 2^roduct hy as many units as would have heen carried from the rejected part of the mul^ff^icandy and one more when the highest figure in the rejected jiart of any product is b or greater than 5 ; and write these partial p)roducts with the lowest figure of each in the same column. III. Add the partial products J and from the right hand of the resxdt j^oint off the required numher of decimal figures. NoTKS. — 1. In ol-taining the number to he carried to each contrncted partial product, it is jrcnerally necessary to multiply (mentally) only one figure at the riirht of the figure above the multiplyino^ fl<;^ure; but when the figures are large, the mtiltiplication should comnienae at least 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 thig CONTRACTED MULTIPLICATION. 1^1 error, the uncertainty may be indicated by the double sign, ±, read, plus, or vnnus, 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 230.45 by 32.46357, retaining 2 decimal places, and 2.563789 by .0347263, retaining decimal places in the product. OPERATION". OPERATION. 236.450 2.563789 75364.23 362 7430. 709350 76914 47290 10255 9458 1795 1419 51 71 15 12 1 2 .089031 7676.02 ± 2. Multiply 36.275 by 4.S678, retaining 1 decimal place in the product. Ans. 158.4 zh. 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 =i=. 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. ^4/^^. 548.07 =i=. 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 ±. 9. Multiply 7jf3 by 3|J§, extending the product to three decimal places. 10. The first satellite of Uranus moves in its orbit 142.8373 + 182 DECIMALS. 1 degrees in 1 day; find how many degrees it will move in 2.52035 days, carrying the answer to two decimal places. A71S, 860.00 degrees. 11. A gallon of distilled water weighs 8.33888 pounds ; how many pounds in 35.8756 gallons? Ans. 299.16 db pounds. 12. One French metre is equal to 1.09356959 English yards; how many yards in 478.7862 metres. Ans. 523.58 db 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. DIWSION. SS3. 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. OPERATION. Analysis. We first divide as f\^7 ^ ^A QAQ r a 1 ^^ whole numbers ; then, since the 09*99 ^ * dividend has 3 decimal places and the divisor 2, we point off 3 — 2 '^ ^"*^ 3= 1 decimal place in the quotient, _Z—-S. (II). The correctness of the work is shown in the proof, where the „„^^„ dividend and divisor are written as PROOF. common fractions. For, when we S_4 3,S 8 y 1 __. 6 4 A 4 - , , , i^oO ^ o3T 10 ^-^ have canceled the denominator of the divisor from the denominator of the dividend, the denominator of the quotient must contain as I DIVISION. ;[33 many ciphers as the number in the dividend exceeds those in the divisor. 234:. Hence the following Rule. Divide as in ichole numherSy and from the right hand of the quotient point off as mani/ 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 prefixin«j 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 phices as tho 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 9.61^8 by 3.46. Ans. 2.78. 2. Divide 46.1975 by 54.35. Ans. .85. 3. Divide .014274 by .061. Aiis. .234. 4. Divide .95£ 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 cir«um- ference make in going a distance of 1 mile or 5280 feet ? 12 134 DECIMALS. CONTRACTED DIVISION. S^o. To obtain a given number of decimal places in tlie 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, (220). 1. Divide 790.755197 by 32.4687, extending the quotient to two decimal places. FIRST CONTRACTED METHOD. COMMON METHOD. S2.4687) 790.755197 ( 649 4 24.35 32.4687 ) 790.7 55198 ( 24.35 649 3 74 1413 141 3 811 129 9 129 8 748 114 115 0639 97 97 4061 17 17 65787 16 16 23435 SECOND CONTRACTED METHOD. 32.4687 ) 790.755197 53.42 1413 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 assu:7iing the four left hand figures of the divisor, we say 324G 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 of the divisor are CONTRACTED DIVISION. I35 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). 3SG. 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 hoiv Tnany figures will he 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 multijplying hy the several 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 =1=. 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 hy .075637, extending the quotient to 3 decimal places. Ans. 11.221 zh. I^Q DECIMALS. 5. Divide 478.325 by 1.43|, extending the quotient to 3 deci- mal places. Ans.SS2M2±:. 6. Divide 8972.436 by 756.3452, extending the quotient to 4 decimal places. 7. Divide 1 by 1.007638, extending tlie quotient to 6 decimal places. Ans. .992425 =h. 8. Find the quotient of .95372843 divided by 4 t. 736546, true to 8 decimal places. 9. Reduce |f ^f to a decimal of 4 places. Ans. .7448 =fc. CIRCULATING DECIMALS. 327. 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. 328. A Finite Decimal is a decimal which extends a limited number of places from the decimal point. 239. An Infinite Decimal is a decimal which extends an unlimited number of places from the decimal point. 230. 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 + . 231. A Eepetend 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 fi«;ure, 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. 232. 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 + . 233. Similar Repetends are those which begin at the same decimal place or order; as .37 and .5, .24 and .375, 1.56 and 24.3. 234. 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 repetends are Similar and Cofiterniiiious when they begin and end at the same deeimal places or orders. I CIKCtJLATING DECIMALS. I37 3«tS. A Pure Circulating Decimal is one which contains no figures but the repetend; as .7, or .704. ^30. 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. 337. The operations in circulating decimals depend upon the following properties. NoTK. — ]. The common fractions referred to are understood to \>q 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 aU 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. For, 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. KoTES. — 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 §j has ^-^ = 5 places. 3. A perfect rcpeteud 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 irive 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. lY. 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 -^ 99 = 74 + J j, or 74^* ; 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 ^ 999 = 74/^V ; and the same is true of any number whatever. Y. 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 ||, then according to the last property, annex- ing two ciphers to the numerator, and reducing to a decimal, we have U = 24jJ4« I^ lil^G manner, carrying the decimal two places farther, .24JJ = .2424^'^ j hence, || — 24. By the same principle, we have I- =.2 ; ^V =- -01 ; i:'^ = -02 ; ^h = -001 ; f ^| = .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. 239 KoTE 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 -f % =* IuoSjjI 5 3f = JM- '-^his is true of every fraction whose denominator terminates with 1, 3, 7, or 9. YI. 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 =- .53G536, or .53G53G53G, 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 sam number of places, or some multiple of the same number, as the given repetend we should not have, in the expansions, the same fgnrea repeated in the sam oi-dnr. REDUCTION. CASE I. 2S8. To reduce a pure circulating decimal to a common fraction. 1. Eeduce .675 to a common fraction. OPERATION. Analysis. Since the repetend has 3 ^•7^ fi75 _ 2 5 places, we take for the denominator of the required fraction the number ex- pressed by three 9's, (237, '^). Hence, KuLE. Omit the j)oints and the decimal sign^ and write the Jigures 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, -^j. 2. E.educe .66 to a common fraction. 3. Reduce .279 to a common fraction. Arts. y\'j. Arts. *A1 TTT* Ans 12 • T3* Ans. 4tV Ans. • If- Ans. Vt"- Ans. 15^^^. 140 DECIMALS. 4. Reduce .423 to a common fraction. 5. Reduce .923076 to a common fraction. 6. Reduce .95121 to a common fraction. 7. Reduce 4.72 to a mixed number. 8. Reduce 2.297 to an improper fraction. 9. Reduce 2.97 to an improper fraction. Note. — According to 237, VF, 2.97 = 2.972. 10. Reduce 15.0 to a mixed number. CASE II. S39. To reduce a mixed circulating decimal to a common fraction. 1. Reduce .0756 to a common fraction. OPERATION. Analysts. Since .756 is equal .0756 = ^^Vo = tV^ **^ ^i^' -^^^^ ^^^^^ ^^ A of Uh 2. Reduce .647 to a common fraction. OPERATION. Analyws. Reducing the finite (*A*j 6 4 1 7 pai't and tke repetend separately (34Q g4 ^ to fractions, we have yVj + ^h- = — - — f-- — . To reduce these fractions to a 900 900 1 • . common clenommator, we must ^^ 040 — d 4 + ^ multiply the terms of the first by 900 9 ; but the numerator, G4, may 647 64 be multiplied by 9 by annexing ^^ 900 ^ cipher and subtracting 04 from 583 . the result, ffivina; — , for the first fraction reduced. The ^^} numerator of the sum of the two .647 given decimal. fractions will therefore be 640 64 finite figures. — 64 + 7 = 583, and supplying 'Too the common denominator, we have IJJ. In the second operation, 583 900 Ans. the intermediate steps are omitted. Hence the followinf): Rule. I. From the given circulathuj decimal subtract the finite party and the remainder will he the required numerator. ^B CIRCULATING DECIMALS. 1^1 ^B 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. i 1. Reduce .57 to a common fraction. Ans, ||. 2, Reduce .048 to a common fraction. Ans. ^y^. 3. Reduce .6472 to a common fraction. I 4. Reduce .6590 to a common fraction. Ans. ||. 5. Reduce .04648 to a common fraction. Ans. -^^^. 6. Reduce .1004 to a common fraction. 7. Reduce .9285714 to a common fraction. Ans. ||. 18. Reduce 5.27 to a common fraction. Ans. ||. 9. Reduce 7.0126 to a mixed number. Ans. 7^1^. 10. Reduce 1.58231707 to an improper fraction. Ans. |^|. 11. Reduce 2.029268 to an improper fraction. I CASE III. S40. 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 ^ ^^ the place of tenths, the .53675 = .53675675675675 v Ans. second at the place of thou- .37234 = .37234723472347 ) 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. AVe therefore expand the given repetends, and place the first point in eadi 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. KuLE. I. Expand the repetendsj 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 mavy places as there are units in the least common mul- tiple of tJie number of places in the severed given repetends. Note. — 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. 241. The processes of adding and subtracting circulating deci- Dials 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 game 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. oPERATiox. Analysis. Since fractions can be 54 = 54444 added only when they have the same c) h: o c}'A'^\h fractional unit, we first make the repe- . "^ *^ . ^ • tends of the given decimals similar and 2.785 = = 2.78527 conterminous. We then add as in finite 6.57214 decimals, observing, how^ever, that the 1 W'hich we carry from the left hand column of the repetends, must also be added to the right hand column ; for this w^ould be required if the repetends were further expanded before adding. CIRCULATING DECIMALS. ^43 2. From 7.4 take 2. 7852. OPERATION. Analysis. Since one fraction can be subtracted 7 J-iii ^J^om another only 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. ^4^. From these principles and illustrations we derive the following Rule. I. Whe7i necessari/^ make the repetends similar and con- terminous. II. To add ] — Proceed as in Jinite decimals^ observing to increase the sum of the right hand column hy as many units as are carried from the left hand column of the repetends. III. To subtract ; — Proceed as in finite decimals, ohserving to diminish the right hand figure of the remainder hy 1, when the repetend in the subtrahend is greater than the repetend of the minuend. lY. Place the points in the residt directly under the points above. Note. — 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, .56t, 7.0o6, 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. 29.745o. 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. 4.3i. 144 DECIMALS. 10. From .99 subtract .433. Ans, .55656. 11. What is the sum of 4.638, 8.318, .016, .54, and .45? Ans. 13|f. 12. From .4 subtract .23. Alls. 7 MULTIPLICATION AND DIVISION. 343. 1. Multiply 2.428571 by .063. OPERATION. 2.428571 = \^ .063 7 -— TTTJ 1'7 V 7 17 TTU — UJi — .154 Atis, Analysis. We first re- duce the multiplicand and multiplier to their equiva- lent fractions, and obtain V"^ and T?ff ; then V X 7 TTUr 2. Divide .475 by .3753. OPERATION. 475 -9-5-5 .475 •3750=375 Analysis. The dividend re- duced to its equivalent common fraction is 4J4, and the divisor ^?|X HIS = 1.26 Ans. reduced to its equivalent com- mon fraction is f ^|J ; and ^J| _i. s " /> ') 10 1 oi ~ -S^^G — TT — ^'^^' 944:. 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. 2. Multiply .0432 by 18. 3. Divide .154 by .2. 4. Divide 4.5724 by .7. 5. Multiply 4.37 by .27: 6. Divide 56.6 by 137. 7. Divide .428571 by .54. 8. Multiply .714285 by .27. 9. Multiply 3.456 by .425. 10. Divide 9.17045 by 3.36. Ans. 2A72, Ans. .7783. Ans. .693. An^. 5.8793. Ans. 1.182. Ans. .41362530. Ans. .7857142. Ans. .194805. Ans. 1.4710037. Alls. 2.72637. 11. Multiply .24 by .57. Ans. .1395775941230486685032. UNITED STATES MONEY. ^45 UNITED STATES MONEY. S4«S. 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. 34:0. 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 he expressed according to the. deci- mal system of notation. II. United States money may he added, suhtracted, 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. 948. 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 it 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 mill or 5 ten-thousandths of a dollar. S40. 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. @oO. 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, $.37i = $.375. 2. Business men frequently write cents as common fractions of a dollar. Thus, $5.19 is also written $5-J^9^, read 5 and ^1^9^ dollars. 3. In business transactions, when the Jinal result of a computation contains 5 mills or more, they are called one cent, and when less 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. A71S. $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. Ans. $48.01 1. 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. Eead the following numbers : $21.18 $10.01 $ .8125 $164.05 $201,201 $15.08J $7.90 $5.37i $96,005 UNITED STATES MONEY. 147 REDUCTION. QCil. 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 oiF two figures from the right; and mills to dollars by pointing off three fii>ures from the ri^ht. 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. 2o9. Since United States Money may be added, subtracted, multiplied and divided in the same manner as decimals, (346, II), it is evident that no separate rules for these operations are required. EXAMPLES FOR PRACTICE. 1. Paid $3475.50 for building a house, $310.20 for painting, $1287.371 for furniture, and $207.12 J 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 $1.75, a pair of gloves for $.87 J, a cravat for $1, and some collars for $.62 J; how much was the cost of all my purchases? 3. Gave $150 for a horse, $175.84 for a carriage, and $62J for a harness, and sold the whole for $390.37}; how much did I gain? Am. $2,035. 4. A man bought a farm for $3800, which was $190. 87J less than he sold it for; how much did he sell it for? 5. A lady bought a dress for $10f, 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. 6. A farmer sold 150 bushels of oats at $.37} a bushel, and 4 cords of wood at $3| a cord. He received in payment 84 pounds of 148 DECIMALS. sugar at 6i cents a pound, 25 pounds of tea at $| a pound, 2 barrels of flour at $5.87 J, and the remainder in cash; how much cash did he receive? Ans. §39.125. 7. A speculator bought 264.5 acres of land for $6726. He afterward sold 126.25 acres for $31} an acre, and the remainder for $33.75 an acre; how much did he g^in 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 J 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 $.06} a yard. How much had he left with which to finish purchasing his stock ? A71S, $6064.62|. 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 lOf cords of wood at $4 J 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 $3f a head, and for the sheep $2|- 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 J 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 i?1936.40 ; and that the money he now had was but | of what he had at the beginning of the month; how much money had he at the beginning of the month ? Ans. $1347.40. 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 J a piece, and gained thereby $1026 ; how much did it cost by the piece ? Ans. $37.20. 19^ A flour merchant paid $3088.25 for some flour. He sold 425 barrels at $Gi 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? A71S. $1909.44. 22. A mechanic earns on an average $1.87i a day, and works 22 days per month. If his necessary expenses are $2 5 J 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 $3.62 J 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 ? Am 69 bushels. 25. A farmer bought a piece of land containing 375 J acres, at $22i per acre, and sold i of it at a profit of $1032|; at what price per acre was the land sold ? Ans. $27.75. 26 If 3i cords of wood cost $11 37i, how much will 204 cords cost? Ans. $65.40f. 27. If I of a hundred pounds of sugar cost $6|, 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 pa^^ment 12i yards of broadcloth at $3i per yard, and the balance in cofl'ee at 12 J 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 J; how much was my profit on the transaction ? Ans. $79.20. 13* 150 DECIMALS. PROBLEMS INVOLVING THE RELATION OF TRICE, COST, AND QUANTITY. PROBLEM I. 3o3. Given, tlie 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, | times the price of 1 unit, etc. Hence, EuLE. Multiply the price of one hi/ the quantift/. PROBLEM II. 254. G iven, 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, Avhich is one of two factors, we have the product and one of two factors given, to find the other factor. Hence, lluLE. Divide the cost hy the quantify. PROBLEM IIL 2*15. Given, the price and the cost, to find the quantity. Analysis. Pteasoning as in Problem II, we find that the cost is the product of two factors, and the price is one of the factors, Hence, PtULE. Divide the cost hy the price. PROBLEM IV. 256. Given, the quantity, and the price of 100 or 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 be 100 times the required result, because the multiplier used is 100 times the true multiplier. For 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 two ways, 1st. By dividing the product by 100 or 1000, as the case may be; 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. 151 Rule. Multiply the price hy the quantity 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 lO'OO. PROBLEM Y. 2o7. To find tlie 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 ton or 1000 pounds. We then have the quantity and the price of 1000 to find the cost. Hence, E-ULE. Divide the price of 1 to7i hy 2, and multiply the quo- tient hy the numher of pounds expressed as thousandths. EXAMPLES IN THE PRECEDING PROBLEMS. 1. What cost 187 barrels of salt, at §1.32 a barrel? Ans. $246.84. 2. What cost 5 firkins of butter, each containing 70^ pounds, at $-^-^g a pound ? Ans. $66.09|. 3. If the board of a family be §501.87-^ for 1 year, how much is it per day ? Ans. §1.37^. 4. At $.10J- a dozen, how many dozen of eggs can be bought for $18.48 ? Ans. 176. 5. ^Vliat is the value of 140 sacks of guano, each sack contain- ing 162 1 pounds, at S17| a ton ? Ans, $201.906.|. 6. Wliat will be the cost oi 3240 peach trees at $161 per hun- dred? Ans. $534.60. 7. At $66.44 a ton, what \A\\ 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 $11 a bushel, how many bushels of wheat can be bought for $37.68| ? Ans. 25^ bushels. 152 DECIMALS. 11. What will be tlie cost of 2172 pounds of plaster, at $3,875 a ton? Ans. $4.208f 12. What cost I of 456 bushels of potatoes at $.37^ a bushel? 13. If 32^ barrels of apples cost $81.25, what is the price per barrel? Ans. $2.50. 14. What must be paid for 24240 feet of timber worth $9.37i per M.? Ans. $227}. 15. At $5| an acre, how many acres of land can be bought for $4234.37i? Ans. 752^. 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 $4| per hun- dred pounds? Ans. $48.37f. 18. What is the value of 5840 pounds of anthracite coai 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 $2 J a yard ? ^^js. $715.87}. 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 $1.62} per C, and 1108 bricks at $4} per M? Ans. $20.69|. 25 At $ J per yard, how many yards of silk may be bought for $151? 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 15} yards of silk cost $27.9, what is the price per yard ? 28. What cost 27860 railroad ties at $125.38 per thousand ? 29. If .7 of a ton of hay cost $134, what is the price of 1 ton ? SO. What is the value of 720 pounds of hay at $12.75 a ton, and 912 pounds of mill feed at $15} a ton ? Ans. $11,658. ACCOUNTS AND BILLS. I53 LEDGER ACCOUNTS. S58. A Ledger is the principal book of accounts kept by mer- chants and accountants. Into it are brought in sununary ibrui 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, (@§j. ao (2-) (3.) (4.) ? 42.17 $ 506.76 §2371. 67 §14763.84 36.24 194 32 4571.84 33276.90 18.42 427.90 1690.50 47061.39 10.71 173.26 2037.69 18242.76 194.30 71.32 5094.46 37364.96 S47.16 39.46 876.54 8410.31 40.00 152.60 679.81 5724.27 12.94 271.78 155.48 '56317.66 86.73 320.00 4930.71 81742.73 271.19 709.08 3104.13 22431.27 103.07 48.50 1987.67 40163.55 500.50 63.41 5142.84 32189.60 7.59 56.00 276.30 7063.21 11.44 410.10 522.71 3451.09 81.92 372.22 3114 60 9200.00 110.10 137.89 1776.82 1807.36 107.09 276.44 7152.91 56768.72 207.16 18.19 9328.42 63024.27 97.20 27.96 472.19 36180 45 21.77 157.16 321.42 90807.08 150.15 94.57 2423.79 28763.81 427.26 177.66 1600.81 87196.75 316.42 327.40 5976.27 4230 61 114.61 1132.16 4318.19 3719.84 81.13 876.57 682,45 1367.92 37.50 179.84 3174.96 8756.47 ACCOUNTS AND BILLS. S^O. A Debtor, in business transactions, is a purchaser, or a person who receives money, goods, or services from another; and ^00. A Creditor is a seller, or a person who parts with money, goods, or services to another. ;j[54 DECIMALS. 36fl, An Account is a registry of debts and credits. Notes. — 1. An ciccount should always contain the names 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. 369. 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. 303. An Account Current is a full copy of an account, giving each item of both debit and credit sides to date. 3154. 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. 36^0 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 Bill of Par- cels ; and an account current having only one side is sometimes called sl Bill of Items. 3©@, In accounts and bills the following abbreviations are in general use : Dr. for debit or debtor ; Cr. for credit or creditor ; \. 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; J lb. tea @ $.75, signi- fies } pound of tea at $.75 per'pound. 2S7. 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 ^\^. or bill is then said to be receipted. 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. Bill : receipted hy clcrh or agent. New York, July 10, 1860. Mr. John C. Smith, Bo't of Hill, Groves & Co., 10 yd. Cassimere, @ $2,85 16 " BIk. Silk, 1.12J 72 " Ticking, .It 42 ^' Bid. Shirting, .16} 12 " Pressed Flannel, .40 24J « Scotch Plaid Prints, .56 $82.03 Rec^d Payment, Hill, Groves & Co., By J. W. Hopkins. (2.) Bill : receipted hy the selling party. Chicago, Sept. 20, 1861. Chase & Kennard, Bo't of McDouGAL, Fenton & Co., 125 pr. Boys' Thick Boots, @ $1.25 275 ^^ ^^ Calf ^' '' 1.75 180 ^' " Kip " '' 1.12J 210 " '' Brogans, ' '' .87J 80 " Women's Fox'd Ggiiters, '' .84 95 '' '' Opera Boots, " .90 175 " " Enameled " " 1.06 8 cases Men's Calf Boots, " 30.50 8 '' Congress Pump Boots, ^^ 35.75 1 '' Drill, 958 yd., " .lOJ 40 gross Silk Buttons, '' .37 i §1828.79 Rec^d Payment y McDouGAL, Fenton & Co. 156 Smith & Perkins, DECIMALS. (3.) BUI : settled hy note. New York, May 4, 1860. Bo't of Kent, Lowber & Co., 40 chests Green Tea, @ $27.60 25 " Black « '' 19.20 16 " Imperial « " 48.10 12 sacks Java Coffee, " 17.75 20 bbl. Coffee Sugar, (A) " 26.30 15 " Crushed " " 31.85 36 boxes Lemons, " 3.87i 42 " Oranges, " 4.12J 25 <^ Raisins, " 2.90 ?2961.60 ec^d Payment^ hy note at 6 mo. Kent, Lowber & Co. (4-) Bill : paid hy draft, and receipted hy Cleric. New Orleans, April 28, 1861. James Carlton & Co. BoH of WiLLARD & Hale. 150 bbl. Canada Flour, @ $6.25 275 " Genesee " " 7.16 170 " Philada. « " 5.87J 326 bu. Wheat, " 1.62i 214 " Corn, « .82 300 " Barley, " .91 500 " Kye, " 1.06 $5413.48 Rec^d Payment, hy Draft on N. T. 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 $ a a " 15 cwt. Eng. Blister Steel, " 10.25 June 21, " 7 doz. Hoes, (Trowel Steel) '^ 7.78 Aug. 10, " 25 " Buckeye Plows, " 8.45 Oct. 3, " 14 Cross-cut Saws, a 16.12} a a " 27 cwt. Bar Lead, a 5.90 u u " 1840 lbs. Chain, i( .09} Cr. $ May 25, By 20 M. Boards, @ $17.60 . July 14, " 50 M. Shingles, " 3.12J a u " 42 M. Plank, " 9.87J Sept. 5, '' Draft on New York, $1000 " 12, " 75 C. Timber, @ 3.10 a a '' 36 C. Scantling, " .87i $ Bal. Due Dodge & Son, §356.61 (6.) Dr. Account Current J another form ; balanced by note. Wm. Richmond •/ !l861 .40 : Jan. V .12K 1 565 25 iiy CI barrels apples, $2.25 '• 70 bushels turnips, .22 " 56 *' dried apples, .87 V^ " 31 drums figs, .68;^ ■ Note at 3 mo. to Bal. Wood & Powell. Boston, Jan. 1, 1861. 158 DECIMALS. 'promiscuous examples. 1. AYliat cost 121 cords of wood @ $4.87i ? Ans, $61.54+. 2. At $.37} per bushel, how many barrels of potatoes, each containing 2} 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 | of a tenth ? Ans. .025. 6. What is the product of 814^^^ X 26|| 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 @ $.87}, the coffee @ $.18f, and the sugar @ $.10}. How many pounds of each could he buy? Ans. 24. 10. If a man travel 13543.47 miles in 365} days, how far does he travel in i of a day ? A71S. 32.445 .miles. 11. Bought 100 barrels of flour @ $5.12}, and 250 bushels of wheat @ $1.06}. 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? ^?is. $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 .0172J| by .03j^g, and obtain a quotient true to 4 decimal places. Ans. .5625=1=. 14. Divide 13.5 by 2} hundredths. Ans. 600. 15. A man agreed to build 59.5 rods of wall; having built 8.5 PROMISCUOUS EXAMPLES. 159 Js in 5 days, how many days will be required to finisli tlie wall at the same rate? A)is. 30 days. 16. A farmer exchanged 28} bushels of oats worth ^.37 J per bushel, and 453 pounds of mill feed worth §.75 per hundred, for 12520 pounds of plaster; how mucb 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? Ans. $101.50. 19. A speculator bought 1200 bushels of corn @ $.56}. He sold 375. J 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 $39.1 8| ; 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 r Ans. ^180. 29. Reduce (tt-^-tt) X 4 of l to a decimal. Ans. .15. \4^ 2|/ ^ ^ 30. If 7.5 tons of hay are worth 375 bushels of potatoes, and 1 bushel of potatoes is worth 8.33|, how much is 1 ton of hay worth? Ans. ?16.GG§. 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, y^^ of it being barley at $.62 J 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 19i 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 @ $1.18}, and 736 bushels of oats @ $.41 ; I had 870 bushels of the wheat floured, and dis- posed of it at a profit of $235.87}, and I sold 528 bushels of the 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 i||, and their difierence is J^l ; 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 year he lost J of what he had at the end of the first year ; the third year he gained | of what he had at the end of the second year, and he then had $28585.70. How much had he gained in the 3 years? Ans. $10594.70. CONTINUED FRACTIONS. 161 CONTINUED FRACTIONS. 368. If we take any fraction in its lowest terms, as ||, and divide botli terms by the numerator, we shall obtain a complex fraction, thus : 13__1 54 ""4 + 2^ 13 Reducing ^^3, the fractional part of the denominator, in the same manner, we have, 13_1 54 ~ 4 + 1 6 + l_ 2 Expressions in this form are called continued fractions. Hence, S69. A Continned Fraction is a fraction whose numerator is 1, and whose denominator is a whole number plus a fraction whose numerator is also 1, ^nd 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, J, i, and J. CASE I. 371. 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+1 for the denominator of the first term of 91]^ the continued fraction. Then in the same -r^ 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 HuLE. I. Divide the greater term, hy the less, and the last divisor hy the last remainder , and so on, till there is no remainder. 14* L 162 CONTINUED FRACTIONS. II. Write 1 yb?' the numerator of each temn of the continued fraction, and the quotients in their order for the respective denom- inators. EXAMPLES FOR PRACTICE. 1. Reduce jT^f^ to a continued fraction. 4 + 1 2 + 1 3 + 1_^ 9* 2. Reduce i|-U to a continued fraction. 8. Reduce |f|f Jf ^^ ^ continued fraction. 4. Reduce -f^j to a continued fraction. CASE II. ^72. 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^ t\YO^ tlireC; or more terms of the continued fraction. S73. 1. Reduce ^^^^3 to a continued fraction^ and find its approximate values. OPERATION. 08 1 r-r ~ ^ , , , the continued fraction. 1G3 4 -f- r 3+jL 2 + 1 5 — _, 1st approx. value. 4 1 •» 3 =: tt;, 2d " r+T 4x3 + 1 13' 1 _1 3 x2 + l _ 3X2 + l _7.3d u 4 + 1 4 + 2 (4X3 + l)X^ + 4 13X2 + 4 30 3+1 3X2+1 2 1 7X5+3 = — -, 4th « 4+1 SO X 5 +13 163' 3T T 2 + 1 5 CONTINUED FRACTIONS. 1(53 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 ^^ 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, ■^Q, is obtained by multiplying the terms of the preceding approxima- tion, j%^, 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 ^ff by the 4th denominator, 5, in the continued fraction, and adding the corresponding terms of -j-\, obtain j^^, the 4th approximate value, which is the same as the original fraction. Hence the following lluLE. I. For the first ajyproxiinate valuCj talze the first term of the continued fraction. II. For the second approximate valuer reduce the complex frac- tion formed hy the first two terms of the continued fraction. III. For each succeeding approximate value , mulflpltj both nu- mrrator and denominator of the last preceding aj)p>roxlmation hy the next denominator in the continued fraction ^ and add to the cor- responding products respectively the numerator and derioininator 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 2d, -Ith, OLh, etc., are less thim the given fraction. EXAMPLES FOR PRACTICE. 1. Find the approximate values of j%\. jLns. jj, 77, 375^, J55. 2. Find the approximate values of -^W. Alii "i 4 5 3 9 8_3 3. What arc the first three approximate values of ^tqW? ^ 4. What are the first five approximate values of f f J ? A 14 9 4 <^ _4_9_ JinS. -3, j-3, ^-g, j'27j? 158- 5. Reduce p to the form of a continued fraction, and find the value of each approximating fraction. 164 COMPOUND NUMBERS. COMPOUND NUMBEKS. ^74. A Compound Number is a concrete number expressed in two or more denominations, (lO). gy^. A Denominate Fraction is a concrete fraction whose integral unit is one of a denomination of some compound number. Thus, I 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. ST6. 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. ^^y. Measure is that by which extent, dimension, capacity or amount is ascertained, determined according to eome fixed standard. Note. — The process by which the extent, dimension, cnpacity, or amount is ascertained, is called Measurincj ; and consists in comparing the thing to be measured with some conventionul 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. I MEASURES OF EXTENSION. ^55 A Solid or Body has three dimensions — length, breadth, and thickness. I. Linear Measure. S79. Linear Measure, also called Long Measure, is ueed 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. 5 J yards, or 16J feet, ** 1 rod, rd. 40 rods *' 1 furlong, fur. 8 furlongs, or 320 rods, '* 1 statute mile, ..mi. UNIT EQUIVALENTS. ft. in. yd. 1 = 12 rd. 1 rrr 3 = 36 fur. 1 = 5J := 16J = 198 mi 1 = 40 =r 220 --= 660 = 7920 1 =r 8 == 320 = 1760 = 5280 = 63360 Scale — ascending, 12, 3, 5 J, 40, 8; descending, 8, 40, 5 J^ 3, 12. i The following denominations are also in use : — 3 barleycorns make 1 inch, {Z\ei;^Tt^^^^^^^ 4 inches " 1 hand, I j^'"^ 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. 1.15 statute miles " 1 geographic mile, | "«^'' '" measuring dis- * ° ^ ' ( tances at sea. 3 geographic *' " 1 league. 60 " * ^^ 1 1 deo-ree I ^^ latitude on a meridian or of GO. 16 statute << " j o | longitude on the equator. 3GQ degrees " the circumference of the earth. NoTKs. — 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 tnble 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 -g-Jjy or ^ tVt^tj ^^ ^^^ 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 09.16 miles, and it gradually decreases toward the poles, where it is 0. surveyors' linear measure. 380. A Gunter'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. ''1 rod, rd. ** 1 chain, . . . ch. ** 1 mile,. . . . mi. 25 4 80 links rods, or chains 66 feet, UNIT EQl mi. 1 ch. 1 80 rd. 1 4 320 1. 1 25 100 8000 in. 7.92 198 792 63360 Scale — ascending, 7.92, 25, 4, 80; descending, 80, 4, 25, 7.92. NoTK. — The denomination, rods, is seldom used in chain measure, distances being taken in chains and links. II. Square Measure. S81. A Square is a figure having four equal sides and four equal corners or right angles. *I8^. 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, Tlie contents or area of a square, or of any other figure having a uniform length and a uniform breadth , is found 1 yd. = 3 ft. ^y m^ultiplying the length hy the hreadth. 1 yd. = 3 ft. 9 " feet 30} <( yards 40 <( rods 4 roods 640 acres MEASURES OF EXTENSION. Ig7 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. 28S. Square Measure is used in computing areas or sur- faces J 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 unit 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. 1 " yard,., .sq. yd. 1 ** rod,. . . .sq. rd. 1 rood, R. 1 acre, . A. 1 square mile,. . . sq. mi. UNIT EQUIVALENTS. sq. yd. 1 = 30}= 1210 = 4840 == 1 = 640 = 2560 = 102400 = 3097600 = 27878400 = 4014489600 Scale— ascending, 144, 9, 30}, 40, 4, 640; descending, 640, 4, 40, 30}, 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. ;N'otes. — 1. Tn estimiiting the painting of moldings, cornices, etc., the mea- suring-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 1^ bricks thick. .3. A thousand shingles are estimated to cover 1 square, being laid 5 inches to the weather. sq. rd. R. 1 = A. 1 = 40 = 1 = 4 = 160 = sq. ft. sq. in. 1 = 144 9 = 1296 272}== 39204 10890 = 1568160 43560 = 6272640 168 COMPOUND NUMBERS. surveyors' square measure. S§ 4. 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.l. gq. ch. 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. S8^. A Cube is a solid, or body, having six equal square sides or faces. S80, 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 ^ X .iittM contain 3x3x1=9 cubic feet ; and the whole cube will contain 3 X 3 X 3 = 27 cubic feet. II ' ' IHHI A solid, or body, may have the three <^ \ lilW dimensions all alike or all different. A body 4 ft. long, 3 ft. wide, and 2 ft. thick it, _ I ya, contains 4 X 3 X 2 = 24 cubic or solid feet. Hence we see that Wr. tei fiiB ■ \ |u/. 1 P .728 cubic inches (cu. in.) 27 cubic feet 40 cubic feet of round timber, or" 50 '^ *' hewn IG cubic feet 8 cord feet, or 128 cubic feet 241- cubic feet MEASURES OF EXTENSION. Jgg The cuhic or solid contents of a hody are found hy multiplymg the length, bread th, and thickness together, QS7. 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. make 1 cubic foot cu. ft. " 1 cubic yard cu. yd. " 1 ton or load T. " 1 cord foot cd. ft. " 1 cord of wood Cd. ii -j I perch of stone ] p i^ I or masonry, J 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. Notes. — 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 I foot in length of such a pile. 4. A perch of stone or of masonry is 16^ feet long, IJ 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 girtj 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, -j^ 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 5^ feet of hewn timber, as estimated for market, are identical. 15 170 COMPOUND NUxMBERS. MEASURES OF CAPACITY. # 288. Capacity signifies extent of room or space.' S89. 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. 290. 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 gills (gi.) make 1 pint, pt. 2 pints *' 1 quart, qt. 4 quarts " 1 gallon, gal. 31 J gallons " 1 barrel, bbl. 2 barrels, or 63 gal. " 1 hogshead, .. hhd. UNIT EQUIVALENTS. pt. gi. qt. 1=4 gal. 1=2= 8 bM. 1 = = 4 = 8 = 32 hhd. 1 = 31J = = 126 = 252 = 1008 1 = 2 = 63 = = 252 = 504 = 2016 Scale — ascending, 4, 2, 4, 31 J, 2; descending, 2, 31J, 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 hnve their capacities in gallons mnrked on them. Several of these denominations are 8till in use in England, (327—330). WEIGHTS, 171 BEER MEASURE. t201. 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. 313 gallons '* 1 barrel, bbl. IJ barrels, or 54 gallons, ** 1 hogshead, .. hhd. UNIT EQUIVALENTS. qt. pt. gal. 1=2 bbl. 1 = 4 = 8 hhd. 1 = 36 == 144 = 288 1 = IJ = 54 = 216 = 432 Scale — ascending, 2, 4, 36, 1}; descending, IJ, 36, 4, 2. This measure is not a standard ; it is rapidly falling into disuse. II. Dry Measure. SO^. 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. pfe. 1=2 bu. 1 = 8 = 16 1 = 4 = 32 = 64 Scale — ascending, 2, 8, 4 ; descending, 4, 8, 2. WEIGHTS. 303. 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. 994. 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 pennyweight, .. pwt. or dwt. 20 pennyweights " 1 ounce, oz. 12 ounces " 1 pound, ...lb. UNIT EQU1YALENT8. pwt. gr. oz. 1 = 24 lb. 1 = 20 = 480 1 = 12 -= 240 = 5760 Scale — ascending, 24, 20, 12; descending, 12, 20, 24. Note. — Troy weight is sometimes called Goldsmiths' Weight, II. Avoirdupois Weight. 39«>. 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 drams (dr.) make 1 ounce, oz. 16 ounces *' 1 pound, lb. 100 lb. " 1 hundred weight, .. cwt. 20 cwt., or 2000 lbs., " 1 ton, T. UNIT EQUIVALENTS. OZ. dr. lb. 1 r= 16 cwt. 1 = 16 256 T. 1 = 100 = 1600 = 25600 1 = = 20 = 2000 = 32000 = 512000 Scale — ascending, 16, 16, 100, 20; descending, 20, 100, 16, 15. 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 Peuu- gylvania mines, and ia the wholesale iron and plaster trade. LONG TON TABLE. 28 lb. make 1 quarter, marked qr. 4 qr. = 112 lb. " 1 hundred weight, '' cwt. 20 cwt. = 2240 lb. " 1 ton, ^ '* T. Scale — ascending, 28, 4, 20; descending, 20, 4, 28. !S90« 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 OP AVOIRDUPOIS POUNDS IN A BUSHEL, As prescribed by statute in the several States named. COMMODITIES. Barley. Beans 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*... Oats Onions Peas Potatoes Kye l^ve M.3al Sali^ Timothy Seed... Wheat Wlieat Bran 48 48 60:60 I4I14 50 52 46 '46 6o'60 25 1 24 56 56 44 1 44 56J56 60 60 54 56 333^ ; 57 60 50 50 45 60 '60 20 '20 J 1 1 he rt c ii i .2 'Z. a. a X 25 1 J 1 1 c c 1 32 46 48 4S 48 60 14 48 48 62 48 46 47 46 45 46 42 42 52 46 50 48 42 48 46 42 60 60 60 64 60 60 60 60 28 28 24 28 28 28 28 33 56 55 55 56 28 28 11 44 56 50 56 50 56 50 52 56 58 5G 56 56 50 56 56 !iO 32 30 30 52 32 32 35 57 30 30 32 60 32 34 32 50 32 30 50 GO 60 60 60 60 60 60 60 60 32 50 56 50 56 56 56 r.o 45 56 56 56 44 56 56 56 "» 56 56 60 60 60 CO 60 20 60 60 60 60 60 60 60 ) 60 28 40 • 60 * In Kentucky, 80 Ibs.of Intuminous coal or 70 lbs. of cannul coal make 1 bushel. * In IVnnsylvania. 80 lbs. coar(«e, 70 lb.«. frround, or 02 Ib.s. fine ealL make I bushel; and in Tliinoi?, 50 Ib.s. common or 56 lb.«!. fine .'^alt make 1 buslu'l. * In Maine, 64 lbs. of rata bajja turnips or beets make 1 bushel. 15* 174 COMPOUND NUMBERS. Notes. — 1. The weight of a barrel of flour is 7 quarters of old, or long ton weiglit. 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 : 5G 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. ^^7. 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 wdiich 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 J. 12 ounces " 1 pound, lb. or lb. UNIT EQUIVALENTS. PC. gr. dr. 1 = 20 oz. 1 = 3 = GO u,. 1 = 8 = 24 = 480 1 = 12 = 9G = 288 = 57G0 Scale — ascending, 20, 3, 8, 12;' descending, 12, 8, 3, 20. apothecaries' fluid measure. ^@§, 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. GO minims, {'^) make 1 fluidrachm, f^. 8 fluidrachms, " 1 fluidounce, f^. IG fluidounces, '* 1 pint, 0. 8 pints, ** 1 gallon, Cong. TIME. 175 UNIT EQUIVALENTS. f^ 1 = 60 0. 1 = 8 = 480 Cong. 1 = 16 = 128 = 7680 1 = 8 = 128 = 2048 = 61440 Scale — ascending, 60, 8, 16, 8 ; descending, 8, 16, 8, 60. MEASURE OF TIME. 2®0. Time is tlie measure of duration. The unit is the day, and tlie table is made up of its divisors and multiples. TABLE. 60 seconds (sec.) make 1 minute, min. 60 minutes, " 1 lioiir, h. 24 hours, " 1 day, da. 7 days, " 1 week, wk. 365 days, *' 1 common year, yr. 366 days, " 1 leap 3^ear, • • . yr. 12 calendar months, '' 1 year, . yr. 100 years, *' 1 century, C. UNIT EQUIVALENTS. min. («ec. h. 1 -= 60 da. 1 == 60 = 3600 ^k. 1 = 24 = 1440 = 86400 1 = 7 :_ 1G8 _ 10080 = 604800 yr. nio. (365 = 8760 = 525600 = 31536000 1 = 12 = |366 = 8784 = 527040 = 31622400 Scale — ascending, 60, 60, 24, 7 ; descending, 7, 24, 60, 60. The calendar year is divided as follows : — No. of month. Season. Names of months. Abbreviations. No of days. ^ w* f 1 January, Jan. 31 2 winter, | pgi^^uary, Feb. 28 or 29 3 [March, Mar. 31 4 Spring, < April, Apr. 30 5 I May, 31 6 [June, Jun. 30 7 Summer, < July, 31 8 I August, ^^S' 31 9 [September, Sept. 30 10 Autumn, ■< October, Oct. 31 11 (November, Nov. 30 12 Winter, 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 astronomi- 170 " COMPOUND NUMBERS. 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. 300. 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 see. 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. IT. 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 6 hours should be disregarded for 3 suc- cessive years, and an entire daj' he added to every fourth year. This day was inserted in the calendar between the 24th and 25th days of February, and is called the iutercalary day. As the Romans counted the days backward from the first day of the following month, the 24th of February was called by them scjcto CmCULAR MEASURE. 177 cnleudaa Jfortit, the sixth before the calends of March. The intercalary day which followed this was called biHsejcto culeiidoH Martii; hence the name hissejitile. 2. In 1582 the error in the calendar as established by Julius CcEsar had in- creased to 10 days; that is, too much time had been reckoned as a year, until the civil year was 10 «lays behind the solar year. To correct this error, Pope Gregory decreed that 10 entire days should be stricken from the calendar, and that the day following the 3d day of October, 1582, should be the 14th. This brought the vernal equinox at March 21 — the date on which it occurred in the year H25, at the time of the Council (»f Nice. 3. The year as established by Julius Caesar is sometimes called the Jnlinn year; and the period of time in which it was in force, namely from 46 years B. C. to 1582, is called the //»/m»* 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 that the day follovring 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 cominences 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 2:!, 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. 301. 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 ^ J^ 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°, " 1 circle, C. M 178 COMPOUND NUMBERS. UNIT EQUIVALENTS. 1 = 60 S 1 = 60 = 3600 c. 1 = 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, sic/ua, is confined exclusively to Astronomy. 3. A degree has no fixed linear extent. When applied to any circle it is always j^-jy 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 " J of a circle. MISCELLANEOUS TABLES. 30S. COUNTING. 12 units or things make 1 dozen. 12 dozen '' 1 gross. 12 gross '' 1 great gross. 20 units " 1 score. 3®3. 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. 30^. COPYING. 72 words make 1 folio or sheet of common law. 90 '* "■ 1 '* "■ '' "■ chancery. GOVERNMENT STANDARDS. 179 GOVERNMENT STANDARDS OF MEASURES AND WEIGHTS. S0&. 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. 3I^T. 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 invariahh 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 same number of times in the same period of duration. Having now the day and the pendulum, the different standards hereafter given have been determined and adopted. STANDARD OF EXTENSION. 300. The JJ. S. standard unit of measiires of extension jV{\ict\\.QV 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 lliiBi 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 TJ, S. standard unit of dry measure is the British Winchester bushel, which is 18 J 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 strinhen 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 19^ 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 597.2067 cubic inches more than a bushel stricken measure. Since 1 peck contains ii5^'* = 637.605 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 Etand:ird bushel was derived, was kept at Winchester, England; hence the name. 4. The wine and drj' measures of the same denomination are of difFerentcapac- ities. The exact and the relative size of each may be readily seen by the fol' 31^. COMPARATIVE TABLE OF MEASURES OF CAPACITY. Cubic in. in Cubic in. in Cubic in. in Cubic in. ia one frallon. one quart. one pint. ouejrill. Wine measure, 231 57| 28 J 7 ^\ Dry measure (Jpk.,).. 2G8J 67i ^^ Sf Note. — The beer gallon of 282 inches is retained in use onl.v by custom. GOVERNMENT STANDARDS. 181 STANDARDS OF WEIGHT. 313. 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 weiglit is the Troy pound of Ihe 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. 31«l* The U. S. Avoirdupois pound is determined from the standard Troy pound, and contains 7000 Troy grains. Hence, the Troy pound is f ^g§ = j4| of an avoirdupois pound. But the Troy ounce contains ^^^^ = 480 grains, and the avoirdupois ounce '''-Jg^ = 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, ihoi only difference in the two tables being in the divisions of the ounce. 3i6« 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. ly. 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. 182 COMPOUND NUMBERS. A wine gallon^ and its subdivisions. A half bushel, and its subdivisions. «li8« State Sealers of AVeights and Measures furnish standard sets of weights and measures to counties and towns. A Count}/ 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 lb. 4. A set of small brass weights, avoirdupois, namely, 4, 2, and I lb., 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., J gal., 1 C[t., 1 pt., J pt., 1 gi.; in a case. 7. A set of dry measures, of copper, namely, J bu., 1 pk., } pk. (or 1 gal.), 2 qt. (or J gal.), 1 qt.; in a case. ENGLISH MEASURES AND WEIGHTS. GOVERNMENT STANDARDS. 310. 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 Sijdcm. 3^0. The Ivvarialle 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 oOlSOo equal parts; then will 10000 of those parts be an imperial inch, 12 whereof make a foot, and 36 whereof make a yard." I ENGLISH MEASURES AND WEIGHTS. 183 STANDARD OF EXTENSION. «|^I, The English Standard Unit of Measures of Extension, whether linear, superficial, or solid, is identical vrith that of the United States, (SI®®). STANDARDS OF CAPACITY. 392. Tlie imjoerial 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 G2° Fahrenheit, the barometer at 30 inches. It contains 277.274 cubic inches. 393. The Imperial Standard Bmlicl 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, w^hich is identical with that of the United States Standard Troy pound of the Mint, (314.) 329S. The Imperial At:oirdupois Pound contains 7000 Troy grains, and the Troy pound 5760. It also is identical with the United States avoirduj^ois pound. TxVBLES. 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 '< 1 11 5 quarters of the standard yard J NoTR. — The cubit wns originally the length of a mnn's forearm and hand; or the distance from the elbow to the end of the middle finger. 184 COMPOUND NUMBERS. 3S8. MEASURES OF CAPACITY. LIQUID MEASURES. 9 old ale gallons make 1 firkin. 4 firkins " 1 barrel of beer. 7J Imperial " " 1 firkin. 52| Imperial gallons or | .. ^ hogshead. G3 wine j 70 Imperial gallons or I ** 1 puncheon or 84 wine " j '* J of a tun. 2 hogsheads, that is | 105 Imperial gallons or v " 1 pipe. 126 wine " ) 2 pipes *' 1 tun. Pipes of wine are of diiferent capacities, as follows : 110 wine gallons make 1 pipe of Madeira. ( Barcelona, 120 '' " ^ " 1 Vidonia, or [ Teneriffe. 130 " " 1 " Sherry. 138 " '' 1 " Port. I4Q u u 1 <' I Bucellas, or I Lisbon. 3^9. 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 5C0 pounds, or i of a ton of 2240 pounds. 330. 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 2210 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 S^»fem, FRENCH MEASURES AND WEIGHTS. 185 33^. 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 mitre, and from this all the standard units of measure and weight are derived and determined. STANDARDS OF EXTENSION. 333. The French Standard Linear Unit is the m^tre. 33z|^ The French Standard Unit of Area is the Are, which is a unit 10 metres square, and contains 100 square metres. 33o* The French Standard Unit of Solidity and Capaciti/ is ohe Litre, which is the cube of the tenth part of the metre. 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 IcUogramme, 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 sea e. 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 niultiples being formed by employing the prefixes deca, (ten), hecto, (hundred), Icilo, (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 make 1 centimetre. 10 centimetres " 1 decimetre. 10 decimetres " 1 metre. 10 metres " 1 decametre. 10 decametres " 1 hectometre. 10 hectometres " 1 kilometre. 10 kilometres " 1 myriametre. Notes. — 1. The metre is equal to 39.3685 inches, the standard rod of brass 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 i 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. 349. 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 cubio inches, or 35.60934 cubic feet. This unit is called a Stere. TABLE. 10 decisteres make 1 stere. 10 steres ** 1 decastere. MONEY AND CURRENCIES. 137 •I41, lY. French Weight. TABLE. 10 milligrammes make 1 centigramme. 10 centigrammes " 1 decigramme. 10 decigrammes " 1 gramme. 10 grammes " 1 decagramme. 10 decagrammes " 1 hectogramme. 10 hectogrammes " 1 kilogramme. 100 kilogrammes " 1 quintal. ir\ • i. 1 n f 1 millier, or 10 quintals 1 i x ^ a ^ [1 ton 01 sea water. Notes. — 1. A gramme is equal to 15.483159 Troy grains. 2. A kilogramme is equal to 2 lb. 8 oz. 3 pwt. 1.159 gr. Troy, or 2 lb. 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. Extonsion, Yd. of 3 ft., or 3(5 in. Same as U. S. Metre. 39.3685 in. r.,n.,n\tv I ^^'"t^ fJ'd., 231 cu. in. Imp'l gal., 277.274 cu. in. Litre, 61.53294 cu. in. ^.ip.icuy, I winch'r bu., 2150.42 cu. in. Irap'l bu., 2218.192 cu. in. Weight, Troy lb., 57 GO gr. Imperial lb., 5760 gr. Gramme, 15.433159 T. gr. NoTKS. — 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 on. in. may be taken as a Winchester bushel, and 2218.2 cu. in. as an Imperial busheL MONEY AND CURRENCIES. 34:3. 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. 34:4:, 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. 34ft5. A Mint is a place in which the coin of a country or government is manufactured. NoTR. — In all civilized countries mints and coinage are under the exclusive direction and control of government. 188 COMPOUND NUMBERS. S43. An Alloy is a metal compounded with another of greater vahie. In coinage, the less vahiable or baser metal is not reckoned of any value. NoTK.— 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. 34: T* 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 gold. 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. 3«10. 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. 3ol« 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^S. Currency is coin, bank bills, treasury notes, and other substitutes for money, employed in trade and commerce. 3«>3. A Circulating Medium is the currency or money of a country or government. 3o4« A Decimal Currency is a currency whose denpmina- tions increase and decrease according to the decimal scale. I. United States Money. 3*]^tS« The currency of the United States is decimal currency, and is sometimes called Federal Money, I MONEY AND CURRENCIES. 189 The unit is the dollar, 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. ^ UNIT EQUIVALENTS. ct. m. d. 1 = 10 $ 1 = 10 = 100 E. 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. S., (United States,) the U being placed upon the S. Coins. The gold coins are the double eagle, eagle, half eagle, quarter eagle, three dollar piece and dollar. The silcer coins are the half and quarter dollar, dime and half dime, and three-cent piece. The nickel coin is the cent. NoTF.s. — 1. The foUowinj^ pieces of gold are in circulation, but.are not legal coin, viz. : the fifty dollar piece, and the half and quarter dollar pieces. 2. The silver dolhir, and the copper cent and half cent, are no longer coined for general circulation. 3. The mill is a denomination used only in computations; it is not a coin. 3«IG* 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 three-cent piece is 3 parts (f) silver, and 1 part (}) copper. The nickel cent is 88 parts copper and 12 parts nickel. STATE CURRENCIES. 3«>7« United States money is reckoned in dollars, dimes, cents, (ind mills, one dollar being uniformly valued in all the States at 100 cents ; but in many of the States money is sometimes reckoned in dollars, shillings, and pence. ^ 190 COMPOUND NUMBEKS. Note. — At the time of the adoption of our decimal currency by Congress, in 1786, the colonial eurrtnci/, or billn of credit, issued by the coh)nies, had depre- ciated in value, and this depreciation, being unequal in the different colonies, gave rise to the different values of the State currencies; this variation continues wherever the denominations of shillings and pence are in use. Georgia Currency. Georgia, South Carolina, $1 = 4s. 8d. = 5Gd. Canada Currency, Canada, Nova Scotia, %1 = 5s. = 60d. New England Currency, New England States, Indiana, Illinois, ] Missouri, Virginia, Kentucky, Tennes- V §1 = 6s. == 72d. see, Mississippi, Texas, J Pennsylvania Currency. New Jersey, Pennsylvania, Delaware, | (&i __ ^g /^^i __ oa^i Maryland, | . . . . s? Neio York Currency. New York, Ohio, Michigan, | $1 = 8s. = 9Gd. North Carolina, j II. Canada Money. 338. The currency of the Canadian provinces is decimal, and the table and denominations are the same as those of the United States money. Note. — The decimal currency was adopted by the Canadian Parliament in 1S58, and the Act took effect in 1859. Previous to the latter year the money of Canada was reckoned in pounds, shillings, and pence, the same as in Eng- land. Coins. The new Canadian coins are of silver and copper. The silver coins are the shilling or 20-cent piece, the dime, and half dime. • The copper coin is the cent. Note. — The 20-cent piece represents the value of the shilling of tho old Canada Currency. 339. Government Standard. The silver coins consist of 925 parts (.925) pure silver and 75 parts (.075) copper. That is, they are .925 fine. NoTR. — The value of the 20-cent piece in United States money is 18§ cents, of the dime 9J cents, and of the half dime 4^ cents. MONEY AND CURRENCIES. 191 360. English or Sterling Money is the currency of Great Britain. i^ 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. «• 1=4 £, or sov. 1 = 12 =^ 48 1 =r 20 = 240 = 960 Scale — 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.)=--id.; 3 fnr. =|d. 2. The old/, the original abbreviation fur 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, «tc. Coins. The gold coins are the sovereign (= £1) and the half sovereign, (= 10s.) The silvei' coins are the crown (= 5s.), the half crown (= 2s. 6d.), the shilling, and the 6 penny piece. The copper coins are the penny, half penny, and farthing. Note. — The guinea (= 21s.) and the half guinea (= 10s. 6d. sterling) are old gold coins, that are still in circulation, but are no longer coined. 301* 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 lb. 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 lb. =» 14 lb. 3 oz. 9 pwt. 1.727 gr. of silver. 192 COMPOUND NUMBERS. TV. French Money. 30^. The currency of France is decimal currency. The unit is the franc, of which the other denominations are divisors. TABLE. 10 millimes make 1 centime. 100 centimes " 1 franc. Scale — ascending, 10, 100; descending, 100,10. Coins. The gold coin is the 20-franc piece, or Louis. The silver coins are the franc and the demi franc. Note. — In France accounts are kept in francs and decimes. A franc is equal to 18.6 cents U. S. money. 363. COMPARATIVE TABLE OF MONEYS. English. U. S. French. U.S. Iqr. = $ .004 1 J Id. = -02;^ Is. = .242 4s. Id. 2xVrqr. = LOO £1 = 4.84 1 millime = 1 centime = 1 franc = $ .000186 .00186 .186 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. 363. Reduction Descending is changing a number of one denomination to another denomination of less unit value; thus, $1 = 10 dimes = 100 cents = 1000 mills. 306. Reduction Ascending is changing a number of one denomination 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. 1 fur. 17 rd. 2 yd. 1 ft. 8. in. to inches. REDUCTION. 193 OPERATION. 3 mi. 1 fur. 17 rd. 2 yd. 1 ft. 8 in. 25 40 fur. 1017 rd. 5} 5087 508} 5595} yd. 3 16787} ft. 12 201458 in. Analysis. Since in 1 mi. there are 8 fur., in 3 miles there are 3X8 fur. == 24 fur., and the 1 fur. in the given number, added, makes 25 fur. in 3 mi. 1 fur. Since in 1 fur. there are 40 rd., in 25 fur. there are 25 X 40 rd. = 1000 rd., and the 17 rd. in the given number added, makes 1017 rd. in 3 mi. 1 fur. 17 rd. Since in 1 rd. there are 5} yd., in 1017 rd. there are 1017 X 5} yd. = 5503} yd., which plus the 2 yd. in the given number = 5595} yd. in 3 mi. 1 fur. 17 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. 1 fur. 17 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 tbj.t we have successively multiplied by the numbers in the descending scale of linear measure from miles to inches, inclu- sive. 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. I. Multiply the JiigJiest denomination of tJie given compound number hy that number of the scale which will reduce it to the next lower denomination, and add to the product the given numbery if any, of that lower denomination, II. Proceed in the same manner with the results obtained in each lower denominationj 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. 2. In £133 6 s. 8d., how many farthings? Ans. 128,000. 3. Change 100 mi. to inches. Ans. 6336000 in. 4. How many rods of fence will inclose a farm IJ miles square ? Ans. 1920 rd. 5. The grey limestone of Central New York weighs 175 lbs. to the cubic foot; what is the weight of a block 8 ft. long and 1 yd. square ? Ans. 6 T. 6 cwt. 6. 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 §.06i a pint; how much was his whole gain ? 9. Eeduce 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 ? A)is. 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? Am. 115,440 d. 15. Reduce $7J to mills. Ani. 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 Sf^ 45iri ^^ minims. 20. Reduce 1 T. 1 P. 1 hhd. to Imperial gallons. Ans. 367 J Imperial gal. 21. How many dollars Canada currency are equal to £126 12s. 6d.? Ans. $506i. 22. How many pint, quart, and two-quart bottles, of each an equal number, may be filled from a hogshead of wine ? Ans. 72. 23. IIow many steps of 2 ft. 9 in. each, will a man take, in walking from Erie to Cleveland, the distance being 95 mi.? PvEDUCTION. 195 BC 24. A grocer bought 12 bbl. of cider at $11 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. 28. If 14 A. be sold from a field containing 50 A., how many gquare rods will the remainder contain ? Ans. 5,760 sq. rd. 29. A man returning from Pike's Peak has 36 lb. 8 oz. of ure gold ; what is its value at $1.04i per pwt. ? Ans. ^9169.60. 30. A person having 8 hhd. of tobacco, each weighing 9 cwt. 42 lb., wishes to put it into boxes containing 48 lb. each ; how many boxes must he obtain ? Ans. 157. 31. A merchant bought 12 bbl. of salt at $li a barrel, and re- tailed it at f of a cent a pound; how much was his whole gain? 32. A physician bought lib lOg of quinine at §2.25 an ounce, and dealt it out in doses of 10 gr. at $.12 J 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. AXALYSIS. OPERATION. 4i gal. X I X f X I Or, 11 = TT gl- V^ To re- duce gallons to gills, we multiply succes- sively by 4, 2, and 4, the numbers in the de- scending scale. And since the given num- ber is a fraction, we indicate the process, as in multiplication of fractions, after which we perform the indicated operations, and ob- tain j\, the answer. Hence, 11 1 4 2 8 = TT g^-^ ^■^«- 196 COMPOUND NUMBERS. KuLE. Multiply the fraction of the higher denomination hy the •^. numhei's in the desc ending scale successively ^ between the given and the required denomination. Note. — Cancellation may be applied wherever practicable. EXAMPLES FOR PRACTICE. 1. Eeduce ^^^ of a lb. Troy to the fraction of a pennyweight. Ans. I pwt. 2. Eeduce ^^^ of a hhd. to the fraction of a pint. 3. Reduce ^ jy^ of a mile to the fraction of a yard. . Ans. I yd. 4. Eeduce -g|^ of a gallon to the fraction of a gill. 5. What part of a dram is -^^^-q of | of | of -^j of S| pounds avoirdupois weight ? Ans, :^2T5 ^^* 6. Eeduce yg^^^j of a dollar to the fraction of a cent. 7. Eeduce ^K of a rod to the fraction of a link. Ans, | 1. 8. Eeduce -^K of a scruple to the fraction of a grain. 9. What fraction of a yard is ^ of ■j\ of a rod ? 10. ^^g of -a week is | of how many days? Ans. 8| da. 11. What fraction of a square rod is j^^g^ of 4| times y^^ of an acre ? Ans, j\ sq. rd. CASE III. jl 369. To reduce a denominate fraction to integers of lower denominations. ,, 1. What is the value of | of a bushel? OPERATION. Analysis. | bu. = | of I bu. X 4 = f pk. = If pk. 4 pk., or 1§ pk.; } pk. = ^ I pk. X 8 = \4 qt. == 4| qt. of 8 qt. = 4| qt. ; and J qt. I qt. X 2 = i pt. = If pt. = I of 2 pt. = 15 pt. The 1 pk. 4 qt. If pt, Ans. "^^*^' 1 P^-' ^ ^*" 1 P*" with the last denominato fraction, | pt., form the answer. Hence, EuLE. I. Multiply the fraction hy 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. EEDUCTIOK. 19T II. Proceed with the fractional i-)artj if any^ as before, until reduced to the denominations required. ^ III. The units of the several denominations j arranged in their ordery ivill he the required residt. i EXAMPLES FOR PRACTICE. 1. Keduce -^j^ of a yard to integers of lower denominations. Ans. 2 ft. 8^ in. o 2. Reduce | of a month to lower denominations. 3. Reduce g^J of a short ton to lower denominations. 4. What is the value of | of a long ton ? Ans. 11 cwt. 12 lb. 7^ oz. ? 5. What is the value of | of 2^ pounds apothecaries' weight ? 6. What is the value of -^^ of an acre ? Ans. 2 R. 6 P. 4 sq. yd. 5 sq. ft. 127/^ sq. in. 7. Reduce | of a mile to integers of lower denominations. 8. What is the value of 4 of a great gross ? Ans. 6 gross 10 doz. 8|. 9. What is the value in geographic miles of -^^ of a great circle? Ans. 12150 mi. 10. What is the value of | of 3| cords of wood ? Ans. 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. G fur. 16 rd. 12. What is the value of || f § ? Ans. 8 f^ 35 rt^. 13. What is the value of -| of a sign ? Ans. 12° 5r 25^". 14. A man having a hogshead of wine, sold -^^ of it; how much remained ? Ans. S3 gal. 3 qt. 1 pt. l-j^^ gi. CASE IV. S7®. 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. OPERATION. Analysis. We first multiply I 125 the given decimal, .125 of a barrel, 31.5 by 31.5 (= 31J) to reduce it to o r%.-,-- 1 callons, and obtain 3.9375 g-allons. 8.93/0 gal. ° .^,. ^1 Q 11 1 1 ^ Omitting the o gallons, we mul- tiply the decimal, .9375 gal., by 3.7500 qt. 4 ^o reduce it to quarts, and obtain _f 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 0-1 1-^ pints. And the decimal part 3 gal. 3 qt. iVt. 2 gi., Ans. ''^ ^^'^ "-f^"'* ^^^^ '""'*'P'y ^'^.^ ^^ reduce it to gills, and obtain 2 gills. The integers of the several denominations, arranged in their order, form the answer. Hence, E-ULE. I. 3Iultip7// the given denominate decimal hy that num- her in the descending scale ichich will reduce it to the next loiver denominatiouy and point off the result as in midtipUcation of decimals, II. Proceed icifh the decimal part of the product in the same manner until reduced to the required denominations. The integers at the left will he the answer required. EXAMPLES FOR rRACTICE. 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. What 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 + far. 7. What is the value of .715° ? Ans. 42' 54". 8. What is the value of 7.88125 acres? Ans. 7 A. 3 E. 21 P. 9. What is the value of .625 of a f^ithom? 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. Ufg 5f3 48 T?i. ^' 12. What is the value of .61 of a tun of wine ? A71S. 1 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 the given number of minutes by 60, because there are ^'j as many hours as minutes, and wo obtain 2645 h. plus a re- mainder of 40 min. AYe next divide the 2645 h. by 24, because there are -^\ as many days as hours, and we find that 2645 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. Eeduce 201458 inches to miles. OPERATION. 12 ) 201458 in. OPERATION. 60)157540 min. 24 ) 2625 h. + 40 min. 7)109 da. -f 9 h. 15 wk. 4- 4 da. 15 wk. 4 da. 9 h. 40 min., Ans. 3)16788 ft. 2 in. 5i or 5.5)5596 yd. 40 ) 1017 rd. 2 yd. 1 ft. 6 in. 8) 25 fur. 17 rd. 3 mi. 1 fur. 3 mi. 1 fur. 17 rd. 2 yd. 1 ft. 8 in., Ans. Analysis. We divide successively by the numbers in the ascending scale of linear measure, in the same manner as in the last pre- ceding operation. But, in dividing the 5596 yd. by 5J or 5.5, we have a ro- 200 COMPOUND NUMBERS. mainder of 2J yd., and this reduced to its equivalent compound num- ber, (369) = 2 yd. 1 ft. G 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 hy that number of the ascending scale which will reduce it to the next higher denomination. II. Divide the quotient hy the next higher number in the scale ; and so proceed to the highest denomination required. The last quotienty with the several remainders annexed in a reversed order ^ will be the answer. Note. — The several corresponding cases in reduction descending and reduo- tion ascending, being opposites, mutually prove each other. EXAMPLES FOR PRACTICE. 1. Reduce 1913551 drams to tons. 2. In 97920 gr. of medicine how many lb. ? Ans. 17 lb. 8. Reduce 1000000 in. to mi. 4. How many acres in a field 120 rd. long and 56 rd. wide ? 6. 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. 4f 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 IJ 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 Tli how many Cong. ? 15. If 135 million Gillott steel pens are manufactured yearly^ how many great-gross will they make ? A7is. 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 ? Ans. G^ 52'. 10. If a man travel at the rate of a minute of distance in 20 minutes of time, how much time would he require to trtivel round the earth ? Ans. SCO 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 86 yr. by rising 45 min. earlier, and retiring 25 min. later, every day, allowing for 9 leap years? Ans. 689 da. 4 h. 30 min. 28. A grocer bought 20 gal. of milk by beer measure, and sold it by wine measure ; how many quarts did he gain ? Ans. 17f 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 mi. 26. In 1 bbl. 1 gal. 2 qt. wine measure, how many beer gal- lons ? Ans. 27-5^. 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 88 ft. wide? Ans. 22^. 29. How many cubic inches in a solid 4 ft. long 8 ft. wide, and 1 ft. 6 in. thick ? 30. How many acres in a field 120 rd. long and 56 rd. wide ? 31. Change 856 dr. apothecaries weight, to Troy weight. 82. A coal dealer bought 175 tons of coal at S8.75 by the long ton, and sold it at §4.50 by the short ton ; how much was his whole gain ? Aiis. $225.75. 38. How many acres of land can be purchased in the city of New York for $78750, at $1.25 a square foot? Ans. 1 A. 56 P. 104 sq. ft. 84. An Ohio farmer sold a load of corn weighing 2402 lb., and a load of wheat weighing 2175 lb. ; for the corn he received S.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. 85. What will be the cost of 54 bu. of corn at 5s. a bushel, New England currency ? OPERATION. 54 X 5 = 270s. 270s. -V- 6 = $45 OPERATION. 3 m Or, i 100 ^ $300 6 25 2 $300 Or, 9 Analysis. Since 1 bu. costs ^^ 5s., 54 bu. cost 54x 5s. = 270s.; -- and since 6s. make $1 N. E. r-p currency, 270 —- G = $45, Ans. ^45 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 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 63 Or, 2 i t $630 88. 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 ? Analysis. The operation in this example is similar to the preceding examples, except that we divide the cost of the apples by the ^rice 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 $lo2 whatever currency. 5 63 2 i $630 OPERATION. 12 11 Analysis. In this ex- ample we first reduce 5 hhd. to quarts by multiply- ing by 63 and 4, and then proceed as in the preceding examples. m KEDUCTION. 203 39. What cost 75 yards of fianncl at 3s. Cd. per yard, New England currency ? Ajis. 813.75. 40. A man in Philadelphia worked 5 weeks at Gs. 4d. a day; how much did his wages amount to ? Ans. $25.33}. 41. A farmer exchanged 2 bushels of beans worth 10s. 6d. per bushel, for two kinds of sugar, the one at lOd. and the other at lid. per pound, taking the same quantity of each kind; how many pounds of sugar did he receive ? Ans. 24 lb. 42. If corn be rated at 6s. lOd. per bushel in Vermont, at what price in the currency of New Jersey must it be sold, in order to gain §7.50 on 54 bushels? CASE IT. 37@. To reduce a denominate fraction from a less to a greater unit. 1. Reduce -fj of a gill to the fraction of a gallon. OPERATION. Analysis. To re- t\ gi- X I X ^ X i = 4^ gal. ^^c«, g'^lls to gallons, we divide successively ^^> by 4, 2, and 4, the 11 ^ numbers in the as- 4 cending scale. And since the given num- ber is a fraction, we 44 1 = -^^j Ans. indicate the process, as in division of frac- tions, after which we perform the indicated operations, and obtain ^^, the answer. Hence, Rule. Divide the fraction of the lower denomination hi/ the numbers in the ascending scale successively/ , between the given and the required denomination. Note. — The operation may frequently be shortened by cancellation. EXAMPLES FOR PRACTICE. 1. Reduce f of a shilling to the fraction of a pound. Ans. £ ^u- 204 COMPOUND NUMBERS. 2. Keduce ^ of a pennyweight to the fraction of a pound Troy. Ans. ^J^ lb. 3. What part of a ton is | of a pound avoirdupois weight ? 4. What fraction of an hour is | of 20 seconds ? 6. What is the fractional difference between g|^ of a hhd. and I of a pt. ? Ans. 35^77^ hhd. 6. ^1^ of I of f of a pint is what fraction of 2 pecks ? Ans. I 7. Reduce | of ^ of /$ of a cord foot to the fraction of a cord. Ans. -^g Cd. 8. What part of an acre is ^^^ of jly of 9^ square rods ? 9. I of 5^ furlongs is ^ of J^ of how many miles? Ans. 12| mi. 10. A block of granite containing | of | of 20^ cubic feet, is what fraction of a perch ? Ans. ^| 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 | of an inch to the fraction of an Ell English. CASE III. S73. 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 lb. Troy = 5760 gr. number is of another, Jf §g lb. = ^5 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 lb. there are 5760 gr. Therefore 1 gr. is 3^'g^^ lb., and 1260 gr. are f^|^ lb. == 3^ lb-> the answer. Hence, Rule. Reduce the given number to its lowest denoniination for the numerator J 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 2 R. 20 P. to the fraction of an acre. Ans. I A. 2. What part of a mile is 6 fur. 26 rd. 3 yd. 2 ft. ? 3. What part of a £ is 18s. 5d. 2/^ far. ? Ans. £if . 4. What part of 21 lb. Apothecaries' weight is 7g 73 29 14 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. ||. 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 ? Ans, 86.30. 11. Reduce 30. of 3 If 3 80 nj, to the fraction of a Cong. 12. What part of a ton of hewn timber is 06 cu ft. 864 cu. in. ? CASE IV. 874. 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. Were- 16 8.0 cu. ft. duce the 8 cu. ft. to the decimal of a cd. 3.5000 cd. ft. 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. = 128 cu. ft. nexing the decimal ^s_fig Cd. = ^5 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., 18 206 COMPOUND NUMBERS. (as in 373), and we shall have j^^g Cd. = -i^g- Cd., which, reduced to its equivalent decimal, equals .4375 Cd., the same as before. Hence, KuLE. Divide the lowest denomination given hy that number in the scale which will reduce it to the next higher denominationy and annex the quotient as a decimal to that higher. Proceed in the same manner witil the whole is reduced to the denomination required.^ Or, Reduce the given numher to aj'raction of the required denomi- natioUy and reduce this fraction to a decimal, EXAMPLES FOR PRACTICE. 1. Eeduce 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 1 R. 11.52 P. to the decimal of an acre. 4. What part of 4 oz. is 2 oz. 16 pwt. 19.2 gr ? Ans. .71. 6. What part of a furlong is 28 rd. 2 yd. 1 ft. 11.04 in. ? 6. Reduce 3|g 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 3| ft. ? Ans. .625 fath. 9. What part of 1\ bushels is .45 of a peck ? Ans. .09. 10. What part of 3 A. 2 R. is 1 R. 11.52 P.? Ans. .092. 11. Reduce | of ^ of 22| lb. to the decimal of a short ton. 12. What part of a f§ is 5 fj 36 T]]^ ? Ans. .7 fg. 13. Reduce 50 gal. 3 qt. 1 pt. to the decimal of a tun. Ans. .20188 + T. ADDITION. S7«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. 376, 1. What is the sum of 50 hhd. 32 gal. 3 qt. 1 pt., 2 hhd. 19 gal. 1 pt., 15 hhd. 46| gal, and 9 hhd. 39 gal. 2^ qt. ? ADDITION. 207 OPERATION. ^lnalysis. Writing the numbers so that hhd. gal. qt. pt. units of the same denomination shall stand 50 32 3 1 in the same column, we add the numbers 2 19 1 of the right hand or lowest denomination, 15 46 1 and find the amount to be 3 pints, which is _^| ^^ ^ ^ 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 1 gal. 3 qt. We write the 3 qt. under the column of quarts, and add Tlie 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., | da., and | h. oPERATiox. Analysis. We -5^^ wk. =r 4 da. 21 h. 36 min. first find the value 3 da. = 14 " 24 min. ^^ ®^^^ fraction in I h. * == 22 " 30 sec. integers^ of less de- nominations, (369), ^ 1'^ 22 30 and then add the Or, resulting or equiva- I da. X 4 = /5wk; lent compound num- I h. X 3j\ X 4 = -5^ wk ; bers. /^ wk. + ^K wk. + ^^^ wk. = j J wk ; ^ Or, we may^ re- 1 J wk. = 5 da. 12 h. 22 min. 30 sec. ^.^^^ *^^ S^'^^^ ^^^^■ 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 arc denominate fractions^ or if any of the denominations are mixed numhei-Sy reduce the frac- tions to integers of lower denominations. II. Vr^rite 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 fail to see that the principles involved in nddinoj compound numbers are the same as those in addition of simple numbers; and that the only difference consists in the different carrying unites. EXAMPLES FQR PRACTICE. (1-) (2-) lb. oz. pvrt. gr. lb. 5 5 9 gf- 10 8 5 1 8 7 7 6 2 13 6 11 7 21 10 16 . 12 1 2 2 3 7 1 19 14 6 12 13 17 5 3 12 15 9 16 2 7 15 20 13 2 1 19 4 1 5 21 66" 11 9 (3.) 5 fur. rd. ft. in. 7 26 11 9 4 16 7 11 86 14 3 1 9 2 8 5 10 1 6 2 5 1 15 13 10 58 4 5 (4.) 2 19 JL. R. P. sq. yd. sq.ft. 140 3 17 27 6 320 1 SO 14 2 111 7 3- 214 2 15 22 7 100 3 6 1 25 1 36 8 104 2 9 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 J in., 2 yd. 2 ft. 6i 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 J A., 7 A. 2 R. 29 J P., 1 A. 3 R. 16.5 P., and | A. ? Ans. 156 A. 39i P. 10. Required the sum of 31 bu. 2 pk., 10| bu., 5 bu. ^ qt., 14 bu. 2.75 pk., and | pk. Ans. 62 bu. 1 pk. 5 qt. IJ pt. SUBTRACTIOK. 209 11. Required the value of 42 yr. 7-J mo. + 10 yr. 3 wk. 5 da. + 9| mo. + 1 wk. 16 h. 40 min. + | mo. + 3| da. Ans. 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 li 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. eh., and 5 sq. ch. 10| P.? Ans. 262 A. 3 sq. ch. 13.8 P. 15. Add 3| Pch. 18 cu. ft., 84.6 cu. ft., | Pch., and |« cu. ft. 16. Add ?3|, $25^, $12|, S2|, and $2.54|. Ans. $47.0725. 17. What is the sum of 3 lb 5 g 4 5 2 9 17 gr., 2 lb 5 5 12 gr., 4 § 2 3 1 9 16 gr. ? Ans. 5 lb 10 g 4 3 2 9 5 gr. 18. AN. Y. farmer received $.60 a bushel for 4 loads of corn ; the first contained 42.4 bu., the second 2866 lb., the third 36i 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 1| T., and the third 2500 pounds; how much did the whole cost? Aiis. $30.20. 20. A man in digging a cellar removed 140| 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. I Analysis. Writing the subtra- hend under the minuend, placing units of the same denomination under each other, we subtract 8 gr. from ~Jq Yi 6 ^^ S^' ^^^ write the remainder, 6 gr., underneath. Since we cannot 18* o OPERATION. •lb. oz. pwt. gr. 18 5 4 14 10 6 10 8 210 COMPOUND NUMBERS. 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 6 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 lb. 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 lb. to the 10 lb. in the subtrahend, we subtract the sum, 11 lb., from the 18 lb. in the minuend, as in simple numbers, and write the remainder, 7 lb., 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. ^^ 1^ ^ 28J gal. 2 qt. But, J gal. = 2 qt., which added to the 2 qt. in the remainder makes ^ !:_i__^ 1 gal., and this added to the 28 gal. makes 4 29 29 gal. ; and the angwer is 4 bar. 29 gal. 8. From | of a rod subtract f of a yard. OPERATION, Analysis. "We first find the value of each I rd. = 4 yd. ft. 4 ' .... , . . . 3 1 9 a o u fraction in integers ot ^ lower denominations, '^ 1 ^ (369), and then sub- Or, tract the less value I yd. X~=l yd. X A = ^\ rd. ; ^^^^ ^^' S^^f ^^- ^J' 0^ ^ *" ^1 ^^ ^ we may reduce the I rd. — 2^^ rd. = f| rd. ; given fractions to frac- II rd. = 3 yd. 1 ft. li in. tions of the 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 KuLE. I. If any of the numbers are denominate fractions, or if any of the denominations are mixed number s, reduce the frac- tions to integers of lower denominations. II. Write the subtrahend under the Tninuend, 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. . I SUBTRACTION. 211 TV. If the numher of any denomination in the siiltrahcnd ex- ceed that of the same denomination in the minuend j add to the mimhcr 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. I EXAMPLES FOR PRACTICE. (1.) (2.) mi. fur. rd. ft. in. A. R. P. From 175 3 27 11 4 820 S 26.4 Take 59 6 10 12 9 150 2 31.86 Rem. 115 5 16 15 1 170 84.54 (3.) (4.) hhd. gal. qt. yr. mo. wk. da. . h. 5 86 3i 45 1 8 17J 2 45 Ij 10 9 1 22 6^ 5. Subtract 15 rd. 10 ft. Si in. from 26 rd. 11 ft. 8 in. Ans. 11 rd. 11 f in. 6. From^l T. 11 cwt. 30 lbs. 6 oz. take 18 cwt. 45 lb. 7. Subtract .659 wk. from 2 wk. 8| da. Ans. 1 wk. 6 da. 5 h. 17 min. 16| sec. 8. From i^-| hhd. take .90625 gal. Ans. 32 gal. 9. From f of 3f A. take 8 R. 12.56 P. 10. Subtract ^% lb. Troy, from 10 lb. 8 oz. 8 pwt. 11. From a pile of wood containing SQ Cd. 4 cd. ft., there was sold 10 Cd. 6 cd. ft. 12 cu. ft. ; how much remained ? 12. From 5 J barrels take ^ of a hogshead. Ans. 4 bbl. 11 gal. 1 qt. 13. Subtract |gj of a day from f of a week. Ans. 4 da. 49 min. 30 sec. 14. From. I of a gross subtract | of a dozen. A^is. 6| dcz. 15. From J of a mile take |.^ of a rod. 16. Subtract 2 A. 8 R. 5.76 P. from 5 A. 1 R. 24.24 P. Ans. 2 A. 2 R. 18.48 P. 212 COMPOUND NUMBERS. 17. Subtract .0625 bu. from | pk. Ans. 4 qt. 18. From the sum of f of 365| da. and | of 5| wk. take 49^ min. Ans. 33 wk. 1 da. 1 h. lOf min. 19. From the sum of § of 3| mi. and 174 ^^-y ^^^^ ^i ^^^• 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 bring ? Ans. $3513.19+. 23. An excavation 58 ft. long, 37 ft. wide, and 6 ft. deep is to be made for a cellar; after 471 cu. yd. 16 cu. ft. 972 cu. in. of earth have been removed, how much more still remains to be taken out ? Ans. 5 cu. yd. 7 cu. ft. 756 cu. in. 24. From the sum of | lb., 4| oz., and 31^ pwt., take the difi*er- ence between | oz. and | pwt. Ans. 1 lb. 3 oz. 8 pwt. 21 gr. 25. From the sum of 5j% A., | of 6| A., -1 E., and ^\ of 2j\ P., take 4 A. 25 P. 12 sq. yd. Ans. 5 A. 3 E. 5 P. 6 sq. yd. 88©. To find the difference in dates. 1. How many years, months, days and hours from 3 o'clock p. M. of June 15, 1852, to 10 o'clock A. M. of Feb. 22, 1860? OPERATION. Analysis. Since the later of two dates yr. mo. da. h. always expresses the greater period of 1860 -J ^2 10 time, we w^rite the later date for a minu- IR 5*^ 6 1.5 15 end and the earlier date for a subtrahend, 7 8 6 19 placing the denominations in the order of the descending scale from left to right, (300, Note 8). We then subtract by the rule for subtraction of compound numbers. When the exact numher of clays is required for any period not exceeding one ordinary year, it may be readily found by the fol- lowing p SUBTRACTION. TABLE, 213 Slioioing tlie number of days from any day of one month to the same day of any other month within one year. FROM ANT DAY OF January February March April May June July August September October November December TO THE SAME DAY OF THE NEXT Jan, Feb. Mar. Apr. May June July Aui,'. Sept. Oct. Nov. Dec. 365 334 306 275 245 214 184 153 122 92 61 31 31 365 337 306 276 245 215 184 153 123 92 62 59 28 365 335 304 273 243 212 181 151 120 90 90 59 31 365 335 304 274 243 212 182 151 121 120 89 61 30 365 334 304 273 242 212 181 151 151 120 92 61 31 365 335 304 273 243 212 182 181 150 122 91 61 30 365 334 303 273 242 212 212 181 153 122 92 61 31 365 334 304 273 243 243 212 184 153 123 92 62 31 365 335 304 274 273 242 214 183 153 122 92 61 30 365 334 304 304 273 245 214 184 153 123 92 61 31 365 335 334 303 275 244 214 183 153 122 91 61 30 365 If the days of the different months are not the same, the num- ber of days of difference should be added when the earlier day belongs to the month /rom which we reckon, and subtracted when it belongs to the month to which we find the time. If the 29th of February is to be included in the time computed, one day must be added to the result. EXAMPLES FOR PRACTICE. 1. "War between England and America was commenced April 19, 1775, and peace was restored January 20, 1783 ] how long did the war continue ? Ans. 7 yr. 9 mo. 1 da. 2. The Pilgrims landed at Plymouth Dec. 22, 1620, and Gen. Washington was born Feb. 22, 1732 ; what was the difference in time between these events ? 3. The first settlement made in the U. S. was at Jamestown, Ya., May 23, 1607; how many years from that time to July 4, 1860 ? 4. How long has a note to run, dated Jan. 30, 1859, and made payable Juno 3, 1861 ? Ans. 2 yr. 4 mo. 3 da. 214 COMPOUND NUMBERS. 5. How many years, months, and days from your birthday to this date ? 6. What length of time elapsed from 16 minutes past 10 o'clock, A. M., July 4, 1855, to 22 minutes before 8 o'clock, p. m., Dec. 12, 1860 ? Alts, 1988 da. 9 h. 22 min. 7. What length of time will elapse from 40 minutes 25 seconds past 12 o'clock, noon, April 21, 1860, to 4 minutes 36 seconds before 5 o'clock, A. M., Jan. 1, 1862 ? 8. How many days from the 4th September, to the 27th of May following ? Ans. 265 da. MULTIPLICATION. 381. 1. Multiply 5 mi. 4 fur. 18 rd. 15 ft. by 6. OPERATION. Analysis. Writinji: the multi- 5 mi. 4 fur. 18 rd. 15 ft. 6 plier under the lowest denomi- nation of the multiplicand, we multiply each denomination in 33 2 33 7i the multiplicand separately in order from lowest to highest, as in simple numbers, and carry from lower denominations to higher, according to the ascending scale of the multiplicand, as in addition of compound numbers. Hence, Rule. I. Write the multiplier under the lowest denomination of the midtipUcand. II. Multiply as in simple numherSj and carry as in addition of compound Jiumhers. Notes. — 1. "When the multiplier is large, and is a composite number, we may ehorten the work by multiplying by the component factors. 2. The multiplier must be an abstract number. 3. If any of the denominations are mixed numbers, they may either be re- duced to integers of lower denominations before multiplying, or they may be multiplied as directed in 193. 4. The multiplication of a denominate fraction is the most readily performed by 193, after vyhich the product may be reduced to integers of lower denomina- tions by 369. J5§2, 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, eacb containing 2 bu. 1 pk. 4 qt. ? I I FIRST OPERATION. 47 = (5 X 9) + 2 2 bu. 1 pk. 4 qt. X 2 5 11 bu. 3 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=(6x 8) — 1 2 bu. 1 pk. 4 qt. X 1 6 14 bu. 1 pk. in 6 bags. 8 114 bu. in 48 bags. 2 ^^ 1 p k. 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 of 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 + 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-) (2-) T. cwt. lb. OZ. mi. fur. rd. ft. 12 15 27 9 14 6 36 14 8 133 6 11 9 102 2 20 8 10 (3.) (-^0 A. R. P. sq. yd. sq. ft. Cd. cd. ft. cu ft. 7 1 33 21 7 6 10 7 13 12 5. Multiply 34 bu. 3 pk. 6 qt. 1 pt. by 14. 6. Multiply 4 lb. 10 oz. 18.7 pwt. by 27. Ans. 132 lb. 7 oz. 4.9. pwt 7. Multiply 9 I 3 5 2 9 13 gr. by 35. 216 COMPOUND NUMBERS. 8. Multiply 5 gal. 2 qt. 1 pt. 3.25 gi. by 96. 9. Multiply 78 A. 3 R. 15 P. 15 sq. yd. by 15f. Ans. 1235 A. 1 E. 2 P. 23} sq. yd. 10. What is 73 times 9 cu. yd. 10 cu. ft. 1424 cu. in.? 11. Multiply 2 lb.. 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 g 4 f 3 25 Tfl by 24. Ans. 22 Cong. 7 0. 13 f g 2 f^. 15. Multiply 3 hhd. 43 gal. 2.6 gi. by 17. 16. Multiply 9 T. 13 cwt. 1 qr. 10.5 lb. by 1.7. Note. — "NVhen 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. t^ Ans. 16 T. 8 cwt. 2 qr. 20.15 lb. 17. If a pipe discharge 2 hhd. 23 gal. 2 qt. 1 pt. of w^ter 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. $1.37} 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 3 how much money did he receive for the grain? Ans. $123.20—. DIVISION. 383. 1. Divide 37 A. 1 R. 16 P. by 8. OPERATION. Analysis. Writing the divisor on the left of the dividend, we divide ^ ; the highest denomination, and obtain 4 2 27 a quotient of 4 A. and a remainder of 5 A. Writing the quotient under the deriomination divided, we reduce the remainder to roods, making 20 R., vdiich added to the 1 R. of the dividend, equals 21 R. Dividing this, we have a quotient of 2 R. and a remainder of 5 R. Writing I DIVISION. 217 the 2 K. under the denomination divided, we reduce the remainder to rods, making 200 P., which added to the 16 P. of the dividend, equals 216 P. Dividing this, we have a quotient of 27 P. and no remainder. 2. Divide 111 bu. 2 pk. 4 qt. by 47. OPERATION. 47)lllbu.2pk.4qt.(2bu.lpk.4qt. a^,,^,,. The 94 divisor being large, 17 bu. rem. and a prime num- _4 ber, we divide by 70 pk. in 17 bu. 2 pk. long division, set- 47 *i^g down the Z^ , whole work of sub- 23 pk. rem. x x- i i o ^ tracting and reduc- I 188 qt. in 23 pk. 4 qt. 188 ing. From these examples and illustrations we derive the following Rule. I. Divide the highest denomination as in simple num- hersj and each succeeding denomination in the same manner, if there he no remainder, II. If there he a remainder after dividing any denomination, reduce it to the next loicer denomination, adding in the given num- her of that denomination, if any, and divide as hefore. III. The several partial quotients will he the quotient required. KoTES. — 1. "When the divisor is large and is b. 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 snme as in simple numbers. 3. The division of a denominate fraction is most readily performed by 195, after which the quotient may be reduced to its equivalent compound number, by 369. H ^ 4 F EXAMPLES FOR PRACTICE. (1-) (2.) £ p. d. far. lb. oz. pwt. gr. 5)62 7 9 3 9)56 6 17 6 Quotient, 12 9 6 3 61b. 3oz. 8pw.l4gr. 19 218 COMPOUND NUMBERS. (3.) (4.) hhd. gal. qt. pt. T. cwt. qr. lb. 12)9 28 2 19 )873 19 2 4 49 2 1' 19 13 2 16 6. Divide 858 A. 1 R. 17 P. 6 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. Aiis. 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. 1 R. 30 P. by 22^. Ans. 341 A. 1 E. 16| P. Note 4. — Observe that 22^ = ^ ; hence, multiply by 2, and divide the result toy 45. 12. Divide 365 da. 6 h. by 240. 13. Divide 3794 cu. yd. 20 cu. ft. 7091 cu. in. by 331. 14. Divide 121 lb. 3g 2^ 19 4 gr. by 13|. 15. Divide 28° 51' 27.76*5'' by 2.754. Ans. 10° 28' 42^. 16. Divide 202 yd. 1 ft. 6| in. by f . Ans. 337 yd. 1 ft. 7i 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. 2 R. 12|| P. 19. A cellar 48 ft. 6 in. long, 24 ft. wide, and 65 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. 121 J. LONGITUDE AND TIME. 384. Since the earth performs one complete revolution on its axis in a day or 24 hours, the sun apj^ears to j^ciss 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 a2'>parcnt motion of the sun^ is as follows: Time. , Longitude. 24 h. = 360° 1 h. or 60 min. = ^^T = 15° = 900^ 1 min. or 60 sec. == ^^° = Vn ^ = 15^ = QOO''^ Isec. = ir = 'i'r' - 15^" Hence, 1 h. of time = 15° of longitude. 1 min. " = 15^ '' ^ 1 sec. '* = 15^/ ** " CASE I. 385. To find the difference of longitude between twc places or meridians, when the difi:erence 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 lirntrs^ minutes^ and seconds, hy 15, according to the rule for midtipUca- tion of compound numbers ; the product will he the difference of longitude in degrees, minuteSj and seconds. Notes. — 1. If one place be in enst, and the other in west lonoritude, the dif- ference of longitude is found by udding 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 at one place, it will he pn^t 12 o'clock at all places east, and hefore 12 at all places west. Hence, if the difference of time between two places, be HuhirHvird from the time at the easterly 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 heing 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 80° 2' west, what is the longitude of Boston? Ans. 71° 3' west. I 220 COMPOUND NUMBERS. 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 ? Ans. 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 ? 6. 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 3 h. 58 min. 23 J 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. 53j| 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. Note .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 1 sec. A. M. at New York; what is the longitude of New York from Greenwich ? Ans, 74° 1' 6" 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 lon«;itude since she set sail? Ans. 18° 3' 48.6". LONGITUDE AND TIME. 221 CASE II. 38G. 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, hi/ 15, according to the rule for division: of compound numbers; the quotient will he the difference of time in hours, minutes, and seconds, EXAMPLES FOR PRACTICE. 1. Washington is 77° 1' and Cincinnati is 84°*24' west longi- tude; what is the difference of time? Ans. 29 min. 32 see. 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, C3° 36' west, at 4 h. 30 min. p. M.; the fact is telegraphed to New York, 74° 1' 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 ? i . Ans. 7 h. 15 min. 11|| sec. A. M. 6. The longitude of Pekin is 118° east, and of Sacramento City 120° west; what is the difference in time? f 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? I 8. What time is it in Baltimore when it is 6 o'cfeck p. M. at Jerusalem? Ans. 10 h. 31 min. 2#;Bec. A.M. 19 * 222 COMPOUXD NUMBEKS. 9. The longitude of Springfield, Mass., is 72° 85' 45" W., and of Galveston, Texas, 94° 46' 34" W.; when it is 20 min. 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' Vv^est; 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 min. 24 sec. p. 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 ? PEOMISCUOUS EXAMPLES IN COMPOUND NUMBERS. 1. In 9 lb. S^ 1^ 29 19 gr. how many grains? 2. IIow much will 3 cwt. 12 lb. of hay cost, at §15^ a ton? 3. In 27 yd. 2 qr. how many Eng. ells? Aas. 22. 4. Heduce §18.945 to sterling money. Ans. £3 18s. 3yyjd. 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? Aus. 384. 7. At 1/6 sterling per yard, how many yards of cloth may be bought for £5 Gs. Gd. ? Ans. 71 yd. 8. In 4 mi. 51 ch. 73 1. how many links? 9. In 22 A. 3 II. 33 sq. rd. 2| sq. yd. how many square yards ? 10. IIow 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 lb. 9 oz., to a jeweler, and ordered them made into tea spoons, each weighing 1 oz. 5 pwt. ; 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|. 14. lleduce | of j\ 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 J ft. long, and 20 ft. wide ? Ans. 58|. / PEOMISCUOUS EXAMPLES. 223 16. If I purchase 15 T. 3 cwt. 3 qr. 24 lb. of English iron, by long ton weight, at 6 cents a pound, and sell the same at ^140 per short ton, how much will I gain by the transaction ? 17. What will be the expense of plastering a room 40 ft. long, 36 J ft. wide, and 22i ft. high, at 18 cents a sq. yd., allowing 1375 sq. ft. for doors, windows, and base board? Ans. $69.78 J. 18. How much tea in 23 chests, each weighing 78 lb. 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 arc 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 J ? 23. If 1 pk. 4 qt. of wheat cost $.72, how much will a bushel cost? Ans. $1.92. 24. IIow many bushels, Indiana standard, in 36244 lbs. 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 IJ feet thick, what will it cost at $1.25 a perch ? Ans, $50.90 jf. 27. The forward wheels of a wagon are 10 ft. 4 in. in circum- ference, and the hind wheels 15 J 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? Aois. 42240. 28. Bought 15 cwt. 22 lb. of rice at $3.75 a cwt, and 7 cwt. 36 lb. 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 -^^ of 7 T. 3 cwt. 26 lb. 30. What is the value in avoirdupois weight of 16 lb. 5 oz. 10 pwt. 13 gr. Troy? Ans. 13 lb. 8 oz. 11.4+dr. 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. 36. Subtract -| bu. + | of f f of 3i qt. from 5 bu. 3|^ qt. ot Ans. 161 pk. 37. What is the difference between f of 5 sq. mi. 250 A. 3 R., and 3i times 456 A. 3 R. 14 P. 25 sq. yd. ? Ans. 2 sq. mi. 254 A. 2 R. 26 P. 24| sq. yd. 38. How many pounds of silver, Troy weight, are equivalent in value to 5.6 lb. of gold by the English government standard ? Ans. 80 lb. 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 384| A., divided it between his two sons, giving to the elder 22 A. 1 R. 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 M 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 694 ^^- corn, worth S46/y. 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 J a square ? / I PROMISCUOUS EXAMPLES. 225 46. If 17 T. 15 cwt. 62} lb. of iron cost $1833.593, how much will 1 ton cost? 47. How many wine gallons will a tank hold, that is 4 ft. long by iij it. wido, and If i't. deep? A)ts. 1871^^^ gal. 48. What will be the cost of 300 bushels of wheat at 9s. 4d. per bushel, 31ichigan currency ? Ans. $350. 49. What will be the cost in Missouri currency? 50. W^hat will be the cost in Delaware currency ? ' 51. What will be the cost in Georgia currency? Ans. $600. 52. What will be the cost in Canada currency? A.ns, $560. ^ 53. Bought the following bill of goods in Boston : /"" 6 J yd. Iriph linen @ 5/4 12 " flannel '' 3/9 8} " calico '' 1/7 9 '' ribbon '' /9 , 4} lb. coffee ^- 1/5 6 1 gal. molasses " 3/8 W^hat was the amount of the bill ? Ans. $21.76 +. 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 lb. 5-^| 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^^/3. 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. 9^ 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? A)ts. 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 3 J ft. deep? 61. Eeduce 3 hhd. 9 gal. 3 qt. wine measure to Imperial gal- lons. Ans, 165.5807+ Imp'l 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 -^^ of it and B the remainder, and the difference between their shares is 15 A. 1 R. 28} P. How much is B^s share? Ans. 38 A. 2 R. 11| P. 64. At 83.40 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 S3 If per acre? / /,, . GQ. Eeduce 9.75 tons of liewn 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, 3-J ft. wide, and 2f ft. deep? ^/?.^. 299i| 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 ? ^^ . , , -"; 'rv ^ * L. >7^ / 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 w^ide, 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. 3 B. 20 P., 6 A, 3 II. 12 P., 9 A. 3 B., and 11 A. 2 B. 32 P. respectively. It is required to divide each piece into the largest sized building lots possible, each lot containing the same area, and an exact num- ber of square rods. How much land will each lot contain ? A71S, 156 P DUODECIMALS. 22T DUODECIMALS. •18 T. Duodecimals are the parts of a unit resulting from con- tinually dividing by 12; as 1, -^^^ -j\^, T72H' ^^^- ^^ 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 : 1^ (inch or prime) is J^ of afoot, or 1 in. linear measure. y^ (second)or y'2 of ^2? " y^^^- of a foot, or 1 '' square '* r^^(third)ory'2of j'aof 3^2-,.. *' jJ^^ofafoot, orl '' cubic TABLE. 12 fourths, [''''), make 1 third V^ 12 thirds '' 1 second, ....... V 12 seconds " Iprime, ... 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 duodecimnl 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. r 8", 15 ft. 3' 5", 17 ft. 9' 7". Ans. 45 ft. 8' 8''. 2. Add 136 ft. 11' 6" 8''', 145 ft. 10' 8" 5'", 160 ft. 9' 5" 5'". Ans, 443 ft. 7' 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' 1". OPERATION. 9 ft. 8' 4 ft. 7' 5 ft. 38 ft. 7' 8' 8" 228 DUODECIMALS. MULTIPLICATION. 3§0« In the multiplication of duodecimals, the product of two dimensions is area, and the product of three dimensions is solidity (282, 286). We observe that V X Ift.^:^^^ of 1ft. =K V X 1 ft. ^ jl:f of 1 ft. = V^. y XV = -iV X ,L of 1 ft. = v. 1// X 1^ = jh X A of 1 ft. = y. Hence, The product of any two orders is of the order denoted by the sum of their indices. 390. 1. Multiply 9 ft. 8' by 4 ft. T, Analysis. Beginning at the right, 8^ X 7^ = 56^^ = 4^ 8^^ ; writing the 8^^ one place to the right, we re- serve the 4^ to be added to the next product. Then, 9 ft. X 7^ + 4^ = 07^ = 5 ft. 7^, which we write in the 44 ft. 3' 8", Ayis, places of feet and primes. Next mul- tiplying by 4 ft., we have 8^ X 4 ft. = 32^ = 2 ft. 8^ ; writing the 8^ in the place of primes, we reserve the 2 ft. to be added to the next product. Then, 9 ft. X 4 ft. + 2 ft. = 38 ft., w^hich we write in the place of feet. Adding the partial pro- ducts, we have 44 ft. y 8^^ for the product required. Hence the Rule. I. Write the several terms of the muliijplier under the corresponding terms of the multiplicand, II. Multiply each term of the multiplicand hy each term of the multiplier y heginning icith the loicest term in each, and call the pro- duct of any two orders, the order denoted hy the sum of their in- dices, carrying 1 for every 12. III. Add the partial products ; their sum will he the required answer. EXAMPLES FOR PRACTICE. 1. How many square feet in a floor 16 ft. 8' wide, and 18 ft. 5' long? 2. How much wood in a pile 4 ft. wide, 3 ft. 8' high, and 23 ft 7Mon 1 filtered in 4 days. Di- ^" " 30 min. viding the quantity fil- - tered in 1 day by 4, 80 " 2 " " 3 Jj " Ans. ^e have the quantity filtered in } da. = 6 h. Dividing the quantity filtered in 6 hours by 12, we have the quan- tity filtered in J h. = 30 min. And the sum of these several results is the entire result required. 75 U 2 '^ " 4 iC 2 '^ 1 " 3 1 ^^ la 7 FOR DrV'ISION. 241 I 2. What will be the cost of 3 lb. 10 oz. 8 pwt. 5} gr. of gold at $15.46 per oz. ? Aiis. $717.52. 3. A man bought 5 cwt. 90 lb. of hay at $.56 per cwt. ; what was the cost? Ans. $3,304. 4. What must be given for 3 bu. 1 pk. 3 qt. of cloverseed, at $4.48 per bushel? Jns. $14.98. 5. A gallon of distilled water weighs 8 lb. 5 oz. 6.74 dr. ; re- quired the weight of 5 gal. 3 qt. 1 pt. 3 gi. Ans. 49 lb. 12 oz. 5.73— dr. 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. lOH-j 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? Am. $4.35 + . 10. What must be paid for 12 J days' labor^ at 5s. 3d. per day, New England currency ? FOR DIVISION. CASE I. 407. "When tlie divisor is an aliquot part of some higher unit. 1. Divide 260 by 3 J, and 1950 by 25. OPERATION. Analysis. Since 3J is J of 10, the next 26|0 19|50 higher unit, we divide 260 by 10 ; and hav- 3 and 4 ing used 3 times our true divisor, we obtain no Wo only i 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 hy a unit of the order next higher than the divisor y hy cutting off ^qures from the right. 21 Q 242 SHOET METHODS. II. Talce as manr/ times this quotient as the divisor is contained times in the next higher unit. EXAMPLES FOR PRACTICE. 1. Divide 63475 by 25. 2. Divide 7856 by 1.25. 3. Divide 516 by 33.3|. 4. Divide 16.7324 by 12J. 5. Divide 1748 by .14f. 6. Divide 576.34 by 1.6f. Ans. 2539. Alls. 6284,S. Ans. 12236 OPERATION. CASE IT. 408. When the right hand figure or figures of the divisor are an aliquot part of 10, 100, 1000, etc. 1. Divide 26923661 by 1233J. Analysis. Since 33J is J of 100, we multiply both dividend and divisor by 3, (117, HI), and we obtain a divisor the component fac- tors of which are 100 and 37. "VYe then divide after the manner of contracted divi- sion, (112). 1233J) 26923661 3_ 3^ 37|00) 80771100 (2183,^715. 67 807 111 . Divide 601387 by 1875. OPERATION. Analysis. Multiplying both dividend and divisor by 4, we ob- tain a new divisor, 7500, having 2 ciphers on the right of it. Multi- plying again by 4, we obtain a new divisor, 30000, having 4 ciphers on the right. Then dividing the new 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 annex the fraction thus formed to the integers of the quotient. 1875) 601387 4 4 7500 ) 2405548 4 4 310000) 96212192 320i|||, Ans, RATIO. 243 Erom these illustrations we derive the following: E-ULE. I. Multiply both dividend and divisor hj a number or numbers that will produce for a new divisor a number ending in a cip)her or ciphers. II. Divide the new dividend by the new divisor. Note. — If the divisor be a whole number, or a finite decimnl, 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. Ans. 2^%%%. 8. Divide 7325 by 433J. 4. Divide 5736 by 431.25. Ans. l^^. 5. Divide 42.75 by 566f. 6. Divide 24409375 by .21875. 7. Divide 785 by 3.14f. . Ans. 249^|. RATIO. 4L09. 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 An'thnu'tical Jiati'o, and the relation expressed by the quotient, Geometrical Itutio. 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. 4:11. 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 \^ -, the ratio of 7 to 5 is ^. 4: IS. 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 are called a couplet, 48 S, A Simple Ratio consists of a single couplet; as 5 : 15. 414. A Compound Eatio is the product of two or more sim- ple ratios. Thus, from the two simple ratios, 5 : 16 and 8 : 2, we 5 : 16 82_2 may form the compound ratio 5x8 : 16x 2, or ^^^ X § = |§ = |. 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- ficquent. Thus, the ratio- of 7 to 9 is 7 : 9 or |, and its reciprocal is |. Note. — The quotient of the second term divided by the first is sometimes called n Direct liatio, and the quotient of the first term divided by the second, jin Inverse or Reciprocal Ilatio. 4115. 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. 41*?. Ratio can exist only between like nunabers, 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 4:18. 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 componndj 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. 4H9. 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). AVe have only to sub- stitute the terms antecedent^ consequent^ and rat'iOy for dluuor, dividend^ and quotient^ and these principles become GENERAL PRINCIPLES OF RATIO. Prin. I. Multiplying the consequent multiplies the ratio ; divi- ding the consequent divides the ratio. Prix. II. Multiplying the antecedent divides the ratio ; dividing the antecedent multiplies the ratio. Prin. III. Multiplying or dividing hoth antecedent and conse^ quent hy the same numhcr does not alter the ratio. 4^0. These three princij)les may be embraced in one GENERAL LAW. A change in the consequent hy multiplication or division produr ces a LIKE change in the ratio ; hut a change in the antecedent produces an OPPOSITE change in the ratio. 421. Since the ratifO 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 '/ jTg — 1 ; or, £8 : 7 as 1 : i ; that is, 28 has the same ratio to 7 that 1 has to }. Ans. J. 2. What part of 42 is 6? 3. What is the ratio of 120 to 80 ? Ans, |. 4. What is the ratio of 8^ to 60 ? Ans, 7. 5. What is the ratio of ^^3 to 26 ? 6. What is the ratio of 7^ to 2^? Ans. f?. 7. What is the ratio of J to -j^^ ? A71S. 44. 8. What is the ratio of 1 mi. to 3 fur. ? 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. 1 E. 20 P. to 6 A. 2 11. 30 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 181° to 45' 30'' ? l^^i 2 of 3 13. What part of -^ is ^—^ ? ^ Ans. ^f 3. 113 93 14. What is the ratio of — / to | of A of -^ ? Ans. q%%%. ■ b^ i'6 15. Find the reciprocal of the ratio of 42 to 28. Ans. 1^. IG. 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? A71S. 1|. 19. If the antecedent be ^ of | and the consequent .75^ what is the ratio ? 20. If the consequent be ^G.12J 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. 3 11. 25 P. and the ratio ||, what is the consequent ? Aiis. 6 A. 2 R. 10 P. PROPERTIES OF PROPORTION. 247 PEOPORTION. 4:22. Proportion is an equality of ratios. Thus, the ratios 5 : 10 arid 6 : 12, each being equal to 2, form a proportion. Note. — When four numbers form a proportion, they are said to be propor- tional. 4:23. Proportion is indicated in three ways : 1st. By a double colon placed between the two ratios; thus, 3 : 4 : : 9 : 12 expresses the proportion between the numbers 3, 4, 9, and 12, and 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, I = ^-^ 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. 42«5. Three numbers are proportional when the first is to the second as the second is to the third. Thus, the numbers 4, G, and 9 are proportional, since 4:6 = 6:9, the ratio of each couplet being |, or IJ. 420. When three numbers are proportional, the second term is called the Mean Proportional between the other two. ^*'^lf. if v,'o Iiavc any proportion, as 3 : 15 =r 4 : 20, Then, indicating this ratio by the second method, we have V = ?• Reducing these fractions to a common denominator, 15 X 4 _ 20 X 3 ~12 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, 62 = 4 X 9. Hence, II. The square of a mean proportional is equal to the product of the other two terms. 428. Since in every proportion the product of the means equals the product of the extremes, (^rSy, I), 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 hy one of the meanSj and the quotient will be the other mean. Or, II. Divide the product of the means hy one of the extremes^ and the quotient will he the otlier 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 : 829.25. Ans. 13^ yd. 4. (?) : 21 A. 3 R. 20 P. : : $1260 : $750. Ans. 36 A. 3 R. 6. 7.50:18 = (?):7yVoz. 6. 7 oz :(?):: £30 : £407 2s. lOf d. Ans. 7 lb. 11 oz. 7. ( ? ) : .15 hhd. : : $2.39 : $.3585. Ans. 1 hhd. 8. 1 T. 7 cwt. 3 qr. 20 lb. : 13 T. 5 cwt. 2 qr. = $9.50 : ( ? ). 9. $175.35 : (?) = I : f Ans. $601.20. 10. (?) : $12^ = 2404 • 149VAV Ans. $20|. 11. I yd. :(?):: $1 : $59.0625. Ans. 40^ yd. SIMPLE PROPORTION. Ojn CAUSE AND EFFECT. ^QQ» Every question in proportion may be considered as a comparison of two cau;es and two cfftrti^. Thus, if 3 dollars as V came will buy 12 pounds as an rffecty G dollars as a can^e wi'.l Duy 24 pounds as an rffect. Or, if 5 horses as a cause consume 10 tons as an effect, 15 horses as a cause will consume oO tons as an ("ffect. Causes and effects in proportion are of two kinds — simple and compound. 4*1^. A Simple Cause or Effect contains but one clement; 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 em- ployed, etc. 43 ^« Since liJce 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 hbid^ (4t'7); 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 PROPOETION. STATEMENT. $ $ yds. 8 : 12 = 86 ; 1st cause. 2d cause. 1st effect. 8 X OPERATION. (?)= 12 X 36 ^4^ X -i^p (?) ^ yds. (?) 2d effect. = 54 yd. FIRST METHOD. 1. If S8 will buy 86 yards of velvet, how many yards may be bought for §12? , Analysis. The re- quh*ed term in this ex- ample is an effect ; and the statement is, $8 is to $12 as 3G yards is to ( ? ), or how many yards. Dividing 12 X 3G, the product of the means, by 8, the given extreme, we have ( ? ) = 54 yards, the re- quired term, (428, II). 2. If 6 horses will draw 10 tons, how many horses will draw 15 tons ? Analysis. In this ex- ample a cause is required ; and the statement is, 6 horses is to ( ? ), or how many horses, as 10 tons is to 15 tons. Dividing 15 X 6, the product of the ex- tremes, by 10, the given mean, we have 9, the re- quired term, (428, I). horses. 6 : 1st cause. STATEMENT. horses. tons. (?) = 10 2d cause. 1st effect. tons. 15 2d effect. OPERATION. 3 40 (?) r 9 horses. 4:S4. Hence the following IluLE. I. Arrange, the terms in the statement so that the causes shall compose one covp)let, and the effects the other, putting (?) in the place of the required term. II. If the required term he an extreme, divide tlie product of the means hy the given extreme; if tlie required term, he a mean, divide the product of the extremes hy the given mean. NoTRS. — 1. If the terms of nny eotiplet be of different denoriiinations, they mu8t he reduced to the same unit value. 2. If the odd term be a compound number, it must be reduced either to its lowe.«t unit, or to a fraction or a decimal of its highest unit. ?>. 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^PROPOHTION. 251 number?, 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 fiictors of the dividend on the right. SECOND METHOD. ^3^. 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 tlirce 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- tiplving this third term, or antecedent, by the ratio of the other two, (421,11). 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 $3G^ what will 5 tons cost? OPERATION. Analysis. In this example, 4 $36 X - = §45 Ans. "tons and 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. AVe 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 2X X - = 12 rods Ans. ^^^ ^^^ ^ ^^^^^ ^^^ ^^^^ ^^^^ 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, 4, and obtain by multiplication, 12 rods, the answer. 4:36. Hence the following Rule. I. With the two given numbers ^ icMch are of the same name or Mnd^ form a ratio greater or less than 1, according as the answer is to he greater or less than the third given number. { II. Multiple/ the third number by this ratio ; 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 rOR PRACTICE. 1. If 12 gallons of wine cost $80, what will 68 gallons cost? /e) / 2. If 9 bushels of wheat make 2 barrels of flour, how many barrels of flour will 100 bushels make ? Aiis. 22|. 8. 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 J bushels of oats cost $8, what will 9i bushels cost? 'i-^ 5. What will 87.5 yards of cloth cost, if 1| yards cost $.42 ? 9.), 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 | of the time ? Ans. 5. 8. If 100 yd. of broadcloth cost $473.07^3, how much will 3.25 yd. cost? f /<:^7 ( V O'O y^ fV 9. If 1 lb. 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 I of a bushel of peaches cost $^|, what part of a bushel can be bought for $3/^ ? " A ns. ^^ bu.' COMPOUND PROPORTION. 253 12. If the annual rent of 46 A. 3 E, 14 P. of land be $374.70, how much will be the rent of 35 A. 2 E. 10 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 ?J y f(^ w^j 14. Two numbers are to each other as 5 to 7}, and the less is 164.5, what is the greater? Ans. 246.75. 15. If 16 head of cattle require 12 A. 3 E. 36 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 60.5 ft. long and 44 ft. wide, if 14.25 sq. yd. cost $34^? ;_ 18. At 6i cents per dozen, w^hat will be the cost of 10 J gross of steel pens? V^^ /^ , S-fh 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. I 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^e 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 efifect. 2d effect. (3= 14 = 40 : (?) Or, 8 X 3 : 9 X 4 = 40 ■ (?) 22 254 PROPORTION. OPERATION. (n-m^.m...... Analysis. In this ex- ample the required term 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 ( ? ) =r 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 cause. 2d cause. 1st effect. 2d effect. f 6 . I (?) ..f 36 . I 1 14 • 1 12 ••! 56 • X 48 54 OPERATION. (?) Analysis. In this example, an element of the second cause is required ; and the state- ment is, 6 compositors 14 hours is to ( ?) com- positors 12 hours as 36 pages of 56 lines each 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 we obtain ( ? ) == 9. 4:3§. From these illustrations we deduce the following Hule.^ I. Of the given tcrmSy 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 hlank term (?) occur in either of the extremeSj divide the j)'^oduct of the means hy the product of the extremes; hut if the hlanh term occur in either mean, divide the product of the extremes hy the product of the means. I^OTKS. — 1. The causes must "he exactly alike in the »innj6erand hhxd of their terms : the same is true of the effects. 2. The same preparation of the terms by reduction is to be observed as in simple proportion. f ■I the L COMPOUND PllOPOETIOK. 255 SECOND METHOD. 4l«l^. 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 odd 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 84320 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, when the rations are 12 ounces per day ? OPERATION, men. days. ounces. §4320 X M X II X If = $4896 BY CANCELLATION. ANALYSIS. In this example there are 4320 three pairs of terms, or couplets, viz., 32 24 men and 24 men, 18 days and 34 days, 15 34 ounces and 12 ounces ; and there is an odd 12 term, $4320, which is of the same kind as ( ) = $4896 Arts, 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 ^4. 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- tion ^5, and we have $4320 X Jf X f^ Next, if it cost $4320 to supply a garrison with rations of 15 ounces, will it cost more, or less, Avhen the rations are 12 ounces? Less; consequently, the multiplier is the proper fraction \Z, and we have $4320 X Jl X ?^ X j^ =$4896, the required term. Hence the following (0 32 18 15 256 PROPORTION. Rule. I. Of the terms com2?osing each couplet form a ratio greater or less than 1, in the same manner as if the aiisicer de- pended on those two and the third or odd term. II. Multiply together the third or odd tei^i and these ralios ; the product will he the ansiver sought. EXAMPLES FOR PRACTICE. 1. If 12 horses plow 11 acres in 5 days, how many horses would plow 33 acres in 18 days? Ans. 10. 2. If 480 bushels of oats will last 24 horses 40 days, how long will 300 bushels last 48 horses,^ at the same rate ? Ans. 12-2 days. 3. If 7 reaping machines can cut 1260 acres in 12 days, in how many days can 16 machines reap 4728 acres ? Ans. 19.7 days. 4. If 144 men in 6 days of 12 hours each, build a wall 200 ft. long, 3 ft. high, and 2 ft. thick, in how many days of 7 hours each can 30 men build a wall 350 ft. long, 6 ft. high, and 3 ft. thick? Ans. 259.2 da. 5. In how many days will 6 persons consume 5 bu. of potatoes, if 3 bu. 3 pk. last 9 persons 22 days ? 6. How many planks lOf ft. long and IJ in. thick, are equiva- lent to 3000 planks 12 ft. 8 in. long and 2$ in. thick? Aus. 6531}. 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? Ans. 1256 bu. 8. If 468 bricks, 8 inches long and 4 inches wide, are 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 17J ft. long, lOJ 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 bbK 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? I PROMISCUOUS EXAMPLES. 257 PROMISCUOUS EXAMPLES IN PROPORTION. 1. If a staff 4 ft. long cast a shadow 7 ft. in length, what is the hight of a tower that casts a shadow of 198 ft. at the same time? Ans. 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 $11.33J? Ans. $7.08+. 3. If 3 cwt. can be carried 660 mi. for $4, how many cwt. can be carried 60 mi. fur $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? Arts. 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 J 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 ? Ans. 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 17i 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 § 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 ? Ai2s. 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* B 258 PROPORTION. 15. What quantity of lining f yd. wide, will it require to line 91 yd. of cloth, 1} yd. wide ? , Ans. 15| yd. IG. 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 IGj'^g cords of wood last as long as Il^^g tons of coal, how many cords of wood will last as long as I5/3 tons of coal? 18. A miller has a bin 8 ft. long, 44 ft. wide, and 2i 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. 7/3 ft. 19. If 4 men in 2 J da3^s, mow 6 J acres of grass, by working 81 hours a day, how many acres will 15 men mow in 3| days, by working 9 hours a day? Ans. 40}^ 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 th% same time ; in how many hours will it be filled ? 22. A merchant in selling groceries sells \^j% 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 lb. of sugar costs $.62 J, and 8 lb. of sugar are worth 5 lb. of coffee, how much will 75 lb. of cofiee 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 859| 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 5 J days of 12 hours each, dig a ditch of 7 degrees of hardness, 232i yd. long, 3f yd. wide, and 2 J yd. deep; in how many days of 9 hours each, will 24 men dig a ditch of 4 degrees of hardness, 387i yd. long, 5^ yd. wide, and 3 J yd deep? Ans. 155 NOTATION. 259 PERCENTAGE. 4:4:0. Per Cent, is a contraction of the Latin phrase per centum, and signifies hy 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. 44^. The character, %, is generally employed in business transactions to represent the words per cent. ; thus G % signifies G per cent. 44^. Since any per cent, is some number of hundredths, it is properly expressed by a decimal fraction; thus 5 per cent. =z b (fo =■ -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. = 6 per cent. =» 6 per cent. = 7 per cent. = 8 per cent. = 10 per cent. = 20 per cent. = 25 per cent. = 50 per cent. = 100 per cent. = 125 per cent. = i per cent. = £ per cent. = 12^ per cent. = Syinlx)ls. 1 % 2 % 4 /. 5 % 6 % 7 % 8 % 10 % 20 % 25 % 50 % 100 % 125 % 1 t/ — 12^ fc Decimals. .01 .02 .04 .05 .06 .07 .08 .10 .20 .25 .50 1.00 1.25 .001 .OOJ .121 Common trac tions. ih = Toff T?ir = oV I'U = o-S 5 lOJS = A 6 JUS == i' 7 = Tffff • TUff » /j 1 ro(5 = iV 2f> TOCT = i ?5 = i 5ft TiJff == i 1 00 150 = I Tft(r foTJ = i f 1 25ff T6^ = i 260 PERCENTAGE, EXAMPLES FOR PRACTICE. 1. Express de 16 EXAMPLES FOR PRACTICE. 1. Express decimally 3 per cent. ; 9 per cent. ; 12 per cent. ; xu per cent.; 23 per cent.; 37 per cent.; 75 per cent.; 125 per cent. ; 184 per cent. ; 205 per cent. 2. Express decimally 15 % ; H % ; 4i % ; 5i % ; 8J % ; 20i % ; 251 % ; 35| % ; 24f % ; 130J %. 3. Express decimally i per cent. ; f per cent. ; i per cent. ; I per cent. ; | per cent. ; /^ per cent. ; j3_6^ per cent. ; If^ per cent. ; 10-1- per cent. 4. Express by common fractions, in their lowest terms, 4 % ; 87J % ; 16| % ; ll^ % ; 42| % ; 45/^ % ; 43/, %. 5. What per cent, is .0725 ? Analysis. .0725 = .071 = 7i ^, Ans. 6. What per cent, is .065? Ans. 6 J %. 7. What per cent, is .14375? Ans, 14i %. 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-j\? Ans. /-j- % 16. What % is .003 J^ ? GENERAL PROBLEMS IN PERCENTAGE. 4:43. In the operations of Percentage there are five parts or elements, namely : Rate per cent., Percentage, Base, Amount, and Difference. 4:44« Hate per Cent., or Rate, is the decimal which denotes how many hundredths of a number are to be taken. Notes. — 1. Such expressions as 6 per cent., and 5 "^.are essentinlly dechnah, the words ^icr cent., or the character % , indicating the decimal denominator. 2. If the decimal bo reduced to a common fraction in its loioent terms, this fraction will still be the equivalent ratef though not the rate^«r cent. PROBLEMS IN PERCENTAGE. 261 I 44d, Percentage is that part of any number wliicli 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- ntage. 1. What is 5 % of 360 ? centage, Rule. Note 1.- f actor 8. Analysis. Since 5 ^ of any number is .05 of that number, (442), we multiply the base, 360, by the rate, .05, and obtain the percentage, 18. Or, since the rate is j^iy = oV* ^^^ h.^^^ 3G0 X 2V == 18, the percentage. Hence the fol- lowing Multiply the hase hy the rate. Percentage is always a product, of which the base and rate ar© the OPERATION, 860 __.05 18.00, Ans, Or, 860 X 2^5 = 18, Ans, 1. What 2 What 3. What 4. What 5. What 6. What 7. What 8. What 9. What 10. What EXAMPLES FOR PRACTICE. s 4 per cent, of 250 ? s 7 per cent, of 3500 ? s 16 per cent, of 324 ? s 12^ per cent, of $5600 ? s 9 % of 785 lbs. ? s 25 % of 960 mi.? s 75 % of 487 bu. ? s38i % of 2757 men? 125 % of 756 ? I % of $2864 ? Ans. 10. Ans. 245. Ans. 51.84. A71S. §700. Ans. 865.25 bu. Ans. $5.91. _1 1 tiU* 262 PERCENTAGE. 11. What is 3| % of $856? Ans. §31.39 12. What is I % of I ? Ans. 13. What is 14f % of di ? 1-1. If the base is §375, and the rate .05, what is the percent- age ? Ans. $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 % + (> % +8 % =Zi %. 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 SO % 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, } % loss; Apr., 2 % gain; May, 2} % gain; June, If % loss; July, 1} % gain; Aug., 1 % loss; Sept., 2| % gain; Oct., 4 % gain; Nov., J % loss; Dec, 3 ^ gain. What were the net profits of his business for the year? Ans. §918. ruoBLE:\i II., 450. Given, the percentage and baso^ to find the rate, 1. What per cent of 360 is 18 ? opERATiox. Analysis. Since the pcrcent- 18 -i- 360 = .05 = 5 % age is always the product of the Qj. base and rate, (449), we divide = .05 = 5 % *' * required ratQ, .05 =? 5 fo. Hence the ,8 1 f).^ P. c/ the given percentage, 18, b}^ the ■3 0^ — -iU — • — /o g.^^j^ i^^g^^ 3QQ^ and. obtain the PROBLEMS IN PERCENTAGE. 263 Rule. Divide the percentage hy the base. EXAMPLES FOR PRACTICE. 1. What per cent, of $720 is $21.60 ? Ans. 3. 2. What per cent, of 1500 lb. is 234 lb. ? 3. What per cent, of 980 rd. is 40 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. 6. What per cent, of 7.85 mi. is 5.495 mi.? * Ans. 70. 7. What per cent, of j% is | ? Ans. 75. 8. What per cent, of ^ is t>^^ ? 9. What per cent, of 560 is 80 ? 10. The base is $578, and the percentage is $26.01 ; what is the rate? Ans. 4i %. 11. The base is $972.24, and the percentage is §145.836; what is the rate ? 12. An editor having 5600 subscribers, lost 448; what was his loss per cent? Ans. 8. 13. A merchant owes $7560, and his assets are $4914; what per cent, of his debts can he pay ? Ans, 65. 14. A man shipped 2600 bushels of grain from Chicago, and 455 bushels were thrown overboard during a gale ; what was the rate per cent, of his loss ? 15. A miller having 720 barrels of flour, sold 288 barrels; what per cent, of his stock remained unsold ? Ans. 60. 16. AVhat per cent, of a number is 30 % of | of it ? 17. The total expenditures, of the General Government, for the year ending June 30, 1858, were $83,751,511.57; the expenses of the War Department were $23,243,822.38, and of the Navy Department, $14,7 12, 610.21. What per cent, of the whole cx- ^jense of government went for armed protection ? ^^B Ans. 45J-, nearly. 18. In the examination of a class, 165 questions were sub- mitted to each of the 5 members; A answered 130 of them, B 125, C 96, D 110, and E 160. What was the standing of the class? Ans. 75.27 %. 264 PERCENTAGE. PROBLEM III. 431. Given, the percentage and rate, to find the base. 1. 18 is 5 % of what number? .^x,,T,.mr^^r Analysis. SmcG tliG pGrceiit- OPERATIOX. . ^ age is always the product of the 18 -^ .Oo = 360, Arts. ^^^^ ^^^ ^^^^^ ^^^^^^ ^^ ^j^j^^ Or, the given percentage, 18, by the 18 -7- 3j^0 = 360, Ans. given rate, .05, or J^, and obtain the base, 360. Hence the Rule. Divide the percentage hy the rate. EXAMPLES FOR PRACTICE. 1. ] 8 is 25 % of what number ? Ans. 72. 2. 54 is 15 % of what number? 3. 17.5 is 2J cjo of what number? Ans. 750. 4. 2.28 is 5 % of what number? 5. 414 is 120 % of what number ? 6 6119 is 105} cjo of what number? Ans, 5800. 7. .43 is 71f ^0 of what number? Am. .6. 8. The percentage is $18.75, and the rate is 2} % ; what is the base ? Ans, $750. 9. The percentage is 31}, and the rate 31} % ; what is the base ? 10. I sold my house for $4578, which was 84 % of its cost; what was the cost ? Ans, $5450. 11. A wool grower sold 3150 head of sheep, and had 30 % of his original flock left ) how many sheep had he at first ? 12. A man drew 40 % of his bank deposits, and expended 13 J % of the money thus drawn in the purchase of a carriage worth $116; how much money had he in bank? Ans. $2175. 13. If $147.56 is 13| % of A's money, and 4f % of A's money is 8 % of B's, how much more money has A than B ? Ans. $461.12}. I PROBLEMS IN PERCENTAGE. 265 4. In a battle 4 % of the army were slain upon the field :|itid 5 % of the remainder died of wounds, in the hospital. The difier- ence between the killed and the mortally wounded was 168 ; how many men were there in the army ? Ans. 21000. Note.— 100 % — 4 ^ = 96 %, left after the battle; and 5 % of 96 "2^ =- 4 J %, the part of the army that died of wounds. 15. A owns f of a prize and B the remainder; after A has taken 40 % of his share, and B 20 % of his share, the remainder is equitably divided between them by giving A $1950 more than B ; what is the value of the prize ? Ans. $7800. PROBLEM IV. . 433. Given, the amount and rate, to find tlie base. 1. What number increased by 5 % of itself is equal to 378 ? OPERATION. Analysis. If any number 1 4- 05 = 1 05 ^® increased by 5 ^ of itself 378 _^ 1.05 = 360, Ans, *be amount will be 1.05 times the number. We therefore di- Or, vide the given amount, 378, by 1 J. 1 *— 2 1 1.05, or |jl, and obtain the base. 37g -^ 2 1 — _ 350 ^^5. 360, which is the number re- quired. Hence the BuLE. Divide the amount hy 1 jplns the rate. Note 1. — The amount is always a product, of which the base is one factor, and 1 plus the rate the other factor. EXAMPLES FOR PRACTICE. 1. What number increased by 15 % of itself is equal to 644 ? Ans. 560. 2. A has ^815.36, which is 4 % more than B has; how much money has B ? Ans. $784. 3. Having increased my stock in trade by 12 % of itself, I find that I have $3800 ; how much had I at first ? 4. In 1860 tiie population of a certain city was 39600, which was an increase of 10 % during the 10 years preceding; what was the population in 1850 ? 23 2(56 PERCENTAGE. 5. Mj crop of wheat this year is 8 ^^ greater than my crop of last year, and I have raised during the two years 5200 bushels; what- was my last year's crop? Ans. 2500 bu. KoTE 2. — 1.00+ 1.08 = 2.08. Hence, 5200 bu. = 2.08 % of last year's crop. 6. The net profits of a nursery in two years were $6970, and the profits the second year were 5 % greater than the profits the first year ; what were the profits each year ? Ans. 1st year, $3400 ; 2d year, $3570. 7. If a number be increased 8 %, and the amount be increased 7 foy the result will be 86.67 ; required the number. KoTE 3. — The whole amount will be 1.08 X 1.07 = 1.1556 times the original number. 8. A produce dealer bought grain by measure, and sold it by weight, thereby gaining 1} % in the number of bushels. He sold at a price 5 % above his buying price, and received $4910.976 for the grain ; required the cost. Ans. $4608. 9. B has 6 %, and C 4 % more money than A, and they all have $11160 ; how much money has A ? Ans. $3600. ' 10. In the erection of a house I paid twice as much for mate- rial as for labor. Had I paid 6 % more for material, and 9 % more lor labor, my house would have cost $1284 ; what was its cost ? Ans. $1200. PROBLEM V. 453. Given, tlie difference and rate, to find the base. 1. What number diminished by 5 % of itself, is equal to 342 ? OPERATION. Analysis. If any number be di- -[ .05 = .95 minished by 5 % of itself, the dif- 342 -f- .95 = 360, Ans. fcrence will be .95 of the number. Qr We therefore divide the given diiFer- 1 ^1^ 1= 19 ence, 342, by .95, or ^^, and obtain 342 -I- ' 5 = 360, Ans. the base, 3G0, which is the, requited number. Hence the '; Hule. Divide the difference r^ 1 minus the rate. XoTF. — The difference is always k product, of which the base is one factor, and 1 minus the rate the other. PROBLEMS IN PERCENTAGE. 2"f57 EXAMPLES FOR PRACTICE. 1. "VYhat number diminished by 10 % of itself fs equal to 504? Atis. 5G0. 2. The rate is 8 ^, and the • difference $4.37; what is the base? 3. After taking a^yay 15 % of a heap of grain, there remained 40 bu. S^ pk. ; how many bushels were there at first ? An.9. 48 bu. 4. Having sold SG % of my land, I have 224 acres left; how much land had I at first ? 5. After paying 65 % of my debts, I find that 62590 will dis- charge the remainder ; how much did I owe in all ? Ans. $7400. 6. A young man having received a fortune, deposited 80 ^^ of it in a bank. He afterward drew 20 % of his deposit, and then had §5760 in bank; what was his entire fortune? Ans. ?9000. 7. A man owning | of a ship, sold 12 % of his share to A, and the remainder to B, at the same rate, for $20020; what was the estimated value of the whole ship ? Ans, $26000. 8. An army which has been twice decimated in battle, now contains only 6480 men ; what was the original number in the army? Avs. 8000. ^ 9. Each of two men, A and B, desired to sell his horse to C. A asked a certain price, and B asked 50 % more. A then re- duced his price 20 %, and B his price 30 %, at which prices C took both horses, paying for them $148 ; what was each man's asking B^? 4^^^ (A, $^0. "^ ^ ' (B, $120. ^ " 10. A buyer expended equal sums of money in the purchase of wheat, corn, and oats. In the sales, he cleared 6 % on the wheat, and 3 % on the corn, but lost 17 % on the oats; the whole amount received was $2336. What sum did he lay out in each kind of grain ? Ans, $800. ^68 PERCENTAGE. APPLICATIONS OF PERCENTAGE. 4.54, The principal applications of Percentage, where time is not considered, are Commission, Stocks, Profit and Loss, Insurance, Taxes, and Duties. And since the five problems in Percentage involve all the essential relations of the parts or elements, we have for the above applications the following General Rule. Note what elements of Percentage are given in the example^ and what elemerd is required ; then apply the spe- cial rule for the corresponding case. COMMISSION. 455« An Agent, Factor, or Broker, is a person who trans- acts business for another. 456. A Commission Merchant is an agent .who buys and sells goods for another. 45 y. Commission is the f&e or compensation of an agent, factor, or commission merchant. 458. A Consignment is a qu«intity of goods sent to one person to be sold on commission for another person. 459. A Consignee is a pei^ptn who receives goods to sell for another; and 46®. A Consignor is a person who sends goods to another to be sold. 461* The Net Proceeds of a sale or collection is the sum left, ?ifter deducting the commission and other charges. NoTB. — A person who is employed in establishino^ mercantile relations between other.s living nt a distance from each other, is called the Correspoudenf of the party in whose behalf he acts. A correspondent is the an the property. The OohcIh ist^ued lor tlje.iiit c;ipit;ii or elaiuis upon corporate I'odies. 4. 'J'he members of an incorporated company are individually liable for the debts and obligations of the coin[»any, to the amount ^ompany was able to declare a dividend of 8 % ; how much scrip had the company issued? Ans. $7086676. 13. Having received a stock dividend of 5 %, I find that I own 504 shares ; how many shares had I at first ? Ans. 480. 14. I received a 6 % dividend on Philadelphia City railroad stock, ajid invested the money in the same stock at 75 %. Mj stock had then increased to $16200; what was the amount of my dividend ?--'\ ]-- 1 ^' ' ; vA ^ f 5 ^ ' ^ ' Ans. $900. .15. A ferry company, whose stock is $28000, pays 5 % divi- dends semi-annually.. The annual expenses of the ferry are $2050 ; what are the gross earnings ? Ans. $5750. STOCKS. 279 STOCK IXTEST:\rEXTS.* 4:SS. The net earnings of a corporation are usually divided among the stockholders, in semi-annual dividends. The income of capital stock is therefore fluctuating, being dependent upon the con- dition of business ; while the income arising from bonds, whether of government or corporations, is fixed, being a certain rate per cent., annually, of the par value, or face of tlie bonds. 489. Federal or United States Securities are of two kinds : viz.. Bonds and Notes. Bonds are of two kinds. First, Those which are payable at a fixed date, and are known and quoted in commercial transactions by the rate of interest they bear, thus, : U. S. 6's, that is, United States Bonds bearing 6 % interest. Second, Those which are payable at a fixed date, but which may be paid at an earlier specified time, as the Government may elect. These are known and quoted in commercial transactions by a cond)i- nation of the two dates, thus : U. S. 5-20's, or a combination of the rate of interest and the two dates, thus : U. S. 6's 5-20 ; that is, bonds bearing G % interest, which are payable in twenty years, bub may be paid in five years, if the Government so elect. When it is necessary, in any transaction, to distinguish from each other different issues which bear the same rate of interest, this is done by adding the year in which they become due, thus : U. S. 5's of 71 ; U. S. 5's of 74 ; U. S. 6's 5-20 of '84 ; U. S. 6's 5-20 of '85. Notes are of two kinds. First, Those payable on demand, without interest, known as United States Legal-tender Notes, or, in common language, "Green Backs." * TJie following eight pages contain/owr pages of new matter, on U.S. Secniitics, Bonds, Treasury Notes, Gold Investments, &c., to meet a necessity which did not exist ut tlie time this book was written. The pupil will find the Cases, Rules, and Operations of the previous editions essentially the same in this, with additional examples, and other matter, which may be used or omitted; so that the present may be used with the preyious editions fuith little or no inconvenience. 280 PERCENTAGE. Second^ Notes payable at a specified time, with interest, known as Treasury Notes. Of these, there are two kinds, — Six-per-cent. Compound-interest Notes, and Notes bearing 7i\ % interest, tlie latter known and quoted in commercial transactions as T.CO's. The nomenclature here explained is the one used in commercial transactions, which involve similar securities of States or corporations. The interest on all bonds is payable in gold. The interest on notes is payable in Legal-tender Notes. When Bonds or Stocks are sold, a revenue stamp must be used equal in value to one cent on each $100, or fraction of $100, of their currency value. If sold by a broker, this is charged to the person for w^hom they are sold. The following are the principal United States Securities : — BONDS. U. S. G's of 1867. U. S. G's of 18G8. U. S. G's of 1880. U. S. G's of 1881. U. S. 5's of 1871. U. S. 5's of 1874. U. S. 6-20's, due in 1882, interest 5 %. U. S. 5-20's, due in 1884, interest G %. U. S. 5-20's, due in 1885, interest G %. U. S. 10-40's, due in 1004, interest 5 %. Pacific Railroad G's of 1895. Pacific Ptailroad G's of 1896. NOTES. Compound-interest Notes of 18G7. Compound-interest Notes of 1868. 7.30 Notes of '1867. 7.30 Notes of 1868. STOCKS. 281 CASE I. 490. To find what income any investment will pro- duce. 1. What income will be obtained by investing $G840 in stock bearing 6 % , and purchased at 95 % ? OPERATION. Analysis. We di- $8840 - .95 = $7200, stock purchased, l'^^'^? i-^^^^^"^; $7200 X .06 = $432, annual income. ^^^^^' \^^^ ^^^* ^^ Si, and obtain $7200, the stock which the investment will purchase, (452). And since the stock bears 6 % interest, we have $7200 X -OG == $432, the annual income obtained by the investment. Hence, Rule. — Fmd how much stock the investment will purchase, and then compute the income at the given rate upon the par value, EXAMPLES FOR PRACTICE. 1. The trustees of a school invested $35374.80 in the U. S. 5 % bonds as a teachers' fund, purchasing the stock at 102J^ if the salary of the Principal be $1000, what sum will be left to pay assistants? Ans. $725. CO. 2. A young man, receiving a legacy of $48000, invested one half in 5 % stock at 95 J %, and the other half in G % stock at 112 %, paying brokerage at J % ; what annual income did he secure from his legacy? Ans. $2530. 3. I have S2300 to invest, and can buy New York Central 6's at 85 %, or N^w York Central 7 's at 95 % ; how much more prof itable will tlj9 latter be than the former, per year ? 4. A owns a farm which rents for $411.45 per annum. If he sell the same for $8229, and invest the proceeds in U. S. 5-20 's of '84, at 105 %, paying \ % brokerage, will his yearly income be increased or duninished, and how much ! Ans. Increased $56.55. 5. A sold $8700 of U. S. 5-20's of '84 at 104 %, paying for necessary revenue stamps, and invested the proceeds in U. S. 10-4 O's at 94 % , brokerage \ % both for selling and buying. Did he gain or lose by the exchange, and how much annually? Ans. $45.62—. 282 PERCENTAGE. CASE II. 491. To find what sum must be invested to obtain a given income. I. What sum must be invested in Virginia 5 per cent, bonds, purchasable at 80 % , to obtain an income of $600 ? OPERATION. Analysis. Since $600 -~ .05 = $12000, stock required. ^^ ^^' ^^^ stock will $1200 X .80 =z $9600, cost or investment, obtain S.05 income, to obtain SGOO will require $G00 -^.05 = $12000, (Case 1). Multiplying the par value of the stocks by the market price of $1, we have $12000 X -80 = S9600, the cost of the required stock, or tho sum to be invested. Hence the IluLE. I. Divide the given income hy the % which the stock pays ; the quotient will he the par value of the stock required. II. Multiply the par value of the stock hy the market value of one dollar of the stock / the product will he the required investment, EXAMPLES FOR PRACTICE. 1. If Missouri State O's are 16 % below par, what sum must bo invested in this stock to obtain an income of $960 ? 2. What sum must I invest in U. S. 5-20's of '82 at 962 %» brokerage ^- %, to secure an annual income of $1500. Ans. $29100. 3. How much must I invest in U. S. 7-30's, at 106 %, that my annual income may be $1752? Ans. $25440. 4. If I sell $15600 U. S. 10-40's at 97 %, and invest a suf- ficient amount of the proceeds in U. S. 5-20's of '85 at 107 % to yield an annual income of $540, and buy a house with the re- mainder, how much will the house cost me ? Ans. $5502. 5. Charles C. Thomson, through his broker, invested a certain sum of money in U. S. 6's 5-20 at 107 %, and twice as much in U. S. 10-40's at 98 J %, brokerage in each case I %. His in- come from both investments was $1674. How much did ho invest in each kind of stock ? Ans. First kind, $10692. Second kind, $21384. STOCKS. 283 CASE III. 492. To find what per cent, the income is of the in- vestment, when stock is purchased at a given price. 1. What per cent, of mj investment shall I secure by purchasing the New York 7 per cents, at 105 % ? Analysis. Since Si of the stock . OPERATION. will cost $1.05, and pay S.07, the in- .07 -r- 1.05 = Gf %. come is ^l^ = 6f % of the invest- ment. Hence the PtULE. Divide the annual rate of mcovfie which the stock hears hy the price of the stock ; the quotient will he the rate vpon the in- vestment. EXAMPLES rOPv PIIACTICE. 1. What per cent, of his money will a man obtain by investing in G per cent, stock at 108 % ? Ans. 5f %. 2. What is the rate of income upon money invested in G per cent, bonds, purchased at a discount of IG % ? Ans. 7-J- %. 3. Panama railroad stock is at a premium of 34|^ %, and the charge for brokerage is IJ % ; what will be the rate of income on an investment in these funds if the stock pays a dividend of %\ % annually? Ans. 6 J %. 4. Which is the better investment, to buy 5's at 70 %, or G's at 80 % ? 5. Which is the more profitable, to buy 8's at 120 %, or 5's a.t 75%? G. What is the rate of income upon money invested in U. S. 7-30's at lOG % ? Ans. GfJ %. 7. Which is the better investment, U. S. 5-20's of '8-1 at 108^ %, or U. S. 10-40's at 98 %, and how much per cent, per annum? ' Ans. U. S. 5-20's, -i%VV %• 8. If a man invest $10000. in U. S. 10-40's at 08 %, and ex- changes them at par for U. S. 7-30's at 102 %, what is his rate of income? ^ 9. What per cent of his money will a man gam by investing in Pacific Ptailroad G's at 105 % ? 284: PERCENTAGE. CASE IV. 493. To find the price at which stock must be pur- chased to obtain a given rate upon the investment. 1. At -wbat price must 6 per cent, stocks he purchased in order to obtain 8 % income on the investment ? OPERATION. Analysis. Since $.06, the in- itio Ar> AQ c>^^ <^ome of SI of the stock, is 8 ^ of 5).UO -r- .UO = Jt?iO. . • 1 /> -^ 1 ^AAn\ the sum paid lor it, we have, (449), $.06 -i- .08 = $75, the purchase price. Hence, HuLE. Divide the annual rate of income which the stock hears hy the rate required on the investment ; the quotient will he the price of the stock, EXAMPLES FOR PRACTICE. 1. What must I pay for Government 5 per cents., that my in- vestment may yield 8 % ? Ans. C2i- %. 2. At what rate of discount must the Vermont G per cent, bonds he purchased that the person investing may secure C^ % upon his money? Ans, 4 %. 3. What rate of premium does 7 per cent, stock bear in the mar- ket when an investment pays G % ? 4. A speculator invested in a Life Insurance Company, and re- ceived a dividend of G %, which was 8J^ % on his investment; at wbat price did he purchase? Ans, 72 %. 5. What must I pay for U. S. 10-40's, that my investment may yield 6 %? Ans, 83^ %. G. What rate of premium docs U. S. G's 5-20 bear in market when an investment pays 5 % ? 7. At wbat rate of discount must U S. 7-30's be purchased, that the investment sball yield 10 % ? 8. What must I pay for government G's of '81, that my invest- ment may yield 7 % ? I STOCKS. 285 GOLD INVESTMENTS. 493 a* Currency is a term used in commercial language, First, To denote the aggregate of Specie and Bills of Exchange, Bank Bills, Treasury Notes, and other substitutes for money employed in buying, selling, and carrying on exchange of commodities between various nations. Second, To denote whatever circulating medium is used in any country as a substitute for the government standard. In this latter sense, the paper circulating medium, when below par, is called Currency, to distinguish it from gold and silver. If, from any cause, the paper medium depreciates in value, as it has done in the United States, gold becomes an object of investment, the same as stocks. In commercial language, gold is represented as rising and falling ; but gold being the standard of value, it cannot vary. The variation is in the medium of circulation substituted for gold ; hence, when gold is said to be at a premium, the currency, or circulating medium, is made the standard, while it is virtually bolow par. CASE I. To change gold into currency. 1. How much currency can be bought for $150 in gold when gold is at 170 % ? OPERATION. Analysis. Since a dollar of gold is nd, $208. PROFIT AND LOSS. 289 * 16. GafTney, Burke & Co. bought a quantity of dry goods for $6840; they sold } of them at 15 per cent, profit, i at 18i per cent., i at 20 per cent., and the remainder at 33 J per cent, profit; how much was- the average gain per cent., and how much the whole gain? Ans. 21| % gain; $1482, entire gain. •17. If I buy a piece of, land, and it increases in value each year at the rate of 50 per cent, on the value of the previous year, for 4 years, and then is worth $12000, how much did it cost ? . 18. A Western merchant bought wheat as follows : 600 bushels of red Southern @ $1.80, 1200 bushels of white Michigan @ $1.62^, and 200 bushels of Chicago spring, @ $1.25. lie shipped the whole to his correspondent in Buffalo, who sold the first two kinds at an advance of 20 % in the price, and the bal- ance at $1.20 per bushel, and deducting from the gross avails his commission at 5 %, and $254.60 for expenses, returned to the consignor the net proceeds. What was the rate of the merchant's gain ? Ans. ^\ ^c 19. A broker buys stock when it is 20 *y^ below par, and sells it when it is 16 % below par; w^hat is his rate of gain ? 20. A man has 5 per cent, stock the market value of which is 78 % ; if he sells it, and takes in exchange 6 % stock at 4 ^ premium, what per cent, of his annual income does he lose? ^ • 21. A machinist sold 24 grain-drills for $125 each. On one half of them he gained 25 per cent., and on the remainder he lost 25 per cent.; did he gain or lose on the whole, and how much? ^ns. Lost $200. B 22. Bought land at $30 an acre ; how much must 1 ask an acre, that I may abate 25 per cent, from my asking price, and still make 20 per cent, on the purchase money ? Am. $48. H^ 23. A salesman asked an advance of 20 per cent, on the cost of some goods, but was obliged to sell at 20 per cent, less than his asking price; did he gain or lose, and how much per cent.? \^ 24. A Southern merchant ships to his agent in IJdston, a quan- tity of sugar consisting of 200 bbl. of New Orleans, each containing 216 lb., purchased at 5 cents per pound, and 560 bbl. of West India, each containing 200 lb., purchased at 5 J cents per pound.- 25 T 290 PERCENTAGE. The agent's account of sales shows a loss of 1 % on the New Or- leans, and a profit of |§ % on the West India sugar ; does the merchant gain or lose on the whole consignment, and what per cent.? Ans. Gains | %. 25. ^A grocer sold a hogshead of molasses for $31.50, which was a reduction of 30 % from the prime cost ; what was the pur- chase price paid per gallon ? 26. A speculator sold stock at a discount of 7| %, and made a profit of 5 % ; at what rate of discount had he purchased the stock? Ans, 12 %. 27. A dry-goods merchant sells delaines for 2^ cents per yard more than they cost, and realizes a profit of 8 % ; what was the cost per yard? Ans. $.31|. 28. If I make a profit of 18| % by selling broadcloth for $.75 per yard above cost, how much must I advance on this price to realize a profit of 31| % ? 29. A speculator gained SO % on | of his investment, and lost 6 % on the remainder, and his net profits were $720. What would have been his profits, had he gained 30 ^ on | and lost 6 % on the remainder? Ans. $405. 30. A man wishing to sell his real estate asked 36 per cent, more than it cost him, but he finally sold it for 16 per cent, less than his asking price. He gained by the transaction $740.48. How much did the estate cost him, what was his asking price, and for how much did he sell it ? Ans, Cost, $5200 ; asking price, $7072; sold for $5940.48. 31. Sold I of a barrel of beef for what the whole barrel cost; what per cent, did I gain on the part sold ? 32. Bought 4 hogsheads of molasses, each containing 84 gal- lons, at $.37 J a gallon, and paid $7.50 for freight and cartage. Allowing 5 per cent, for leakage and waste, 4 per cent, of the sales for bad debts, and 1 per cent, of the remainder for collecting, for how much per gallon must I sell it to make a net gain of 25 per cent, on the whole cost? Ans. $.55 + . INSURANCE. 291 INSURANCE. 4©C5. Insurance is security guaranteed by one party to ano- ther, against loss, damage, or risk. It is of two kinds; insurance on property, and insurance on life. 40?^. The Insurer or Underwriter is the party taking the risk. 498. The Insured or Assured is the party protected. 400. The Policy is the written contract between the parties. ^00. Premium is the sum paid for insurance. It is always a certain per cent, of the sum insured, varying according to the degree or nature of risk assumed, and payable annually or at stated intervals. NoTKS. — 1. Insurance business is generally conducted by joint stock compa- nies, though Siunetimes by individuals. 2. A 3/ntnal luHurauce company is one in which each person insured is enti- tled to a share in the profits of the concern. 3. The act of insuring is sometimes called taking a risk. FIRE AND MARINE INSURANCE. '" ^01. Insurance on property is of two kinds; Fire Insurance and Marine Insurance. Fire Insurance is security against loss of property by fire. Marine Insurance is security against the loss of vessel or cavgo by the casualties of navigation. ^O^. The Sum Covered by insurance is the difference be- tween the sum insured and the premium paid. Notes. — 1. As security against fraud, most insurance companies take risks at not more than two- thirds of the full value of the property insured. 2. When insured property suffers damage less than the amount of the policy, the insurers are required to pay only the estimated loss. ^03, The calculations in insurance are based upon tho fol- lowing relations : I. Premium is perrenta//e (44:3). 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 mj stock of goods to the amount of ?5760 at li % ? 292 PERCENTAGE. OPERATION. Analysis. According to $5760 X .0125 == $72, Ans. Prob. I, (449), we multiply $5760, the base of premium, by .0125, the rate, and obtain $72, the premium. 2. For what sum must a granary be insured* at 2 % in order to cover the loss of the wheat, valued at $1617 ? OPERATION. Analysis. According to 1.00 .02 = .98 P^^ob. V, (453), we divide the $1617 ~- .98 = $1650, Ans, sum to be covered, $1617, which is difference, by 1 minus the rate of premium, and obtain $1650, the base of premium, or the sum to be insured. Proof. $1650 X .02 = $33, premium ; $1650 — $33 = $1617, the sum covered. 3. What must be paid for an insurance of $5860 at 1 J % ? 4. What is the premium of $860 at } % ? Am. $4.30. 5. What is the premium for an insurance of $3500 on my house and barn, at IJ % ? Ans. $43.75. 6. A fishing craft, insured for $10000 at 2J %, was totally wrecked ; how much of the loss was covered ? A^is. $9775. 7. A hotel valued at $10000 has been insured for $6000 at li %, $5.50 being charged for the policy and the survey of the premises ; if it should be destroyed by fire, what loss would the owner suffer? Ans. $4080.50. 8. A merchant whose stock in trade is worth $12000, gets the goods insured for | of their value, at J % ; if in a conflagration he saves only $2000 of the stock, what actual loss will he sustain ? 9. If I take a risk of $36000 at 2J %, and re-insure J of it at 3 %, what is my balance of the premium? Ans. $360. 10. I pay $12 for an insurance of $800 ; what is the rate of premium? A71S. \\ ^q. 11. A trader got a shipment of 500 barrels of flour insured for 80 ^c of its cost, at 3} %, paying $107.25 premium; at what price per barrel did he purchase the flour? Ans. $8.25. 12. The Astor Insurance Company took a risk of $16000, for a premium of $280 ; what was the rate of insurance ? 13. A whaling merchant gets his vessel insured for $20000 in INSURANCE. 293 the Gallatin Company, at f %, and for $30000 in the Howard Company, at J % ; what rate of premium does he pay on the whole insurance? Ans. | %. 14. If it cost $46.75 to insure a store for J of its value, at 1| ^, what is the store worth? Ans. $6800. 15. For what sum must I get my library insured at H %, to cover a loss of $7910 ? Ans. $8000. 16. What will be the premium for insuring at 2| %, to cover $27320? Ans. $680. 17. A shipment of pork was insured at 4f %, to cover | of its value. The premium paid was $122.50; what was the pork worth? Ans. $4480. 18. A gentleman obtained an insurance on his house for f of its value, at li % annually. After paying 5 instalments of pre- mium, the house was destroyed by fire, in congcqueiice of which he suffered a loss of $2940 ; what was the value of the lum^e ? Ans. $960.0. 19. A man's property is insured at 2 J % payable annually; in how many years will the premium equal the policy ? 20. A company took a risk at 2} %, and re-insured | of it in another company at 2 J %. The premium received exceeded the premium paid by $72. What was the amount of the risk? 21. The Commercial Insurance Company issued a policy of insurance on an East India merchantman for f of the estimated value of ship and cargo, at 4} % , and immediately re-insured J of the risk in the Manhattan Company, at 3 %. During the out- ward voyage the ship was wrecked, and the Manhattan Company lost $1350 more than the Commercial Company; what did the owners lose? ' Ans. $40590. LIFE INSURANCE. ^©-4. Life Insurance is a contract in which a company agrees to pay a certain sum of money on the death of an individual, in consideration of an immediate payment, or of an annual premium paid for a term of years^ or during the life of the insured. The 25* 294 PERCENTAGE. policy may be made payable to the heirs of the insured, or assured, person, or to any one whom he may designate. ^^e>. The policies issued by life insurance companies are of various kinds, the principal of which are as follows : 1st. Term policies, payable on the death of the insured, if the death occurs during a specified term of years ; these require the payment of an annual premium till the policy matures or expires. 2d. Life policies, payable on the death of the insured, the annual premium to continue during life. 8d. Life policies, payable on the death of the insured, the annual premium to cease at a given age. 4th. Endowment assurance policies, payable to the assured person on his attaining a given age, or to his heirs if his death occurs before that age, annual premium being required till the policy matures. NoTK. — The premium on tlie firpt nrd peconri cln?ses of policies innj he dis- cliar it will be seen that the legal rate of interest in -^2 States is 6 per cent. This is a sufficient reason for introducing the following brief method into this work : Analysis. At 6 J/^ per annum the interest on $1 For 12 months is $.06. *' 2monihs(i% = Jofl2mo.) " .01. '* 1 month, or 30 days (y^ of 12 mo.) " .0OJ = $.005 (^j of $.06). " 6 days (I of 30 da.) *' .001. ** 1 " (i of 6 da. = jV of 30 da.) " .OOOJ. Hence we conclude that, 1st. The interest on $1 is $.005 per month, or §.01 for every 2 months; 2d. The interest on §1 is g.OOOi per day, or ?.001 for every 6 days. From these principles we deduce the lluLE. I. To find the rate: — Call ever?/ year §.06, every 2 montlu $.01, every 6 days $.001, and any less number of days sixths of 1 niill. II. To find the interest : — Multiply the jprincijyal by the rate. Notes. — 1. To find the interest at any other rate ^ by this method, first find it at 6 ^ot and then increase or c'lmin'sh the result by as many sixths of itself as the given rate is units greater or lc*i«s than G ^. Thus, for 7 J^ add ^, fur 4 ^ subtract J, etc. 2. The interest of $10 for 6 days, or of $1 for 60 days, is $.01. Therefore, if tho principal be less than^ $10 and the time less than 6 days, or the principal less than $1 and the time less than 60 days, the interest will be less than $.01, and may be disregarded. 3. Since the interest of $1 for 60 days is $.01, the interest of $1 for any num- * This method of finding the interest on $1 by inspection was first published in The Scholar's Arithmetic, by Daniel Adams, M. D., in ISOl, and from its simplicity it has come into very general use. SIMPLE INTEREST. 311 ber of days is as many cents as 60 is contained times in the number of days. Therefore, if any principal be multiplied by the number of days in any given number of months and dnys, and the product divided by 60, the result will be the interest in cents. That is, Mnltiplji the pritio'jxrl bij the uiinihvr of dnj/H, divide the pntdiict by 60, aiid point off two decimal. pfficen in the quotient. The result will be the interest iu the name denominulion as the principal, EXAMPLES FOR PRACTICE. What is the interest on the following sums for the times given, at 6 per cent. ? 1. $325 ror 3 years. Ans. $58.50, 2. $1600 tor 1 yr. 3 mo. Ans. $120. 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 at 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. S1.228+. 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 l*yr. 3 mo. 17. $647,654 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. I 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. $j600 for 3 yr. 2 mo. 17 da. Ans, $154.266f. 312 PERCENTAGE. 26. $1700 for 28 da. Ans. $10-58-. On the following at 10 per cent.? 27. §3045.20 for 7 mo. 15 da. ' ' Ans, §190.32-f . 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 6} per cent. ? 34. What is the interest of $970.99 for 6 mo., at 5i 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 4i per cent.? Ans. $1,353+. 37. What is the amount of $145.20 for 1 yr. 9 mo. 27 da., at 12i per cent.? ^«s. $178.32375. 38. What is the amount of $215.34 for 4 yr. 6 mo., at 3 J per cent.? Ans. $U9.256+. 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 4i 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 % ? 42. If a loan of $3155.49 be made Aug. 15, 18§8, at 6 per cent., what amount will be due May 1, 1866, no interest having been paid? 43. How mucb 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 in 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 J. 47. May 10, 1859, I borrowed $3840, with which I purchased Sour at $5.70 a barrel. June 21, 1860, 1 sold the flour for $6.62} 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 to 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? A71S. $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 hi« business, and loans his money at 7i %. How much does he lose in 2 yr. 5 mo. 10 da. by the change ? Ans. S2535.86f. 52. A speculator wishing to purchase a tract of land containing 450 acres at $27.50 an acre, borrows th3 money at 5} per cent. At the end of 4 yr. il mo. 20 da. he sells | 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} 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 ? 1' 27 814 PERCENTAGE. TARTIAL PAYMENTS OR INDORSEMENTS. «541:9. A Partial Payment is payment in part of a note, bond, or oilier obligation. 41 4S. 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. ^•J.3. 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 ot 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 balance of the principal remaining due. III. '^ If the payment be less than the interest the surplus of interest miist 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 Kule. I. Find the amount of the given principal to tlie tirm of Hie first iiaymentj and if this payment exceed the interest then diie^ TMhiract it from the amount obtained, and treat the remainder as a new j)ri'nrj'pal. II. But if the interest he 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 oj the 'payments from the amount of the principal ; the remainder will form a new principal ^ with which proceed as he/ore, EXAMPLES FOR PRACTICE. flOOO. 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 Sura 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. $^^0^- Richmond, Va., Oct. 15, 1859. 2. One year after date we promise to pay Jatnes Peterson, or order, twelve hundred dollars, for value received, with interest. Wilder k Son. Indorsed as follows: Oct. 15, 1860, $1000; April 15, 1861, §200. How much remained due Oct. 15, 1861 ? Ans, ?82.56. ^^50//^. Boston, June 10, 1855. 3. Eighteen months after date I promise to pay Crosby, Nich- ols & Co., or order, eight hundred fifty and j'y^ 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 ? $2^j¥Tr Savannah, Ga., Sept. 4, 1860. 4. Six months after date I promise to pay John Rogers, or order, three hundred eighty-four and -j^^^^ dollars, for value re- ceived, with interest. \Vm. Jenkins. This note was settled Jan. 1, 1862, one payment of $126.50 having been made Oct. 20, 1861 ; Ibow much was due at the time of Settlement ? ^^^"'^- New Orleans, March 6, 1857. 5 On demand we promise to pay Evans & Hart, or order, three thousand four hundred seventy-five dollafrs, 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? PARTIAL PAYMENTS. 317 g5^Q» 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, $10; Nov. 14, 1861, $8; April 1, 1862, $12; May 1, 1862, $30. How much was due Sept. 16, 1862? Ans. $455.57+. 544. Connecticut Rule. I. Payments made one year or more from the time the interest ■ commenced J or from another payment ^ and 'payments, less lluin the interest due, are treated according to the United Sfates rule. II. Payments exceeding the interest due and made within one year from the time interest commenced , or from a former paymc7it, shall draw interest for the balance of the year^ p)rovidf'd the interval does not extend hey and the settlment^ and the amount mwt be sub- tracted from the amount of the principal for otie year; the re- maijider will be the new principal. III. If the year extend beyond the settlemrnt^ 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. •145. A note containing «- promise to pay interest annually is not considered in law a contract for any thing more than simple interest on the principal. Fir 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. Fiiid the amount of each payment from the time it icas made to the time of settlement. III. Subtract the sum of the amounts of the payments from the amount of the pi'incipal ; the rc-niainder will be the sum due. _NoTE. — This rule is in quite extensive use among merchants and others. 27* 318 PERCENTAGE. S4rG« 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 he/ore 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 he 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^f interest a part of a year, accordingly, EXAMPLES FOR PRACTICE. ^IQQQ - New Haven, Conn, Feb. 1, 185G. 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. $2000. Burlington, Yt., 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 ? Ans. $831.58. 4. How much, computing by the New Hampshire rule? By the United States rule ? . f N. H. rule, $833.21; ""'' 1 U. S. " $831.90. SAVINGS BANK ACCOUNTS. 319 SAVINGS BANK ACCOUNTS. 547. Saving's 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. 548. It is the custom of all savings hanks 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. 540. A savings bank furnishes each depositor with r. book, in which is recorded from time to time the sums deposited and the sums drawn out. The Dr. side of such an account shews 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 amount due on any depositor's account, we have the following Rule. At the end of each term^ add to the hcdance of the account one term^s interest on the smallest balance on deposit at any one time durincj that term j the final balance thus obtained loill be the sum due. Notes. — 1. It will be seen that by this rule no interest is allowod for money on deposit during a partial term, whether the period be the first or tlio last part of the term. 2. An exception to this general rule occurs in the practice of some of tlio savings banks of New York city. In these, the interest term is G moiitlis, and the depositor is allowed not only the full term's interest on the smallest. Inlanco, but a half term's interest on any deposit, or portion of a deposit made during the first 3 months of the term, and not dratcn out duriuij ant/ subsequent 2)art 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, tho terms commencing Jan. 1, April 1, July 1, and Oct. 1 ? Dr. Savings Bank in account with James Taylor. Cr. 1858, Jan. 12,... S75 1858, March 5, ^-30 " May 10, 150 " Aug. 16, 50 . '^ Sept 1, 20 " Dec. 1, 48 1859, Feb. 16, 130 320 PERCENTAGE. OPERATION. Deposit, Jan. 12, 1858, $75 Draft, March 5, " _30 Balance, Apr. 1, 1860, $45 Deposit, May 10, 1858, 150 Int. on $45, for 3 mo 68 Balance, July 1, 1860, $1-95.68 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 Note.- Jan. 1. Bal. due after Apr. 1, 1860, $253.50 ^n^. -In the following examples the terms commence with the year, or on 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 ? Dr. Bay State Savings Institution, in account with Jane Ladd. Cr. 1860. Jan. 3 «( 8 i< 20 Feb. 20 27 March 6 29 April May 25 7 30 July Aug. 28 3 u 26 To cash, " check, " cash, " check, " cash, " draft, H ii " cash, " check, *' cash, 5 75 13 45 7 60 16 45 8 40 14 65 7 98 3 49 26 60 45 79 15 68 18 45 4 60 1860. Jan. Feb. March April June Aug. 28 7 20 11 3 12 20 17 By check, " draft, " check, k, 5 00 8 48 10 00 12 76 3 96 10 48 , 17 48 ^, 5 64 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 Detroit Savings Institution, in account tvith E. L. SeJclen, Cr 1S<,9. J n. 1 M.-.rch 12 June 20 Aujr. 3 isco. Jan. 25 To cash, }^.9. 47 ro May 1-24 30 Oct. 13) 6(5 Nov. 08 75 1 l»ec. ICO 80 1 12 J By check, If) 28 50 ! 36 25 i 78 36 j 48 12 50 Ans. S423.22. 4. How mnch 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, 1858. 1858. Junu 4 To cash, 175 S.'pt. 14 Nov. 1 u a 150 1859. 185S). July 25 Feb. 24 '' draft. 200 Dec. 3 Sept. 10 « check, 66 By check, 120 80 ' Ans. $337.02. 5. Interest at 5 %, allowed according to Note 2, how much was due, Jan. 1, 1860, on the book of a savings bank in the city of New York, having the following entries ? Dr. Sixpenny Savings Bank, in account with William Gallup, 1858. 1 1858. Jau. 1 To check, 30 60 Sept. 16 By check. March 17 a .i 26 38 1 1859. Aujr. 1 " cash, 84 72 1 Jan. 27 « (t Ib59. March 1 « (( June 11 " draft, 60 00 Nov. 1(5 " cubh, 40 78 Ans, §179.10. COMPOUND INTEREST. •>«^0« Compound Interest is interest on both principal and interest, when the interest is not paid when due. NoTK. — The ."imple interest m ly be addled to the principal nnnnally, Rp»ni- jniiiually, or quarterly, as the partiea may agree -, but the taking of compui nd interest in not legal. 1 What is the compound interest of $640 for 4 years, ^i 5 per cent. ? V 822 PERCENTAGE. OPERATION. $610 Principal for 1st year, $340 X 1.05 = $372 '' " 2d '' $872 X 1.05 = $705.60 " " 3d " $705.60 X 1.05 = $740.88__ " " 4th '' §740.88 X 1.05 ==: S777.924 Amount « 4 years, 640. Given principal, $137,924 Compound interest. This illustration is sufficient to establish the folloAving KuLE. I. Find the amount of the given 'principal at the given rate for one year^ and make it the principal for the second year, II. Find the amount of this new principal, and mahe it the principal for the third year, and so continue to do for the given number of yrars, III. Subtract the given principal from the last amount ; the re- mainder will be the compound interest. Notes. — 1. When the interest is payable femi-annually or quarterly, find the amount of the given principal for the first interval, and make it the principal for the second interval, proceeding in all respects as when the interest is payable yearly. 2. When the time contains years, months, and days, find the amount for the years, upon which compute the interest for the months and days, and add it to the last amount, before subtracting. EXAMPLES FOR PRACTICE. 1. What is the compound interest of $750 for 4 years at 6 per cent.? Ans. $196.86- 2. What will $250 amount to in 3 years at 7 per cent, compound interest? Ans. $306.26. 3. At 7 per cent, interest, compounded semi-annually, what debt will $1475.50 discharge in 2i years? Ans. $1752.43. 4. Find the compound interest of $376 for 3 yr. 8 mo. 15 da.;, at 6 per cent, per annum. J^ns. $90.84. ^«il« A more expeditious method of computing compound interest than the preceding is by the use of the compound interest table on the following page. COMPOUND INTEREST. 823 TABLE, Showing the amount of $1, or <£!, at 2J, 3, 3J, 4, 5, G, 7, and 8 '^.er cent., compound interest, for any number of years from 1 to 40. Years 1 2Kpe'-ct. 3 percent. sT^perct. 4 per cent. 5 per cent. 6 per cent. 7 per cent. 8 per cent. 1.025000 1.030000 1.035000 1.040000 1.050000 l.OGOOOO 1.070000 l.OSOOOO 2 1.050625 1.060900 1 071225 1.081600 1.102500 1.123600 1.1449U0 1.166-100 3 1.076891 1.092727 1.108718 1.124864 1.157625 1.191016 1.225043 1.259712 4 1.103813 1.125509 1.147523 1.169859 1.215506 1.262477 1.310796 1.3G0489 5 1.131408 1.159274 1.187686 1.216653 1.276282 1.338226 1.402552 1.469328 6 1.159693 1.194052 1.229255 1.265319 1.340096 1.418519 1.500730 1.586874 7 1.188686 1.229874 1.272279 1.315932 1.407100 1.503630 1.605782 1.713824 8 1.218403 1.266770 1.316809 1.368569 1.477455 1.593848 1.718186 1.850930 9 1.248863 1.304773 1.362897 1.423312 1.551328 1.689479 1.8G84.>9 1.999005 TkJ 1.280085 1.343916 1.410599 1.480244 1.628885 1.790848 1.967151 2.158925 11 1.312087 1.384234 1.459970 1.539454 1.710339 1.898299 2.104852 2.331639 12 1.344889 1.425761 1.511069 1.601032 1.795856 2.012197 2.252192 2.518170 13 1.378511 1.468534 1.563956 1.665074 1.885649 2.132928 2.409845 2.719624 14 1.412974 1.512590 1.618695 1.731676 1.979932 2.260904 2.578534 2.937194 15 1.448298 1.557967 1.675349 1.800944 2.078928 2.396558 2.759032 3.172169 16 1.484506 1.604706 1.733986 1.872981 2.182875 2.540352 2.952164 3.425943 17 1.521618 1.652848 1.794676 1.947901 2.292018 2.692773 3.15S815 3.700018 18 1.559659 1.702433 1.857489 2.025817 2.406619 2.854339 3.379932 3.99C020 19 1.598650 1.753506 1.922501 2.106849 2.526950 3.025600 3.616528 4.315701 20 1.63S616 1.806111 1.989789 2.191123 2.653298 3.207136 3.8G9CS5 4.660957 21 1.679582 1.860295 2.059431 2.278768 2.785963 3.399564 4.140562 5.033834 22 1.721571 1.916103 2.131512 2.369919 2.925261 3.603537 4.4-30402 5.43C540 23 1.764611 1.973587 2.206114 2.464716 3.071524 3.819750 4.740530 5.871464 24 1.808726 2.032794 2.283328 2.563304 3.225100 4.048935 5.072.307 6.341181 25 1.853944 2.093778 2.363245 2.665836 3.386355 4.291871 5.427433 6.848475 26 1.900293 2.156591 2.445959 2.772470 3.555673 4.549383 5.807353 7.396353 27 1.947800 2.221289 2.531567 2.883369 3.733456 4.822346 6.21C868 7.98s^062 28 1.996495 2.287928 2.620172 2.99S703 3.920129 5.111687 6.648838 8.627100 29 2.046407 2.356566 2.711878 3.118651 4.116136 5.418388 7.114257 9.317275 30 2.097568 2.427262 2.806794 3.243398 4.321942 5.743491 7.612255 10.062657 31 2.150007 2.500080 2.905031 3.373133 4.538040 6.088101 8.145113 10.867C69 32 2.203757 2..575083 3.006708 3.508059 4.764942 6.453387 8.715271 11.737083 33 2.258S51 2.652335 3.111942 3.648381 5.003189 6,840590 9.325340 12.076050 34 2.315322 2.731905 3 220860 3.794316 5.253348 7.251025 9.97S114 13.690134 35 2.373205 2.813862 3.333590 3.946089 5.516015 7.686087 10.676582 14.785344 36 2.432535 2.898278 3.450266 4.103933 5.791816 8.147252 11.423942 15.968172 37 2.493349 2.985227 3.571025 4.2GS090 6.081407 8.6360S7 12.223618 17.245026 38 2.555682 3.074783 3.696011 4.438S13 6.-385477 9.154252 13.079271 1 8.625276 39 2.619.=i74 3.167027 3.825372 4.616366 6704751 9.703508 13.994820 20.115298 40 2 685064 3.262038 3.959260 4.801021 7.039989 10.2'«5718 14.974458 21.724522 324 PERCENTAGE. EXAMPLES FOR PRACTICE. 1. What is the amount of $300 for 4 years at 6 per cent, com- pound interest,payable semi-annually ? OPERATION. Analysis. The amount of $1 at 6 per cent., ^1 26677 compound interest payable semi-annually, is 300 the same as the amount of $1 at 3 per cent., £^^0 OtToO compound interest payable annually. We ^ ' therefore take, from the table, the amount of $1 for 8 years at 3 per cent., and multiply this amount by the given principal. 2. What is the amount of $536.75 for 12 yr. at 8 per cent, com- pound interest ? ^ws. $1351.63. 3. What sum placed at simple interest for 2 yr. 9 mo. 12 da., at 7 per cent., will amount to the same as $1275, placed at com- pound interest for the same time and at the same rate, payable semi-annually? Ans. $1292.51 — . 4. At 8 per cent, interest compounded quarterly, how much will $1840 amount to in 1 yr. 10 mo. 20 da. ? Ans. $2137.06. 5. A father at his death left $15000 for the benefit of his only son, who was 12 yr. 7 mo. 12 da. old when the money was de- posited; the same was to be paid to him when he should be 21 years of age, together with 7 per cent, interest compou'uded semi- annually. How much was the amount paid him ? 6. What sum of money will amount to $2902.263 in 20 years, at 7 % compound interest ? Ans. $750. PROBLEMS IN INTEREST. PROBLEM I. SS3. Given, the time, rate per cent., and interest, to find the principal. 1. T7hat sum of money will gainr $87.42 in 4 years, at 6 per cent. ? OPERATION. Analysis. Since $.24 $.24, interest of $1 for 4 years. ^« *^^ interest of $1 for 4 $87.42 — - .24 = $364.25, Ans, years at 6 per cent., $87.42 must be the interest of as PROBLEMS IN INTEREST. 325 many dollars, for the same time and at the same rate, as $.24 is con-* tained times in $87.42. Dividing, we obtain $364.25, the required principal. Hence the Rule. Divide the given interest hy the interest of %\ for the given time at the given rate, EXAMPLES FOR PRACTICE. 1. What sum of money, invested at 6 J per cent., will produce $279,825 in 1 yr. 6 mo.? Ans. $2870. 2. What sum will produce $63.75 interest in 6 mo. 24 da. at 7 J per cent. ? 3. What sum will produce $12 J interest in 10 days at 10 per cent.? Alls. $4500. 4. What sum must be invested in real estate paying 12 i per cent, profit in rents, to give an income of $3125 ? 5. What is the value of a house and lot that pays a profit of 9i per cent, by renting it at $30 per month ? 6. What sum of money, put at interest 6 yr. 5 mo. 11 da.^ at 7 per cent., will gain $3159.14 ? Ans. $7000. 7. What sum of money will produce $69.67 in 2 yr. 9 mo. at 6 ^0 compound interest ? Ans. $400. 8. What principal at 6 % compound interest will produce $124.1624 in 1 yr. 6 mo. 15 da. ? Ans. $1314.583. PROBLEM II. ^o3. Given, the time, rate per cent., and amount, to find the principaL 1. What sum of money in 2 years 6 months, at 7 per cent., will araount.te $136,535? OPERATION. Analysis. Since $1,175, amount of $1 for 2 yr. 6 mo. ^1-175 istheamount $136,535 - 1.175 = $116.20, Ans. ^^ ^ V"' ^ ^f' ^ months, at 7 per cent., $136,535 must be the amount of as many dollars, for the same time and at the same rate, as $1,175 is contained times in $136,535. Dividing, w© obtain $116.20, the required principal. Hence the 2^ 326 PERCENTAGE. Rxjle. Divide flic given amoitnt hy the amount q/* §1 for the given time at the given rate. EXAMPLES FOR TRACTICE. 1. \Yhat principal in 2 yr. 3 mo. 10 da.^ at 5 per cent., will amount to $1893 61^ ? Ans, $1700. 2. A note which had run 3 yr. 5 mo. 12 da. amounted to $081,448, at 6 per cent. ; how much was the face of the note ? 3. What sum put at interest at 3 J per cent., for 10 yr. 2 mo., will amount to $15660? 4. "What is the interest of that sum for 2 yr. 8 mo. 29 da., at 7 per cent., which at the same time and rate, will amount to $1568.97? Ans. $253,057+. 5. What is the interest of that sum for 243 days at 8 per cent., which at the same time and rate, will amount to $11119.70 ? 6. What principal in 4 years at 6 per cent, compound interest, will amount to $8644.62 ? Ans. $6847.34. 7. What sum put at compound interest will amount to $26772.96, in 10 yr. 5 mo., at 6 per cent. ? Ans. $14585.24. PROBLEM III. 554. Given, the principal, time, and interest, to find the rate per cent. 1. I received $315 for 3 years* interest on a mortgage of $1500; what was the rate per cent. ? OPERATION. Analysis. Since ^15^00 $^^ ^s the interest on 3 the mortgage for 3 TTTT77 . » r. ^ ^ years at 1 per cent., $45.00, int. for 3 yr. at 1 %. Ig^g ^„^, {^ ^^^^ .^_ $315 ~ $45 = 7 %, A)is. terest on the mortgage for the same time, at as many times 1 per cent, as $45 is contained times in $315. Divid- ing, and we obtain 7, the required rate per cent. Hence the Rule. Divide the given interest hy the interest on the principal for the given time at 1 per cent. PROBLEMS IN INTEREST. 327 EXAMPLES FOR PRACTICE. 1. If I loan $750 at simple interest, and at the end of 1 yr. 3 mo. receive $796,874, what is the rate per cent. ? Ans. 5. 2 If I pay $10.58 for the use of $1700, 28 days, what is the rate of interest? Ans. 8-hper cent. 3. Borrowed $600, and at the end of 9 yr. 6 mo. returned $356.50 ; what was the rate per cent. ? 4. A man invests $7266.28, which gives him an annual income of $744.7937; what rate of interest does he receive ? 5. If C buys stock at 30 per cent, discount, and every "^6 months receives a dividend of 4 per cent., what annual rate of interest does he receive ? Ans. 11| per cent. 6. At what rate per annum of simple interest will any sum of money double itself in 4, 6, 8, and 10 years, respectively ? 7. At what rate per annum of simple interest will any sum triple itself in 2, 5, 7, 12, and 20 years, respectively ? 8. A house that rents for $760.50 per annum, cost $7800 : what % does it pay on the investment? Ans. 9| per cent. 9. I invest $35680 in a business that pays me a profit of $223 a month ; what annual rate of interest do I receive ? Ans. 7i 5^. PROBLEM IV. 535. Given, the principal, interest, and rate, to find the time. 1. In what time will $924 gain $151,536, at 6 per cent.? OPERATION. ^ Analysis. Since 4924 $55.44 is the interest .06 of $924 for 1 year at $5144, int. of $924 for 1 yr. at 6 X. ^ ^7 ^^'^^ ^^^^^^^^ ' J /«> jjmst \yQ the interest $151,536 ^ $55.44 = 2.73 ^f the same sum, at 2.73 yr. = 2 yr. 8 mo. 24 da., Ans. the same rate per I cent., for as many irs as $55.44 is contained times in $151,536, which is 2.73 times, ducing the mixed decimal to its equivalent compound number, have 2 years 8 months 24 days, the required time. Hence the S28 PERCENTAGE. Rule. Divide the given interest hi/ the intereft on the jirincijial for 1 year ; the quotient will he 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, 4}, 6, 7, and 10 per cent. ? How long to quadruple itself? ^^^ J To double itself at 3 %, 33J yr ^^* ( 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-t-1000^$1.171353, the amount of $1 for the required time. From the table, $1, in 2 years, will amount to $1.123G ; hence $1.171353— $1.1236=$.047753, the interest which must accrue on $1.1230 for the fraction of a year; and $1.1230 X .00 = :i^.0G741G ; $.047753 ~ $.007410 = .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. 556, Discount is an abatement or aliowance made for the payment of a debt before it is due ^^7. 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 becomes 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 Si. * present worth of $1.33775 $642.12 ^ 1.33775 =« §180 for the given time at the $642.12, given sum. given rate of interest, the 480. present worth. present worth of $G42.12 $162~12^ discount. ^"^^^ ^^ ^^ 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 hy the amount of $1 for the given rate and time; the quotient will he 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 worthy discount, and debt, are equivalent to principal^ interest, and amount. Hence, when the time, rate per cent., and amount are given, the principal may be found by Prob. II, (663); 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 eaoh payment {separately. Their sum will be the presewt worth of the several payments, and this sura 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*f 86, 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 mon^y, 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. ? Ans. Gained $238.10. 5. B bought a house and lot April 1, 1860, for which he waa to pay $1470 m the fourth day of the following September, and 330 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, 1860, discounting at the rate of 7 % ? Ans. $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 ^o per annum dis- count? Ans. $7.11^. BANKING. S58. A Bank is a corporation chartered by law for the pur- pose of receiving and loaning money, and furnishing a paper circulation. ^^9. A Promissory Note is a written or printed engagement to pay a certain sum either on demand or at a specified time. S60. 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 hanJc of issue ; one which lends money, a bank of discount ; and one which takes chart^e of money belonging^ to other parties, a hank of deposit. Some banks perform two and some all of these duties. ^61. 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. 831 S63. The Holder is the owner. ^04: A Negotiable Note is one which may be bought and sold, or negotiated. It is made payable to the hearer or to the order of the payee. «i6o« Indorsing a note by a payee or holder is the act of writing his name on its back. Notes. — 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 ^iven should be written out in words. ^06. The Face of a note is the sum made payable by the note. •5G7. 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 wxpira- tion of the time mentioned in them. 5@9* Notes may contain a promise of interest, which will be reckoned from the date of the note, unless some other time be specified. NoTK. — A note is on interest from the day it is due, even though no mention be mude 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 mdorsers of its non-payment. Notes. — 1. The fulhire to protest a note on the tliird day of grace releases the iu- 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. «>73« Bank Discount is an allowance made to a bank for the payment of a note before it becomes due. ■ 3g2 PERCENTAGE. ^73. The Proceeds of a note is the sum received for it when discounted, and is equal to the face of the note less the discount. ^74:, 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. ^73. 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. NoTB. — Many banks take only true discount. CASE I. S7&. Given, the face of a note, to find the discount and the proceeds. KuLE. I. Compute the interest on tlie face of the note for three dey^ more than the specified time ; the residt will be the discmint, II. Subtract the discount from the face of the note; the re- mainder will be the proceeds. NoTKS. — 1. When a note is on interest, payable nt a future specified time, the omomit is the face of the note, or the sura made payable, and must be made the b;i.«jis 01 discount. 2. To indicate the maturity of a note or draft, a vertical line ( | ) is used, with the day at which the note is nominally due on the left, and the date of maturity (Ml tk« wght; thug, Jan. "^ | jq. 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 $884.50 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 K. 25 P. for $27.59 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 I 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 j^j^^^ dollars at the Exchange Bank, for value received. John Demar^st. Discounted Aug. 26, at 7%. Ans. Due Nov. ^^ | 09; term of discount 95 days; preoec^, $1926.20. $1066yY^ . Baltimore, April 19, 1859. 6. Ninety days after date we promise to pay to the order of King & Dodge one thousand sixty-six and -^j^j^ dollars at the Citi* zens' Bank, for value received. Case & Sons. Discounted May 8, at 6 %. Ans. Due July » » | , , ; term of discount, 74 da. ; proceeds, $1058.59. $784^. . 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 -^^^^ dol* lars at the Traders' Bank. Wm. Hamilton. Discounted July 5, at 8 ^. ■ 834 PERCENTAGfB. Sl845^<\j. Chicago, Jan. 31, 1862. 8. One inontli after date we jointly and severally agree to pay to W. H. Willis, or order, one thousand eight hundred forty-five and -f^Q dollars at the Marine Bank. Payson & 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 difierence 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 $~^795, proceeds of $1. ^^^* ^^^^' *^^^' '^"^^ $750 -- .9795 = $765,696, Ans. ^-^^^^ is *h^ P^^ 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 hy the proceeds of $iybr the time and rate mentioned ', the quotient will he 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 7A % ; 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. Ana. $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 jmonth? Ans. $593.70. 8. What must be the face of a note which, when discounted at a broker's for 110 days at 1 5^ 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. $100 as the face of the $100, base. note, the discount for 33 1.65, discount for 33 days. days at IJ^ a month will $98.35, proceeds. ^^ ^^'^^ ^"^ ^^^ proceeds $1.65 -f-. 090154 J =18AV^%,^7is. $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 Rule. I. Find the discount and the proceeds of $1 or $100 for the time the note has to run. II. Divide the discount hy 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. ^H^ %• 336 PERCENTAGE. 2. A note payable in 2 months is discounted at 2 % a month; "what rate of interest is paid? Ans. 2b-^^^^ %. 3. When a note ^payable in 90 days is discounted at IJ % a month, what rate of interest is paid? Ans, 18yf Jf %• 4. What rate of interest corresponds to 5, 6, 7, 10, 12 % dis- count on a note running 10 months without grace ? 5. What rate of interest does a man pay who has a 60 day note discounted at |, 1, 2, 2^, 3 % a month ? CASE IV. 579. Given, tke rate of interest, to find the corres- ponding rate of bank discount. 1 A broker buys 60 day notes at such a discount that his money earns him 2 ^ a month; what is his rate ^ of discount? OPERATION. Analysis. If we assume 60 da. -f 3 da. = 63 da. ^1^^ a,s the proceeds of a $100 base. note, the interest for 63 days 4.20, interest for 63 da. at 24 per cent, will be $4.20, il0r20, amount " ^^ ^"^ *^^ ^"'^""^ ^^ ^^^^ ^^ $4.20 -f- .18235 = 23^^^- %, Ans. the note will be $104.20. We then have $104.20 the prin- cipal, $4.20 the interest, and 63 days the time, to find the rate per cent., which we do by (549) as in the last case. He^oe the Rule. I. Find the interest and the amount o/%\ or §100 /or the time the note has to run, II. Divide the interest hy the interest on the amount at 1 per cent, for the same time. EXAMPLES FOR PRACTICE. 1. W^hat rates of bank discount on 30 day notes correspond to 5, 6, 7, 10 per cent, interest? 2. At what rate should a 3 months^ note be diseounted to pro- duce 8 % interest? Ans. Vjlf ? %. 3. At what rates should 60 day notes be discounted to pay to a •broker 1, li, 2, 2} % a month? 4. At what rate must a note payable 18 months hence, without grace, be discounted to produce 7 % interest? Ans. 6/^*^ %• EXCHANGE. 837 EXCHANGE. 580. Exchange is a method of remitting money from one place to another, or of making payments by written orders. 581. A Bill of Exchange is a written request or order upon one person to pay a certain sum to amother person, or to his order, at a specified time. 58S. A Sight Draft or Bill is one requiring payment to be made " at sight,'' which means, at the time of its presentation to the person ordered to pay. In other bills, the time specified is usually a certain number of days '' after sight/' There are always three partiQ3 to a transaction in exchange, and usually four : I58S. The Drawer or Maker Is the person who signs the order or bill ; 584. The Drawee is the person to whom ,the order is ad- dressed ; 585. The Payee is the person to whom the money is ordered to be paid ] and 586. The Buyer or Remitter is the person who purchases the bill. He may be himself the payee, or the bill may be drawn in favor of any other person. 587. The Indorsement of a bill is the writing upon its back, by which the payee relinquishes his title, and transfers the pay- ment to another. The payee may indorse in blank by writing his name only, which makes the bill payable to the hearer , and con- sequently transferable like a bank note ; or he may accompany his signature by a special order to pay to another person, who in his turn may transfer the title in like manner. Indorscrs become sep- arately responsible for the amount of the bill, in case the drawee fails to make payment. A bill made payable to the hearer is transferable without indorsement. 588. The Acceptance of a bill is the promise which the drawee makes when the bill is presented to him to pay it at ma- turity; this obligation is usually acknowledged by writing* thp word " Accepted," with his signature, across the face of the bill* 29 w 838 PERCENTAGE. Notes. — I. In this country, and in Great Britain, three days of grace are al- lowed for the payment of a bill of exchange, after the time specitied has expired. In regard to grace on siglit hills, however, custom is variable ; in New York. Penn^ivlvania, Virginia, and some other States, no grace is allowed on sight bills, 2. ^V'hen a bill is protested for non-acceptance, the drawer is obligated to pay it immediately, even though the specified time has not expired. Exchange is of two kinds — Domestic and Foreign. 58l>. Domestic or Inland Exchange relates to remittances made between different places of the same country. Note. — An Inland Bill of Exchange is commonly called a Draft. •5@0o Foreign Exchange relates to remittances made between different countries. *5IJ1. A Set of Exchange consists of three copies of the same bill, made in foreign exchanges, and sent by different conveyances to provide against miscarriage; when one has been paid, the others are void. ^1^^. The Face of a bill of exchange is the sum ordered to be paid ; it is usually expressed in the currency of the place on which the draft is made. 593. The Par of Exchange is the estimated value of the coins of one country as compared with those of another, and is either intrindc or commercial. t594. The Intrinsic Par of Exchange is the comparative value of the coins of different countries, as determined by their weight and purity. Q^^» The Commercial Par of Exchange is the comparative value of the coins of different countries, as determined by their nominal or market price. Note. — The intrinsic par is always the same while the coins remain un- • changed; but the commercial par, being determined by commercial usage, is fluctuating. ^9G. The Conrse of Exchange is the current price paid in one place for bills of exchange on another place. This price varies, according to the relative conditions of trade and commercial credit at the two places between which exchange is made. Thus, if Boston is largely indebted to Paris, bills of exchange on Paris will bear a high price in Boston. When the course of exchange between two places is unfavor- EXCHANGE. 839 able to drawing or rcinitting, the disadvantage is sometimes avoided, by means of a circuitous exchange on intermediate places between which the course is favorable. DIRECT EXCHANGE. ^97. Direct Exchange is confined to the two places between which the money is to be remitted. j 598. There are always two methods of transmitting money ' between two places. Thus, if A is to receive money from B, 1st. A may draw on B, and sell the draft; 2d. B may remit a draft, made in favor of A. Note. — One person is said to draw on another person, when he is the maker of a draft addressed to that person. CASE I. 599. To compute domestic exchange. The course of exchange for inland bills, or drafts, is always ex- pressed by the rate of premium or discount. Drafts on time, however, are subject to hank discount , like notes of hand, for the term of credit given. Hence, their cost is affected by both the course of exchange and the discount /or time. 1, What will be the cost of the following draft, exchange on Boston being in Pittsburgh at 2} ^ premium ? $600. Pittsburgh, June 12, 18G0. Sixty days after sight, pay to William Barnard, or order, six hundred dollars, value received, and charge the same to our account. To the Suffolk Bank, Boston. Thomas Bauer & Co. OPERATION. $1 4- $.0225 == $1.0225, course of exchange. .0105 , bank discount of $1, (63 da.) $1,012, cost of exchange for $1. S600 X 1.012 = $607.20,^728. NALTSis. From $1.0225, the course of exchange, we subtract $.0105, the bank discount of $1 for the specified time, and obtain $1,012, the cost of exchange for $1 ; then $G00 X 1.012 = $G07.20, the eost of exchange for $600. 340 PERCENTAGE. 2. A commission merchant in Detroit wishes to remit to his employer in St. Louis, $512.36 by draft at 60 days ; what is the face of the draft which he can purchase with this sum, exchange being at 2 i % discount ? OPERATION. $1 — S.025 = ?.975, course of exchange. .01225, discount of §1. $.96275, cost of exchange for $1. ?512.36 ~ .96275 = §532.18+, Ans. Analysis. From $.975, the course of exchange, we subtract $.01225, the bank discount of $1 for the specified time, at the legal rate in Detroit, and obtain $.96275, the cost of exchange for $1 ; and the face of the draft that will cost $512.36, will be as many dollars as $.90275 is contained times in 512.36, which is 532.18+, times. Hence we have the following Rule. I. To find the cost of a draft, the face being given Multiply the face cf the draft hy the cost of exchange for $1. II. To find the face of a draft, the cost Doing given. — Divide the given cost hy the cost of exchange for $1. Note. — The cost of exchange for $i may always be found, by subtracting from the course of exchange the bank discount (at the legal rate where the draft is made), for the specified time. Foi sight drafts, the course of exchange is tho cost of $1. EXAMPLES FOR PRACTICE. 1. What must be paid in New York for a draft on Boston, at 80 days, for §5400, exchange being at J % premium ? Ans. $5392.35. 2. What is the cost of sight exchange on New Orleans, for $3000, at 3} % discount? 3. What must be paid in Philadelphia for a draft on St. Paul drawn at 90 days, for $4800, the course of exchange being lOlf % ? Ans. $4791.60. 4.. A sight draft was purchased for $550.62, exchange being at a premium of 3^ ^ ; what was the face ? 5. An agent in Syracuse, N. Y., having $1324.74 due his em- ployer, is instructed to. remit the same by a draft drawn at 30 days; what will be the face of the draft, exchange being at If % premium? Ans. $1310.22—. EXCHANGE. 841 6. My agent in Charleston, S. C, sells a house and lot for $7500, on commission of IJ %, and remits to me the proceeds in a draft purchased at J % premium ; what sum do I receive from the sale of my property ? 7. A man in Hartford, Conn., has $4800 due him in Baltimore; how much more will he realize by making a draft for this sum on Baltimore and selling it at J % discount, than by having a draft on Hartford remitted to him, purchased in Baltimore for this sum at I % premium? Ans, $11.73 + . 8. The Merchants^ Bank of New York having declared a divid- end of 6} %, a stockholder in Cincinnati drew on the bank for the sum due him, and sold the draft at a premium of If %, thus real- izing $508.75 from his dividend; how many shares did he own ? 9. Sight exchange on New Orleans for $5000 cost $5075; what was the course of exchange ? Ans. IJ % premium. 10. A man in Buffalo purchased a draft on St. Paul, Minn., for $5320, drawn at 60 days, paying $5141.78; what was the course of exchange ? -4»s. 2i % discount. CASE IT. ®00. To compute foreign exchange. 001. The following standards of the decimal currency of the United States were established April 2, 1792. Coins. Weight. Fineness. Gold eagle, 270 grains, 91Gf thousandths. Silver dollar, 416 '' Copper cent, 264 " In 1834, the eagle was reduced in weight to 258 grains, and in 1837 its fineness was fixed at 900 thousandths pure, which is likewise the present standard of purity for all the U. S. gold and silver coins. In 1837, also, the silver dollar was reduced in weight to 412.5 grains. In 1853, the silver half dollar was reduced in weight to 192 grains, and the smaller silver coins proportionally. NoTn. — The object of the change in the silver coinage of the United States, made in 1853, was to prevent its exportation bj raiding the nominal value of silver above its foreign market value. ®0^» The intrinsic par of exchange between the United States and different countries, is given in the following 29* 342 PERCENTAGE. TABLE OP FOREIGN COINS AND MONEY. Crown, Baden " Bavaria " England " France " Geneva " Portugal " Tuscany " Wurtemberg • " Zurich Dollar, Argentine Republic " Bolivia " Chili " Columbia " Mexico " Norway " Peru '' Spain " Sweden Doubloon, Bolivia " Columbia (Bogota)... " " (Popayau). « Chili (since 1835) " " (before 1835).. .. •* La Plata '* Mexico (average) « Peru (Cuzco) " " (Lima) « Spain Drachma, Greece Ducat, Austria " Bavaria..... " Cologne *' Hamburg '• Hungary " Netherlands., " Saxony " Sweden " Wurtemberg Florin, Austria " Bavaria " Hanover " Italy Gold. Silver. Gold. Silver. Gold. Silv £>euouiiiations. 8 reals. 8 " 100 cents. 8 reals. 8 " 6 marks. 8 reals. 10 " (old). 6 marks. 4 gilders 12 marks. 6C kreutzers. 60 « 60 groshen. 12 soldi. i 1.077 1.157 1.072 1.151 1.100 1.181 1.100 1.181 .960 1.031 5.813 1.050 1.128 1.070 1.148 .960 1.031 1.016 1.091 1.011 1.086 1.011 1.080 1.022 1.09S 1.005 1.079 1.051 1.129 1.005 1.079 1.003 1.077 1.059 1.136 15.580 15.617 15.390 15.060 15.570 14.060 15.534 15.534 15.551 15 570 166 .177 2.278 2.274 2.250 2.257 2.2S1 2.269 2.264 2.267 2Ji36 .485 .521 .395 .425 .547 .587 .181 .194 EXCHANGE. 343 TABLE OP FOREIGIT COINS AND MONEY — Continued. Florin, Mecklenburg " Prussia and Poland. " Tuscany Franc, Belgium '• France Frederick d'or, Denmark Gilder, Baden " Darmstadt " Demerara " Frankfort " Netherlands " Wurtemberg Ghersh, Tripoli Guinea, England Lira, Lombardy " Leghorn « Milan " Yenice Livre, Genoa " Leghorn " Switzerland Mark banco, Hamburg " current, " Milree, Azores " Brazil...' " Madeira « Portugal Mohur, Ilindostan Ounce, Naples Pagoda, Madras , Piaster, Tunis " Turkey Pi.«tareen, Spain Pistole, Spain Pound, British ProTinces.... Ileal, plate, Spain " vellon, " " Egypt Rix dollar, Austria " '• Batavia , « « Denmark Silver. Gold. Silver. Gold. Silver. Gold. Silver. Lower Denoiniuations. 30 groshen. 12 soldi. 100 centimes. 100 " 60 kreutzers. 60 " ' 20 stivers. 60 kreutzers. 20 stivers. 60 kreutzers. 100 paras. 21 shillings. 20 soldi. 20 « 20 " 100 centimes. 20 soldi. 20 " 100 centessini. 16 skilliugs 10 » loco reis. 1000 " 1000 " 1000 " 16 rupees. 3 ducats, 42 fan am s. 16 carobas. 100 aspers. 4 reals vellon. 20 shillings. 34 marvedis. 34 " 20 piasters. 120 kreutzers. 4S stivers. 9u ^killings. 1?° i'-iS 5.^?. ML pm .541 .571 .227 .244 .202 .281 .ISO .200 .180 .200 3.932 .397 .420 .397 .426 .202 .282 .397 .426 .400 .436 .395 .423 .105 .112 5:059 .162 .173 .162 .173 .162 .173 .162 .173 .186 .L98 .102 .173 .273 .292 .350 .375 .2S5 .3-J5 .830 .800 .830 .890 1.000 1074 1.120 1.203 7.109 2.485 1.840 .124 .133 .020 Xj7 .028 .211 3.904 4.016 .097 .104 .0{8 .051 .DCS l.OiO .971 1.043 ,7S2 .840 1 l.Oil ..,.0 844 PERCENTAGE. TABLE OF FOREIGN COINS AND MONEY — Continued. Rigsbank dollar Rix dollar, Norway Rouble, Russia Rupee, India Ruspone, Tuscany Sequin, Tuscany Scudo, Milan " Naples " Rome « Sicily : Sovereign, Great Britain. Thaler, Brunswick " Hanover " Hesse-Cassel " Prussia " Saxony " Bremen Tale, China " Japan , Tomaun, Persia Utchlik, Tripoli Yirmilik, Turkey , Silver. Gold. Silver. Gold. Silver. Gold. Silver. Gold. Lower Denominations. -1: « ~ bo :i5|i 48 skillings. .526 .565 96 « 1.051 1.129 100 copecks. .754 .806 16 annas. 6.925 2.301 .445 .477 117 soldi. .973 1.045 12 carlini. .950 1.021 1.006 1.080 12 tari. .985 1.058 20 Bhillings. 4.861 30 groschen. .692 .743 30 « .694 .735 30 « .687 .738 SO « .692 .743 30 « .694 .735 72 grotes. .788 .846 10 mace, 100 -» candarines. S 1.480 1.590 10 mace, 100 \ candarines. i .760 .800 100 mamvodis. 2.233 120 paras. .149 .160 20 piasters. .877 Notes. — 1. The standard value of gold a.s compared with silver in the United States, is as 15.407 to 1 in the coinage of 1792, as 15.988 to 1 in the coinage of . 1837, nnd as 14.922 to 1 in the coinage of 1853. 2. The relative values of gold and silver differ in the coinage of different coun- tries. In England the ratio is 14.288 to 1 j in France it is 15.5 to 1 ; in Ham- burg it is 15 to 1. 3. In the present gold coinage of the United States, a Troy ounce of pure gold is equal to $20,672, and of standard gold to $18,605. In the present silver coinage of the United States, a Troy ounce of pure silver is equal to $1,388, and of standard silver to $1.25. 603. It will be seen by the table that the par of exchange between the United States and Great Britain is £1 = §4.861. Previous to the changes in the U. S. coinage^ made in 1834 and in 1837, the par of exchange was £1 = S4.44|, or £9 = §40, which is called the old par of excfiange. By the new par of ex- EXCHANGE. 345 change, sterling money is worth about 9| % more than by the old par. @04:. The course of exchange on England is usually given with reference to the old par of exchange. Hence, when sterling money is really at par ^ according to present standards, it is quoted in the market at 9| % premium. 60S. The course of exchange between different countries may be expressed either by the rate per cent, above or below par, or by giving the sum of money in one country which is equal to a certain sum in another country. In the latter case, the exchange requires simply a reduction of currencies ; in the former, it requires both a reduction of currencies and a computation of percentage. 1. What will be the cost in Boston of the following bill of ex' change on Liverpool, at 9 J % premium? ^^^2- Boston, June 16, 1860. 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 k Son, ) Liverpool y England. j Analysis. Since OPERATION. exchange on Liver- £9 = §40 X 1 .095, course of exchange, pool is at 9 J % pre- $40 X 1.095 mium, £9 will co-^t ^^ = 9 ' '''' '^ ^^' $40 X 1.095, (603) : Aoo $40x1.095 ^oino 4A A and £1 will therefore 432 X ^^ =§2102.40,^.... ^^^ ^ ^,,5 cost — g . Multiplying the face of the bill, £432, by the cost of exchange of £1, we obtain $2102.40, the required cost of the bill. 2. What is the face of a bill on London, that may be purchased in New York for $2768.70, exchange being at 10 % premium in favor of London ? 346 PERCENTAGE. OPERATION. £9 = $40 X 1.10, course of exchange, £1 = — '- — J cost of £1, $2768.70 -^ \ ' =£bQQ 6s. 6d., Ans. y 40 X 1.10 Analysis. We divide $2768.70, the given cost, by g— ^ — , the cost of exchange for £1, and obtain £566 6s. 6d., the face. 3. What cost, in Hamburg, a bill on New Orleans for $4500, the course bf exchange being 1 mark = $.365 ? OPERATION. Analysis. Since ^l=-\%%^ of a maik, cost of a unit. exchange for §1 will $4500 xVg00^12328 marks 12 skillin^s. ^^«* ^^ Hamburg iJ^V of a mark, a bill for $4500 will cost 4500 X'||3«= 12328 marks 12 skillings. G06. From these illustrations we derive the following Rule. I. To find the cost of a bill, the face being given. — Multi2)h/ the face hy the co$t of a unit of the currency in which the hill is expressed. II. To find the face of a bill, the cost being given. — Divide the given cost by the cost of a unit of the currency in which the hill is to he expressed, EXAMPLES FOR PRACTICE. 1. What is the cost in Portland of a bill on Manchester, Ene:., f^ £325 3s. 9d., at Of % premium? Ans, $1586.19 + . 2. What must be paid in Charleston for a bill of exchange on Paris for 6000 francs, at 18f cents per franc? 3. What cost in Bostc^^ bill on St. Petersburg for 3000 roubles at IJ % premium, the |^ar of exchtinge being $.754 for 1 rouble? 4. What will be the cost in Naples of a bill of exchange on New York fdi- $831.12, at the rate of $.96 for 1 scudo? Ans. 865 scudi 9 carlini. 5. A draft on Philadelphia cost £125 in Birmingham, Eng., exchange being at 8 % premium for sterling; required the face of the draft. EXCHANGE. 34:7 6. An agent in Boston, having $7530.30 due liis emploj^er 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 being at 11 % premium? An^^ £1527 12s. Gj^d. 7. A merchant in Cincinnati has 9087 gilders 10 stivers due him in Amsterdam^ and requests the remittance by draft; what sum will he receive, exchange on U. S. being in Amsterdam at 2 J gilders 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 pur- chased for this sum, at the rate of 64. 5d. sterling per milree, what sum in the currency of Portugal will the agent receive ? Am. 9302 milreeo 325 Jf reis. 9. A draft on Dublin for £360 cost $1736; what was the course of exchange? A'lu. 8 J % premium. 10.* A merchant in Baltimore, having receiv(?d an importation of Madeira wine invoiced at 1500 milrees, 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 930 rcis = §1 : how much would the merchant have saved, by remitting a draft on Madeira, purchased at $1,065 per 1 milree ? Am. $15.40. 11. An importer received a quantity of Leghorn hats, invoiced at 25256 lire 16 soldi which was paid in U. S. gold coin, ex- ported at a cost of 3 % for transportation and insurance, the price of fine gold in Leghorn being 131 lire per ounce Troy. How much less would the goods have cost in store, had payment been made by draft on Leghorn, purchased at the rate of 16 cents per lira? Am. §64.0 1. NoTK. — In U.S. gold coiqage, $10 contains 258 X .9 = 232.2 grains of//ie gold, (601). 12. When silver is worth in England 67d. per oz. fine, what sum of money in the U. S. silver coinage of 1853 is equal to 20 shillings, or £1 sterling? Am. $1,975. 13. At what rate of premium is Prussian coin, when S88.23 in U S. silver coinage of 1837 is paid for 125 thalers ? Am. 2 %. 348 PERCENTAGE. ARBITRATED EXCHANGE. 607, Arbitration of Exchange is the process of computing exchange between two places by means of one or more interme- diate exchanges. Notes. — 1. When there is only one interraediate exchange, the process is called Simple Arbitration ; when there are two or more interaiediato exchanges, the process is called Comjoound Arbitration. 2. The arbitrated price is generally either greater or less than the price of direct exchanges; and the object of arbitration is to ascertain the best route for making drafts or remittances. 608. 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 ; 8d. B may draw on A, and remit to C. 1. A man in Albany, N. Y., paid a demand in Paris of 5400 francs, by remitting to Amsterdam at the rate of 21 cents for 10 stivers, and thence to Paris at the rate of 28 stivers for 3 francs ; how much Federal money was required ? OPERATION. Analysis. We are to deter- $ f ? ^ = 5400 francs. mine how much Federal money 3 francs = 28 stivers. is equal to 5400 francs, and the 10 stivers = 821. question may bo represented ( ?) = ?51058.40, Arts. "^"« = ^ ( - 5400 francs. Now Or, 10 5400 since 3 francs = 28 stivers, and 10 stivers = $.21, we know that if the required sum be multi- ^^ plied successively by 3 francs LLl! and 10 stivers, the result will be '. () = §1058.40, Ans. equal to the product of 5400 francs by 28 stivers and $.21 successively, (Ax. 3). Canceling the units of currency, 1 franc, 1 stiver, and $1, and also the equal numerical factors, we have (?) = $1058.40, the sum required. EXCHANGE. 849 Or, since the course of exchange betTveen Amsterdam and Paris gives 1 franc = ^j stivers, and the course between Albany and Am- sterdam gives 1 stiver = f J cents, vre multiply the 5400 francs by ^^ and f J- successively, using the vertical line and cancellation, and obtain $1058.40, 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 Lon- don, who remits the proceeds to Naples, reserving his commission of ^ % on the draft sent. If exchange on London is 9 % in Boston, and the rate between London and Naples is £1 for 5 scudi, how much does the man realize from his bequest ? oPERATioi Analysis. AVe make (?) scudi = $8720 the statement as in the $40 X 1.09 = £9 first example, according £1 = 5 scudi. to the given rates of ex- 1.005 change. Then, since the (?) = 8955 scudi 3 carlini. ^g^^* '\ *^ ^^^^«* i /^ commission on the face of the draft before the purchase, vre place 1.005 on the left as a divisor, (159), and obtain by cancellation 8955 scudi 3 carlini as the proceeds of the exchange. Note. — Since the par of exchange on Ennrland is £9 = $40, the course of ex- change will always be £9 = $40 X 1 plus the rate of exchange. 3. A merchant in Chicago directs his agent in Albany to draw upon Baltimore at 1 % discount, for $1200 due from the sales of produce ; he then draws upon the Albany agent at 2 % pre- mium, 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 (?) C. = 1200 B. given rates of exchange, 100 dol- 100 B. = 99 A. lars in Baltimore is equal to 99 100 A. = 102 a dollars in Albany; and 100 dol- .995 lars in Albany is equal to 102 /'?\__ §1205.70 Ans. dollars in Chicago ; and since the unit of currency is the same in <3ach place, being $1, we represent its exchange value in each town 350 PERCENTAGE.* "by the initial letter, and make the statement as in the other exam- ples. Then, since the agent is to reserve J % commission from the avails of his draft, we place 1 — .005 = .995 on the right as a mul- tiplier, and obtain by cancellation { ? ) = $1205.70, the answer. From these principles and illustrations we have the following Rule. I. Represent the required sum hy ( 'i ), 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 op^posite 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. lY. Divide the product of the numbers on the right hy the j^^'od- uct of the numhers on the left, cancelling equal factors ', the result will he the answer. Notes. — 1. Commission for drawing is commission on the sale of a draft; commission for remitting is commission on the purchase price of a draft. 2. The above method is sometimes called the Chain liulef or Conjoined Pro- portion. EXAMPLES FOR PRACTICE. 1. A gentleman in Philadelphia wishes to deposit $5000 in a bank at Stockholm, by remitting to Liverpool and thence to Stock- holm ; if exchange on Liverpool is at 10 % premium in Phila- delphia, and the course between Liverpool and Stockholm is 6 roubles 48 copecks per £1, how much money will the man have in bank at Stockholm, allowing the agent at Liverpool J % for remitting? A7is^ 6610 roubles 74 copecks. 2. When exchange at New York on Paris is 5 francs 16 cen- times per $1, and at Paris on Hamburg 2i francs per marc banco, what will be the arbitrated price in New York of 7680 marc bancos of Hamburg? Ans. $3162.79. 3. A man in Cleveland wishes to draw on New Orleans for a bank stock dividend of $750, and exchange direct on New Or- leans is li % discount; how much will he save by drawing on . . EXCHANGE. 35I his agent in New York at 1^ % premium, allowing his agent ta draw on New Orleans at 1 % discount, brokerage at ^ % '/ 4. A gentleman in Boston drew on Wurtemberg for GO 00 gild- ers at $.415 per gilder; how much more would he have received if he had ordered remittance to London, and thence to New York, exchange at AYurtemberg on London being 11 J gilders per £1, and at London on New York 9 J %, in favor of sterling, broker- age at IJ % in London for remitting? Ans. $67.66. 5. If at Philadelphia exchange on Liverpool is at 9| % pre- mium, and at Liverpool on Paris 26 francs 8'6 centimes per £1 ; what is the arbitrated course of exchange between Philadelphia and Paris, through Liverpool ? Ans, 1 franc = $.181. 6. An American resident of Amsterdam wishing to obtain funds from the U. S. to the amount of $6400, directs his agent in London to draw on the U. S. and remit the proceeds to him in a draft on Amsterdam, exchange on the U. S. be'ing at 8 % in favor of London, and the course between London and Amsterdam being 18d. per gilder. If the agent charges commission at J % both for drawing and remitting, how much better is this arbitra- tion than to draw directly on the U. S. at 40 cents per gilder ? 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. 0. Now, if exchange on N. 0. is at i (fo discount in N. Y., and the agent's commission for re- mitting is i %, how much does the stock cost in Pittsburgh? Ans. $5606.08. 8. A banker in New York remits $3000 to Liverpool, by arbi- tration, as follows : first to Paris at 5 francs 40 centimes per $1 ; thence to Hamburg at 185 francs per 100 marcs; thence to Am- sterdam at 85 stivers per 2 marcs; thence to Liverpool at 220 stivers per £1 sterling. How much sterling money will he have in bank at Liverpool, and what will be his gain over direct ex- change at 10 % premium ? . ( Proceeds in Liverpool, £696 lis. 2d. I Gain by arbitration^ £82 18s. 5d. 852 PERCENTAaB. 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. 615. 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 term of credit. In averaging, there are two kinds of equations, Simple and Compound. ' 61^. 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 (late. 1. In settling with a creditor On the first day of April, I find that I owe him $12 due in 5 months, 815 due in 2 months, and $18 due in 10 months ; at what time may T pay the whole amount? EQUATION OF PAYMENTS. 353 OPERATION. Analysis. The $12 X 5 = GO whole amount to be I 15 X 2 = 30 paid, as seen in the ope- 18x10 = 180 ration, is $45 ; and we L^K ©T^ 270 ^^® *^ ^^^ ^^^ ^^^S ^* B 270 --45 = 6 mo., average credit, shall be withheld, or IB Apr. 1, + 6 mo. = Oct. 1, Ans. what term of credit it I^P 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 $60 for 1 mo., because 12 X 5 = GO 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 G months on $45, because 45 X 6 = 270 x 1. Hence the following Rule. I. Multiply each payment hy its term of creditj and divide the sum of the products hy the sum of the payments ; the quotient loill he the average term of credit, II. Add the average term of credit to the date at which all the credits begin ; the result will he the equated time of payment. Notes. — 1. The periods of time used as raultipliers must all be of the same deuomination, 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 duo 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 ? Ans Term of credit, 62 da. ; time of payment, Mar. 3, 1860. 2. A man purchased real estate, and agreed to pay i of the price in 3 mo., } 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 ? ^ 30* X 854 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 oficrs 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* "WTien 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 FIRST OPERATION. SECOND OPERATION. Due Da, Items. Prod. Jan. 12 i Mar. 16 1 64 Apr. 20 99 400 600 375 38400 37125 1375 75525 Due. Da. 99 35 Items. Prod. Jan. 12 Mar. 16 Apr. 20 400 600 375 1375 39600 21000 60600 Ans 75525 -^ 1375 == 55 da. 55 da. after Jan. 12, or Mar. 7. •!: Ansig < 60600 -f 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 earliest maturity, Jan. 12, for a focal date, and find the difi'erence in days between this date and each of the others ; thus, from Jan. 12 to Mar, I 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 lias 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 dj^-aw 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 Lecomes due, hy adding to the date of each transaction the term of credit y if any he specifiedj 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 hy the corresponding number of days in the preceding columii, writing the products in a fourth column. IV. Divide the sum of the products hy the sum of the items. TJie quotient will he 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 cents are given, it is generally suflficient to take only dollars in the multiplicand, rejecting the cents when less than 50, and car- ryinj^ 1 to the dollars, if the cents are more than 50. 2. Months in any terms of credit are understood to be calendar months; the time must therefore be carried forward to the same day of the month in which the term of credit expires. 356 percentage. examples ror practice. 1. James Gordon, 1860. To Henry Lancey, Dr. Mar. 4. To 100 yd. Cassimere, @ $2 50, S250 " 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 ? Arts. Apr. 12, 1860. 2. I sell goods to A at different times, and 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 Tf 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.I7. " " " 537 Ans. Mar. 10, 1861. 4. I have 4 notes, as follows : the first for $350, due Aug. 16, 1859 • the second for $250, due Oct. 15, 1859 ; the third for $300, 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 ? I m EQUATION OF PAYxMENTS. COMPOUND EQUATIONS. C19. 1. Average the following account. John Lyman. 357 Or. 1860. 1860. June 12 To Mdse. 530 GO June 24 By draft at 30 da. 480 00 Sept. 12 li u 428 00 Aug. 20 " cat?h, 280 00 pet. 28 " Sundries, 440 00 Oct. 8 a ii 140 00 OPERATION. Dr. Due. Da. Items. Products. Due. Da. Items. Products. June 12 Sept. 12 Oct. 28 138 46 530 428 440 73140 19688 July 27 Aug. 20 Oct. 8 93 09 20 480 230 140 44640 15870 2800 1398 850 92828 63310 850 63310 ' Balances, 548 29518 29518 -T- 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 botJi 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. Cliarles Derby. 1859. 1859. Jan. 21 To Mdse. H2 00 Jan. 1 By cash, 84 00 Mar. 5 " u 145 00 Feb. 4 40 00 " 22 " " 194 00 Mar. .30 " " 12 00 OPERATION. Cr. Due. Da. Items. Products. Due. Da. Items. Products. Jan. 21 Mar. 5 « 22 68 25 8 32 145 194 2176 3625 1552 Jan. 1 Feb. 4 Mar. 30 88 54 84 40 12 7392 2160 371. 136 7353 136 9552 7353 Balance of account, 235 Balance of products, 2199 2199 -T- 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. 6S0* From these examples we derive the following lluLE. 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 find 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 doie TOWARD the other dates when both EQUATION OF PAYMENTS. 359 ^^ktes when the balances are on opposite sides of the account, ^^^J^'oTES. — 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. A. Z. Armour, Dr. Cr. 1859. 1859. Feb. 12 To Mdse. 85 75 March 15 By bal. old acc't. 97 36 25 (( u 36 24 April 17 " cash, 56 (K) April 16 (( « 174 96 May 25 a u 25 00 Mav 20 « « 94 78 June 8 " sundri'es, 94 75 OPERATION. Dr. Cr. Due. Da. Items. Int. Due. Da. Items. Int. 1.3S .40 .06 Feb. 12 " 25 April 16 Ma^ 20 116 103 53 19 85.75 36.24 17496 94.78 1.66 .62 1 1.55 .30 March 15 April 17 May 25 June 8 85 62 14 97.36 56.00 25.00 94.75 391.73 273.11 4.13 1.93 273.11 1.93 Balances, 118.62 2.20 Int on $118.62 for 1 da. = $.0198. 2.20---.0198=lll da.; June 8—111 da.=:.Feb. 17,1859, Ans. Analysis. Taking the latest maturity, June 8, for the focal date, we find the interest of each item, at 6 ^, 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. l7, 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. 860 PEllCENTAQE. 2. What is the balance of the following account, ^.and when is it due ? ■■ ^ Thomas Lardner, Dr. Cr. 1860. 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 u 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. 1869. Jan. 1 To Mdse. on 3 mo. 54 i April 1 By cash, 50 •00 Feb. 12 " " '^ 2 " 28 May 16 " draft, at 30 da. 30 00 March 16 " Sundries, 95 75 June 12 ii a 125 00 June 25 " Mdse. 26 32 " 20 " cash, 150 00 Ans. Aug. 28, 1859. Oliver Waimcright. Dr. Cr. 1858. 1858. Jan. 1 To Mdse. 36 72 Jan. 10 By fcash, 98 72 Feb. 1 U ii 48 25 " 21 it a ' 25 84 March 17 (( ii 72 36 March 23 " sundries, 15 17 April 1 ii il 98 48 April 6 a a ' 8 90 If the above account were settled April 6, 1858, by draft on time, how many days' credit should be given ? Ans. 20 Qa. 6. 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. Bide, and then average. 6. A man owes $684, payable Aug. 12, and $4:68, payable Oct. 15. If he pay $839; Aug. 1, what will be the equated time for the payment of the balance ? Ans. Dec. 15. 7o 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^^ug. 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. 861 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 ? Ans. 1st, June 28; 2d, Aug. 27. 9. When shall a note be made payable, to balance the followino- account ? James Tyler, Dr. Cr. 1«59. 1859. June 12 To Mdflo. on 3 mo. 530 84 Sept. 14 By caFh, 436 no *' 20 " " f' " 236 48 '• 25 i. u 320 00 « 30 a a « « V39 56 Oct. 3 (( il 560 00 July 6 « a u a 273 44 " n li u 370 00 " 16 " <; a « 194 78 Nov. 16 a « 840 00 " 29 u u a it 636 42 *• 24 ACCOUNT SALES. ©SI. 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. 03S. 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. 6S3. Storage is a charge made for keeping the goods, and may be reckoned by the week or month, on each article or piece. G24. Primage is an allowance paid by a shipper or consignor of goods to the master and sailors of a vessel, for loading it. OS«i. 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. — Multiple/ each article or parcel hy the time it is in store j and multiply the sum, of the pro- ducts hy the rate per unit ; the result will he the storage. II. To find when the net proceeds are due. — Average the sales alone ^ and the result will he the date to he given to the com,mission and guaranty ; then make the sales the Cr. sidcj and the charges the Dr. side, and average the entire account hy 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. 18 GO April 15 May 5 June 28 April 1 " 1 « 1 June 28 Sold 32 Pipes, 4160 gal. @ $1.05, on 30 days,. " 40 '' 6240 " @ 1.02, cash, «• 28 « 3650 " @ LOO, ♦' 6344 3650 100 CHARGES. To Freight and Primage, $136.76 " Wharfage and Cartage 48.54 " huty Bonds, at CO days 3207.07 " Storage from April 1, viz. : On 32 Pipes, 2 wk.s 64 wks. " 40 " 5 " ... 200 " - " 28 « 13 « ... 364 « 100 « equal to 628 «@6c 37.68 « Commission on S13362.80. at 23^ % 334.07 " Guaranty on $i368, at 21^ % 109.20 3873 00 80 00 80 What are the net proceeds of the ahove 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 interval, reduced to weeks, hetween Apr. 1, when the pipes were received into store, and the date of sale. Every fraction of a wecli 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. 363 day, freight $175.48, cartage $56.25, and cooperage $8.87. He sold out the shipment as follows: June 3, 200 bbl. @ $6.25; June 30, 850 bbl. @ $6.50; July 29, 400 bbl. @ $3.12 J; Aug. 6, 50 bbl. @ $6.00. Required the net proceeds, and the date when they shall be accredited to the owner, allowing commission at 3 J %, and storage at 2 cents per week per bbl. Ans. Net proceeds, $5614.28 ; due, July 10. SETTLEMENT OF ACCOUNTS CURRENT. ©26. To find the cash balance of an account current, at any given date. /. Burns in account current with Tyler dh Co, Dr. Cr. 1860. 1860. Feb. 25 To Mdse. on 3 mo. 360 75 March 1 By cash on acct. 250 00 March 20 «4 U it 3 i< 240 56 April 20 " accept, at 30 da. 300 00 April 26 « a « 3 « 875 24 June 12 " Sundries, 375 00 June 24 (( (( (( 2 « 235 26 « 27 " cash on acct. 400 00 Required the cash value of the above account, July 1, 1860, interest at 6 %. OPERATION. ur. \JT. Due. Da. Items. Int. Cash val Due. Da. Items. Int. Cash val. May 25 June 20 July 26 Aug. 24 37 11 25 54 360.75 + 2.22 240.56 -f- .44 875.24 — 3.65 235.25—2.12 362.97 241.00 871.59 233.13 March 1 May 20 June 12 " 27 122 42 19 4 250.00 + 5.08 300.00 + 2-10 375.00 + 1.19 400.00+ .27 265.08 302.10 376.19 400.27 1333.64 1708.69 ^ $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 ■ 864 PARTNERSHIP. E-ULE. I. Find the number of days intervening between each maturity and the day of settlement. II. Compute the interest on each item for the 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 a u 250 48 Feb. 3 " cash, 486 57 Feb. 13 « u 400 00 March 26 U (( 1250 78 March 16 (( » 750 00 April 20 (( U 756 36 April 25 " " 200 00 May 12 a it 248 79 Ans. ^1196.67. 2. What is the cash balance of the following account on Dec. 31, at 7 per cent. ? James Hanson. Dr. Cr. 1859. 1859. 8«pt. 3 To Sandriea, 478 36 Sepk. 17 By Sundries, 96 54 Oct. 2 " Mdse. on 3 mo. 256 37 « 20 " cash on acct. 200 00 " 21 " " " 3 " 375 . 26 f oa Oct. S (i a u 325 00 Nov. 12 « « « 3 « 80 Nov. 17 « (( (( 50 00 Dec. 15 " Sundries, 148 1 it Dec. 27 « « « 84 00 PAETNERSHIP. 037* 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. 6^8. The Partners are the individuals thus associated. Note. — The terms Capital or Stock, Dividendj and Assessment, have the same ?ignificafcion in Partnership as in Stocks. PARTNERSHIP. 365 CASE I. 6S9. 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- ^5 ^ 5^^ A^s part of the stock. ^^«^?^« t¥A =t2^{ the \^Q0_ 7* JVa ic u u u capital, and B furnishes |96 X A = $40 A's share of the gain. t^Vo = h of t^e capi- $90 X Ss = $56, B's '^ '' " '' *^^- ^""^ '^^'^ ^^'^ man's share of the pro- fit or loss will have the same ratio to the whole profit or loss as his part of the capital has to the whole capital, A will have /g of the $9G, and B y\ of the $96, for their respective shares of the profits. We may also regard the whole capital as the first cause^ and each man's share of the capital as the second cause, the whole profit or loss as the first effect, and each man's share of the profit or loss as the second effect, and solve by proportion thus : 1st cause. 2d cause. 1st efifect. 2d effect. $1200 : : $500 = $96 : ( ? ) = $40, A's gain, $1200 : $700 = $96 : (?) = $56, B's " Hence we have the following BuLE. Multiple/ the whole projli or loss hy the ratio of the whole capital to each man's share of the capital. Or, The whole capital is to each man^s share of the capital as the whole profit or loss is to each man's share of the profit or loss» EXAMPLES FOR PRACTICE. 1. Three men engage in trade; A puts in $6470, B $3780, and C $9860, and they gain $7890. What is each partner's share of the profit? Ans. A% $2538.453 ; B's, $1483.053; C's, $3868^.493. 2. B and C buy pork to the amount of $1847.50, of which B pays $739, and C the remainder. They gain $375 ; what is each" one's share of the gain ? ' 31* 366 , PARTNERSHIP. 3. A, B, and C form a company for the manufacture of woolen cloths. A puts in $10000, B $12800, and C $3200. C is allowed $1500 a year for personal attention to the business; their ex- penses for labor, clerk hire, and other incidentals for 1 year are $3400, and their receipts auring the same time are $9400. What is A^s, B's, and C's income respectively from the business ? 4. Four persons rent a farm of 115 A. 32 P. at $3.75 an acre. A puts on 144, B 160, C 192, and D 324 sheep; how much rent ought each to pay ? 5. Three persons gain $2640, of which B is to have $6 as often as C $4, and as often as D $2 ; how much is each one's share ? 6. Six persons are to share among them $6300 ; A is to have ^ of it, B i, C |, D is to have as much as A and C together, and the remainder is to be divided between E and F in the ratio of 3 to 5. How much does each receive ? Arts. A, $900; B, $1260; C, $1400; D, $2300; E, $165; F, $275. 7. Two persons find a watch worth $90, and agree to divide the value of it in the ratio of | to | ; how much is each one's share ? Note. — If the fractions be reduced to a common denominator, they will be to each other as their numerators, (418, III). 8. A father divides his estate worth $5463.80 between his two sons giving the elder J more than the younger ; how much is each son's share J* Ans. Elder, $2892.60; younger, $2571.20. 9. Three men trade in company. A furnishes $8000, and B $12000 Their gam is $1680, of which C's share is $840; required, C's stock, and A's and B's gain. Ans, C's stock, $20,000. 10. Four persons engage in the lumber trade, and invest jointly $22500; at the expiration of a certain time, A's share of the gain is $2000, B's $2800.75, C's $1685.25, and D's $1014; how much capital did each put in ? Ans. I> put in $3042. 11. A legacy of $30,000 was left to four heirs in the propor- tion of ^, I, I, and 5, respectively; how much was the share of each? 12. Three men purchase a piece of land for $1200, of which sum C pays $500. They seli it so as to gain a certain sum^ of PARTNERSHIP. 367 which A takes $71.27, and B $142.54; how much do A and B pay, and what is C's share of the gain ? Ans, C's gain, $152.72 1. 13. Three persons enter into partnership for the manufacture of coal oil, with a joint capital of $18840. A puts in $3 as often as B puts in $5, and as often as C puts in $7. Their annual gain is equal to C's stock; how much is each partner's gain? 14. Ay B, and C are employed to do a piece of work for $26.45. A and B together are supposed to do | of the work, A and C -j^^, and B and C ^^, and are paid proportionally; how much must each receive? Ans, A, $11.50; B, $575; C, $9.20. CASE ir. 030. To find each partner's share of the profit or loss when their capital is employed for unequal periods of time. It is evident that the respective shares of profit and loss will depend equally upon two conditions, viz.: the amount of capital invested by each, and the time it is employed. Hence they will be proportional to the products of these two elements. 1. Two men form a partnership ; A puts in $320 for 5 months, and B $400 for 6 months. They lose $140 ; what is each man's share of the loss ? OPERATION. $320 X 5 = $1600, A^s capital for 1 mo. $400 X 6 = $2400, B's '' '' " $4000, entire '^ '^ '^ $lg-Q-0 = |, A's share in the partnership $140 X I = $56, A^« loss- $140 X I = $84, B's loss. Analysis The use of $320 for 5 months is the same as the use of 5 times $320, or $1G00, for 1 month ; and the use of $400 for 6 months is the same as the use of 6 times $400, or $2400, for 1 month ; hence the use ot the entire capital is the same as the use of $1000 + $2400 ==r $4000 for 1 month. A^s interest in the partnership is therefore \l\\ = §, and he will suffer § of the loss, or $140 X f — $56 .• and 368^ PARTNERSHIP. B's interest in the partnership is f ^ J§ = |, and he will suffer | of the loss, or $140 X §-=$84. We may also solve by proportion, the causes being compounded of the two elements, capital and time • thus : 84000 : 81600 = $140 : (?) = $56, A^s loss, $4000 : $2400 = $140 : (?) = $84, B's loss. Hence the following Rule. 3/wZ^/jjZ^ each maiis cajntal hy tlie time it is employed in tradey and add tlie products. Then multiply the entire profit or loss hy the ratio of eacji product to the sum of the products ; the results will he the respective shares of profit or loss of each part- ner. Or, Midtiply each man^s capital hy the time it is emptloycd in trade, and regard each product as his capital, and the sum of the p7'o^ ducts as the entire cap>ital, and solve ly proportion, as in Case I. EXAMPLES FOR PRACTICE. 1. A, 11, and C enter into partnership. A puts in $357 for 5 months, B $371 for 7 months, and C $154 for 11 months, and they gain $347.20; how much is each one's share? Ans. A^s $102; B's $148.40; C's $96.80. 2. Three men hire a pasture for $55.50. A put in 5 cows, 12 weeks; B, 4 cows, 10 weeks; and C, 6 cows, 8 weeks; how much ought eacii to pay? Ans. A $22.50; B $15; C $18. 3. B commenced business with a capital of $15000. Three months afterward C entered into partnership with, him, and .put in 125 acres of land. At the close of the year their profits were $4500, of which C was entitled to $1800 ; what was the value of the land per acre ? 4. A and B engaged in trade. A put in $4200 at first, and 9 months afterward $200 more. B put in at first $1500, and at the end of 6 months took out $500. At the end of 16 months their gain was $772.20 ; how much is the share of each ? 5. Four companies of men worked on a railroad. In the first company there were 30 men who worked 12 days, 9 hours a day; in the second, there were 32 men who worked 15 days, 10 hours PARTNERSHIP. 369 a day; in the Ihird, there were 28 men who worked 18 days, 11 hours a day; and in the fourth, there were 20 men who worked 15 days, 12 hours a day. The entire amount paid to all the com- panies was $1500; how much were the wages of each company? 6. A and B are partners. A's capital is to B's as 5 to 8 ; at the end of 4 months A withdraws J of his capital, and B | of his ; at the end of the year their whole gain is $4000 ; how much be- longs to each ? Ans. A, $17141 ; B, $2285^; 7. B, C, and D form a manufacturing company, with capitals of $15800, $25000, and $30000 respectively. After 4 months B draws out $1200, and in 2 months more he draws out $1500 more, and 4 months afterward puts in $1000. C draws out $2000 at the end of 6 months, and $1500 more 4 months afterward, and a month later puts in $800. D puts in $1800 at the end of 7 months, and 3 months after draws out $5000. If their gain at the end of 18 months be $15000, how much should each receive? Ans. B, $3228.07; C, $5258.15; D, $6513.78. 8. The joint stock of a company was $5400, which was doubled at the end of the year. A put i for J of a year, B | for J a year, and C the remainder for one year. How much is each one^s share of the entire stock at the end of the year ? 9. Three men engage in merchandising. A^s money was in 10 months, for which he received $456 of the profits ; B's was in 8 months, for whioh he received $343.20 of the profits; and C's was in 12 months, for which he received $750 of the profits. Their whole capital invested was $14345 ; how much was the capital of each? An$, A's, $4332; B's, $4075.50; C's,' $5937.50. 10. Three men take an interest in a coal mine. B invests his capital for 4 months, and claims -j'^ of the profits; C's capital is in 8 months ; and D invests $6000 for 6 months, and claims | of the profits ; how much did B and C put in ? 11. A, B, and C engage in manufacturing shoes. A puts in $1920 for 6 months; B, a sum not specified for 12 months; and C, $1280 for a time not specified. A received $2400 for his stock and profits, B $4800 for his, and C $2080 for his. Required, B's stock, and C's time ? Y \ 870 ALLIGATION. ALLIGATION. 631. Alligation treats of mixing or compounding two or more ingredients of different values or qualities. 63S. The Mean Price or ftnality is the average price or quality of the ingredients, or the price or quality of a unit of the mixture. CASE I. 633. To find the mean price or quality of a mixture, when the quantity and price of the several ingredients are given. Note. — The process of finding the mean or average price of several ingredi- ents is called AUxyation Medial. 1. A produce dealer mixed together 84 bushels of oats worth $.30 a bushel, 60 bushels of oats worth $.38 a bushel, and 56 bushels of oats worth $ 40 a bushel ; required, the mean price. OPERATION. Analysis. The worth of 84 $.30 X 84 = $25.20 bushels @ $.30 is $25.20, of .38 X 60 = 22.80 60 bushels @ $.38 is $22.80, .40 X 56 = 22.40 and of 56 bushels @ $.40 is 200 ^ ^70 40 $22.40 ; and we have in the ;; whole compound 84 + 60 + 56 $.3520, Ans, —200 bushels, worth $25.20+ $22.80 + $22.40 = $70.40. One bushel of the mixture is therefore worth $70.40 -h 200 = $.352. Hence the following Rule. Find the entire cost or value of the xngredientSy and divide it hy the sum of the simples. EXAMPLES FOR PRACTICE. 1. A grocer mixed 4 lb. of tea at $.60 with 3 lb. at $.70, 1 lb. at $1.10, and 2 lb. at $1.20; how much is 1 lb. of the mixture worth? -^/is. $.80. 2. A dealer in liquors would mix 14 gal. of water with 12 gal. of wine at $.75, 24 gal. at $.90, and 16 gal. at $1.10; how muck is a gallon of the mixture worth ? Ans. $.73^^-5. ALLIGATION. 371 3. If 3 lb. 6 oz. of gold 23 carats fine be compounded with ^4 lb. 8 oz. 21 carats, 3 lb. 9 oz. 20 carats, and 2 lb. 2 oz. of alloy, what is the fineness of the composition ? Ans, 18 carats. 4. A grain dealer mixes 15 bu. of wheat, at $1.20 with 5 bu. at $1.10, 5 bu. at S.90, and 10 bu. at $.70 ; what will be his gain per bushel if he sell the compound at $1.25. 5. A merchant sold 17 lb. of sugar at 5 cts. a pound, 51 lb. at 8 cts., 68 lb. at 10 cts., 17 lb. at 12 cts., and thereby gained on the whole 33 J per cent; how much was the average cost per pound ? 6. A drover bought 42 sheep at $2.70 per head, 48 at $2.85, and 65 at $3.24 ; at what average price per head must he sell them to gain 20 per cent.? Ans, $3.567^ f. 7. A surveyor took 10 sets of observations with an instrument, for the measurement of an angle, with the following results : 1st, 36° 17' 25.4"; 2d, 36° 17' 24.5"; 3d, 36° 17' 27.8"; 4th, 36° 17' 26.9"; 5th, 36° 17' 25.4"; 6th, 36° 17' 24.7"; 7th, 36° 17' 24.2"; 8th, 36° 17' 26.3"; 9th, 36° 17' 25.8"; 10th, 36° 17' 26.7". What is the average of these measurements ? Ans, 36° 17' 25.77''. 8. Three trials were made with chronometers to determine the difi'erence of time between two places; the first trial gave 37 min. 54.16 sec, the second 37 min. 55.56 sec, and the third 37 min. 54.82 sec Owing to the favorable conditions of the third trjal, it is entitled to twice the degree of reliance to be placed upon either of the others ; what should be taken as the difference of longitude between the two places, according to these observations? Ans, 9° 28' 42.6". CASE IL 634. To find the proportional quantity to be used of each ingredient, when the mean price and the prices of the several simples are given. Note. — The process of finding the quantities to be used in any required mix- ture is commonly called Alliyation Alternate. 1. A farmer would mix oats worth 3 shillings a bushel with peas worth 8 shillings a bushel, to make a compound worth 5 shil- lings a bushel ; what quantities of each may he take ? 372 ALLIGATION. oPERiTiON. Analysis. If a mixture, in any pro- ^ o I , I Q \ portions, of oats worth 3 shillings a 5 -< o ? 9 [■ Ans. bushel and peas worth 8 shillings, be ^ ^ ^ priced at 5 shillings, there will be a gain on the oats, the ingredient worth less than the mean price, and a loss on the peas, the ingredient worth moi^e than the mean price ; and if we take such quantities of each that the gain and loss shall each be 1 shilling, the unit of value, the result will be the required mixture. By selling 1 bushel of oats worth 3 shillings for 5 shil- lings, there will be a gain of 5 — 3 = 2 shillings, and to gain 1 shil- ling would require J of a bushel ; hence we place i opposite the 3. By selling 1 bushel of peas worth 8 shillings for 5 shillings, there will be a loss of 8 — 5 = 3 shillings, and to lose 1 shilling will require J of a bushel ; hence we write J opposite the 8. Therefore, } bushel of oats to :J of a bushel of peas are the propoy^tional quantities for the required mixture. It is evident that the gain and loss will be equal, if we take any number of times these proportional terms for the mix- ture. We may therefore multiply the fractions J and J by 6, the least common multiple of their denominators, and obtain the integers 3 and 2 for the proportional terms (418,111); that is, we may take, for the mixture, 3 bushels of oats to 2 bushels of peas. 2. What relative quantities of sugar at 7 cents, 8 cents, 11 cents, and 14 cents per pound, will produce a mixture worth 10 cents per pound ? OPERATION. Analysis. To preserve the equality of gains and losses, we must compare two prices or sim- ples, one greater and one less than 10 ^ -n -, o r» the mean rate, and treat each pair or couplet as a separate example. Thus, comparing the simples whose prices are 7 cents and 14 cents, we find that, to gain 1 cent, J of a pound at 7 cents must be taken, and, to lose 1 cent, J of a pound at 14 cents must be taken ; and com- paring the simples the prices of which are 8 cents and 11 cents, we find that J pound at 8 cents must be taken to gain 1 cent, and 1 pound At 11 cents must be taken to lose 1 cent. These proportional terms are 11 14 1 2 3 4 5 T 4 4 i 1 1 1 2 2 i 3 3 10^ 1 2 3 7 T 1 8 J 4 11 1 3 14 } 2 ALLIGATION. 373 "written in columns 1 and 2 We now reduce these couplets separately to integers, as in the last example, writing the results in columns 3 and 4 ; and arranging all the terms in column 5, 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 fo^ 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 for either term is the difi'erence 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 difi'erence between 14 and the mean rate 10 ; and the proportional number corresponding to the latter simple is 3, which is the difi'erence 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 difi'erence 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 EuLE. 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 mean ratCy 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 m 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- tijyly 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. NoTKS. — 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 mul- 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 $5i, $6, and $71 per barrel, must be sold, to realize an average price of $6i per barrel? OPERATION. Analysis. Comparing the r ^1 4 14 A first price with the third, we ob- 0J. J g 4 j 12 12 ^^^^ *^® couplet J to t; and com- (74ffj2 2 4 paring the second price with the third, we obtain the couplet 4 to •J. 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 kjnd 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 lb. of each of the first and third kinds, 2 lb. of the second, and 5 lb. 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 ? J (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 per pint, will make a mixture worth $.45 a pint ? 6. What relative quantities of silver | pure, | pure, and ^^^ pure, will make a mixture | pure ? Ans. 3 lb. J pure, 3 lb. | pure, and 20 lb. j% pura. CASE III. 63«>. "When two or more of the quantities are es- quired to be in a certain proportion. 1- A farmer having oats worth $.30, corn worth $.60, and wheat ins. < ALLIGATION. 375 r 30 50 ) 60 (no 1 2 3 4 5 6 ^ 7u 1 3 6 7 t'd 2 2 e^i 1 2 2 worth $1.10 per bushel, desires to form a mixture worth §.50 per bushel, which shall contain equal parts of corn and wheat ; in what proportion shall the ingredients be taken ? OPERATION. Analysis. We first obtain 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 IlULE. I. Compare the given prices, and obtain the proportional terms by couplets, as in Case IL 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 Ihve, till a set of pro- portional terms is obtained, answeri7ig 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. 3 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 ? J, ( For the first mixture, 1, 1 and 2. 1 For the second mixture, 1, 4 and 4. 376 ALLIGATION. OPERATION. r28 1 58 ] 441 ' 3'tT 1 T4 7 5 140 100 3*5 .'s 6 2 8 160 CASE IV. 636. Wlien the quantity of one of the simples is limited. 1. A miller has oats worth ^.28, corn worth 8.44, 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 ? Analysis. By Case 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 ' J*^ =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 o. oats, and 8 X 20 = 160 bushels of barley. Hence the following HuLE. Find the proportional quantities by Case II or Case III. Divide ihe^ given quantity hy the proportional quantity of this ingredient J and multiply each of the other proportional quan- tities hy 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 $i a yard, that the average price of the whole may be $1^? Ans, 12 yards at $1J 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 lbs. of the third kind; hoT? many pounds of each of the first two kinds did he take ? ALLIGATION. 377 OPERATION. $.48 X 18 = § 8.64 .52 X 8 = 4.16 .85 X J = 3.40 30 ) SI 6.20 Mean price of the ) ^ ^j_ given simples J ( <■ 54 1114. 2 6 30 84. 108 ^■4 5 5 25 { Ll44 gV 1 1 5 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 ? Analysis. Of the given quantities there are 18 -{- 8 + 4 = 30 bushels, whose mean or average price we find by Case I to be $.54. We are therefore required to mix 30 bushels of grain worth $.54 per bushel, with rye at $1.08, and wheat at $1.44, to make a compound worth $.84 per bushel. Pro- 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 y and find their mean price hy Case I; then consider this mixture as a single ingredient whose quantify and price are known, andfitid the quantities of the other ingredients hy Case IV, EXAMPLES FOR PRACTICE. 1. A gentleman bought 7 yards of cloth @ $2.20, and 7 yards @ $2 j 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 $J a bushel. How many oats worth $} 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, $.45, and $.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, 1 o o ^ r r» we find the proportional 1 2 3 4 5 6 ^ / XI. quantities to lorm tne (SO TU tV 2 13 36 mixture to be 3 lb. at 40 } 45 I 2 2 24 $.30, 2 lb. at $.45, and (OO ^jj 1 1 12 1 lb. at $.00. Adding ~' ~ these proportional quanti- ^ '^ ties, we find that they would form a mixture of 6 pounds. And since the required mixture Is y ^ 12 times 6 pounds, we multiply each of the proportional terms by 12, and obtain for the required quantities, 36 lb. at $.30, 24 lb. at $.45, and 1 2 lb. at $.60. Hence the following Rule. Fi7id the proportional numbers as in Case IT or Case III, Divide the given quantify hy the sum of the proportional quantities, and multiply each of the proportional quantities hy 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 lb, that shall be worth 20 cts. a pound ? Ans. 40 lb. at 8 cts., 40 lb. at 16 cts., and 160 lb. at 24 cts. 2. A man bought calves, sheep, and lambs, 154 in all, for $154. He paid $3 J for each calf, $li for each sheep, and $J 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, and to the boys $} a week ; how many were there of each ? Ans, 30 men, 5 women, and 20 boys. INVOLUTION- 379 INVOLUTION. OSf^* A Power is the product arising from multiplying a number by itself, or repeating it any number of times as a factor. ^ G4L0^ Involution is the process of raising a number to a given power. 649. The Square of a number is its second power. 643. The Cube of a number is its third power. 643* In the process of involution, we observe, I. 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 mora 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 ? Analysis. We multiply 6 by it- self, and this pro- duct by 6, and so on, until 6 has 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, (11). 2. What is the 6th power of 12 ? Analysis. Wg find the cube of ^-^ = i^"^^ the second power, which must be 144» = 2985984, Ans. .^e 6th power, (III). 644. Hence for the involution of numbers we have the fol* lowing OPERATION. x6 X 6 X 6 X Or 6 = 7776, Ana. 6 X 6 = 6^ = 36 86 X 6 = 6' = 216 X 6' =:: 6^ = 216 X 36 = 7776, Ans. 880 INVOLUTION. KuLE, I. Multiply the given number hi/ itself in continued mulHplicution, till it has been taken as tnany times as a factor as there are v.vits in the exponent of the required power. Or, IL Multipl}/ 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, NoTKS. — 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. Eaise 16| to the 4th power. Ans. 79659|f I.' 5. Eaise 2 to the 20th power. Ans. 1048576, 6. Kaise .4378565 to the 8th power, reserving 5 decimals. \ Ans. .00135 -t: 7. Raise 1.052578 to the 6th power, reserving 4 decimals. Ans. 1.3600 db. 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 (1)3. 11. (;2|/. 12 4.G» X 25' 13. (6|y — 7.25*. 15. I of (1)3 of (^^y, Note.— Cancel like powers of the same factor. 16. 7«-f-3.08. 17 (4^ X 5« X 12«) -^ (4^ X 10* X 32). Ans. If 3. Ans. mfil. Ans. 1520875. 14. (8J/ X 2.5^ Ans. 5|. Am, 1200 EVOLUTION. 381 E^^OLUTION. 64:«>. A Root is a factor repeated to produce a power; thus, in the expression 7x7x7 = 34.3, 7 is the root from which the power, 343, is produced. 64®. 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. 04T. 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. 6«S0. 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. 6«dl* The names of roots are derived from the corresponding powers, and are denoted by the indices of the radical sign. Thus, •s/lOO denotes the square root of 100; \^1U0 denotes the cube roo^of 100; v^ 1 00 denotes the /oi^r^/i root of 100; etc. 6^S« 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. 6«S3« Fractional exponents are also used to denote both invo- lution and 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, 7 denotes the cube root of the second power of 7, and is the same as >/V) so also 7^ = \/7^ 6«>4. In extracting any root of a number, any figure or 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 oyder, or the 254 as tens of the first order. 882 EVOLUTION. SQUARE ROOT. 653. 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 8 X 8 = 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. 6«56« The relative number of places in a given number and its square root is shrwn in the following illustrations. Roots. Squares. i 1 9 81 99 98,01 999 99,80,01 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 off a numher into two-figure 'periods^ commencwg at the right hand, the number of fall periods avd the left hand full or partial period loill indicate the numher of places in the square roof. 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 ro lated in local value as follows : 20002 ^ 4 00 00 00 23002 = 5 29 00 00 23402 == ^ 47 56 00 23452 ^ 5 49 90 25 : Hence, II. The square of the first figure of the root is contained xchoUy in the first period of the power ; the square of the first two figures Roots. Squares. 1 1 10 1,00 100 1,00.00 1000 1,00,00,00 I 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 we 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 part separately : thus, 40 + 3 40 4- 3 120 + 9 1600 + 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 hy the units J 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, lY. If the square of the tens he subtracted from the entire square, the remainder will he equal to tiolce the tens plus the units multiplied hy the units. 1. What is the square root of 5405778576 ? OPERA noN. Analysis. Pointing ofi" the 5405778576 ( 73524 gi^^n number into periods of 49 two figures each, the 5 periods show that there will be 5 fig- ures in the root, (I). Since the square of . the first figure of the root is always contained wholly in the first period of the power, (II), we seek for the iT^^yTTi cooi-t^ greatest square in the first pe" 14/044 588176 ^. . m w i, fi a \. 588176 nod, 54, which we find by trial to be 49, and we place 143 505 429 1465 7677 7325 14702 35285 29404 884 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) j)^us the second fiigt. re (units), multiplied hy the second figure, (IV). Now if we could divide this remainder by tioice 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 w^e therefore write twice the first figure, or 14, at the left of 505 for a tried 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 '^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 of the root is contained wholly in the first three periods, (II) we bring down the third period, 77. to the remainder, 7G, and obtain for a new dividend 7677, which must contain at least ticice ihe two figures already found plus the third, mxdtiplied hy the third, (lY). 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 in the former steps, and thus continue till the entire root, 73524, is obtained. 6«5T. From these principles and illustrations we derive the following Rule. 1. Point off the given numher into periods of two figures each, counting from units place toicard the left and right, II. Find ihe greatest square numher in the left hand p)criod, and write its root for the first figure in the root ; subtract the square numher from the left hand period, and to the remainder hring dawn the next period for a dividend. SQUARE ROOT. §§5 III. At the left of the dividend write twice the fir^t figure of the root, for a trial divisor ; divide the dividend, exclusive of its right hand figure, hy the trial divisor, and write the quotient for a trial figure in the root, Y\ . Annex the trial figure of the root to the trial divisor for a complete divisor ; midtipli/ the complete divisor hy the trial figure in the root, subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. V. Multiply the last figure of the last complete divisor by 2 and add the product ^o 10 times the previous divisor, for a new trial divisor, with which proceed as before. Notes. — 1. If at any time the product be greater than the dividend, diminish the trial figure of the root, and correct the erroneous work. 2. If a cipher occur in the root, annex a cipher to the trial divisor, and another period to the dividend, and proceed as before. 3. If there is a remainder after all the periods have been brought down, annex periods oi ciphers, and continue the root to as many decimal places as are required. 4. The decimal points in the work may be omitted, care being taken to point off in the root according to- the number of decimal periods used. 5. The square root of a common fraction may be obtained by extracting the souare roots of the numerator and denominator separately, provided the terms are perfect squares; otherwise, the fraction may first be reduced to a decimal. 6. Mixed numbers may be reduced to the decimal form before extracting the root ; or, if the denominator of the fraction is a perfect square, to an improper fraction. 7. The popil will acquire greater facility, and secure greater accuracy, by keeping units of like order under each other, and each divisor opposite the correspoading dividend, as shown in the operation. EXAMPLES FOR PRACTICE. 1. What is the square root of 315844 ? Ans. 562. 2. What is the square root of 152399025 ? Ans. 12345. 3. What is the square root of 56280004 ? Of 597 ? ' 4. What is the square root of 10795.21 ? Ans. 103.9. 5. What is the square root of 58.14061 ? Ans. 7.62i. Find the values of the following expressions : 6. v/. 01)00316969. A7is. .00563. 7. v/3858.07694409"64. Ans. 62.11342. 8. n/|. Ans. .745355+. 9. \/9^225 — 63504. 10. \/. 126736— \/.045369. 11. ^\n X ^im^ Ans. {3. 12. v/8P~x 625^ x~2^ Ans. 202500. 33 z 886 EVOLUTION. 48 400 384 562 1600 1124 5648 47600 45184 5656 2416* 2262 566 154 113 CONTRACTED METHOD. 6*18. 1. Find the square root of 8; correct to 6 decimal places. OPERATION. Analysis. Extracting the square 12.8284274-, Ans, root in the usual way until we have 8 000006 obtained the 4 places, 2.828, the 4 corresponding remainder is 2416, and the next trial divisor, with the cipher omitted, is 5656. We now omit to bring down a period of ciphers to the remainder, thus contracting the dividend 2 places ; and we contract the divisor an equal number of places by omitting to annex the trial figure of the root, and regarding the right hand figure, 6, as a rejected or re- dundant figure. We now divide as __ ,-- in contracted division of decimals, ^Q (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+, contains as manT/ places as there are places in the periods used. Hence the following KuLE. I. If necessary^ annex periods of ciphers to the given number J and assume as many figures as tJiere are places requirecl in the root ; then proceed in the usual manner until all the assumed figures have been employ ed, omitting the remaining figures^ if any, II. Form the next trial divisor as usualy but omit to annex to it the trial figure of the root j reject one figure from the right to form each subsequent divisor ^ and in multiplying regard the right hand fi.gvre of each contracted divisor as redundant NoTKR. — 1. If the rejected figure is 5 or more, increase the next left hand fj^'uie by 1. 2. Alwii3''s take full periods, both of decimals and integers. EXAMPLES FOR PRACTICE. 1. Find the square root of 82 correct to the seventh decimal place. Ans. 5.6568542 + . CUBE ROOT. 387 2. Find the square root of 12 correct to the Beventh decimal place. Ans. 3.4G41016+. 8. Find the square root of 3286.9835 correct to the fourth decimal place. Ans. 57.3322 -f-. 4. Find the square root of .5 correct to the sixth decimal place. Ans. .745355+. 5. Find the square root of 6^ correct to the sixth decimal place. Ans. 2.563479 + . 6. Find the square root of 1.06^ correct to the sixth decimal place. Ans, 1.156817+ . 3 7. Find the value of 1.0125^ 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. 6«S9« 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 ; and 4th. The factors of these combinations. 600* The relative number of places in a given number and its cube, is shown in the following illustrations : Roots. Cubes. 1 9 99 999 1 729 907,299 997,002,999 Roots. Cubes. 1 10 100 1000 1 1,000 1,000,000 1,000,000,000 From these examples, we perceive, 3g8 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 : 6000» = 125 000 000 000 6400» = 157 4G4 000 000 5420* =159 220 088 000 5423» = 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, 542 == 502 -f- 2 X 50 X 4 + 42 50 + 4 502 X 4 + 2 X 50 X 42 + 43 508+2x50^x4+ 50X4^ 543= 503+3 X 50^x4+3 X 50 X 42 + 43 = 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 ten^ and units is equal to the cube of the tern, 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 EOOT. 38& parts, considered as one number, may therefore be separated into two factors, thus, (3 X 502 + 3 X 50 X 4 + 42) X 4 Hence, TV. If the cube of the tens he subtracted from the entire cube^ the remainder will be composed of two factors^ one of which will be three times the square of the tens plus three times the tens multipUed by the units plus the square of the units ; and the other j the units, 1. What is the cube root of 145780726447 ? OPERATION. 145780726447 ( 5263, Ans. I II 125 152 304 1566 9396 7500 20780 7804 15608 811200 5172726 820596 4923576 83002800 249150447 15783 47349 83050149 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 powor, (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), ^Zz/5 the square of the second ; and the other, tho second figure (IV). Now if we could divide this dividend by the first of these factors, the quotient would be the other fuctor, or the second 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 tho square of the first figure, regarded as tens, or 50^ 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, (lY). But as 50 X 3 X 2 -f 2^ r= (50 X 3 -f- 2) X 2, we annex the second figure, 2, to three times the first figure, 5, and thus obtain 50 X 3 -f 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 tliret 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 hy 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 = (502 X 3) + (50 X 3 X 2) + 22 304 = 50 X 3 X 2 + 22 4= 22 8n2^ (502 + lOO'x 2 +22) X 3 = 522 X 3 This number is obtained in the operation without re-writing the parts, 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. 6G1* From these principles and illustrations we deduce the following Rule. I. Point off the given number info periods of three figures each, counting from units" place toward the left and right. II. Find the greatest cube that does not exceed the left hand period J and icrite its root for the first figure in the required root; subtract the cube from the left hand period^ and to the remainder bring 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. lY. Annex the trial figure to three times the former figure, and write the result in a column marhed 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 ; sidttract the product from the dividend, and to the remainder bring dozen the next period for a new dividend. YI. Add the square of the last figure of the root, the last term in column II, and the complete divisor together, and annex two ciphers^ for a new trial divisor j with which obtain another trial figure in the root. 392 EVOLUTION. YII. Multiply the unit figure of the last term in column I hy 3, and annex the trial figure of the root for the next term of column I; multiply this result hy the trial figure of the root for the next term of column II ; add this term do the trial divisor for a complete divisor, icith which proceed as hefore. Notes. — 1. If at any time the product be greater than the dividend, diminish the trial figure of the root, and correct the erroneous work. 2. If a cipher occur in the root, annex two more ciphers to the trial divisor, nnd another period to the dividend; then proceed as before with column I, an* nexlng both cipher and trial figure. EXAMPLES FOR PRACTICE, 1. What is the cube root of 389017 ? Ans, 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? A7is. .053. Find the ^'alues of the following expressions. 8. ^12'2615327232y Ans. 4968. 9. ^n7i34^W^ Ans. 8. 10. Va¥30¥'? Ans. 1156. • ^^I)D5 ^ ^ 3TT9 * ^'^^' B5' 12. How much does the sum of the cube roots of 50 and 31 exceed the cube root of their sum? Ans, 2.4986 +. CONTRACTED METHOD. G63. In applying contracted decimal division to the e -tac- tion of the cube root of numbers, we observe, 1st. For each new figure in the root, the terms in the operati' a extend to the right 3 places in the column of dividends, 2 placv.s 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 II OPERATION. |5.73 879355dz, Ans. 189.000000 125 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 fcH* the next figure of the root. To complete the divisor, we obtain a correc- tion, 1375, contracted 2 places by omitting to annex 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 157 1099 5139 7500 8599 64 000 60193 1713 974700 979839 3 807000 2 939517 1719 1375 984987 986362 867483* 789090 17 12 98774 98786 78393 69150 9880 9243 8892 988 351 296 99 55 50 10 5 5 894 ETOLUTION. 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 db, 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 , amiex ciphers to the given number j 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 tinal divisor, III In completing the contracted divisor Sy omit at first to annex the trial figure of the root to the term in column I, and reject 2 2ilaces in forming each succeeding term in this column. lY. In multiply'mg, 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 disrej2;arded in forming the new trial divisors. 2. Employ oxAy full periods in the number. EXAMPLES FOR PRACTICE. 1. Find the cube root of 24, correct to 7 decimal places. Ans. 2.8844992 ±. 2. Find the cube root of 12000.812161, correct to 9 decimal places. Ans. 22.894801334 db. 8. Find the cube root of .171467, correct to 9 decimal places. Ans. .555554730 ±. 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 ±. 6. Find the value of v^l" to 6 places. Ans. .941035 ±. 7. Find the value of ^.571428 to 9 places. Ans. .829826686 ±. ROOTS OF ANY DEGREE. 395. 8. Find the value of VlU8G74325^ to 7 places. Ans, 1.057023 zfc. 5 9. Eind the value of 1.053 to 7 places. Ans. 1.084715 ±. ROOTS OP ANY DEGREE. 063. 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. 6G4. 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, (64:3, 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 \/ 2176782336 = 46657 therefore extract the square root v^ 46656 = 36 Ans, ^^ *he given number, and the cube root of this result, and ob- Or . ' tain 36, which must be the 6tli %^2176782336 = 1296 root required. Or, we first find \/l296 = 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 KuLE. Separate the index of the required root iiito its prime factors, and extract successively the roots indicated hy the several factors obtained ; the final result will he the required root, EXAMPLES FOR PRACTICE. 1. What is the 6th root of 6321363049 ? Ans. 43. 2. What is the 4th root of 5636405776 ? Am, 274. 396 EVOLUTION. 3. What is the 8th root of 1099511627776 ? Ans, 82. 4. What is the 6th root of 25632972850442049 ? Ans. 543. 5. What is the 9th root of 1.577635 ? Ans. 1.051963 +. Note. — Extract the cube root of the cube root by the contracted method, carrying the root iu each operation to 6 decimal places only. 6. What is the 12th root of 16.3939 ? Ans. 1.2624 + . 7. What is the 18th root of 104.9617 ? Am. 1.2950+. CASE II. G&S. 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 ^oQg ^ 2.59-f fi^^ ^y ^^s® I' tl^e 6t^^ ^3Qg __ 2.044- ^^^*» ^^^ ^^^^ *^^® ^'^'^ 2.59-f- 2.04 = 4.63 root of 308 ; and since 4.63 ^ 2 = 2.31, assumed root. the 7th root must be 2.316 =151.93 less than the former 308 ~ 151.93 = 2.0272+ and greater than the ?;.'L79'-+7'-'?9ATr''l^^^' • ^- latter, we take the ave- 15.8872 -r- 7 = 2.2596, 1st approximation. ^ ^^ , ^-^ rage of the two, or one fo^f 1^''r7°if '% 9^^ir,, half of theirsums,2.31, 308 — 13b.b748 = 2.253452-f- , ,, ., ,, \ 2.2696 X 6 + 2.253452 = 15.871052 ^^^ ^^^^ ^* *^® assumed 15.871052 -^ 7 = 2.267293, 2d approx. '^oot. We next raise the assumed root, 2.31, to the 6th power, and divide the given number, 308, by 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. 897 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 2.2G06, and obtain 2.253452-f- 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 hy trial wine number nearJy equal to the re- quired rootj and call this 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 odd as many times the assumed root as there (ire 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 of the required root contains the factors, 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 5tii root of this result. EXAMPLES FOR PRACTICE. 1. What is the 20ih root of 617 ? OPERATION. 20 = 2 X 2 X 5. v/617 = 24.839485+. v^^ ^4.839485 = 4.983923+. s/IWd^rS = L378206 + . ^ws. 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 ? 34 898 SQUARE AND CUBE ROOTS. APPLICATIONS OF THE SQUARE AND CUBE ROOTS. 666. An Angle is the opening between two lines ^ that meet each other. A 667. A Right Angle is an angle formed by two lines perpendicular to each other. Thus, B A C is a right angle. 6685 If an angle is less than a right angle, it is acute ; if greater than a right angle, it is obtuse. Thus, the angle .on the right of the ^ line C B is acute, and the angle on the left of C B is obtuse. ^ 669. Parallel Lines are lines hav- a ing the same direction, as A and B. b 670. A Triangle is a figure having three sides and three angles, as A B C. 671. A Right- Angled Triangle is a triangle having one right angle, as at C. 672. The Hypotenuse Is the side opposite the ^ right angle, as A B. 673. The Base of a triangle is the side on which it is sup- posed to stand, as A C. 674:. The Altitude of a triangle is the perpendicular distance from the base, or the base produced, to the angle opposite, as C B. Note. — The altitude of a right-angled triangle is the side called the perpen- dicular. 67o. A Sq[ijare is a figure having four equal sides and four right angles 676. A Rectangle or Parallelogram is a figure having four right angles, and its opposite sides equal. 677. A Diagonal is a line drawn through a figure, joining two opposite angles, as A C. APPLICATIONS. 399 I G78. A Circle is a figure bounded by one uniform curved line. 679. The Circumference of a circle is the curved line bounding it. 080. The Diameter of a circle is a straight line passing through the center, and terminating in the circumference. 681, A Semi-Circle is one half of a circle. 685. A Prism is a solid whose bases or ends are any similar, equal, and parallel plane figures; and whose sides are parallelograms. 683. A Parallelepiped is a solid bounded by six parallelograms, the opposite ones of which are parallel and equal to each other. Or, it is a prism whose base is a parallelogram. 684:, A Cube is a solid bounded by six equal squares. The cube is sometimes called a Right PruTYi, 083. A Sphere or Globe is a solid bounded by a single curved surface, which in every part is equally distant from a point within called its center. 686. The Diameter of a sphere is a straight line passing through its center, and terminating at its surface. 687. A Hemisphere is one half of a globe or sphere. 688. Similar Figures and Similar Solids are such as have their like dimensions proportional. PROBLEM I. 689. To find eitlier side of a right-angled triangle, the other two sides being given. Let us take any right-angled triangle, as ABC, and form the equare, A E D C, on the hypotenuse. Now take a portion, ABC, of this square, and move it as on a hinge at A, until the points B and C 400 SQUARE AND CUBE ROOTS. D 1 F \ \ ^ G ^ are brought to the positions of H and E, respectively. Take also another portion, D F C, and move it as on a hinge at D, until the points F and C are brought to the positions of G and E, respectively. Then the figure formed by the parts thus moved and the remain- ing part will be composed of two new squares, one on A B, the base of the triangle, and one on _ D F, which is equal to the per- pendicular of the triangle. Hence, The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. From this property we derive the following Rule. I. To find the hypotenuse ; — Add the squares of the two sides J and extract the square root of the sum. II. To find either of the shorter sides ; — Subtract the square of the given side from the square of the hypotenuse j and extract the square root of the remainder. EXAMPLES FOR PRACTICE. 1. The top of a tower standing 22 feet from the shore of a river, is 75 feet above the water, and 256 feet in a straight line from the opposite shore 3 required the width of the river. Ans. 222.76 ft. 2. Two ships set sail from the same port, and one sails due east 50 leagues, the other due north 84 leagues ; how far are they apart ? 3. A ladder 50 ft. long will reach a window 30 ft. from the ground on one side of the street, and without moving the foot, will reach a window 40 ft. high on the other side ] what is the breadth of the street? 4. What is the distance through a cubical block, measured from one corner to the opposite diagonal corner, the side of the cube being 6 feet? ' Ans. 10.39 ft ^ ICATIONS. 401 PROBLEM II. 690. To find the side of a square equal in area to a ''given rectangle. Note. — This case, arithmetically considered, requires us to find a mean pro- portional between two given numbers. The product of the sides of the rectangle will be the area which the square is to contain ; hence EuLE. 3Iultiply the sides of the rectangle together, and extract the square root of the product. EXAMPLES FOR PRACTICE. 1. There is a field whose length is 208 rods, and whose breadth is 13 rods ; what is the length of the side of a square lot contain- ing an equal area ? Ans. 52 rods. 2. If it cost $312 to inclose a farm 216 rods long and 24 rods wide, how much less will it cost to inclose a square farm of equal area with the same kind of fence ? 3. What is the mean proportional between 12 and 588 ? Ans, 84. 4. A and B traded together. A put in $540 for 480 days, and received J of the gain ; and the number of dollars which B put in was equal to the number of days it was employed in trade. What was B's capital ? Ans. $720. PROBLEM III. 691. To find the two sides of a rectangle, the area and the ratio of the sides being given. Note. — This case, arithmetically considered, requires us to find two numbers whose product and ratio are given If we multiply together the terms of the given ratio, the product will be the area of a rectangle similar in form to the rectangle whose sides are required. Now we perceive, by the accompanying figures, that multiplying both sides of any rect- angle by 2, 3, 4, eta, multiplies the area by the 34* 2a 402 SQUARE AND CUBE HOOTS. I j I squares of these numbers, or 4, 9, 16, etc. If, therefore, we divide the given area by the rectangle of the terms pro- portional to the required sides, the quo- tient will be the square of that number which must be multiplied into these pro- portional terms to produce the required sides. Hence the following E.ULE. I. Divide the given area hy the product of the terms proportional to the sideSy and extract the square root of the quotient. II. Multiply the root thus obtained hy each proportional term ; the products will he the corresponding sides. EXAMPLES FOR PRACTICE. 1. The sides of a rectangle containing 432 square feet are as 4 to 3 ; required the length and breadth. Ans. Length, 24 feet; breadth, 18 feet. 2. Separate 23 into two factors which shall be to each other as 2 to 3. Ans. 3.91578 + ; 5.87367 +• 3. It is required to lay out 283 A. 2 E. 27 P. of land in the form of a rectangle whose length shall be 3 times the width; what will be the dimensions ? Note. — The proportional terms are 3 : 1. Ans. 369 rods; 123 rods. PROBLEM IV. 69^. To find the radius, diameter, or circumference of a circle, the ratio of its area to a known circle being given. All examples of this class relating to circles, may be solved by means of the following property : — The areas of two circles are to each other as the squares of their radiij diameters^ or circumferences. Note. — This property of the circle is only a particular cnse of a more general principle, viz. : That the areas of similar figures are to each other as the squares of their like dimensions. This principle is rigidly demonstrated in Geometry, but cannot be easily proved here. iffpLICATIONS- 403 EXAMPLES FOR PRACTICE. 1. The radius of a circle containing 28.2744 sq. ft., is 6 ft. ; what is the radius of a circle containing 175.7150 sq. ft. ? 28.2744 : 175.7150 = 6=^ : () = 225, square of radius re- quired. Hence, v^225 = 15, Ans. 2. If it cost $75 to inclose a circular pond containing a cer- tain area, how much will it cost at the same rate to inclose an- other, containing 5 times the area of the first? Ans, $167.70. 3. If a cistern 6 feet in diameter hold 80 barrels of water, what must be the diameter of a cistern of the same depth to hold 1200 barrels ? 4. If a pipe 1.5 in. in diameter will fill a cistern in 5 h., what must be the diameter of a pipe that will fill the same cistern in 55 min. 6 sec. ? Am. 3.5 in. PROBLEM V. 693. To find the side of a cube, the solid contents being given. Note. — This case, arithmetically considered, requires us to separate a number into three equal factors. The solid contents of a cube are found by cubing the length of one side; hence. Rule. Extract the aihe root of the given contents. EXAMPLES FOR PRACTICE. 1. What must be the length of the side of a cubical bin that shall contain the same quantity as one that is 24 ft. long, 18 ft. wide, and 4 ft. deep ? Ans. 12 ft. 2. What must be the length of the side of a cubical bin that will contain 150 bushels ? 3. What must be the depth of a cubical cistern that will hold 200 bbl. of water ? 4. How many sq. ft. in the surface of a cube whose solidity is 79507 cu. ft. ? Ans. 11094. 404 SQUARE AND CUBE ROOTS. PROBLEM VI. 694. To find the three dimensions of a parallelo- piped, the solid contents and the ratio of the dimen- sions being given. Note 1. ^ This case, arithmetically considered, requires us to separate a num- ber into three factors, proportional to three given numbers. The three dimensions will be like multiples of the proportional terms, (691) ; the product of the three dimensions, or the solid contents, will therefore contain the product of the three propor- tional terms, and the cube of the common ratio which the pro- portional terms respectively bear to the corresponding dimensions, and no other factor. Hence the EuLE. I. Divide the given contents hy the product of the terms proportional to the three dimensions^ and extract the cube root of the quotient. II. Multiply the root thus obtained hy each proportional term; the products will he the corresponding sides. Note 2. — The dimensions are supposed to be taken in a direction perpen- dicular to the faces of a solid, and to each other. EXAMPLES FOR PRACTICE. 1. A pile of bricks in the form of a parallelepiped contains 8000 cu. ft , and the length, breadth, and thickness, are to ^ach other as 4, 8, and 2, respectively; what are. the dimensions of the pile? Ans. 10, 15, and 20 ft. 2. Three numbers are to each other as 2, 5, and 7, and their continued product is 4480 ; required the numbers. Ans. 8, 20, and 28. 8. Separate 100 into three factors which shall be to each other as 2, 2 J, and 8. Ans. 3.76414 + ; 4.70518 + ; 5.64622—. 4. A person wishes to construct a bin that shall be of equal width and depth, and the length three times the width, and that shall contain 450 bushels of grain ? what must be its dimensions ? PROMISCUOUS EXAMPLES. 405 PROMISCUOUS EXAMPLES. 1. There is a park containing an area of 10 A. 2 R. 20 P., and the breadth is equal to f of the length. If two men start from one corner and travel at the rate of 3 miles per hour, one going by the walk around the park, and the other taking the diagonal path through the park, how much sooner will the latter reach the opposite corner than the former? Ans. 1 min. 29.3 sec. 2. What is the length of one side of a square piece of land con- taining 40 acres? A7is. 80 rd. 3. The ground situated between two parallel streets is laid out into equal rectangular lots whose front measure is 44 per cent, greater than the depth. Now, if the streets were 20 feet further apart, the ground could be laid out into square lots of the same area as the rectangular. What is the distance between the streets ? Ans. 100 feet. 4. How much less will it cost to fence 40 acres of land in the form of a square, than in the form of a rectangle of which the breadth is i the length, the price per rod being $1.40 ? Ans. $112. 5. If a cistern 6 feet in diameter holds 80 barrels of water, how much water will be contained in a cistern of the same depth and 1 8 feet in diameter ? 6. What is the length of the side of a square which contains the same area as a rectangle 5i by 7 feet ? Ans. 6 ft. 2.4 + in. 7. What is the length of the side of a square which can just be inclosed within a circle 42 inches in diameter ? Ans. 29.7 — in. 8. If it costs $75 to inclose a circular fish pond containing 3 A. 86 P., how much will it cost to inclose another containing 17 A. HOP.? Ans. $167.70. Note. — It is proved in Geometry that all similar solids are to each other as the cubes of their like dimensions. Hence, any dimension may be found by proportion^ when its ratio to the corresponding dimension of a known similar solid is given. 9. What is the length of the side of a cubical vessel that con- tains J as much as one whose side is 6 ft. ? Ans. 3 ft. 406 SERIES. 10. How many globes 4 in. in diameter are equal in volume to one. 12 in. in diameter? 11. If an ox that weighs 900 lb. girt 6.5 ft., what is the weight of an ox that girts 8 ft. ? Ans. 1677 lb. 14 -f oz. 12. If a cable 3 in. in circumference supports a weight of 2500 lb., what must be the circumference of a cable that will support 4960 1b.? 13. If a stack of hay 4 feet high contain 4 tons, how high must a similar stack be to contain 20 tons ? SERIES. G93. A Series is a succession of numbers 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. G90« 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 de- pends 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^, etc.; the term preceding 64 is 64 -^-4. Hence the series, as far as formed, is 16, 64, 256, 1024. G97. 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. 698. 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. 090. 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. PROGRESSIONS. 407 700. An Arithmetical or Eqnidifferent 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. 701. 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. 702. 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. 703* 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, i, J, etc., is a geo- metrical progression in which the common multiplier is J. 704:. The Ratio in a geometrical progression is the common multiplier. 705. 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. 408 SERIES. GENERAL PROBLEMS IN ARITHMETICAL PROGRESSION. PROBLEM I. 706. 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 hy the number of terms less 1 / add the product to the given term if it he 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 |, the common difference f , and the number of terms 20, what is the last term ? Ans. 1^1, ARITHMETICAL PROGRESSION. 40g PROBLEM II. 7®7. 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, (706), we have the following ^ Rule. Divide the difference of (he extremes h^tJie 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 tlie number of terms is 8 ; what is the common difference? Ans, .00714285. 4. If the extremes are and 2 J, and the number of terms is 18, what is the common difference ? PROBLEM HI. 708. 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, (706), we have tho following Rule. Divide the difference of the extremes hy 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 ? Ana. 15. 2. The extremes are J and 20, and the common difference is G} ; find the number of terms. 35 410 SEKIES. 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. ^ / ,/- ^ , / 709w Given, Ijie extremes and nuniber of terms, to iind the sum ol the series. # Let VIS 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 + 14 = 40, once the sum, 14 -^11 -J- 8+ 5+ 2 = 40 , " '' *' IG + IG + IG + 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, w^hich is tivice the sum of the series ; and 80 -7- 2 = 40, the sum of the series. Hence EuLE. Mulliphj the sum of the extremes hy the number of terms, and divide the product hi/ 2. a/yi,.'>-^ ANNUITIES. ^ — 0^S^ fi*- -+- 719* 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. yS©. A Certain Annuity is one whose period of continu- ance is definite or fixed. ySfl. 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. 722. A Perpetual Annuity or Perpetuity is one which con- tinues forever. 723. Each of these kinds is subject, in reference to its com- mencement, to the three following conditions : 1st. It may he deferred^ i. e., it is not to be entered upon until after a certain period of time. 2d. It may he 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. Sd. It may he in possession^ i. e., it is to be entered upon at once. ANNUITIES. 417 724. 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. 725. 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, TV. All the payments being settled at one time, each 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. 726. 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 S20 + UO X 17 = $21.70, first term, l^^^ month's wages, ^20 + $21.70 ^^^' ^^ *^^ ^^^^ term; ^ X 18 = 375.30, sum. the number of months, 18, is the number of 2b 418 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 706 we find the first term, which is the amount of the first month's wages for 17 months ; and by 709 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 $1860x2— $210x12 ^-.^^ J ., $210, the amount of Y^ = ^1^0, deposit. the first deposit, is 21Q IQO t^6 fi^st term ; 12, -jj" = 10 %, rate. 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 709 we find the last term to be $100, which is the annual deposit ; and by 707 we find the common difi'erence to be $10, which is the annual rate % . 3. What is the amount of an annuity of $150 for o J years, pay- able quarterly, at IJ per cent, per quarter? Aiis. $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. 727. An Annuity at compound interest constitutes a geomet- rical progression whose first term is the annuity itself; the common multiplier is one plus the rate per cent, for one interval expressed decimally 5 the number of terms is the number of intervals for which the annuity is taken; 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. PPtOMISCUOUS EXAMPLES. 4I9 73S. The Present Value of an Annuity is sucli 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 hy K^5\ the/ii find the present value of this amount h[^ 5*^^^ talcing out the anioimt o/Sl, or divisor^ from ^•51, Notes. — 1. The present value of a rcverpionary annuity is that principal which will amount, at the time the reverb-ion expiree, to what will theu be the prOi>ei)t value of the annuity. 2. The present value of a perpetuity is a sum whose interest equals the an- nuity. "^SO. Questions in Annuities at compound interest can be solved by the rules of Geometrical Progression. PROMISCUOUS EXAMPLES 1^ SERIES. 1. Allowing G per cent, compound interest on an annuity of $200 which is in arrears 20 years^ what is its present amount ? Ans. $7857.11. 2. Find the annuity whose amount for 25 years is 81G459.85, 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. 82791.18. 4. What is the present value of a reversionary lease of $100, commencing 14 years hence, and to continue 20 years, coujpound interest at 5 per cent.? Ans. §629.420. 5. Find the sum of 21 terms of the series, 5, 4|, 4 J, 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 scries, 30, 15, 7 J, etc. 8. The first term of a geometrical progression is 2, the last term 512, and common multiplier 4; find the sum of the scries. Ans. 682. 420 SEllIES. 9. The distance between two places is 360 miles. In liow 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 7; find the common ratio. Ans. 5. 11. An annual pension of S500 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 difi*erence 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 yeany payments, mak- ing the first payment $2, the hist $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 others 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. Ans. $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 numl)er 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 ; I PROMISCUOUS EXAMPLES. 421 the distance from the starting place to the first station was 5 miles, and to the last 25 miles; 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,UlS2M + . 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 of a reversionary perpetuity of $100, commencing 30 years h(^nce, allowing 5 per cent, compound interest? Ans. $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 7 % simple inte- rest, and the sum left the other at 6 % compound interest, paya- ble semi-annually, and the amount of each boy's money will be $2000 when he is 21 years old. What is the sum left Jo 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. 86 422 MISCELLANEOUS EXAMPLES. MISCELLANEOUS EXAMPLES. 1. How many thousand shingles will cover both sides of a roof 36 ft. long, and wiiose rafters are 18 ft. in length ? 2. From f of -^ of i of 70 miles, subtract .73 of 1 mi. 3 fur. 3. What number is that from which if 7J be subtracted, f of the remainder is 91^? Ans. 144 J. 4. What part of 4 is | of 6? Ans. |. 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, i, and J of itself equals 125 ? 7. What is the hour, when the time past uoon 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. @ $7i; how much was 1 cwt. of the mixture worth? 9. If $240 gain $5.84 in 4 mo. 2G da., what is the rate fo ? J^ns. 6. 10. If 24 men, in 189 da., working 10 h. a day, dig a trench 33 J yd, long, 2| yd. deep, and 5j- yd. wide; how many hours a day must 217 men work, to dig a trench 23 i yd. long, 2 J yd. deep, and 3| yd. wide, in 5} days? Ans. 10 h. 11. What is the difference between the interest and the discount of $450 at 5 per cent., for 6 jr. 10 mo.? 12. A younger brother received $G300, which was i as much as his elder brother received; how much did both receive? 13. Reduce .7, .88, .727, .91325 to their equivalent common fractions. 14. A person by selling a lot of goods for $438, loses 10 ^c ; how much should the goods have been sold for, to gain 12J ^? 15. For what sum must a note be drawn at 4 mo., that the proceeds ^ of it, when discounted at bank at 7 per cent., shall be $875.50? ^9 ^^/ 16. Three persons engaged in trade with a joint capital of $2128; A's capital was in trade 5 mo., B's 8 mo., and C's 12 mo.; A's share of the 2;ain was $228, B's $266.40, and C's $330. What was the capital of each? Ans. A's, $912 ; B's, $666; C's, ^^555. crG Sep.. ., — ^__, ^.. _ , ^ J,. , ^ .. @ $.40. When was the a|c due per average? Ans. iNov. 8. 18. A B and C can do a job of work in 12 da., C can do it in 24 da., and A in 34 da. ; in what time can B do it alone? Ans. 81^ da. 19. If a man travel 7 mi. the first day, and 51 mi. the last, increas- ing his journey 4 mi. each day, how many days will he travel, and how far? Ans. 12 da., and 348 mi. MISCELLANEOUS EXAMPLES. 423 20. What is the difference between the true and bank discount of $2500, payable in 90 days at 7 per cent. ? Ans. 5;>2.21. 21. Which is the more advantageous, to buy flour at $5 a bbl. on 6 mo., or $4.87 J cash, money being worth 7 ^ ? Ans. At $5 on G mo. 22. Sold J of a lot of lumber for what | of it cost ; what fc was gained on the part sold? Ans. 25 4,. 23. If $500 gain $50 in 1 yr., in what time will $900 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. 42-^^-. 25. The sum of two numbers is 365, and their difference is .0675 ; what are the numbers ? 26. If the interest of $445,621 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 IJ ^, and remitted the proceeds by draft. What was the face of the draft, exchange being ^ fc discount? Ans. $12629.28+. ^ 28. A man in Chicago haa 5000 francs due him on'account in Paris. He can draw on Paris for this amount, and negotiate the bill at 19| 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 6J ^, at 105^, to '^^^roduce an income of $1000?/^^ jl ... . ., -^ns. «?16153.84. 30. A person exchanges 250 shares of 6 per cent, stock, at 70 fo, for stock bearing 8 per cent., at 120 ^ ; what is the difference in his income? Ans. $333. 33J. 31. If f of A^s money equals f of B's, and f of B's equals 'j of O's, and the interest of all their money for 4 yr. 8 mo. at 6 ^ is $15190, how much money has each ? Ans. A has $18859.44+ ; B, $16763.95+ ; C, $18626.61. 32. A boy 14 years old is left an annuity of $250, which is de- posited in a savings bank at 6 ^, interest payable semi-annually; how much will he be worth when of age? Ans, $2104.227. 33. If a boy buys peaches at the rate of 5 for 2 cents, and sells them at the rate of 4 for 3 cents, how many must he buy and sell to mal^e a profit of $4.20 ? / J^ ^^ 34. What fo in advance of the cost must a mercnant mark his goods, so that, after allowing 5 j^ of his sales for bad debts, an ave- rage credit of 6 months, and 7 ^ of the cost of the goods for his ex- penses, he may make a clear gain of 12^] ^ on the first cost of the goods, money bein^ worth 6 ^ ? Ans. 29.56 •{- ^ . < 424 MISCELLANEOUS EXAMPLES. 35. Four men contracted to do a certain job of work for $8600; the first employed 28 laborers 20 da., 10 h. a day; the second, 25 Laborers 15 da., 12 h. a day; the third, 18 hiborers 25 da., 11 h. a day; and the fourth, 15 laborers 24 da., 8 h. a day. How much should each contractor receive ? Ans. Ist, $2686; 2d, $2158.39; 3d, $2374.24; 4th, $1381.37. 36. If I exchange 75 railroad bonds of $500 each, at 36 % below f/^JCpar, for bank ^ck at 5 % premium, how many shares of $100 each will I receive? Ans. 2281^. . 37. A trader has bought merchandise as follows : July 3, $35.26 ; July 4, $48.65, on 30 da. ; Aug. 17, $6.48 ; Sept. 12, $50. What is due on the account Oct. 12, interest at 9 % ? Ans. 142.60. 38. A farmer sold 34 bu. of corn, and 56 bu. of barley for $63.10, receiving 35 cents a bushel more for the barley than for the corn ; what was the price of each per bushel? 39. A speculator purchased a quantity of flour, Sept. 1 ; Oct. 1 its value had increased 25 % ; Nov. 1 its value was 30 % more than Oct. 1; Deo. 1 he sold it for 15 % less than its value Nov. 1, receiving in payment a 6 months' note, which he got discounted at a bank, at 7 %, receiving $12950 on it. How much was his profit on the flour? Ans. $3228.51. 40. A flour merchant bought 120 bbl. of flour for $660, paying $5.75 for first quality and $5 for second quality ; how many barrels were first quality? ')/ Ans. 80. 41. Two mechanics work together ; for 15 days' work of the first and 8 days' work of the second they receive $61, and for 6 days' work of the first and 10 days' work of the second they receive $38 ; how much does each man earn ? Ans. 1st, $63 ; 2d, $36. 42. The duty, at 15 %, on Rio cofl'ee, in bags weighing 180 lbs. gross, and invoiced at $.12J per pound, was $961. 87J, tare having been allowed at 5 % ; how many bags were imported ? A7is. 300. 43. A dairyman took some butter to market, for which he received $49, receiving as many cents a pound as there were pounds ; how many pounds were there ? Ans. 70 lb. 44. A mechanic received $2 a day for his labor, and paid $4 a week for his board ; at the expiration of 10 weeks he had saved $72 ; how many days did he work, and how many was he idle ? 45. To what would $250, deposited in a savings bank, amount in 10 yr., interest being allowed sem^-|innually at 6 % per annum ? 46. How much water is there in^a mixture of 100 gal. of wine and water, worth $1 per gal., if 100 gal. of the wine cost $i^^? ^^ 47. If a pipe 3 in. in diameter will discharge a cermm quantity of water in 2 h., in what time will 3 two-inch pipes discharge 3 times the quantity ? Ans. 4 h. 30 min. 48. Wm. Jones & Co. become insolvent and owe $8100. Their assets amount to $4981.50. What per cent, of their indebtedness can MISCELLANEOUS EXAMPLES. 425 they pay, allowing the assignees 2J ^ on the amount distributed for their services ? Ans, GO per cent. 49. Shipped a car load of fat cattle to Boston, and offered them for sale at 25 per cent. adAance on the cost ; but the market being dull I ^ sold for 14 per cent, less than my asking price, and gained thereby /^^ $170. How much did the cattle cost ; for how much did they sell ; and what was my asking price? Ans. Cost $22G6.6Gf ; sold for $2436.66f ; asking price, $2833.33 J. /V^ 50. What must be the dimensions of a cubical cistern to hold 2000 \ 51. A man died leaving $5000 to be divided between his three sons, aged 13, 15, and IG yr. G mo., respectively, in such a proportion that the share of each being put at simple interest at 6 %, should amount to the same sum when they should arrive at the age of 21. How much was each one's share? <^. >^^ Ans. Youngest, $1536.76+ ; second, $1672.36+ ; oldest, $1790.88 + . 52. A vessel having sailed due south and due east on alternate days, was found, after a certain time, to be 118.794 miles south-east of the place of starting ; what distance had she sailed ? Ans. 1G8 miles. 53. Imported 4 pipes of Madeira wine, at $2.15, a gallon, and paid \- $57.60 freight, and a duty of 24 per cent. I sold the whole for $1980 ; i' what was my gain ^ ? 54. If 34J bu. of corn are equal in value to 1 7 bu. wheat, 9 bu. of wheat to 59 J bu. of oats, and 6 bu. of oats to 42 lb. of flour, how many bushels of corn will purchase 5 bbl. of flour ? y^ Aiis. 42|J|. /y 55. If stock bought at 8 % discount will pay 7 ^ on tbo it.^-. ct„ P^ ment, at what rate should it be bought to pay 10 fo 2 '^^ 5G. A merchant in New York gave $2000 for a bill of cxciiaii-f .a X400 to remit to Liverpool ; what was the rate in favor of Engfand? 57. A, B, and C start from the same point, to travel around a lake 84 miles in circumference. A travels 7 miles, and B 21 miles a day in the same direction, and C 14 miles in an opposite direction. In how many days will they all meet? Ans. 12. 58. The exact solar year is greater than 365 days by i^i^^ of a day ; find approximately how often leap year should come, or one day be added to the common year, in order to keep the calendar right ? Ans. Once in every 4 yr. ; 7 times in every 28 yr. ; 8 times in every S3 yr. ; 31 times in every 128 yr. ; or 163 times in every 673 3^r. , 59. A gentleman purchases a farm for $10000, which he sells after a certain number of years for $14071, making on the investment 5 fo compound interest. He now invests his money in a perpetuity, which is in reversion 11 years from the date of purchasing the farm. Al- lowing 6 fo compound interest for the use of money, find the annuity and the length of time he owns the farm. y Ans. Annuity, $1065.85 : owned the farm 7 yr. 426 MISCELLANEOUS EXAMPLES. . What will I gain % by purchasing goods on 6 mo., and selling immediately for cash at cost, money being worth 7 % ? 4/«^. GO. them Gl. What sum must a man save annually, commencing at 21 years of age, to be worth $30000 when he is 50 years old, his savings being invested at 5 % compound interest? -f. Ans. $481.37. 62. Three persons are to share $10000 'in the ratio of 3, 4, and 5, but the first dyin^: it is required to divide the whole sum equitably be- tween the other two. What are the shares of the other two ? Ans. $4444f, and $5555f. G3. If 50 bbl. of flour in Chicago are worth 125 yd. of cloth in New York, and 80 yd. of cloth in New York are worth G bales of cotton in Charleston, and 13 bales of cotton in Charleston are worth 3 J hhd. of sugar in New Orleans, how many hhd. of sugar in New Orleans are vrorth 1500 bbl. of flour in Chicago? Ans. 75 ,^4^. G4. Seven men all start together to travel the same way round an island 120 miles in circumference, and continue to travel until they all come together again. They travel 5, G}, 7J, S^, 9J, 10} and 11} miles a day respectively. In how many days will they all be together again ? Aiis. 1440 da. G5. There are two clocks which keep perfect time when their pen- dulums beat seconds. The first loses 20 seconds a da}^ and the second gains 15 seconds a day. If the two pendulums beat together when both dials indicate precisely 12 o'clock, what time does each clock show when the pendulums next beat in concert? Ans. The first shows 41 min. 8 sec. past 12 ; and the second 41 min. 9 sec. past 12. G6. If a body put in motion move J of an inch the first second of time, 1 in. the second sec, 3 in. the third, and so continue to increase in geometrical ratio, how far would it move in 30 seconds ? A71S. 541o9u730fMi mi. G7. If stock bought at 5 fo premium will pay 6 ^ on the invest- ment, what fo will it pay if bought at 15 fo discount ? Ans. Ty^ %* G8. If G apples and 7 peaches cost 33 cts., and 10 apples and 8 peaches cost 44 cts., what is the price of one of each ? A71S, Apples, 2 cts. ; peaches, 3 cts. GO. A gentleman in dividing his estate among his sons gave A $9 as often as B $5, and C $3 as often as B $7. . C's share was 5538G2.50 ; what was the value of the whole estate? '}( Ans. $£1,097.50. 70. A farmer sold IG bu. of corn and 20'bu. of rye for $30, and 24 bu. of corn and 10 bu. of rye for $27. How much per bushel did he receive* for each ? ^/ Ans. Corn, $.75 ; rye, $.90. 71. A drover sold some oxen at $28, cows at $17, and sheep at $7.50 per head, and received $749 for the lot. There were twice as many cows as oxen, and three times as many sheep as cows. How many were there of each kind ? /Jy ^'^f .\.' i 72. For what sum must a vessel, valued at $25000, be insured, so MISCELLANEOUS EXAMPLES. 427 that in case of its loss, the owners may recover both the value of the vessel and the premium of 24 f.W2^9p^V?g. .^v^, /%>m^^i 73. A boy hired to a mechani^for 20 weeKs,oav condition that he should receive $20 and a coat. At the end of 12 weeks the boy quit work, when it was found that he was entitled to $9 and the coat ; what' was the value of the coat? y! Ans. ^l.bO. 74. An irreo-ular piece of land, containing 540 A. 36 P., is ex- chan'o-ed for a "square piece containing the same area; what is the length of one of its sides ? If divided into 42 equal squares, what wiUbe the length of the side of each? -/ 75 What will be the difference in the expense of fencing two fields of 25 acres each, one square, and the other in the form of a rectangle, whose length is'lwice its breadth, the fence costing $.62i a rod? /^ Ans. $9.59+. 76. At what time between 5 and 6 o'clock are the hour and minute hands of a watch exactly together ? ^ v 77. A general, forming his army into a square, had 284 men re- maining ; but increasing each side by one man, he wanted 25 men to complete the square. How many men had he? / Ans. 24000. 78. Divide $3618 among 3 persons, so that the share of the first to ^that of the second shall be as 7 to 9, and of the first to the third as 3 to 4. ♦ Ans. $1008, $1296, $1344. ^ 79. If a lot of land, in the form of an oblong or rectangle, contains 6 A. 3 E. 12 P., and its length is to its width as 21 to 13, what are its dimensions ; and how many rods of fence will be required to in- close it? Ans. to last, 136 rd. of fence. f\6/ \ 80. Five persons are employed to build a house. A, B, C, and D /^can build it in 13 days ; A, B, C, and E in 15 days ; A, B, D, and E in 12 days; A, C, D, and E in 19 days; and B, C, D, and E in 14 days. In how many days can all together build it ; and which cna could do the work alone in the shortest time ? Ans. llf*oVj\ da. ; B in shortest time. 81. Divide $500 among 3 persons, in such a manner that the share of the second may be J greater than that of the first, and the share of the third J greater than that of the second. Ans. 1st, $105/g; 2d, $157 jj; 3d, $236] J. v^ 82. A and B engage in trade ; A puts in $5000, and at the end ofj 4 mo. takes out a certain sum. B puts in $2500, and at the end of 5 mo. puts in $3000 more. At the end of the year A's gain is $1066 §, 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 ^ 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 fc 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 ? ^8 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 tiian the marked price, ahd still make 20 % prorit? Aas. $LOGf. 86. A merchant bought 20 pieces of cloth, each piece containing 25 yd. at %\% 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 G ^^ ? 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 shoiild 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 ^ ? Ans, To oldest son, $22080.28 ; to second son, $1^329.51 ;., 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 ^ ? 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 paynrients which fell due after Jan. 1, 1864, being deducted; what sum canceled the bond? J.ns. $2972.54+. 92. A gentleman desires to set out a rectangular orchard of 864 trees, so placed that the number of rows shall be to the number of trees in a row, as 3 to 2. If the trees are 7 yards apart, how much ground will tlie orchard occupy ? Ans, 39445 sq. yd, V 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 ^o, when he sold the stock at a premium of 11 ^. Money being worth 7 ^0 compound interest, how much did he gain? Ans, $137.31. THE METRIC SYSTEM WEIGHTS AND MEASURES.* INTRODUCTION. Tlie metric system of weights and measures — so called, because the metre is the unit from which the other units of the system are derived — had its origin in France during the Eevolution, a time when all regard for institutions of the past was repudiated. In the year 1790, the French government resolved to introduce a new Fystem ; and, in order that it might be received with general favor, other countries were invited to join with it in the choice of new units. In response to this invitation, a large number of scientific men, com- missioned by various countries, met in Paris, in consultation with the principal men of France. In the year 1791, a commission, nomi- nated by the Academy of Sciences, was appointed by the Government to prepare the new system. The first work of the commission was to select a standard of lengths from which the system of units adopted might at any time be restored if from any cause the original unit should be lost. A quadrant of the earth's meridian was chosen as the standard, and the ten-millionth part of it taken as iha unit cf lengths, which was called a metre. In 1795, this standard and a provisional metre whose length was determined from measurements * 31. McYiCAR, A.M., Principal of the State Normal and Training School at Brockport, N.Y., a most thorough and critical scholar as well as teacher, prepared this article, which contains many practical improvements in Notation, Nomencla- ture, and Applications, not before presented to the public. . Entered, according to Act of Congress, in the year 1867, by D. W. Fisn, A. M., in the Clerk'tt Office of the District Ck>urt of the United States for the Southern District of New York. (429) 430 THE METRIC SYSTEM. of the earth's meridian, "which had already been made, was adopted by the government. In the meantune, two eminent astronomers, Mechain and Delambre, were engaged in determining the exact length of the arc of the meri- dian between Dunkirk in the north of France, and Barcelona in Spain. At a later period, Liot and Arago measured the prolonga- tion of the same meridian as far as the island of Formentara. From these measurements, together with one formerly made in Peru, they deduced, as they supposed, the exact distance from the equator to the pole, which differed slightly from the standard assumed in 1795. In 1790, a law was passed changing the length of the metre adopted in 1795 so as to conform with this diSerence. The metre thus de- termined was marked by two very fine parallel lines drawn on a pla- tinum bar, and deposited for preservation in the national archives. While a part of the commission were engaged in establishing the exact length of the metre, other members pursued a course of inves- tigation for the purpose of determining a unit of weights, which would sustain an invariable relation to the unit of lengths. As the result of their investigations, the weight of a cube of pure water whose edge was one-hundredth part of a metre was the unit chosen. The water was weighed in a vacuum, at a temperature of 4° C, or 39.2° F., which was supposed to be the temperature of greatest density. This weight was called a gramme ^ and a piece of platinum weighing one thousand grammes was deposited as the standard of weights in th^ national archives. Had the work of the commission ended in determining these standards of lengths and weights, their labor would have been futile. For, while the conception of basing their system upon an absolute standard in nature was good, the execution proved a failure. Later in vesti stations have shown that the metre is less than the ten-millionth part of the earth's meridian ; consequently the metric system of weights and measures is referable not to an invariable standard in nature, but to the platinum metre deposited in the national archives of France. The great benefits which result from the labors of tlie commission arise from the adoption of the decimal scale cf units, and a simple yet general and expressive nomenclature. The amount of THE METRIC SYSTEM. 431 time and money nsed in carrying on exchanges between different coun- tries, Yrliich would be saved by the universal adoption of this system, is incalculable. The system was declared obligatory throughout the whole of France after Xov. 2, 1801 ; but, owing to the prejudices of the people in favor of established customs, and the confusion con- eequent upon the use cf the new measures, the Government, in 1812, adopted a compromise, in the systeme usuile, whose principal units were the new ones, while the divisions and names were nearly those formerly in use, ascending commonly in the ratios of two, three, four, eight, or twelve. In 1837, the government abolished this system, and enacted a law attaching a penalty to the use of any other than the metric system after Jan. 1, 1841. Since that time, the system has been adopted by Spain, Belgium, and Portugal, to the exclusion of other weights and measures. In Holland, other weights are used only in compounding medicines. In 18G4, the system was legalized in Great Britain ; and its use, either as a whole or in some of its parts, has been authorized in Greece, Italy, Norway, Sweden, Mexico, Guatemala, Venezuela, Ecuador, United States of Columbia, Brazil Chili, San Salvador, and Argentine Republic. In 1866, Congress authorized the metric system in the United States by passing the fol- lowing; bills and resolution ; — ■ An Act to atttiiorize the tjse of tiie Metric System of Weights AND Measures. Be it enacted hy the Senate and House of Representatives of the United Stauts 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 contract or dealing, or pleading in any court, shall be deemed invalid, or liable to objection, be- cause the weights or measures expressed or referred to therein arc weights or measures of the Metric System. Section 2. And he it farther enacted, That the tables in the schedule hereto annexed shall be recognized in the construction of contracts, and in all legal proceedings, as establishmg, 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 be lawfully used for computing, determining, and expressing in customary weights and measures, the weights and measures of the Metric System. 432 THE METRIC SYSTEM. A Bill to authorize the Use in Post Offices of the Weights OF THE Denomination of Grammes. Be it enacted by the Senate and House of Representatives of the United States of America in Congress assembled^ That the Postmaster General be, and he is hereby, authorized and directed to furnish to the post-offices exchanging mails with foreign countries, and to such other offices as he shall think expe- dient, postal balances denominated in grammes of the metric system ; and, until otherwise provided by law, one-half ounce avoirdupois shall be deemed and taken for postal purposes as the equivalent of fifteen grammes of the metric weights, and so adopted in progression; and the rates of postage shall be applied accordingly. Joint Resolution to enable the Secretary of the Treasury TO furnish to each State one set of the Standard Weights AND JVIeASURES of THE MeTRIO SySTEM. Be it resolved by the Senate and House of Representatives of the United States of America in Congress ossejnbled, That the Secretary of the Treasury be, and he is hereby, authorized and directed to furnish to each State, to be delivered to the governor thereof, one set of the standard weights and measures of the metric system, for the use of the States respectively. TABLES AUTHORIZED BY CONGRESS. MEASURES OF LENGTHS. Metric Denominations and Values. Equivalents in Denominations in use. Myriametre, . . . 10,000 metres, 6.2137 miles. Kilometre, .... 1,000 metres, 0.62 137 miles, or 3280 feet, 10 inches. Hectometre , . . . 100 metres, 328 feet and 1 inch. Decametre, . . . 10 metres, .. 393.7 inches. Metre, 1 metre, 39.37 inches. Decimetre, .... ^Q- of a metre, . . 3.937 inches. Centimetre, . . . ■ji^ of a metre, . . 0.3937 inch. Millimetre, T (tW ^^ ^ ^^^^^' • • 0.0394 inch. MEASURES OF SURFACES. Metric Denominations and Values. Equivalents in Denominations in use. Hectare, Are, Centiare, 10,000 square metres, 100 square metres, 1 square metre, 2.471 acres. 119.6 square yards. 1550 square inches. THE METRIC SYSTEM. 433 MEASURES OF CAPACITY. Metric Denominations and Values. Equivalents in Denominations in use. Names. No. of litres. Cubic Measure. Dry Measure. Liquid or wine measure. Kilolitre, or stere. Hectolitre, Decalitre, Litre, lOOOjl cubic metre, 100| jl^ of a cubic metre, . . . 10 10 cubic decimetres,. . . 1 1 cubic decimetre 1.308 cubic yd. 2 bu. 3.35 pk... 9.08 quarts,.... 0.908 quart, . . . 6.1022 cubic in. 0.6102 cubic in. 0.061 cubic in. . 264.17 gallon. 26.417 gallon. 2.6417 gallon. 1.0567 quart. 0.845 gill. 0.338 fluid oz. 0.27 fluid dr. Decilitre, Centilitre, Millilitre, -j\j- of a cubic decimetre, 10 cubic centimetres, . . 1 cubic centimetre, WEIGHTS. Metric Denominations and Values. Equivalents in De- nominations in use. Names. Number of grammes. Weight of what quantity of water at maximum density. Avoirdupois weight. Millier, or toimeau, . Quintal, 1,000,000 100,000 10,000 1,000 100 10 1 iV iJo TOTT^ 1 cubic metre, 2204.6 pounds. 220.46 pounds. 22.046 pounds. 2.2046 pounds. 8.5274 ounces. 0.3527 ounce. 15.432 grains. 0.5432 gi-ain. 0.1543 grain. 0.0154 gi-ain. 1 hectolitre, Myriagramme, Kilogramme, or kilo. Hectogramme, Decagramme, Gramme, 10 litres, 1 litre, 1 decilitre, 10 cubic centimetres, 1 cubic centimetre, 1-10 of a cubic centimetre, 10 cubic millimetres, 1 cubic millimetre, Decigramme, Centigramme, Milligramme, Note. — The spelling in the above tables is not the same as in the tables in the schedule annexed to the report of the committee of the House of llepresentatives on weights and measures. The change is not made to indicate any preference for any standard upon this subject ; but to carry out what the author believes to be an essential condition to the utility and success of the system. As remarked by a distinguished senator when the tables were adopted by Congress, '"'-The names are cosmopolitan ;^'' and to re- tain this character fully, the spelling must also he cosmopolitan. The French introduced the nomenclature and spelling ; and, so long as the names remain unchanged, the spelling should be retained. 434 THE METRIC SYSTEM. NOMENCLATURE AND TABLES. t| Tbcre are eight kinds of quantities for wbich tables are usnally constructed; viz., Lengths, Surfaces, Volumes or Solids, Capacities, Weights, Values, Times, and Angles or Arcs. The table for Times is the same in the metric as in the ordinary system. The table for Angles is constructed upon a centesimal scale. The tables for the other six kinds of quantities are constructed upon a decimal scale. In each of the tables for Lengths, Surfaces, Volumes, Capacities, and Weights, there are eight denominations of units, — one principal and seven derivative. The principal units are the metre, which is the base of the system, and those derived directly from it. The two following tabular views present the facts regarding the principal and derivative units, which should be fixed in the memory. " 1. Principal unit of Lengths. 2. The- base of the metric system, and nearly one ten-millionth part of a quadrant of the earth's meridian. 3. Equivalent, 39.3708 inches. 1. Principal unit of surfaces. 2. A square whose side is ten metres. 3. Equivalent, 119.6 square yards. 1. Principal unit of volumes or solids. 2. A cube whose edge is one metre. 3. Equivalent, 1.308 cubic yards. 1. Principal unit of capacities. 2. A vessel whose volume is equal to a cube whose edge is one-tenth of a metre. 3. Equivalent, .908 quart dry measure, or 1.0567 quarts wine measure. f 1. Principal unit of weights. 2. The weight of a cube of pure water whose edge is .01 of a metre. 3. The water must be weighed in a vacuum 4° C, or 39.2° F. ^ 4. Equivalent, 15.432 grains. «2 H o P5 p-l I. Metre, . II. Are, in. Stere, IV. Litre, . . ^ L V. Gramme, THE METRIC SYSTEM. 435 ft o o o IS? fa >5 c *^ as ^^ |2l O g I M o 2 o. 1. Three orders of smaller units, or submultiples of each kind, are formed by dividing each of the principal units into tenths, hundredths, and thousandths. 2. Four orders of larger units, or multiples of each kind, are formed by considering as a unit ten times, one hundred times, one thousand times, and ten thousand times, each of the principal units. " The names of derivative units are formed by attaching a prefix to the name of the princi- pal unit from which they are derived, which indicates their relation to the principal unit. 1. Millesimus, one thousandth, contracted Milli. Example, Millilitre = j^ViJ ^^ ^ ^i^^^ '? 8 millilitres = j^%jj of a litre. 2. Centesimus, one hundredth, contracted centi. Ux., Sentiare = jo (j ^^ ^^ ^^^i ^ centiares = yj^j of an are. 3. Decimus, tenth, contracted deci. jSJx., De- cimetre = ^ metre ; 3 decimetres = j\ metre. 1. Deca, ten. Example, Decametre = 10 metres ; 5 decametres = 50 metres. 2. Hecaton, one hundred, contracted hecto. Ex,, Hectolitre = 100 litres ; 7 hectolitres = 700 litres. 3. Kilioi, one thousand, contracted kilo. Ex. Kilogramme = 1000 grammes. 4. Myria, ten thousand. Ex., Myriastere = 10,000 steres; 3 myriasteres =30,000 stores. 5. The a in deca and myra, and the o in hecto and kilo, are dropped when prefixed to are. ^ The tables being constructed upon a decimal scale, ten units of a lower order make one of the next higher, thus: 10 millimetres = 1 centimetre; 10 centimetres = 1 decimetre ; 10 decimetres = 1 metre ; 10 me- tres == 1 decametre, &c. to CO S 9 •9 S o fl U 03 3. O 2 -( g s 436 THE METRIC SYSTEM. The facts in tbe preceding views being mastered, tlie tables can be constructed bj the pupil at sight. For example : The names of the derivative units are formed by attaching the seven prefixes, in their order, to the principal units of the tables. The order of progression being ten, the table of capacities will be written thus : — • 10 Millilitres = 1 Centilitre. 10 Litres = 1 Decalitre. 10 Centilitres = 1 Decilitre. 10 Decalitres = 1 Hectolitre. 10 Decilitres = 1 Litre. 10 Hectolitres = 1 Kilolitre. 10 Kilolitres = 1 Myrialitre. All the tables peculiar to the Metric System are presented together in a convenient form in the two following tables : — TABLE OF SUBMULTIPLES AND PRINCIPAL UNITS. Names of Units. T* ROX TINT" T A TT OV Symbols. PREFIX. BASE. JL £h\J^^ \J XI Vy JL^V X X\^^^ f p Metre Miir-e-mee'-ter 3M 10 Milli- Are Miir-e-are A Equal Stere MilI'-©-ster sS 1 Centi- Litre Miir-e-li'-ter gL . Gramme Mill^-e-gram 8^ r Metre Sent^-e-mee'-ter 2^ 10 Centi- Are Sent'-e-are 2^ Equal Stere Sent'-e-ster sS 1 Deci- Litre Sent'-e-li'-ter ^ . Gramme Sent^-e-gram 2<5 - Metre Des'-e-mee'-tcr jM le Deci- Are Des'-e-are 1^ Equal Stere Des^-e-ster iS 1 Principal Unit. Litre Des^-e-li'-ter iL - Gramme Des'-e-gram 6 - Metre Mee'ter M 10 Principal Units Are Are A Equal Stere Ster S 1 Deca- Litre Li'-ter L - Gramme Gram G THE METRIC SYSTEM. TABLE OF MULTIPLES. 437 Namks of Units. T*ROXrTVrTATTOV PREFIX. BASE. X XW^i O ■'■I v/X^x X ±\J^ . r Metre Dek^-a-mee-ter 'M 10 Deca- Are Dek'-are >A Equal -< Store Dek'-a-ster 's 1 Hecto- Litre Dek'-a-li'-ter 'l . Gramme Dek'-a-gram 'G - Metre Hec^-to-mee-ter 'm 10 Hecto. Are Hec'-tare 'a Equal - Stere Hec'-to-ster 's I Kilo- Litre Hec'-to-li'-ter 'l * Gramme Hec'-to-gram 'Gr r Metre Kiir-o-mee-ter \l 10 Kilo- Are Kiir-are \ Equal - Stere KilK-o-ster ■ 's 1 Myria- Litre Kill'-o-li^-ter 'l . Gramme Kill'-o-gram =G - Metre Mir'-e-a-mee-ter *M Are Mir^-e-are *A Myria- - Stere Mir'-e-a-ster *S Litre Mir^-e-a-li'-ter *h ^ Gramme Mir'-e-a-gram 'G ABBREVIATED NOMENCLATUUE. To secure the fullest advantage to business men by the universal adoption of the new system of weights and measures, it is necessary that the names used should be short and easy to write and pronounce, that they should express clearly the relation of the different denomi- nations of the same table to each other, and that they should be identical in all languages. The last two of these requirements would be secured by the uni- versal use of the nomenclature adopted by the French. It is cosmo- politan in its character : it belongs to their language no more than to any other.*- The former, however, is not secured. It is evident to all, that, for business purposes, the long names of the metric system are inconvenient, and that to shorten them would prove a great 438 THE METRIC SYSTEM, advantage. Efforts have been made to introduce short names; but these efforts have invariably sacrificed their universal and expres- sive character, which is of more importance to the business world than their shortness. The only true course which seems to be open, is to abbreviate the names already introduced, in such a way as to retain their peculiar charapteristics. To secure this, the following plan of abbreviation is suggested : — First. Let the prefixes be abbreviated thus : Myr, kil, hect, dec, des, cent, mil. Second. Let the initial letter of the names of the five principal units be used, instead of the names themselves, thus : For metre, use a capital M ; for are, use a capital A ; for store, a capital S ; for litre, a capital L ; and, for gramme, a capital Gr. Third. For the names of multiples and sub-multiples, attach to these initial capital letters the abbreviated prefixes, thus ; Kil M, pro- nounced kill-em'^ ; Kil S, pronounced kill-ess', &c. By this method of abbreviation, the elements of the original terms are retained in such a form that each part is clearly indicated. The capital letter used after the prefix will always point to the base-word of which it is the initial, although the pronunciation is changed. TABLES WITH ABBREVIATED NOMENCLATURE. MEASUEES OF LENGTHS. Written. Pronounced. 10 Mil M, Mill-em', make 1 Cent M. 10 Cent M, Cent-em', 1 Des M. 10 Des M, Des-em' 1 M. 10 M, Em 1 Dec M. 10 Dee M, Dek-em', 1 Hect M. 10 Hect M, Hect-em', 1 Kil M. 10 Kil M, Kill-em', 1 Myr M.* Jlyr M, Mir-em'. THE METRIC SYSTEM. 439 MEASURES OF SURFACES. Written. Pronounced. 10 Mil A, Mill-ii', make 1 Cent A 10 Cent A, Cent-a', 1 Des A. 10 Des A, Des-a', 1 A. 10 A, A, IDccA. 10 Dec A, Dek-a', 1 Hect A 10 Hect A, Hect-a', 1 Kil A. 10 Kil A, KiU-a', 1 Myr A. Myr A, Mir-a'. MEASURES OF VOLUMES, OR SOLIDS. Written. Pronounced. 10 Mil S, Mill-ess', make 1 Cent S. 10 Cent S, Cent-ess', 1 Des S. 10 Des S, Des-ess', IS. 10 S, Ess, 1 Dec S. 10 Dec S, Dek-ess', 1 Hect S. 10 Hect S, Hect-ess', 1 Kil S. 10 Kil S, Kill-ess', 1 Myr S. Myr S, Mir-ess'. MEASURES OF CAPACITY. Written. Pronounced. 10 Mil L, Mill-eir, make 1 Cent L. 10 Cent L, Cent-eir, 1 Des L. 10 Des L, Dess-ell' IL. 10 L, Ell, 1 Dec L. 10 Dec L, Dek-ell', 1 Hect L. 10 Hect L, Hect-eir, 1 Kil L. 10 Kil L, Kill-eir, 1 Myr L. Myr L, Mir-ell'. 440 THE METRIC SYSTEM. MEASURES OF WEIGHTS. Written. Pronounced. 10 Mil G, Mill-gee', make 1 Cent G. 10 Cent G, Cent-gee', 1 Des G. 10 Des G, Des-gee', IG. 10 G, Gee, 1 Dec G. 10 Dec G, Dek-gee', 1 Hect G. 10 Hect G, Hect-gee', 1 Kil G. lOKilG, Kill-gee', 1 Myr G. MyrG, Mir-gee'. NOTATION AND NUMERATION. In the practical application of the metric system, it is not always convenient to use the principal units as the unit of number. For example : Should the gramme, the principal unit of weight, be used as the unit of number, in the grocery or any similar business, small quantities would be expressed by inconveniently large numbers. Example : 386 lbs. are expressed by 175,000 grammes. To avoid this inconvenience, the higher denominations aie used as the unit of number when large quantities are measured. No general system of notation is yet agreed upon. The same quantity is written in various ways by different authors. Example ; 42 metres, 8 decimetres, and 5 centimetres, are written m cm 42.85 M. 42? 85. 42.85. M 42 85. &c. Inasmuch as the principal units of measure are not always used as the unit of number, it is important that a system of notation be adopt- ed, which will apply equally well to both principal and derivative units. It is believed that the system given below, while simple and con- venient, expresses clearly the relation of the unit of number to the principal unit of measure ; and, hence, has an advantage over any contractions of the names of the derivative units or arbitrary signs which might be adopted. THE METRIC SYSTEM. 441 GENERAL PRINCIPLES OF NOTATION. I. The scale in the metric system being decimal, the consecutive denominations are expressed by the consecutive orders of units in a number. Thus, 78G42.358 metres is an expression for 7 myria- metres, 8 kilometres, 6 hectometres, 4 decametres, 2 metres, 3 deci- metres, 5 centimetres, 8 millimetres. II. Whichever one of the eight denominations of units of measure is used as the unit of a number, the higher denominations are ex- pressed as tens, hundreds, and so on ; and the lower as tenths, hun- dredths, and so on. Example : 784.56 decametres. Here the unit of the number is a decametre ; consequently the tens and hundreds are, respectively, hectometres and kilometres, and the tenths and hundredths are metres and decimetres. From these principles and illustrations, we derive the following rule for notation : — Rule. Write the consecutive denominations in their order, com- mencing with the higher, and placing a cipher wherever a denomi- nation is omitted, and the decimal point after the denomination which is the unit of the number. KULES FOU INDICxVTING THE DENOMINATION. KuLE I. When a principal unit of measure is the unit of Clum- ber, place the initial letter of the unit used before the number, thus : M 342.5. Read, three hundred and forty-two and five-tenths metres ; or^ 3 hectometres, 4 decametres, 2 metres, 5 decimetres, EXAMPLES FOR PRACTICE. Write the numbers which represent the following quantities, con- sidering the principal unit of measure the unit of number. 1. Seven myriametres, 4 hectometres, three decametres, and eight centimetres. Ans. M 70430.08. 2. Thirty-four kilometres and forty-three millimetres. Ans, M 34000.043. 3. Eighty-seven hectogrammes and fifty-nine centigrammes. Ans: G 8700.59. 442 THE METRIC SYSTEM. 4. Thirty-two myriagrammes, forty-eigbt decagrammes, five milli- grammes. Ans. a 320480.005. 5. Three hundred and two kilares, eight hundred and seven cen- tiares. Ans. G 302008.07. 6. Four myrialitres, sixty-two decalitres, live millilitres. Ans, L 40620,005. 7. Four hundred and thirty-three kilosteres, nine hundred and eighty four hectosteres, seven thousand two hundred and three centi- steres. A?is. S 53147203. EuLE II. When a multiple of a principal unit of measure is the unit of number ; — First, Place before the number the initial letter of the principal unit from ivhich the multiple is derived. Second, Indicate the order of multiple used by a small figure placed to the left and above the letter prefixed to the number. (See symbols in table of multiples.) Example. 42.5 kilometres, is written ^M42.5. The M before the number indicates that the metre is the unit of measure from which the unit of the number is derived. The small 3 indicates that the third order of multiple, or kilometre, is the unit of number. EXAMPLES FOR PRACTICE. Write the numbers which represent the following quantities, con- sidering the denomination named as the unit of number : — Unit of Number, Kilogramme, 1. 43 myriagrammes, 7 decagrammes, 5 grammes. Ans. ^G 430.075. 2. 8 kilogrammes and 3 centigrammes. Ans. ^G 8.00003. 3. 736 hectogrammes, 243 centigrammes, and 4 milligrammes. Ans. ^G 73.602434. 4. 2009 hectogrammes and 3 centigrammes. Ans. ^G 200.90003. Ufiit of Number , Decalitre. 5. 254 litres and 43 milUlitres. Ans. ^L 25.4043. THE METPJC SYSTEM. 443 6. 364 mjrialitres, 47 litres, 384 millilitres. ^;is. 1L3G4004.7384. 7. 243 decalitres, 47 centilitres. Ans, ^L 243.047. Unit of Number, Second Order of 3Iultiples. 8. 23 myriametres, 72 millimetres. Ans. ^31 2300.00072. 9. -4000007 steres and 2 millisteres. Ans, ^S 40000.07002. 10. 3 kilares and 43 centiares. Ans, ^A 30.0042. Unit of Numher, Myriametre. 11. 3 hectometres and 2 centimetres. Ans. "^M .030002. 12. 5 millimetres. Ans. ^xM .0000005. 13. 3 decametres and 2 centimetres. Ans. ^M .003002. Rule III. When a submultiple of a principal unit of measure is the unit of number ; — First, Place before the number the initial letter of the principal unit from which the submultiple is derived. Second, Indicate the order of submultiple used by a small figure placed to the left and below the letter prefixed to the number. (See symbols in table of submultiples.) EXAMPLES FOR PRACTICE. Write the numbers which represent the following quantities, con- sidering the denomination named as the unit of number. Unit of Number, Millimetre. 1. 32 decametres and 2 decimetres. Ans. gM 320200. 2. 7002 hectometres. Ans. .M 700200000. 3. 7 myriametres and 5 metres. Ans. gM 70005000. 4. 3 kilometres and 2 decametres. Ans. 3M 3020000. Unit of Number, Second Order of Submultiples. 5. 5 kilogrammes and 9 grammes. Ans. gCr 500900. 6. 302 myriasteres, 5 decasteres, and 3 centisteres. Ans. 2S 302005003. 7. 4009 kilolitres and 5 litres. Ans. 2L 400900500. 8. 2 hectares and 2 centiares. Ans. 2 A 20002. 444 THE METRIC SYSTEM. Unit of Number f Decilitre. 9. 3002 hectolitres and 4 millilitres. Ans, iL 3002000.04. 10. 6 mjrialitres and 1 decalitre. Ans. iL 600.100. 11. .404 millilitres. Ans. iL .00004. DEDUCTION. Rule for Reduction Descending. Multiply the given quantity by the number of the required denomination which makes a unit of the given denomination. Since the multiplier is always 10, 100, 1000, &c., the operation is performed by removing the decimal point as many places to the right as there are ci23hers in the multiplier, annexing ciphers when necessary. examples for PRACTICE. 1. RcMluce ^M 32.58 to milHmetres. 2. Reduce ^M 5 to decimetres. 3. Reduce G402 to milhgrammes. 4. Reduce ^ A 42.3 to centiares. 5. Reduce "L 93.2 to decilitres. 6. Reduce *S 895 to decasteres. 7. Reduce ^A 903.2 to mllliares. 8. Reduce ^G 539 to centii!:rammes. Rule for Reduction Ascending. Divide the given quantity by the number of its own denomination which makes a unit of the required denomination. Since the divisor is always 10, 100, 1000, &c., the operation is performed by removing the decimal point as many places to the left as there are ciphers in the divisor, prefixing ciphers when necessary. 1. Reduce gA 5 to myrlares. 2. Reduce 3M 403 to kilometres. 3. Reduce iS 42.3 to hectosteres. 4. Reduce 3 A 7.2 to decares. examples for practice. 5. Reduce 3G 3 to kilogrammes. 6. Reduce 2L5.7 to hectoUtres. 7. Reduce 3M 9 to myriametres. 8. Reduce 2S47.3 to decasteres. MEASURES OF SURFACES. RELATIONS OF UNITS OF SURFACE TO UNITS OF LENGTH.. Decimilliare == One square decimetre = 100 square centimetres. _ f -^^ square decimetres, or a plane figure whose ~ 1 length is one metre and breadth one decimetre. Centiaro = One square miCtre = 100 square decimetres. THE METRIC SYSTEM. 445 . _ f 10 square metres, or a plane figure wbose length is one • ^ ~ ^ decametre and breadth one metre. Are = One square decametre = 100 square metres. j^ ^ ^ _ ( 10 square decametres, or a plane figure whose length (is one hectometre and breadth one decametre. Hectare = One square hectometre = 100 square decametres. y^., (10 square hectometres, or a plane figure whose length 1 is one kilometre and breadth one hectometre. Myriare = One square kilometre = 100 square hectometres. NUMERAL EXPRESSION FOR SURFACE. The contents of a plane figure is expressed numerically by giving the number of times it contains some given area, which is assumed as the unit of surface. The following illustrations will show how the. various denomina- tions of the table are used in numerical expressions of surface : — ILLUSTRATION FIRST. r, o B 1 Length 6 metres. It will be seen, by examining this figure, that the lines drawn parallel to the sides, at the supposed distance of a metre from each other, divide the surface into square metres, and that there are as many rows of square metres as there are metres in the breadth, each row containing as many square metres as there are metres in the length. Hence the number of square metres in the area of the figure is equal to the product of the two numbers which indicate the length and breadth ; and A 0.18 is a numerical expression for its contents. 446 THE UETEIC STSTEM. ILLUSTRATION SECOND. - •S o 5 u 'O'S^ ''^ a In this figure, the lines drawn parallel to the sides divide the figure into 36 milliares, or oblongs, whose length is one metre and breadth one decimetre. It is evident that ten of these oblongs put together will constitute a centiare, or square metre. Hence the ex- pression, 36 milliares, may be written 3.6 centiares; and read, three and six tenths centiares, or three centiares and six milliares. By reducing the length to decimetres, the numerical expression of the contents will be, by Illustration First, 60 x 6, or 360 decimiliiares or square decimetres. ILLUSTRATION THIRD. Length 1 decametre, 2 metres, and 1 decimetre. o u a « QJ 'O »-l n a O) B 03 o « -O 1H .d 'O a t W Declare. /■ Declare. 1 — Milliares. Decimilliare. In this figure, we have illustrated the relations of different denomi- nations of units in expressing the contents of a given surface^ THE METRIC SYSTEM. 417 In the following analysis, each part of the contents is presented separately, as it would be obtained by multiplying the length by the breadth. The learner should carefully note each part, and analyze a sufficient number of examples to fix the principles in the mind. ANALYSIS. Jj ( One decimetre = 1 decimilliare = A 0.0001 One decimetre x \ Two metres = 2 miliiares = A 0.002 ( One decametre = 10 miliiares = 1 ceatiare = A 0.01 ( One decimetre =r 2 miliiares = A 0.002 Two metres x \ Two metres = 4 centiares = A 0.04 ( One decametre = 2 declares = A 0.2 {One decimetre = 10 milliare = 1 ccntiare = A 0.01 Two metres = 2 declares == A 0.2 One decametre = 1 are or square metre = A 1. X = 1—) ^M 1.21 X 'M 1.21 = A 1.4641 From these illustrations, we derive the following rule for finding a numerical expression for a given surface of utiiform length and breadth : — Rule. Reduce the length and breadth to the same denomination ; find the product of the two dimensions after reduction, and point off as many decimal places in this product as there are decimal places in the two dimensions. The unit of the numerical expression thus found will be a decimil- liare when the unit of length is a decimetre, a ccntiare when the unit of length is a metre, an are when the unit of length is a deca- metre, a hectare when the unit of length is a hectometre, and a myriare when the unit of length is a kilometre. EXAMPLES FOR PRACTICE. 1. How many ares in a floor M 1.25 long, and M 8.7 wide ? Ans. A. 10875. 2. How many centiares, how many kilares, and how many hec- tares in the same floor? Ans. gA 10.875. 3. How many ares in a board M 5.32 by 2M 47. ? Ans. A. 025004. 4. How many miliiares, how many myriares, and hectares in the same board ? 5. How many metres of a carpet nine decimetres wide will cover 44& THE METRIC SYSTEM. a floor six metres long and five and four-tenths metres wide ? and what would be the cost of the carpet, at $2.50 a centiare ? Ans. M 36. $93. 6. In a farm consisting of four fields of the following dimensions, how many hectares ? First field, length M 342, breadth M 273 ; second field, length M 634, breadth M 350 ; third field, length M 450, breadth M 329 ; fourth field, length 31 730, breadth M 632.7. Ans. 2A 92.5187. 7. A pile of lumber was found to contain 150 boards M 4 long and iM4. wide, 225 boards M 6.2 long and gM 52. wide, and 642 boards M 5.2 long and 2M 43 wide. How much was it worth, at $42. per are, face measure. Ans. $1008.38 -f- 8. How many bricks iM2.2 X iM 1.1 would pave a side- walk M 842.6 long and M 2.2 wide? and what would be the whole cost at 82 cents per centiare. Ans. 76600 bricks. $1520.05 -\-. MEASUHES OF VOLUMES, OU SOLIDS. RELATIONS OF U2^ITS OF VOLUMES TO UNITS OF LENGTHS. Millistere = A cubic decimetre = 1000 cubic centimetres. r 10 cubic decimetres, or a volume, or solid, whose Centistere = ■< length is one metre, and breadth and thickness one 1 decimetre. r 10 centisteres = 100 cubic decimetres, or a volume Decistere = -< whose length and breadth is one metre, and thick- C ness one decimetre. ^^ _ f ^ ^^^^ metre =10 decisteres = 100 centisteres = \ 1000 millisteres or cubic decimetres. _ (10 cubic metres, or a volume whose leno;th is one Decastere ^^^ ■% * . (. decametre, and breadth and thickness one metre. ( 10 decasteres = 100 cubic metres, or a volume whose Hectostere = -< length and breath is one decametre, and thickness ( one metre. Kilostere = A cubic decametre = 1000 cubic metres. C 10 kilosteres, or a volume whose length is one hecto- Myriastere = ■< metre, and breadth and thickness each one doca- ( metre. THE METRIC SYSTEM. 449 NUMERICAL EXPRESSION FOR VOLUME, OR SOLIDITY. The solidity, or contents, of a volume is expressed numerically by giving the number of times it contains some given solid as the unit of volume. The following illustrations will show how the various denominations of the table are used in numerical expressions of volume. Millistere, or Cubic Decimetre, 10 millisteres, placed side by side, make a volume whose length is one metre, and breadth and thickness each one decimetre, thus, — 10 centistere, placed side by side, make a volume whose length and breadth is each one metre, and thickness one decimetre, thus, — Decistre = 10 Centisteres = 100 Millisteres, 10 decisteres, placed face to face, make a cube whose edge is one metre, thus, — Stere =10 Decisterea = 100 Centisteres = 1000 Millisteres, From these illustrations, it is evident that the contents of a cubic metre may be expressed numerically, as S 1, iS 10, 2S 100, ^S 1000. 450 THE METRIC SYSTEM. The following figures illustrate the use of the same four denominations in expressing the contents of a cubic volume whose edge is one metre and one decimetre. The sur- face of one face of the volume contains one centiare, two milliares, and one deci- milliare, thus, — Centiare. Milliure. F .^^ Taking a slab of the face one decimetre thick, thus, — and we have one decistere, two centisteres, and one millistere. But the volume is eleven deci- metres thick ; therefore we have eleven such slabs, or eleven times one decistere, two centisteres, and one millistere. r 11 millisteres = 1 centistere and 1 millistere = S 0.011 = < 22 centisteres = 2 decisteres and 2 centisteres = S 0.22 (11 decisteres = 1 stere and 1 decistere = S 1.1 Ml.l X Ml.l X Ml.l = S1.331 From these illustrations, we derive the following rule for finding a numerical expression for a given volume of uniform length, breadth, and thickness : — Rule, deduce the length, breadth, and thickness to the same denomination ; find the product of the three dimensions, after re- duction, and point off as many decimal places in this product as there are decimal places in the three dimensions. The unit of the numerical expression thus found will be a millistere when the unit of length is a decimetre, a stere when the unit of length is a metre, a kilostere^ when the unit of length is a decametre. EXAMPLES FOR PRACTICE. 1. How many steres in a wall twenty-four metres long, eight and five-tenth metres high, and fifty-two centimetres thick ? And what would be the cost of building it, at $4.25 a stere ? Ans, S 106.08. Cost, $450.84. THE METRIC SYSTEM. 451 2. What would be the cost of a pile of wood fifteen and seven- tenths metres long, three metres high, and seven and fifty- two hun- dredths metres wide, at $1.50 a store? Ans. §531.29. 3. What would be the cost of excavating a cellar eighteen and three-tenths metres long, ten and seventy-three hundredths metres wide^ and three and four-tenths metres deep, at 15 cents per store ? A71S, $100.14+. 4. How deep must a box be, whose surface is thirty-two milliares, to contain seven and thirty-six hundredths stores? Ans. iM 23. 5. How many stores in five sticks of timber of the following di- mensions : First, jM 5.2 by jM 7.3, and M 13 long; second, 2M 43. by 2M 65, and M 17.5 long; third, iM 5.3 by iM 3.7, and M 15.42 long; fourth, 2M 39 by gM 56, and M 14 long; fifth, iM 4.52 by iM 3.78, and M 15 long. Ans. S 18.470352. 6. What must be the height of a load of wood, M 3.2 long and M 1.1 wide, to contain S 4.0128. A^is. M 1.14. MEASUREMENT OF ANGLES. In the ordinary or sexagesimal system, a right-angle, which is used as the measure of all plane angles, is divided into 90 equal parts, called degrees; a degree is divided into 60 equal parts, called minutes ; and a minute into 60 equal parts, called seconds. In the centesimal or French system, a right-angle is divided into 100 equal parts, called grades ; a grade into 100 equal parts, called minutes ; and a minute into 100 equal parts, called seconds. The former is called the sexagesimal system, on account of the occurrence of the number sixty in forming the subdivisions of a de- gree ; and the latter centesimal, on account of the occurrence of the number one hundred. Grades, minutes, and seconds are usually written thus : 35^ 42^ 24^^ ; read, thirty-five grades, forty-two minutes, twenty-four seconds. Since the scale is centesimal, minutes may be expressed as hun- dredths, and seconds as ten-thousandths ; hence any number of grades, minutes, and seconds may be expressed decimally thus : 73^ 4569 ; read, seventy-three grades, forty-five minutes, sixty-nine seconds. 452 THE METRIC SYSTEM. In a rlght-angie, there are 100 grades, or 90 degrees ; hence, for every 10 grades there are 9 degrees. Dividing the 10 grades into 9 equal parts or degrees, each part will contain 1-J- grades; therefore a degree s equal to 1 J grades. Hence, in any number of grades there are as many degrees as 1^ is contained times in the given number of grades ; and, conversely, in any number of degrees there are 1^ times as many grades as there are degrees. Hence the fol- lowing rules : — TO CHAIs^GE THE CENTESIMAL MEASURE TO THE SEXAGESIMAL. Rule. Express the mmutes and seconds as a decimal of a grade/ divide hyl^x the quotient will express the number of de- grees and decimals of a degree in the given number of grades, min- utes, and seconds, EXAMPLES. Change the following quantities from the centesimal measure to the sexagesimal. 1. 25» 34^ 42^\ Ans. 22° 48' 35.208'^ 2. 57'93\ Ans, 3ri6.932^ 3. 83" 13^ 87^\ Ans, 74° 49' 29.388''. 4. 3G^ 98^ 15^^ Ans. 33° 17' .06". 5. 14^15^60^ Am, 12° 44' 25.44". 6. 90^ 90^ 90^\ Ans. 81° 49' 5.16". 7. 18^ 50^ 25^\ Ans, 16° 39' 8.1". TO CHANGE THE SEXAGESIMAL MEASURE TO THE CENTESIMAL. Rule. Reduce the minutes and seconds to a decimal of a de- gree / multiply the degrees and decimal of a degree by 1^: the pro- duct is the number of grades, minutes, and seconds in the given number of degrees, minutes, and seconds, examples. Change the following quantities from the sexagesimal measure to the centesimal. 1. 3G° 18' 27". Ans. 40^ 31^ 16 J^\ 2. 56' 54". Ans. 1« 5^ 37^y^ 3. 27° 36' 45". 4. 189° 15' 20". 5. C3° 14' 58". 6. 147° 24' 48". 7. 117° 36' 54'. THE METRIC SYSTEM. 453 Ans. 30« G8^ 5f . Ans. 210^ 28^ 39| V\ Ans. 70^ 27' 71|f\ Ans. 168^ 79' 25|f'\ Ans. 130« 68' 33^''. TO CHANGE THE METRIC TO THE COMMON SYSTEM. Rule. Reduce the given quantity to the denomination of the principal unit of the table ; multiply hy the equivalent , and reduce the product to the required denomination, 1. ^M 3.6, how many feet? OPERATION. Analysis. — The metre is ^M 3.6 X 1000 = M 3600 *^e principal unit of the table ; 39.37 in. X 3600 == 141732 in. ^^"^^ we- reduce the kilome- 141732 in. -- 12 in. = 11811 ft. *^^' ^ "^J^'^^' . ^^""'^ ^^^l^ are 39.37 inches in a metre, in 3600 metres there are 3600 times 39.37 inches; and since there are 1 2 inches in a foot, there are as many feet as 1 2 inches is contained times in 141732 inches. Therefore ^M 3.6 is equal to 11811 feet. EXAMPLES FOR PRACTICE. 2. How many feet in 472 centimetres ? Ans, 15.4855 J ft. 3. How many cubic feet in 2 kilosteres? Ans. 70632 cu. ft. 4. How many gallons, wine measure, in 32^5 decilitres? Ans. 8 gals. 2.343— qts. 5. How many gallons in 108.24 litres ? Ans. 28.594 -|- gals. 6. How many bushels in 3262 kilolitres ? Ans, 92559.25 bush. 7. How many square yards in 436 ares ? Ans. 52145.6 sq. yds. 8. In 942325 centilitres, how many bushels ? Ans. 267.3847 + bush. 9. In 436 myriagrammes, how many pounds ? Ans, 9611.9314 lbs. 454 THE METRIC SYSTEM. TO CHANGE FROM THE COMMON TO THE METRIC SYSTEM. Rule. Reduce the given quantity to the denomination in which the equivalent of the principal unit of the metric table is expressed; divide by this equivalent, and reduce the quotient to the required denomination. 1. In 10 lbs. 4 oz. liow many myriagrammes ? OPERATION. Analysis. — 10 lbs. 4 oz. = 10.25 lbs. The gramme, 10.25 lbs. X 7000 z= 71750 gr. *^^ principal 71750 gr. -^15.432 gr. = G 4649.43— ""^^ ^^ *^® G4649.43 — --10000 = ^G. 464943— Ans, *''^^^' '' ^^" pressed in grains; hence we reduce the pounds and ounces to grains. 15.432 grains make one gramme; hence there are as many grammes in 71750 grains as 15.432 grains is contained times in 71750 grains. And since there are 10000 grammes in a myriagramme, dividing G 4649.43 — by 10000 will give the myriagrammes in 10 pounds 4 ounces. Therefore, 10 lbs. 4 oz. is equal to ^G 464943 — examples for practice. 2. In 6172.8 pounds, how many decagrammes? Ans, ^G 280000. 3. How many hectares in 2392 square yards? Ans. ^A .2. 4. How many ares in a square mile ? Ans. A 25899.665552—. 5. How many millisteres in 18924 cubic yards? Ans. sS 14467889.9082 +. 6. In 892 grains, how many hectogrammes ? ^715. 2G. 578019. 7. In 2 miles, 6 furlongs, 39 rods, and 5 yards, how many kilometres? Ans. ^M 4.626416 +. 8. Bought 454 bush, wheat at $3, and sold the same at $8.75 per hectolitre ; how many hectolitres did I sell ? Did I gain or lose, and how much ? Ans. 2L 160. Gain, $38. THE METRIC SYSTEM. 455 MISCELLANEOUS EXAIVIPLES. Required the footings of the following bills : — (1.) New York, April 23, 1867. W. J. Milne, M 122 Broadclotli, " 320 Bid. Shirting, . " 230 White Flannel, '* 206.5 Ticking, •' 107.9 Blk. Silk, Rec'd Payment, BoH of L. CooLEY & Son. @ $6.00 .35 .30 .31 *' 2^ Ans, $1235.975 L. CooLEY & Son. (2.) Chas. D. McLean, Buffalo, May 1, 1867. Bo't of Wm. Benedict. each «G 30.5 @ $ 2.50 ^^ 40.00 tt .32 (t .38 it .50 Ans, $4951.00 Wm. Benedict. 40 chests Tea, 12 sacks Java Coffee, 25 bbls. Coffee Sugar, each ^G 110 10 ** Crushed '* '' ^G 95 30 boxes Raisins, " ^G 12 JRec^d Payment, 3. A man bought a lot of land ^M 40 long and ^M 20 wide, and sold one-third of it. How many ares had he left, and what was the cost of the lot, at $100 per acre? Ans. to first, A 53333.33J. Ans. to second, $197685.95. 4. A farmer sold ^L 540 of wheat at $6, and invested the pro- ceeds in coal at $8 per ton. How many myriagrammes of coal did he purchase? Ans. ^G 36741.835147 +. 5. What will be the cost of a pile of wood M 42.5 long, M 2. high, M 1.9 wide, at $2 per stere ? Ans. $323. 456 THE METFJC SYSTEM. 6. How many metres of shirting, at $.25 per metre, must be given in exchange for ^L 300 oats, at $1.20 per hectolitre? Ans. M 1440. 7. A grocer buys butter at $.28 per lb., and sells it at $.G0 per kilogramme. Does he gain or lose, and what per cent. ? Ans. Lost 2{4|- %. 8. A bin of wheat measures M 5 square, and M 2.5 deep. How many hectolitres will it contain, and what will be the cost of the wheat, at $2 per bushel? A?is. ^L 625. $3546.875. 9. What price per pound is equivalent to $2.50 per ^G? Ans. $11.34. 10. A merchant bought M 240 of silk at $2, and sold it at $1.95 per yard. Did he gain or lose, and how much ? Ans. Gain $31.81. 11. Find the measure of 1^ 5^^ in decimals of a degree. Ans. .00945. - 12. A merchant shipped to France 50 bbls. of coffee sugar, each containing 250 lbs., paying $2 per hundred for transportation. Ho sold the sugar at $.34 per kilogramme, and invested the proceeds in broadcloth at $4 per metre How many yards of broadcloth did he purchase? ^ws. 458.71 -f- yds. 13. The difference between two angles is 10 grades, and their sum is 45'** Find each angle. Ans. 18° and 27°. 14. Determine the number of degrees in the unit of angular measure when an angle of 66 § grades is represented by 20. Alls. 3^. 15. How many centiares of plastering in a house containing six rooms of the following dimensions, deducting one-twelfth for doors, windows, and base ? and what would be the cost of the work at 38 cents per centiare? First room, M 6.2 X M 4.7; second room, M 4.52 X M4 ; third room, M 6 X M 5.2 ; fourth room, M 382 X M3.82; fifth room, M7 X M6.2; sixth room, M4.5 X M 4.25r Height of each room, M 3.8. Ans. gA 562.039 +. $213.57 +. iiJ p /(■■■■■ ^ -^y • ^Ji t • ^ q,^ ■, 'I ^ /^ /^ ^>^^ /c- ////l^^^ 'U-ij 1 . 1^4- - I 44 ■ /r-j. H w /'V^ ' ' ..L s i -k '^ I ^^ Lj^, -Z^.