BURNHAM'S ARITHMETIC, REVISED. A NEW SYSTEM OF AR I T M E T I C, ON AN IMPROVED PLAN: EMBRACING THE RULES OF THREE, SINGLE AND DOUBLE, DIRECT AND INVERSE; BARTER LOSS AND GAIN; REDUCTION; MULTIPLICATION AND DIVISION OF FRACTIONS: EXCHANGE OF CUIRRIENCIEB; INTERESIqTAIA) &LLt. PROPORTIONAL QUESTIONS IN ONE RULE APPLICABLE TO THE WHOLE THE PROCESS GREATLY SIMPLIFIED AND ABRIDGED. BY CHARLES G. BUTRNHAM, A.M. KEENE, N. H.: PUBLISHIED-BY S. & G. S. WOODWARD 1857. Entered according to Act of Congress, in the year 1857, by C. G. BURNHAM, In the Clerk's Office of the District Court of the District of Massachusetts. PREFACE. HE who writes a book in this age merely for the sake cf being a bookmaker, will find that he has written for other times than these; and his fame will be like one of those " second sights," Laving existence only in the mind of him who sees it. Every invention, every thing new, every book, from the clild's primer to the most profoundly scientific text-book, must be tested by a comparison with others, professing, each and all, to be the best extant. 1or will any production gain for its author that for which he labored, unless it finally proves to be what it professes. Improvement is the "charmed word" of the age. It rings hourly in the ear of the multitude. The strong wind bears it onward, and the gentle zephyr wafts its echo. He who has already written his name far above his competitors, now seeks to outdo himself; and the tyro fancies, that he can begin where the best have ended, and run his race alone.' The author of this series of Aritlhmetics cherishes none of these fancies, having already received satisfactory compensation for all his toil, in imparting, from time to time, to those he has had the pleasure to instruct, the improvements here embodied. But should it be found, when the decisive test has been applied, that he has said some things not before said, which may be of benefit to teachers, and the cause of education generally, his pleasure will not be less, because he had ventured to indulge some slight anticipation of the fact. If the experience of some twenty-five years in teaching, has not failed to discover to him the real wants of our schools, then it will be found that his series of Arithmetics is adapted to meet those wants, and is in some measure suited to the spirit of the age in which we live. The Cancelling Arithmetic, published in 1837, was the first work known to the author, which, to any considerable extent illustrated, and practically applied the principle of cancelling. Although it is true that the principle is coextensive with the science of numbers, for no question in Simple Division can be solved without employing it, still Division was not explained as embodying the whole of it, nor was the principle so applied and illustrated as to simplify Division. The mode of writing numbers for the convenience of cancelling, in connection with the ordinary mode, affords a variety of illustrations interesting' and useful, both to teachers and scholars. The application of the cancelling principle is not, however, the only peculiar characteristic of this work. It aims throughout, by the connection of its subjects, and illustration of principle, to impress upon the mind of the scholar the truth, that he will never discover nor need a new principle beyond the simple rules. Hence the first object is to make the scholar thoroughly acquainted with those rules. One thing at a time, 4 PREFACE. and in its time, is the plan. The simple rules are presented in their order singly; then in contrast; then a review of the whole, to exercise the judgment of the scholar. Fractions are introduced as the result of division, or rather as division implied. They are made to occupy the same position, and are illustrated and solved the same as whole numbers. The same numbers are again written in the fractional form, and the scholar is enabled to perceive, at a single view, that a change of position, and of names, is a matter of convenience and not of necessity. In the ordinary mode of presenting fractions, the idea is not precluded from the mind of the scholar, that new positions and new names do not necessarily introduce new principles. The result is, that he perceives no connection between the present and the past, and consequently the subject is ever new, and new difficulties aresconstantly arising. A new form of notation, and new names being introduced, it is in vain to insist that no new principle is employed, so long as the subject is but imperfectly illustrated, and the scholar does not perceive that the change is not a matter of necessity. It is one thing to gain the assent of the pupil to a truth, and it is often quite another to give him a practical understanding of it. It is a fact too little realized, that much time is consumed in going over ground, from which no practical knowledge is gained. Not that the studies themselves are not practical, but they are not pursued in a practical manner. The scholar may be often informed that a fraction is the result of division; that the fractional form of writing numbers is division implied; and that numerator is the same as dividend, and denominator is the same as divisor; and yet difficulties will arise which did not occur in whole numbers. Whereas, a practical knowledge of this fact would enable him to solve most questions, in fractions, with the same facility as in whole numbers; nor would he find any necessity for some half dozen rules; which he is usually required to commit to memory. When the simple rules are thoroughly understood, the pupil may be introduced to the subject of fractions, in a manner similar to the following, at the blackboard. If we divide 2 by 2, the quotient is a unit or 1, 212==1, for the dividend is just equal to the divisor. Vere we required to divide 1 by 2, we should meet with a difficulty, for the dividend is less than the divisor and consequently will not contain it; we must therefore employ a new form of notation, 21=t. We write the divisor under the dividend, and'give a new name to the expression; we call it a fraction, which means a part of a thing. The quotient usually shows how many times the dividend contains the divisor. If the quotient is 2, the dividend contains the divisor twice; if 3, three times. But here the quotient is a fraction, less than a unit, or 1, which shows that the dividend is only a part of the divisor. But what part? The same part the quotient is of a unit. But what part is the quotient of a unit? It will now be convenient to introduce new names, in order to value the fraction. You perceive, that the number which we employed as divisor, we have written under the line, and the number employed as divi dend, is above the line. If our divisor be 2, our quotient is one-half of the dividend; if our divisor be 3, the quotient is one-third of the dividend. Thus it is plain, that in whole numbers, the divisor gives name to the quotient. The same is true when we imply division and write the numbers in the form of a fraction. Our divisor in this example is 2. and PREFACE. 5 our quotient is one-half of the dividend: it is also one-half of a unit. The unit is divided into two parts; our quotient is now denominated; we therefore call the figure below the line, denominator, or namer, because it gives name to the parts into which the unit is divided. Thus we have our fraction named, or denominated; but what is its value? It is halves, but how many halves does it contain? Evidently one, which the figure above the line shows. We have now the fraction denominated or named, and numbered. Its denomination is halves, and their number is one. Making use of the figure above and below the line in one expression, we call the fraction one half, or one-half. Thus you perceive that numerator is the same as dividend, and denomiLator the same as divisor. And, as in division multiplying the dividend increased the quotient, so in fractions, multiplying numerator increases the value of the fraction. Thus: 211 1X2=2 x2-2 221 2 2 Let the scholar write numbers in this manner, side by side, and be exercised, as in division, by multiplying dividend and divisor, numerator and denominator, employing the language of division and the language of fractions, until he is practically familiar with the fact that the principle employed in fractions and whole numbers is the same. Whenever new names are introduced, and new positions employed, let the different forms be written side by side, and extra exercises be given, until the scholar clearly perceives the unity of the principle. (See example under Art. 147.) In Decimal Fractions, also, the points in which they are like whole numbers and common fractions, and points in which they differ, are distinctly brought out as the scholar proceeds, and then, at the close, those points are presented in one general view. In Proportion, new names and new positions are again employed. Let the same pains be taken to contrast the new positions with the former, and to explain the new terms introduced. TO TEACHERS. It cannot be expected that a School Arithmetic, limited in size as it must be, should exhaust its subjects, or give all those illustrations which might be both interesting and useful. The most it can do upon any one subject is to give a single illustration of a principle, a formula of a particular mode of teaching. And that text-book is the best, which by its connection of thought and subjects, and illustration of principle, interests both teacher and scholar, and incites the teacher to invent new modes for himself. Teachers are here presented with an Arithmetic which is the result of much experience in teaching and effort at improvement. It has been the purpose and aim of the author to prepare a work which should accord with the spirit of the age, and be adapted to the schoolroom. It is not expected, nor is it desirable, that the teacher should be confined to the forms laid down in the book. They are designed simply to open the subject-to serve as hints to something better. The peculiar mode of stating questions for the convenience of cancelling and for illustrating fractions as whole numbers, teachers can 1* 6 PREFACE. adopt, or apply the principle of cancelling to the ordinary mode of statement. It will be well to employ both modes, as together they open a wider field for illustration. It is sometimes remarked of the cancelling system, that it is good as far as it goes. The same may be said of arithmetic; for the principle is inseparable from it. It is the only principle by which any question in division can be performed. Wherever it cannot be applied, the numbers must be written in the form of a fraction. When the question involves multiplication and division, it will generally be found to be a great saving of labor, to write down all those numbers which are to be factors of the dividend and divisor, before proceeding to the operation. The eye will then detect at a glance equal factors, and they can be excluded from the operation. The teacher will bear in mind the importance of giving general illustrations of arithmetical principles, whenever it can be done, as its tendency is to enlarge the views of the pupil and to give importance to the study. For example, let simple division be illustrated not only arithmetically, but on general principles. Let it be required to divide 16 by 8, and it may be done and illustrated in the following manner:8)16=&X2 —. 8=2 Ans. Now substitute the letter a for 8, and the letter b for 2, and read the question thus: divide ab by a.'a) b b Ans. Here, as before, we exclude from the dividend a factor equal to the divisor. But this latter process is algebraic; hence the scholar's views are extended, and he perceives at once, and for the first time, the connection between arithmetic and algebra. Formulas are also given to aid the less experienced teacher, and also to bring out more prominently arithmetical principles. MANNER OF RECITATION. Promptness and dispatch are characteristics of our times, and young men must be educated in reference to them. There is no place, perhaps, better calculated to train a scholar to think and act with precision and energy, than at the blackboard. When a scholar is called out fiom his class to solve a question, let him quickly, and with gentlemanly mien endeavoring to be self-possessed, take'lis stand at the board, read his question distinctly, and with the same reference to rhetorical notation as though he were called out on purpose for the reading of the question. Then let him state his question, giving the reasons for each step as he proceeds; or let him state and solve his question, then return to the commencement, and illustrate the principle, and give the reason for each step in the.solution. Then let him pause at the board a moment, for his teacher to propose such questions as he may think proper. A brief view has now been given of the plan and mode of teaching arithmetic adopted in this system. It is confidently believed, from the long experience the author has had in teaching, that the mode here DEFINIT' ONS. 7 adopted for prusenting the subject of aritimnctic, will be found better calculated to induce a fondness for the study; that it unfolds more of the science, and brings out principles more clearly than any other system now before the public. With these views the author submits the work to the candid perusal of all who are interested in the progress of knowledge. CHARLES G. BURNHAM. DANVILLE, VT., Oct. 18, 1849. DEFINITIONS, AXIOMS, AND SIGNS. DEFINITIONS. A definition is what is meant by a word or phrase. The language of a definition should be so plain as not to be capable of misapprehension. 1. Quantity is any tiling which may be multiplied, divided, and measured. 2. Magnitude is that species of quantity which is extended; i. e. which has one or more of the three dimensions-length, breadth, and thickness. A line is a magnitude, because it has length. 3. Mathematics is the science of quantity. 4. Aritlmnetic is the science of numbers. 5. Algebra is a method of computing by letters and othensymbols. 6. Geometry treats of lines, surfaces, and solids. Aritlmetic, Algebra, and Geometry are those parts of mathematics, on which all the others are founded. 7. A Demonstration is a course of reasoning which establishes a truth. 8. A Proposition is any thing proposed: if to be proved or demonstrated, it is called a Theorem; if to be done, it is called a Problem. 9. A plus quantity is a quantity to be added, and has this sign + before it; thus, +6. 10. A mninus quantity is a quantity to be subtracted, and has this sign - before it; thus, - 6. 11. An Equation is a proposition expressing equality between one quantity; or set of quantities, and another, or between different expressions for the same quantity; thus, 5=3+2. 12. A member of an equation is the quantity or quantities on one side of the sign of equality. Os. —For definitions of terms in more common use in this work, see Art. 54, or Part I. 8 AXIOMS-SIGNS. AXIOMS An axiom is a-self-evident proposition. 1. Things which are equal to the same thing are equal to each other. 2. If equals be added to equals, the wholes will be equal 3. If equals be taken from equals, the remainders will be equal 4. If equals be added to unequals, the wholes will be unequal. 5. If equals be taken from unequals, the remainders will be unequal 6. Things which are double of equal things are equal to each other. 7. Tlings which are halves of the same thing, are equal to each other 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. SIGNS. - Equality is denoted by two horizontal lines. + Addition: as 4-3=-7; which signifies that 4 added to 3 equals 7. X Multiplication: as 4 X3=12; which signifies that 4 multiplied by 3 equals 12. - Subtraction: as 4-3=1; which signifies that 3 taken from 4 leaves 1. ), 214, 4 Division: as, 2)4(2, and 4+- 2=2, and 4=2, and 214=2. In either case it signifies that 4 divided by 2 equals 2.:::: Proportion: as, 2 4: 6: 612; which is read, 2 is to 4 as 6 is to 12. Vinculun: as 4+3=-7; which is read, the sum of 4 and 3 equals 7, and 4-3~:l, is read, the difference of 4 and 3 equals 1. / Radical sign: placed before a number denotes that the square root is to be taken. 42 implies that 4 is to be raised to the second power. 43 implies that 4 is to be raised to the third power. n implies the third root. CONTENTS. PAGE PAG Simple Numbers. Multiplication of Whole Numbers by Notation............................ 11 Fractions....................... 74 Numeration...................... 12 Division of Whole Numbers by FracAddition............................ 19 tions.......................... 75 Subtraction................... 22 Multiplication and Division of Whole Addition and Subtraction 2........... 26 Numbers by Fractions........ 77 Practical Questions in Addition and Sub- Multiplication of Fractions by Fractraction.....................27 tions................. 78 Multiplication....................... 28 Division of Fractions by Fractions..., 79 Multiplication and Division Table..... 29 Multiplication and Division of FracDivision....................... 37 tions......... 81 Short Division.............. 39 Promiscuous Examples............. 82 Long Division............... 42 Addition of Fractions................ 83 Multiplication and Division.......... 9 To find a Common Denominator....... 85 Multiplication and Division by Cancel- Subtraction of Fractions.......... 86 ling............................. 49 To find the Least Common Multiple.. 87 Supplement to the Four Fundamental Decimal Fractions.................. 90 Rules........................ 52 l)w Decimal Fractions are produced. 92 Exercises in the use of the Signs..... 54 Addition of Decimals................ 93 Ratio, or the Relation of Numbers..... 553 Subtraction of Decimals............. 96 Fractions-their Origin............... 55 Multiplication of Decimals............ 97 Common or Vulgar Fractions......... 58 Division of Decimals........ 93 Definitions.............. 60 Federal Money..................... 100 To reduce a Compound Fraction to a Addition of Federal Money.......... 101 Simple.......................... 63 Subtraction of Federal Money....... 102 To change any given Fraction to an Multiplication of Federal Money.... 103 Equivalent, having a given Denom- Division of Federal Money........... 104 inator........................... 64 Supplement to Decimal Fractions and To reduce a Whole Number to an Federal Money.................. 104 Equivalent Fraction, having a given Bills of Parcels...................... 105 Denominator..................-.. 64 Compound Numbers........ 106 To reduce Improper Fractions to Mixed Tables of Compound Numbers....... 106 Numbers, and the reverse........ 65 Reduction......................... 112 To reduce a Fraction to its Lowest Reduction Descending and Ascending 113 Terms................ 6 Supplement to Reduction of Whole To reduce a Complex Fraction to a Sim- Numbers...................... 117 ple, and the reverse...-.... 68 Reduction of Fractions -............ 118 General Rule for the Multiplication and Reduction Descending and Ascending 121 Division of Fractions by Whole Comparison of Numbers and QuantiNumbers, Whole Numbers by Frac- ties............................ 122 tions, Fractions by Fractions..... 69 To reduce Fractions to Integers of Multiplication of Fractions by Whole lower Denominations, and the reNumbers.. ve......... 70 verse.......................... 123 Division of Fractions by Whole Num- Reduction of Vulgar Fractions to Decibers............................. 72 mal............................ 125 Multiplication and Division of Fractions To reduce a Decimal Fraction to a Vulby Whole Numbers............. 73 gar................ 126 10 CONTENlTS. PAGE PAOG To reduce Integers of different Denomi- Supplement to Interest, Discount, etc. 216 nations to a Decimal Fraction, and Equation of Payments............... 218 the reverse.................. 127 Fellowship.............. 220 Inspection.......................... 128 Assessment of Taxes................ 222 Compound Addition.............. 129 Double Fellowship............ 224 Compound Subtraction.............. 133 Involution................. 225 To reduce Longitude to Time........ 136 Evolution...................... 227 Compound Addition and Subtraction. 137 Extraction of the Square Root........ 228 Compound Multiplication and Division 138 Extraction of the Cube Root......... 235 Supplement to Compound Numbers.. 142 Extraction of Roots in General...... 242 Supplement to Fractions............ 145 Arithmetical Progression....... 243 Circulating Decimals................ 147 Geometrical Progression........ 247 Ratio and Pl portion................ 149 Compound Interest by Progression... 250 Direct Proportion................... 156 Annuities at Compound Interest..... 252 Inverse Proportion.................. 167 Permutation................... 256 Compound Proportion............... 169 Single Position.................. 257 Supplement to Proportion.............. 176 Double Position................... 259 Exchange................... 178 Alligation Medial............,262 Domestic Exchange............... 179 Alligation Alternate.... 263 Foreign Exchange................. 181 Duodecimals........................ 267 Bills of Exchange................. 185 Miscellaneous Rules............ 271 Reduction of Currencies............... 188 Mensuration of Surfaces.............. 271 Percentage.................. 190 Mensuration of Solids........ 282 Interest............................ 191 Gauging. 284 Interest by Cancelling............... 196 Measuring Grain................ 285 Time, Rate, and Amount given, to find Tonnage of Vessels.. 286 the Principal....-............. 197 Mechanical Powers................ 287 Time, Rate, and Interest given, to find The Lever......................... 287 the Principal.............. 198'Te Wheel and Axle............... 288 Principal, Interest, and Time given, to The Pulley..................... 289 find the Rate per cent........... 198 The Screw...... 290 Principal, Rate, and Interest given, to The Wedge......................... 291 find the Time............. 199 Mathematical Problems.............. 292 To find the Interest on Notes, etc., Levelling.......-...... 297 when Partial Payments have been Philosophical Problems............. 298 made.................. 199 Table of Specific Gravities........... 302 Commission, Brokerage, and Insurance 202 Astronomical Problems.............. 303 Compound Interest............. 203 Of Balls and Shells................. 304 Discount................ 206 Piling of Balls.................... 305 Bank Discount..................... 208 To find the Weight of Cattle.......... 306 Loss and Gain............ 210 Miscellaneous Questions............ 307 Stock......................'. - 213 Book-keeping -.................... 317 Barter............ -.....'....- 214 Notes, Receipts, etc................. 322 ARITHMETIC. Art. 1.-ARITHMETIC is the science of numbers. It explains their properties, and teaches how to apply them to practical purposes. Art. 2.-The principal, or fundamental rules, are, Notation, Numeration, Addition, Subtraction, Multiplication, and Division. These are called fundamental rules, because all questions in Arithmetic are solved by one or more of them. Art. 3.-Notation is the expressing of any number or quantity by figures; thus, 1 one; 2 two; 3 three; 4 four; 5 five; 6 six; 7 seven; 8 eight; 9 nine; 0 cipher. The first nine figures are sometimes called digits, from the Latin word digitus, which means a finger. In the early stages of society people counted by their. fingers; they were also formerly all called ciphers-hence the art of Arithmetic was called ciphering. Art. 4 —There are two methods of Notation-the Arabic, as above, and the Roman, which is expressed by the following seven letters of the alphabet: I, V, X, L, C, D, M. 1 2 3 4 5 6 8 9 10 20 30 40 60 1, II, III, IV, V, VI, VII, VIII, IX, X, XX, XXX, XL, L, 6 0 80 9 100 00 00 1000. LX, LXX, LXXX, XC, C, D, M. Art. 5,-When a letter of less, is placed before one of a greater value, it diminishes the value of the greater, by the value of itself-thus, X signifies ten, but IX is only nine. When a letter of less, is placed after one of greater value, it increases the value of the greater by the value of itself. This method is seldom used except in numbering chapters, sections, etc. QUESTIONS.-1. What is Arithmetic? 2. What are the principal, or fundamental rules? 3. Why so called? 4. What is Notation? 5. What are the first nine figures sometimes called? 6. What were they all formerly called? 7. How many methods of Notation, and what are they? 8. How many are the Arabic characters, or figures? 9. By what is the Roman method expressed? 10. How is a letter affected when one of less value is placed before it? 11. flow when one of less value is placed after it? 12. For what is the Roman method of Notation principally used? 12'NUMERATION.-TABLE. NUM ERATION. Art. 6.-Numeration teaches to express in words the value of any number represented by figures. Thus, 365 is read, three hundred and sixty-five. Art. 7. —Figures have a simple and relative value. When a figure stands alone its value is simply so many units, or ones; as, 2 two; 3 three; 4 four. Their relative value is derived. from the place they occupy when joined together, or from their distance from the unit's place. Thus, 2 and 3 express their own value; simply so many units; but they are made to express either 23 or 32; that is, either three units and two tens, or two units and three tens. Hence it appears that the first, or right-hand place, always expresses so many units; it is therefore called the unit's place; the second, the place of tens, expressing always as many tens as the figure contains units. The third place is hundreds; the fourth, thousands, as may be seen by the following TABLE. a. 0 8, X ^4, 8 2 1 F3 2 5 3. i D' o &11' c i ~. * us: *... * -.1 On.': -: -'. 2 1 TT enty-one.... 2 1 irtee hundred and 21.. 4, 3 2 1 Four tlhousand land 321. *.. 5 4, 3 2 1 Fifty-four thousand and 321. 6 5 4, 3 2 1 Six hundred and fifty-four thousand 321. 7, 5 4, 3 2 1 Seven millions 654 thousand 321. 8 7, 6 5 4, 3 2 1 87 millions 654 thousand 321. 9 8 7, 6 5 4, 3 2 1 987 millions 654 thousand 321. QUESTIONS. —13. What is Numeration? 14. What is the value of a figure standing alune? 15. From what is their relative value derived? 16. What does the first, or right-hand figure, always express, and what is it called? 17. What are the second, third, and fourth places called? 18. What is the value of the cipher, when standing alone, or at the left hand of another figure? 19. What effect has it when placed at the right of another figure? NUMERATION TABLES. 13 Art. 8.-The cipher, when standing alone, or at the left hand of another figure, signifies nothing, as 05, 005, is five in either case, because it still occupies the unit's place. But when placed at the right hand of another figure, it increases its value in a tenfold ratio, by removing the figure farther from the unit's place. This may be seen by the followingTABLE II. 0' Nothing. 20 Twenty. 200 i Two hundred. 2,000 iTwo thousand. 20,000 Twenty thousand. 200,000 Two hundred thousand. 2,000,000 iTwo millions. Art. 9.-To know the value of any number offigures. RULE.-1. Numerate from the right hand to the left, by saying units, tens, hundreds, &c., as in the Table. 2. To the simple value of each figure join the name of its place, reading from the left hand to the r'gh'. TABLE III. - I 0 I 10 I 0 I 4o g..~ Q ^> Q 0 ^c i 0? 0* ^ ff E 783 694 542 987 5621714 923 610 782 184 542 3651987 963 ) n a. o oq e E -a I 1 1 The first division of the foregoing Table is according to the French method, into periods of three figures each: the name of the period is superadded. The second division is according to the English method, into periods of six figures each. The name of each period is subjoined. The two divisions of the QUESTIONS.-20. How may the value of any number be found? 21. What are the two methods ofnumeration in the third table? 22. In what respect do they differ? 2 14 EXERCISES [N NUMERATION. Table agree for the first nine figures-beyond that they assume different names. The principles of Notation in both are the same. In the former method the names, units, tens, hundreds, are repeated in each period; in the latter method, thousands, tens of thousands, hundreds of thousands, are repeated with the name of the period. If the sum be not expressed in figures, it is necessary to know the method of notation employed. Art. 10.-Let the scholar point the following numbers into periods, and read them. 3445 67891 983452 5437643 67821356 436543897 5678923412 96754329876 1234678901263 Art. 11.-Express the following numbers in figures. I. Twenty-three. 2. Thirty —ve. 3. One hundred and twenty. 4. One hundred and twenty-six. 5. Ten thousand three hundred and twenty. 6. Four millions four thousand and four. 7. One hundred and seventeen millions, one hundred and two. 8. Three lillions, three millions, seventeen thousand and ten. 9. One hundred billions, onehundred thousand, two hundred and fifty. 10. Twenty billions and twenty. 11. Seven billions, seven thousand and seventeen. 12. One hundred-and seven billions, twenty-seven thousand and one. 13. Five hundred and four trillions, two billions, ten millions, ten thousand and ten. 14. Forty-five trillions, forty billions and thirteen. 15. Two millions, two thousand, three hundred and three. 16. Thirty quadrillions, fifty millions, four thousand, three hundred and forty-eight. EXERCISES IN NUMERATION. 15 17. Four hundred and foul quadrillions, seven hundred and seven thousand, two hundred and two. 18. Four quintillions, thirty-five quadrillions, three trillions, two billions, twenty-seven millions, three hundred and forty thousand, four hundred and seventeen. ADDITION TABLE. Art. 12.-Signs.-A cross + is the sign of addition. It shows that the numbers between which t is placed are to be added. Two parallel horizontal lines = signify equality. Thus: 3+4==7 is read, 3 added to 4, or 3 plus 4 (plus is a Latin word, which signifies more) is equal to 7. The following TABLE may be read thus: 2 and 0 are two; 2 and I are 3, &c. 2+0= 2 3+0= 3 4+0= 4 -5+0= 5 2+1= 3 3+1= 4 4+1= 5 5+1= 6 2+2= 4 3+2= 5 4+2= 6 5+2= 7 2+3= 5 3+3= 6 4+3= 7.5+3= 8 2+4- 6 3+4= 7 4+4= 8 5+4= 9 2+5= 7 3+5= 8 4+5= 9 5+5=10 2+6= 8 3+6= 9 4+6=10 5+6=11 2+7= 9 3+7=10 4+7=11 5+7-=12 2+8=10 3+8=11 4+8=12 5+8=13 2+9=11 3+9=12 4+9-=13 5+9=14 6+0= 6 7+0= 7 8+0= 8 9+0= 9 6+1= 7 7+1= 8 8+1= 9 9+1=10 6+2= 8 7+2= 9 8+2=10 9+2=11 6+3= 9 7+3=10 8+3=11 9+3=12 6+4=10 7+4=11 8+4=12 9+4=13 6+5=11 7+5=12 8+5=13 9+5=14 6+6=12 7+6=13 8+6=14 9+6=15 6'+7=13 7+7=14 8+7=15 9+'7=16 6+8=14 7+8=15 8+8=16 9+8=17 6+9=15 7-9=16 8+9-=17 9+9=1.8 QUESTIONS.-Two and 0-how many? 2. Two and 1 —how many? 3. Two and 2 — how many? The scholar should be questioned in this manner, until he is familiar with the above table. 16 - EXERCISES IN NUMERATION. The scholar should be well versed in Notation and Numeration, before proceeding to the following questions. EXERCISES. Art. 13. —1. If John has 6 apples, and his brother gives him 3 more, how many will he have? 2. James being on a visit at his uncle's, one of his cousins gave him 3 walnuts, another 4, and his uncle gave him 9; how many did he receive? 3. Samuel bought a book for 15 cents, and a slate for 17; how many cents did he give for both? 4. If a boy pay 15 cents for a book, 10 for a knife, and 6 for a dozen of apples, how many cents does he pay in all? 5. If an inkstand cost 10-cents, an orange 5, a lemon 3, and a dozen of quills 14 cents, what is the cost of the whole? 6. A man bought of a drover 3 sheep and a cow; for one of the sheep he paid 4 dollars, for the other two he paid 3 dollars apiece, for the cow he paid 20 dollars; how many dollars did he pay for the whole? 7. Joseph bought a sled for 25 cents, a yoke for 12 cents, and a whip for 6 cents; what did the whole cost him? 8. If I pay 6 dollars for a hat, 8 for a cap, 4 for a vest, and 14 for a coat, what do I pay for the whole? 9 If I owe one man 6 dollars, another 8, another 12, another 20, how much do I owe in all? 10. The scholars in a certain school are divided into 4 classes; in the first class there are 10 scholars, in the second 12, in the third 9, and in the fourth 14; how many in all? 11. If from my library I lend to one man 5 books, to another 10, to another 8, to another 12, to another 20, how many do I lend in all? 12. In my garden there are 6 apple-trees, 8 pear:trees, 10 peach-trees, 18 plum-trees; how many trees are there in all? 13. In a certain school 10 study music, 12 French, 14 Spanish; how many are there in all these studies? 14. Eliza had 4 finger-rings, Mary had 10, and Susan had 7; how many had they in all? 15. A certain man had 4 boarders; for two he received 3 dollars each per week, for one 2 dollars, for another 5; how much did he receive per week for the whole? 16. A young lady bought two dresses; for one she paid 7 dollars, for the other 9 dollars; how much did she pay for both' ADDITION. 17 ADDITION. Art, 14.-ADDITION is the putting together of two or more numbers so as to make but one. The number thus obtained, is called their sumn or amount. Art. 15 —SIMPLE ADDITION is the putting together of two or more numbers of the same kind. OBS.-It is called Simple Addition, because the numbers are all of one denomination; that is, all dollars, or all cents. When the numbers are pounds, shillings, pence, &c., the denominations are different. If one man owe me 25 dollars, another 22,-to find the amount of what both owe, I write the sums in the following manner, units under units, tens under tens, and add them together, thus: Tens. Units. 25 or thus, 2 + 5 22 2+ 2 47 amount. 4 + 7 = 47. Illustration.-Beginning at the right hand, or unit's place, I say 2 and 5 are 7; then, in the second place, or place of tens, I say 2 and 2 are 4-which is 4 tens, or 40. 2. A man has three fields; one contains 31 acres, another 25, another 42; how many acres are there in all? Operation. Illuzstration.-Having written the numbers O 31 according to the directions, units under units, 25 tens under tens, &c., we begin at the right hand 42 to add, and find the amount to be 8 units, which we place under the column of units. The nas. 98 amount of the second column, or column of tens, we find to be 9 tens, or 90. The answer, then, is 9 tens and 8 units, or 98. 3. What will a carriage, horse and harness cost, if the carriage cost 102 dollars, the horse 80 dollars, and the harness 16 dollars? Ans. 198. 4. If a wagon cost 78 dollars, and a yoke of oxen 96 dollars, what will be the cost of both? QUESTIONS.-1. What is Addition? 2. What is Simple Addition? 3. How are the numbers to be added written? 4. By what number do you carry? 5. Why? 6. What is the number called arising from the operation? 7. What is the sign of Addition? 8. Sign of Equality? 9. Sign of Subtraction? 10. What does plus signify? 2 * 18 ADDITION-RULE. In the preceding examples, the numbers, when added, have been less than 10, and, of course, have required but one figure to express them. In the last example it will be seen that the numbers in the unit column, when added, amount to more than 10, and in the column of tens, the amount is more than tenthat is, ten tens. Let the student write the numbers to be added on the blackboard, and illustrate in the following manner: Seventy-eiglit equals seven tens plus eight units, and ninetysix equals nine tens plus six units. Operation 1st, 2d, 3d. Writing the numbers, -Tens. n units under units, and Tels. Units. Tens. Units. 7 + 8= 7 + 8 = 78 tens under tens, and 9 + 6 9 + 6 = 96 adding, we have sixteen 16 + 614 = + 4 1= 6 tens plus fourteen units, 16 +q 14 -- 17 - 4 - - 17^4 Ans. but fourteen units equal one ten plus four units; the left-hand figure of the units therefore belongs in the column of tens. It will be seen by this operation, that what is called carrying for ten, is simply adding numbers to the column where they belong. In practice, numbers are written as in operation 3d, and a part of the operation is carried on in the mind. The same may be illustrated, thus: 78 Placing the numbers as before directed, and 96 adding the right-hand column, we find it amounts to 14 units, or I ten and 4 units. The next 14 column amounts to 16 tens, or 100 and 6 tens, which, when added, make 4 units, 7 tens,-and 174 Ans. 100, or 174, the answer. From the foregoing it is evident, that one in the column of tens is equal to ten in the column of units, and one in the column of hundreds is equal to ten in the column of tens. This is the reason why we carry for 10 rather than any other number. RULE. Add each column, beginning at the right hand, and set down the amount directly under the column, if it be less than 10; but if it be 10 or more, set down the right-hand figure, and add the left to the next column. Under the last left-hand column set down the whole amount. This is the same as carrying one for every ten. QUESTIONS.-11. What is the rulv for Simple Addition? 12. How is Addition proved? ADDITION-RULE AND EXAMPLES. 19 Proof.-Perform the addition downwards, and if this last amount corresponds with the sum total, the work is supposed to be right. The following method may be adopted when the scholar has become acquainted with the rule of Division: Add the digits in the top line, and from their sum reject the nines, and write down the excess at the right hand, directly even with the figures in the line. Proceed in the same manner with each proposed line of figures. Then add the several remainders, or excess of nines, and from their sum reject the nines and write down the excess. Add the digits in the sum total, and reject the. nines, and the excess, if the work be right, will be the same as the excess last obtained. Thus, The sum of the digits in the sum total 3497 m 5 is 16, and the excess above 9 is 7; the 6512 5 same as the last excess. 8295 ~ 6 18304 M 7Ex. 9's. This method of proof depends upon a property of the number nine, which belongs to no other digit but 3, which is a factor of 9; -namely, that any number divided by 9 will leave the same remainder as the sum of its digits divided by 9. This peculiar property of the number 9 grows out of the decimal relation of place. Were the ratio of increase any other than ten fold, it would belong to the number next before the ratio. E. g. Were the ratio five fold, it would belong to 4; were it six, it would belong to 5, &c. It will be observed, that if we remove a digit from the units' place to the tens, we increase its value by as many 9's as there are units in the digit. Thus: 2 units, 20 units, or two 9's plus 2 units.If we remove the digit still further to the left, we increase the number of 9's ten fold at each remove. Thus: 200=22 nines plus q units, and 2000-222 nines plus 2 units. Hence it appears that whatever be the number of 9's expressed by the digit, the digit itself is always a remainder; and since the proof depends not upon the number of 9's, but upon the remainders, it is plain that if we add any number of columns of units upwards, and then add the several lines from right to left, we shall have the sum of the same digits. If therefore we cast out the nines from each sum, the remainders, if the work is right, must be the same. OBS.-This method of proof, although not a demonstration, is, nevertheless, very satisfactory, for it presents an entirely different combination of digits; nor should we be likely to drop a 9, or a multiple of 9. 20 EXERCISES IN ADDITION. EXERC[SES. Art. 16.-1. If a man pay 1496 dollars for a house, 734 dollars for a lot of land, 300 dollars for railroad stock, and 145 dollars for a share in a bridge, how much does he expend in the whole? Ans. 2675 dollars. 2. A man sold plank to the amount of 834 dollars; boards to the amount of 376; shingles to the amount of 400; timber 621; two masts, Qne for 30 and the other for 50 dollars; what was the amount of the whole? Ans. 2311 dollars. 3. A merchant, on settling his accounts, finds himself in debt to A. $100; to B. 60; to C. 78; to D. 80; to E. 447; how much does he owe in all? Ans. 765 dollars. 4. From the creation of the world to the Christian era was 4004 years; from that time to the Declaration of American Independence was 1776, and 64 years since that period. How many years since the Creation? Ans. 5844 years. 5. A man by his will left his two sons 1450 dollars each; his four daughters 1200 each; to his wife 1500; to various charitable objects, 1834; what was the value of his estate? Ans. 11034 dollars. 6. If 1889 figures cover one side of a slate, how many will it take to cover both sides of 4 slates'? Ans. 15112. 7. Bonaparte was born in the year 1769; lived 52 years. In what year did he die? Ans. 1821. 8. General Jackson took the Presidential chair in 1829; occupied it 8 years. In what year did his course terminate? Ans. 1837. 9. George Washington was born in the year 1732. He lived 67 years. In what year did he die? Ans. 1799. 10. The distance from New York to Rahway, N. J., is 20 miles, from Rahway to New Brunswick 12 miles, from New Brunswick to Princeton 18 miles, from Princeton to Trenton 12 miles, from Trenton to Bristol 10 miles, from Bristol to Philadelphia 20 miles. What is the distance from New York to Philadelphia? Ans. 92 miles. 11. Lafayette was born in the year 1757. He died at the age of 78. In what year did he die? Ans. 1835. 12. A man sold five oxen, each weighing 864 pounds; how much did they all weigh? Ans. 4320. 13. How many times does a common clock strike in 24 hours? Ans. 156. 14, A gentleman left his two sons each 1480 dollars; his SIMPLE ADDITION. 21 only daughter 1500 dollars, and his wife 200 more than all his children; what was the wife's portion, and what was the value of the whole estate? Ans. Wife's portion, 4660 dolls. Whole estate, 9120d 15. There are two numbers the less is 1768; their difference is 961; what is the larger number? Ans. 2729. 16. From Boston to Providence it is 40 miles; from Providence to New York 198 miles; from New York to Philadelphia 92 miles; from Philadelphia to Wilmington 28 miles from Wilmington to Baltimore 72 miles; from Baltimore to Richmond 110 miles; from Richmond to Raleigh 155 miles; from Raleigh to Charleston 256 miles; from Charleston to Savannah 113 miles; from Savannah to New Orleans 713 miles. How many miles from Boston to New Orleans, passing through the above places? Ans. 1777 miles. 17. A man bought five firkins of butter; one firkin contained 150 pounds, another 60, another 75, another 98, another 125. How much did they all contain? Ans. 508. 18. There were five churches erected, one in -, which cost 16,500 dollars, two in ~, which cost 18,350 dollars each, one in -, which cost 19,386 dollars, and one in -, which cost 12,640 dollars. How much was the expense of the whole? Ans. 85,226 dollars. When the columns to be added are long, the following method will be found convenient. Begin to add with the unit figure, as usual; and for every ten, place a dot against that figure which makes ten, or more than ten, and add the excess to the figure above it; and thus proceed to the top of the column. Write the excess of ten at the foot of the column added; then count the dots, and as many as they are, so many carry to the next left-hand column. 27687 3978 3.4.9.5 78989 2129 6 7 8.9. 87896 9723 2.3.1 6. 98988 1320 7 6 3 4 65769 9621 2 3.16. 75645 8732 5.5 6.7. 1256 2148 30265 QUESTIoN. —When the columns to be added are long, how may you proceed? 22 SUBTRACTION.-TABLE. SUBTRACTION. Art. 17.-1. John's father gave him 6 apples. He gave his brother 4 of them. How many had he left? 2. Joseph bought sixpence worth of candies, and ninepence worth of hazel-nuts. How much more did he give for the hazel-nuts than for the candies? 3. Henry was 10 years old when his mother died; his sister was 6. How much older was Henry than his sister? SUBTRACTION TABLE. Art. 18.-Signs. A short horizontal line - signifies subtraction. Thus: 7-4=3, is read: 7 minus 4 (minus is a Latin word, which signifies less) equals 3. 2-2= 0 3-3= 0 4-4= 0 5-5= 0 3 —2= 1 4-3= 1 5-4= 1 6-5= 1 4-2= 2 5-3= 2 6-4= 2 7-5= 2 5-2= 3 6-3= 3 7-4= 3 8-5= 3 6-2= 4 7-3= 4 8-4= 4 9-5= 4 7-2= 5 8-3= 5 9-4= 5 10-5= 5 8-2- 6 9 —3- 6 10-4= 6 11-5= 6 9-2='7 10-3= 7 11 -4= 7 12-5= 7 10-2= 8 11-3= 8 12-4= 8 13-5= 8 11-2- 9 12-3= 9 13-4= -9 14-5- 9 12-2=10 13-3=10 14-4=10 15-5=10 6-6= 0 7- 7= 0 8-8= 0 9-9= 0'7-6= 1 8-7- 1 9-8= 1 10-9= 1 8-6= 2 9-7= 2 10-8= 2 11-9= 2 9-6= 3 10-7- 3 11-8= 3 12-9= 3 10-6= 4 11-7= 4 12-8= 4 13-9= 4 11-6= 5 12-7= 5 13-8= 5 14-9= 5 12-6= 6 13-7= 6 14-8= 6 15-9= 6 13-6= 7 14-7=- 7 15-8= 7 16-9= 7 14-6= 8 15-7= 8 16-8= 8 17-9= 8 15-~6= 9 16-7= 9 17-8= 9 18-9= 9 16-6=10 17-7=-10 18-8=10 19-9=10 QUESTIONS. —1. Two from 2-how many? 2. Two from 3-how many? 3. Two from 4-how many? 4. Two from 5-how many? In this manner the scholar should be questioned, until he is familiar with the above Table. SIMPLE SUBTRACTION.-ITS OBJECT. 23 Art. 19.-Susan had 6 frocks; 4 of them she burnt. How many had she left? Four from (, and 2 remain; 6, the larger number, is called the Minuend, because it is the number to be diminished, or made less; 4 is called the Subtrahend, because it is the number to be subtracted; 2, the difference, is called the Remainder, because it is the number left after subtraction. The process of finding the difference between two numbers, is called Subtraction. Art.'20.-SIMPLE SUBTRACTION teaches to find the difference between two numbers of the same name or kind. (OBs. Art. 15.) The object in subtraction is to take the whole subtrahend from the whole minuend. Whenever the numbers are small, the operation may be performed in the mind; but when they are large, it is better to write them down, and subtract a part at a time. Thus, from 252 subtract 161. First Operation. Second Operation. Hunds. Tens. Units. Hunds. Tens. Units. 252 = 2 + 5- 2= I + 15 + 2 161 = 1 + 6 + =1 + 6 + 1 Ans. 91. =9 + 1 Ans. We first subtract I unit from 2, and write down the remainder. Then, because 6 tens, the next left-hand figure of the subtrahend, cannot be taken from 5, the figure above it, we take one from the next left-hand figure of the minuend, or place of hundreds, equal to 10 tens, which, added to 5 tens, makes 15 tens. Second Operation. Then, 6 from 15, and 9, or 9 tens remain. Subtracting tens from tens, the remainder is tens. Now, because we have taken 1 from 2, in the place of hundreds, and added it to the place of tens, we call the 2,as it really is,-1, and say, 1 from 1, and 0 remains; or, which is the same thing, we may add I to the lower figure, and say, 2 from 2 and 0 remains. Thus it appears that what is sometimes called borrowing ten is really making a new division of the minuend. This may be illustrated in the following manQUESTIONS.-1. What is Subtraction? 2. What does Simple Subtraction teach? 3. Why is it called simple? 4. How many numbers are required to perform the operation? 5. Which is the minuend? 6. Why called minuend? 7. Which is the subtrahend? 8. Why called subtrahend? 9. What is the remainder? 10. Why called remainder? 24 SIMPLE SUBTRACTION. ner: Suppose a man have 252 bushels of grain in 3 boxes; il the first, 200 bushels; in the second, 50; in the third, 2. He sells to A. 100 bushels; to B. 60; to C. 1. How many has he left? 252=200+50+2 He may take one bushel from the 161=100+60+1 smallest box, but the 60 bushels cannot be taken from the 50 in the second box; he therefore takes 100 bushels from the largest box, and adds it to the 50, in the smaller. Thus: 100+150+2=252 He can now take 60 from 100+ 60+1-161 150, or 6 tens from 15 90+1= 91- A9ns. tens, and 9 tens remain. Then, because he has taken 100 bushels from the largest box, there remains but 100; therefore, 100 from 100, or I from 1, and nothing remains. Whenever, therefore, the lower figure exceeds the upper, we take 1 from the next left-hand column of the upper line, calling it 10, because 1 in the left-hand column is equal to 10 in the right, and add it to the upper figure. OBs.-We take from the left-hand place, because the right can never contain enough. We take but 1, because I is always sufficient. 2. From seven thousand and five, take six thousand seven hundred and forty-six. Operation. In this example the 6 units of the subtrahend'7-006000+ 990+15 cannot be taken from the 5 units of the minuend. Ans. 259- 250 + 9 We must, therefore, borrow, or rather make another division of the minuend; but as the second and third places contain ciphers, we must go to the thousand's place. From the one thousand, which we borrow, we take ten units, and add them to the 5 units of the minuend. The remainder, nine hundreds and nine tens, now occupy the second and third places, instead of the ciphers, and we say 4 from 9, and 5 remain; 7 from 9, and 2 remain. From the foregoing we derive the following RULE. Place the numbers, the less under the greater; units under units, tens under tens, hundreds under hundreds, etc. Begin at the right hand, or unit's place, and take each figure in the lower line from the-one above it, and set down the remainder. If either EX AMI LES IN SUBTIIACTION. 25 9f the lower fyiures be grieater than the one above it, suppose ten to be added to the upper figure, subtract the lowerfigurefronm it, land set down the dctffercnce, observitn to carry one to the next left-hand figure of the subtrahend,-or suppose the next left-hand figure of the minuend to be diminished by one. If the next figure of the minuend be a cipher, call it 9. Proof.-Add the remainder and lower line together. If the work be right, the amount will correspond with the upper line. EXAMPLES. From 39070 From 506789 From 67023491 Take 28931 Take 467898 Take 57216532 EXERCISES. Art. 21.-1. From four hundred and seventy-nine, take three hundred and seventy-five. 2. Take twenty-five thousand nine hundred and twentythree, from forty-four thousand five hundred and twenty. 3. What number must be subtracted from 2081 that the remainder may be 1104? Alns. 977. 4. From thirty-four thousand, take seventeen thousand and ninety-one. 5. John's incle gave him 20 cents. He lost 5 of them; how many had he left? 6. A western hunter met with 45 buffaloes in one drove, and killed all but 18; how many did he kill? Ans. 27. 7. The Arabian, or Indian method of notation, was first known in England about the year 1150. How long is it since to the present year, 1849? Ans. 699. 8. The mariner's compass was invented about the year 1302. How long before that period was the Arabian method of notation known in England? Ans. 152 years. 9. The first settlement in New England was made at Plymouth, by the Puritans, in the year 1620. How long is it since that time to the year 1837? Ans. 217 years. QUESTIONS.-11. What is the rule for Simple Subtraction? 12. How do you prove Subtraction? 13. When the lower figure exceeds the upper, what is to be done? 14. What do you call it? 15, Why? 16. Suppose the next left-hand figure be a cipher, what is to be done? 3 26 SIMPLE SUBTRACTION. 10. Gunpowder was invented in the year 1320. How long was it after the invention of the mariner's compass? Ans. 18 years. 11. Virginia contains 64000 square miles; New York contains 46000. What is the difference? Ans. 18000. 12. The library of Dartmouth College contains 12,800 volumes; Harvard University contains 34,600 volumes. How many does one contain more than the other? Ans. 21,800. 13. Dartmouth College was incorporated in the year 1769; Harvard University in the year 1638. What is the difference in time? AAns. 131 years. 14. The population of the state of New York in 1820 was 1,372,812; in the year 1835 it was 1,616,482. How many years between these two periods, and how much was the increase? Ans. 15 years. 243,670 increase. 15. An officer, with a company of 102 soldiers, was met by a party of Indians, who killed all his army but 17 men. How many were killed? Ans. 85. 16. A merchant bought 40 tuns of wine, containing 10080 gallons, which cost him 2410 dollars. He sold 28 tuns, containing 7056 gallons, for 1814 dollars. How many gallons had he left, and how much money did he want to make up the first cost? Ans. 3024 galls. ~ 596 dolls ADDITION. SUBTRACTION. Art. 22.-1. If a harness is Art. 23.-2. If a horse cost worth 18 dollars, and the horse is 86 dollars, and the harness 18 dol. worth 68 dollars more than the lars, how much more than the har' harness, what is the value of the ness did the horse cost? horse? 3. Ifa merchant have 1734 4. A merchant having 8322 yards of cloth, after selling 6588 yards, sells 6588 yards; how yards, how many had he at first? many has he left? 5. What is the amount of 6. What number must be added 2269+8625? to 8625 to make 10894? 7. Dr. Franklin was born in 8. How many years before the 1706, 93 years before the death death of Washington, in 1799, of Washington. In what year was the birth of Franklin? did Washington die? ADDITION AND SUBTRACTION. 27 9. The mariner's compass was 10. Sir Isaac Newton was born invented in 1302; Sir Isaac New- in 1642; the mariner's compass ton was born 340 years after. In was invented 340 years before. what year was he born?' In what year was it invented? 11. If a piece of land be bought 12. If a piece of land sell for for 550 dollars, and sold for 250 800 dollars, which is 250 more more than it cost, for how much than it cost, what was the first is it sold? cost? 13; Peter the Great died in 14. Peter the Great died in the 1725,112 years before the inde- year 1725. How many years pendence of Texas was acknowl- from that period to the acknowledged by the United States. In edging of the independence of what year was the independence Texas, in 1837? acknowledged? 15. Supposing a man to be born 16. Supposing a man to be 98 in the year 1738; lived 98 years; years old in the year 1836, in in what year did he die? what year was he born? 17. Columbus first sailed for 18. The independence of AmerAmerica in the year 1492; the ica was declared in the year 1776; independence of America was de- 284 years before, Columbus first dared 284 years after. In what sailed for America; in what year year was it declared? did he sail? 19. Washington was born in 20. Washington was born in 1732; was 67 years old when he 1732, died in 1799; how old was died; in what year did he die? he when he died? 21. Noah's flood happened about 22. Noah's flood happened about the year of the world 1656; the the year of the world 1656; the birth of Christ was about 2348 birth of Christ was about 4004; years after; in what year was he how long was the flood before the born? birth of Christ? PRACTICAL QUESTIONS IN ADDITION AND SUBTRACTION. Art. 24. —1. Add 900, 400, and 752; subtract from their sum 647. Ans. 1405. 2. Charles had 18 peaches. He gave his mother 6 and his sister 4. How many had he left? Ans. 8. 3. A man buys at one store 84 eggs, at another 4 dozen, at another 3 dozen. As he returns, he sells 5 dozen. How many did he purchase, and how many had he left when he arrived at home? Ans. I168. A \108. 4. James owes A. 20 cents; B. 30; C. 40; D. 50. John owes A. 18 cents; B. 23; C. 35; D. 47. How much do they both owe, and which owes the most? 28 2SEXAMPLES IN MULTIPLICATION. 5. A merchant owes A. 1300 dollars; B. 1900 dollars; C, 2500 dollars. He is worth 3500 ddlars. How much does he owe, and how much more than he is worth? 6. A man borrowed of his friend, at one time 100 dollars; at another, 150 dollars; at another, 175 dollars. He paid 275 dollars. How much did he borrow, and how much does he still owe?. Borrowed 425. Afs. \Owes 150. 7. On a certain farm there are 750 apple-trees, 425 peartrees, 1000 peach-trees, and 389 plum-trees. What is the amount of the whole, and how many more apple-trees than pear-trees, and how many more peach than plum? 8. A man left to his wife 2500 dollars; to his four sons 900 dollars each; to his three daughters 450 each. What was the amount of property left, and how much more was left to the mother than to the daughters, and to the sons more than to the mother? 9. There were five important events in the course of 215 years, viz:-1. The invention of the mariner's compass. 2. The invention of gunpowder. 3. The art of printing. 4. The discovery of America. 5. The reformation. The last was accomplished A. D. 1517; the third, 77 years before; the second, 18 years after the first, and the fourth 172 years after the second. The question is, In what year did each happen? n. J Mariner's compass in 1302; gunpowder in 1320; printing, 1440; discovery of America, 1492. MITULTIPLIICATION. Art. 25.-1. If Mary give 4 cents for one picture-book, how much must she give for 2 books? how much for 4? how much for 5? 2. If one dWen of eggs cost 10 cents, how many cents will two dozen cost? how many will 4? 5? 6? 3. If one share in a library cost 5 dollars, how much will three shares cost? how much 4? how much 5? how much 6? 4. If a picture-frame cost 12 dollars, what will 4 cost? what will 6? what will 7? what will 8? what will 9? MULTIPLICATION AND DIVISION TABLE. 29 5. Four men bought a piece of land, each paying 12 dollars. What did they all pay? 6. If a horse can trot 11 miles in one hour, how many miles can he travel in 8 hours.? how many in 10? 12? 7. If a bushel of wheat cost 2 dollars, how many dollars will 8 bushels cost? how many will 9? how many will 10? how many will 11? how many will 12? 8. If a man receive 4 shillings for a day's work, how much will he receive for a week's work? 9. A pound of sugar is worth 8 cents. What are 6-7-8 -9-10-11-12 pounds worth? 10. John bought a writing-book for 6 cents. What will 2 cost? what will 4? what will 6? what will 8? Art. 26.-The scholar should commit to memory the following Table before proceeding any further. MULTIPLICATION AND DIVISION TABLE TWICE 3 TIMES 4 TIMES 5 TIMES 6 TIMES 7 TIMES 1 make 2 1 make 3 1make 4 1make 5 1 make 6 1 make 7 2 4 2 6 2 8 2 102 122 14 3 6 3 9 3 12 3 15 3 18 3 21 4 8 4 32 4 16 4 20 4 24 4 28 5 10 5 15 5 20 5 25 5 30 5 35 6 12 6 186 24 6 306 366 42 7 14 7 21 7 287 357 427 49 8 168 248 328 40.8 488 56 9 18 9 27 9 36 9 45 9 54 9 63 10 2010 3010 4010 5010 60 10 70 11 22 11 3311 4411 5511 6611 77 12 24 12 36 12 48 12 60 12 72112 84 8 TIMES 9 TIMES 10 TIMES 11 TIMES 12 TIMES 1 make 8 1 make 9 1 make 10 1 make 11 1 make 12 2 16 2 18 2 20 2 22 2 24 3 24 3 27 3 30 3 33 3 36 4 32 4 36 4 40 4 44 4 48 5 40 5 45 5 50 5 55 5 60 6 48 6 54 6 60 6 66 6 72 7 56 7 63 7 70 7 77 7 84 8 64 8 72 8 80 8 88 8 96 9 72 9 81 9 90 9 99 9 108 10 80 10 90 10 100 10 110 10 120 11 88 11 99 11 110 11 121 11 132 12 96 12 108 12 120 12 132 12 144 3* 30 MULTIPLICATION. OBS —The student may be required to write out the table, as an ex. ercise, up to 24 times 24, rand commit it to memory. 11. If a man pay 85 dollars for a carriage, what must he pay for 5 carriages? The answer may be obtained by setting down 85 five times, and adding them up, thus: 85 It will be seen, by examining this operation, that 85 the product of five times five units is two tens and 85 five units,-and five times eight tens is 40 tens. 85 The answer, then, is 40 tens, 2 tens and 5 units, or 85 425. 425 This method would be tedious when a number is to be many times repeated, and can be solved much easier by multiplication, thus: 85 Instead of setting down 85, five times, we write 5 5, the multiplier, under the unit figure of the num425 ber to be multiplied; then say, 5 times 5 are 25, setting down 5, the excess of tens; and reserving in the mind, 2, the number of tens, we say, 5 times 8 are 40; adding the two tens which we reserved from the unit column, we set down 42. The answer, then, is 42 tens and 5 units, or 425. Art. 27.-From the above we derive the following defin. tions: 1st. Multiplication is the concise method of performing many additions. 2d. Multiplication consists;n repeating a given number a required number of times. OBs. 1.-It is always true of multiplication, that it can be performed by addition; but it is not always true that addition can be performed by multiplication: it is only the case when a number is to be repeated. OBs. 2.-The word factor signifies an agent, or doer: it is derived from the Latin wordfactum, which signifies a deed, or thing done. A person employed to do business for another, is called an agent, or factor. Hence, when two numbers are employed as multipliers, or as the means of obtainig a product, they are calledfactors. (See def Art. 54.) 12. If a share in a bridge is worth 142 dollars, how many dollars are 6 shares worth? ILLUST'iATIONS OF MULTIPLICATIUN. 31 Operation. HIunds. T.ens. Units. Iunds. Tens. Units. 112 or thus, 1 4 + 2 1 - 4+2 =142 6 6 6 6 12= 2x6 6 + 24 + 12 + 5 + 2 = 852 240= 40x 6 600=100x6 Ans. 852=142X6 13. If one man receive 164 dollars for a year's labor, what ought 32 men to receive, for the same time? Operation. Since we cannot conveniently 164 multiply by a larger number 32 than 12 collectively, it will be 328= 2 times the mult. necessary, in this example, to 492 =30 times the mult. adopt a new mode of operation. 492 =-30 times the mult. The multiplier consists of 2 units 5248_32 times the mult. and 3 tens. We first multiply each figure of the multiplicand by the units of the multiplier. Units into units give units; units into tens give tens; units into hundreds give hundreds. We next multiply each figure of the multiplicand, beginning with the units, by the tens of the multiplier, observing to set 2, the first figure of the product, in the place of tens, because it is the product of tens. We next multiply the 6 tens of the multiplicand by the 3 tens of the multiplier, carrying one for every ten as in Addition, and set the product in the place of hundreds. The product of tens into tens is hundreds. Lastly, we multiply the hundreds of the multiplicand by the tens of the multiplier, and set the product in the place of thousands. The product of tens into hundreds is thousands. Adding together the several products, we have 5248, the answer. The above illustration may be better understood by setting the product of each figure of the multiplier into each figure of the multiplicand down by itself; thus, Hunds. Tens. Units. Iunds. Tens. Units. 1 + 6 +4 = 1 6 + 4 3 + 2 3 + 2 2 + 12- +8 3 + 2 +8 5000 3 + 18 +12 4+ 9 + 2 200 3 -20 + 24+8 5 +2+4+8= 40 8 5248 Ans. 32 RULE.-METHOD OF PROOF. From the foregoing examples we derive the following RULE. I. Place the multiplier directly under the multiplicand, units under units, tens under tens, etc., then draw a line underneath. II. Whlen the multiplier is 12, or less than 12, begin at the right hand of the multiplicand, and multiply each figure contained in it by the multiplier, setting down the numbers, and carrying as in Addition. III. When the multiplier is greater than 12, write down the figures, as before directed, and multiply the multiplicand by each figure in the multiplier, commencing with the unit figure; observing to place each figure in the product directly under the figure by which you multiply. In this way proceed, and the sum of the products will be the answer. There are three methods of proving Multiplication. First-Make the multiplicand and multiplier change places, and multiply the latter by the former, in the same manner as before; if the latter product be the same as the former, the work is supposed to be right. Second-Cast the 9's out of the product, or answer, and set down the remainder. Cast the 9's out of the sum of the two factors; multiply the two remainders together, and cast the 9's out of the product. The last remainder, if the work is right, will be equal to the first. OBS. 3.-The four remainders may be set within the four angular spaces of a cross, as in the following example. Third.-Multiplication may be proved by Division. The product divided by either of the factors will give the other. OBs. 4.-The second and third methods can be deferred until the scholar becomes acquainted with Division. QUESTIoNs.-1. What is Multiplication? 2. How many numbers are required to perform the operation? 3. What is the number to be multiplied, called? 4. What is the number by which you multiply, called? 5. Taken together, what are they called? 6. Why called factors? 7. What is the answer called? 8. How many figures are there in the multiplier of the 13th question? 9. By which do we multiply first? 10. What is the product of units multiplied into units? 11. Of tens into tens? 12. 01 hundreds into hundreds? 13. How are the numbers placed in Multiplication? 14. How do you proceed when the multiplier is 12, or les than 12? 15. When the multiplier is greater than 12? EXERCISES IN MULTIPLICATION. 33 (14) EXAMPLES. 3542 96 Proof. 21252 3 31878 340032 (15) (16) (17) (18) 3467 124567 54678901 34567892 6 8i 341 4567 (19) (20) (21) (22) 679834 678345 126789123 908764584768 12 23 27678 632976 Multiply 24 by 2; then double the multiplier; then double the multiplicand; then double the product. 1st. 2d. 3d. 24 24x2= 48 2X2= 4 2 48x2- 96 = 96 What effect upon the product has multiplying the multiplier? What effect upon the product has multiplying the multiplicand? 23. What will 432 barrels of flour cost, at 14 dollars a barrel? Ans. 6048 dollars. 24. How many rods in 84 miles, there being 320 rods in a mile? Ans. 26880 rods. 25. What will be the cost of 6328 thousands of boards, at 18 dollars per thousand? Ans. 113904 dollars. 26. How many dollars would a man count in 12 days, if he count 42000 in one day? Ans. 504000. How would you solve the above question by Addition? 27. What will 64 cows cost, at 16 dollars apiece? Ans. 1024 dollars. The multiplier, 16, in the last example, is a number which can be formed by the multiplication of two numbers-thus: 34 RULE.-EXERCISES IN MULTIPLICATION. 4x4=16: or 8x2=16. Any number thus produced is called a composite number. The numbers thus multiplied are called component parts. Sixteen, then, is a composite number, and 4 and 4, or 8 and 2 are the component parts of 16. Thus, taking the above question, 64 X 4=256, the price of 64 cows at 4 dollafs eache; and this product multiplied by 4 gives 1024, the price at 16 dollars each, because 4 times 4 are 16. The same result will be produced if we multiply 64 by 8 and 2. Art. 28.-When the multiplier is a composite number. RULE. _Multiply first by one of the component parts, and that product by the other, and so on, if there be more than two; the last product will be the answer. 28. What is the product of 78 multiplied by 25? 78 5 It will be seen that 5 and 5 are the 390 component parts of 25. 5 1950 Ans. 29. There are 365 days in a year. If a man live 48 years, how many days does he live? Ans. 17520. 30. Multiply 7684 by 112. 8X'7x2=112. 31. Multiply 8410736 by 56. 32. Multiply 17548671 by 81. 33. Multiply 998673214 by 1864. 34. Multiply 99998887777 by 445566. 35. Multiply 88900236789456 by 77889123. 36. If it take 142 stones to build a rod of wall, how many will it take to build 10 rods? It will be seen by this example, that the answer 142 is obtained by annexing a cipher to the multipli- 10 cand; annexing a cipher, therefore, to any number 1420 multiplies that number by 10. Therefore, Art. 29.-To multiply by 10, 100, 1000, or by 1, with any number of ciphers, we have this QUBSTIOS. —16. What is a composite number? 17. What are the numbers called which form a composite number? 18. What is the rule for multiplying by a compoite numrber? EXERCISES IN MULTIPLICATION. 35 RULE. Annex as many ciphers to the multiplicand as there are ciphers at the right hand of the multiplier, and it will give the answer required. 37. Multiply 142 by 100. Ans. 14200. 38. Multiply 864 by 1000. Ans. 864000. 39. Multiply 999 by 100000. Ans. 99900000. 40. Multiply 2400 by 2200. Ans. 5280000. OBs.-A significant figure is one which has value in itself The nine digits are significant figures. Operation. 2400 In this example, we multiply by the signifi- 200 cant figures only, placing as many ciphers at 220 the right hand of the product as there are 48 ciphers in the multiplier and multiplicand. 48 5280000 41. What is the product of 68400 multiplied by 18000? Ans. 1231200000. Art. 30. —When there are ciphers between the significant figures of the multiplier. RULE. Reject the ciphers and multiply by the significant figures, observing to place the first product of each figure directly unw that by which you multiply. Operation. 42. Multiply 2008 by 604. 2008 604 8032 12048 1212832 43. Multiply 8624 by 108. 44. Multiply 340824 by 909. 45. Multiply 5678902 by 770901. 46. What will 412 hogsheads of molasses cost, at 31 dollars per hhd.? Ans. 12772 dollars. QUESTIONS.-19. When the multiplier is 10, 100, 1000, &c., how may you proceed? 20. When there are ciphers at the right hand of the multiplier and multiplicand? 21. How do you proceed when there are ciphers between the significant figures? 22. What arc siguificant figur;es' 36 nxEXTERICISE S IN MIUI,'T.IF'lIATiON. 47. What number is Ihat of which 8, 9, 1, are factors? A.ns. 792. 48. If 80 men dig a canal in 94 days, how many men could dig the same in one day?. Ans. 7520. 49. How many shillings ought 7520 men to receive for one day's work, at 5 shillings each per day? Ans. 37600. 50. A merchant bought 28 boxes of sugar, each weighing 235 lbs., at 8 cents per lb. How many cents did they cost?'Ans. 52640. 51. How many shillings will 89 cords of wood cost, at 15 shillings per cord? Ans. 1335. 52. A merchant bought 15 pieces of cloth, each piece containing 27 yards, at 7 dollars per yard. How much did he pay for the whole? Ans. 2835 dollars. 53. If a ship sail 12 miles per hour, how far will she sail in 12 days? Ans. 3456 miles. 54. If a man hoe 3 rows of corn, 28 hills each, in I hour, how many hills will he hoe in 12 days, working 8 hours in a day? Ans. 8064. 55. What will 50 firkins of butter cost, weighing 54 lbs. each, at 14 cents per lb.? Ats. 37800. 56. A man has 9 piles of wood, 16 cords in a pile. What is it worth at 7 dollars per cord? Ans. 1008 dollars. Irt. 3, —When the multiplier is 9, or any number of 9's. RULE. Annex as many cip)hes to the multiplicand-as there are 9's in the multiplier, and from it subtract the giren multiplicand. EXAMPLES. 57. Multiply 162 by 9. Operation 1st. Operation 2d. 1620 162 162 9 Ans. 1458 1458 Ans. The reason of this process is evident; annexing a cipher to the multiplicand, multiplies it by 10, which is repeating it once more than is required. uKsTnnONs.-23. How do you proceed when the multiplier is 9? 24. How is this expiaineJ? 25. How is Multiplication proved? DIVISION. 37 On the same principle we may multiply by 8, by annexing a cipher, and subtracting twice the multiplicand; or by 98, by annexing two ciphers, and subtracting twice the multiplicand. 58. Multiply-3452 by 99. 59. Multiply 46784 by 999. 60. Multiply 576213 by 98. Art. 32 —To multiply by any number between 10 and 20. RULE. Annex a cipher to the multiplicand, which is to multiply it by 10, and to this result add the product of the given multiplicand into the right-hand figure of the multiplier; their sum will be the answer required. EXAMPLES. Operation. 61. Multiply 562 by 11. 5620 562 6182 Ans. Operation. 62. Multiply 789 by 12. 7890 1578 9468 Ans. Operation. 63. Multiply 843 by 19. 8430 7587 16017 Ans. DIVISION. Art. 33. —1. John has 5 oranges given him. He keeps one himself, and divides the others equally between his two sisters. How many did each receive? 2. Samuel divided 9 walnuts equally between three boys. How many did each receive? 3. James's father gave him 8 butternuts to divide equally between himself and his three brothers. How many did each receive? 4 38 D viSr IN. 4. If Harriet gave 24 cents for 6 pictures, what did she pay for? 5. Mary divided 36 cents equally between 6 poor children. What did each receive? 6. If 8 yards of cloth cost 5a cents, what did one cost? 7. How many yards of broadcloth can be bought for 72 dollars, at the rate of 8 dollars per yard? 8. How many bushels of apples can be bought for 100 cents, at 25 cents per bushel? 9. How many barrels of flour may be bought for 77 dollars, at 7 dollars per barrel? 10. In how many hours will a man travc. 48 miles, at the rate of 4 miles per hour? 11. Eighty cords of wood arc piled in 8 different piles. How many cords in each pile? 12. A farmer sold wool to the amount of 81 dollars, for 9 shillings a fleece. How many fleeces did he sell? 13. How many thousands of boards may be bought for 144 dollars, at 12 dollars per thousand? 14. How many books may be bought for 84 cents, at 7 cents apiece? 15. If 1G apples be divided equally between 4 boys, how many does each receive? It is evident, that as many times as 4 is contained in 16, or, as many times as it can be subtracted from it, so many apples each boy will receive. Operation 16 4 12 4 8 4 We find, by trial, that 4 is contained in 16 four 4 times, which is the number of apples each boy is to 4 receive. 0 16. If 4 boys receive 4 apples each, how many do they all receive? It is plain that 4 boys will receive 4 times as many as one; therefore, if one boy receive 4 apples, 4 boys will receive 4 x4=16. DIVISION.-THE SUBJECT ILLUSTRATED. 39 Art. 34.-From the foregoing we derive the following defi& nitions: 1. Division is a concise method of performing many subtractions, or, the reverse of Multiplication. 2. Division consists in finding how many times one number contains another. As in Multiplication two numbers are required to perform the operation, so in Division. The number to be divided is called the Dividend; the number by which you divide is called the Divisor. T^he Dividend is to be regarded as the product of two factors, of which the Divisor is one, and the other is sought, which is the Quotient after division. The Divisor and Quotient multiplied together produce the Dividend. Thus, it appears that Division and Multiplication mutually prove each other. OBS.-All questions in Division may be performed by Subtraction; but all questions in Subtraction cannot be performed by Division.-When a number is to be divided into equal parts, the operation may be performed by Division. Art. 35.-When theoDivisor is not greater than 12, the process of operation may be carried on in the mind, and the Quotient only be written down. This process is called SHORT DIVISION. 17. If 336 dollars be divided equally among 3 men, how many dollars will each receive? Illustration.-To subtract 3 from 336 as many times as would be necessary to give each man his share, would be long and tedious; but, by Short Division the operation becomes simple: 336 is 3 hundreds, 3 tens, and 6 units. It will be perceived, that if 300 be divided into 3 equal parts, one of these parts will be 100; and 3 tens divided in like manner become 1 ten; and 6 units divided by 3 become 2 units. The answer, then, is 1 hundred, 1 ten, and 2 units, or 112. Operation. 300 divided by 3 gives 100, 30 di- 300 3=100 vided by 3 gives 10, and 6 di- 30 3== 10 vided by 3 gives 2: Then, 6 3= 2 1004-10+2=112 Ans. 336 3-112 Ans. QUEsTONS. —1. What is Division? 2. How many numbers are required to perform the operation? 3. What are they called? 4. How is the dividend to be regarded? 5. What are the two factors which, multiplied together, will produce the dividend 6. How is Division proved? 40 RULE IN DIVISION. hun. tens. units. Or thus: 3)3+3+6 1+1+2=112 Ans. By carrying on the process partly in the mind, the oreration may be made still shorter, thus: 3)336 Proof. 112 Ans. 112 X 3=336 18. If 4 shares of bank stock cost 456 dollars, what will I share cost? 4)45(6 114 Ans. We first set down the number to be divided, or the dividend; at the left of this number, place the number by which we divide, or the divisor. Taking the first left-hand figure, or hundreds, we find how many times the divisor is contained in it. The number of times, or 1, we place directly under the divided figure. We next divide the tens of the dividend: 4 is contained in 5 once, and 1 ten over, which also must be divided. If this ten be added to the unit, it will make I ten and 6 units-equal to 16 units: 4, then, is contained in 16, foul times, which we place in the column of units; and in 4o6, 114 times. 19. Six brothers received a legacy of 1512 dollars. What was the share of each? 6)1512 252 Ans. In this question the divisor is not contained in the first lefthand figure of the dividend. We, therefore, take the next figure, 5, which with the 1 makes 15 hundred; 6 is contained in 1500, 200 times, and 30 tens, or 300 over. The 3 added to the 1 ten in the next column, is 31 tens; 6 is contained in 31, five times, or in 31 tens, 50 times and 1 ten over, which, with the two units, makes 12 units: this, divided by C, is 2 units. The answer, then, is 200+50+2=252. From the foregoing, we derive the following RULE. Write the divisor at the left-hand of the dividend, with a line drawn between them. QUESTIONS..-7. How is the process partly carried on in Short Division? S. When do you work by Short Division? 9. What is the method of procedure in the 18th questin? EXERCISES IN DIVISION. 41 Find how many times the divisor is contained in the first left-,and figure or figures of the dividend, setting the result directly under the divided figure or figures. The remainder, if there be any, carry to the next figure, calling it so many tens. Find how many times the divisor is contained in this dividend, and set it down as before; and so continue to do until the figures in the dividend are all divided. EXERCISES. Art, 36.-1. Paid 150 dollars for six tons of hay. How much.was it a ton? Ans. 25 dollars. 2. In a certain town there are 1280 inhabitants. The average number in each family is 8. How many families are there-? Ans. 160. 3. How many yards of cloth can be bought for 1155 dollars, at 7 dollars per yard? Ans. 165. 4. If man labor one month for 12 dollars, how many months will he labor for 1008 dollars? and how many years, allowing 12 months to a year? Ans. 84 months-7 years. Art. 37. — hen the divisor is a composite number, and greater than 12. 1. If 15 horses consume 2550 bushels of oats in one year, how many will one horse consume? It will be seen that 15 is a composite number, produced by the multiplication of 5 and 3, thus: 5 X3= 15. As Division is the reverse of Multiplication, it is evi- Operation. dent that when the divisor is a composite 5)2550 number, we may divide, first, by one of 3)510 the component parts, and that quotient by 170 Ans. the other. For example: Suppose 30 apples to be divided equally between 15 boys. In the first place, we divide the whole number by 5. If there were only 5 boys, they would receive 6 apples each; but as there are 3 times 5 boys, they can have only one-third as many as 5 boys would have. 2. How many days would it take a man to travel from Boston to New York, travelling at the rate of 30 miles a day, the distance being 240 miles? Ans. 8 days. QUESTIONS.-10. What is the rule for Short Division? 11. When the divisor is a composite number, how may you proceed? 12. What is the first step in the 1st example? Second step? 4 42 LONG DIVISION EXPLAINED. 3. If a horse travel 2184 miles in 42 days, how many miles will he travel in one day? Ans. 52 miles. 4. If 112 barrels of flour cost 612 dollars, what will 1 barrel cost? Ams. 6 dollars. Art. 38. — Whe' the divisor is net a composite number, and is greater than 12. 1. If 2624 bushels of corn be divided equally among 41 men, how many bushels will each receive? As 41 is greater than 12, and not a composite number, the operation must be performed by the whole divisor at once. This process is called LONG DIVISION. Operation. Setting down the numbers as before 4262Otion. directed, and taking 41 for the divisor, we 24( find that it is not contained in the 1st figure, nor in the 1st and 2d taken to- 14 gether, but it is contained in the first three 164 figures, 6 times; that is, 41 is contained in 262 tens, 60 times and something'over. To find what this remainder is, we find the product of 6 times 41, which we place under the three figures employed in the dividend, and, subtracting it therefrom, we find the remainder to be 16, which is 16 tens. We next bring down the 4 units of the dividend, and place them at the right hand of the 16 tens, which make 164 to be divided by 41, which is contained in it 4 times. The answer, then, is 64. By examining the work of the last question, it will be seen, that it is the same as Short Division, only that the operation is all set down, instead of being carried on in the mind. For example-divide 868 by 7, Long Division. Operation.. 7)868(124 7 16 14 28 28 By Short Division, we say, 7 in 8, once, and 1 over; 7 in 16, twice, and 2 ovr; 7 in 28, 4 times, and no remainder, RULE IN DIVISION. —EXERCISES. 43 In Long Division, we say, 7 in 8 once, and place 1 for the first figure in the quotient; we then multiply 7 by this quotient figure, and place the result under the 8, and subtracting it we find the difference to be 1, to which we bring down the next figure for a new dividend, and proceed as before. From the preceding explanations is deduced the following RULE. Place the divisor at the left hand of the dividend. Draw a line at the right and left of the dividend, and take as many figures of the dividend as will contain the divisor one or more times. Place the number of times at the right hand of the dividend, for the first figure of the quotient. MIultiply the divisor by this quotient figure, and place the result under the divided figures; find the difference between them, by subtraction, anid to this diference bring down the next figure of the dividend, and divide as before; so continue to do until all the figures of the dividen&dare brought down. Should it be necessary to bring down more than one figure to contain the divisor, a cipher must be annexed to the quotient. OBss.-The number of figures of the dividend we assume at any one tep is a matter of convenience. EXERCISES. Art. 39.-2. A man raised 6996 bushels of potatoes on 33 acres How many did he raise per acre? Ans. 212 bushels. 3. How many years in 32485 days, if 365 days make a year? Ans. 89 years. 4. A legacy of 15808 dollars was left to a certain number of men, giving them 832 dollars each. How many men were there? Ans. 19 men. 5. How many pounds in 11520 farthings, there being 960 farthings in one pound? Ans. 12 pounds. 6. How many hogsheads in 49896 pints, if 504 pints make one hogshead? Ans. 99 hogsheads. 7. There are 8 furlongs in one mile. How many miles in 123 furlongs? Operation. 8)123 15- Ans. QuzsTiONs. —13. What Is the difereace between Long and Short Division? 14. Rule for Long Division? 44 EXERCISES IN LONG DIVISIoA. OBs. —For the illustrations of this and the following questions, see Fractions. By dividing 123 by 8, we find the quotient to be 15 miles, and there is a remainder of 3 furlongs. As it takes 8 furlongs to make a mile, it is evident that I furlong is 1 of a mile, and 3 furlongs are 8 of a mile, and 8 furlongs are s, equal one mile. The answer, then, is 15 miles and 3, which we place at the right of the quotient. We have, then, when there is a remainder in division, this RULE. Place it at the right hand of the quotient, as the numerator of afraction, and under it place the divisor, as a denominator. A number like this is called a mixed number: thus, 153 is a mixed number. To prove this last question, we multiply the quotient into the divisor, and add to the product the numerator of the fraction, or the remainder, thus: 153 8 123 That the remainder, as the numerator of a fraction, is a part of the quotient, will appear from the following: 120+3 is the whole dividend; 8, the divisor, is contained in 120 units 15 times; it is also contained in three units A of a time. Since, therefore, 120+3 is the whole dividend, it follows that 15+1- is the whole quotient. 8. What is the quotient of 1832 divided by 16? Ans. 1148. 9. There are 320 rods in a mile. How many miles in 66327 rods? Ans. 207 8-. 10. How many miles from Boston to Providence, the distance being 12800 rods? Ans. 40 miles. 11. If 10 shares in a factory be worth 2220 dollars, what is one share worth? We have seen, that annexing a cipher Operation. 1 * -' 10)22210 to any number is the same as multiplying 2220ns. by 10. To remove a cipher, therefore, 222 Ans. from the right of any number, is dividing QUESTIONS.-15. When there is a remainder, after dividing, what is to be done with it? 16. How do you know it is apart of the quotient? Illustrate. 17. What is a number like 15* called? 18. When there are ciphers at the right hand of the divisor, how may you proceed? 19. How must the figures cut off from the right hand of the dividend be placed? EXAMPLES IX DIVISION. 45 that number by 10. To remove a cipher from divisor and iividend, is dividing both by 10. Therefore, Art. 40.-When the divisor is 10, 100, 00, 0, or 1, with any number of ciphers, we have the following RULE. Cut of those ciphers from the divisor, and a corresponding number of figuresfrom the right of the dividend. The figures on the left will be the quotient, and those on the right, a remainder. 12. Divide 18986421 by 10000. Opera tion. In this example, the divisor is not a factor of the whole dividend, but 110000)18986421 of a part only. 10,000 will divide Ans. 1898,6421 18980000, but is not contained in 642Z. The quotient, therefore, is 1898, and 6421 units are left undivided, which are a remainder. 13. Divide 3330 by 30. Operation 1st. Operation 2d. 30)3330(111 Ans. 310)33310 30 Ans. 111 33 In Operation 1st we reject a factor from 30 the dividend equal to the whole divisor; 30 but the dividend may be separated into the 30 the factors 3, 10, and 111; 3 and 10 are also factors of the divisor. By cutting off the cipher from the dividend, the process of dividing by 10 is performed. —Operation 2d. We have now only to divide by 3, the other factor of the divisor, and the factor 3 is rejected from the dividend. The remaining factor, 111, is the quotient. 14. Divide 342871 by 7000. Operation. In this example, cutting off three fig71000)3421871 ures from the dividend is dividing by 71000: 1000 is contained in 342,000, 48,6871 three hundred and forty-two times, and there is a remainder of 871 units. 7 is contained in 342, fortyeight times, and 6 remain, which is 6000, because it was taken from the place of thousands, and therefore must be prefixed to the first remainder. Proof. 48X 7000+6000+871=342871. 46 CANCELLING OPERATION. Art. 41.-To divide by any number whose right-hand.fig. ures are ciphers. RULE. Cut off the ciphers, and figures of the dividend, as before directed, and divide the remaining figures of the dividend by the remaining figures of the divisor. To the right hand of the remainder bring down the figures cut off from the dividend. 15. What is the quotient of 421998 divided by 8400? 16. What is the quotient of 406224 divided by 9600? 17. What is the quotient of 7864234 divided by 67200? Art. 42.-From the foregoing it is manifest, that in the process of division, a factor is rejected from the dividend equal to the divisor. Upon the same principle —f equal factors be rejected from divisor and dividend, the value of the quotient will not be altered. 18. Divide 16 by 8. Operation 1st. 8)16=$X2 -=2 Ans. 2 Ans. For convenience' sake, we will draw the line between divisor and dividend straight instead of curved, and write the factors one over the other, those of the dividend on the right, and those of the divisor on the left, and draw a line through those factors which are rejected or cancelled. Thus: Operation 2d. 2 4=2 Ans. The dividend, 16, is resolved into the factors 4 and 4; and 8, the divisor, into the factors 4 and 2. If we strike out the factor 4 from each, we have 4 22, as before. Again, we may separate the dividend into the factors 4, 2, and 2, and strike out the factors 4 and 2 from each side of the line. Thus: Operation 3d. It is evident, since to reject equal factors 4 A from divisor and dividend does not affect the quotient, that to multiply divisor and dividend 2 Ans. by the same quantity would not affect the quotient. If 16 contain 8 twice, the double of 16 would contain thedouble of 8 twice, - - - -- - 16 X2-8 x 2 —2 answer. QUESTIONS.-20. In the process of division, what factor is rejected from the dividend. 21. What effect upon the quotient has rejecting equal factors tfom divisor and divi dead? CANCELLING OPERATIONS. 47 This expression is read, The product of 16 into 2, divided by the product of 8 into 2, equals 2. Os..-Let the scholar now be called upon to illustrate, by a variety o( examples, the principle employed in division, in the following manner taking the above question:Teacher. What do you infer from the process in operation 1 st? Scholar. That in the process of division we reject from the dividend a factor equal to the divisor.- T What do you infer from the process in operation 2d? S. That to reject equal factors from divisor and dividend does not affect the quotient. T. What from operation 3d? S. The same as from operation 1st. In the following questions let the student be required to separate divisor and dividend into their prime factors, and write them down, as before directed. Then let him reject an equal factor from each, and perform the operation with the remaining; then let him reject another, and perform the operation with the remaining, and so on until all the factors of the divisor are rejected. 3. Divide 72 by 12. 72 —12=2x2X3X2X3 2 x2x3=6 Ans. Or thus: Operation 1st. Operation 2d. Operation 3J. 22 2 3 3 33 3 2 2 2 3 3 3 6 36=6 Ans. 3 18=6 Ans. 6 Ans. OBS. 2-In the following questions, let the teacher proceed thus: Teacher. What are the factors of 84? Scholar. 7 and 12. T. Which is prime? S. 7.-T. Write it on the right of the line.-What are the factors of 12? S. 2 and 6.-T. Which is prime? S. 2.-T. Write it under the 7.-What are the factors of 6? S. 3 and 2. T. Are they prime? S. They are. T. Write them down. i. Divide 84 by 21. Ans. 4. 5. Divide 108 by 18. Ans. 6. 6. Divide 112 by 28. Ans. 4. 7. Divide 224 by 56. Ans. 4. 8. Divide 336 by 16. Ans. 21.. 9. Divide 96 by 8. 96 8=12 Ans. In solving this question, into what factors must the dividend be separated? Ans. 8 and 12. 48 ILLUSTRATIONS. How do you know? Ans. Because 8, the divisor, is one factor, and therefore 12 must be the other; for 12 x 8=96. 10. Divide 72 by 6. 72 6=12 Ans. Into what factors is the dividend separated in this example? Ans. 6 and 12. Why not 8 and 9? They are also factors of 72.-Ans. Because neither is like the divisor. Let the teacher propose similar inquiries in regard to the following exercises: 11. Divide 84 by 12. 12. Divide 108 by 9. 13. Divide 121 by 11. 14. Divide 132 by 12. Let the student read the following forms of implying division, and write others similar: 16-8=2 56- 7=8 24 12160=5 6 7= 72 6)36=6 9 818=6 12)84=7 OBS. 3.-For the reading of the following forms, see Art. 48. Art. 3. —Illustration of general principles.-What effect up16-8 _ 2 on the quotient has multiplying the dividend? W_ _ hat effect upon the quotient has multi16 x 2 8 =4 plying the divisor? What would be the effect of multiplying 16 8 X2=1 both divisor and dividend? Illustrate. _What effect upon the quotient has dividing 16 - 2 8 =1 the dividend? What effect upon the quotient has dividing 6 8 2=4 the divisor? What effect upon the quotient has dividing both the dividend and divisor? Illustrate. Art, 44.-From the foregoing illustrations, the following principles are manifest. The larger the dividend, with a given divisor, the larger the quotient; and the less the dividend, with a given divisor, the less the quotient. Therefore, To multiply the dividend is the same as to multiply the quotient, and to divide the dividend is the same as to divide the quotient. To divide the divisor is the same as to multiply the dividend, and to multiply t' a divisor is the same as to divide the dividend. MULTIPLICATION AND DIVISION, 49 IULTIPLICATION. DIVISION. Art. 45.-I. What will 1574 Art. 46.-2. If 1574 yards yards of cloth cost, at 12 dollars of cloth cost 18888 dollars, what per yard? will 1 yard cost? 3. How many inches in 56541 4. How many feet in 678492 feet? inches, if 12 inches make 1 foot? 5. If a man travel 38 miles in 6. If a man travel 608 miles in a day, how many will he travel in 16 days, how many will he travel 16 days? in I day? 7. If 60 minutes make 1 hour. 8. In 784201560 minutes, how how many minutes in 13070026 many hours? hours? 9. How many hours in 336 10. In 80641hours, how many days? days? 11. The quotient of two num- 12. The product of two numbers is 46; the divisor 14; what bers is 644; the multiplier 14? is the dividend? what is the multiplicand? 13. The quotient of two nnm- 14. The product of two numbers is 72; the divisor 84; what bers is 6048; the multiplicand 84; is the dividend? what is the multiplier? 15. How many pounds of flour 16. A man has 125440 pounds may be put into 640 barrels, each of flour to be put into barrels, containing 196 pounds? colining 196 pounds each. How many barrels must he have? 17. What will 24 oxen cost, at 18. If 24 oxen cost 1104 dol46 dollars each? lars, what do they cost apiece? 19. If a carriage wheel turn 20. If a carriage wheel turn round 340 times in a mile, how round 81600 times between Bosmany times will it turn in going ton and New York, and turn 340 from Boston to New York, it times in a mile, what is the disbeing 240 miles? tance? 21. If 33 men do a piece of 22. If 1 man do a piece of work in 24 days, in what time work in 792 days, in what time will 1 man do it? will 33 men do it? 23. In 7894 feet, how many 24. How many feet in 284184 barley-corns? barley-corns? MULTIPLICATION AND DIVISION, BY CANCELLING. Art. 47.-The operation of questions, involving Multiplication and Division, may be greatly abridged by the following RULE. I. Draw a perpendicular line, and place dividends and numbers to be multiplied for dividends, on the right, and divisors on tes left hand. 5 50 EXERCISES IN CANCELLING. OBS. 1.-The perpendicular line is the same as the curve line in Division, separating divisors from dividends. II. If there be two equal numbers on each side of the line, cross them out, and omit them in the operation. Thus: Multiply 8 by 9, and divide by 8. Operation. As 8 is found on both sides of the line, cross 0$ - them both, and 9, remaining on the right, is the $9 Ans. answer. The- principle upon which this RULE proceeds is that of cancelling, or rejecting equal factors from dividends and divisors. Thus, taking the above example, 8 and 9 are the factors of'2: 8 x 9=72. The quotient of 72 divided by 8, is 9, one of its factors; the other fa'ctor, 8, equal to the divisor, is rejected. III. If a number on one side of the line will divide a number on the other side, without a remainder, erase both numbers, and substitute for the larger the number of times it contains the smaller. Multiply the remainders together, on the right, for a dividend, and the remainders on the left, for a divisor. Thus: Multiply 6 by 3, and divide by 2. e 3 In this example the divisor, 2, is not the saie 3 as either figure of the giidend, but it is a factor 9 Ans. of one of them, 2x3=6. We may, therefole, cross 2 and 6, since the divisor, 2, cancels one of the factors of 6, the dividend, and write 3, the other factor, against 6 as the quotient. The remainders on the right multiply together, 3X 3= 9, and 18. 2=9, the answer, as before. When there is no remainder on either side of the line, and the numbers are all cancelled, the answer is 1: that is, the right-hand side contains the left-hand, once. OBs. 2.-A stroke drawn through any number denotes its being cancelled; and any number which takes its place may be set alongside of it. 3. Multiply 8 by 5, and divide the product by 3; multiply the quotient by 18, and divide the product by 9; multiply again by 9, and divide the product by 6; multiply the quoQUESTIONS.-1. What is the rule for Multiplication and Division by cancelling? 2. First, second, and third steps? 3. Is the answer affected by striking out equals on each side of the line? 4. Why not? 5. What is done with remainders 6 When there is no number left on either side of the line, what is the answer? CANCELLING OPERATIONS. 51 tient by 24, and divide the product by 12; multiply the quotient by 2, and divide the product by 4. Operations. 8 Or thus: 8x5=40 a5 $5 40 An. 40 =1 40 Ans. 40 Ans. Having stated the question, according to thie foregoing RULE, we proceed.to cancel, or cross equals on each side of -the perpendicular line. In the first place, 9 is found on each side of the line. We therefore cross them both; for 9 is contained in 9 once, and multiplying any number by 1 does not alter its value. Secondly: 3 and 6, on the left hand of the line, multiplied together make 18-equal to 18, on the right hand of the line, which may be crossed out. Again: 4 and 12, on the left, multiplied together, are 48, equal to the numbers 2 and 24 on the right multiplied together, and may be crossed out. The numbers now are all cancelled, except the 5 and 8, on the right, which, multiplied together, give 40, the answer. 4. A boy gathered 16 nuts under each of 4 trees, and divided them equally between himself and 7 schoolmates. How many did each receive? Operation In this example, it is evident, that had ^ a 2 - the boy gathered but 16 nuts, there would 4X92-8 have been but 2 apiece; but as he gath_ 4 x28 ered the same number under each tree, the 8 Ans. 16 must be multiplied by 4; and as there were 8 to share them, the product of 16 multiplied by 4 must be divided by 8. 5. Multiply 20 by 5, and divide by 6; multiply by 7 and divide by 14; multiply this again by 6, and divide by 10, and multiply by 12. Ans. 60. 6. Multiply 120 by 40, divide by 400, multiply by 20, divide by 30, multiply by 250, divide by 50, multiply by 300, divide by 500, and give the answer. Ans. 24. QUESTION.-7. Give the reason for placing 16 and 4, in Example 4, on the right of the line, and 8 on the left. 52 SUPPLEMErT TO THE FUNDAMENTAL RULES. SUPPLEMENT TO THE FOUR FUNDAMENTAL RULES OF ARITHMETIC, VIZ: ADDITION, SUBTRACTION, MULTIPLICATION, AND DIVISION. EXERCISES. 1. A man purchased a farm for 6720 dollars; sold it for 199 dollars more than he gave. For how much did he sell it? Ans. 6919 dollars. 2. Suppose a tree. broken by the wind 39 feet from the ground, and the part broken off to be 56 feet in length. How high was the tree? Ans. 95 feet. 3. A merchant having 784 bushels of salt, sold 99 bushels. How many had he left? Ans. 685 bushels. 4. A man left his estate, valued at 8956 dollars, to his wife and daughters, giving his wife 4688 dollars. How much did the daughters receive? Ans. 4268.dollars. 5. Sir Isaac Newton was born in the year 1642, and died in the year 1727. What was his age? Ans. 85 years. 6. The greater of two numbers is 624; their difference is 89. What is the less number? Ans. 535. 7. What will 58 yards of broadcloth cost, at 4 dollars per yard? Ans. 232 dollars. 8. Bought 122 bushels of wheat, at 2 dollars a bushel; 8 oxen for 27 dollars each; 4 cows, 16 dollars each, and a wagon for 60 dollars. How much was paid for the whole, and how much more for the wheat and oxen than for the cows and wagon? Ans 584. 336. 9. The factors of a certain number are the difference between 1632 and 1700, and between 94 and 5 dozen. What is that number? Ans. 2312. 10. How many barrels of flour may be bought for 6721 dollars, at 13 dollars per barrel? Ans. 517 barrels. 11. Paid 57600 cents for eggs, paying at the rate of 12 cents a dozen. How many dozen did I buy? Ans. 4800 dozen. 12. What wll 168 firkins of butter cost, at 29 dollars a firkin? Ans. 4872 dollars. SUPPLEMENT TO THE FUNDAMENTAL RULES. 53 13. A man bought at vendue the following articles, viz.:A colt for 18 dollars; a horse for four times as much as the colt; a wagon for 8 dollars less than the cost of the horse; 4 cows for 4 dollars more than the cost of the wagon; 12 sheep, at 3 dollars each; a plough for 5 dollars; a ton of hay for 16 dollars; and a pair of oxen for four times the cost of the hay. Now, supposing he sells the whole for 527 dollars, how much does he gain; and if with the gain he pays 4 men, to whom he is in debt, equal sums, what does each receive?.ns. 46 dollars. 14. How many square feet in a board 12 feet long, and 2 feet wide? It is evident that a board 12 feet long and 1 foot wide would contain 12 square feet; then a board of the same length and 2 feet in width would contain twice as many feet. The answer, then, is 12 X 2=- 24 feet. 15. How many feet in length is a board which contains 24 square feet, and is 2 feet in width? Ans. 12 feet. It is evident that this question is the reverse of the preceding. Then, 24 2=12. 16. How many square feet of boards in a log which will make 26 boards, 15 feet in length and 3 feet in width? Ans. 1170. 17. How many square feet will it take for the floor of a hall, 40 feet long, 22 in width, allowing 24 feet for waste? Ans. 904 feet. 18. What is the width of a house which is 42 feet long, and the length and width multiplied make 1260 feet? Ans. 30 feet. 19. Supposing it take 60 yards of carpeting to cover the floor of a room 15 feet in width, what is the length of the room, and how much will be the cost of the carpeting, at 1 dollar 50 cents per yard? Ans 12 yds. in length. ~ 9000 cents. 20. How much money will a man lay up in a year of 52 weeks, if he lay up 25 cents a day, Sundays excepted? Ans. 7800 cents. 21. What is the difference between 7 times 35, and 7 times 5 and 30? Ans. 180. 22. How many days, months and years wilt a man be in travelling around the globe, it being 25000 miles, at the rate of 5 miles per hour, 10 hours in a day? 5* 54 SUPPLEMENT TO THE FUNDAMENTAL RULES. 23. The less of two numbers is 432; the difference between them is 17i. What is the greater? Ans. 607. 24. The remainder of a sum in Division is 423; the quotient 423; the divisor is the sum of both and 19 more. What, then, was the number to be diviaed? Ans. 366318. 25. What number, multiplied by 72084, will produce 5190048? Ans. 72. 26. The remainder of a sum in Division is 244; the quotient 1269; the divisor is twice the sum of the remainder, less 32. What was the sum divided? Ans. 578908. 27. What is that number, which, being divided by 7, the quotient resulting multiplied by 3, that product divided by 6, from the quotient 20 be subtracted, to the remainder 30 added, and half the sum shall make 35? Ans. 700. Art 48, —Exercises in tke use of the signs. 1. Write 9, plus 3, minus 7, plus 4. 9+3 —7+4-9 Ans. 2. Write the sum of 9 plus 3, minus the sum of 7 plus 4. 9+3-7+4=1,ins. 3. Write the sum of the products of 8 into 7, and 9 into 4. 8x7+9x4=92 Ans. 4. Write the product of the sum of 8 and 7 into the sum of 9 and 4. 8+7x9+4=195 Ans. 5. Write the difference of the products of 8 into 7, and 9 into 4. 8X7-9X 4=20 Ans. 6. Write the product of the difference of 8 and 7, and 9 and 4. 8-7 X9-4=5 Ans. 7. Write the sum of the difference of 9 and 3, and 7 and 4. 9-3+7-4=-9 Ans. 8. Write the product of 16 into 2, divided by 8. 16x2 —8=4 Ans. 9. Write 16 divided by the product of 8 into 2. 16- 8x2=l1 Ans. 10. Write the quotient of 16 divided by 2, divided by 8. 16-2~ 8=1 Ans. 11. Write 16divided by the quotient of-8 divided by 2. 16-8 —2=4 Ans. RELATION OF NUMBERS. 55 Art. 49. —Let the scholar write and perform the following questions, as the preceding. 1. What is the product of the sum of 16 and 12 into the sum of 9 and 10? Ans. 532. 2. What is the sum of the products of 7 into 11, and 5 into 8? Ans. 117. 3. What is the difference of the products of 9 into 12, and 7 into 9? -Ans. 45. 4. Divide the sum of 5 and 19, by the sum of 3 and 5. Ans. 3. 5. Divide the product of 7 into 10, by the product of 5 into 7. Ans. 2. 6. Divide the product of 8 into 16, by the sum of 9 and 7. Ans. 8. 7. Divide the sum of 15 and 17, by the product of 4 into 2. Ans. 4. RATIO, OR RELATION OF NUMBERS. Art. 50.-The ratio, or relation of one number to another, is found by division. It is the quotient afing from dividing one number by another. Thus the ratio of 8 to 4 is 2; 8-4 =2. The quotient shows that the dividend is twice as large as the divisor. Instead,. therefore, of the word quotient, we might use the word ratio. EXAMPLES. 1. What is the ratio of 25 to 5? Operation. 5)25 5 Ans. 2. What is the ratio of 30 to 6? Ans. 5. 3. What is the ratio of 56 to 7? Ans. 8. 4. What is the ratio of 144 to 12? Ans. 12. 5. What is the ratio of 6 to 7? Ans. 6 6. What is the ratio of 7 to 8? Ans.. When the dividend is less than the divisor, the ratio is expressed by writing the divisor under the dividend. Art. 51.-Since no new principle is ever discovered or needed in arithmetical operations not embraced in the simple rules, it is important that the student should understand these QUESTION.-1. What is ratio? 56 RELATION OF NUMBERS. rules, in all their varied applications. New names, and a new mode of writing and solving questions, naturally suggest the idea of new principles. Hence the beginner, in Fractions, is generally perplexed; to avoid this, fractions are written, and the various operations are explained, in the following exercises, as in whole numbers. Operation. The quotient of 2 divided 1. Divide 2 by 2. 2)2=1 Ans. by 2, is a unit, or 1. Operation. 2. Divide 1 by 2. 2)1- Ans. The quotient of 1 divided 2 by 2, is something less than a unit, and is called a fraction. A fraction is, therefore, the result of division. The terms of the division, which were dividend and divisor, are now the terms of the fraction, and assume the new names, "Numerator and Denominator." The scholar will, therefore, bear in mind, that numerator is the same as dividend, and denominator the same as divisor. The fraction 1 as a quotient, expresses the relation of dividend to divisor; that is, it shows that the dijdend was half as large as the divisor. But it may still be regarded as division implied-the numerator may be considered as a whole number, and the expression read, 1 divided by 2. 3. Multiply 4, or 1 divided by 2, by 2. Operation. It is evident that twice 4, or twice 1 divided "1 by 2, is equal to unity, or 1. To multiply the dividend is the same as to divide the divisor. Ans. Twice I divided by 2, is multiplication and division of whole numbers, and requires no new rule. 4. Divide 4, or 1 divided by 2, by 2. Operation. To obtain one half of any number, we 2 1 divide it by 2. But our dividend is a number 21 already divided; the operation, therefore, is a 411- Ans. repetition of division, and consequently no new rule is necessary.. 5. Multiply A by 2. That is, multiply 3 divided by 4, by 2 divided by 3. RELATIONS OF FRACTIONS. 67 Operation. Were it required to multiply 3 by 2, 2 4f3 we should write 3 and 2 under each $ A other as now. But in this example, 211 =l Ans. the numbers to be multiplied are divided numbers. The operation, therefore, involves multiplication and division of whole numbers — rules with which the student is already familiar. It will be perceived that the only difference between multiplication of wale numbers by whole numbers, and the multiplication of fractions by fractions, is, that in the former case, we have no divisor, in the latter we have, viz., the denominators of the fractions. 6. Divide A by 2. That is, divide 3 divided by 4, by 2 divided by 3. Operation. Were we required to divide 3 by 2, we 41j3 should write the 3 and 2 as we have now 21 3 done,-the 3 in the place of dividends, and S19=1l. Ans. the 2 in the place of divisors, thus, 213. But the 3 is already a divided number; therefore we have another divisor, or a factor to introduce into the divisor, written thus, 213. Illustration. -To multiply divisor is the same as to divide dividend. But 2, the divisor, is also a divided number. We have, therefore, another factor to introduce into the dividend, written thus, 213. Illustration.-To multiply dividend is the same as to divide divisor. * It will be seen, by this mode of writing fractions, that the numerators occupy the same position, that whole numbers would occupy standing in the place of fractions. Having thus disposed of the numerators of fractions, it is easy to recollect that their denominators are not to occupy the same side of the line. The terms of the fraction thus disposed of, we may apply the language of whole numbers to the statement; thus-Divide 3 by 2: multiply the quotient by 3, and divide the product by 4. Thus it appears that whole numbers may be written as fractions, and fractions as whole numbers, and the same principle of illustration employed. Fractions will hereafter be written and illustrated in both forms. QUCSTION-.-2. What is a fraction the result of? 3. Does division always result is a fraction? 4. When does it? 5. Can division be performed when the dividend is less than the divisor? 6. How, then, does division result in a fraetio 7. What i the value of a fraction? 58 FRACTIONS. FRACTIONS. Art. 52.-A fraction is part of a thing. The word fraction is derived from the Latin word frango, which signifies to break. When, therefore, any thing is broken into parts, those parts are called fractions. If a stick be broken into parts, each part becomes a fraction of the whole. The method of expressing whole numbers has been shown in Notation. Thus, the characters, 1, 2, when Wvritten alone express their own value; that is, 1 unit, 2 units, &c.;.but, when taken together, they express either 12 or 21. To express one half of a unit, or 1, we make use of the same figures, thus:. The unit is here divided into two parts, and one of those parts is here expressed. If a thing be divided into two equal parts, these parts are called halves; if into three equal parts, they are called thirds; if into four, fourths, or quarters, &c. The equal parts of a thing are expressed thus: i read, one half, or 1 divided by 2; 4 - one third, or 1 divided by 3; r - one fourth, or 1 divided by 4; I - two halves, equal 1, or 2 divided by 2; 2 - two thirds, or 2 divided by 3; 3- three thirds, equal I, or 3 divided by 3; 3 - three fourths, or 3 divided by 4; - four fourths, equal 1, or 4 divided by 4. From the nature of division, The greater the dividend with a given divisor, the greater the quotient, and the less the dividend with a given divisor, the less the quotient. If the quotient of 2 divided by 2, be 1, then the quotient of 1 divided by 2 must be one half of 1. Whenever, therefore, the dividend is less than the divisor, the quotient will be less than a unit. A man divides an acre of land into 4 equal parts; each part is one fourth of the whole. The quotient of 1 divided by 4 is one fourth, (l-.) It is evident, since 4 will contain 4, one time, that 1 will contain 4 one fourth of a time, and 2, two fourths (2), and 3, three fourths (3), and 4, four fourths (4), equal to one time. When a less number is to be divided by a greater, the division is performed by writing the divisor under the dividend, and drawing a line between them. Thus we have, at one view, in a fractional expression, the divisor, dividend, and quo EXAMPLES IN FRACTIONS. 59 tient. As a quotient, it expresses the ratio of dividend to di. visor. The divisor, or figure below the line, is called the denominator, because it gives name to the parts, or shows into how miany parts the unit is divided. If the denominator be 2, the unit is divided into two parts; if 3, three parts; if 4, four parts, etc. The figure above the line is called the numerator, because it numbers the parts, or shows how many parts are contained in the fraction. If the numerator be 2, the fraction contains two phrts; if 3, three parts; if 4, four parts, etc. A fraction is also the result of division, when the dividend is greater than the divisor, but will not contain it without a remainder. OBS.-The idea connected with the numerator and denominator of a fraction, may be familiarly illustrated thus: Suppose I have a box of 12 oranges, labelled on the outside, thus: 1 12 Oranges. The number above the line shows how many tlings are contained in the box, and the word below the line shows what kind of things they are. If we write, instead of the word "Oranges," the figure i, it would then show that the box contained 12 things, or units. Again, should we write the numbers, 2, 3, or 4, instead of the word "Oranges," they would show, not what kind of things were in the box, but into what parts the things were divided. Thus: EXAMPLE. Art. 53.-In the question, What is the quotient of 123 divided by 8? were the question, How many miles in 123 furlongs? (Art. 39,) 8 would still be the divisor. As 8 furlongs are equal to 1 mile, (8 - 8= 1,) I furlong is equal to one eighth of a mile, (1 8=!,) and 2 furlongs to 2, and 3 furlongs to. Increasing the numerator is only repeating the units to be divided; and as 123 units are to be divided by 8, we may write the whole in the form of a fraction, thus: 123. The question now is, HIow-nany miles in 123 eighths of a mile? The same is required as at the first, viz., the quotient of 123 divided by 8. QUESTIONS.-1. What are Fractions? 2. What is their origin? 3. If a thing be divided into two equal parts, what are those parts called? 4. Into four equal parts? To what does the dividing figure give name? 6. How are fractions expressed? 7. hat is the figure below the line called? 8. What is the figure above the line called? Wrh t do - it show? 10. What does the denominator show? 11. The numerator? 60 ILLUSTRATION OF FRACTIONS. Operation. We find by trial, that 15 is not the exact 8)123(15 quotient of 123 divided by 8, and 16 would be 120 too large; therefore, the true quotient is between those numbers, and must be expressed by a fraction. Having divided 120, the greatest number of units contained in the dividend of which 8 is a factor, we have 3 units of the dividend left, as a remainder, each of which must be divided by 8. 1~-8=-, and 3 8 =-, which evidently is the fraction required to express the exact quotient; for 120q-3 equals the dividend, and 120+3 - 815 -l =158. All questions in Division might be written in the form of a fraction, and to all fractional expressions the language employed in Simple Division might be applied, viz: What is the quotient of - divided by -? Hence it appears1. That the value of afraction is the quotient of the numerator divided by the denominator. 2. If the numerator be less than the denominator, the value of the fraction is less than a unit, or 1. 3. If the numerator be equal to the denominator, the value of the fraction is equal to a unit, or 1. 4. If the numerator be greater than the denominator, the value of the fraction is greater than a unit, or 1 DEFINITIONS. Art. 54 —Fractions are of two kinds; Vulgar, or common, and Decimal. They differ in the form of expression and mode of operation. In Decimal Fractions, the unit, or integer, is divided into 10, 100, 1000, etc., equal parts; or the denominator is always 1, with as many ciphers annexed as the numerator has places. In Vulgar Fractions, the integer may be divided into any number of parts; and the denominator being always expressed, may be any thing but 1 with a cipher or ciphers annexed. Vulgar Fractions are either proper, improper, compound, or mixed. QUESTIONS.-12. Fractions aredf how many kinds, and what are they? 13. In what do they differ? 14. How is the unit divided in Decimal Fractions? 15. What is always the denominator? 16. In Vulgar Fractions, how is the integer divided? 17. What may the denominator be? 18. How are Vulgar Fractions subdivided? DEFINITION OF FRACTIONS. 61 1. A Proper Fraction is one whose numerator is less than the denominator, as 2, 3, i, etc. 2. An Improper Fraction is one whose numerator is equal to, or greater than the denominator, as a, 8, 9, etc. 3. A Compound Fraction is a frction of a fractionf, as of ] of 5, etc. 4. A Mixed Number is a whole number and fraction written together, as 2X, 14-, 15.5, etc. 5. A Complex Fraction is one which has a fraction either in its numerator or denominator, or in both of them, thus: 6- 8 31 2 8' 9 51' 5' 6. A Common Divisor, or Common Measure of two or more numbers, is a number which will divide each of them without a remainder. 4 is the common measure of 12 7. The Greatest Common Di wor of two or more numbers, is the greatest number which will divide those numbers without a remainder. Thus, 12 is the greatest common measure of 12 8. Two or more fractions are said to have a common denom. inator, when the denominator of each is the same. 9. A Common Multiple of two or more numbers, is a number which may be divided by each of those. numbers without a remainder. 10. The Least Common Multiple of two or more numbers, is the least number which may be divided by those numbers, without a remainder. Thus 8 is the least common multiple of 8, 4, and 2. 11. A Prime Number is that which can be measured only by itself or a unit. 12. Two numbers are prime to each other when a unit is the only number that will measure both of them. Thus, 3 and 5 are prime to each other. 13. A Prime Factor of a number is a prime number that will measure it; and all the prime factors of a number are all QUESTIONS.-19. What is a proper fraction? o0. An improper fraction? 21. A compound fraction? 22. A mixed number? 23. What is a mixed fraction? 24. What is a common divisor, or common measure cf two numbers? 25. The greatest common divisor? 26. When are fractions said tc havelcommon denominator? 27. What is meant by a common multiple of two or more numbers? 28. The least common multiple? 29. What is a prime number? 30. What is a perfect number? 31. What is meant by the terms of a fraction? 32. When is a fraction said to be in its lowest terms? 6 62 EXERCISES IN FRACTIONS. the prime numbers that will measure it. Thus, 3 is a prime factor of 21, and 3 and 7 are all the prime factors of 21. 14. A Component Factor of a number is a composite number that will measure it; and all the component factors of a number are all the composite numbers that will measure it. Thus, 4 is a component factor of 12, and all the component factors of 12 are 4,,6, and 12. 15. An Aliquot Part of any number, is such a part of it as, being taken a certain number of times, will exactly make that number. 16. A Pefect Number is equal to the sum of all its aliquot parts. The smallest perfect number is 6, whose aliquot parts are, 3, 2, 1; and 3+2+1=6. The next perfect number is 28, the next 496, and the next 8128. Only ten perfect numbers are yet known. 17. The numerator and denominator of a fraction, taken together, are called the terms of the fraction. 18. A fraction is said to be in its lowest terms, when no number greater than 1, or unity, will divide the terms of the fraction without a remainder. EXERCISES. Art, 55 —1. If I divide an apple into 8 parts, by what fiaction will one of those parts be expressed? 2 of those parts? 3,4,5,6,7,8'? Ans. 1 2 3 4 5 6 7 8 F' F' F' A,, F, f. 2. If 1 be divided by 4, what will be the quotient? if by 6, what? if by7? if by 8? if by 9? if by 10? i by 11? if by 12? Ans. 1, 1I 11 1 1 1 4', 6, 7',', 9, 10, T' T-. 3. If 2 be divided by 4, what will be the quotient? if by 6? by 7? by 8? by 13? 2 6 ns,, 2 4. If a bushel be divided equally among 4 persons, what part of a bushel does each receive? 5. If 2 bushels of apples be divided equally among 4 persons, what will each receive? 6. If a bushel of corn be divided into four parts, what are those parts called? if into 5? into 6? into 7? 8? 9? 10? 7. If I give away 6 quarts of nuts, what part of a peck is it? if 7? if 8? if 90 8. How many of the four last questions are proper fractions? Are any improper, and which are they? EXERCISES IN FRACTIONS. 63 9. How is the quotient of 4 divided by 3 expressed? What rs the expression called? 10. If I divide an apple into halves, and give away a of a, what part of the apple do I give way? What is the expression, 7 of I called? In the foregoing question, the unit is divided into two equal parts, and each part is 2 of the unit. A division is again made of one of these parts into two other equal parts, and each part is 1 of -, or lo of the unit first divided. The expression, 2, as it respects the unit of which it is a part, is a fraction, (see definition, Art. 52;) but as it respects itself, or a subsequent division, it is to be regarded as itself a unit, and may be divided into halves, or any number of parts. A quarter, or l of a thing, is a whole quarter; and is made up of as many parts as the thing of which it is a part. It is, therefore, in relation to a division already made, that an expression is to be regarded as a fiaction. As it respects itself, or a subsequent division, it is to be considered a unit. Example. —A yard may be divided into 3 equal parts, or feet. A foot, when spoken of in relation to the yard, is ~; but 3 of a yard is one foot, and may be divided into 12 equal parts, or inches, and each inch is l-I of a foot, or -J of - of a yard. The inch may be divided into 3 equal parts, or barley-corns, and each barley-corn is itself a unit of less value, and it is also a fraction of a unit of a higher value; that is, 1 barloy-corn is ~ of ~- of I of a yard. That the terms, unit, and fraction, are merely relative, may be seen by the following formula: yd. yd. ft.. f ft. in. in. in.'bar; 1 +3=~=1, and 1 12==1=-, and 1 3=1=1. Art. 56. To reduce a compound fraction to a simple one. 1. Reduce 2 of - to a simple fraction. If we multiply the denominator of I by 3, we obtain onethird of 4. If we multiply this numerator by 2, we obtain twothirds. Hence the RULE. Multiply the numerators together, and the denominators, having cancelled all the equal factors in the numerators and denominators. Oper ~ 3'a 1 1 2%1i 3 x 6 or 1 = Ans. 64 EXERCISES IN FRACTIONS. 2. Reduce - of 3 of 4 to a simple fraction. Ans. 2, 3. Reduce of I of 3 of - to a simple fraction. Ans. a. Art. 57~.-To change any given fraction to an equivalent fraction, which shall have any required denominator. Change I to an equivalent fraction whose denominator shall be 6. In this example, the unit is already divided into thirds, and we wish to divide it into 6ths: We have, therefore, simply to reduce thirds to sixths. 2 sixths make a third, for the unit is divided into twice as many parts, and therefore the parts are one-half as large. Hence the RULE. Divide the reofired denominator by the denominator of the given fraction, and multiply the quotient by the numerator. The product will be the required numerator. Art. 58.-To reduce a whole number to an equivalent fraction, having a.given denominator. 1. Reduce 8 to a fraction whose denominator shall be 4. As in 1 unit there are 4 fourths, so in 8 units there must be 8 X-4=32 fourths, expressed thus: 32; therefore the RULE. Multiply the whole number by the given denominator, and set the product over the given denominator. 2. Reduce 16 to a fraction whose denominator shall be 7. Ans. 11. 3. Reduce 40 to a fraction whose denominator shall be 9. Ans. 36. 4. Reduce 129 to a fraction whose denominator shall be 21. Ans. fr 5. Reduce 339 to a fraction whose denominator shall be 39. Ans. 13221 A whole number may be expressed fractionally, by writing 1 under it for a denominator. Thus 2 may be written, 2; and read 2 ones; T 3 " it 3 ones; 4 " " 4 " 4 ones; As the expression, -, is equal to 2, and 3 to 3, the value QUESTION.-HOW is a whole number reduced to an equivalent fraction, having i given denominator? EXERCISES IN FRACTIONS. 65 of a number is not affected by writing 1 under it, as a denominator. To reduce improper fractions to mixed numbers, and mixed numbers to improper fractions. IMPROPER FRACTIONS. MIXED NUMBERS Art. 59. —1. Change 6X to a Art. 60.-2. Change 163 to whole, or mixed number. an improper fraction. 4)67 As the denominator of The scholar will per163 a fraction denotes the 163 ceive, that the mixed number of parts into 4 number, 163, was the which the unit is divided, it is evi- 67 quotient, in the last dent that'1 contains as many - question, of 67 divided units, or wholes, as 4 is contained by 4; and let it be retimes. in 67, which we find, by membered that a mixed number is trial, to be 16 times and X of a the quotient of a division whose time. Hence, divisor is the denominator of the To reduce an improper frac- fraction; therefore, tion to a whole, or mixed num- To reduce a mixed number to ber, we have this an improper fraction, we have this RULE. RULE. Divide the numerator by the de- Multiply the whole nqmber by nominator, and the quotient will be the denominator of the fraction, the whole number; the remainder, and to the product add the numerif any, written over the denomina- ator; under the result, place the lor, must be placed at the right denominator of thefraction. hand of the quotient. EXAMPLES. EXAMPLES. 3. Change 42 to a whole or 4. Change 5] to an improper mixed number. fraction. 5. In 59 how many wholes? 6. In 6, how many ninths? 7. In?-9 of a week, how many 8. In 10 9 weeks, how many weeks? 15ths? 9. Change 223- to a whole or 10. Change 159- to an immixed number. proper fraction. 11. In 347 of a day, how many 12. In 202s-5 days, how many days? 16ths? 13. In 46334 of a year, how 14. In 12652 years, how many many years? 39ths? 15. In ~ of a cent, how many 16. Change 3~ cents to the cents? fraction of a cent. 17. Change 27 to a whole or 18. Change 567 to an improper mixed number. fraction. 19. In H- of a minute, how 20. In 72 minutes, how many many minutes? 7ths? 6* 66 EXERCISES IN FRACTIONS. Art. 61.-To reduce a fraction to its lowest terms. 1. Reduce 4 to its lowest terms. If 4 bushels were divided equally between two persons, it is evident that one person would receive ~ of 4 bushels, or 2 bushels; so if 4 of a bushel be divided equally between two persons, one person will receive one half of -, or 8 of a bushel. Dividing the numerator by 2, we take one half of those parts which are contained in the fraction, while the value of each part remains the same. Therefore, To divide the numerator diminishes the value of thefraction. If we divide the denominator of 4 by 2, the fraction becomes 4. In this expression the unit is divided into half as many parts as at the first, and consequently, these parts are twice as large. It is evident, therefore, that To divide the denominator of afraction, the numerator remaining the same, increases its value. If we divide the terms of the fraction by 2, it becomes, which is equal to 2, or, for in either case the numerator is one half of the denominator. Hence it appears, that the value of a fraction is not affected by dividing or multiplying both the numerator and denominator by the same number. (See Art. 43.) To reducea fraction to its lowest terms, we have this RULE. Divide both the numerator and denominator by any number that will divide both without a remainder; and so continue to do.until no number greater than 1 will divide them. 2. Reduce 4' to its lowest terms. \ 84 16)12 210( 30 A 3. Reduce 16 to its lowest terms. Ans. -. 4. Reduce -y32 10,0 83,, to their lowest terms. Art. 62.-Were the greatest number known which would divide the terms of the fraction, a simple division would at once reduce the fraction; but, as this is not the case, the greatest divisor may be found by the following QUzsTIoNs.-1. What is the rule for reducing an improper fraction to a whole ot mixed number? 2. For reducing a mixed number to an improper fraction? ILLUSTRATION OF FRACTIONS 67 RULE. lDivide the denominator by the numerator, or the larger numPer by the less, and if there. be no remainder, the numerator, or the less number, will be that divisor; but if there be a remainder, divide the last divisor by the last remainder, and thus proceed until there be no remainder; and the last divisor will be the greatest common measure sought. The above Rule may be illustrated in the following manner: Suppose I have two lines, and wish to obtain a third which shall be an exact measure of the two. I first apply the shorter line to the longer, and find it contains it twice, but not three times. The remainder I now apply to the shorter line, and find it contains it twice and no remainder. Therefore, this last divisor is the third line sought, and is the exact measure of the other two. _ Suppose the longer line to be 40 feet and the shorter 16 feet, - and we are required to find the greatest common measure, or divisor, of 16 and 40, or to reduce 16-40 to its lowest terms, we should proceed in the following manner: O_ r emr ios. It is evident that 16 is the greatest number rem Option. that will divide 16 without a remainder; and 32 would 16 divide 40 without a remainder, it would be the greatest common measure of the 8)16(2 terms of the fraction. But we find, by trial, 16 that 16 is contained in 40 twice and 8 remainder; hence 16 is not the common measure. Dividing 16, the last divisor, by 8, the remainder, we find it contains it twice and no remainder; therefore 8 is the greatest common divisor'of the terms of the fraction. For, if 8 will divide 8, it will also twice 8, which is 16, and five times 8, which is 40. 6. Reduce s 2 to its lowest terms. Ans. 1. 7. Reduce 4?68- to its lowest terms. Ans. -7. 8. Reduce 36 -1 to its lowest,erms. Ans. i. 9. Reduce 2-1 6 to its lowest terms. Ans.. If it be required to find fie greatest common measure of more than two numbers, find the greatest common measure of two of them, as before; then, of that common measure, and of one of the other numbers, and so on through the whole. The common measure last found will be the one sought. QuzSTIoNS.-3. What is the rule for reducing a fraction to its lowest terms? 4. Were the greatest number known which would divide the terms of the fraction, how might you proceed? 5. When this is not the case, how may the greatest divisor be fiund? 6. How is the common measulreof more than two numbers found? 68 ILLUSTRATION OF FRACTIONS. 10. What is the greatest common measure of 48 and 192? Ans. 48. 11. Reduce 48s to its lowest terms. Ans. ~. 12. What is the greatest common measure of 35, 42, 63? Ans. 7. Art. 63 —To reduce a complex fraction to a simple fraction, 1. Reduce 5- to a simple fraction. To multiply numerator is the same as to divide denominator, 2 2 x 4 (Art. 44;) therefore, --— _ 5-; and to multiply denominator 3 3 2x4 is the same as to divide numerator; therefore =3X =T85. Hence the RULE. If the numerator be whole or mixed numbers, reduce them to improper fractions. Then multiply the numerator of each fraction by the denominator of the other; the product will be the fraction required. 2. Reduce 57 to a simple fraction. Ans. 7.. Reduce to a simple fraction. ns. 203. Reduce - to a simple fraction. Ans. 2. 8 4. Reduce 8j to a simple fraction. Ans. 9. Art. 6. —To change a simple fraction to a complex. 1. Change 4 to a complex fraction. 4 X2 2 ~24=2 -X2 Ans. Hencs the 5 2 RULE. If the numerator or denominator, or both, be a composite number, separate them into factors, and transfer one or more from numerator to denominator, and from denominator to numerator, observing that a factor transferred, becomes a divisor. 7 3 2. Change 2 to a complex fraction. Ans. - or —. The answer depen n the factor tansferred OBs.-The answer dependsolpon the factor transferred. ILLUSTRATION OF FRACTIONS. 69 3. Change 2 to a complex fraction. 4. Change I! to a complex fraction. Art. 65.-The following, if made familiar, will aid the scholar in cancelling. 1. Any number ending with an even number, or cipher, is divisible, or can be divided, by 2. 2. Any number ending with 5, or 0, is divisible by 5. 3. If the right-hand place of any number be 0, the whoe is divisible by 10; if there be two ciphers, it is divisible by 100; if 3 ciphers, by 1000, and so on, which is only cutting off those ciphers. 4. If the two right-hand figures of any numbers be divisible by 4, the whole is divisible by 4; and if the three right-hand figures be divisible by 8, the whole is divisible by 8, and so on. 5. If the sum of the digits in any number be divisible by 3 or by 9, the whole is divisible by 3 or 9. 6. If the right-hand digit be even, and the sum of all the digits be divisible by 6, then the whole will be divisible by 6. 7. A number is divisible by 11, when the sum of the 1st, 3d, 5th, etc., or all the odd places, is equal to the sum of the 2d, 4th, 6th, etc., or of all the even places of digits. 8. If a number cannot be divided by some quantity less than itself, that number is a prime, and cannot be divided by any number whatever. Art, 66 —To multiply and divide fractions by whole numbers, whole numbers by fractions, fractions:b fractions. RULE. Draw a perpendicular line, and write numerators, in all cases, as you would a whole number standing in the place of thefraction; viz., the numerators of fractions to be multiplied or divided on the right of the line, and their denominators on the left. The question thus stated, equals on each side of the line may be crossed, as cancelling each other. (See Art. 47.) When no two numbers remain, one on each side of the line, capable of being divided by any one figure, multiply the figures on the right of the line, for a numerator, or dividend, and those on the left, for a denominator, or divisor, and the result will be the answer in the lowest terms of the fraction. 70 ILILUSTRATION OF FRACTIONS. Multiplication of Fractions by Whole ~Numbers. Art. 67.-1. If a man receive 4 of a dollar for 1 lay's work, what will he receive for 2 days' work? It is evident, if a man receive - of a dollar for 1 day's work, that he would receive, for 2 days' work, twice as much, or 2=- Multiplying the numerator by 2, the denominator remaining the same, we have twice the number of parts, while the value of each part remains the same. Dividing the denominator by 2, the numerator remaining the same, we have the same number of parts, while the value of each part is twice as great. Hence, to multiply the numerator of a fraction is the same, in effect, as to divide the denominator. If the numerator of - be multiplied by 2, it becomes — 1. If the denominator be divided by 2, it becomes I= 1. Therefore, to multiply a fraction by a whole number, we have the following RULE. Multiply the numerator, or divide the denominator, and the result will be the answer required. 2. If a pound of lead cost I of a dollar, how much must 16 pounds cost? Operation. 1 10 1 10 Or thus: -X=- = Ans. i Ans. 1 1 1 It is evident, if one pound cost 1 dollar divided by 16, that 16 pounds would cost 16 dollars divided by 16, equal to 1 dollar. Therefore-A fraction is multiplied into a quantity equal to its denominator, by cancelling or removing the denominator. 3. If a pound of iron cost g of a dollar, how much will 9 pounds cost? Operation. 2 Id 1 0 Or thus: X -- n. 2 1- Ans. QuzsTIoNs.-7. What is the rule for the multiplication and division of fractions, ctc, by cancelling? 8. When the question is stated. what is the method of procedure? 9. When no two numbers are left, one on each side of the line, capable of being divided by any one figure, what is to be done? 10. How do you multiply a fraction by a whole number? 11. Why, in example 2, are 16 and the numerator of the fraction placed on the right of the line? ILLUSTRATION OF FRACTIONS. 71 If.1 pound cost I dollar, 9 pounds would cost 1 x 9=9 dollars; but the cost of 1 pound is 1 dollar divided by 18: therefore, 9 dollars, the cost of 9 pounds, must be divided by 18. By the rule already given, the numerator of the fraction, with 9, its multiplier, is placed x)n the right of the line, and 18, the divisor, on the left. 9 and 2 are factors of 18; therefore, cross 9 and 18, and write 2, the remaining factor, in the place of 18. The answer, then, is, 1 divided by 2; or, -. (See Art. 66.) On the principle above stated, A fraction may be multiplied into any factor in its denominator, by cancelling that factor. 4. If a pound of lead cost T of a dollar, how much will 8 pounds cost? If the cost of 1 pound be -- of a dollar, 8 pounds will cost y x 8 -^ = of a dollar. Making the horizontal line, which separates the numerator of the fraction from the denominator, perpendicular, it will be seen that the numerator occupies-the place of dividends, (the right of the line,) and the denominator the place of divisors, (the left of the line,) thus: 2 1 1 Ial the latter mode the question is resolved into this. Mul2 1=1 Ans. tiplv 1 by 8, and divide by 16; therefore, the numerator of the fraction and 8, its multiplier, occupy the right of the line, and 16, the divisor, the left. It is to be remembered, that the numerator of afraction, in all cases, is to be disposed of as a whole number, without regard to its dt nominator. On whichever side of the line the numerator falls, the denominator must be placed on the opposite side. 5. What will 8 bushels of apples cost, at a a dollar per bushel? Ans. $4. OBs.-This character ($) placed before any number, shows that it is dollars. 6. If one man can plant 4 of an acre in one day, how much could 12 men plant in the same time? Ans. 9 acres. 7. If 1 barrel of fish cost 6- dollars, what will 9 barrels cost? Ans. 56-. OBS.-Mixed numbers must be reduced to improper fractions. 8. If 1 chest of tea cost $251, what will 15 cost? A is. $3781. 72 ILLUSTRA'TION OF' FRACTIONS. 9. If a man can walk 29^ miles in I day, how far could he walk in 30 days? Ans. 885 miles. 10. What will 600 pounds of cotton cost, if 1 pound cost 9~ cents? Ans. $57. OBs.-Dividing any number of cents by 100, reduces them to dollars. Diviszon of Fractions by Whole Numbers. Art. 68 —1. If a man receive - of a dollar for two days' work, what does he receive per day? We have seen, that a fraction is multiplied either by multiplying its numerator, or dividing its denominator; then, as Division is the reverse of Multiplication, the reverse of the rule for Multiplication will be the rule for Division. If I divide the dollar into 4 parts, or quarters, and pay a man 4, or quarter, it is the same as though I should divide it into 8 parts, or half quarters, and pay him -, or 2 half quarters=. Multiplying the denominator by 2, the numerator remaining the same, is dividing the unit into twice as many parts, and consequently the value of each part is diminished by one half. Dividing the numerator by 2, the denominator remaining the same, is taking half as many parts, while the value of each part is the same. Therefore, to divide a fraction by a whole number, we have this RU LE. Divide the numerator of the fraction by the whole number, when it can be done without a remainder; otherwise, multiply the denominator. 2. If 8 pounds of lead cost I of a dollar, what does it cost per pound? Operation. Were the cost of 8 pounds 1 dollar, 2 1 the cost of 1 pound would be the quo8 tient of 1 divided by 8. Regarding the 16 1- I Ans. numerator as expressing the cost of the 16 lead, without reference to the denominator, we place it on the right, as a dividend, and 8, the number of pounds, on the left, as a divisor; but the cost of 8 pounds is 1 divided by 2; therefore, write 2 also on the left. As no QUESTIONS.-12. How do you divide a fraction by a whole number? 13. Why, in example 2d, are 8 and 2, the denominator of a fraction, placed on the left 9 14. M1ltt plying the denominator of a fraction, is the same as what?i EXERCISES IN FRACTIONS. 73 reduction can be made, the 2 and 8 are to be multiplied together, for a divisor or denominator: (multiplying the denominator divides the fraction.) The answer, then, is 1 divided by 16, or H. Or thus, - =T -Ans. 2x8 6 3. If a pair of oxen plough, in 4 days, 2d of a field, what part do they plough in one day? Ans. -7' 4. If 6 men earn 8 of a guinea in I day, what part of a guinea does 1 man earn in the same time? Ans. 3. 5. If 16 hats cost $64~, what does 1 hat cost? Ans. $4. 6. What will 1 pipe of molasses cost, if 14 pipes cost $707 —? - Ans. $50~. 7. What will pound of rice cost, if 50 pounds cost $150-1? Ans. $3. Muttiplication and Division of Fractions by Whole Numbers. MULTIPLICATION. DIVISION. Art. 69.-1. If a dollar will Art. 70. —2. If 9 dollars will buy -l of an acre of land, how buy' of an acre, how much will much will 9 dollars buy? 1 dollar buy? 3. If a man travel 2 of a mile 4. If a man travel 8 of a mile in 1 minute, how far will he travel in 12 minutes, how far will he in 12 minutes? travel in I minute? 5. If a man consume -5 of a 6. If 7 men consume 5 of a barrel of flour in 1 month, what barrel of flour in 1 month, how will 7 men consume in the same much will 1 man consume? time? 7. If 3 of a box of glass cost 8. If 21 dollars will buy 41 1 dollar, how many boxes will 21 boxes of glass, how much will 1 dollars buy? dollar buy? 9. If a pound of chocolate cost 10. If 7 pounds of chocolate 25 of a dollar, how much will 7 cost X of a dollar, what will I pounds cost? pound cost? 11. If a man can do -L of a 12. If a man can do ofa piece piece of work in one day, how of work in 8 days, how much can much could he do in 8 days? he do in I day? 13. What will 16 yards of cloth 14. If 16 yards of cloth cost 12 cost, at 3 of a dollar per yard? of a dollar, what will 1 yard cost? 15. What will 40 yards of car- 16. If 40 yards of carpeting peting cost, at ~ of a dollar pir cost \ — of a dollar, what will 1 yard? yard cost? 74 EXERCISES IN FRACTIONS. 17. If 1 pint of wine cost X of 18. If 12 quarts of wine cost a dollar, how much will 12 quarts 13 dollar, what is it per pint? cost? 19. Multiply; by 11. 20. Divide 9 by 11. Art. 71, —Multiplication of whole numbers by fractions. 1. If a barrel of flour cost $9, how much will 2 of a barrel cost? -Had the cost of 2 barrels been required, the price of 1 barrel being given, we should multiply the price of I barrel by the number of barrels. The same is now to be done, that is, the price of one barrel is to be multiplied by the part or parts of a barrel taken. To multiply by 1, is to repeat the units of the multiplicand once. To multiply by 2, is to repeat the units of the multiplicarnd twice. To multiply by I of 1, is to repeat one-half of the units of the multiplicand once. The product of any number- multiplied by a fraction is proportionally as much less than the multiplicand as the multiplier is less than the unit, or. I. Therefore, if we multiply 9 by ~ of. the product will be 2 of 9, or 6. Hence, it appears, that, to multiply by a fraction, is to repeat such a part of the multiplicand as the fraction is part of a unit. If 9 dollars be the cost of I barrel, then the quotient of 9 divided by 3 will be 3 dollars, the cost of ~ of a barrel, and 3 x 2=6, the cost of 2 Thus it appears, that the only difference between multiplying by a whole number and a fraction, is, that in the last case the multiplier is a number divided; now to divide the multiplicand or the product is the same as to divide the multiplier. Hence, to multiply a whole number by a fraction,.RULE. Multiply the whole number by the numerator of the fraction, and divide the product by the denominator; or divide the whole number bythe denominator of the fraction, and multiply the quotient by the numerator. OBs.-As multiplying by a fraction is repeating a part only of the multiplicand,we divide by the denominator of the fraction, to obtain that part of the multiplicand to be repeated, and multiply by the numerator to repeat that part. Thus, to multiply by i, we divide by 3, and obtain one-third of the multiplicand, which is to be repeated twice, or multiplied by 2, the numerator. The same principle is applicable to the multiplication of fractions by fractions. EXERCISES IN FRACTIONS. 75 2. What will 16 yards of cloth cost, at I of a dollar per yard? Were the cost of I yard 3 dollars, the cost of 16 yards would be 16 times 3 dollars; but the cost of one yard is 4 of 3 dollars; therefore the cost of 16 yards will be - of 16 times 3 dollars. Operation 1st. Operation 2d. 4 3 4 rs 4 3X 1212 1 12 Ans. 2 Exclding equal factors from divisor and dividend, or from numerator and denominator, does not affect the result. In this example we exclude the factor 4, and the remaining factors 3 and 4 multiplied together, being factors of the dividend or numerator, give 12, the answer. 3. What will 40 yards of carpeting cost, at - of a dollar per yard? Ans. $35. 4. What will 64 bushels of oats cost, at 3 of a dollar per bushel? Ans. $24. 5. What will 24 bushels of corn cost, at 5- of a dollar pei bushel? Ans. $15. 6. Multiply 21 by 3., by q, by ~. Ans. 9, 18, 7. 7. What is the product of 324 multiplied by 1? 8. What will 9 pounds of tea cost, at 27 of a dollar per pound? Ans. $4-. 9. What will 56 pounds of butter cost, at 2 of a dollar per pound? Ans. $14. 10. What will 124 pounds of sugar cost, at 2 of a dollar per pound? Ans. $154. 11. Multiply 32 by,4, by, by 8, by -. Ans. 8, 8, 12, 28. 12. Multiply 224 by 5-. Ans. 4. Division of Whiole lumbers by Fractions. Art 72. —1. If 2 of a barrel of flour cost $6, what will be the cost of 1 barrel? QUESTIONS.-15. What is the product of any number multiplied by 1? 16. How many times greater than the multiplicand is the product of a multiplier greater than 1? 17. How much less than the multiplicand is the product, when the multiplier is less than 1? 18. What is the product of a whole number multiplied by a fraction? 19, Rule? 76 ILLUSTRATION OF FRACTIONS. As this question is the reverse of question 1st, in the preceding section, the reverse of the rule there given will be the rule for solving this question. Had 6 dollars been the cost of 2 barrels, we should divide the price by the number of barrels. The same is now to be done; that is, the price is to be divided by the parts of a barrel taken. The quotient of any number divided by a unit, or 1, is the same as the dividend. The quotient of any number divided by a greater number than 1, is as many times less than the dividend, as the divisor is times greater than a unit, or 1. On the same principle, if the-divisor be less than a urt, or 1, the quotient will be greater than the dividend. If the divisor be -, the quotient will be 2, or twice the dividend. If the divisor be -, the quotient will be 3, or three halves of the dividend. Therefore, if we divide 6 by 2 of 1, the quotient will be 3 of 6, or 9. It is evident, that 6 will contain - of 2 three times as often as it will contain 2. Again: if 6 dollars be the cost of 2 of a barrel, the quotient of 6 divided by 2 will be 3 dollars, the cost of 1 of a barrel, and 3 x 3 =9 dollars, the cost of = I barrel. Thus it appears, that the difference between dividing by a whole number, or by a fraction, is, that in the latter case the divisor is a number divided; for, To multiply the dividend or the quotient, is the same as to divide the divisor. Hence the RULE. Divide the dividend by the numerator of the fraction, and multiply the quotient by the denominator; or multiply by the denominator, and divide the product by the numerator. Operation. Regarding the numerator of the 3 X 3 x3=9 Ans. fraction as a whole number, we say, -3 if 2 barrels cost 6 dollars, the quotient of 6 divided by 2 will be the cost of I barrel; therefore, place 6, the dividend, on the right, and 2, the divisor, on the left of the line. But the 2 barrels is 2 divided by 3; therefore, place 3, the divisor, or denominator, on the opposite side of the line, as a multiplier. To multiply the dividend, or quotient, is the same as to divide the divisor. We then say, 2 in 6, three times; cross 6 and 2, and multiply 3 into 3. 3 X 3$9 Ans. OBS-Were the foregoing question, How many times will 6 bushels ILI.USTRATION OF FRACTIONS. 77 contain 2 pecks? we should multiply the dividend by 4, to bring it into pecks, or fourths of a bushel. 6 bushels equal 24 pecks, or 14 of a bushel, and 2 pecks equal 2 of a bushel. So, to divide 6 units by -, we multiply the dividend by 3, the denominator of the fraction, to bring it into thirds. The same is true in division of fractions by fractions. 2. If - of a ton cost $15, how much will 1 ton cost? Ans. $18. 3. If - of an acre be worth $12, what is 1 acre worth? Ans. $32. 4. If -I of the number of rows in a corn-field be 30, what is the whole number? Ans. 40. 5. Divide 20 by -, ~ 1. Ans. 40, 80, 100. 6. If a pound of tea cost -3 of a dollar, how many lbs. may be bought for $60? Ans. 100 lbs. 7. In what time. can a man build 7 rods of wall, if he build 0- of a rod in an hlour? Ans. 17- hours. 8. At - of a dollar for building one rod of stone wall, how many rods may be built for $69? Alns. 80-. 9. At $33 per yard, how many yards may be bought for $80? Ans. 21- yards. 10. If 1- bushel of wheat sow an acre of land, how many...ies will 12 bushels sow? Ans. 9 acres. 11. How many times is 7 contained in 56? Ans. 64. 12. How many times is Zs contained in 21? Ans. 43. lMultiplication and Division of W/hole Numbers by Fractions. IMULTIPLICATION. DIVISION. A'rt 3. —1. If a man can A it, 74 -2. If a man earn earn $16 in a month, how mrrch 6 in $ of a month, how much can he earn in - of a month? can he earn in a month? 3. If a man lay up $84 in a 4. If a man in - of a year lay year, how much would he lay up up 60, how much would he lay in - of a year? up in a year? 5. If the price of a horse be 6. If -- of the value of a horse $75, what would be the price of be 125, what is the whole value? a horse worth y a-s much? 7. If a house be worth $672, 8. If - of a house be worth how much is 19 worth? $378, what is the whole worth? 9. If a farm be worth $840, 10. If of a firm be worth what is I of it worth? $525, what is the whole worth? QUESTIONS-20. How much greater than the dividend is the quotient of a whole number divided by a fracion? 21. How do you divide a whole number by a frac tion? P. t. 78 ILLUSTRATION OF FRACTIONS. 11. Multiply 156 by. 12. Divide 117 by {. 13. A man has 360 apple-trees 14. 40 is v of what number? in two orchards; in the smaller, there is y of the whole. How many are there in the smaller? 15. A ship and cargo are val- 16. If A of a ship and cargo ued at;$100,000, the ship at 2 are valued at $5,000, what is the of the whole. What is the value value of the whole? of the ship? Multiplication of Fractions by Fractions. Art. 75 —We have seen, that to multiply a whole number by a fraction, is to repeat such a part of the multiplicand as the multiplier is part of a unit. If the multiplicand be a fraction, the principle is the same.' 1. If a bushel of corn be worth i of a dollar, how much is a of a bushel worth? Operation. Were the cost of 2 bushels required, we I1 should multiply the price by the quanti-. 2 ty; but as the quantity is less tharr 1 3 1- S. bushel, we multiply by the parts taken; _ 4 n- e (see Art. 71.) The qestion then is, How much is 1 of? To multiply a whole number by a, we take. of the multiplicand. The same is now to be done: _ of 2=. If the numerators be multiplied together, and also the denominators, we have the answer: Thus, X — = —. Multiplyingthe denominator of 2 by 2, is dividing the fraction by 2. We thus obtain that part of the multiplicand to be repeated, or multiplied by the numerator of A-. (See Art. 71, Obs.) Therefore, to multiply a fraction by a fraction, we have this RULE. Multiply the numerators together for a new numerator, and their denominatorsfor a new denominator. EXAMPLE. 1. A man owning i of a ship, sold 5 of his share: I of 3 is how much? QUESTIONS.-1. How do you multiply a fraction by a fraction? 2. Why, in example 1st, are the numerators of the f1ictios placed on the right of the line, and the denominators on the eft? ILLUSTRATION OF FRACTIONS. 79 Operation. As the numerators of the fractions are to 5 ^ be multiplied together for a new numerator, 2 A 413 or dividend, they are placed on the right 1013=3- Ans. of the line, and the denominators, which are to be multiplied for a new denominator, or divisor, are placed on the left of the line. The numl)ers are cancelled and multiplied as in preceding examples. Or thus: 2 3 The factor 2 in' the numer2 4 5 ator, cancels 2, one of the factors of 4, in the denominator. 2. What is the product of - x -; of 7 x 3? Ans. 1, MEW. 3. Multiply * by i, and 6 by 8. 4. A boy having I of a dollar, gave i of it for toys; what did the toys cost him? Ans. 4 of a dollar. 5. At - of a dollar per yard, what will 9 of a yard cost? 6. At - of a dollar per pound, what will - of a pound of tea cost? Ans. 21 of a dollar. 7. At A of a dollar a pound, what will 6 of a pound of coffee cost? Ans. I of a dollar. 8. At 2- dollars per bushel, what will 6 —T bushels of wheat cost? Ans. $13P. 9. If a house lot be worth 100-5 dollars, what is -5 of the lot worth? Ans. $4U^._ 10. If a flock of sheep be worth 75-1 dollars, what is 1 of the flock worth? Ans. $185. Division of Fractions by Fractions. 1. If a bushel of corn cost - of a dollar, how many bushels may be bought for 3 of a dollar? It is evident that I will contain I twice: -2 - -==-2, Ans. In this example, both dividend and divisor are divided by 3. It has been shown, that to divide the dividend is the same as to divide the quotient, and to divide the divisor is the same as to multiply the quotient; and also, that to multiply and divide any number by the same quantity does not affect its value. Art. 76,-Therefore, when the denominator of dividend and divisor are alike, divide the numerator of the dividend by thk numerator of the divisor, and the quotient will be the answer. 80 ILLUSTRATION OF FRACTIONS. Operation. Or thus:'2 2 $2= 2 Ans. 1_2 A ns. 2. If a bushel of corn cost 3 of a dollar, how much may be bought for 3 of a dollar? Were the money to be expended 3 dollars, and the price 3 dollars, the answer wouldbe 1 bushel, 3-3 = 1. But suppose the price 3 dollars, and the money to be expended 3 dollars divided by 7, equal 3. The answer would then be 3 = =1 of a bushel. To divide the quotient is the same as to divide the dividend. But the price is not 3 dollars, but 3 dollars divided by 4. To multiply the quotient is the same as to divide the divisor. Therefore, the true answer is, 4 divided by 7 equal 4. Again, 3=- Y, and =_ X 3. To multiply dividend and divisor by the same quantity does not affect the quotient. Art. 77 —Therefore, when the numerators of divisor and dividend are alike, RU LE. Divide the denominator of the divisor by the denominator of the dividend. Art. 78.-3. If a bushel of oats cost 3 of a dollar, how many bushels may be bought for A9 of a dollar? Operation: 8)-96 (3=1_l Ans. 4. If a bushel of rye cost -5 of a dollar, how many bushels may be bought for } of a dollar? In this example, the terms of the dividend cannot be divided by the corresponding terms of the divisor without a remainder; but to multiply the numerator is the same as to divide the denominator. Hence the RULE. Divide the terms of the-dividend by the corresponding terms of the divisor, when it can be done without a remainder; otherwise, invert the divisor, and proceed as in Multiplication. Operaticpn. ^By the rule already given for placing the numerators of raactions as whole 91 numbers, the numerator of -, the ~~ dividend, is placed on the right of the 45156=I- — As. line, and the numerator of, the ILLUSTRATION OF FRACTIONS. 81 divisor, on the left. This is the same as inverting the divisor. *9 DX -45 Ans. Division of fractions by fractions may be variously illustrated. 1. To multiply the dividend is the same as to divide the divisor, (see Art. 43.) To multiply the denominator divides the fraction. Therefore, I X 8 - 5 = X =5 6= 1- - Ans. 2. Multiplying the numerator of 7 by 8'reduces it to eighths. Multiplying the denominator by 5 divides the fraction, (see Art. 72, Obs.) 1. If a bushel of potatoes cost -2 of a dollar, how many bushels may be bought for - of a dollar? Ans.. 2. If - of a bushel of apples cost 5 of a dollar, how much will 1 bushel cost'? Ans. Yh. 3. How many bushels of rye, at } of a dollar per bushel, may be bought for 2 of a dollar? Ans. 3 of a bush. 4. If r of a ton of hay cost - of a dollar, what does it cost per ton? Ans. $9IL. 5. If 4~ pounds of tea cost 3{ dollars, what is it per lb.? Ans. 1 3 of a dollar. 6. If 1 of a dollar buy 1 pound of tea, how much will 3I dollars buy? Ans. 41 pounds. 7. Divide 17~ by 71 and 183 by 2. Ans. 2~; 56-. Multiplication and Division of Fractions. MULTIPLICATION. DIVISION. Art. 79,-1. A man owning Art. 80.-2. A man sold. & of a house, sold f of his share. of a house, which was - of his What part of the house did he share. What part of the house sell? did he own? 3. If a bushel of salt cost 19 4. If I of a bushel of salt cost of a dollar, what will ~ of a bush- lW of a dollar, what does it cost el cost? per bushel? 5. If a peck of coal cost 5 of 6. If 3 of a peck of coal cost a dollar, what will 3 of a peck 15 of a dollar, what will one cost? peck cost? 7. If 1 cord of wood cost 2 of 8. If 7 of a cord of wood cost a dollar, how much will V of a 31 dollars, how much is it per cord cost? cord? 9. If 1. foot of hammered stone 10. If yl of a foot of hammered cost s5 of a dollar, what will 2- stone cost I of a dollar, what will of a foot cost? one foot cost? 82 EXERCISES IN FRACTIONS. The simple rule may now be repeated for solving any question which may arise in Multiplication and Division of Fractions by Whole Numbers-Multiplication and Division of Whole Numbers by Fractions-Multiplication and Division of Fractions by Fractions. RULE. Plii all those numbers which are to be multiplied together for a numerator, or diviq4nd, on the right of the perpendicular line, and those numbers which are to be multiplied togetherfor a denominator, or divisor, on the left of the line, and proceed to cancel, as before directed. PROMISCUOUS EXAMPLES. Art. 81.-1. A man owning -a of - of o of 4 of a ship, sold 7 of - of - of his share. What part of the ship did he sell? Thus:'1 1 2 T: Orthus: X - x x $x $ 2 2X —Ans. ~8 $ 4 85 W $ 315 5 A Fractions connected by the word of, 1/ are called compodif fiactions. They X $ are reduced to simple fractions, by mul$ $ tiplying all the numerators together for 3 2 a new numerator, and all the denomi15 2=-5 Ans. nators for a new denominator. By cancelling, the process of multiplying and reducing the fraction is performed at once. 2. Reduce 7 of - of. of 2 of 56 of 87 to a simple fraction. Ans. 25 84 3. A man owning 1 of 5 of 7 of 9 of a factory, sold - of - of 2 of his share. What part of the factory did he sell? Ans. 7. 4. What simple fraction is equivalent to 7 o f 2, o f of of A of I of 9, of 6 of 18, of T of 2. Ans. 17-. 5. Multiply 3 by 4. 6. Multiply ~ by 6. 7. Multiply I by I. 8. Divide 4- by 31. 9. Divide I of I by 5 of - of 9. Ans. -. EXERCISES IN FRACTIONS. 83 10. Divide of 9 by - of 7. Ans. 11. 161 122 IHaving reduced the terms of 11. Divide by. divisor and dividend to improper 58 s fractions, it will be found that the numerators and denominators themselves become fractions. Thus, the numerator of the dividend, 163 =49, and the denOminator, 18 =73 3 4 The dividend now: assumes this form: 49 The denomina3 tors may be removed from the - terms of the frac4 tion, and the process illustrated in the following manner. The numerator is 49 divided by 3. To multiply the denominator is the same as to divide the numerator. 49 To multiply the numerator is the same as to 3 49 divide the denominator. Therefore73 73 X3 3 49 3 X 3 49X4 196 4 4'13 X 3x= - 73><3 219 Let the scholar reduce the divisor, and illustrate in a similar mannner. 7 6- 4 3 55 12. Divide 8 — by -5 -. Ans. 64 8 II - 5 n 7 64 13. Divide 1 of4 f of o - of 18 by 4 of 5 of - of 12; multiply by I of 2 of 5 of 2; divide by 6 of 7 of 2 of 6. Ans. -. 14. A man who owns 2 of a farm, sells 1 of 4 of 3 of - of his half. What part of the farm does he sell? Ans. -. 15. Multiply 12 by I of 3, divide by - of 1, multiply by 2 of 6, divide by - of 14, multiply by I of 18, divide by - of 27. Ans. 9. Addition of Fractions. Fractions are added on the same principle as whole numbers. As tens can only be added to units, and pounds to shillings, by first reducing the higher denomination to the lower, 84 ILLUSTRATION OF FRACTIONS. so fractions of different denominations, or which have different denominators, can only be added by first reducing them to the same. Art. 82.-Fractions which have a common denominator may be added by the following RULE. Add their numerators, and write their sum over the denominator. Operation. 1. Add 3 and 1 8+IL4 Ans. 2. Add I+2+ +3 +T. Ans. - T Art. 83.-Addition of fractions whose denominators are different, and one is a multiple of each of the others. 1. Add - and -. In this example sixths is the lowest denomination mentioned; thirds must, therefore, be reduced to sixths. That is, 2 must be reduced to an equivalent fraction, whose denominator is 6. (See Art. 58.) OBs.-That fraction is of the lowest denomination whose denominator is the largest. To ascertain how many of the smaller fractions make one of the larger, we divide the denominator of the smaller by the denominator of the larger; 6 contains 3 twice, or - contains 1 twice. That is, 2 sixths make x. Were the nllmerator a unity, we should now have the fraction required. But since it is greater, if we multiply it by 2, we have -, an equivalent fraction. Hence the RULE Divide the denominator of that fraction whose denominator is a multiple of each of the other denominators, first by the denominator of one of the other fractions; multiply its numerator into the quotient, and write the product over the denominator thus divided; and so continue to do, until'the fractions are all reduced to the same denomination, or to a common denominator. 2. Add +-TA+. Oper ttion. 24-12X2-4 Then 4_4 + 4 Ans 24-i- Gxl=4 The T+-2 4 9T ILLUSTRATION OF FRACTIONS. 85 3. dd, i, T 6 Ans. o. 4. Add 1, 1, 2 Ans. 1. Art. 84.-Addition of fractions when no denominator is a multiple of each of the other denominators. Add 2, 4, 2. In this example we have no common denominator, or denomination given, to which each of the fractions may be reduced. If we multiply all the denominators together, we shall obtain a common multiple of all the denominators, or a denomination to which each of the fractions may be reduced. 3X4X5==60. We have now to reduce each of the given fractions to 60ths. This may be done by the foregoing Rule. It will be observed, that the quotient of 60 divided by any one of the denominators is the product of all the other denominators. Hence'the product of all the other denominators shows how many 60ths make one inithe denomination of the fraction to be reduced; we may, therefore, adopt the common RULE. Multiply all the denominators together for a common denominator, and each numerator into all the denominators except its own for a new numerator. Operation 1st. Denominators, 3X 4 X 5-60 cor. denominator. Then, 60-3,x2=40) 60 — 4x 1-=15 > new numierators. 60-5 X2==24) Operation 2d. Denominators, 3 X 4 x 5=60 corn. denominator. 1st numerator, 2x4x5=40 2d " 1x3X5=15 new numerators. 3d " 2x3x4=24) Then A 1- + 2 4 - 9 = 19 Ans. EXAMPLES. 1. Reduce 4, i, -, and 3- to fractions having a common denominator. Ans. 6 44, 16 2 344, 304 2. Reduce 2, -,, and A to fractions having a common denominator. Ans. 36, 96 9, 6930 8316 3024 3. Add together i, 4, and - Ans. 2J24-4 = 21 OBs. 1.-Reduce the fractions to a common denominator, find new numer ators, and add them together. -- 86 ILLUSTRATION OF FRACTIONS. 4. Add 1 I, 3 and - n. 27. 5. Add together 6 of - and 1 of 3. Ans.. — 5. OBs. 2.-Compound fractions must be reduced to simple fractions. 6. Add ~ of 96 and 7 of 14' together. Ans. 4411 3 s 2 toget.8 Y 1 1 7. Add together ~ of I and 2 of 1. Ans. 112. 8. Add together 6 and 7 of -9 and 4 of 1 and 71 Ans.'143 21. OBS. 3.-Mixed numbers may be reduced to improper fiactions, or the Lrctional parts may be reduced to a common denominator, and added as ih, he foregoing examples. If their sum amount to an integer, add it to ti, whole numbers. 9. Add together 143 and 163. Operation. 42 3 —1 and 2 -8; then 1 +T= 11 2 -4 - I y — Y- -2iv. 16 s 1a ^s We find the common denominator to be 1'T2 As. 12, and the new numerators to be 9 and 8, which when added are 7 =1 -l Write the - under the fractions, and carry 1 to the whole numbers. 10. Add together 173, 18k, 19. Ans. 55.1 11. A grocer sold the following parcels of sugar, viz: 162 lbs., 191, 133, 20-, 251, 301, and 111 lbs. How many 4 8 1' 6' 4 pounds did he sell in all? Ans. 13643. Subtraction of Fractions. RULE. Art. 85 — Prepare the.fractions as in Addition, and subtract the less numerator from the greater, and under the difference write the denominator. EXERCISES, 1. From - take t. Ans. 1. 2. From T take?. Ans. 2. 3. From 17 take 1. Ans. 6. 4. From 2 take a. Ans. -4-. QUESTION.-What i the rule for the subtraction of fractions? ILLUSTRATION OF FRACTIONS. 87 5. From 6- take. Ans. 56-. 6. From 5 of 1 take 5 of 2. Ans. 14. 7. Add together 4 and 5, and from their sum subtract 2 of 9 2 An s. 1 -4-. 8. A. owns I of of av e B. n of How much 3 4 8 75 greater is A.'s share than B.'s? Ans. 5. 9. Subtract 131 from 152. 15i 3 2_- 9 8 Having reduced the fractions 139-2 I to a common denominator, and ~ ii A. found new numerators, as in Ad- T- 2 n dition, we have 2 to be taken from 2. We therefore borrow a unit, and say 9 from {2 and add 3, the remainder, to 8, the numerator of the subtrahend. 3+8=1-, which we write under the fractions, and carry 1 to 13, the whole number, which makes 14; and 14 from 15, and 1 remains.. The answer, then, is 1I. 10. A man bought a horse for I of 2- of $150, and sold him for i of 7 of: of $60. Did he gain or lose, and how much? Ans. $40 gain. To find the least Common Multiple. Art. 86.-The common denominator found by the preceding rule, is a common multiple of the denominators of the given fractions; for every product must be divisible by all its factors; but it was not the least common multiple. 1. What is the least common multiple of 4, 6, 8, 10? 4 X 6 X 8 X 10=1920. 1920 is evidently a common Operation. multiple of 4, 6, 8, and 10, be2)4, 6, 8, 10 cause they are its factors; but it 2)2, 3, 4, 5 is not 1he least common multiple. 1, 3 x 2 x 5x 2 x 2-=120. We find, also, that each of these numbers is a multiple of 2, or that 2 is a prime factor of each. Dividing by 2, we find the other factors, which are 2, 3, 4, 5. Again: As the quotient, 4, is a multiple of 2, we may substitute for it 2, one of its factors; and as we employ the other factor for a divisor, we erase the other quotient, 2. We now have 3, 2, 5, undivided numbers, which are prime factors of the dividends 6, 8, 10; the other prime factors are the divisors. If now we multiply to 88 ILLUSTRATION OF FRACTIONS. gether these undivided numbers and the divisors, we shall have a combination of all the prime factors of each dividend, and consequently it must be divisible by them. Thus the divisors 2 and 2 are the factors of 4, the first dividend, 2 x 2=4. If 4 be divisible by 4, then 2, 3, and 5 times 4 must be divisible "by 4; also 3, the first undivided number, is a factor of 6, the seond dividend; and 2, the first divisor, is the other factor, 3x2=6. If 6 be divisible by 6, then 2 and 5 times 6 must be divisible by 6. The same may be said of 2 and 5, the other undivided numbers. Hence it appears, that the product of the continued multiplication of the remainders and divisors is divisible by the several dividends; and by examining the operation, it will be found to be the least number which can be divisible by them; for all repetition of prime factors beyond what is necessary to produce each dividend, is avoided. Therefore, to find the least common multiple of two or more numbers, we have the following RULE. Write the numbers in a horizontal line; divide them by the least prime number that will measure two or more of them; write the quotients and undivided numbers in a horizontal line under the given numbers; divide the numbers in this second line, in the same manner. Thus continue to divide until the quotients and undivided numbers are all prime to each other. The product of the continued multiplication of the divisors and undivided numbers will be the least common multiple required. OBs. l.-We divide by any number that will divide two or more of the numbers, to find first the least common measure of two or more. EXERCISES. 2. What is the least common multiple of 3, 4, 9 and 12?'Ans. 36. 3. What is the least number which can be divided by 7, 8, 10, and 12, without a remainder? Ans. 840. 4. What is the least common multiple of 7, 14, 28, 35? Ans. 140. 5. What is the least number which can be divided by the nine digits without a remainder? Ans. 2520. QUESTIONS.-1. How is the least common multiple of two or more numbers found? 2. Why do you divide by any number that will divide two or more without a remainder? ILLUSTRATION OF FRACTIONS. 89 6. Reduce, 2, 5, to equivalent fractions having the least common denominator. 214 5 6 60 being the least common mul-2x5x3x260 tiple of 4, 5, ana 6, it is, therefore, the least common denominator of the fractions 4, 2, 5. The remaining part of the process is performed by the Rule under Art. 83. Still other illustrations may be given. The value of the fraction -, is three-fourths of a unit. That is, the unit is divided into 4 parts, and the fraction expresses - of them. If we divide the unit into 60 parts, and wish to express the same part of a unit, we must take -- of 60. If we divide 60 by 4, we have onefourth; if we multiply one-fourth by 3, we have three-fourths, 60-. 4=15, and 15X3=45. Now 3 is.three-fourths of 4, and 45 is three-fourths of 60; therefore 4=_, an equivalent fraction. Operation. 60 4x3=45 60 —5X2=24 Then 45 24 0 Ans. 60-6 X5-=50 OBs. 2.-By this process the fractions are all reduced to the same denomination. 7. What is the least common denominator of 1, 5, and -? Ans. 20 10 24 1 s, io~-, i,!wo 8. Reduce I, -9, -7, to fractions having the least common denominator.. Ans. 352 324 231 ~~3denominator. 2n. Y76 31V 196, 9. Reduce a, 2, l, and I, to fractions having the least common denominator. Ans. -2 64 56, 46O 2 10. A merchant buys 5 pieces of cloth. The first contains 403 yards; the second, 27~; the third, 347; the fourth, 43 -; and the fifth, 39- yards. How many were there in the whole? Ans. 185T-. 11. Which is the greater fraction, I - or1. Ans. ^ is greater by T-4. OBs. 8. —If the denominator of either of the given fractions be a multiple of each of the other denominators, it will be the least common denominator. QUESTION.-3. How are fractions of different denominators reduced to equivalent fractions having the same denominator? 8* 90 DECIMAL FRACTIONS. 12. Reduce;, and -7 to equivalent fractions having the least common denominator. Multiplying the first by 4 and the second by 2, we have the answer required. ixF4= x2 2=1 Ans. h], l27. 34 2- 3 2 13. Which is the greater fraction, 5y or 0? Dividing the terms of -24 by 2, we have -, and 1- - -o=3-3S' the Answer. Fractions whose denominators are 10, 100, or 1000, etc. form a very important class of fractions, and will be treated under a separate head, called DECIMAL FRACTIONS. Art. 87.-The term decimal signifies tenth. It is derived from the Latin word decem, which signifies ten. It is, therefore, applied to all fractions whose denominator is 10, or 1, with any number of ciphers. If a dollar be divided into ten parts, one of these parts, being worth ten cents, is one tenth of a dollar. If the dollar be divided into one hundred parts, one of these parts is the one hundredth part of a dollar. It is, nevertheless, a decimal fraction, because 100 is the product of 10's. The same may be said of a thousand, or ten thousand. A fraction is always known to be decimal, if its denominator be ten, a hundred, or a thousand. The denominator of a decimal fraction is not always expressed, but it can always be ascertained by the numerator. If it contains but one figure, the denominator is ten; if two, it is a hundred, etc. It is always one, with as many ciphers annexed as the numerator has places. When the denominator is not expressed, the fraction is distinguished from a whole number by a period placed at the left of it. [The period is called the separatrix.] Example:.5,.50, is read five tenths, fifty hjdredths, as though they were written -5, 5-0. If the nmerator have not so many places as the denominator has ciphers, supply the defect by prefixing ciphers, thus: for —, write.05; T-Ao, write.005. Ciphers placed at the right hand of a decimal do not affect its value, QUESTIONS.-4. What does the term decimal signify? 5. From what derived? 6. To what applied? 7. How is a fraction known to be decimal? 8. Is the denominator always expressed? 9. How, then, can it be known? DECIMAL FRACTIONS 91 as 5 -, 0% are the same in value; for while the addition of the cipher indicates a division into parts ten times smaller than the preceding, it makes the decimal express ten times as many parts. Thus, 5 tenths denotes 5 parts of a unit which is divided into 10 parts; and 50 hundredths denotes 50 parts of a unit which is divided into 100 parts. It is, therefore, plain, that the value is not altered, since 5 is half of 10, and 50 is half of 100. The value of a decimal depends upon its distance from the unit's place. As whole numbers increase from the unit's place towards the left in a tenfold proportion, so decimals, in the same ratio, decrease from the unit's place towards the right hand; as will appear from the following T B L E. o 2 1 2 3 4 5 6 4.-4 o. C n'" o #12 - 0 6 54 32 123 456 7 a whole number or decimal, takes its value from the unit's place. If it be in the first place on the right of units, it is tenths; if in the second, it is hundredths, etc. Consequently, every decimal will have for its denominator 1, with as many ciphers as the decimal is places distant from the unit's place; thus, 2 in the Table is2; 3 is - 4; 4 is, etc. Art. 88.-The manner in which decimal fractions are produced, and the relation they bear to whole numbers, may be seen by the following formula: Qvxzoxs.-10. On what does the value of a decimal depend? 11. In what proportion do decimals decrease from the unit's place towards the right? 12. From what does each decimal figure take its value? 13. What is the value of the first figure on the right of units 14. What effect have ciphers placed at the right hand of a decimal? 15. What effect at the left? 92 DECIMAL FRACTIONS. 1000 10 =100 NOTE.-Tt wiU be noticed that the deci100+ 10 ~ 10 i mal point is the same as the straight line U100 1U 0 =10( used Art. 40. Divide I by 10. 10 -10 = 1 Operation. 110)11=.l Conceive the straight line employed to cut 1. — 10~i o.1 of figures from the right of the divisor and L- *.'10= -Tt.01 dividend to be contracted to a point, and.'i i'\,\.0 you have the origin and use of the sepaT<.~ —10 TT —.001 ratrix. Thus it appears, that from any given place in whole numbers to any given place in decimals, is a regular descending series, formed by a uniform divisor. The right-hand place is the quotient of the left divided by 10. The first decimal place is the quotient of the unit's place divided by 10, and is called the tenth's place. The decimal point, therefore, occupies a positior between the unit's place ad its quotient. The second place is the quotient of the tenth's place divided by 10, or the unit's place divided by i00, and is called the hundredth's place. Thus decimal fractions, like whole numbers, have a local value, and are. subject to the same law of increase from the right hand towards the left. As 1, in the place of tens, is equal to 10 in the unit's place, so 1 in the place of units is equal to 10 in the place of tenths. From this circumstance, we may know the value of those parts of the unit contained in the numerator, although the denominator be not expressed. This proety of a decimal fraction also distinguishes it from a vulgar ction, for there is no place on either side of the unit where the numerator of a vulgar fraction can be placed, which will give name to the fraction; its denominator must, therefore, always be expressed. Although ciphers placed at the right hand of a decimal fraction do not affect its value, yet, placed at the left, they diminish it in a tenfold proportion, by removing the significant figure so much farther from the unit's place. Thus,.5.05.005 express different values, viz.-.5 is -,.05 is o,.005 is 0 5. Write denominators to the following decimals:.5;.25;.026;.3245;.56783;.789024. Write the following without their denominators. 1. Twenty-five hundredths. Ans..25. 2. Four hundred and fifty-two thousandths. Ans..452. 3. Five hundred and sixty ten thousandths. 4. Sixty-two hundred thousandths. 5. Forty-five millionths. 6. Eighty-seven billionths. DECIMAL FRACTIONS. 93 7. Ninety-eight trillionths. 8. Twenty-five, and four thousandths. As whole numbers are written, units under units, tens under tens, from right to left, so decimals are written tenths under tenths, from left to right. EXAMPLES. 1. Write 2 tenths; 3 hundredths; 4 thousandths; 6 ten thousandths..2.03.004.0006 2. Write twenty-nine thousandths; three hundred and fourteen thousandths; five ten thousandths, and sixty-seven millionths..029.314.0005.000067 3. Write five tenths; five hundredths; fifty thousandths, and forty-nine; one hundred thousandths, and sixteen thousandths. 4. Write forty-five and five tenths; six hundred and fortyfive and four thousandths; twenty-nine and four thousandths; sixty-seven and forty-seven thousandths. 5. Write four hundred and fifty-three, and fifty-seven ten thousandths; five thousand and five hundredths; twenty-four and three millionths; thirty-six and eighty-two billionths. ADDITION OF DECIMALS. Art. 89.-1. Write one hundred and one tenth; twenty and two hundredths; five units and five thousandths, and add them together. Operation. As whole numbers can only be added 100.1 by writing them in their proper places and 20.02 uniting those of the same name; so deci5.005 mals, when written tenths in the place of 125.125 Ans. tenths, hundredths in the place of hundredths, etc., are added by uniting those 94 DECIMAL FRACTIONS. of the same name or denomination. The amount, both in decimals and whole numbers, takes its name from the lowest, or right-hand place of the numbers added: thus, 1 hundred, 2 tens and 5 units, when added, are read 125 units; and one tenth, 2 hundredths and five thousandths, when added, are read, 125 thousandths. Decimal fractions may also be added and illustrated in the same manner as vulgar fractions. 2. Add two and five tenths; four and six hundredths; seven and three thousandths. 2.5 =, and 4.06-406, and 7.003=- 7. Then x x<100-=25, and 06X 10-o06 OBs.-Multiplying the terms of a fraction by the same quantity does not alter its value. (See Art. 61.) The fractions added: 2500 4060 _ 7O00313O 5 L3 63 153 Ans. f 00bo To6-iS- -r 1 60 -d T 0 1000 The same, added by decimal fractions: 2.5 = 2.500 4.06 = 4.060 7.003= 7.003 13.563=13.563 Ans. From the foregoing it is evident, that decimal fractions are reduced to a common denominator by writing tenths in the place of tenths, and hundredths in the place of hundredths, and supposing those decimal places, which are deficient, to be supplied by ciphers. Applying the decimal point to the amount, is equivalent to dividing itby its own denominator, which we have seen is the denominator of the lowest of the given decimals, or that decimal'whose denominator is the largest. But the decimal places in the numerator of a decimal fraction, are equal to the number of ciphers in its denominator, the denominator being understood; therefore, addition of decimals may be performed by the following QUEsTIONS —16. How is the first decimal place produced? 17. The second, third, &c.? 18. How are decimals to be added, written? 19. From what does the amount take its name? 20. Applying the decimal point is equal to what? 21. How are decimal fractions reduced to a common denominator? DECIMAL FRACTIONS. 95 RULE. Place the numbers, tenths under tenths, hundredths under hundredths, etc.; or, so that the decimal points may stand directly under each other. Add as in whole numbers; observing to point off as many places for decimals in the amount, as will be equal to the greatest number of decimals in any of the given numbers. EXAMPLES. (1) (2) (3) 2345.6 Add 459.51 79.01 1987.51 371.62 891.67 3456.712 129.03 137.79 21098.6543 1271.007 1239.812 16723.24567 1090.215 2671.927 65431.002001 3321.382 4. Add thirty-five and four tenths; five hundred twenty. nine and seven millionths; sixty-nine, four hundred and sixtythree thousandths; two hundred, sixteen and two hundredths; seventy-seven, nine hundred and two tenths. Ans. 1827.083007. 5. Add forty-nine and sixty-seven hundredths; six hundred seventy-nine, two hundred seventy-five thousandths; one thousand four hundred, fifty-five thousandths, nine hundred and ninety-nine millionths. 6. Add 249.39; 6712.9123; 6.3219; 2739.235; 5.671; 723.2674; 926.679; 72.601. 7. Add.7+9.2+.321+279. +4.67+349.2+3.956. 8. Purchased of one man 325.5 lbs. of beef; of another, 175.75; of another, 178.028; what was the amount? 9. I receive of A. $183.25; of B. $138.89; of C. $372.218; of D. $88.99; of E. $137.29; what is the amount of the whole? 10. Add $59.67; $158.355; $375.752; $167.375; $567.756. 96 DECIMAL FRACTIONS. SUBTRACTION OF DECIMALS. Art. 90.-1. From three and two tenths, take one and five tenths. Operation. 3.2 1.5 1.7 Ans. Because five tenths cannot be taken from two tenths, we borrow I from the unit's place, which, reduced to tenths, equals 10 tenths;'o TO 12 then 1 -6 = =.7. Lastly, 1 from 2, and 1 remains. Again, three and two tenths is the quotient of 32 divided by 10, (see definition of a mixed number, Art. 54;) therefore, 3.2=32, and 1.5=15; then 2 15 = 1 l7 Ans., as before. Pointing off the remainder is dividing it by its own denominator. Hence the RULE. Write the numbers and point the result, as in Addition of Decimals, and subtract as in whole numbers. (2) (3) (4) (5) From 429.67 87.02 359.76 2029.5 Take 319.76 59.2 126.571 1718.279 109.91 311.221 6. From two hundred and sixty-nine and three tenths, take fifty-seven and thirty-nine hundredths. Ans. 211.91. 7. Take twenty-four thousandths from nine hundredths. Ans..066. 8. Take sixty-five millionths from five tenths. 9. From three hundred seventy-five thousand and three tenths, take two hundred forty-nine and thirty-nine one hundred thousandths. Ans. 374751.29961. 10. From 361.2 take 276.75. 11. From 456.35 take 27.356. 12. From 5678.0002 take 3980.96715. QUESToNS. —22. How are decimals to be subtracted, written? 23. How can ftle tenths be taken from two tenths? 24. What is done with the unit borrowed? DECIMAL FRACTIONS. 97 MULTIPLICATION OP DECIMALS. Art, 91 —1. Multiply three and five tenths by five tenths. Operation. 3.5.5 1.75 Answer. The product of tenths into tenths is hundredths: -~ x -= =o=.25. The product of tenths into units is tenths: 3 X = —= 1.5. The sum of the product,.25+1.5=1.75 Ans. Again, 3.5= 3 and 5 x 5= 1 7 5 =1.75 Ans., as before. The value of the product is the quotient of its numerator divided by the denominator. Hence the figures cut off from the right of the numerator are equal to the ciphers in the denominator; but the ciphers in the denominator of the product, it will be perceived, are equal to the decimal places in both factors; therefore the multiplication of decimals may be performed by the following RULE. Multiply as in whole numbers, and point of as many places for decimals in the product as there are decimal places in both factors. If there are not so many places, supply the defect by prefixing ciphers. EXAMPLES. 2. Multiply five hundredths by five tenths. Operation..05.5.025 Answer. The product of tenths into hundredths is thousandths. In this example, the tenth's place in the product is wanting; we must, therefore, supply it by prefixing a cipher. QuEsnTIoNs.-25. What is the product of tenths into units? 26. Of tenths into tenths? 27. What is the rule for the multiplication of fractions? 28 What is the value of the product? 98 DECIMAL FRACrIONS. 3. Multiply 49.5 by 3.2. 3.2 99.0 1485. 158.40 4. Multiply 569.39 by 27.05. 5. Multiply 6.791 by 2.67. 6. Multiply 549.05 by 35.25. 7. Multiply six hundred and seventy-five by twenty-sevon and three tenths. 8. Multiply sixty-seven thousand by three hundredths. 9. Multiply 34.56 by 1.3. 10. Multiply 674.49 by 37.16. 11. Multiply 5648 by 6.78. 12. Multiply 7864 by 467. 13. Multiply fifty-seven and three tenths by twenty-nine. 14. Multiply thirty-seven thousand by three hundredths. 15. Multiply fifty thousand and seven tenths by four hun. dredths. DIVISION OF DECIMALS. Art. 92. —1. Divide twenty-five hundredths by five tenths. By Vulgar Fractions By Decimal Fractions. Operation. Operation. Proof. )5 (_5=. 5 A ns..5).25.5 When no remainder will arise.5 Ans..5 from the division, the terms of the.25 dividend may be divided by th WVVe have seen that the decimal corresponding terms of the divi-plcs in the product of any two sor. (See Art. 78.) It will b factors are equal to the decimal seen that the decimal point im- paces in both those factors. Tie plies a division of the numerator divisor and quotient are factors of of the quotient by its own denon- t dividend; therefore the d ciinator. m;al places in th quotient and divisor, taken together, must be equal to the decimal places in the dividend. Hence the RULE. Divide as in whole numbers, and point off so many placesfor decimals in the quotient, that the decimal places in the quotient DECIMAL FRACTIONS. 99 and divisor, taken together, shall equal the decimal places in the dividend; or, so many as the decimal places in the dividend exceed those of the divisor. If there are not so many, supply the deficiency by prefixing ciphers. OBS. 1.-The above rule may be illustrated by reference to the operation of the preceding question by Vulgar Fractions, thus: the ciphers in the denominator of the divisor and quotient are equal to the ciphers in the denominator of the dividend; but the decimal places in the numerator of a decimal fraction are equal to the ciphers in its denominator; therefore the decimal places in the numerator of the quotient and divisor, taken together, must be equal to the decimal places in the numerator of the dividend. 2. Divide five tenths by twenty-five hundredths. Operation..5 = 5 = — 0=.50; then.25).50(2 Answer..50 OBS. 2.-Annexing a cipher to a decimal fraction multiplies the terms of the fraction by 10, and, therelore, does not'alter the value. (See Art. 61) Whenever the decimal places il the divisor exceed those of the dividend, annex a cipher or ciphers to the dividend; this reduces it to the denomination of the divisor. 3. Divide three hundred and sixty-nine thousandths by nine. Operation. 9).369.041 Answer. The necessity of prefixing a cipher to the quotient will be more readily seen by the following: 3o~' —9_=y-4o. If we remove the denominator of the quotient, and prefix the decimal point to the numerator, it will then be 41 hundredths, which is not its true value; but, by placing a cipher between the decimal point and the left-hand figure, the right-hand figure of the quotient will be made to occupy the thousandths' place, which will denominate the parts into which the unit is divided, or show their true value. Prefixing a cipher, therefore, divides the fraction, by multiplying its denominator. 4. Divide 36.72 by 18. 5. Divide 21.7 by 7. 18)36.72(2.04 7)21.7 36 3.1. —----— __ 3.1 72 72 Quis-roNs.-29. What is the rule for the division of decimals? 30. How is the quotient pointed off? 31. Illustrate the rule. 100 FEDERAL MONEY. 6. Divide 2.17 by 7. Ans..31. 7. Divide.217 by.7. Ans..31. 8. Divide.217 by 7. Ans..031. 9. Divide one hundred and seventeen and nine tenths by nine tenths. Ans. 131. 10. Divide four hundred fifty-six and three hundred thirtythree thousandths by three hundredths. 11. If three hundred fifty pounds of beef cost twelve dollars twenty-five hundredths, what cost one pound? Ans..035. 12. If 565.05 pounds cost 25.42725 dollars, what will one pound cost? Ans..045. Art, 93.-From the foregoing it appears that decimal fractions are like whole numbers in the following particulars: 1. The figures that compose them have an appropriate place to occupy, from which they take their value. 2. They take their name from the lowest right-hand place. 3. They increase in value from the right-hand place. 4. They can only be added by being first reduced to the lowest denomination. 5. They are reduced by writing them in their proper place. They are unlike whole numbers in the following particulars: 1. They diminish in value from the unit's place. 2. A cipher, placed at the left hand, diminishes their value. 3. They may be written and treated as common fractions. FEDERAL MONEY. Art. 94.-FEDERAL MONEY is the coin of the United States. Its denominations are eagles, dollars, dimes, cents, and mills. From the above examples and illustrations in Decimal Fractions, we have seen that a decimal is the division of the unit into tens, and that from the unit's place towards the right hand it decreases in a tenfold proportion. If we examine the denominations of Federal Money, we shall find that all bear a decimal relation to the dollar, which is considered the unit. This will be seen by the following ADDITION OF FEDERAL MONEY. 101 TABLE. 10 Mills =1 Cent. 10 Cents =1 Dime. 10 Dimes =-1 Dollar. 10 Dollars=1 Eagle. OBs.-The eagle is a gold coin, the dollar and dime are silver coins, the cent is a copper coin. The mill is only imaginary, there being no coin of that denomination. The dime being 1 tenth of a dollar, it occupies the first, or right-hand place from the dollar; thus, 0.1. The cent, being 1 tenth of a dime, and consequently 1 hundredth of a dollar, occupies the second place, or place of hundredths; thus, 0.01. The mill, being 1 tenth of a cent, and consequently 1 thousandth of a dollar, occupies the third place, or place of thousandths; D. D. C. M. thus, 0.001. Placing them together, 1 1 1 1. This may be read, one dollar, one dime, one cent, and one mill; or, one dollar, eleven cents, and one mill-as eleven cents is equal to one dime and one cent. The same may be said of eagles and dollars; thus, 25 dollars may be read, 2 eagles and 5 dollars, since 20 dollars are equal to 2 eagles. Write 4 eagles, 5 dolE. D. D. C. M. lars, 8 dimes, 3 cents, 5 mills-4 5 8 3 5. This may be read, 4 eagles, 5 dollars, 8 dimes, 3 cents, and 5 mills; or, 45 dollars, 83 cents, and 5 mills. Hence, it is evident that the denominations in Federal Money are dollars and decimals of a dollar, and may be treated as Decimal Fractions. Federal Money is denoted by this character ($) placed before the figure. ADDITION OF FEDERAL MONEY. RULE. Write the denominations, add and point the result as in Addition of Decimals. EXAMPLES. Art. 95.-1. If I buy a bushel of wheat for $2.25; a bushel of corn for *1.32; four yards of cloth for $14.285; how much do I pay for the whole? QUzsTINSs.-1. What is Federal Money? 2. What are its denominations? 9* 102 SUBTRACTION OF FEDERAL MONEY. 2.25 OBs.-The scholar will do well to 1.32 turn now to the rule for reducing a 14.285 vulgar fraction to a decimaL $17.855 Ans. 2. Bought 8 yards of cloth for $16.254; a pair of shoes for 874 cents; a hat for $4.33; a whip for 42 cents; a knife for 37- cents. How much did I pay for the whole? Ans. $22.255. 8. Bought a cart for $17.62; a wagon, $62~; a plough, $7.48; 4 rakes, $1.26; 3 hoes, $2.15; a pitchfork, 87 cents. How much did the whole cost? Ans. $91.88. 4. Purchased a barrel of flour for $9.25; 4 pounds of tea, $2.08; 2 gallons of molasses, 64 cents; 3 pounds of raisins, 371 cents; 9 pounds of sugar, $1.211; 8 yards of calico, $2.234. What is the amount of the whole? Ans. $15.795. 5. Add forty dollars, sixty-seven cents and three mills; six hundred seventy-nine dollars, twenty-five cents and seven mills; one thousand and four dollars, five cents, and five mills; nine hundred, ninety-nine dollars, thirty-nine cents and nine mills. Ans. $2723.384. SUBTRACTION OF FEDERAL MONEY. RULE. Write the numbers, subtract and point the result as in Subtraction of Decimals. EXAMPLES. Art. 96 —1. A man bought 50 bushels of wheat for $125.50; sold it for $145.75. How much did he gain? Ans. $20.25. 2. Bought 26 bushels of oats for $8.49; sold the same for $8.94. How muct did I gain? Ans. $0.45. 3. Purchased a horse for $92; lost on the sale of him, $15.25. For how much did I sell him? Ans. $76.75. 4. Bought 2 barrels of flour for $22.50; but, it being damaged, I am willing to sell it at $4.25 less. What must I receive for it? Ans. $18.25, MULTIPLICATION OF FEDERAL MONEY. 103 5. Bought 8 yards of cloth for $36; gave a $50 bill. What must I receive in change? Ans. $14. 6. Subtract 1 mill from $333. Ans. $332.999. 7. Subtract half of a cent from $100,000. 8. Bought a wood lot for $879; sold the same for $1000.81. How much did I gain? 9. If a man's wages in a year amount to $1434, and he spends $928.45, how much does he save at the end of the year? 10. How much must be added to $32.50 to make $1000? MULITIPLICATION OF FEDERAL MONEY. RULE. Write the numbers, and point the product as in Multiplication of Decimals. EXAMPLES. Art. 97.-1. How much will six pairs of shoes cost, at 11.37~ a pair? Operation. 1.375 6 Ans. *8.250 It will be seen that the operation is the same as in simple numbers. The product will always be in the lowest denomination of the given sum, until distinguished by points. 2. What will 9 sheep cost, at $3.75 each? Ans. $33.75. 3. How much must be paid for 45 bushels of corn, at $1.37 per bushel? 4. What will 38 pounds of sugar cost, at 13- cents per pound? Ans. $5.13. 5. What will 3 dozen hats cost, at $4.75 each? Ans. $11l. 6. What will 75 dozen eggs cost, at 15~ cents a dozen? Ans. $11.625.'. How much will a man spend in a year, if he spend 12i cents a day? 8. What will 55 yards of broadcloth cost, at $3.87r per yard? 104 DECIMAL FRACTIONS AND FEDERAL MONEY. DIVISION OF FEDERAL MONEY. RUL E. Write the numbers, and point the quotient as in Division of Decimals. EXAMPLES. Art. 98.-1. Bought 8 bushels of wheat for $17.92. How much was it per bushel? Ans. $2.24. 2. Bought 9 pounds of tea for $3.37~. What was it ner pound? Operation. 9)3.375.375 Ans. 3. Bought tea to the amount of $3.374, at 374 cents per lb. What quantity did I buy? Ans. 9 lbs. 4. Bought 141 bushels of corn for $21.75. How much was it per bushel? Ans. $1.50. 5. If a man pay $38.4371 for 201 casks of lime, how much was it per cask? Ans. $1.871. 6. Bought 6 yoke of oxen for $450. What was paid for each ox? Ans. $37.50. SUPPLEMENT TO DECIMAL FRACTIONS AND FEDERAL MONEY. Art. 99 —1. Purchased 49.5 pounds of butter of A., at 12- cents per pound; 37.51 pounds of B., at 183- cents per pound; 155.05 pounds of C., at 20 cents per pound. How many pounds did I buy, and what was the cost of the whole? Ans. $44.23+. Ans. 242.06 pounds. 2. When butter is worth 18 cents 4 mills per pound, how many pounds can be bought for $671.60? Ans. 3650 pounds. 3. At 9 mills per yard, how many yards of tape can be bought for 45 dollars, 81 cents, 9 mills? Ans. 5091 yds. 4. If 5091 yards of tape be worth $45.819, what is 1 yard worth? Ans. 9 mills. BILLS OF PARCELS. 105 5. What will 629.21 feet of boards cost, at $20.18 per thousand? 6. What will 36 bushels 9 tenths of corn amount to, at 1 dollar 5 tenths per bushel? Ans. $55.35. 7. If corn be worth 3 and 5 tenths as much as potatoes, which are worth 25 hundredths of a dollar per bushel, and rye 5 tenths more than corn, and wheat 2 and 4 tenths more than rye, what is the value of wheat? Ans. $3.15. 8. Bo ht 4 cords of wood for $12.28; 15 pounds of beef for $1.2 IHow much do I pay for the whole, and how much more for the wood than for the beef? 9. Bought 28 bushels of potatoes, at 28 cents abushel.; 451 bushels of apples at $1.124 per bushel. How much did the whole cost, and how much more did the apples cost than the potatoes? A. i 59.02'7 whole cost. $43.347l difference. BILLS OF PARCELS. Art. 100 —It is customary for the merchant, when he delivers goods, to give also a bill of the articles, and their prices, with the amount cast up. Such bills are called Bills of Parcels. Concord, May 18, 1837. Mr. JOHN WORTKY, Bought of PETER TRUSTRUM, 5! bushels of oats, at $0.63 per bushel................$3.465 1A bushels of wheat, at $1.50 per bushel............... 18.750 71 cords of woodat $3.45 per cord................... 25.875 Received Payment, $48.090 PETER TRUSTRUM. Mr. BENJ. SAVAGE, r. BEN. SAVAGE, Bougnt of JOSEPH EASY, 121 yards of broadcloth, at $3.87 per yard............ 55 casks of nails, at $5.50 per cask................. 112 pounds of iron, at 91 cents per pound.............. 16 pounds of steel, at 18 cents per pound.............. 25 pounds of lead, at 9~ cents per pound.............. I hogshead of sugar, (8 cwt.) at $9.24 per cwt....... 2 boxes of glass, at $7.50 per box............ $191.810 106 COMPOUND NUMBERS. —MONEY. COMPOUND NUMBERS. Art. 101.-ALL preceding numbers have been simple; that is, numbers whose sum may be expressed by a certain number of units of one and the same kind, as 256. By reference to Notation, it will be seen, that this expression is 2 hundreds, 5 tens, and 6 units, which, instead of being written separately, are expressed as two hundred. and fifty-six units, which are said to be of the same denomination. But if a menz lave 10 pounds and 2 shillings, he cannot add them so tPo make 12 pounds, nor 12 shillings, but they must be expressed separately. So if a man travel 3 miles and 25 rods, the sum is neither 28 miles, nor 28 rods; but they must, in like manner, be expressed, the miles and the rods each by themselves. So of feet and inches, barrels, quarts, and pints. These are called different denominations. Hence, compound numbers are those which treat of quantities consisting of different denominations. TABLES OF COMPOUND NUMBERS. MONEY. Art. 102.-Federal Money. 10 mills make.................................1 cent, ct. 10 cents ".................................1 dime, d. 10 dimes "..1 dollar, dol. 10 dollars ".................................1 eagle, e. The above denominations of Federal Money are authorized by tli laws of the United States; but, in the transaction of business in New England, we seldom hear any of them named but dollars and cents. " A coin is a piece of money stamped, and having legal value. The coms of the United States are, three of gold; the eagle, half-eagle, and quarter-eagle; five of silver, the dollar, half-dollar, quarter-dollar, dime, and half-dime; and two of copper, the cent and half-cent. Of the small foreign coins current in the United States, the most common are the New England four-pence-halfpenny, or New York sixpence, worth 6*,cents; and the New England ninepence, or New York shilling, worth 12j cents. The value of the several denominations of English money is different in different places. A dollar is reckoned at 4s. 6d. in England; 5s. in Canada; 6s. in New England, Virginia, and Kentucky; 8s. in New York, Ohio, and North Carolina; 7s. 6d. in Pennsylvania, New Jersey, Delaware, and Maryland; and 4s. 8d. in South Carolina and Georgia." Art. 103.-English Money. 4 farthings, qrs., make.......................1 penny, d. 12 pence "........................1 shilling, s. 20 shillings "........................1 pound, 1. or ~ TIME.-WEIGHTS. 107 Art. 104, —Time. Time is the measure of duration or existence. 60 seconds, s., make................................. 1 minute, m 60 minutes "................................1 hour, h. 24 hours "................................1 day, d. 7 days ".......................1 week, w. 3654 days, or 365 d. 6 h., or 52 w., make..........1 year. yr. "The year is commonly divided into 12 months, as in the following table, called calendar months: Months. Days. M. D. M. D. M. D. 1 January, 31 4 April, 3017 July, 31110 October, 31 2 February, 28 5 May, 3i 8 August, 31 11 November, 80 3 March, N4 6 June, 3q 9 September, 80112 December, 31 Another day is added to February every fourth year, making 2q days in that month, and 366 in the year. Such years are called Bissextile, or Leap Year. To know whether any year. is a common or leap year, divide it by 4; if nothing remain, it is leap year; but if 1, 2, or 3 remain, it is 1st, 2d, or 3d after leap year. The number of days in the several months may be called to mind by the following verse: Thirty days hath September, April, June, and November; All the rest have thirty-one, Excepting February alole, Which hath twenty-eight, nay more, Hath twenty-nine one year in four. The true solar year consists of 365 days, 5h. 48m. 57s., or nearly 3651 days. A common year is 365 (lays, andl one day id added in leap year to make up the loss of i of a day in each of the preceding years. This method of reckoning was ordered by Julius Caesar, 40 years before the birth of Christ, and is called the Julian Account, or Old Style. But, as the true year falls 11mn. 3s. short of 3054 days, the addition of a day every fourth year was too much by 44ei. 12s. This amounted to one day in about 130 years. To correct this error, Pope Gregory, in 1582, ordered that 10 days should be struck out of the calendar, by calling the bth of October the 15th; and to prevent its recurrence, he ordered that each succeeding century divisible by 4, as 16 hundred, 20 hundred, and 24 hundred, should be leap years, but that the centuries not divisible by 4, as 17 hundred, 18 hun.Ired1 a:nd 19 hlnlredl, should be common years. This reckoning is called the Gregorian, or New Style. The New Style differs now 12 days from the Old Style." Art. 105. —Troy Weight. Troy weight is used in weighing gold, silver, platina, diamonds, and other precious stones. The standard Troy pound of the United States, is the weight of 22.794377 cubic inches of distilled water, weighed in air. 24 grains, grs., make................. pennyweight, pwt. 20 pennyweights ".................1 ounce, oz. 12 ouices ".............. 1 pound lb. )08 WEIGHTS AND MNEASURES. Art. 106.-~A potheccaries' Weight. This weight is used only by apothecaries and physicians in compounding medicines. 20 grains, grs. make............................... scruple, 3. 3 scruples "............................. 1 dram, 3. 8 drams'-..............................1 ounce 3. 12 ounces *'................................1 pound, lb. "The original standard of all our weights was a corn of wheat taken from the middle of the ear, and well dried. These were called grains, and 82 of them made one pennyweight. But it was afterwards thought sufficient to divide this same pennyweight into 24 equal parts, still calling the parts grains, and these are the basis of the table of Troy weight, by which are weighed gold, silver, and jewelry. Apothecaries' weight is the same as Troy weight, only having different divisions between grains and ounces. Apothecaries make use of this weight in compounding their medicines, but they buy and sell their drugs by Avoirdupois weight. In buying and selling coarse and drossy articles, it became customary to allow a greater weight than that used for small and precious articles, and this custom at length established Avoirdupois, or common weight, by which all articles are weighed, with the following exceptions. Avoirdupois weight is about one-sixth part more than Troy weight,-the former being 7000 grains, and the latter 5750 grains. In buying and selling by the hundred weight, 28 pounds have been called a quarter, 112 pounds,a cwt.; but this practice of grossing, as it is called, is now pretty generally laid aside, and 25 pounds are considered a quarter, and 4 quarters (100 pounds) a hundred weight." Art. 107.-Avoirdupois, or Common Weight. This weight is used in weighing most kinds of merchandise, and all metals, except silver and gold. 16 drams make..........................1 ounce, oz. 16 ounces "..........................1 pound, lb. 28 pounds..........................1 quarter, qr. 4 quarters, " I hundred, cwt. 20 hundred, "........................ ton, ton. MEASURES. Art. 108 — Linear Measure. This measure is used in measuring distances, lengths, breadths, heighths, and depths. 3 barleycorns,bar., make................ inch, in. 12 inches "...............1 foot, ft. 3 feet "................1 yard, yd. 5J yards, or 161 feet, "...............1 rod, or pole, rd. 40 rods ".............1 furlong, fur. MEASURES. 109 8 furlongs make..................... mile, n. 3 miles "......................1 league, lea. 69' miles ".....................1 degree, deg. 360 degrees "..................... circle of the earth. T,92 inches.................. link, lk. 25 links "................... rod, 4 rods, or 100 li"................... chain, Cem. 80 chains "............... mile, mi. "The original standard of English Long Measure, was a barleycorn taken from the middle of the ear, and well dried. Three of these in length were called an inch, and then the others as in the table. Long Measure is employed for denoting the distance of places, and for measuring where length only is concerned. When measure is applied to surface where both length and breadth are concerned, it is called Square Measure. A square inch is a square measuring an inch on every side. The table of Square Measure is made from that of Long Measure, by multiplying the several numbers of the latter into themselves. Thus, 12 inches are a foot in length; a square foot, then, is a square which measures 1 foot, or 12 inches on every side, and contains 12X12=144 square inches. Three feet in length make a yard. A square yard is a square ~ — ~ measuring 3 feet on each side; but such a square contains _ (see figure) nine (3 X 3=9) square feet; and when we say that a surface contains so many square feet, or square yards, _ ~ ve mean that the surface is equal to such a number of squares, meaning a foot, or a yard, on each side." Art. 109 — Clolh Afeasure. This measure is used for measuring cloth, and other goods which are sold by the yard or ell. 24 inches make........................ nail, na. 4 nails "........................I quarter, qr. 4 quarters........................1 yard, yd. 3 quarters "........................1 ellFlemish, E. Fl. o quarters ".......................1 ell English, E. E. 6 quarters "........................ ell French, E. Fr. 37.2 inches "........................1 ell Scotch, E. S. Art. 11. — Square Mfeasure. This measure is used in measuring all kinds of surfaces, such as land, paving, flooring, plastering, and every thing wilich has length and breadth. Gunter's chain, used by surveyors in measuring land, also in measur. ing distances, is 4 rods, or 66 feet in length, and is composed of 100 links. 144 inches make................... square foot, ft. 9 feet ".......................1 square yard, yd. 304 yards "........................1 square rod, rd. 2724 feet "........................1 square rod, rd. 40 rods ".......................1 rood, ro. 4 roods "......................1 acre, acr. 10 110 MEASURES. 640 acres make...............1 square mile, mi. 10 square chains.......... acre, acr. 6400 chains "..............1 square mile, mni. Art. 11l. — Solid, or Cubic Measure. Cubic Measure is used in measuring solids, or any thing that has the dimensions, length, breadth, and thickness. 1728 inches make.........................1 for t, ft. 27 feet ".........................1 yard, yd. 40 feet of round timber, or I t ton. 50 feet of hewn timber,.-1 ton, t 128 feet........................1 cord, cor. A perch of stone is equal to 24 cubic feet, used by masons in measuring stone walls/ A square of earth is equal to 216 cubic feet. Art. 112.- Wine Measure. Wine Measure is used in measuring wine and all spirituous liquors, except porter, ale, and beer. 4 gills make....................1 pint, Ipt. 2 pints "................... quart, qt. 4 quarts ".....................1 gallon, gal. 31~ gallons ".1 barrel, bar. 63 gallons..................1 hogshead, hhd. 2 hogsheads ".................... pipe, p. 2 pipes "........1... tun, t. OBs.-The wine gallon contains 231 cubic inches. The hogshead is used only in estimating the contents of cisterns, wells or large bodies of water. Th'e common gallon is 231 cubic inches. A gallon of milk, or malt liquor, is 282 cubic inches. "Four pounds of Troy Weight of wheat, gathered fiom the middle of the ear, and well dried, were called one gallon; and this was the original standard of all Englih measures, both liquid and dry; and this was the same as the present wine gallon. But, in time, it became customary to use a larger measure in selling cheap liquors; and this custom at length established the Beer Measure, which bears about the same proportion to Wine Measure that Avoirdupois does to Troy Weight. The Dry Measure was also made.rger than the Wine Measure, and was at length established at about a mean between Wine and Beer Measure. The statute bushel for measuring coal, ashes, and lime, in Vermont, contains 38 quarts, or 2553.6 cubic inches" REMARKS.-When measure is applied to magnitudes which have length, breadth, and thickness, it is called solid, or cubic measure. A solid inch is a body or block, having six sides, each of which.is an inch square, and the number of inches in a 4olid foot is equal to the number of such blocks that would be required to make a pile a foot square and a foot high. Now it would require 144 blocks to cover a square foot one inch high. Hence to raise the pile 12 inches high would require 12 times 144-1728 blocks MEASURES 111 or iches. In like manner it would require 9 solid blocks, a foot each way, to cover a square yard to the height of one foot, and 3 times 9-=2, to raise it 3 feet, or make one solid yard. A cord of wood is sometimes called 8 feet. In this case, four feet in length, four in breadth, and one in height=16 solid feet; or 8 feet in length, 4 in breadth, and 6 inches in height, a foot; that is, 8 of a cord is called one foot, 2, two feet, etc. In measuring lands, roads, etc., the distances are usually taken in chains and links. In ordinary business, feet and inches are the most common measures. By forty feet of round timber, in the table of solid measure, is meant so much round timber as will make forty feet after it is squared. Art. 113 —Ale, or Beer Measure. This measure is used in measuring porter, ale, beer, milk, and water. 2 pints,pts. make.........................1 quart, t. 4 quarts ".........................1 gallon, gal. 36 gallons ".........................1 barrel, bar. 54 gallons ".........................1 hogshead, hhd. 2 hogsheads......................... 1 butt, butt. 2 butts ".........................1 tml, tu. DBS.-The ale gatlon contains 282 cubic or solid inhe3. Art. 1144 —Dry Measure. This measure is used in measuring grain, fruit, seeds, roots, salt, sand, oysters, coal, etc. 2 pints,pts. make..................1 quart, qt. 4 quarts ".........................1 gallon, gal. 8 quarts ".........................I peck, pk. 4 pecks ".........................1 bushel, bu. 8 bushels..........................1 quarter, gr. 4 quarters ".........................1 chaldron, ch. OBs.-A gallon dry measure contains 268- cubic inches. A Winchester bushel is 18 inches in diameter, 8 inches deep, and contains 215 6 cubic inches. The coal bushel must be 194 inches in diameter; and 36 bushels, heaped up, make a London chaldron of coal, the weight of which is 3156 lbs. Avoirdupois. Art. 115.-Circular Measure. Circular Measure is used for measurirfg circles, latitude and longitude, and in computing the revolution of the earth and other planets round the sun.'60 seconds, "make.........................1 minute, 60 minutes "..................... 1 degree,' 30 degrees ".......................1 sign, s 12 signs, or 360............................1 circle. "Every circle, without regard to its size, is supposed to be divided into 860 equal parts, called degrees, and these again to be subdivided into minutes and seconds; so that the absolute quantity expressed by any of these denominations must always depend upon the size of the circle. A 112 REDUCTION. degree on the circumference of a circle, whose radius or semidiameter is 68 miles, is one mile; if the radius is 58 rods, the length of the degree is one rod. In this measure are reckoned latitude, longitude, and the planetary motions." Art. 116.-Miscellaneous. 12 units, or things, make...............1 dozen, doz. 12 dozen "...............1 gross, gro. 12 gross, or 144 doz.".............. greatgross, G.gro 20 things "..............I score. 24 sheets of paper ".....I...... quire, 20 quires ".............. ream. 112 pounds............ quintal 6 points 1.............. line. 12 lines "..............1 inch. 4 inches ".............. hand. 6 feet "..............1 fathom. Os..-Points and lines are applied to measuring pendulums of clocks; hands, to measuning horses; fathoms, to measuring depths of the sea. Books. When a sheet is folded into two leaves, it is called Folio. When folded into 4 leaves, it is called Quarto. When folded into 8 leaves, it is called Octavo. When folded into 12 leaves, it is called Duodecimo, or 12eno When folded into 18 leaves, it is called 18mo. When folded into 24, it is called 24mo. When folded into 48, it is called 48mo. A gallon of train oil weighs........................... pound A stone of butcher's meat weig................... 8 " A gallon of molasses I................... 11 " A stone of iron ".................. 14 " A firkin of butter...................56 " A fother of lead ".................. 19 cwt. A barrel of flour ".................. 196 pounds. A " of ]ork or beef "...................200 " A " of soap "...................256 A quintal of fish "...................112 " REDUCTI ON. Art. 117.-REDUCTIoN is the changing of numbers from one denomination to another, without altering their value. 1. In ~6 7s. 8d. 2qrs., how many farthings? QUESTIONS.-1. What is Reduction? 2. By what numbers do you multiply in Example lst? and why? REDUCT'ION. 113 OP)eration. It is plain, that if in one pound ~6 7s. 8d. 2tqrs. there are 20s., in 6 pounds there 20 are 6 times as many, or 120s.; 227 and in.~0 7s. there are 127s. 12 Again, if in one shilling there are ^~1532 ~ 12d., in 127s. there are 127 times 14 12, which, with the 8d. added, 4-~ ~ equals 1532d. Lastly, it is evi6130qrs. Ans. dent that, if in 1 penny there are 4 farthings, in 1532d. there are 4 times the number of farthings, or 6130, the 2qrs. in the given question being added. Hence it appears, that in any given number of pounds there are 20 times as many shillings as pounds, 12 times as many pence as shillings, and 4 times as many farthings as pence. This process is called Reduction Descending, because higher denominations are brought into lower. 2. In 6130 farthings, how many pounds? Operation. It will be seen that this question 4)6130 is the reverse of the former; and 12)1532 2qrs. as 4 farthings are equal to 1 penny, 8210 )191.7 8 so the number of pence in 6130qrs. 210)1217 8 will equal the number of times it ~6 7s. 8d. 2qrs. Ans. contains 4, or 1532d. and 2qrs. over. Again, as 12d. are equal to 1 shilling, so in 1532d. the number of shillings will equal the number of times it contains 12, or 127s. and 8d. over. Lastly, since 20s. is equal to 1 pound, 127s. must equal 6 pounds and 7s., because 20 is contained in 127s. 6 times, and 7s. over. It is always to be remembered, that the remainder is of the same denomination as the dividend, whatever may be the divisor. This latter process is called Reduction Ascending, because lower denominations are brought into higher. By these examples, it appears that Reduction Ascending and Descending mutually prove each other. From the preceding operations we derive the following Rules: Art. 118. Art. 119. Reduction Descending. Reduction Ascending. RULE. RULE. Multiply the highest denomina- Divide the denomination given tion given by that number which by that number which expresses expresses how many it takes of the how many of that denomination it 10* 114 EXAMPLES IN REDUCTION. next lower to make one of the takes to make one of the next highhigher, observing to add the next er. Divide farthings by as many lower denomination to the product. farthings as it takes to make a Multiply pounds by w at makes a penny; pence by as many pence pound; shillings by what makes as it takes to make a shilling, and a shilling, and so on, until you so on, until you have reduced it to have reduced it to the denomina- the denomination required. tion sought in the question. English Money. EXAMPLES. EXAMPLES. 3. In 624 pounds, how many 4. In 599040 farthings, how farthings? many pounds? 5. If 6 shillings make a dollar, 6. In 2603 dollars, how many how many dollars in ~780 18s.? pounds and shillings? 7. How many pence in ~24 8. How many pounds in 5963 16s. 1d.? pence? 9. How many guineas, 28s. 10. How many pounds in 35 each, in ~49? guineas? Troy Weight. 11. In 18 Ibs. 11 oz. 6 pwt. 18 12. In 109122 grains, how grs., how many grains? many pounds? 13. Bought jewelry weighing 14. If I pay $437.60 for jew1 lb. 10 oz. 15 pwt. 20 grains. elry, at the rate of $0.04 per grain, Paid $0.04 per grain. What did what was the weight? I pay? Avoirdupois Weight. 15. In 5 tons, 16 cwt. 3 qrs. 16. Reduce 3350762 drs. to 12 lbs. 14 oz. 10 drs., how many tons. drams? 17. What will be the cost of 18. Bought lead to the amount 40 tons of lead, at $0.12 per of $10752, at $0.12 per pound. pound? What was the weight? 19. What must I pay for 20 20. How much hay can be tons, 18 cwt. of hay, at the rate bought for $2340.80, at 5 cts. per of 5 cts. per pound? pound? Apothecaries' Weight. 21. Reduce 6tr 103 73 29 16 22. In 39836 grains, how many grs. to grains. pounds? 23. Reduce 211b 113 33 19 13 24. In 126453 grains, how grs. to grains, many pounds? QUESTIONS.-3. What is the process called? 4. By what numbers do you divide In question 2? and why? 5. What is this process called? 6. What is the rule for Reduction Ascending? 7. For Reduction Descending? EXAMPLES IN REDUCTION. 115 Long Measure. 25. Reduce 640 degrees to 26. In 234854400 feet, how feet. many degrees? To reduce degrees to statute To divide by 16-, first reduce miles, multiply first by 69, and 161 to halves, and the dividend alfor the I in the multiplier take so to halves. Thus, 16=-33 half of the multiplicand. Thus: feet; and 234854400=469708800 2)640 degrees. half feet, which, divided by 33 69- gives 142338 rods. 5760 3840 320 44480 miles. 27. How many barleycorns 28. How many degrees in will reach across the Indian 594475200 barleycorns? Ocean, it being 45 degrees? 29. A teamster, after travelling 30. If a teamster receive for 20 miles, met a man who offered his load $264.00, being paid 5 him as many 5 cent pieces for his cents for each revolution of the load, as his larger wheel had turn- larger wagon-wheel, the circumed round times since he cor- ference being 20 feet, how far menced his journey, the wheel had he travelled? being 20 feet in circumference. How much did he get for his load? 31. How many barleyccrns 32. In 4755801600 barleywill reach round the globe, it corns, how many degrees? being 360 degrees? Land, or Square Measure. 33. If in a county there are 34. How many square miles 1200 square miles, how many in 122880000 square rods? square rods are there in this county? 35. How many square inches 36. In 2697235200 square inchin 430 square acres? es, how many square acres? Solid, or Cubic Measure. 37. How many solid inches in 38. Reduce 2211840 solid inch10 cords of wood? es to cords. 39. In 40 cords, how many 40. In 320 cord feet, how many cord feet? cords? 41. How many solid inches in 42. In 3732480 solid inches of a pile of timber containing 54 round timber, how many tons? tons? 116 EXAMPLES IN REDUCTION Cloth Measure. 43. Reduce 6324 yds. 3 qrs. 44. How many yards in 101199 3 na. to nails. nails? 45. What will 54 yds. 3 qrs. 46. If I buy cloth to the amount 0 na. of cloth cost, at 12- cts. an of $246.371, at the rate of 124 inch? cts. per inch, how many yards do I buy? WVine Measure. 47. In 5 tuns, how many 48. In 5040 quartse how many quarts? tuns? 49. What will be the cost of 50. How much wine can be 8 hhds. of wine, at $0.05 per bought for $201.60, at b cents pint? per pint? Ale, or Beer Measure. 51. What will 40 hhds. of beer 52. How much ber -can be cost, at $0.02 per pint? bought for $345.60, at 2 cents per pint? Dry Measure. 53. Reduce 8 ch. 10 bush. 3 54. In 19135 pints, how many pks. 7 qts. 1 pt. to pints. chaldrons? 55. How many pints in 30 56. In 1920 pints, how many bushels? bushels? Circular Mleasure. 57. In 6 s. 28 deg. 10 m. 8 sec., 58. Reduce 749408 seconds to how many seconds? signs. 59. Reduce 7 s. 21~ 30' 29" 60. In 833429 seconds, how to seconds. many signs? Time. 61. How many minutes in 10 62. In 5303925 minutes, how yrs. 30 da. 18 hrs. 45 m.? many years? 63. How many seconds from 64. In 219326400 seconds, how May 1, 1830, to April 12, 1837, many years? inclusive? 65. Alexander the Great as- 66. In 1140807240 minutes, cended the throne 332 years be- how many years? fore the Christian era. How many minutes from that period to 1837? MISCELLANEOUS EXAMPLES. 117 SUPPLEMENT TO REDUCTION OF WHOLE NUMBERS. Art. 120.-1. How many dollars in ~480 18s.? Ans. $1603. 2. In ~332 16s. 8d. how many ninepences? Ans. 8875 and 5d. 3. How many times will a regular clock strike in 400 years? Ans. 22776000. 4. A man sold four trees standing in the forest, measuring as follows: 6 tons, 5 tons, 41 tons, 3 tons, at 12- cts. per foot. What was their value? Ans. $92.50. 5. A man buys 20 tons of hay, at 45 cts. per cwt. He pays a man 62 cts. a day for himself, and 50 cts. for his team. It takes him 6 days to cart it. How much does the hay cost him? Ans. $186.75. 6. How many plank one foot wide will it take to cover a bridge 60 rods in length, and 2 rods wide, and what will it.cost, at 20 cts. per hundred feet? Ans. 65.340 lank. Ans. $65.34 cost. 7. If a boy be paid for wheeling a bushel of apples over said bridge 1 mill for every revolution of the wheel, which is 5 feet in circumference, how much does he receive? Ans. 19 cts. 8 m. 8. What will 2 tons of molasses amount to, at 6: cents a pint? Ans. $252. 9. A Vermonter, being in Boston, 80 miles from home, sold his dog and returned. At 6 o'clock the following night, the dog left his new master, and at 6 o'clock the next morning stood at the door of his former master. How many steps did he take of 8 inches each? 10. What will the plastering of a room, 15 feet square, the walls 9 feet high, amount to at 23 cents a square yard, deducting for 2 doors, 7 feet by 3, and 2 windows, 5 feet by 3? Ans. $17.71. 11. How much time would a person lose in 20 years, by lying in bed half an hour later every day than he ought? Ans. 152 days, 4 hours, 30 minutes. 12. How many cords of wood would a man draw in 6 weeks, drawing 4 loads a day, and 61 cord feet at a load? Ans. 117 cords. 13. A merchant failing in trade, owes A. ~15 7s. 9d.; B. ~69 11s. 6d.; C. ~102 16s. 11d.; D. ~41 19s. IOd.; 118 REDUCTION OF FRACTIONS E. ~139 1I7s. 5d. His whole estate is valued at ~300. How much does he owe more than he is worth? lAns. ~69 13s. 5d. 14. How many shingles will cover the roof of a factory 100 feet in length, one side of the roof being 40 feet in width, if 4 shingles in width cover 2 feet in length, and 2 courses make a foot? Ans. 32,000. 15. How many boxes, each 12 lbs., can be filled from a hogshead of sugar containing 7- cwt.? Ans. 70. 16. In 46 bales of cloth, each containing 24 pieces, and each piece 42 ells Flemish, how many yards? Ans. 34,776 yards. 17. The sun travels through 6 signs of the zodiac in half a year. How many degrees, minutes, and seconds? Ans. 180 deg., 10,800 m., 648,000 sec. 18. How many English crowns, at 6s. 8d. each, in 10 English guineas, at 28s. each, and 24 pistoles, at 22s. each? Ans. 121c. Is. 4d. 19. The forward wheels of a wagon are 14~ feet in circumference, and the hind wheels 15- feet. How many more times will the forward wheels turn round than the hind wheels, in running from Concord to Boston, the distance being 60 miles? Ans. 1.734, rejecting fractions. REDUCTION OF FRACTIONS. Art. 121.-1. Reduce 21g of a pound to the fraction of a penny. We have seen that integers of a higher denomination are brought into integers of lower, by multiplication, (see Art. 117;) and also that fractions are multiplied either by multiplying the numerator, or dividing the denominator. (See Art. 67.) Pounds are reduced to shillings by multiplying by 20, and shillings to pence by multiplying by 12. Therefore, to reduce 2s8 of a pound to the fraction of a penny, multiply the fraction by 20 and 12, thus: Or thus: 1 2If X20 X 12 = 22 - = 5= Ans. 6 8!f 5 65= Ans. REDUCTION OF FRACTIONS. 119 As the numerator of the fraction is to be multiplied, place it with its multipliers on the right of the line, and 288, the divisor, on the left. Cross 288 and 12, and write 24 in the place of the larger number; 4 is contained in 20 five times, and in 24 six times. The answer, then, is 5 divided by 6, or 6, in the lowest terms of the fraction. OBs.-It will be seen, that the only difference between reducing 1 pound to pence, and, of a pound, is, that in the latter case the multiplicand is a number divided; consequently, having multiplied the numerator as we should a whole number, we divide the product by the denominator. To divide the product is the same as to divide the multiplicand. Art. 122.-To change fractions of a higher denomination into fractions of a lower denomination, we have the followingRULE. Miultiply the numerator of the fraction, or divide the denominator by all the denominations between it and that denomination into which it is to be reduced, including the lower denomination. EXAMPLES. 1. Reduce -3- of a pound to the fraction of a penny. Ans. 3 2. Reduce -1g-6 of a pound to the fraction of a farthing. -ins. 2. 3. What part of a pound is -x cwt.? Ans. -. 4. Reduce -I- of a yard to the fraction of a nail. Ans. -. 5. Reduce Tio- of a pound to the fraction of a farthing. Art. 123 —It has been shown (see Art. 117) that Reduction Ascending is the reverse of Reduc0 tOperatn. tion Descending, and also, (Art. 68,) I that a fraction is divided, either by di1920,20 -viding its numerator or multiplying its 2 x1 denominator. Farthings are reduced to ___ ___ pence -by dividing by 4, and pence to 211=1 Ans. shillings by dividing by 12; shillings to pounds by dividing by 20. ThereforeTo reduce - of a farthing to the fraction of a pound, divide the fraction by.4, 12, and 20. I What part of a pound is - of a farthing? 120 REDUCTION OF FRACTIONS. Operations. 1 Or thus: 2 I X4X12 X 20=T9?6so 4 12 20 1920 1-=T-92 Ans. Art. 124.-To change fractions of lower, into fractions of higher denominations, we have, then, this RULE. lMultiply the denominator of the fraction by all the denominations between it and that into which it is to be reduced, and write the product under the numerator of the given fraction. 2. Reduce - of a penny to the fraction of a pound. Operations. Or thus: 6!$ xl2x20= =5 121 5 X 1 2 X 20 = r h= 4 12. 4 d _____ 288 1 — Ans. As the denominator of the fraction is to be multiplied for a divisor, place it with its multipliers on the left of the line. Then, by cancelling, 5 on the right of the line, and 20 on the left, equal 4: therefore cross 5 and 20, and write 4 on the left. Multiply the remaining numbers on the left together, for a divisor. We have, then, the answer in the lowest terms of the fraction, 2 8 3. Reduce A of a penny to the fraction of a pound. Ans. ~,-0. 4. Reduce 2 of a farthing to the fraction of a pound. Ans. 60.. W at part of a cwt. is - of a pound? Ans. 1 5. What part of a cwt. is 2 of a pound? Ans. g-. 6. Reduce F of a nail to the fraction of a yard. Ans. -. QUESTION.-1. What would be your multipliers in reducing the fraction of a pound to the fraction of a penny? REDUCTION OF FRACTIONS. 121 REDUCTION ASCENDING AND DESCENDING. EXAMPLES. DESCENDING. ASCENDING. Art. 125.-1. Reduce T-4 Art. 126.-2. Reduce 3 of of a pound to the fraction of a a farthing to the fraction of a farthing. pound. 3. Reduce A-T of a pound to 4. Reduce; of a penny to the the fraction of a penny. fraction of a pound. 5. Reduce o —g of a guinea to 6. Reduce of a penny to th the fraction of a penny. fraction of a guinea. 7. Reduce 4 of a guinea to the 8. What fraction of a guinea fraction of a pound. is 4 of a pound? Guinea 1 $s. 4 ~ I l0s. 4 5 s.0 1~ s.$ 1 guinea. 5 4= 4 Ans. 7 4=- Ans. 9. Reduce 30- of a guinea to 10. What part of a guinea is the fraction of a penny., 4 of a penny 11. Reduce I of a shilling to 12. What part of a shilling is the fraction of a farthing. - of a farthing? 13. Reduce -5, of a pound 14. What part of a pound Troy Troy to the fraction of an ounce. is 5 of an ounce? 15. Reduce 8 of a hhd. of 16. What fraction of a hhd. is wine to the fraction of a quart. a of a quart? 17, What fraction of a rod is 18. What fraction of an acre is T-BT of an acre? 4y - of a rod? 19. Reduce' of a mile to 20. Reduce 54- of a rod to the the fiaction of a rod. fraction of a mile. 21. Reduce -1 4v- of a degree 22. Reduce 3 of a foot to the to the fraction of a foot. fraction of a degree. 23. Reduce' of a bushel to 24. Reduce' of a gill to the the fraction of a gill. fraction of a bushel. 25. Reduce'3- of a tun to 26. Reduce I of a gill to the the fraction of a gill. fraction of a tun. 27. Reduce -L of 1 of 4 pounds 28. -?- of a penny is -- of 3 to the fraction of a penny. of how many pounds? 29. -7 of a pound is { of what -0 of 1 penny is,-1 of what fraction of 7 guineas? fraction of 4 pounds? - of a pound is -A of what 8_0 of a penny is - of what fracfraction of 7 guineas? tion of 4 pounds? 7 of a pound is I of; of how 30. Reduce I of A- of 7 guin. many guineas? eas to the fraction of a pound. 11 122 REDUCTION OF FRACTIONS. 31. Reduce f2-is of a week 32. Reduce 2 of a second to to the fraction of a second. the fraction of a week. 33. Reduce Ige of a year to 34. Reduce X of an hour to the the fraction of an hour. fraction of a year. Comparison of Numbers and Quantities. Art. 127.-We compare quantities.rnd numbers, to ascertain what part the one is of the other. Things compared must be of the same kind, or those properties compared must be alike..We do not compare rods with hours, nor minutes with days, bt rods with rods, and minutes with minutes. (See Art. 166.) The terms quotient, ratio, value of the fraction, each expresses what part the dividend is of the divisor. What is the quotient of 6 divided by 3? What is the ratio of 6 to 3? 3 Ans. 2. What is the value of the fraction 9? i What is one-third of 6? A How many times greater than 3 is 6? What part of 3 is 6? j The answer to each question is the same, and obtained by the same process. The quotient expresses what part the dividend is of the divisor. The quotient, with unity over it, expresses what part the divisor is of the dividend, =_-, and 6-2-. That is, 6 is twice as large as 3, and 3 is half as large as 6. The numerator of a fraction is the same part of the denominator, that the fraction is of unity. 1. What part of 4 is 3? 2. What part of 3 is 4? Ans. X. Ans. 4. 3. What part of 1 dollar is 1 4. What part of X1 is is.? shilling? Ans.. Ans. -. 5. What part of 1 yard is 1 6. What part of 2 yards is 1 quarter? Ans. 4. quarter? Ans. -. RULE If the numbers consist of different denominations, reduce them to the same, and write that number which the question requires to be a part, as the numerator of a fraction, and the other as a denominator. REDUCTION OF FRACTIONS. 123 7. What part of 2 yards and 2 8. What part of 2 yards and 2 qrs. is I qr.? qrs. is 1 yard and 1 qr.? yds. qrs. yds. qrs. 2+2=10 qrs. Ans. L 12+21 5 qrs. Ans. -1 — 2+2 —10 qrs. I 0i~ 9. What part of 8 dollars is 2 10. What part of 20 miles is dollars and 25 cts.? Ans. 8 mi. 6 furlongs Ans. 11. When wood is worth 4 dol- 12. When corn is worth 3 of a lars a cord, what part of a cord dollar a bushel, what part of a can be purchased for 2 dollars? bushel can be bought for ~ of a Ans. 2-. dollar? 13. What part of 5 is 3? 7 _7. 3_7 _4 7 A Ans. 8 4 _ _ 3-614. What part of 7- is 51? 15. What part of 7s. 6d. is Ans. 8. 4s. 2d.? Ans. a. For a fuller illustration of the subject, see Art. 129. To reduce Fractions to integers of lower denominations, and the reverse. Art. 128,-1. What is the Art. 129.-2. Reduce 13s. value of of a pound? 4d. to the fraction of a pound. - of a pound reduced to the In ~1 there are 240d. 1 penny fraction of a shilling is 3 X20=- then is A-u of a pound, and 160d., 43- of a shifng —which, reduced the number of pence in 13s. 4d., to a mixed number, (Art. 59,) is is!6 of a pound, or 160 times as 13s. The Iofa shilling reduced much as 1 penny. Therefore, to to the fraction of a penny, is reduce integers of lower, to fracX 12=a2 —=4d. Hence, to re- tions of a higher denomination, duce fractions of one denomina- we have this tion to integers of a lower, we have this RULE. RULE. Multiply the numerator (f the Reduce the given numbers to given fraction by that number the lowest denomination mentioned, which expresses how many of the for a numerator, and an integer next lower denomination make 1 of the denomination required to of that denomination in which the the same denomination for a de. fraction is given, and divide the nominator, and they will form the product by the denominator of the fraction required. fraction. If there be a remainder, proceed as before, until it is reduced to the lowest denomination. If there be still a remainder, place it at the right of the last answer. 124 REDUCTION OF FRACTIONS. OBs.-Let question 2d be written upon the blackboard, in the following manner, and illustrated. Teacher. Express by writing upon the blckboard, what part of a pound is 138s. Scholar. 13is. is,~ of a pound, which equals -. T. What kind of fractions are those you have written? S. Complex T Remove the denominator of the numerator, and illustrate. 40 40 & - =-. To multiply the denominator is the same as to divide a 60 the numerator, for, to multiply the divisor is the same as to divide the dividend. T' Remove the denominator of the fraction, and illustrate. 40 S - =. To divide the numerator divides the fraction, for, to divide 2af the dividend divides the quotient. Let other questions be written ar.d illustrated in a similar manner. 3. What is the value of I of a 4. Reduce 7d. 2qrs. to the fracshilling? tion of a shilling. Operation. Operation. 2 $ 5 Integer. d q grs. 1 I 3 Is. 7 2 2 15=7~ Ans. 12 4 12 6)30(5 Or thus: 4 48(8Ans. 5 x/ 3 15 7 Ans X- =' — Ans. 2 $ 1 2 Or, we may reduce the farthings to the fractions of a penny, and reduce the whole to an improper fraction. Thus: 7=1 2I5 5 4 _ 1 8 5= — Ans. Oss.-It will be recollected that it was said (Art. 60) that a mixed number is the quotient of a division, whose divisor was the denominator of the fraction. Consequently, example 10 is the quotient of a division whose divisor was 11. We have therefore only to multiply quotient and divisor together, or to reduce a mixed number to an improper fraction, and we have the fraction of a cwt., which we divide by 20, to reduce it to the fraction of a ton. Thus: QcEsTinos.-1. Rule for reducing fractions to integers of lower denominations? 2. Rule foi reducing integers of lower denominations to a fraction of a higher? REDUCTION OF FRACTIONS. 125 Operation. If there should be no fraction in cwt. qr. lbs. oz. dr. the question, the lowest denomination 16 1 12 11 10-2 may be reduced to a fraction of the 11 higher. In example 12, the 36 minutes may be reduced to the fraction 11 1A0 0 0 0 0 of an hour; thus, ~3=_. We then ~0 9 have 9} hours, a mixed number. 11 9=19 Ans. 5. What is the value of 4 of a 6. Reduce 6 furlongs, 26 po. mile? 11 ft. to the fraction of a mile. 7. What is the value of 4 of a 8. Reduce 8 mi. 5 fur. 20 po. degree? to the fraction of a degree. 9. What is the value of -1 of 10. Reduce 16 cwt. 1 qr. 12 lbs. a ton? 11 oz. 10O2- dr. to the fraction of a ton. 11. What is the value of 4 of a 12. Reduce 3 w. 1 da. 9 hr. month? 36 m. to the fraction of a month. 13. What is the value of 4 of a 14. Reduce 8 oz. 11 pwt. 102 pound Troy? grs. to the fraction of a pound. 15. What is the value of f of 16. Reduce 3 roods, 13 rods, 90 an acre? feet, 108 in. to the fraction of an acre. 17. What is the value of I of a 18. Reduce 3 qrs. 2 na. to the yard of cloth? fraction of a. yard. 19. What is the value of I of a 20. Reduce 4s. 6d. to the fracdollar in shillings? tion of a dollar. 21. Whatis the value of ~ of a 22. Reduce 11 cwt. 0 qr. 12 lbs. ton? 7 oz. 1- drs. to the fraction of a ton. 23. What is the value of 7 of a 24. Reduce 49 gals. to the hogshead? fraction of a hogshead. Reduction of Vulgar Fractions to Decimal. Art. 130.-1. Reduce 4 to a decimal fraction. In this example, 4 being a proper fraction, the numerator will not contain the denominator; but, by annexing a cipher, which reduces it to tenths, we can divide by the denominator. In 1 unit there are 10 tenths, but the example is one half of a unit; therefore, one half of 10 tenths, which is 5 tenths, will be the answer. Hence the RULE. I. Annex a cipher, or ciphers, to the numerator, and divide by the denominator. 11* 126 REDUCTION OF FRACTIONS. II. If there be a remainder, a cipher, or ciphers, may be annered, and the process of division carried on until there be no remainder, or the quotient is sufficiently exact. The decimal places in the quotient must be equal to thenumber of ciphers annexed to the numerator. If, after division, the quotient does not contain so many, supply the deficiency by prefixing ciphers. 2. Reduce 4 to a decimal fraction. Ans. q5. 3. Reduce 4, 4, and 4, to decimal fractions. Ans..625,.125,.6. 4. Reduce,, I, and ~, to decimal fractions. Ans..875,.375,.75,.25,.5. 5. Reduce I of 2 of 3 to a decimal fraction. Operation. 4 $ 4 1.001.25 Ans. 6. Reduce 4 of 6 of 4, divided by * of of, to a decimal fraction. Ans. 2.25. 7. Reduce I to a decimal fraction. Ans..04. 8. Reduce 126- to a decimal fraction. Ans..632071+. Art. 131.-To reduce a decimal fraction to a vulgar. RULE. Write down the given decimal, as a numerator, andfor a denominator, write 1, with as many ciphers annexed as there are figures in the numerator, and then reduce the fraction to its lowest terms. (See Art. 61.) 1. Reduce.25 to a vulgar fraction. Operation. 25 I 25 1- 00 IAns. QuzSTIONs.-1. Rule for reducing a vulgar fraction to a decimal? 2. How many decimal places must there be in the quotient? 3. If the quotient does not contain a sufficient number of figures, what is to be done? REDUCTION OF FRACTIONS. 127 2. Reduce.125 to a vulgar fraction. Ans. 1. 3. Reduce.45 to a vulgar fraction; Ans. t. 4. Reduce.24 to a vulgar fraction. Ans. 26. 5. Reduce.945 to a vulgar fraction. Ans. 1 To reduce Integers of different denominations to a Decimal Fraction of a higher denomination, and the reverse. Art,. 132,-1. Reduce 4 Art. 133 —-2. Reduce.375 pence 2 farthings to the decimal of a shilling to integers of lower of a shilling. denominations. Operation 2 farthings is As this question is the reverse Operation. T 4 2.0 of a penny; then, of the former, and as the decimal, 121 4500 by the rule for re-.375, was obtained by dividing -.7 ducing vulgar frac- the integers, it is plain, that the *5 tions to decimal, integers may be obtained by mulwe have -=.5, or - of a penny. tiplying the decimal by the same This, placed at the right of 4 numbers. pence, 4.5, and divided by 12, the Operation. number of pence in a shilling, or.375 because 4 pence is -1 of a shil- 12 ling, gives.375 of a shilling. 4.500 Hence the 4 2.000 Hence the RULE. RULE. Place the numbers one above Multiply the given decimal by another, the highest denomination that number which expresses how at the bottom. Divide the lowest many of the next lower denomidenomination by that number nation it takes to make one of that which expresses how many of that in which the decimal is given; it takes to make 1 of the next higher observing to point off as many denomination, writing the quo- places in the product, for decimals, tient at the right of the next higher as there are figures in the given denomination; and so proceed decimal; and so proceed through until the whole shall be reduced to all the denominations; and the the required decimal. several numbers at the left of the Tdecimal points uwill be the answer OBs.-Integers of different denomina- deciml points ill be the answer tion may be reduced to a decimal of a required. hik!r, by reducing the given numbers to the lowest denomination mentioned for a numerator, and the integer, to which the given numbers are to be reduced, to the same denomination for a denominator, OBs.-Pointing off the product is the and dividing the numerator by the denom- same as dividing by the denominator of inator, i the decimal. 128 REDUCTION OF FRACTIONS. 3. Reduce 8s. 4d. 2qrs. to the 4. Reduce.41875 to integers decimal of a pound. of lower denominations. Operation. Operation. 4 2.0.41875 12 4.500 2 21C 8.373o10 s. 8.375000.41873 12 d. 4.50000 4 qrs. 2.00000 5. Reduce 4 oz. 4 pwts. to the 6. What is the value of.35 of decimal of a pound. a pound Troy? 7. Reduce 4 oz. 8 drs. to the 8. What is the value of.28125 decimal of a pound. of a pound? 9. Reduce 2 cwt. 2 qrs. to the 10. What is the value of.125 decimal of a ton. of a ton? 11. Reduce 3 qrs. 2 na. to the 12. What is the value of.875 decimal of a yard. of a yard? 13. Reduce 20 h. 16 ti. 48 sec. 14. What is the value of.845 to the decimal of a day. of a day? 15. Reduce 21s. 6d. to the 16. What is the value of.875 decimal of a guinea. of a guinea? Art. 134.-To riduce shillings, pence, and farthings to the decimal of a pound, by inspection. 1. Reduce 7s. 8d. 2qrs. to the decimal of a pound? One shilling is -- of a pound: therefore, two shillings is -^0, or T. Having, therefore, any number of shillings given, if we take one half the even number, they will be reduced at once to the decimal of a pound. If there is an odd shilling, it is the same as yOs of a pound: 2 —.05. The-fiarthing, which is 9,6 of a pound, is made to occupy the 1000ths place. But s is greater than r- - by 4o; there will, therefore, be a loss of 2i —0 on every farthing; but if we add one to the number, when they exceed 12 and do not exceed 36, and two when they exceed 36, the expression will be nearly so many 1000lths of a pound. 2. Reduce 4s. 6d. to the decimal of a pound. Operation..2 half of the even shillings..024 farthings in 6d..001 for excess of 12...225 Ans. COMPOUND ADDITION. 129 If we call the farthings in 6d. T4C, there will be a loss of 2400o=o0o0 if we add 1 to the 1000ths place, we have, in this instance, precisely the decimal required. 3. Reduce 7s. 8Id. to the decimal of a pound. Operation..3 half of the even shillings..05 for the odd shilling..034 farthings in 8,d..001 for excess of 12..385 Ans. 4. Find by inspection the decimal expression of 18s. 3-d., and 17s. 8~d. Ans. ~.914 and ~.885. 5. Reduce to a decimal by inspection the following sums, and add them together, viz:-15s. 3d.; 8s. 11d.; 10s. 61d. Is. 8Wd.; 21d. Ans. 1.832. Decimals may be reduced back to shillings, pence, and farthings, by reversing the above process. Double the left-hand figure, or tenths, for the shillings;*if the second figure be 5, or greater than 5, deduct 5 from it, and add 1 to the shillings. Then consider the second and third figures so many farthings; if they exceed 12, deduct 1; if they exceed 36, deduct 2. 6. Find by inspection the value of ~.385. 7. Find by inspection the value of ~.927. Ans. 18s. 6d. 2qrs. 8. Find by inspection the value of ~.491, and ~.984. Ans. 9s. 9d. 3qrs.; 19s. 8d. lqr. COMPOUND ADDITION. Art. 135.-1. A boy bought a slate for 4d. and a book for 8d.. What did both cost? Ans. 1s. 2. If I buy a book for 2s. 4d., another for 4s. 8d., what do I pay for both? Ans. 7s. 3. If a boy pay 4s. 8d. for a sled, and 5s. for a wagon, what does he pay for both? Ans. 9s. 8d. 4. How many shillings in 4d. 8d. 9d. 3d. 6d.? Ans. 28. 6d. 130 COMPOUND ADDITION. 5. How many pounds are 8s. 7s. 4,. 3s. 9s. 5s.? Ans. ~1 16s. 6. How many yards are 3 feet, 4 feet, 5 feet, 6 feet? Ans. 6 yards. 7. Bought two pieces of cloth; one 10 yards, 1 foot; the other 12 yards, 2 feet. What was the length of both pieces? Ans. 23 yards. 8. What is the amount of ~1 4s. 2d. 3qrs., and ~10 8s. 3d. and 2qrs.? Ans. ~11 12s. 6d. lqr. 9. Add ~4 5s. 6d. 3qrs., and ~5 17s. 7d. 2qrs. ~ s. d. qrs. 4 5 6 3 In adding the first column, or col5 17 7 2 umn of farthings, we find the amount 10 3 2 1 Ans. to be 5 farthings. Now as 4 farthings are equal to 1 penny, we write the 1 farthing over, in the line of farthings, and carry the 1 penny to the column of pence. One to 7 is 8, and 6 are 14d. In 14d. there is 1 shilling and 2d. over, which we write in the column of pence, carrying the Is. to the column of shillings. One added to 17 is 18, and 5 are 23. In 23s. there is- ~1 and 3s. over, which we write in the column of shillings, and carry I to the column of pounds. Had the numbers to be added in the question been simple numbers, we should have had none to carry, because 5, in the column of units, is not equal to 1 in the column of tens. Again, had 14 been in the column of tens, we should have written 4 and carried 1. Lastly, had 23 been in the column of hundreds, we should have written 3, and carried 2, because 23 in the right-hand column, is equal to 2 in the left, and 3 remain; or, 23 hundred is equal to 2 thousand, and 3 hundreds remain. Art, 136. —From the foregoing questions and illustration we derive the following definition and rules. COMPOUND ADDITION is the adding of numbers of different denominations. By different denominations is meant a different name-as shillings, pence, farthings, etc. Were' the numbers given to be added, all pence, or all farthings, there would be but one denomination. RULE. I. TVrite numbers of the same denomination directly under each other, pounds under pounds, shillings under shillings, etc. COMPOUND ADDITION. 131 II. Begin to add at the right-hand column, observing to carry one for as many in that column as make one in the next left-hand column. Proof-The same as in addition of simple numbers. EXAMPLES. 1. Bought 4 books at the following prices, viz., ~1 4s. 6d.; ~2 3s. 8d.; ~2 19s. lild.; 2s. 3d. 2qrs. To what did they amount? Ans. ~6 10s. 4d. 2qrs. 2. Add the following numbers: ~46 26s. 7d. 3qrs.; ~49 18s. 5d. lqr.; ~57 17s. 9d. 2qrs.; ~102 19s. 10d. lqr. Ans. ~258 2s. 8d. 3qrs. 3. Add $286 12 cts. 6-m.; $347 20 cts. 4 m.; $119 18 cts.' m.; $542 93 cts. 9 m.; $314 89 cts. 1 m. Ans. $1610 34 cts. 7 m. 4. Add 45 lbs. 9 oz. 15 pwt. 18grs.; 90 lbs. 6 oz. 16 pwt. 23 grs.; 30 lbs. 10 oz. 11 pwt. 6 grs.; 85 lbs. 11 oz. 13 pwt. 4 grs.; 91 lbs. 7 oz. 7 pwt. 23 grs. AVOIRDUPOIS WEIGHT. Ton cwt. qr. lbs. oz. dr. Ton cwt. qr. lbs. oz. 40 18 2 156 14 15 19 16 23 3 13 80 19 3 17 13 12 14 13 1 19 12 67 11 1 12 9 7 29 11 2 12 11 79 17 1 23 15 13 39 17 1 16 9 93 13 2 26 10 11 47 19 2 19 15 APOTIIECARIES' WEIGHT. tb 5 3 9 gr. tI b 3 9 gr. 29 10 7 2 17 99 11 3 1 19 25 11 6 1 13 102 9 7 2 5 37 8 4 1 9 81 4 6 1 13 71 5 3 2 1.1 120 7 3 2 18 89 0 5 1 10 341 6 1 3 16 132 COMPOUND ADDY rION CLOTH MEASURE. Yd. yr. na. E. E. q. na. E. Fr. qr. na. 125 3 2 176 4 3 69 3 2 300 1 3 57 3 1 76 5 3 159 2 2 102 1 2 57 4 1 260 1 1 69 2 2 89 2 1 1910 3 3 267 4 1 97 1 2 357 4 2 179 2 3 88 3 3 WINE MEASURE. Tun hhd. gal. qt. pt. Tun hhd. gal. qt. pt. 66 2 57 2 1 86 3 39 3 2 79 3 60 3 0 121 2 51 1 2 88 1 49 1 1 67 1 19 0 3 91 2 38 2 —- 1 76 1 29 1 2 72 3 20 1 0 167 2 38 2 1 61 1 39 - 1 129 0 31 1 2 ALE AND BEER MEASURE. Hld. gal. qt. pt. Hhd. gal. qt. pt. 102 21 2 1 171 29 1 1 201 39 3 0 169 49 3 0 310 42 2 0 289 38 1 1 412 38 1 1 169 42 1 1 121 39 2 1 128 31 2 1 16. Add 49 bushels, 3 pecks, 4 quarts, 1 pint; 39 bu 1 pk. 5 qt. 1 pt.; 59 bu. 2 pk. 3 qt. 0 pt.; 40 bu.'7 pk. 2 qt. 1 pt.; 150 bu. 0 pk. 6 qt. 1 pt.; 69 bu. 1 pk. 2 qt. Opt. 17. Add 360 degrees, 15 miles, 6 furlongs, 16 poles, 13 feet, 6 inches; 240 deg. 19 m. 5 fur. 29 p. 11 ft. 5 in. 2 b.; 159 deg. 51 m. 7 fur. 32 p. 14 ft. I in. 2 b.; 201 deg. 63 m. 3 fur. 15 p. 12 ft. 9 in. 2 b. 18. Add 971 miles, 6 furlongs, 11 poles, 3 yards, 1 foot; 239 m. 5 fur. 9 p. 2 yd. 2 ft.; 269 m. 7 fur. 31 p. I yd. 2 ft.: 67 m. 6 fur. 9 p. 2 yd. 2 ft.; 691 m. 5 fur. 8 p. 2 yd. 2 ft. COMPOUND SUBTRACTION. 133 19. Add 69 acres, 2 roods, 1 rod; 76 acr. 3 ro. 39 rd.; 88 acr. 1 ro. 32 rd.; 150 acr. 3 ro. 29 rd. 20. Add 150 years, 221 days, 13 hours, 31 minutes, 29 seconds; 230 yr. 300 d. 23 h. 49 m. 59 s.; 191 yr. 149 d. 21 h. 39 m. 23 s.; 359 yr. 75 d. 23 h. 59 m. 19 s. COMPOUND SUBTRACTION. Art. 137.-1. If a picture-book cost 4d. and a spellingbook lid., how much more does one cost than the other? 2. James bought a book for 9d. arid sold it for Is. How much did he gain by the bargain? 3. From 2s. 6d., take Is. 8d. 4. From 8s. 9d. 3qrs., take 6s. 8d. 2qrs. 5. From 4 qts., take 3 pts. 6. If a bushel of rye be worth 7s. 6d., and a bushel of corn 6s. 4d., how much more is the rye worth than the corn? 7. How much more is wheat worth at 9s. 8d. per bushel, than corn at 7s. 6d. per bushel? 8. How much more is 2 bushels 2 pecks, than 1 bushel 3 pecks? 9. From ~29 9s. 6d. 3qrs., take ~23 1Os. 7d. 2qrs. Operation. In this example, we write the ~9 a. 6 r. difference between 2 and 3 far29 9 6 3 o 23 10 7 2 th2gs in the line of farthings, and. ~ proceed to the column of pence; 5 18 11 1 we carry none, because we borrowed none-but 7d. from 6d. cannot be obtained; we therefore borrow as many pence as make a shilling, and say, 7 from 12-the remainder 5, we add to 6, in the upper line, and write 11 in the column of pence. We now carry I to the column of shillings, which is equal to the 12 pence we borrowed, and say, 11 from 9, which cannot be obtained; again we must borrow as many of the denomination we have to subtract as make one of the next higher, which is 20s., and say, 11 from 20, and 9 remain, which added to 9 in the upper QUESTIONS.-1. What does Compound Subtraction teach? 2. Rule? 3. If the number in the upper line be less than the one standing under it, how may you proceed? 4. Why do you carry 1 to the next left-hand column? 12 134 COMPOUND SUBTRACTION. line_ is 18, which must be written in the column of shillings. Las[ly, the 20s. which we borrowed, we pay by carrying 1 to the line of pounds, which must be subtracted as in simple subtraction. Hence, Art. 138.-COMPOUND SUBTRACTION teaches to find the difference between two compound sums, or quantities. RULE. I. Write the less number under the greater, so that numzbers of the same denomination may stand directly under each other. II. Begin to subtract with the lowest denomination, and take the lower line from the one above it; proceed in this way with all the denominations. 11I. Should the number in the upper line be less than the one standing under it, borrow as many units as make 1 in the ne.^t higher denomination. IV. From the units borrowed, subtract the lower number, and to the difference add the upper number; write their sum under the.figures subtracted, observing to carry I to the next left-hand column. Proof-The same as Simple Subtraction. EXAMPLES. TROY WEIGHT.:Lbs. oz. pwt. g. Lbs. oz. pwt. gr. 91 10 19 21 39 1 14 20 87 11 15 19 37 11 15 19 AVOIRDUPOIS WEIGHT. Ton cuwt. qr. lbs. oz. dr. Ton cwt. qr. lbs. 122 1t 3 22 13 12 39 11 14 20 110 13 2 23 14 13 37 9 15 19 APOTIIECARIES* WEIGHT. lf 5 3 9 gr. 3tb gr. 21 10 7 2 16 33 9 6 1 13 19 9 6 1 17 29 7 7 0 14 COMPOUND SUBTRACTION. 135 WINE MEASURE. Gal. qt. pt. gi. Hhd. gal. qt. pt. gi. 17 2 29 3 600 3 59 2 0 69 3 49 2 459 3 47 3 1 ALE AND BEER MEASURE. HAhd. ga. qt. t.. tihd. gal. qt. pt. 981 49 1 1 1000 37 3 0 392 51 3 0 999 49 2 1 11. From 31 tuns, 3 hhd. 15 gal., take 29 tuns, 2 hhd. 26 gal. 12. From 39 yds. 3 qr. 2 na., take 27 yds. 2 qr. 3 na. 13. From 127 E. E. 3 ql. 2 na., take 121 E. E. 4 qr. 3 na. 14. From 247 E. Fl. 0 qr. 2 na., take 159 E. Fl. 2 qr. I na. 15. From 671 E. Fl. 4 qr. 3 na., take 582 E. Fl. 5 qr. 2 na. 10. From 971 mi. 6 fur. 11 p. 3 yds. 1 ft., take 439 mi. 5 fur. 12 p. 4 yds. 2 ft. 17. From 69 acr. 2 ro. 31 rd., take 49 acr. 3 ro. 37 rd. 18. From 150 yrs. 221 d. 13 h. 31 m. 29 s., take 130 yrs. 129 d. 14 h. 39 m. 41 s. 19. From 2600'15 mi. 5 fur. 16 p. 13 ft. take 1500 17 m. -3 fur. 17 p. 12 ft. 20. From 240~ 49' 31" take 159~ 59' 41". 21. From 9s. 21~ 31' 42" take 7s. 22~ 36' 37". 22. A note dated Feb. 3d, 1826, was paid March 12th, 1837. How long was it from the first date until it was paid? The time from one date to another may be found by subtracting the former date from the latter, observing to number the months in their order; thus, January, 1st month; February, 2d month, etc. A. D., 1837 3d mo. 12th day. A. D., 1826 2d mo. 3d day. Ans. 11 ys. 1 mo. 9 days. Ons.-The month, in casting interest, is reckoned 30 days. 23. What is the time from June 3d, 1835, to July 15th, 1837? Ans. 2 yrs. 1 m. 12 d. 136 COMPOUND SUBTRACTION. 24. The latitude of a certain place is 42~ 50' north; that of another place is 39~ 37'; what is the difference of latitude? Ans. 3013'. 25. What is the difference of longitude between 390 40', and 290 49' west? Ans. 90 51'. As every circle,-whether greater or less, is divided into 360 equal parts, or degrees, consequently,'the circle described by the revolution of the earth on its axis every 24 hours, contains 360 equal parts, or degrees; and as 360 degrees are described in 24 hours, it is plain that in 1 hour, I of 360, or 15 degrees, would be described; and, also, if 15 degrees be described in 1 hour, or 60 minutes, it is equally plain that 1 degree would be described in i of 60 minutes, or in 4 minutes, and 1 minute of a degree in 4 seconds. Hence, Art. 139 —To reduce longitude to time, we have the following RULE. Multiply the longitude in degrees and minutes by 4, and we have the time in minutes and seconds. EXAMPLE. Reduce 140 15' to time. 140 15' 4 57' 0" Ans. Art. 140.-To find the difference of time between any two places, having the time of one place given, and their difference of longitude. RULE Reduce the longitude to time, and add it to the given time, if the longitude of the place whose time is required be east of the place whose time is given; and subtract it, if the longitude be west. OBs. —The reason of this is, because the farther we go east, the later is it in the day; and the farther west, the earlier in the day. That is, when it is 12 o'clock, at noon, in London, 15 degrees east of London it would be 1 o'clock, P. M.; and 16. degrees west of London it would be but 11 o'clock, A. M. QUESTIONS.-5. How is the time from one date to another found? 6. How many degrees in a circle? 7. How many degrees does the earth describe in one hour, in its revolution round the sun? 8. In one minute? 9. In one second? 10. What is the rule for finding the difference of time between two places, the longitude being known? COMPOUND ADDITION AND SUBTRACTION. 137 26. When it is 12 o'clock in London, what is the hour in Boston, 70 degrees west longitude from London? Ans. 7 o'clock 20 m. 27. When it is 12 o'clock in Boston, what is the time in London, lon. 70 deg. cast? Ans. 4 o'clock 40 m. COM0POUND ADDITION AND SUBTRACTION. ADDITION. SUBTRACTION, Art. 111.-1. A man bought Art. 112.-2. Ifa pair of oxen a horse for ~32 10s. and a pair and a horse cost ~57 Is. 6d. 2qrs., of oxen for;~24 lls. 61id. How and the horse cost ~32 10s., what mulch (lid both cost? was the cost of the oxen? 3. If I plirchase a farm for 4. I' I sell a farm for ~1150 ~1092 4s. 81., for how much 4s. 4d., and gain ~57 19s. 8d. by must I sell it to gain;~57 19s. the bargain, what did the farm 8d.? cost? 5. A pipe of wine sprang a- 6. From a pipe of wine conleak, and 31 gal. 1 qt. 1 pt. were taininig 118 gallons there leaked lost, and there remained 86 gal. out 31 gal. 1 qt. 1 pt. How many 2 qts. 1 pt. How many gallons rezmained? were there at first? 7. There was a silver tankard 8. If the weight of a silver which weighed 4 lbs. 3 oz., the tankard and lid be 4 lbs. 9 oz. 4 lid weighed 6 oz. 4 pwt. 6 grs. pwt. 6 grs., and the lid alone weigh How much did both weigh? 6 oz. 4 pwt. 6 grs., what was the weight of the tankard? 9. A merchant bought a quan- 10. A merchant bought 17 cwt. tity of sugar; sold 9 cwt. 3 qrs. 2 qrs. 14 lbs. of sugar; sold 9 25 lbs.; had 7 cwt. 2 qrs. 17 lbs. cwt. 3 qrs. 25 lbs. How much left. How much did he buy? had he left? 11. From a piece of cloth were 12. If from a piece of cloth consold 6 yds. 2 qrs., and there re- taining 39 yds. 2 na., were sold 6 mained 32 yds. 2 qrs. 2 na. How yds. 2 qrs., how many remained? much was there at first? 13. A farmer has two mowing 14. A farmer has two mowing fields; one contains 08 acres, 3 fields, containing 31 acres, 1 ro. ro., the other 12 acres, 2 ro. 24 24 rds.; one contains 12 acres, 2 rds. How many acres in both? ro. 24 rds. How many acres does the other contain? 15. A note dated July 20, 1834, 16. A note dated July 20,1834, was paid in 9 mo. 46d. At what was paid June 6, 1835. How time was it paid? long was it on interest? 12* 138 COMPOUND MULTIPLICATION AND DIVISION. COMPOUND MULTIPLICATION AND DIVISION. MULTIPLICATION. DIVISION. Art. 143.-1. If a bushel of Art. 144.-2. If 2 bushels oats cost 3s. 6d., how much will of oats cost 7s., how much are two bushels cost? they per bushel? 3. How much must be paid for 4. If 4 books cost 17s., what 4 books, at 4s. 3d. each? will 1 book cost? 5. What will 5 yds. of cloth 6. If 5 yds. of cloth cost 18s. cost at 3s. 8d. per yard? 4d., how much is it per yard? 7. How much beer in 8 bottles, 8. If 8 bottles contain 22 qts., each containing 2 qts. 1 pt. 2 how much does 1 contain? gi.? 9. If 1 gallon of molasses cost 10. If 8 gallons of molasses 2s. 8d. 3qrs., what will 8 gallons cost ~1 Is. 10d., what cost 1 galcost? Ion? s. d.qr. E. s. d. qr. If the price of 2 8 If one gallon cost 8)1 1 10 gallons be di2s. 8d. 3qrs.,itisev- 0 2 8 3 videdinto 8 parts, ~1 1 10 0 identthat 8 gallons it is evident that one of these will cost 8 times as much. We parts would be the price of one begin to multiply with the lowest gallon. Thus, 1 pound divided denomination, which is farthings. by 8, gives a cipher as a quotient 8 times 3qrs. are 24qrs.=6d. Oqr. figure, which must be written Place a cipher in the column under the column of pounds, and of farthings, and.proceed to mul- 1 pound remains, which must be tiply the column of pence, reserv- reduced to shillings: 1 X 20=20s., ing the 6d. found in 24qrs. to be and Is. added=21s. Dividing added; 8 times 8d. are 64d., and 21g. by 8, we have 2 as a quotient 6d. added are 70d.=5s. 10d. figure, and 5s. remainder, which Write the 10d. in the column of reduced to pence, 5 X 12=60, and pence, and reserve the 5s. to be 10d. added=70d., which divided added to the column of shillings. by 8=8d. and 6d. over; reduce La'stly, 8 times 2s. are 16s., and 6d. to qrs., 6X4=24qrs., divided 5s. added are 21s.=~1 Is. and by 8=3qrs.; we have then, ~0 10d., the answer. 2s. 8d. 3qrs., the answer. Art. 145.-COMPOUND MUL- Art. 146. —COMPOUND DivTIPLICATION is when the multipli- SION is wln the dividend consists cand. consists of different denomi- of different denominations. nations. RULE. RULE. Multiply the price by the quan- Divide the price by the quantity. When the quantity does not tity. When the quantity does not COMPOUND MULTIPLICATION AND DIVISION. 139 exceed 12, set down the price of exceed 12, divide by the whole quanone yard, one pound, or one gal- tity at once. Divide the highest ion, etc., and the quantity under denomination by the divisor; then, the lowest for a multiplier, observ- multiply the remainder, if any, by ing to carry as in Compound that number whichexpresses how Addition. many of the next lower denomination make one of that, adding to the product the next lower denomination; divide this sum by the given divisor, and so proceed. EXAMPLES. EXAMPLES. 1. What will 9 yds. of cloth 2, If 9 yards of cloth cost ~2 cost at 5s. 6d. per yard? 9s. 6d., what will 1 yard cost? 3. What will 8 cwt. of cheese 4. If 8 cwt. of cheese cost ~12 cost, at ~1 1O0s. 5d. per cwt? 3s. 4d., what is it per cwt.? 5. What will 24 yards of cloth 6. If 24 yards of cloth cost ~18 cost at 15s. 3d. per yard? 6s. what will one yard cost? ~ s. d. ~. d. 15 3. Whe the multi- 6)18 6 0 When the'di6 plier is greater 4) 3 1 visor is greater 4 11 6 than 12, and is a A hanl2,andisa 4 composite number, 0 3 Ans. composite num18 6 0 Ans. multiply by its ber, divide by its component parts, component parts, as in the last example, 6 X 4=24. 7. What is the weight of 56 8. If 56 casks of raisins weigh casks of raisins, each weighing 1 90 cwt., what is the weight of one cwt. 2 qrs. 12 lbs.? cask? 9. How much will 66 acres of 10. If 66 acres of land cost land come to, at ~7 9s. 6d. per ~493 7s., what will I acre cost? acre? ~11. What will 108 boxes of su- 12. If 108 boxes of sugar weigh gar weigh, each weighing 2 cwt. 256 cwt. 2 qrs., what is the weight I qr. 14 lbs.? of one box? 13. What will 112 yds. of cloth 14. If 112 yards of cloth cost cost, at ~1 10s. 6d. per yard? ~170 16s., what is it per yard? 15. How much cloth will be 16. If it take 476 yds. 1 qr. 3 required to make 121 coats, if, to na. to make 121 coats, how much make one, it requires 3 yds. 3 qrs. will it require to make one? 3 na.? 17. What is the value of 336 18. If 336 yards of cloth cost QUESTIONS.-1. What is Compound Multiplication? 2. Compound Division? 3. Rule for Compound Multiplication? 4. Rule for Compound Division? 5. How do you proceed when the multiplier is a composite number 6. When the divisor is a composite number? 7. How do you proceed when the multiplier is greater than 12, and not a composite number? 8. How, when the divisor is not a composite number? 140 COMPOUND MULTIPLICATION AND DIVISION. yards of cloth, at 2s. 5d per ~40 12s., what is the cost of one yard? yard? 19. What will 153 barrels of 20. If 153 barrels of sugar sugar weigh, each barrel weigh- weigh 516 cwt. 1 qr. 14 lbs., what ing 3 cwt. I qr. 14 lbs.? is the weight of one barrel? As 153 is not a composite num- When the divisor is not a comber, we will first find the weight posile number, divide by the whole of 100, then of 50, then of 3; the divisor at once, after the manner of several products added will be the Long Division. answer. Thus: Thus, taking the last question, cwt. qr. lbs. cwt. qr. lbs cwt. qr. ibs. 3 1 14X3= 10 0 14 153)516 1 14(3 cwt. 10 459 33 3 0X5=168 3 0 57 The divisor, 10 4 153, is con337 2 0 weight of 100. 153)229(1 qr. tail d in 516 168 3 0 weight of 50. 153 thr times, 10 0 14 weight of 3. 76 ther is a remainder of 516 1 14 weight of 153. 28 57- Tat is 622 if 153 barrels The above may be given in the 152 wscihl 516 form of a rule. C form of a rule. 153)2142(14lb3. cwt., 1 barrel 153 wei' hs3 cwt.,,When the mulliplier is not a L_ r s3 ct,7' composite number, and is hun- 612 and 57 remain dreds, multiply by 10, and this 612 part, ofi acr wht~~. ITmst arts of a cwt. product by 10, which winl give the product by 10, which gil a ive the and must be reduced to quarters, product of 100, and this by the the next lower (enomination; number of hundreds. For tens, therefore, mltily 57 by, and multiply the product of ten by the o e t ld the one qllanumber )f tens, foia units, multi- to the product add th one qarply the multiplicand. The several er, and d w he 1 qr. as products added will be the answer by 153. e now v qr. sought. the qlotient, and a remainder of ^~sought.'76, which must bo reduced to pounds by multiplying it by 28, and adding the 14 lbs. to the product. Again, dividing by 153, we have 14 lbs. as the quotient. The several quotients, 3 cwt. 1 qr. 14 lbs., are the answer. 21. How much will a man 22. If a man in one year spend spend in a year, if he spend 4d. a ~6 Is. 8d. how much will he day? spend in a day? 23. What is the value of 1900 24. If 1900 yards of linen cost yards of linen at 5s. 8 d. per ~542 5s. 10d., what will one yard yard 7 cost? COMPOUND MULTIPLICATION AND DIVISION. 141 25. What will 68 hogsheads 26. If 68 hhds. of lime cost of lime cost, at ~l Is. 6Gd. per ~73 3s. 5d., what is it per hhd. t hhd.? 27. What is the value of 26 28. If 26 yards of silk be worth yards of silk, at 9s. 6Id. per yard? ~12 8s. ld., what will 1 yard be worth? 29. How many gallons of beer 30. If 14 bottles of beer conin 14 bottles, each containing 3 tain 12 gal. 2 qts. 1 pt. 2 gills, qts. 1 pt. 1 gill? how much does 1 bottle contain? 31. What is the weight of 6 32. If 6 chests of tea weigh chests of tea, each weighing 3 21 cwt. 1 qr. 26 lbs., what is the cwt. 2 qrs. 9 lbs.? weight of I chest? 33. How many acres in 9 fields, 34. If in 9 fields there are 113 each containing 12 acr. 2 ro. 25 acr. 3 ro. 25 rds., how many in rds.? I field? 35. How many cords of wood 36. If 37 piles of wood contain in 37 piles, each containing 8 304 cords and 12 ft., how much cords, 28 ft.? in 1 pile? 37. How much will 17 casks 38. If 17 casks of nails weigh of nails weigh, each weighing 27 cwt. 3 qrs. 23 lbs. 3 oz., what 1 cwt. 2 qrs. 16 Ibs. 3 oz.? will 1 cask weigh? 39. How many bushels of ap- 40. If 125 barrels contain 425 ples can be put into 125 barrels, bush. 3 pks. 1 qt., how much each containing 3 bu. 1 pk. 5 qts.? does 1 contain? 41. If a ship sail 2 deg. 30 m. 42. If a ship sail 75 deg. 5 m. 10 sec. in 1 day, how far will she in 30 days, how far will she sail sail in 30 days? in 1 day? 43. If 3 men build 14 rds. 8 44. If 3 men build 376 rods, feet of wall in one day, how much 10 feet, in 26 days, how much do will they build in 26 days? they build in I day? 45. If 1 yard of cloth cost 46. Bought 229 yards of cloth ~2 2s. 6d., what will 229 yards for ~486 12s. 6d.; what did it cost? cost per yard? 47. The moon passes through 48. If the moon pass through 1 sign of the zodiac in 2 days, 12 signs of the zodiac in 27 days, 6 h. 38 m. 34 sec. In what time 7 h. 42 m. 48 sec., in what time does it pass through 12 signs? does it pass through 1 sign? 49. If one gallon of molasses 50. If 1000 gallons of molasses cost 4s. 2d. 2qrs., what will 1000 cost ~210 8s. 4d., what is it per gallons cost? gallon? 51. If 1 pound of tea cost 8s. 52. If 108 pounds of tea cost 5d. 2qrs., what will 108 lbs. cost? ~45 13s. 6d, what will 1 pound cost? 53. If 1 quintal of fish cost 54. If 345 quintals of fish cost 23s. 9d., what will 345 quintals ~409 13s. 9d., what was it per cost? quintal? Art. 147.-A concise view of the application of the prinm 142 SUPPLEMENT TO COMPOUND NUMBERS. ciple employed in the addition of simple numbers to compound numbers and fractions:Add 2 tens and 2 units. Add ~2 and 2 shillings. Operation. Operation. tens. units. ~ s. 2 + 2 2+ 2 10 20 22 Ans. 42 An.s. Add 1 and. Oper tion. 2= —2, and - + — I = Ans. In each case it appears that the numbers to be added'mnst be reduced to the lowest denomination mentioned; and also, th(t they are reduced by multiplying the higher by that nutmber which expresses how many of the lower make one of the higher. SUPPLEMENT TO COMPOUND NUMBERS. Art. 148,-1. What is the weight of two pieces of gold, one weighing 1 lb. 0 oz. 6 pwt. 4 rts.; the other, 2 ibs. 3 oz. 8 pwt. 16 grs.? lns. 3 lbs. 3 oz. 14 pwt. 20 ors. 2. A man has one wedge of gold, weighing 25 lbs. 3 oz. 12 pwt., mnd another weighing 1 lb. 11 oz. 12 pwt.i f irs. What is the weight of the two? Ans. 27 lbs. 3 oz. 4 pwt. 7 grs. 3. A silversmith had a quantity of silver, weighing 21 lbs. 9 oz. After refining it by melting", it weighed 15 lbs. 10 oz. 11 pwt. 19 grs. What was lost by melting? Ans. a lbs. 10 oz. 8 pwt. 5 grs. 4. What is the sum and difference of 3 lbs. 10 oz., and 2 lbs. 11 oz. 7 pwt. 4 grs.? An. Sum: 6 lbs. 9 oz. 7 pwt. 4 grs. Difference: 10 oz. 12 pwt. 20 grs. 5. What will 13 lbs. of coffee cost, at is. 2d. 3qrs. per pound? Ans. 15s. l d. 3qrs. SUPPLEMENT TO COMPOUND NUMBERS. 143 6. What will 47 yards of cloth cost, at 17s. 9d. per yard? Ans. ~41 14s. 3d. 7. How much will 10 cwt. of lead cost, at 7d. per lb.? Ans. ~32 13s. 4d. 8. What is the value of 7 cwt. of sugar, at 41d. per lb.? Ans. ~1 5 10s. 4d. 9. What is the weight of 4 hogsheads of sugar, each weigh. ing 7 cwt. 3 qrs. 19 lbs.? Ans. 31 cwt. 2 qrs. 20 lbs. 10. Bought 1~ doz. large silver spoons, each weighing 3 oz. 5 pwt.; two doz. teaspoons, each weighing 15 pwt. 14 grs.; thjree silver cups, each weighing 9 oz. 7 pwt.; two silver tankards, each 21 oz. 15 pwt.; 6 silver porringers, each 11 oz. 18 pwt. What is the weight of the whole? Ans. 18 lbs. 4 oz. 3 Pwt. 11. If 6 ells cost ~5 7ts. 6d., what will 1 ell cost? Ans. 19s. 7d. 12. What must a man spend per month, to spend ~17 14s. 3d. in a year'? Ans. ~1 9s. 61d. 13. If 8 cwt. of cocoa cost ~15 17s. 4d., what is it per pound? Ans. 4d. lqr. 14. If 132 bushels of oats cost ~20 12s. 6d., what is the cost of one bushel? Ans. 3s. ld. 2qrs. 15. If 147 bushels of corn cost ~47 12s. 5d., what does it cost per bushel? Ans. 6s. 5d. 3qrs. 16. If I'acre produce 152- bushels of oats, how much will a square rod produce? Ans. 3 pks. 6 qts. 1 pt. 17. How much wood in 11 piles, each containing 120 cords, 7 cord-feet, 11 solid feet? Ans. 1330 cords, 4 cord-feet, 9 solid ft. 18. Multiply ~86 12s. 6el. by 9; divide the product by 6; multiply the quotient by 4; divide the product by 12, and give the result? 19. If it take a printer 297 h. 59 m. 24 sec. to set 108 pages, how long will it take him to set I page? Ans. 2 h. 45m. 33 sec. 20. A person wishes to draw a pipe of wine into bottles, containing a quart, 2 quarts, 1 pint, ~ pint, of each an equal number. How many must he have? Art. 149, —When it is required to find how many times several quantities, each an equal number, may be had in a given quantity 144 SUPPLEMENT TO COMPOUND NUMBERS. RULE. Reduce the given quantity to the lowest denomination men" tioned, for a dividend, and each of the other quantities to the same denomination, and add etem together for a divisor. The quotient will be the answer. 1 quart = 8 gills. The I pipe reduced to 2 quarts 16 gills. gills equals 4032 gills, and 1 pints = 6 gills. 4032 —32=126 bottles, the ~ pint = 2 gills. Answer, 32 gills. 21. How many bushel, half bushel, and peck baskets, of each an equal number, will it take to contain 175 bushels? 22. There are four fields, one containing 10 acres, 2 roods; another 9 acres; another 11 acres, 3 roods; another 6 acres, 3 roods, 30 rods. How many shares, of 65 rods each? Ans. 94. 23. A man left $1043.28 to be divided as follows: His wife is to have two thirds; of the other third, his sister is to have one-half, and the remainder is to be divided between two nephews and nine distant relatives. To one nephew he gives 3 shares, to the other 2, to each of the relatives I share each. What is the share of each respectively? Wife,.$695.52 Sister, 173.88 Ans, Nephew, 37.26 do. 24.84 Relatives, 12.42 24. What will 156 acres of land cost, at ~5 6s. 9d. 2qrs. per acre? Ans. ~832 19s. 6d. 25. A. values a piece of land at $120, B. at 8100, C. at $110. What is the average judgment? A. 1 $120 The average is found by dividing the sum B. 1 100 of the several judgments by the number of C. 1 110 judges. 3 ) 330 $110 Ans. 26. Two gentlemen wished to exchange vehicles. One was a gig, the other was a wagon; but not being able to agree as QUESTION —. What is the rule for the 20th example? SUPPLEMENT TO FRACTIONS. 145 to the conditions, referred the matter to A., B. and C., who decided as follows: A. said the owner of the gig should pay the owner of the wagon $20, and B. said he should pay $15; but C. said the owner of the wagon should pay the owner of the gig $10. What is the average judgment? Ans. The owner of the gig must pay $8-. In cases where the judgment of the referees is part on one side of the question, and part on the other, subtract one side from the other, and divide the remainder by the number of referees, and the quotient will be the answer. 27. A. and B. wish to exchange watches, but cannot agree upon the difference. They refer the matter to C., D. and E., and agree to abide by their decision. C. gives his opinion that A. should give B. $3. D. thinks the difference in B.'s favor is $4; but E. takes the. other side of justice, and says B. should pay A. at least $1. What is the average judgment? Ans. A. must pay $2. SUPPLEMENT TO FRACTTONS. Art. 150.-A factor may be tramfierred from t/he numerator of a fraction to its denominator, and from its denominator to its numerator, without altering the value of the fraction. Thus, =_ —2.2 = - -. Let it be required to separate the terms of 4.- into their prime factors, and transfer the factor 2 from numerator to denominator, and the factor 5 from denominator 2 2 On whna principle is the factor 2 transferred from numerator to denominator, and the factor 5 from denominator to numerator? On what page is the principle first illustrated? Repeat the language. Separate the terms of the following fractions into their prime factors, transfer as above, and illustrate. 6 14 9 10 12 12' 28' 27' 48' 25' EXAMPLES. 1. What is the sum of - and? Ans. 1- =1i. 13 146 SUPPLEMUNT' TO FRACTIONS. 2. If a manll cceive - a- of a dolllr for 1 day's w-ork,; 4or another, and I for alnolher, how much does he receive for the 3 days' work? Ais. 121. 3. If I receive v of a cord of bark from one mnl;, and 6 from another, what part of a cord do I receive from both? Ans. 1 -8. 4. Add 5, G1 28, 14-. Ans. 15s-. 5. From - take 1. Ans. 8. 6. From 12- take 10. Ans. 15. 7. Bought a piece of land contaiining 491- acres. Sold 213 acres. How many were there left? Ans. 28-. 8. From -1- of a d-la tlke -0 of a minute. 40 A1n. 5 h. 59l m. 9. From 78 of of4 take ^ of- of. of S. 2s. V f 4 5' 10. A man had 3 bags of mone, oe, ontaiing in,l1 450 lbs.; in the first bag he had 230127 lbs., in the second 1001-}. How many were there in the third? Ans. 1199. 1. From 1I of a pound take 1 of - of - of 7 of 8 of 8 shillings. Ins. ~ 2 - 12. If a bushel of corn cost 6 of a dollar, what will - of a bushel cost? Ans. $y. 13. If I of a bushel of corn cost i of a dollar, how much must be paid for 1 bushel? Ans. -^. 14. If 9 of a bogslhead cost ~4-3-4, what wiill be the cost of 6a of a hhd.? Ans. ~1i 70 41 15. Multiply 30 by 41 of 2 of a of -, and divide the product by of 8 of of 21-. Ans. 0. 16. Divide } of- - byx of - Ans. 1 8 3.3 4' 17. Reduce 8 of a pound, avoirdupois, to the fraction of a cwt. Ans. 12 6.18. Multiply - of a day, reduced to minutes, by the fraction L. Ans. 315 m. 19. Reduce 3s. 6d. to the fraction of a pound. Ans.'~-7. 20. Reduce I g cwt. to thle fraction of a pound, avoirdupois. Ans.. 9. 21. What is the value of 4 4 of a dollar? Ans. 5s. 7Id. 22. What is the value of 2 of a Julian year? Ans. 257 d. 19 h. 45 m. 521 sec. 23. What is the value of:9r of a g-iaeal? Ans. 18s. CIRCULATING DECIMALS. 147 24. Reduce 4 cwt. 2 qrs. 12 lbs. 14 oz. 123 drs. to the fraction of a ton. Ans. -. 25. Reduce 16 h. 36 m. 55-3 s. to the fraction of a day. Ans. -3. 26. Reduce 2 qrs. 9 lbs. 10 oz. 72! d's. to the fraction of a cwt. iAns. 7 27. If 100 oranges cost 10As., how many hundred may be bought for 105 s.? Ans. 10. 28. HIow much will - cwt. cost, at 153s. per cwt.? iAns. 3s. 11~1d. 29. If a of a yard cost 18d., what will 1 yard cost? Ans. 2s 30. If 3 of v of I of a ship be worth - of 6 of 9 of the 7 5 8 9 7'T cargo, valued at $36,000, what is the value of the ship? Ans. $45,000. CIRCULATING DECOIMAILS. Art. 151 — CIRCULA'ING, or RECURRING DECIMALS, are those that consist of a repetition of a number of digits, as.646464, etc.,.4127127127, etc.; in fact, every decimal that is not finite is a circulating decimal, or is such, that if continued far enough, the same figures will again recur; but it is only those of which the periods of circulation consist of a few figures, that generally receive the definition of Circulating Decimals. When the circulation consists of the same digit repeated, it is called a Simple Circulate, and is distinguished by a point placed over it; thus,.11l, etc.=.1;.333=.3, etc. When the period of circulation consists of more than one digit, it is called a Compound Circulate, and is distinguished by a point over the first and last repeating figure; thus,.234234234, etc.=.234. A Mixed Circulate is that which has other figures in it that are not repeated, as.7848484, etc., and these are represented thus,.784. As all operations, as multiplication, division, etc., of these numbers may be performed by the same rules which are give~ for common decimals, and as but few cases occur in whicb those rules are not to be preferred, some rules only will be 148 %aCULATI'C DEC(iMALS. given for the reduction of circulating decimals to vulg3ar fi'ra tions, leaving the student to apply the rui:'s. If, however, he should wish to pursue the subject farther, he can find his curiosity amply gratified by consulting the following authors: Brown, Cunn, Mhalcolm, Emerson, Donn, and particularly Henry Clarke; also, Dr. Wallis, all of whom have treated at some length the theory of Ciiculating Decimals. REDUCTION OF CIRCULATING DECIMALS. Art. 152.-To reduce a simple, or compound circulate, to its equivalent fraction. R U LE. Take the given decimal, considered as a whZole numnber, for the numerator; and as many 9's as there are places in the circulalte, for the denominator. TVhten there ore aniy integral figures in the circulate, as many ciphers must be annexed to the numerator as the highest place in the repetend is distant from the decimalI point. EXAMPLES. 1. The circulate-.6= 6= 2 2. 3- 3- 4.3 9 3. - -.09 = 9 4. - - 2.063=2 —- 2 15. - - 1.62= 1. Art. 153.-To reduce a mixed circulate to its equivalent fraction. RULE Subt act the finite part of the expression, considered as a whole number, from the whole mixed repetend, taken in the same manner for the numerator; and to as many 9's as there are repeating places in the circulate, annex as many ciphers as there are finite decimal places for a denominator; thus1. The circulate-.138= 1-o 3- 2-, 5 2. Again-2.418 =24-=2 4 - 2 4 OBs.-This rule, as it is not of great practical utility, may be passed over until the review. RATIO AXN) P.aOORTIOa X. 149 RATIO AND PROPORTION. Art. 151. —We arrive at a knowledge of particular quantities by comparing them with other quantities, which are either equal to, or greater or less than those which are the objects of inquiry. We may inquire, how much greater one quantity is than another; or how many times the one contains the other. The answer to either of these inquiries is termed a ratio of the two quantities. One is called arithmetical, and the other geometrical ratio. Art. 155,-Arithatetical ratio is the diference between two quantities. Thus, the arithmetical ratio of 6 to 3 is 3. It is sometimes expressed by two points placed horizontally between the two quantities; thus, 6. 3=3, which is the same as 6 —3=3. Art. 156.-Geometrical ratio is the quotient arising fporn dividing one iuantity by anzoter. Thus, the ratio of 6 to 3 is -, or 2. Geometrical ratio is expressed by two points placed one over the other, between the two quantities compared; thus 6: 3=2. If the ratio is not specified, it is always understood to be geometrical. The two quantities taken together, are called a couplet. The number which is compared, being placed first, is called the antecedent, and that with which it is compared, the consequent. Of these three, the antecedent, the consequent, and the ratio, any two being given, the other may be found. EXAMPLES. 1. If the antecedent be 16, and the consequent 4, what is the ratio? Ans. 4. 2. If the antecedent be 18, and the ratio 3, what is the consequent? Ans. 6. Art. 157.-Inverse, or reciprocal ratio, is the ratio of the reciprocals of two quantities. OBs.-The reciprocal of any quantity is a unit divided by that quantity. Thus, the reciprocal of 4 is 4, the reciprocal of 3 is ~. The reciprocal ratio of 6 to 3 is I to I; that is, - ~, which QUESTIONS.-1. What is Ratio? 2. What is arithmetical ratio? 3. What is geometrical? 4. What is compound ratio? 150 RATIO AND PROPORTION. is equal to -. Hence, a reciprocal ratio is expressed by inverting the terms of the couplet. The reciprocal ratio of antecedent to consequent, is the direct ratio of consequent to antecedent. The direct ratio of 6 to 3 is =2. The reciprocal ratio of 6 to 3 is -A= -. Art. 158 — Compound ratio is the ratio of the products of the corresponding terms of two or more simple ratios. Thus, the ratio of 9: 3 is 3. And the ratio of 6: 2 is 3. 54: 6 —'9. OBS. 1. —A compound ratio is not different in its nature from a simple ratio. The term compound is used merely to denote the origin of the ratio. Art. 159.-In a series of ratios, if the consequent of each preceding couplet is the antecedent of the following one, the ratio of the first antecedent to the last consequent is equal to that which is compounded of all the intervening ratios. Thus, 12: 6 6: 18 18: 3 3: 4 Art. 160.-If we multiply all the antecedents together, and all the consequents together, it will be found that the ratio of the products of the antecedents to the product of the consequents, is equal to the ratio of 12, the first antecedent, to 4, the last consequent, which is 142=3. OBS. 2.-Rejecting all the antecedents but the first, and all the consequents but the last, is cancellilg equal factors from dividends and divisors. (See Art. 42.) Art. 161. — f, in the several couplets, the ratios are equal, the sum of all the antecedents has the 3,ame ratio to the sum of all the consequents, which any one of the antecedents has to its consequent. Thus, 12: 6=2 OB. 3. —It will be observed, in 10: 5 =2 this example, that the terms of the 8: 4=2 ratio are not used as factors. The ratio is, therefore, not a compound 6'' 3=2 ratio. 36: 18-2 It has already been shown (Art. 44) that to multiply the dividend with a given divisor, is the same as to multiply the RATIO AND PROPORTION. 151 quotient, and to multiply the divisor with a given dividend; is the same as to divide the quotient. In Fractions the same principle was recognised, with this difference only in the mode of expression; we substituted numerator for dividend, and denominator for divisor. We shall now substitute antecedent for numerator or dividend, consequent for denominator or divisor, and ratio for value of the fr-ction. Art. 162. —To multiply the antecedent, or to divide the consequent, is the same as to multiiply the ratio. (See Art. 44.) Thus the ratio of 12: 6 is 2 Multiply the antecedent by 2, the ratio of 24: 6 is 4 Divide the consequent by 2, the ratio of 12: 3 is 4 Art. 163.-To divide the antecedent, or to multiply the consequent, is the same as to divide the ratio. Thus, the ratio of 8: 4 is 2 Divide the antecedent by 2, the ratio of 4: 4 is 1 Multiply the consequent by 2, the ratio of 8: 8 is 1 Art. 161,-To multiply both antecedent and consequent by the same quantity, does not affect the ratio. Thus, the ratio of 6: 3=2 Multiply both terms by 3, 18: 9 —2 the same ratio. Divide both terms by 3, 2 1 =2 The ratio of two fractions, which have a common denominator, is the ratio of their numerators. (See Art. 76.) Thus, 2. 1=. The direct ratio of two fractions, which have a common numerator, is the reciprocal ratio of their denominators. (See Art. 77.) Thus, 3: 3=. Art. 165. — A factor may be transferred from antecedent to consequent, and from consequent to antecedent, without altering the ratio; observing, that when a factor is transferred, it becomes a divisor, and when a divisor is transferred, it becomes a factor. (See Art. 150.) Thus, the ratio of 16: 2 X 42 same io. Transferring the factor 2, 6: 4 =2 Art. 166.-It may be observed, in regard to ratio, that it exists only between quantities of the same nature, or, the things compared must be so far alike that one may be said to be larger or smaller than the other. For example, a rod can. 152 RATIO AND PROPORTION. not be said to be longer than an hour, nor can there be a comparison between them in any respect, for there is no common property. But a rod can be said to be longer than a foot, for it is made up of feet. There may be, however, a relation between the numbers which stand for quantities of a dissimilar nature. Thus, the ratio of 16 to 8 is 2. Now, 16 may stand for rods, and 8 for hours, which things bear no relation to each other. The subject of ratio is of incalculable importance, since it lies at the foundation of all arithmetical investigation. The practical nature of ratio will be seen by the following example. 1. If 6 yards of cloth c(st 30 dollars, what will 12 yards cost? The ratio of 12: 6 is 2, which shows that 12 is twice as large as 6. It is, therefore, plain that the cost of 12 yards will be as much greater than the cost of 6 yards, as 12 is greater than 6. Therefore, 30 x 2=60, the cost of 12 yards. Again, if we know the price of 1 yard, we can repeat this price 12 times, and thus obtain the price of 12 yards. If 6 yards cost 30 dollars, it is evident that one-sixth of 30 will be the cost of 1 yard. Although, strictly speaking, there is no relation between the cost and the number of yards, yet the ratio of 30 to 6, considered as numbers merely, is a number which will represent the cost of 1 yard. Therefore, 30: 6 =5, the cost of 1 yard, and 5 x 12=60, the cost of 12 yards, as before. If we now compare the cost of the second with the cost of the first piece, we shall find that the ratio is equal to the ratio of the length of the second piece, to the length of the first piece. Thus, 12: 6=2, and 60: 30=2. When two or more couplets of numbers have equal ratios, these numbers are said to be proportionals. Hence, (Art. 167,) Proportion is an equality of ratios. Arithmetical Proportion is an equality of arithmetical ratios, and Geometrical Proportion is an equality of geometrical ratios. Proportion may be expressed, either by the common sign of equality, or by four points placed between the couplets. Thus8 * 6=4. 2, or 8. 6:: 4 2, arithmetical proportion. 12: 6=8: 4, or 12: 6:: 8: 4, geometrical proportion. The latter is read,-the ratio of 12 to 6 equals the ratio of 8 to 4, or 12 is to 6 as 8 is to 4. The first and last terms are called the extremes, and the others the means. RATIO AND PROPvA-rlOI. 153 Alrt 167.-The number of terms must be at least four, for the equality is between the ratios of the couplets; and each couplet must have an antecedent and consequent. There may be, however, a proportion between three quantities; for one of the quantities may he repeated, so as to form the two terms. Thus, 6:12:: 12: 24. Art. 168.-If four numbers are in geometrical proportion, the product of the extremes is equal to the product of the means. Thus, 12: 8::15: 10, for 12 X 10=8 X 15. Art. 169.-By multiplying the extremes and means together, a proportion is reduced to an equation. When the product of any two numbers is equal to the product of any other two, the numbers may be formed into a proportion by taking the factors on one side of the equation for the extremes, and those on the other for the means. Thus, 4 x3=6 6x2. Making 4 and 3 constitute the extremes, and 6 and 2 the means, we have the following proportion; 4: 2:: 6: 3. Form proportions of the following equations: Ox 8= 4X12 3x12- 4X 9 4x 7=14X 2.8X 9=-12x 6 Art. 170 —In compounding proportion, equal factors may be rejected from antecedents and consequents. Thus: 12: 4::9: $ 4: S::$: 0 $:20:: 1: 15 12:20::9:15 Art. 171 —If the corresponding terms of two or more ranks of proportional quantities be multiplied together, the products will be proportional. Thus: 12: 4:: 6:2 10: 5:: 8:4 120:20::48: 8 Art. 172.-If the terms in one rank of proportionals be divided by the corresponding terms in another rank, the quotients will be proportional. Ts 12: 6::18: 9 Thu 6:2:: 9:3 Then, 6:.::. ~:: -T 154 RATI'O AN7D PROPORTION. Art. 173.-If to or from th.e terms of any proportion, there be addted or subtracted the corresponding terms of any other proportion, having tlh3 same ratio, their sums or remainders will be proportional. Thus, 14:'7::16: 8 4:2 6 3 4:2:: 6: 3 18:9:: 22: 11 10: 5::10: 5 Art. 174.-A factor may be transferred from one mean to the other, or from one extreme to the other, without altering the ratio, 1: 8:: 12: 6. The scl'olar may be exercised upon the foregoing proposition, in the followin g manner' eatcler. lWhat are the factors of the antecedent of the first couplet? cll., 1ar. 4 an1 4..1 Transfer one of these to the consequent, and illustrate. S. 4:: 12: 6. To divide antecedent and consequent by the same quantity, does not affect the ratio. Y' How does it appear that the antecedent has been divided? S. We have removed from it the factor, 4. Removing a factor from any quantity, divides by that factor. r ]What are the factors of the antecedent of the second couplet? S. 4 and 3. T, Transfer the 4 to the consequent, and illustrate. S. 4::: 3:.'7' Remove the denominators from the consequents. f. 4: 8:: 3:6. T. What effect upon the consequents has removing the denominators? T1 Is there a proportion between the four following numbers? 4: 2:: 6:3. Illustrate. S. 4 X32 X6. T. Remove the factor, 4, from the left of the equation, and illustrate. 2 X 6 To divide both members of tl^3 equation by the same 4.' 3 -quantity, does not affect the equation. T Do the four following numbers, 16 8:: 12: 6, constitute a proportion? S. They do. T. HowI do vou know? S. Thle ratios between the couplets are equal.'T Divi de the consequents by 2, and will there then be a proportion? S. There will. 7' Are not the ratios affected? S. Thlev are.'. Wliy then is not the proportion destroyed? S. The ratimo are still equal. 7' In what, therefore, does proportion consist? IRATI1j AND) R OF I TON. 155 S. In equality of r.atios. T. IlowA (ldo you ascertain when tlie ratios;re equal? S. By dividillg the antecedents iy tle cons;eqiuetlt s, or by dividingl consequents by antcccdents, or by multiplying the extremes tiogether and the means togetlher.'. How do you reduce a proportion to an equation? ilowv ( you form a proportion fiom an equation Rteduce to an eqluation tle iflowl-ilg proportion; 5:10::4:8. Add 2 to each member oi' tle equation. Is the equation affected? Why not? Add 2 to one nmember, anld 3 to to te other. Is the equation now affected? Repeat the axiom. Let the teacher multiply exercises of this kind. Art. 175.,Inverse, or reciprocal proportion, is an cqu:al.ity between a direct and reciprocal ratio. Thus, 4:2::' -. That is, 4 is to 2 as 3 is to 6 reciprocally. Sometimes the order of the terms is inverted, without writing them in the form of a fraction. Thus, 4: 2:: 3: 6, inversely. In this case the first is to the second as the fourth is to the third. We have seen that a factor may be removed fiom antecedent to consequent, and the reverse, and the proportion still be preserved. Art. 176'-The terms of the proportion mray also be changed, provided that the equality of the ratios be not affected. Thuls, 12: 8::15:10 12:15:: 8:10 8:10::12:15 8: 12::10: 15 10: 8::15:12 10:15:: 8:12 15:10::12: 8 15:12:: 10: 8 In all these chalnges the product of the extremes will be found equal to the product of the means. If, therefore, we have the product of the extremes, and one of the means, it is easy to find the other. We can, therefore, find any one term of the proportion when we know the other three, for the term sought must be one of the extremes, or one of the means. The operation by which, three terms being given, a fourth proportional is found, is called the " Rule of'Three," or " Rule of Proportion." Tie.lre must always be three terms or numbers given, two of wh-ich are of the same kind, alid the other of the kind of the answer required. 156 RATIO AND PROPFOTION. SIMPLE PROPORTION, OR RULE OF THREE. Art. 177.-Proportion is of two kinds, Direct and Inverse. Proportion is direct, when the ratios are in the order in which the question is proposed; Inverse, whien one of the ratios is inverted. A question is known to belong to Direct Proportion when more requires more, or less requires less. More requires more, when the second term is greater than the first, and requires that the fourth be greater than the third. Less requires less, when the second term is less than the first, and requires that the fourth be less than the third. 1. If 3 men build 12 rods of wall in a given time, how many rods will 6 men build in the same time? In this question the ratios are in the order in which the question is l rcposed, 3: 12:: 6: the answer. M.ore requires more: for, tvidently, 6 men will perform more labor in the same tine than 3 men. We may employ the same numbers in the proposal of a different que.;tion, and the ratios will be inverted. 2. If 3 men perform a certain amount of labor in 12 days, in how many days would 6 men perform the same? In this question more requires less: for G men would require less time to perf)orm the s:mne atmount of labor than 3 men: 3 is to 6 reciprocally as 12 is to the answer,::: 12: the answer. Of tlhe three terms given in Proportion, two are called the terms of conditi),, and one the term of demanzd. Thus, 3 men, 12 days, are the terms of condition, and 6 men the terrm of' demand. It may be observed that, in Proportional questions, the term of demand is the only term which presents any difficulty. The two other terms simply state a fact, or the condition upon which the conclusion rests, and are to be employed as the means of solving the difficulty. The answer to the question proposed, in Direct Proportion, depends upon the ratio of the term of demand to that term of the condition, which is of the QtESTIONS. —5. What is Proportion? G. VWhat are the first and last terms called? 7 lHaviln the extremes and one of the means given. lhow nmy the other mean be found? 8. What is the Rule of Three? 9. How are the ratios? 10. What is meant b: the order in whici the question is proposed? 11. -low is a question known to belon; to the Rule of Three Direct? Illustrate. 12. 1Lale lor staling the question? RATIO AND PROPORTION. 157 same name or kind. In Inverse Proportion, the answer depends not upon the ratio of demand to condition, but of condition to demand. In either case, this ratio multiplied into that term, which is of the same name or kind as tile answer required, gives the desired result. Solution of Question 1st. Soluti.on of Question 2d. IX 12=24 Ans. X 126- Ans. The ratio of 6 men to 3 rren The ratio of 3 men to 6 men, expresses how much more labor expresses how much less time ( 6 men can perform in a given mein would require than 3 men, time, than 3 men, — 2. They to perform the same amount of would perform twice as much. labor. The first step in solving a ques- If 3 men will perform a certain tion is to find the ratio; the sec- amount of labor in 12 days, 6 ond, to multiply. If 3 men build men will perform the same in - 12 rods, then 6 men will build of 12-6 days. Also, if it takes 3 of 12=24 rods. 3 men 12 days to perform a certain amount of labor, it will take 1 man 3 times as lonf, 3X 12= 36, and 6 men one-sixth as long as 1 man, 36- 6 = 6 days. 3. If 5 tons of hay cost 10 4. If 7 men reap a field cf dollars, how many dollars will 20 grain in 14 days, how many men tons cost? can reap the same in 21 days? Does this question belong to Does this question belong to Direct or Inverse Proportion? Direct or Inverse Proportion? How do you know? How do you know? What is meant by more re- What is meant by more requiring more? Illustrate by ex- quiring less? ample 3d. How (do you know when more Which is the term of demand? requires less? Which are the terms of con- Are the ratios direct or roeipdition? Statf the question ac- rocal? cording to the example given. What do you mean by a recipWhat is the first step in the rocal ratio? solution? What is the second? Is the ratio that of the demand Which term presents the diffi- to the condition, or of the condiculty? tion to the demand? State the How are the other terms to be question, and illustrate. employed? Art. 178 —It has been stated that, in Proportion, we have either the two extremes and one of the means, or the two means and one of the extremes given, to find the other. 14 158 RATIO AND PROPORTION. 1. The extremes in a proportion 2. Tile extremes in a proporare 12 and 4, one of the means 6. tion are 12 and 4, and one of the What is tho other mcan? moans 8. What is the other mean? 3. The moans in a proportion 4. The means in a proportion are 8 and 6, and ono of the ex- are 8 and 6, and one of the extremes is 4. What is the other i tremes is 12. What is the other extreme? I extreme? 5. If a man travel in 3 days 6. The extremes in a propor30 miles, how many miles can he tion are 12 and 9, and one of the travel in 9 days? means is 4. What is the other mean? Operation..-X 60-180 Ans. If he travel 60 miles in 3 days, 7. The extremes in a proporthe ratio of 9 to 3 shows how tion are 12 and 9, and one of'the many more miles he could travel means is 27. What is the other in 9 days than in 3 days. mean? From the.preceding illustrations of ratio and proportion, we derive the following RULE. I. WVrite that term of the condition which is of the same name or kind as the fourth term, or answer required, for the third term'. If the Proportion be Direct, write that number which e.rjzesses the demand for the second, and the remaining term of the condition for the first. If the Proportion be Inverse, write the number' which expresses the demand, for the first, and the remaining, termn of the condition? fbr the second term. II. M~ultiply the second and third terms together, qnd divide the product by the first. III. If the first or second terms consist of different denominations, reduce both to the lowest mentioned. If the third term consist of different denominations, reduce it to the lowest, or the lower to afraction of the highest. IV. Since the second and third terms are factors of a dividend, and the first term is the divisor, any factor common to the first and second or first and third terms, may be rejected. OBS.-For the statement of a question proportionally, the foregoing is the rule, but for practice, the following has many advantages: 1st, It is more convenient for cancelling; 2d, We avoid all fractions in the operation; 3d, We can often avoid the labor of the reduction of the terms. RATIO AND PROPORTION. 159 RULE FOR CANCELLING. Draw a pe7eendicular line, and place the sign of the answer on the left, (if the answer is to be in dollars, this $ is called the sign; if pounds, this ~,) and the term which expresses the demand on the right of the line, and that term of the condition, which is of the same name or kind, on the left, and the remaining term of the condition, on the right. Reduce, cancel, multiply, and divide, as before directed. If we do not wish to consider whether the proportion be direct or inverse, the following may be adopted:RUL E. Write that number which is of the same name or kind as the answer to the question, for the third term of the proportion. If the answer is greater than the third term, write the greater of the remaining terms for the second, and the less for the first, but if the answer is less than the third term, write the less of the remaining terms for the second term, and the remcaining for the first. Reduce the terms, cancel, multiply, and divide, as before directed. EXAMPLE. Art. 179.-If 3 yards of cloth cost 18 dollars, what will 9 yard s cost? Operation 1st. Operation 2d. 3 6 $:9:: 18 $: $:: 9 3 6 54 Ans. 54 Ans. Having stated the question and solved it, the student will give the following illustration at the black-board. This question belongs to Direct Proportion; more requires more-the third term is greater than the first, and requires that the fourth be greater than the second. Having placed the name of the answer for the third term, we write 9, the number expressing the demand, for the second term, and 3, the remaining term of the condition, for the first. The answer depends upon the ratio of demand to condition. The ratio of 9 to 3 shows how much more 9 yards will cost than 3 yards; but 3 18::9: the answer. The ratio of 18: 3 expresses the 160 RATIO AND PROPORTION. cost of one yard. Hence, by analysis, if 3 yards cost 18 dollars, one yard will cost 18 = 6, and 9 yards will cost 6x9=54. Teacier. In what does Proportion consist? Y2 How mnuy ratios are there? Scholar. Two. T. Iow many circumstances in the given question affect the answer S. One. YT What is it? S. The difference between the number of men in the demand and condition. Operation 3d. 81 3 $ 18 54 Mns. This mode of slating the question will admit of the same genera illustration as the otller. Buit a variety of illustrations may be employed. The following may be adopted. We place the siign of the answer on the left of the line, and first dispose of the term of demand, because it is the telrm which presents the difficulty. We place it on the right of the line, because it is to be a dividend. It is also properly the antecedent in the last couplet, whose consequent is the term sought. We therefore place that term which is of the slnme n1ame, on the left, for a divisor, and close the statement tby placing the name of tle answer on the right. The first step il the solution is to filn tie ratio of demand to condition, which we fiond to be 3. The second, to multiply the ratio into the name of the answevr. We thus obtain 54 dollars, the answer. Y' Do we necessarily, in the solutioi, first find the ratio of demand to condmitiou.? S. We do not. T. Why not? S. We illnty reject from either factor of the dividend, a factor equal to the divisor: for 18 X 3 = 9 X 6 = 54. i' If we reIjet the factor 3 fronm the divisor, and fronm 18, one of the factors of the dividend, how will the question read? If we reject the factor 3 fionm the divisor, and fiom 9', one of the factors of the dividend, how will the question then read? T Prove y our answer to be right, and also show the connection between antece-dent rand con:sequent. N. We imay sutb-itute in the place of the sign of the answer, the answer itself; we shall then have the means on the right, and the extremes on the left of the line. We coiinmence with the answer, or last consequent, anl! dratw a line to its antecedent, or the demand, thence to the left, coinnecting it with the condition, or firsi antece- 9 dent, thelice to its consequenIt on the right, then back \ to the point of commencemert, The statement then 3_p8 reads, consequent, antecedent, antecedent, consequent. Thus antecedents alre connected with antecedents, and consequents with consequents. RATIO AND PROPORTION. 161 EXAMPLES. 1. If 12 yards of cloth cost $48, what will 4 yards cost? ilns. $16. 2. If 4 bushels of wheat cost $8, how much will 16 bushels cost?. ns. $32. 3. If a man earn $24 in 12 days, how much does he earn in 6 days? Ans. $12. 4. If 8 yards of cloth cost $12, how much will 10 yards cost? Ans. $15. 5. If 10 yards of cloth cost $15, how much will 8 yards cost? Ans. $12. 6. If 6 acres of land are bought for $180, for how much may 15 acres be bought? Ans. $450. 7. If 15 acres of land cost $450, what will 6 acres cost? Ans. $180. 8. If 18 yards of cloth cost $36, what will 20 yards cost? Ans. $40. 9. If 7 men be paid $82 for a certain amount of labor, what ought 25 men to receive at the same rate? Ans. $30. OBs. 1.-Mixed numbers must be reduced to improper fractions, and the numerators placed on that side of the line where the whole numbers would be placed. Let it be remembered, that the numerator of a fraction always occupies the same side of the line which a whole number would occupy, standing in the place of the fraction. (See question 14, below.) 10. If 2 horses plough 51 acres in a day, how many acres would 18 horses plough in the same time? Ans. 46 acres. 11. If 31 dollars will buy 6- yards of cloth, how many yards will $40 buy? Ans. 81 yards. 12. If $123 buy 4- yards of cloth, how many yards will $174 buy? Ans. 57 yards. 13. If 1 4 of a bushel of wheat cost 25-, how much will 60 bushels cost? Ans. $115-. 14. If 4 of a yard of cloth cost - of a dollar, how much will I- of a yard cost? Ans. $2~. Read this question thus: What will 7 yards cost, if 1 yard cost 2 dollars? QUEzTIONS.-13. When either of the terms is a compound quantity, what is the rule? 14. Analyze question'2d. 15. Rule for mixed numbers? 1i. On which side of the line it the numerator of a fraction to be placed " 14':' 162 RATIO AND PROPORTION. Statement. Now write each denominator on the side 7 of the line opposite its numerator, thus: ~1572 0Operation. $7 1~ 3 3 7 = 22 Ans. 15. If 3 horses consume 4;- tons of hay in 4 months, how many tons will 22 horses consume in the same time? Ans. 33 tons. 16. If 1 yard of ribbon cost 8 pence, how many dollars will 2 yards cost? Ans. $8. Operation. ) operation OBs. 2.-When the answer is required in a How many $ I7 yd. different denomination from that given in yd. 1 8 d. the supposition, follow the tables from the d. T 1I s. denomination given to the denomination res. 1 $ quired. In the last example, the price of 1 yard is 8 pence. The answer is required in $8 Ans. dollars; therefore, continue the statement by saying, 12 pence make 1 shilling, and 6 shillings make 1 dollar, the denomination required. Then 6 times 12 on the left cancels 72 on the right; 8 being the only number left on the right of the line, and there being no number on the left greater than 1, 8 is the answer in dollars. 17. If 1 pint cost 10d., what will 3 hhds. cost in pounds? Operation. How many ~ - $ hhds. Redung In this example hhd. 1 6- 3 gal. iiogsheads the first.and third gal. 1 4 qts. to Pit, terms are of difsee Red. qt. 1 g pts. Descend. ferent denominapt. i-^- 0 d. ) Reducing tions, and the secd. It 1 s. pence to. ond term is difsee Red. s. I0' \ I ~ ) Ascend. ferent from the ~63 Ans. answer sought; therefore, follow the tables until you find the name of the answer required, observing to commence each successive step on the left with the QUESTIONS.-17. Repeat the Ohs. under question 16. 18. What is the ratio of each tern in the demand to each corresponding term in the supposition, in question 16? RATIO AND PROPORTION. 163 denomination last placed on the right, thus: IIow many ~ will 3 hhds. cost? The facility of passing from one If I hhd. is 63 galls. denomination to another will be and 1 gal. is 4 qts. readily seen in the statement of and 1 qt. is 2 pts. this question; and also that the and 1 pt. is 10 pence. process of Reduction Descend- and 12 d. is 1 s. ing and Ascending employed in and 20 s. is 1 ~ the ordinary mode, is rendered unnecessary by cancelling; thus, 3 times 4, on the right, are equal to 12 on the left, and 2 times 10 on the right are equal to 20 on the left; 63 being the only number left greater than 1, and standing on the right, it is the answer in pounds. 18. If 3 hhds. cost ~63, what will I pint cost in pence? Ans. IOd. Statement. Having made the stateHow many pence, d. -1 pt. ment, the connection bepts. 2_ -1' I qt. tween the numbers may qts. A 1 gal. be shown by connecting, gals. 63 — 1 hhd. with a continued line, hhds. 3 f:63~ first, the sign of the an~1 =-X__20s. swer on the left with the s.l - \1 2d. demand on the right, and the demand with a number of the same name on the left, and this again with its equal, or its value, (as 3 hhds. with ~63, its price,) on the right; and thus on from left to right alternately, until you come to the name of the answer, on the right; then returning with the line to the sign of the answer where you commenced. 19. If 1 yard of cloth cost 13{ shillings, how much will 12 ells English cost in pounds? Ans. ~10. Operation. 13=4- l 20. If 12 ells English cost How many ~ It Ell E. ~10, what will 1 yard cost? EllE., I $ qrs. Ans. 131s. Qrs., g 1 yd. 3 Qrs., 4 1 yd. l21. If 1 pint of rye cost 2 Yd., 1 f0s. 10 pence, what will 7 chaldrons Shil, $0ji cost in guineas? Shil, As~1 Ans. 851 guineas. I 11~ Ans. 161 RATIO AND PROPORTION. 22. If 1 pint cost 10d., what will 3 hhds. cost in dollars? Ans. $210. 23. If I nail cost 3 farthings, how many pounds will 40 yards cost? Ans. ~2. 24. If 7 chaldrons cost 85- guineas, what will 1 pint cost in pence? Ans. 2d. 25. How many pounds will 3 tons of lead cost, at 2 farthings per ounce? How many guineas? How many dollars? Ans. ~224; 160 guineas; $7463. 26. What will 20 dozen pairs of gloves cost, at 4s. 6d. per pair'? Ans. ~54. OBS. 3. —4 shillings 6 pence may be reduced to pence=54 pence; or thus, 4 shillings==s. 27. How many yards of cloth may be bought for ~32 10s., at 12s. Gd. per ell English? Ans. 65 yds. 28. If X of a yard cost - of a shilling, how many guineas will 42 yards cost? Ans. 5 guineas. 29. If 96 lbs. of red lead cost ~3 12s., what is 1 lb. worth? Ans. 9d. 30. What is the value of 11 cord of bark, if 41 cords be worth $20.25? Ans. $5.620. 31. If 6 yards cost $3, what will 91 yards cost? Ans. $4.269+. 32. How many miles will a man travel in 8 of a day, if in - of a day he travel 5120 rods? Ans. 20 miles. 33. If - of a barrel of flour serve 3 men 1 day, how much will be sufficient to serve 402 men the same length of time? Ans. 7l- barrels. 34. A man owning I of a coal mine, sold A of his share for $36000. What was the value of the mine? Ans. $80000. 35. What will 8 bales of cloth cost, each bale containing 12 pieces, and each piece 27 yards, at $54 per piece, and what will be the cost of 1 yard? Ans. The whole $5184, and I yd. $2. 36. If $100 gain $6 in one year, how much will $450 gain in the same time? Ans. $27. 37. If 17 tons 12 cwt. of iron cost ~165, what will be the cost of 2 cwt.? Ans. 18s. 9d. 38. If 112 lbs. of beef cost 18s. 8d., what is 1 lb. worth? Ans. 2d. 39. If a man travel 200 miles in 15 days, what is the average distance for every 3 days? Ans. 40 m. RATIO AND PROPORTION. 165 40. If - of a yard cost I of a pound, what will 60 yardscost in dollars? Ans. $240. 41. If 30 horses consume 70 bushels of oats in 4 weeks, how many bushels will 9 horses consume in the same time? Ans. 21 bushels. 42. A merchant bought a number of bales of velvet, each containing 129~17 yds., at the rate of $7 for 5 yds. Sold the same at the rate of $11 for 7 yds., and gained $200 by the bargain. How many bales were there? Ans. 9 bales. Operation. Yds. in a bale. FIRST STATEMENT. 129 7- 35 How many $ 5 yds. Sold 5 yds. for'$7. yds. 7 11 $ Bought 5 yds. for $7. 7 55 Q$7. Gained on 5 yds. $1. SECOND STATEMENT. Bales]200$ a$ $ 7 $ yds. yds. $Wl00 9 1 bale. 9 bales Ans. Bs.4.-In the above question it is necessary to make two statements. First, to find the gain on 5 yards. We then say, on how many bales is $200 gained, if 6 of a dollar is gained on 5 yards, and 3 5 0 yards make 1 bale? 43. If 500 men consume 102 barrels of flour in 9 months, how many barrels will 365 men consume in the same time? Operation 1st. Operation 2d.'7500 7500 102 54 -7 10254 73- 73 7 — 73 57500 How manvbbls. 30$ men $ 500: 365:: ~ en7 73 Men b00 Then, 5 x 73=365 bbls. The first and second terms are now equal, and may be rejected. 7 bbls. Ans The answer is therefore 75. A divisor transferred from one antecedent to the other, becomes a multiplier. 166 RATIO AND PROPORTION. 44. A merchant bartered 5S cwt. of sugar, at 6d. per lb., for tea at 8-5-s. per lb. How much tea did he receive? Operation 1st. Operation 2d. 5 3 cvt. 53 of 1 2 = 5 93 5= 53 cwt. 6 3 _ 2-7 6 ~2 64 = 7 - 4- 4 69 of 2d 828 4 4 85 6 8 1 8- 88 = 98. 27 X 5 936 _ 1 6. 7 Tea. 5 3 suoar. Invert the divisor. cWt. 1 tIlbs. 7 X 8 X 53=2968 8 160272:: l 1 b., d. 828 36 4 30 hl-. 3 ls. 4968 160272 s. 69 8 2484 8 lIb. tea. 29808)1282176(432- Ans. 69 2968-43 Anw. 45. If I buy 3- lbs. of sugar for 25 cents, what part of a ton car.I buy for 6'! Operation 1st. lbs. ton. 3 7 of -of 1 of 1 -7 2 2 228 4 20 4480 2 Operation 2d. ^ 1lbs. 31 7 4 82 8 1L-24 Ton.O1.0~ 20- cts.:1.7 lbs. Xi 3 2 4480 lbs. U 1 qr. 25: 600:: qrs. 4 1 cwt. 7 4480 cwt. 20 1 ton. 25)4200(168 80 3="-' Ans. 25 170 150 200 200 7) 814801 510 80 46. If 2 lbs. of sugar cost { of a dollar, what will 100 lbs. of coffee cost, if 8 lbs. of sugar are worth 5 lbs. of coffee? 9o INVERSE PROPORTION. 167 FIRST STATEMENT. Operation 1st. Operation 2d. cof. sug. cof. Wh:at will SI 0A coffee 20 0: 8:: 100 20 co(.fe ^ sugar 8 sugar l1B 160 lbs. sugar. ____A SECOND STATEMENT. O As.. sug. sug. cts. f: 00:: 25 80 80 $20.00 AIns. 47. If I lb. less by 1 cost 13W., what will 14 lbs. less by 5 of 2 lbs. cost? Ans. ~4 9s. O39d. 48. A merchant failing in trade owes 16000. His property amounts to $2400. What does his creditor receive to whom he owes $500, and what does he pay on the dollar? ns Creditor, $200. ns On dollar, 40 cts. 49. Bought 2 yd. of cloth for; sold I of 1 of I for s. Did I make or lose? A2ns. M Ide 10 cents +. 50. If 40 yards of cloth cost $32, what will I ell English cost? Ans. $1. 51. A mercer bought 32 pieces of silk, each piece containing 24- yds., at 6s. 6d. per yd. What did the whole cost him? lns. ~27 13s. 7d. 52. If -5 of a gallon cost ~5, what will I of a tun cost? Ans. ~140. 53.. If 21- yds. of cloth cost 60 cents, what will 1251- yds. cost? A ls.''30.18. 54. If A of a cord of wood is worth $6, what is 40 cords worth? _ns. $270. INVERSE PROPORTION. (Foithe rule, see Art. 178.) EXAMPLES. Art. 180.-1. If 4 men build a wall in 20 days, in how many days could 8 men build the same wall? 168 INVERSE PROPORTION. Operation. If 4 men can build a wall in 20 Inhow many ( days, 8 men would build the same days. d in 4 - 1 of the time. of 20, or 1dym. s 2s 0 20 multiplied by the ratio of 4 to 12 days Ails. 81s 0 days, the answer. 2. If 6 men mow a field in 21 days, in how many days would 9 men mow the same? Ans. 14 days. 3. If it take 9 men 14 days to mow a field, how long would it take 6 men to mow the same field? Ans. 21 days. 4. If it take 6 men 21 days to mow a field, how many men would mow the same in 14 days? Ans. 9 men. 5. If a man perform a journey in 6 days, when the days are 16 hours long, in how many days can he perform the same when the days are 12 hours long? Ans. 8 days. 6. If 1 cwt. be transported 150 miles for 1 guinea, how far can 6 cwt. be carried for the same money? Ans. 25 miles. 7. How many yards of carpeting, - yard in width, will cover a room 30 feet long and 20 feet wide? Ans. 133- yds. 8. What must be the length of a garden, 16 rods in breadth, to contain 2 acres? Ans. 20 rods. 9. How many yards of lining, A yard wide, will it take to line a cloak 44 yds. long and 1-1 yd. wide? Ans.'7~ yds. 10. If I lend a friend $200 six months, how long ought he to lend me $1000 to repay the kindness, allowing the month to be 30 days? Ans. 36 days. 11. Suppose 800 men were placed in a garrison, with provision sufficient to last them 2 months, how many must depart that the provision may last them 5 months? Ans. 480 men. 12. A ship's company of 15 persons is supposed to have bread to last their voyage, allowing each person 8 ounces per day. They pick up a crew of 5 persons in distress, whom they permit to share their daily allowance with them. What will be the allowance of each person? Ans. 6 ounces. 13. When wheat is sold at 93 cts. per bushel, the penny loaf weighs 12 ounces. What must it weigh when wheat is $1.24 per bushel? Ans. 9 ounces. 14. How many yards of cloth, 1~ yd. in width, are equal in measure to 30 yds. 1 ell English in width? Ans. 25 yds. 15. How long must a board, 4~ inches in breadth, be, to contain a square foot? Ans. 32 inches. COT )iFOUNI) PR'OlTlu I II'. 169 16. A certain buildcing Ywas raised in 8 rmonths by 120 workmen. How many workmen could hLave (lone tlhe same amount of labor in 2 months? 4A8i. 480 men. 17. How much in length that is 16 rods in width will it take to make an acre??-isa. 10 rods. 18. There is a cistern having a pipe which will elmpty it in 6 hours. How many pipes of the same capacity will empty it in 20 minutes? Ans. 18 pipes. 19. If 30 men can perform a piece of work in I1 (days, how many men will accomplish another piece of work, 4 times as large, in a fifth part of the time? 1Ans. 600 men. COMPOUND PROPORTION. Art. 181.- When a proportion is formed by the combination of two or more simple proportions, it is called Compound Proportion, or Double Rule of Three. 1. If 8 men consume 24 bushels of wheat in 5 months, how many bushels will 4 men consume in 15 months? In this question, the number of bushels consumed depends on two circumstances-the number of men, and the time. We may consider the circumstances separately, and solve the question by two statements in the Single Rule of Three. First, the number of men. If 8 men consume 24 bushels in 5 months, how many bushels will 4 men consume in the same time? Operation 1st. 2) 2:::24(12 Ans. Secondly, the time. If 4 men consume 12 bushels in 5 months, how many bushels will the same number of men consume in 15 months? Operation 2d. 3 $::: 12 3 36 Ans. QUESTIo.N. —.L How is Compound Proportion formed? 170 COMPOIUND PROPORTION. The first operation is in Simple Pr'oportion, because we employed but one simple ratio as a multiplier upon 24 bushels, the name of the answer, viz., the ratio of - men to 8 men. The second operation is also in Simple Proportion, for the same reason. The' ratio employed is the ratio of 15 months to 5 months. We may now unite these two statement.s in one, applying the rule already given in Simple Proportion. Thus, 2$: 4:: )___ Orthus: b. 4 3 2 24x3 3 2=-36 Ans. 12'$ 3 $: $:: J iU4 12'36 Ans. Here we have two terms of demand, viz., 4 men and 15 months; and two terms of condition, 8 men and 5 months. The ratio of 4 to 8 is 1-, and the ratio of 15 to 5 is 3. If we multiply these two simple ratios together, we have a compound ratio, which, multiplied into 24 bushels, gives the answer. X 3 — and 24 x- = —36, the answer. It is the use of a compound ratio whlich constitutes Compound Proportion. All questions in Compound Proportion may be solved by two or more statements in Simple Proportion, or they may be analyzed thus: If 8 men consume 24 bushels in 5 months, 1 man would consume 8 of 21=-3 bushels, and in I month -1 of 3= — of 1 buslil. Tilen 4 men would consume 4 times 5 — xin 1 montll, and in 15 months 15 times'-=36 bushels, the answer-. Art, 182,-Co1mpound Proporton telahes's teles o solf e by one stattcmenit (questions whlich uwoll rejuire two or more y re by Simple Proportion. Ors. 1.-The student should be required, first, to solve the question by analysis, then by proportion. 2. If a man build 27 rods of wall in 3 days, when the days are 12 hours long, how many rods can he build in 9 days, when the days are 16 hours long? If a man in 3 days build 27 rods, in one day he would build - of 27=9 rods. If in one day, 12 hours long, he build 9 rods, in one hour he would build 9g- of a rod, and in 16 hours, X 16 = -4. = l 12 rods. If in 1 day, 16 hours long, he build QUESTIONS.-1. What does Compound Proportion teach? 2. What constitutes Compotnd Proportion? COMPOUND PROPORTION. 171 12 rods, in 9 days he would build 12 x9=108 rods, the answer. In this example, 3 days, 12 hours long, are equal to 12 x 3= 36 hours; and 9 days, 16 hours long, are equal to 16 x 9=144 hours. We have, then, this proportion; 36 h.: 144 h.:: 27 rds.: 108 rods, for 1-44=4, and 10 84. The ratio of the time in the demand, to the time in the supposition, is the same as the ratio of the term sought to the rods in the conditions of the question. That is, the ratio of 144 hours to 36 hours, expresses how many more rods can be built in 9 days, 16 hours longr, than in 3 days, 12 hours long. It will be perceived, that the ratio of the time in the demand, to the time in the supposition, is the product of two simple ratios. It is a ratio of the ratio of days to days, and hours to hours, (a ratio produced by the multiplication of simple ratios is called a compound ratio.) Thus, if a man in 3 days, 12 hours long, build 27 rods of wall, the amount of wall built in 9 days, (the days being of equal length,)'is expressed by the ratio of 9 to 3, 3 -; that is, he could build 3 times the number of rods, 27 x3=81 rods; but the days are 16 hours long; this circumstance, again, affects the result, and is expressed by the ratio of 16 to 12, 1=1I-; that is, the amount of labor performed in 16 hours is greater by - than the labor performed in 12 hours; I of 81 rods -27, and 27+81 =-108 rods. If we multiply these simple ratios, 3 X 11 =4, we have a compound ratio, the same as above, and 27 x4= 108 rods, the same answer. The ratios of the days to days, and hours to hours, may be expressed thus. If, in 3 days, 12 hours long, 27 rods of wall are built, how many rods days. hours. can be built in 9 of 1"? of =6-1-4=4, and 27 x4-108, the answer. OBs. 2.-The teacher will now call upon some member of the class, to select the terms, and form first a simple proportion, and then a compound, in the following manner, and illustrate as he proceeds; 3: 9:: 27: This statement involves Simple Proportion. The days are considered of equal length. There is but one circumstance that affects the answer, viz.: the difference in the number of the days. This affects it in a threefold ratio. The ratio of 9: 3=3, which shows how many more rods could be built in 9 days, than in 3 days. But the days are not of equal length. This circumstance must also be considered. We will therefore introduce into the statement another simple ratio. The ratio of 16 to 12, wlich shows how many more rods could be built in 16 hours, than in 12. 172 COMPOUND PROPORTION. 4 - 4 Thus, $ I1 ~:: J~4 108 Ans. Multiplying antecedents and consequents by antecedents and consequents, the ratios are compounded, and thus the question becomes Compound Proportion. T'I How does it appear that antecedents and consequents have been multiplied? S. Intlroducing a factor, multiplies by that factor, and cancelling equal- flctors fromn antecedent;s and consequents, does not affect the ratios. In this example, the factors are all cancelled but 4 and 1. 4 is therefore the compound ratio.: D)oes this question involve Direct or Inverse Proportion? IHow do you know? How many simple ratios are there? Upon how many circumstances does the answer depend? What are they? In what ratio does the first circumistance affect the answer? The second? In what both combined? What is this ratio called? What is a compound ratio Does it differ, in itself considered, fiom a simple ratio? Question 2.-If a man build 2.7 rods of wall in 3 days, when the days nre 12 hours long, in how many days can he build 108 rods, when the days are 16 hours long? 3 X3 X3=-9 ns. 4 A.: n::, Let the student state and illustrate tlis question, in the following manner: This question involves Inverse Proportion-more requires less. It would require a less numlber of days to perform the same amount of labor, when they are 16 hours long, than \when they are 12. The number of the days will be inversely as their length. We therefore place the 16 hours in the demand, for the first term, and the 12 hours in the condition for the second. We reject the factor 4, which is common to 16, the antecedent, and to 12, the consequent. Then 4 X 27=108, which is an antecedent and consequent, and therefore may be rejected. The ratios are n(ow compounded, and the proportion reads, 1: ~:: 3: 9 the answer. T1 By what rule are tle foregoing questions stated? S. By the rule given for Simple Proportion. 3. If 3 men can build 360 rods of wall in 24 days, how many rods can 8 men build in 27 days? 4. How many men will it take to build a wall, 75 rods long, 8 feet high, 3 feet thick, in 6 days, working 9 hours each day, if 20 men can build a wall 100 rods long, 6 feet high, 4 feet thick, in 12 days, working 12 hours each day? COMPOUND PROPORTION. 173 Statement. 100) 6: 81:: 4 3 } 20 men. 6 12 ~:: 9 12 J Opera tion. Iow many men? ji; 5 X8=40 men Ans. a Ga$ 1;'0 men. 40 men Ans. 5. If 3 of a yard of cloth, 8 — yd. wide, cost ~j, wlat is the value of 8 yard, 13 yard wide, of the same quality? 6. If a man travel 240 miles in 12 days, when the days are 12 hours longc, how far can he travel in 27 days, when the days are 16 hours long? Art. 1S. —In Proportion, both Simple and Compound, the terms in the supposition and demand may be distinguished by cause and effect, or producing and produced terms. That which causes any thing, or produces an effect, as men, time, length, breadth, depth, etc., may be denominated a producing term. Thus, in the foregoing question, among the terms of supposition, one man, 12 days, 12 hours long, are the joint cause, oi 2producing tl'ms, and miles the effect, or produced term. Among the terms of demand, 27 days, 16 hours long, are the joint cause, or the producinq terms, and the rods required are the effect, or the produced term. In all questions in Proportion the answer required will be either cause or effect, (a producing or produced term.) Hence, Art. 181.-When the term required is a produced term, Draw a perpendicular line, and place all the terms of demand on the right of the line, and all the corresponding terms of the condition on the left, closing the statement by placing the name of the answer on the right. QUESTIONS.-1. What is meant by producing and produced terms? 2. Rule, when the trm required is a produced term? 15* 174 COMPOUND PROPORTION. OBs. 1.-All questions under the foregoing head will be found to be in Direct Proportion. 7. If 4 students spend ~19 in 3 months, how many pounds will 8 students spend in 9 months? Ans. ~114. Operation. How many ~? 8 students. l Producing terms Producing terms of j Students 4 9 months. of the demand. the supposition. Months 3 ~19 Produced term of the (I19 supposition. I~114 Ans. 8. If 7 men can reap 84 acres in 12 days, 12 hours long, how many acres can 20 men reap in 5 days, 14 hours long? Ans. 1161 acres. 9. If 8 reapers receive ~3 4s. for 4 days' work, how much ought 20 reapers to receive for 15 days' work? Ans. ~30. 10. If 3 men receive ~8.9 for 19.5 days' labor, how much autght 20 men to receive for 100.25 days' labor? Ans. ~305 Os. 8d. lqr. 11. If 20 cwt. may be carried 80 miles for $35, how much will it cost to transport 40 cwt. 100 miles? Ans. $87~. 12. If the fieight of 9. hhds. of sugar, each weighing 12 cwt., 20 leagues, cost ~16, what must be paid for the freight of 50 tierces, each weighing 21- cwt., 100 leagues? OBs. 2.-IIundred weight and distance are the producing terms, and the money received, the produced term. 13. If 2 — yards of cloth, 13 yards wide, cost ~33, how much will 361- yards, 11 yards wide, cost? Ans. ~54. 14. If 24 bushels of wheat be consumed by 8 persons in 5 months, how many persons will consume 36 bushels in 15 months? Operation. How many persons? 13I bushels 4. $ bushels k41 months. $ men 1X $ persons. 4 persons Ans. Art. 185 —When the term required is a producing term, Draw a perpendicular line, and place the produced term of the demand and the produciny terms of the supposition on the COIMPOUND PROPORTIO)N. 175'igsht, and the'rCintainti term's ov2 the left of the line, and procced as before. OBS.-This head will be found to correspond with Inverse Proportion, and may be applied to both Simple and Compound. 15. If 6 men build a wall, 20 feet long, 6 feet high, and 4 feet thick, in 16 days, working 12 hours in a day, in how many days will 24 men build a wall 200 feet long, 8 feet high, and 6 feet thick, working 10 hours in a day? Ans. 96 days. Operation. How many days? W l Produced terms of deProduced terms of,0 8A mand. supposition.' O 6 ~nProducing terms of v; A Producing terms of Producing terms of 4 {4 12 supposition. demand. 10 supposition. 96 days Ans. 16. If 3 men, in 24 days, 9 hours long, can dig 328 rods of trench, 6 feet wide and 4 feet deep, how many men will it take to dig a trench 984 rods long, 9 feet wide, and 8 feet deep, in 27 days, when the days are 12 hours long? Ans. 18 men. 17. If a man can travel 240 miles in 16 days, when the days are 14 hours long, how many days will it take him to travel 720 miles, when the days are 12 hours long? Ans. 56 days. 18. If 98 lbs. of bread he sufficient to serve 7 men 14 days, how many days will 63 lbs. serve 21 men? Ans. 3 days. 19. If 40 men in 15 days, 12 hours long, build a wall 200 feet long, 12 feet high, and 5 feet thick, how many hours long must the day be, that 20 men, in 12 days, may build a wall 100 feet long, 10 feet high, and 6 feet thick? Ans. 15 hours. 20. If 20 men in 12 days, 15 hours long, can build a wall 100 feet long, 10 feet high, and 6 feet thick, in how many days, of' 12 hours long, can 40 men build a wall 200 feet long, 12 feet high, and 5 feet thick? Ans. 15 days. 21. If 16 compositors set 150 pages of types, each page QUESTION.-3. Rule, when the tear required is a producing term? 176 SUPPLEMENT TO PROPORTION. consisting of 48 lines, and each line of 50 letters, in 8 days, 10 hours long, how many compositors will be required to set 500 pages of 72 lines each, and 45 letters in a line, in 6 days, 8 hours long? Ans. 120 compositors. SUPPLEMENT TO THE RULES OF PROPORTION. EXAMPLES. Art. 186.-1. If I can hire 30 horses pastured 7 weeks for ~6, how many weeks may 5 horses be pastured for ~4 5s. 8dd.? Ans. 30 weeks. 2. If 8 men build 48 rods of fence in I day, how many rods will 24 men build in the same time? Adzs. 144 rods. 3. If 24 men build 144 rods of fence in 1 day, how many rods will 8 men build in the same time? iJns. 48 rods. 4. How many men will it require to build 144 rods of fence in 1 day, if 8 men build 48 rods in the same time? Ans. 24 men. 5. If 20 men build a mill in 160 days, in how many days could 25 men build the same mill? Ans. 128 days. 6. If 25 men build a mill in 128 days, how many men will build the same mill in 160 days? A2ns. 20 men. 7. How many cords of wood may be bought for ~40, if 22 cords cost $G? lns. 531 cords. 8. How many tbushels of wheat may be bougaht for ~ 40, if 1 bushel of wheat be worth 2 bushels of rye, and 4 bushels of rye be worth 5 bushels of corn, and 8 bushels of corn be worth 16-bushels of oats, and 1 bushel of oats be wo:-th 2 shillings? Ans. 80 bushels. 9. If 18 cords of oak wood be worth 26-2 cords of hemlock, and a cord of hemlock 14- shillings, how much will 4- cords of oak cost in cents, and how many guineas? An. 1600 cents. Ans 38} guineas. 10. How many stoves may be bought for 672 shillings, if 8 fire-frames are worth 3- stoves, and 16 fire-frames are worth 32 guineas? Ans. 5 stoves. 11. How much will $450 gain in a year, if $100 in the same time gain $8? Ans. $36. 12. A merchant owning 2 of a vessel, sold - of his share for $934. What was the value of the ship? Ans. $3736, SUPPLEMENT TO PROPORTION. 177 13. How many guineas will 8 yards of cloth cost, if 7 of a yard cost 5 of a dollar? Ans. 1 2 guineas. 14. If 1 of a pound of sugar cost -7- of a shilling, what will 3- of a pound cost? Ans. 4d. 23651- qrs. 15. If 2- lbs. of tobacco cost 4s. 6d., how much will 180 lbs. cost in dollars? Ans. $54. 16. If when wheat is 4s. 6d. per bushel, the penny loaf weigh 12 oz., what ought it to weigh when wheat is $.50 per bushel? Ans. 18 oz. 17. How many yards of cloth, 3- yd. wide, are equal in measure to 30 yds. 1~ yds. in width? Ans. 60 yards. 18. If 40 bushels of grain will pay a debt when the price is 60 cents per bushel, how many bushels will it take when the price is $1.20? Ans. 20 bushels. 19. How far may 30 cwt. be transported for $8, if 2} cwt. be carried 180 miles for the same money? Ans. 15 miles. 20. How many yards, of A yd. wide, will it take to line 850 suits of clothes, each suit to contain 3-1 yards of cloth, 13 yds. in width? Ans. 6941 yds. 2 qrs. 21- nails. 21. If 30 horses consume 600 bushels of oats in 8 months, how much will each horse consume per day? Ans. 2- quarts. 22. A man owns 2 of a ship, which ship is valued at X of the ship and cargo-the latter worth 896000. What is the value of I of his share? Ans. $4000. 23. If 12 men consume I of 2 of 3 of 4 of 30 bushels of wheat, in a of 14 of 6 of -Y of 20 months, how much will 4 men consume in - of 7 of 1 6 of 3 of 14 of 40 months? Ans. 12 bushels. 24. If 4 men spend - of of 7 of' of ~30 in 7- of -3 of 26 of 1 of 9 days, how many dollars will 21 men spend in 3 of 14 of of - of 45 days? ins. 8420. IL 5 7 12 25. If a man travel 336 miles in 14 days, when the days are 18 hours long, in how many days will he travel 672 miles, the days being 12 hours long? Ans. 42 days. 26. If a man travel 240 miles in 12 days, when the days are 12 hours long, in how many days will he travel 720 miles, when the days are 16 hours long? Ans. 27 days. 27. If a family of 9 persons spend $450 in 7 months, how much would be sufficient to maintain them 8 months, if 5 persons more were added to the family? Ans. $800. 28. What is the value of I grain of gold, if 17 lbs. be worth ~10224? Ans. 2 1-s. 178 rXCHANGE. EXCHANGE. Art. 187.-EXCHANGE is the act of paying or receiving the money of one country for its equivalent in the money of another country, by means of Bills of Exchanye. It comprehends both the reduction of moneys and the negotiation of bills. It determines the comparative value of the currencies of different nations, and shows how foreign debts may be dischalrg ad, and remittances made from one country to another, without tlhe risk, trouble, or expense of transporting specie or bullion. When the United States were British colonies, the stelilng value'of the pound was the same in nll the colonies; but the legislatures of the different colonies emitted bills of credit, which afterwards depreciated in their value-in some states more, and in others less. Art. 18-8.-The following table exhibits the number of shillings in a dollar in each of the states. TABLE I. To exzchanpefrom Now Eng. Penn'a, N. Jer- New York Soulth Carexctatange from s and sy, Del.ware, and North oinla andl to Virginia. and Maryland. Carolina. Georgia. New England Dollar Add x 7 r-' Add Add -. States and Va. 6s. Od. a 3' and, 9 Pennsylvania, Subtract New Jersey, Dela- Dollar'is.6d. Add I. 3. ware and Maryland' New York and Subtract Subtract Dollar x 7 North Carolina. X -.. 8s. Od. and 12 South Carolina 5 and xby 12 Dollar and Georgia. Ad.T 3:-. - by 7. 4s. 8d. OBS.-The value of a dollar in any state is found either opposite to that state, or under it in the table. As the number of shillings in a dollar is different in different states, the value of the dollar being the same, it follows, that the value of the shilling is different; and as the number of shillings in a pound is the same, the value of the pound must EXCHANGE. 179 differ in the ratio of the shillings. That is, if 6 shillings in New England, and 8 shillings in New York, make a dollar, then a pound in New York is to a pound in New England, as 8 is to 6; =-4; or ~4 in New York are equal to ~3 in New England. Art. 189,-The relative value of the pound in different states may be seen by the following Table. TABLE II. New Jersey, PNew Jersey 15=16 New York and North Carolina. Pennsylvania, Delaware, and 5= 4 New England and Virginia. Maryland, j 45 = 28 South Carolina and Georgia. New England 4= 5 New Jersey, Pennsylvania, &c. States and 3= 4 New York and North Carolina. Virginia, 9= 7 South Carolina and Georgia. New ork n 16 =15 New Jersey, Pennsylvania, &c. N. C o a 4-= 3 New England and Virginia.. Carol, 5 12= 7 South Carolina and Georgia. S. C )and 28=4o New Jersey, Pennsylvania, &c. S. Carolina and *aGo inn 7 — N12 New York and North Carolina. Georgia, ) 7= 9 New E ghland and Virrinia. Art. 199.-The following table shows the value of pounds, shillings, and pence in each of the United States, according to their respective currencies: TABLE III. New Jersey, Pennsylvania, New York and N. En!rand States S. Carolina and Delaware, and North Carolina. and Virginia. Georgia. Maryland. ~ 3=$8. ~ 2=$5. ~ 3=$10. ~ 7=$30. s. 3=40cts. s. 2=25 cts. s. 3=50 cts. s. 7=1.50 cts. d. 9-10 cts. d. 24=25 cts. d. 18=25 cts. d. 14=25 cts. DOMESTIC EXCHANGE. Art. 191.-To reduce the currency of one state to that of another. 180 EXCHANG E. IRULE. Draw a perpendicularl line, anl place the demand of the question on the right, and the suppos,tion on the left, as in the Rule of Three Direct. EXSA MPLES. 1. What sum in Georgia, is equal to ~1800 New England currency? OQcrtion. In stating the quesHow many Georgia ~ ~X00 N. E. tion, say how many If NT. E. ~p 7 G. 200 pounds Georgia cur~ 1400 dAs. rency are equal to ~1800 New England, if ~9 New England arc equal to ~7 in Georgia. (See Table II.) 2. How many pounds in New York currency will ~240 New Jersey currency marke? Ans. ~256. 3. What sum in Virginia is equal to ~375 16s. 9d. New York currency? Ans. ~281 17s. 6d. 3qrs. 4. What is the value in New Jersey currency of a bill of exchange for ~375 10s., on a cotton dealer in Georgia? Ans. ~603 ns. 7d. 3qrs. 5. A manufacturer in Massachusetts sends to Georgia a lot of shoes, which amount to ~420 7s. What is the value in New England currency? Ans. ~540 9s. 6. A manufacturer in New Jersey consigns to his agent in Charleston a quantity of ready-made clothing, vhlich, when sold, and the charges deducted, amounted to ~532 1 ls. What is the value in N. Jersey currency? Ans. ~855 17s. 8d+. 7. EXCHANGE FOR ~320 10s. 6d. Boston, July 26th, 1837. Twelve days after sight, please pay to PETER FINCH, or order, three hundred twenty pounds, ten shillings and sixpence, value received, and place the same to the account of your Ob't Servant, To PETER FiNCH. IsAAc WATERFORD, JR. What sum in New York currency will discharge this bill? Ans. ~427 7s. 4d. QUESTIONS.-]. Vhat does Exchange teach? 2. Why is the value of the pound different in different states? 3. Is the value of a dollar everywhere the same' 4. Is tllo value of a shilling the same? 5. What is Domestic Exchange? 6. Rule of statement? EXCi AN Gu. 181 S. What sum in South Carolinat currency is equal to ~429 1s. 3d. in New England? Ans. ~833 18s. lid. 2qrs. 9. What sum in Pennsylvani-a is equal to ~259 15s. 9d. Georgia currency? Ans. ~417 1Os. 3d. 2qrs. 10. Philadelphia, Jiue 1st, 1837. EXCHANGE FOR ~240 lOs. PENNSYLVANIA CURRENCY. Sixteen days after sight, pay to GEOIRGE STMPSON', or order, two hundred and forty pounds, ten shiliings, Pensylvania currency, as per advice from Yours, etc., To THOMAS SMART, Merchant, N. Y. JosiuH LITTLE. What sum in New York currency will discharge the above bill? Ans. ~256 10s. 8d. I'OKhEIGN EXCHANGE.* Art. 19,.-.lI foreCignl coins, by a late act of Congress of the United States, are prohibited being a lawful tender. The gold coins of Great Britain and Portugal, of their present standard, are valued at the rate of 100 cents for every 27 ralins, or 88- cents per dwt. The gold coins of France, of their present standard, are to be valued at the rate of 100 cts. for 27-1 grains, or 87- cts. per dwt. The gold coins of Spain, of their present standard, are to be valued at the rate of 100 cts. for 28-5 grains, or 84 cts. per dwt. In England, Ireland, and the English West India islands, accounts are kept in pounds, shillins,, pence, and farthings; though_ the intrinsic vailue,,in eac plac, is not the same. Exchange is said to be at par between two countries, or states, when the money given in one is equivalent in value to that received for it in another. The course of exchalenge is fluctuating, being above or below par, according to the occurrences of trade, or the demand for money. A Bill of Exchanqe is a written order for the payment of a certain sum of money, at an appointed time. It is a mtercanttile contract, in which four persons are mo4.tl concerned, viz.: * lis rule mayv be omitted tmtil the review. 10 182 LXCIANG, First. The driawer, waho receives the value, and is also called the make, and seller of the bill. Second. The debtor in a distant place is one on whom the bill is drawn, and who is called the drawcee. He is also called tle acceptor, after he accepts the bill, which is an engagement to pay it when due. Third. The person who gives the value for the bill, and is called the buyer, taker, and remitter. Fourth. The person to whom the bill is ordered to be paid, who is called the payee, and who may, by endorsement, pass it to any other person. Art. 193 —The following tables show the par value of forvign money in the United States. TABLE IV. Coins current in the United States, with their Sterling and Federal value. Stn d- Sterlin N Yor.kN. Jors'y S. CaroAM0ES OF COIN. ard moneyof N.E. ad N. Penn., iinaand Federal weight reat Brit- States. Caroli- IDel.,adl Geor- value. weight ain. na. Maryl'd. gia. Gold. wt.g.~ s. d. ~ s. d.~. s. d.~. s.dl. ~C s. d.$ e. mn. AJohannes, 18 013 11 0 4 16 06 8 06 004 0 016 00 0 Ahalf Johann. 9 01 16 0 2 8 03 4 0 0 0 0 8000 A Doubloon, 16 213 6 0 4 8 05 16 05 12 6 310 014 93 3 AMoidore, 6 1811 7 01 16 012 8 02 5 011 8 0 6 00 0 AnEnlg.Guin. 6 611 1 0 1 801117 01 150119 4 66 A French do. 5 51 1 0 1 761 160 11 1 5 4 600 A Span. Pistole 4 6'0 16 0 1 2 01 9 0 l 8 0018 0 3 7 3 A French do. 4 41 016 0 1 2011 801 7 6 017 6 3 667 Silver. Crown, - g lislaorFr. }118 00 5 0 680 8 90 8 30 5 0 1 100 Doll.of Spain Sweden, or 17 60 4 60 600 800 6 4 8 000 Denmark, ) Eng Shil 3 180 0 141 19 0 0 1 1 0 2 2 ling, APistareen, 3 1110 0 10}0 1 20 1 70 1 60 011 0 20 0 All other gold coins, of equal fineness, at 89 cents per dwt., and silver at $1.11 cents per oz. QUESTIONS.-1. What is foreign exchange? 2. When is exchange said to be at par? 3. What ia the meaning of par? Answer: it is a Latin word, which sigiifie, eyptat. EXCHANGE. 183 TABLE V. Art. 194. — alue of Foreign Coins in Federal Money, as estab. lished by a late Act of Congress. d. c. m. Pound Sterling*...................................... 4 44 4 Pound, of Ireland..................................... 4 10 0 Pagoda, of India....................................... 1 94 0 Tale, of China..................................... 1 48 0 Mill-ree, of Portugal.................................. 1 24 0 Ruble, of Russia.................................0... 66 0 Rupee, of Bengal................................. 0 65 6 Guilder, of the United Netherlands............... 0 39 0 Mark Banco, of Hamburgh........................... 0 33 5 Livre Tournois, of France............................ 0 18 5 Real Plate, of Spain.............................. 0 10 0 * ~1 sterling before 1832 was $4.444; since that time $4.80. TABLE VI. Art. 195.- Moneys of different countries. FRANCE. 12 derniers=1 sol. 20 sols = 1 livre=18- cts. 3 livres =1 crown. OB. —The above is according to the old system; the present method of keeping accounts in France is in francs and centimes, or hundredth parts, thus: 10 centimes= 1 decime. 10 decimes =1 franc=$.1873125. The 5-franc piece weighs 25 grammes, or 386.1 grains Troy, and is equal in value to $.9365625. SPAIN. The money of Spain is of two kinds; one is called vellon, the other plate money. Accounts are most generally kept in rials and marvadies vellon. 4 marvadies vellon, or 1 quarta. 2 — marvadies of plate, 8~ quartas, or 8 quartas, = 1 rial vellon. 34 marvadies vellon, 15 rials vellon, - - =1 peso, or current dollar. 16 quartas, or 1 rial of plate=0 cts. 34 marvadies of plate, 184 TABLE OF FOREIGN COINS 8 rials of plate,- - =1 piastre- 80 cts. 10 rials of plate, - - =1 dollar=$1.00. 5 piastres, - - - 1 Spanish pistole=$4. ITALY. 12 derniers-=l sol. 20 sols = 1 livre. 5 livres =l piece of eight at Genoa. 6 livres =I piece of eight at Leghorn. 6 solidi -1 gross. 24 grosses =1 ducat. PORTUGAL. 400 reas —l crusado. 1000 reas=-1 milleaa-=,1.24. The reas and millreas are imaginary pieces of money; the real moneys of Portugal are as follows: Silver. 1 crusado =400 reas =50 cts. 12 vintin piece=280 reas -30 cts. 5 do. =100 reas —12 cts. 2; do. 50 reas — 6- cts. Gold. 1 double johannes-25 millreas, 600 reas $3 2. I single do. =12 do. 800 leas- ^16. half do. = 6 do. 400 r-e-s=88. quarter do. = 3 do. 200 Ireas:- 4. eighth do. = 1 do. 600 I as-_ 12. festoon, or - =I 800 reas= 1. I moidore = 4 do. 800 reas -- 6. HOLLAND. 8 pennings = groat - - - ct. 2 groats =1 stiver - - - 22d., or 2 ets. 6 stivers = I shilling - - = 2 cts. 20 stivers -1 florin, or guilder =40 cts. 2- florins =1 rix dollar - - $1. 00. 6 florins =1 pound Flemish =$2.40. 5 guilders =1 ducat - - - -$2.00. BILLS OF EXCHANGE. 185 DENMARK. 16 schillings=1 mark =$0.33-. 3 marks =1 rix dollar=$1.00. 61 marks =1 ducat =$2.08-. RUSSIA. 3 copecs=l altima. 10 do. =1 grivena. 50 do. =1 politin. 2 politin = 1 ruble= 75 cts. 2 rubles =1 ducat. CHINA. 10 caxa =1 candareen=$.0148. 10 candareens=l mace =$.148. 10 mace =1 tale =$1.48. BARBARY. 10 aspers =1 rial =$0.121. 2 rials =1 double -.25 cts. 4 doubles=l dollar =$1.00. 24 medins = 1 chequin =.75 cts. 32 do. =1 dollar =$1.00. 180 aspers =1 zequin ==2.25. 15 doubles=l pistole =$3.75. TURKEY. 3 aspers=l para =$0.00652925. 40 paras =1 piastre =$0.26117647. BILLS OF EXCHANGE. Art. 196,-To find the value of bills of exchange above par. EXAMPLES. 1. What is the value of a bill of exchange for $860, at 5 per cent. above par? Operation. What value $ 860 $ If $100 105 $ $903 Ans. qGls 186 EXERCISES IN EXCIHANGE. 2. A. of Boston, is indebted to C. of London, ~1000. How miuch sterling must be remitted, exchange being 50 per cent.? Ans. ~1500. 3. B. of New York, is indebted to D. of Liverpool, ~650 sterling, to discharge which he purchases a bill at 3 per cent. above par. How many dollars does he give for it? Ans. $2975.258. Art. 197.-To find the value of bills of exchange below par. 1. What sum sterling money is equal to ~340 Os. 4d. Massachusetts currency, exchange 40 per cent.? Ans. ~204 3s. 9d. 2qrs. Reduce the shillings and pence to the decimal of a pound by inspection. Opra tioz. 340.317 I 01 60 204.1902~ —204 3s. 9d. 2qrs. Ans. 2. D. in Philadelphia, owes E. in London, ~600 sterling, to discharge which he purchases a bill at 3 per cent. below par. How many dollars must he give? Ans. $2586.666+-. (a) EXCHANGE FOR. ~540 8s. 9d. STERLING. Boston,. At thirty days' si.lht, pay to TIMOTHY DICES, or order, five hundred and forty poutds, eight shillings, and ninepence, value received, and place the same to the account of To JOHN Jou.ISON, Merchant, Liverpool. JAMES STRIKER What is the value of this bill in Pennsylvania currency, exchange at 56 per cent.? 4. A. of Cork, draws upon B. of London, for ~870 12s. 4d. Irish, exchange at 8 per cent. How much sterling will discharge this bill? (5.) EXCHANGE FOR 2446 LIVRES, 6 SOLS, 4 DERNIERS. Thirty days after sight of this my second of exchange, first of same tenor and date not paid, pay to TITUS TRUE, or order, two thousand four hundred forty-six livres, six sols, four derniers, value received, and place the same to my account. PETER J. TUTTLE. To TRUSTRAM CROGcER, Merclhat, Paris. EXERCISES IN EXCHANGE. 187 How much sterling is the aLove bill, and how much in Pennsylvania currency? 6. A merchant in Toulon is indebted to a merchant in Boston 8462 francs, 20 centimes. What is the amount in federal money? 7. A. of Albany, buys a draft on C. of Paris, of 8846 francs, 34 centimes, for $1600. What is the rate of exchange? 8. Q. of Barcelona, is indebted to P. of New York, 925 piastres, 3 rials, -24 nmarvadies plate. I-ow much, in federal money, is Q. chargl-ed in P.'s book? 21ns. $740.37. 9. C. of Ireland, remits to D. of London, ~345 10s. Irish. With how much stealing must C. be credited, exchange being 8 per cent.? ilns. ~319 18s. Id. 34-qrs. 10. A bill for 3625 pesos, 4 rials, 31 marvadies, being remitted to Cadiz, what sum New Jersey money is equal to it, at 7s. Gd. per peso? Ans. ~1359 9s. 11d. Iqr. 11. A Virginia merchant shipped tobacco to Norway worth ~1673 18s., Virginia currency. How many rix dollars, at 6s. each, must he receive? Ans. $5579.666. 12. A. in Philadelphia, owes B. of Amsterdam, $750. How many guilders is it, at 40 cts. per guilder? Ans. 1875. 13. What sum must be paid in Savannah for an invoice of goods charged at 490 florins, 15 stivers, allowing the exchange at 40 cents per florin, and freight and duties 30 per cent.? Ars. 255.19. 14. A merchant in Philadelphia receives of a merchant in Amsterdam an invoice of goods, amounting to 12340 florins, 19 stivers, 12 pennings. How much must be remitted, in Pennsylvania currency, to discharge the bill, at 35-d. per florin, and what sum in sterling, exchange at 38s. 6d. Flemish per pound sterling? Ans. ~941 12s. Od. Oqrs. sterling. 15. In 16745 marks, how many dollars, allowing 331 cents per mark? Ans. $5581.666. 16. In 2045 piastres, 9 rials plate, how many dollars? Ans. $1636.90. 17. What will 8400 arsheens of ravens duck cost, at 15 rubles for 45 arsheens, in rubles, and also in federal money? Ans. 2800 rubles; $2100. 18. A. of Bordeaux, draws on B. of Liverpool, for 1400 crowns, at 55d. sterling per crown; for the value of which B. draws again on A. at 56d. sterling per crown; besides com 188 REDUCTION OF CURRENCIES. mission of ~ per cent. What did A. gain or lose by this transaction? Ans. Gained 18 crowns. 19. In $1820, how many pagodas of India? 20. In $605, how many rupees of Bengal? 21. How many dollars in 4678 tales, 8 mace, 7 candareens? 22. In 8000 aspers, how many dollars? Ans. $100. 23. A merchant in Philadelphia imported from England 700 ells of cloth, at 5 shillings sterling per ell. The cost of transportation and duty, on the whole amount, was 35 per cent., the exchange at par. For how many cents must 1 yard be sold in Philadelphia, to gain 12~ per cent.? Ans. 135 cents. REDUCTION OF CURRENCIES. Art. 198.-REDUCTION OF CURRENCIES teaches to reduce pounds, shillings, pence, &c., to federal money, and the reverse. RULE. Reduce the dollar to the fraction of a pound; and, if there be shillings, pence, andfarthings, in the given sum, reduce them to the decimal of a pound, by inspection, and proceed as in the Rule of Three. 1. In ~63, New England and 2. In $210, how many pounds, Virginia currency, how many New England and Virginia curdollars? rency? Operation. Operation. How many $10 ~ 21 How many ~ $210 If ~$ 10 If $1 3~ 1_ $ __1 $210 An. 63 Ans. If $1 is ~L,3, then it is evident If ~- is $1, then the product that the quotient of ~1 divided by of ~a3 multiplied by any number ~ would be the number of dollars of dollars, will be the number of in 1 pound, and so of any number pounds required. of pounds. QUESTIONS. —1. What does Reduction of Currencies teach? 2. Rule for reducing pounds, shillings, and pence, to Federal Money? 3. Rule for reducing Federal Money to pounds, shillings, pence, etc.? REDUCTION OF CURRENCIES. 189 3. In ~240 10s. Virginia, &c., 4. In $801.666+ how many currency, how many dollars,cents, pounds and shillings, New Engand mills? land, &c., currency? 5. Reduce ~210 15s. Virginia, 6. Reduce $702.50 to Vir&c., currency, to federal money. ginia, &c., currency. 7. What sum in federal money 8. What sum in New England is equal to ~300 10s. 6d., New and Virginia currency is equal to England and Virginia currency? $1001.75? 9. Reduce ~380 9s. North Car- 10. Reduce $951.125 to North olina currency, to federal money. Carolina currency. II. Reduce ~67 5s. 3d., New 12. Reduce $179.366+ to New Jersey currency, to federal money. Jersey currency. 13.,What sum in federal mon- 14. What sum in South Caroev is equal to ~67 us. 3d., South lina and Georgia currency is equal Carolina and Georgia currency? to $288.265? 15. Reduce ~102 17s. 4d., 16. Reduce $440.858 to South South Carolina, &.c., crrency, to Carolina, &c., currency. federal money. 17. Reduce~630 6s.4 d.,Penn- 18. Reduce $1680.85 to Pennsylvania currency, to federal mon- sylvania, &c., currency. cy. 19. Reduce ~600 10s. 6d. 20. Reduce $2669 to sterling sterling, to federal money, the money. dollar being 4s. 6d. 21. What sum in federal money 22. What sum in Canada and is equal to ~126 14s. Canada and Nova Scotia is equal to $506.80? Nova Scotia currency, the dollar being 5s.? 23. Reduce ~346 16s., New 24. Reduce $867 to New York York, &c., currency, to federal and North Carolina currency. money. 25. Reduce ~125 7s. 9d., Ma- 26. Reduce $334.365+ to Maryland, &c., currency, to federal ryland, &c., currency. money. 27. Reduce ~501 3s. 9d., Mas- 28. Reduce $1670.623 to Massachusetts currency, to federal sachusetts currency. money. 29. Reduce 450d., New Jersey, 30. Reduce 500 cents to New &c., currency, to cents. (See Jersey currency. Table III.) Operation. Operation. How many cents? 450 pence. Howmany pence? 500 d. 9 10 cents. cents 10 9d. 500 cts. Azis. 450 pence. 31. Reduce 540 pence, New 32. Reduce 750 cents to pence, England currency, to federal mon- New England currency. ey. 190 PERCENTAGE. PERCENTAGE. Art. 199,-The consideration of profit and loss adds the chief interest to all business operations. It is necessary, therefore, that there should be some standard by which all should agree to make their estimates: 100 has been adopted, and hence gain and loss are said to be so much per centum; that is, so much by the hundred. The gain or loss per centum is called percentage. The individual who makes 20 per cent. profit on his goods, makes a high percentage; and he who makes but 4 per cent. makes a low percentage. Since, therefore, 100 denominates or is the denominator of the gain, it is plain the gain itself will be the numerator. If the gain be 5 per cent., it would be expressed thus: -o=.05. 1 per cent. equals T-Q =.01 2 per cent. equals x-}2 =.02 3 per cent. equals --- =.03 4 per cent. equals yo- =.04 5 per cent. equals Tg - =.05 6 per cent. equals X6 =.06 Whatever, therefore, be the amount of capital invested, the gain or loss will be so many hundredths of the capital, to be added to, or subtracted from it. The gain or loss on any sum is to be calculated by the rule for the multiplication of decimals. 1. What is 1 per cent. of 20 dollars? Ans. 20 cents. 2. What is 2 per cent. of 40 dollars? Ans. 80 cents. 3. What is 3 per cent. of 50 dollars? Ans. $1.50. 4. What is 4 per cent. of 75 dollars? Ans. $2.00. 5. What is 5 per cent. of 90 dollars? Ans. $4.50. 6. What is 6 per cent. of 100 dollars? Ans. $6.00. 7. What is 7 per cent. of 250 dollars? Ans. $17.50. 8. What is 8 per cent. of 375 dollars? Ans. $30.00. Under the general head Percentage, may be reckoned Interest, Discount, Insurance, Commission, Loss and Gain. SIMPLE INTEREST. 191 INTEREST. Art. 200.-INTEREST is a premium paid, or an allowance made by the borrower to the lender, for the use of a certain sum of money. The money lent, upon'which interest is to be received, is called the principal. The premium paid for the use of the principal, is called the interest. The sum paid on $100, or 100 cents, or ~100 per annum, is called the rate per cent., or per centum. (Per centum signifies by the hundred; per annum, by the year.) The principal and interest added together, is called the amount. OBS.-The rate of interest established by law in the New England states, is 6 per cent. In New York the legal interest is 7 per cent. In England it is 5 per cent. When the rate is not mentioned in this work, 6 per cent. is understood. Interest is either simple or compound. SIMPLE INTEREST. Art. 201. —SIMPLE INTEREST is that which arises from the principal only. What is the interest of $12 for 1 year, at 6 per cent? Operation. If the interest of 81 for 1 year be six $12 cents, or 6o- of a dollar, then the in~.6 terest of.12 would be 12 times.06, or.72 Ans..06 x 12=72 cents. QUESTIoss. —1. What is interest? 2. What do you understand by the principal? 3. What do you understand by the rate per cent.? 4. What does per cent. signify? What, per annum? 5. What is the amount? 6. What is the legal rate of interest in New England? 7. Whatin New York? 8. What is simple interest? 9. Rule to obtain the interest for one year? 10. Why is the rate per cent. written so many hundredths of a dollar? ]192 SI S[MPLE INTERL-EST. 03s. 1.-The rate pelr ce:t. i.s;written as (o) mian liun -redths of a rdollar thus 6 per ceit, i w \rit.te.i (.06; 7 iper celt;.' r cent.,. 053 2 per cent...02. It is evident that the rate per cent. inmist be w-ritten so many hund(redths, because, beiln, so many cents on every hunldred cents, it is so many 100ths of a dollar. EXAIMPLES. 1. Whalt is the interest of.,30, for 2 years, at 5 per cent.? Opteration. If the interest of $1, for I year, at $30.05 per cent. be 5 cents, then the inter-.05 est of. I30. will be.05 x 30 1.50, and 1io fior 2 years, 51.50 X 2 3.00. Ience, 2 to compute tlhe. interest for 1 or more ~ yea-rs, we have the following RULE. fultiply the prit:cipal byT/ the rate, e.rressed as the decimal of a doll(r, anl the p};(rodlct will be the interest for onze y/ear. jiVhc the time is m)ore thal. onet year, multiply the inzterest fir one year by the?nutmbber of years. 2. What is the interest of -$43, for I year, at 6 per cent.? A s.:2.70. 3. What is the interest of $22.25 for 1 year, alt 3 per cent.?.AReduccd to (a decimal. 0 - Y =. 0)Oera. tion,. Opleration. 822.2 O thus: ): 22.25.055.052 11 125 11 1i2 11125 1112 ~1.2'237 5 81.2237 0(ns. 2.-The deciimals below Inills are not regarded in the answer in this, or the following (iqestions. For pointing the product, see Jluitip!icaatiAL of 1)<" a!N. qf t ),?ci, c,,./.. 4. \v at is the interest of'62.75, for 2 years, at, 3 per cent. Ans. ~3.76 5.. WVi,,at is- the interest of $535.42, for 4 years, at 2 per *^;.... - h iCr~tof115.075, f va,, at 7-. cc:.t".t, 0,:r cent.? a~t 81? a.t 9'at 12 per cont. SIMPLE'INTEREST. 193 7. What is the interest of $450.50, for 3 years, at 6 per cent.?' Jls.;81.09. In the preceding examples, the interest h:s teen; computed for 1 or more yea.; b.ut it is often neceSssiary o calculate the interest for montls indcl days. Now, as t1he iIterest on 1, at.06 per cent., for 1 year, or 1'2 months, is 0 (cents, it is ev ident that it amoulls to Ihilf a c cnt a month, or 1I lialf cents a yearl on a dollar. If, thcl''cfore, we ml;tit)l ay numb er o' dotllrs by half the nuumbr of moti.l;, we s;ll h! htv e ine ite'test for the time in cents. A..i as 1 onth is 30 das, d the interest for 1 nIonth is cent, or. 5 mills, ~or 1 dayt i \ ould he -1 -'of a. mill. if', tlhe fi-c're, wet m1ult iply by'- of the days, we have the interest in rmills; or.; we many reduce the days to the fraction of a monthD, and multiply by half the fraction. 8. What is the interest of 840, for 1 aearr, 6 montls, and 5 days? Opera tion.. If the interest of 81 for 12 months s-=_ & 1-)P40 be.06 cents, ti:e interest for 6 monthls.090(3 will be.03 cents, and foir 5 days. -I of 360 a mill; therefoewe," thle interlest of;il 20 for 12 mos.,6 rnos. and 5 days, will be 13.06+.03 -1-.)00 —-.09)5OO!=t.he samt $36~-33 Ans. as one half the months, and one sixth Q,3.633 Ans. i of the days. Hence the RULE. Art. 202. —When there are months and days in the given time — ultiply bby 1rif the 7numtber of months in the whole time, and one sixth of the cldrs. If there be coa odd month, call it 30 days, to.which add the odd days,, f any; atd, dividin them by 6, writi the quotient in the pJlace of miells, in the multiplier. Oa. 1.-If tlie interest is required foli a number of years, multiply the interest for 1 year by the number of years, and compute the interest for the months anld tday;, as above tirected. EX AM PL ES. 9. What is the interest.of'275,, for 2 years, 5 months, and 6 days? Ans. $40.15. QUESTIONS. —. Wlrhat is ite rule f)or poiltiing off the product? 12. What is the rule for computing interest for riionths andl days? 13. Why do we multiply by or6 half the month, and one sixth of the days? 17 194 SIMPLE INTEREST. 10. What is the interest of $749.605, for 3 years, 7 months, and 15 days? 11. What is the interest of $342, for 1 month, 15 days? Operation. 2)342.0071 OBs.2.-As there is no even number of months, we supply the two first decimal 2,394 places with ciphers, as a guide in pointing off 171 the product. $2.565 Ans. 12. What is the interest of $678.59, for 1 year, 3 months, and 11 days? 13. What is the amount of $678.59, on interest, for 1 year, 3 months, and 11 days? Ans. $730.728. OBS. 3.-The amount is the principal and interest added together. 14. What is the interest of $600, for 27 days? 15. What is the amount of $750.60, on interest, for 18 mos. and 18 days? 16. What is the amount of $1000, on interest, for 4 years and 6 months? Ans. $1270. 17. A note for $450, on interest, was dated January 1st, i835. What was due, principal and interest, March 16th, 1837? Ans. $509.625. yrs. mo. d. 1837 3 16 1835 1 1 2 2 15 time. 18. A note for $60.50, on interest, was dated Dec. 20, 1834. What was there due, principal and interest, Jan. 28, 1837? Ans. $68.143. 19. What is the amount of $879.30, on interest, 2 years, 5 months, and 19 days? Ans. $1009.582. 20. What is the interest of $375 for 7 days? Ans. $.437. 21. What is the interest of $89.285, for 1 year, 7 months, and 29 days? Ans. $8.913. 22. What is the interest of $336 for 5 months and 16 days? Ans. $9.296. 23. What is the amount of $1844.48, on interest 2 months and 21 days? Ans. $1869.38. 24. What is the amount of $2731.50, on interest 3 years, 9 months, and 26 days? Ans. $3357.924. SIMPLE INTEREti. 195 25. What is the amount of $1764, on interest from June 14, 1829, to July 14, 1837? Ans. $2619.54. 26. What is the interest of ~240 8s. 6]d., for I year? Operation. Reduce the shillings, pence, and far~240.428 things, to the decimal of a pound by in6 spection, (see Art. 134;) then proceed 14.42568= as in Federal Money. The interest will ~14 8s. 6d. Ans. be in pounds and decimal parts, which must be reduced to shillings. 27. What is the interest of ~379 15s., for 1 year and 6 months? Ans. ~34 3s. 61d. 28. What is the interest of ~416 12s. 6d., for 10 months? Ans. ~20 16s. 7.d. 29. What is the interest of ~427 13s. 9d. 2qrs., for 1 year and 8 months? Ans. ~42 15s. 41d. 30. What is the interest of ~129 7s. 3d. 3qrs., for 3 years, 7 months, and 5 days? Ans. ~27 18s. 6dd. 31. What is the amount of ~320 10s. 6d., on interest for 2 years, 6 months, and 15 days? Ans. ~369 8s. l'd. 32. What is the interest of ~430 7s. 8d. 3qrs., for 4 years, 3 months, and 20 days? Ans. ~111 3s. 7]d. Art. 203. —When the rate of interest is any other than six per cent., and the time consists of years, months, and days, RULE. Find the interest first for 6 per cent., and then for I per cent., and multiply the interest at 1 per cent. by the given rate, and the product will be the answer. EXAMPLES. 33. What is the interest of $680, for 1 year and 6 months, at 7 per cent.? Ans. $71.40. Operation. $680.09 6)61.20 interest at 6 per cent. 10.20 interest at 1 per cent. 7 $71.40 interest at 7 per cent. QUISTIONS.-14. What is the rule for computinginterest on pounds, shillings, pence, etc.? 15. Rule, when the rate of interest is any other than 6 per cent.? 196 INTEREST BY CANCELLING. 34. What is the interest of'336.40, fo: 2 ye:irs, 8 months, and 3 days, at 3 per cent.? Ah4s. $26.9P6 35. What is til interest of $556.36, for 3 years, at 1 per cet.? Ans. $16.69. OnB. —Tl interest of any sum at 1 per cent., for 1 year, is the principal itself, with the separatr x moved twot fiures towards the left; therefore, to obtain the interest at 1 per cent., for any number of years, we lhav\e only to multiply by t.he nlmber of years. 36. \W at is the interest of i0.50 cents, for 5 years, 5 months, and 10 days, at 9 per' cent. ils. ^.'274. 37. WhlIt is the ansount of 81000, on interest for 5 years and 7 nonthlls, at 7- per cent.? Ans. $1418.75. 38. What is the interest of 1 569.20, for 1 year, at I per cent.? An.. 815.692. INTEREST BY CAN.CELLING. RULE. State the question, as in Direct Proportion, by placing the term^s of de'man. on the rig/ht, and the terms of suapposition on the left. EXAMPrLES. Art. 2W, —1. What is the interest of`500, for 3 years, at 6 per cent.? OBS. 1.-1The terlms of supposition in Interest are not expressed, being always 100 and 1 year. The foregoing question may be expressed thus: What is the interest of $500, for 3 years, if the interest of $100 for 1 year be 86? Operation. What interest?;500 $ If $10013 years. Year 16 $ 1890 Ans. 2. What is the interest of $720, for 1 year and 6 months, at 6 per cent.? Ans. 864.80. QUErSTION.-16. What is the interest of any sum for 1 year at I p) r cent.? EXAMPLES IN INTEREST. 197 OBS. 2. —When the given time is months, weeks, or days, either less or greater than a year, reduce it to the lowest denomination, and 1 year, the time in the supposition, to the same denomination. 3. What is the interest of $642.255, for 2 years and 6 months? Ans. $96.338. 4: What is the interest of $1000.68, for 2 months and 15 days? Ans. $12.508. 5. What is the interest of $440, for 4 years, at 4 per cent.? Ans. $70.40. 6. What is the interest of $60.10, for 5 years, at 5 per cent.? Ans. $15.025. 7. What is the interest of $160, for 36 days, at 7 pei cent.? Ans. $1.12. 8. What is the amount of $780, for 3 years and 4 months, at 3 per cent.? Ans. $858. OBs. 3.-If we multiply the amount of $1 for the given time, by the given principal, the result will be the same as adding the principal to the interest. Thus, the amount of $1 for 3 years and 4 months, at 3 per cent., is $1.10, which, multiplied by $780, gives $858, the answer. Art. 205.- When time, rate, and amount are given, to find the principal. 1. What principal will amount to $858, in 3 years and 4 months, at 3 per cent.? The student will perceive, that this question is the reverse of question 8th, preceding, and also that 858 is there a product, of which 1.10, the amount of $1 for the given time, is a factor; therefore, if we divide 858 by 1.10, we shall obtain the other factor, or the principal required; 858- 1.10=$780, the answer. Hence the RULE. Divide the given amount by the amount of $1 for the given time, and the quotient will be the answer. EXAMPLES. 2. What principal will amount to $778.10, in 4 years and 3 months, at 6 per cent.? Ans. $620. 3. What principal will amount to $650, in 6 years, at 5 per cent.? 1)7* 198 EXAMPLES IN INTEREST. Operation by cancelling. What principal. I650 amount. Amount, $1.30 $1 principal. i500 Ans. 4. What principal will amount to $738.40, in 7 years? Ans. $520. Art. 206 —When time, rate, and interest are given, to find the principal. 1. What principal will gain $27.52, in 1 year, 5 months, and G days? We have seen, that the interest of a given principal for a given time, is the product of the interest of $1 for the same length of time, and the principal; therefore, if we divide $27.52 by.086, the interest of $1 for the given time, we shall obtain the principal required, as before. Hence the RULE. Divide the given interest by the interest of $1 for the given time, and the quotient will be the answer. 2. What principal will gain $19 in 4 months, at 6 per cent.? Ans. $950. 3. What principal will gain $1500 in 5 years, at 6 per cent.? Ans. $5000. Art, 207.-When principal, interest, and time are given, to find the rate per cent. 1. If $50 in 6 months gain 81.50, what is the rate per cent.? If the interest of $50, at 1 per cent., be 25 cents, then the quotient of $1.50, the whole interest, divided by 25 cents, will be the rate per cent. required. 1.50 25=6 per cent., the answer. Hence the RULE. Divide the given interest by the interest on the given principal, at 1 per cent. for the given time, and the quotient will be the answer. 2. If $300 gain $12 in 8 months, what is the rate'per cent.? EXA^MPIES IN INTEREST. 199 Operation by cancelling. What interest. 100 $ If $300 12 m. m. 8 12 $ int. lAns. 6 per cent. 3. If $740 gain 827.75 in 9 months, what is the rate per cent.? Ans. 5 per cent. 4. If $1000 gain $75 in 6 months, what is the rate per cent.? Ans. 15 per cent. Art. 208 —When principal, rate, and interest are given, to find the time. 1. In what time will $300 gain $12, at 0 per cent.? Operation. Having found the interest In what time. 12$ Int. of $300 for I year, the quesInt. $18 1 year. tion may be expressed thus: 3 2 8 months. In what time will $12 interest be gained, if $18 be gained in 1 year? It is evident, that the ratio of the interest for I year, is to the given interest, as 1 year is to the time required. Hence the RULE. Divide the given interest by the interest of the given principal for 1 year, and the quotient will be the answer. 2. In what time will $240 gain $4.80, at 6 per cent.? Ans. 4 months. 3. In what time will $600 amount to $645, at 5 per cent.? Ans. I year and 6 months. 4. In what time will $375 gain $28.12~, at 6 per cent.? Ans. 1 year and 3 months. 5. The interest on a note of $225, at 4 per cent., was $11.40. What was the time? Ans. 1 year, 3 months, 6 days. rARTIAL PAYMENTS. Art. 209.-W- hen notes are paid within one year from the time they become due, it has been the usual custom to find the amount of the principal from the time it became due, until the time of settlement, and then to find the amount of each endorsement, from the time it was paid, until settlement, and to subtract their sum from the amount of the principal. 200 EXAMPLES IN INTEREST. EXAMPLES. Boston, January, 1, 1841. For value received, I promise to pay SAMUEL FULTON, or order, tw< hundred and fifty dollars and forty cents, in three months, with interest afterwards. ELIHU JONES. On the back of this note were the following endorsements: March 15, 1841, received one hundred and fifty dollars. June 10, 1841, received forty-five dollars. The balance on the note was paid January 1st, 1842. How much was the balance? First payment, $150 2d payment, $45 Principal, $250.40 Int. 9 m. 16 d..15 Int. 6 m. 21 d. 1.507 Int. 9 mI. 11.268 $157.15 $46.507 261.668 157.15 203.657 Amount of payments, $203.657 Balance, $58.011 Concord, Sept. 1, 1840. For value received, I promise to pay JOHN FOSTER & CO., or order, one thousand dollars, on demand, with interest. STEPHEN PAYWELL. On this note are the following endorsements: March 1, 1841, received two hundred dollars..April 6, 1841, received one hundred and fifty dollars. July 5, 1841, received two hundred and forty dollars. What was there due at the time of settlement, which was August 15, 1841? Ans. $457.042. If settlement is not made till more than a year has elapsed after the commencement of interest, the preceding mode of computing interest, when partial payments have been made, is not in strict conformity with law. The methods of computing interest on notes and bonds differ in different places. The United States Court, and the courts of several of the states, have established a general rule for the computation of interest, when partial payments have been made. The following is, in substance, the RULE. Compute the interest up to the time of the first payment; and if the payment exceed the interest, deduct the excess from the principal, and cast the interest on the remainder up to the second payment, and so on. If the payment be less than the interest, cast the interest up to the time when the sum of the payments shall exceed the interest; then deduct the excess from the principal, and proceed as before. EXAMPLES IN INTEREST. 201 When a note is given specifying interest annually, simple interest is cast on the note to the time of final settlement; and also simple interest on the several sums of interest from the time they became due to the time of final settlement. $3184.25. (1.) For value received, I promise to pay JAMES LARNED, or order, three thousand seven hundred eighty-four dollars and twenty-five cents, with interest. July 10, 1826. Ji FI. On this note were the following endorsements: Jan. 16, 1827, received $148.21 Aug. 11, 1827, " 50.00 Dec. 24, 1828, " 2789.25 Feb. 12, 1830, " 1000.00 What was due Dec. 14, 1830? Ans. $464.867. The first principal, on interest from July 10, 1826, $3784.25 Interest to Jan. 16, 1827, time of the first payment, (6 months, 6 days,)................ 117.311 $3901.561 Payment exceeding the interest, Jan. 16....... 148.21 Remainder for a new principal................ $3753.351 Interest from Jan. 16, 1827, to Dec. 24, 1828, (1 year, 11 months, 8 days,)............... 436.639 4189.990 Payment, Aug. 11, less than the interest, $50.00 Payment, Dec. 24, exceeds the interest, 2789.25 Sum of the payments, 2839.250 Remainder for a new principal................ $1350.740 Interest from Dec. 24, 1828, to Feb. 12, 1830, (1 year, 1 month, 18 days,)................ 91.850 1442.590 Payment, Feb. 12, exceeds the interest,........ 1000.000 Remainder for a new principal................ $442.590 Interest from Feb. 12, 1830, to Dec. 14, 1830, (10 months, 2 days,)..................... 22.277 Balance due Dec. 14, 1830.................. $464.867 Whalt would have been due on the foregoing note at the time of final settlement, had annual interest been specified? 202 COMMISSION, BROKERAGE, AND INSURANCE. (2.) $6420.50. For value received, I promise to pay THOMAS TERRIL, or order, six thousand four hundred twenty dollars and fifty cents, with interest. SAMUEL FGLISH. May 4, 1830. On this note were the following endorsements: March 4, 1831, received 840.00 Dec. 1, 1831, " 200.00 Feb. 10, 1832, " 5000.00 June 28, 1833, " 1534.25 What was the sum due March 1, 1834? Ans. t500.784. 3. A.'s note of $374.62 was given Jan. 1, 1834, on interest after 90 days. June 4, 1836, hle paid 6320. What was due August 15, 1837? Ans. $110.942. 4. B.'s note of $654.32 was given Dec. 12, 1831, on which was endorsed the interest for 18 months and 4 days. What was due on settlement, Nov. 20, 1833? Ans. 8671.114. COMMISSION, BROKERAGE, AND INSURANCE. Art. 210.-CoMMISSION and BROKERAGE are compensations of so much per cent. to factors and brokers, for their respective services in buying and selling goods, etc. INSURANCE is an exemption from hazard, obtained by the payment of a certain sum, which is generally so much per cent. on the estimated value of the property insured. Premium is the sum paid by the owner of the property, for the insurance. Policy is the name given to the instrument, or writing, by which the contract of indemnity is effected between the insurer and insured. The Policy should always cover a sum equal to the estimated value of the property insured, together with the premium: that is, a policy to secure the payment of $100, at 3 per cent., must be made out for $103. QUESTIONs.-1. What are Commission and Brokerage? 2. What is Insurance? 3. What is a Premnlium? 4. What is a Policy' 5. What sum should the Policy cover? 6. Give the example. COMPOUND INTEREST. 203 RULE. Method of operation the same as in Simple Interest. EXAMPLES. 1. If a factor purchase goods to the amount of,1800, and I allow him i per cent. for his services, what must I pay him? Operation. 500 i000 9 *2 43 2 -27 13 Ans. 2. What commission must a factor receive for selling groids to the amount of $864.78, at 4} per cent? Ans. 8$38.915. 3. What is th1 commission on $3784.22, at 121 per cent? Ans..5473.027. 4. A factor buys goods to the amount of $1200. What will be his commission, at 11 per cent.? A rs. 18. 5. What is the brokerage on $9798.671, at 5 per cent. 3? Ans.,563.423. 6. The value of a certain ship and cargo is 5'0000. What is the insurance, at 15 per cent.? Ans. $7,500. 7. What is the duty on 4 boxes of tea, each weighing I cwt. 2 qrs. 14 lbs., at 1 cent per lb.? Ans. $10.92. 8. What may a broker demand on $1000 at 3 per cent.? Anrs. 830. 9. What will be the premium for insuring a ship and cargo, valued at,57840, at 3- per cent.? Ans. $2024.40. 10. What may a broker demand on ~320 10s. 6d., at 4s. 3d. per cent.? Ans. ~68 2s. 2d. 3qrs. OBs.-The above example is not reduced to decimals by Inspection. 11. What will be the premium for insurance on property to the amount of'9248.28, at ~ per cent.? at 3 per cent.? at } per cent..? at? per cent. at -j per cent.? at 4 per cent.? COMPOUND INTEREST. Art. 211.-Co MPOUND INTEREST iS interest upon interest, or that which arises from making the interest a part of the principal, whenever it becomes due. 204 COMPOUND INTEREST. R U LE. Find the amount of the given prin2cipal.for the first year, or the first stated tihne fbr the interest to becom-e due, by simple interest, and make the amount the principal, for the next year, or stated period; and so on to the last. From the last amount, subtract the given principal, and, the renainder will be the compound interest required. EXAMPLES. 1. What is the compound interest of $200, for 3 years, at 6 per cent.? Operation. $200, first principal..06 12.00 interest. ) be 0 0 ~ ~ ip V.to be added. 200 principal. 212, amount, or principal, for 2d year..06 12.72, compound interest, 2d year. to be 212 principal, " added. 224.72, amount, or principal, for 3d year..06 13.4832, compound interest, 3d year. to be 224.72 principal, " added. 238.2032, amount. 200 first principal, subtracted. $38.2032, the compound interest required. 2. What is the compound interest on a note of $325, on interest 5 years? Ans. $109.92. 3. What is the compound interest of *680, for 4 years? Ans. $178.479. 4. What is the compound interest of $500, for 4 years, at 7 per cent. per annum? 5. What is the compound interest of 8470, for 5 years, at 5 per cent. per annum? 6. To what sum will $478 amount, in 3 years, at 6 per cent., compound interest? QUESTIONS.1.. What is Compoud Intereset? 2 tWhat is the Rule? COM.PO'UND 1NTEI'RE ST. 205 TABLE, Showizgui ihe airoutnt of il, or ~1, for any n;Ui;,ir o/'years, not exceeding 30 years, at the rates of 5 and G 2jsr cent. comp,)ound interest. Years. 5 per cent. 6 per cent. Years. 5 per cent. 6 per cent. 1 1.05 1.06 16 2.18287+ 2. 5 0351! 2 1.1025 1.1236 17 2.29201+ 12 2.t977- 13 1.15762+ 1.19101+ 18 2.40661-i- 2. 851t3+ 4 1.21550+ 1.262247+ 19 2.52695-4- 3 02559 + 5 1.27628+ 1.33822+ 20 2.65329-. 3. 2)07 3 6 1.34009+ 1.41851+ 21 2.78596 3. v39 6-+ 7 1.40710+ 1.50363+ 22 2.92526 — 3.60353 - 8 1.47745+- 1.5934M l 23 3.07152+ 3.81974+ 9 1.55132+ 1.68947+ 24 3.22509-+ 4.04893+ 10 1.62889+ 1.79084+ 25 3.36355+ 4.293187+ 11 1.71033+ 1.89829+ 26.55562+- 4.54938 + 12 1.79585+ 2.01219+ 27 3.73345-{- 4.82234+ 13 1.88564+ 2.13292+ 28 3.92012+ 5.11168+ 14 1.97993- 2.26090+ 29 4.116131 + 5.41838415 2.07892-f 2.39655 1 30 4.32194+ 5.74349+ OBs. 1.-Although the decimals, in the preced(ing nunbers, are car ied to five places, yet four are generally sufficient for most business operations. 7. What is the compound interest of G650, for 6 years, at 6 per cent.? Ans. $272.031. By the foregoing table we find the amount of $1 for 6 years to be $1.41851; which, multiplied by $650, gives $922;031, theamountof $650 for 6years,ana $922.031-650 —=272.031, the interest required. 8. What is the compound interest of 8350 for 2 years and 6 months? Ans.'55.0 57. Oss. 2.-When there are months and (lays, first filnd the aion!llt for ihe vears, and on this amount cast the interest for the months and das; this, added to the amount, will give the answer. 9. What is the compound interest of f135, for 3 years, 6 months, and 6 days? Ans. $30.77. 10. What is the compound interest of G 78.25, for 12 years and 6 months, at 5 per cent.? Ans. 5 50.236. 11. What is the compound interest of 5079.75, for 20 years?-for 30 years? QUESTION.-3. What is the Rule, when there are emouths and days? 1 206 DISCOUNT. 12. What is the amount of a note of $150, for 4 years, at 6 per cent., compound interest? Ans. $189.37. 13. Thp amount of a certain note, at compound interest for 4 years, was 1189.37+. What was the principal? Thlis question, it will be perceived, is the reverse of the last. If the amount required is obtained by multiplying the amount of.1 for the given time by the given p incip.il, then it follows, that if we divide the given amount by the amount of ^$1 for the given time, we shall obtain the required principal. 14. What is the amount of $597.75, for 20 years, at 6 per cent., compound interest? Arts. $1917.01. 15. Wait is the amount of $1350, for 3 years, at 5 per cent., comnpound interest? Aa.s'. $156 2.793. 16. \Vhat is the amount of a note for;150, for 2 years, compound interest, the interest becomiing due at the end of every 3 months? Ans. $168.967. 17. Whtl is the compound interest of ~240 10s. 60., for 2 yea's, at 6 per cent.? Ans. ~29 14.s. 61. 3(r-s. 18. Whilt is the amount of ~450, for 3 years, at 5 per cent., compouniid interest? Ans. ~520 18s. 7d. 19. What is the amount of ~256 10s. for 7 years, at 6 per cent., compound interest? Ans. ~385 13s. 71d. DISCOUNT. Art. 212.-DIsco NT is an allowance made for the payment of money before it becomes due.'he present worth/, of a debt due at any future period, is so much money as, being put on interest, at a given rate per cent., will amount to the debt, when it becomes due. 1. A. holds B.'s note for 8106, due in 1 year: What is the present wo.'th of the note, discouniing at 6 per cent.? It is evident, that if B. pays A. $106 now, at the end of the year, when the note becomes due, A. will have the interest of $106 more than is his due; therefore, B. ought to pay him such a sum, as, being put o, interest, woVld amount to 8106 QLCs'ros.;Q.-1. Wihat i di'sc4;Ant1. What i wrst.Ort? EXERCISES IN DISCOUNT. 207 at tie end of the year. If we divide 106 by the amount of $1 for 1 year, we shall have the principal, or that sum which being put on interest at the usual rate per cent., will amount to the debt when it becomes due. (See Art. 205.) $106 -$1.06= $100, the present worth of $106 due a year hence. From the above we derive the following RULE. Art. 213 —To find the present worth-Divide the given sum by the amount of $1 for the given time, and the quotient will be the PRIESENT WORrTI. The present worth, subtracted from the dtbt, will leave the discount. 2. What is the present worth of $246.21, payable in 2 years and 8 months, discounting at 6 per cent.? Ans. $212.25. 3. How much ready money will purchase a note of -$1719.04, due 6 years hence, discounting at 6 per cent.? Ans. $1264. 4. Suppose I owe a note of $416, to be paid in 4 years and 2 months, and wish to pay it now, what must be discounted for present payment? Ans. $83.20. 5. How much ready money will purchase a note of $37.165, due 5 years, 1 month, and 18 days hence, discounting at 6 per cent.? Ans. L28.413+. 6. What is the present worth of a note of $840, payable one half in 10 months, the other half in 20 months, discount, 6 per cent. per annum? Ans. $781.818. 7. What is the present worth of $1500, due 40 years hence, discount, 12 per cent. per annum? Ans. $258.62 -. 8. What is the discount of $420, due in 1 year and 6 months, at 6 per cent.? Ans. 834.679+. 9. What is the discount of $109.86, foc 1 year, at 6 per cent.? Ans.,6.219. 10. Bought goods to the amount of $1909.34, at four months' credit. How much ready money must I pay, discounting at 3~ per cent.? Ans. $1887.322. 11. What is the present worth of ~4000, payable in 9 months, at 43 per cent. discount? Ans. ~3862 8s. Od. 2qrs. + QUESaIONS.-3. Rule for finding the present worth? For finding the discount? 4 What is the difference between interest and discount 1 208 EXERCISES IN DISCOUNT. The foregoingr is the co'rrect method of reckoning discoun, yet tie usual method in practice is to compute the interest for the time, and deduct it from the given sum. The interest thus found is called the ciscount. The difference between interest and discount, on a small sum, for a short time, is intonsiderable; but the difference becomes very considerable when the sum is large and the time long for which the discount is to be made. 12. What is the difference between the interest and discount of 8100 for 1 nonth, at 6 per cent.? Ans. 21 mills, nearly. 13. What is the difference between the interest and discount of:t649, for 3 years, at 6 per cent.? Ans. $17.82. Art, 211, —Bank discount is the same as simple interest \Yhen a note is discounted at a bank, the interest is computed on the sum fromi the date of the note to the time when it becomes due, including three days of grace, and deducted as discount. Thlus, if a note of 8100 be discounted for 30 days, the interest is computed for 33 days. Custom has allowed to the borrower 3 days after the day on which the note becomes due, called days of grace; and as payment is generally withheld until the third day, it is justice that interest should be paid for these davs. If the payment of a note cannot conveniently be made at the proper time, the note may be taken up, if the bank allow the indulgence, by a new note, which must be presented on the day of discount immediately preceding the day on which the note would have become due, paying at the same time the discount, or interest, as before stated. Thus the borrower loses the discount on his note from the day on which he replaces it by another to the day on which it would have been to be paid. The discount of any sum discounted for 30, 60, or 90 days, is found by multiplying by - of the days. (See Art. 202.) EXAMPLES. 14. What is the bank discount 15. What is the bank discount an a note of $714, for 30 days, at on a note of $1692, for 60 days, 6 per cent.? at 6 per cent.? QVES'rION.-5. VWhat is the usual method( in practice? EXERCISES IN DISCOUNT. 209 Operation. Operation. 2)714 2)1692.0051.0 10 3570 16920 357 846 $3.927 Ans. $17.766 Ans. 16. What is the bank discount on a note of $784, for 90 days, at 6 per cent.? Operation. 2)784.0151 3920 784 392 $12.152 Ans. 17. What is the bank discount on a note of $53, for 30 days? Ains. $.291+. 18. What is the bank discount on a note of $1092, for 30 days? Ans..OG06. 19. What is the bank discount on a note of 12049, for 30 days? Ans. 11.269. 20. A.'s note of $561, for 60 days, is discounted at the bank, at 6 per cent. What ready money does he receive? Ans. $555.109. 21. B.'s draft for $150, drawn at 15 days' sight, is cashed at the bank, at 3 per cent. discount. How much money does he receive? Ans. $149.812+. 22. What is the bank discount on a note of $340, for 90 days, at 6 per cent.? Ans. $5.27. 23. What is the bank discount on a note of $632.75, for 90 days, at 6 per cent.? Ans, $9.807. When a note is offered at the bank for discount, one or two endorsers are generally required; and the note is presented in one of the following forms: QUESTIONs.-6. What is bank discount? 7. What is meant by days of grace? 8. flow is discount found for 30,60, and 90 days? 9. When a note is offered at a banik for discount, what is required? 18, 210 LOSS AND GAIN. $500. CONCORD, July 4th, 1849. For value received, we, the subscribers, jointly and severally promise to pay the President, Directors, and Company of the New England Bank, or order, five hundred dollars, at said bank, on demand, with interest after sixty days. When a note, called business paper, is offered for discount, it is generally made in the following form: $350. BOSTON, August 6th, 1849. Three months after date, I promise to pay to the order of Mr. JOHN SAVAGE, at the Commonwealthl Bank, three hundred and fifty dollars, value received. A. B. In order to negotiate this note to an individual, or to procure a discount of it at a bank, the said Savage should endorse his name upon the back of the note, and such other names of endorsers should be procured as may be required; in which case, the promiser, or payer, A. B., is first liable for the note, and the note should be demanded of him, when it becomes due. If not paid, immediate notice should be given to the endorsers of the note; and on such demand and notice, the endorsers become liable for payment of the note; otherwise they are not holden. The promiser, or payer of a note, is the individual who signs it. The promisee, or payee, is the person to whom the note is payable. When a note is endorsed, the promisee, or payee, is always an endorser. LOSS AND GAIN. Art. 215. —Loss AND GAIN teach to find what is gained or lost in the purchase and sale of goods; and also to regulate the price, so as to gain or lose, at a certain rate per cent. 1. If I purchase goods to the amount of $50, and sell the same for $60, what do I gain per cent.? QUESTIONS. —10. What is the form of a note payable to the president, directors, &c., of a bank? 11. What is the form of a note calledl hsiness paper? 12. Who is the p)romiser of a note? 13. Who the promisee? LOSS AND CAIN. 211 It is evident that the gain on $1 would be ~- as much as on $50. Since, then, the gain on O50 is $10, or the gain is l of the cost, then $10-.50- 20 cts. on a dollar, or 20 per cent., the Answzer. Hence the RULIE. When the prices at which goods are bought and sold are given, to find the gain or loss per cent.: Divide the gain or loss, found by subtraction, by the cost of the article. EXAMPLES. 2. A merchant bought goods to the amount of $500, and sold the same for $700. What did he gain per cent.? The question may be thus expressed, as in the Rule of Three: What is the gain on $100, if on $500 the gain be $200? Opuera tion. What gain?.'00 If m000 j0 40 -40 per cent. Ans. 3. A merchant purchased goods to the amount of $342.25,.. gains on the sale $41.07. What is the gain per cent.? Ans. 12 per cent. 4. Bought flour to the amount of $840. Sold the same for $907.20. What do I gain per cent? Ans. 8 per cent. 5. Suppose a merchant purchase goods to the amount of $1000, and sell them for $910, what is the loss per cent.? Ans. 9 per cent. 6. Bought fur caps for $7 apiece; sold them for $7.25. What was the whole gain in laying out $630, and what was the gain per cent.? As. Whole gain, $22.50. Gain per cent., 3.57+. 7. What is the whole loss, and what is the loss per cent., in laying out $70 for hats, at $1.75 each, and selling them for 25 cents apiece less than cost? Whole loss, $10. Loss per cent., 142. QUESTIONS.-1. What is Loss and Gain? 2. How is the gain or loss per cent. found? 3. Having the gain or loss per cent, how is the price found at which an article is bought or sold' 212 LOSS AND GAIN. 8. Bought 100 yards of cloth, at $6.72 per yd., and sold the same for $8.40. What did I gain per cent? Ans. 25 per cent. Art. 216,-When the gain or loss per cent. is given, to find the-price at which the goods are bought and sold. RULE. If the per cent. be gain, add it to 100; if the per cent. be loss, subtract it from 100, and proceed as in the Rule of Three. EXAMPLES. 9. A merchant sold cloth, which cost $6.72 per yard, at 25 per cent. profit. For how much did he sell the cloth per yard? (See Interest, Obs. 3, Art. 204.) Operation. How many $iO./ $ 168 4 $ x00 10 $ 5.8.40 Ans. 10. A merchant sold cloth at $8.40 per yard, and gained 25 per cent. What was the first cost? Operation. How many $18.40 $ I $!x00 $ 4 {j$6.72 Ans. 11. If 1 tun of wine cost ~40, for how much must it be sold to gain 6- per cent.? Ans. ~42 10s. 12. Sold 10 yards of cloth for ~4 16s., and gained 10 per cent. What was the prime cost per yard? Ans. 8s. 8 -id. 13. Bought 7 tuns of wine, at $61.20 per hhd.; sold at 18 cents a pint. What was the whole gain, and how much per cent.? ns. Whole gain, $826.50. Gain per cent., $48.235. 14. Purchased 40 gallons of molasses, at 3s. per gallon. By accident, 6 gallons leaked out. At what rate must I sell the remainder per gallon to gain 10 per cent. upon the first cost, and give 8 months' credit?, Ans. 4s. Od. I qr.15. If I sell a pound of silk for $12.72, and gain 81.20, how much should I gain in selling a bale which cost $1 152? Ans. $120. 16. Bought 300 lbs. of coffee, at 4s. 2d. per lb., ready money. STOCK. 213 and sold the same for 5s. per lb., payable in 8 months. How much was gained upon the whole, and how much per cent. Ans. ~9 128. 3d. 2qrs.+ * 1513 per cent. 17. Bought 50 yards of broadcloth, at 85 per yard, which I purpose to sell at 25 per cent. profit, ready money; but if I sell it on credit, I must have 5 per cent. extra. How must I sell it per yard, at 6 months, to make both these gains? Ans. $6.695. 18. If by selling tea at 57 cents per lb. I lose 3 cents, what is the loss per cent.? Ans. 5 per cent. 19. A merchant purchases 180 casks of raisins, at 16s. per cask; sells the same at 28s. per cwt., and gains 25 per cent. What is the weight of each cask? Ans. 80 lbs. 20. What will be the gain in selling $500 worth of flour, at 8 per cent. advance? Ans. $40. 21. Bought 1000 bushels of corn, for $1922.25. For how much must it be sold to gain 15 per cent.? Ans. $2210.587. 22. Bought 80 reams of paper, at $2.50 per ream. For how much must the whole be sold to lose 5 per cent.? Ans. $190. 23. A merchant bought 500 yards of broadcloth for 82125. For how much must he sell the whole to lose 10 per cent.? Ans. $1912.50. 24. If I buy 45 bushels of salt, at 95 cents per bushel, for how much must it be sold per bushel to gain 20 per cent.? Ans. $1.14. 25. Bought 64 bushels of wheat, at $1.75 per bushel. For how much per bushel must I sell it to lose 3 per cent.? Anrs. 1.697. S TO CK. Art. 217.-STOCK is a general name for capital employed in trade, manufactures, insurance, banking, etc. Also, for money loaned to government, or property in a public debt. The CAPITAL STOCK of a corpany, or corporation, is the whole amount originally invested by such company, or corpoQUESTIOSS.-1. What is stock? 2. What is the capital stock of a company or corporation? 214 BAfTER. ration, which sum is divided into shares, and each holder receives a certificate of the number of shares to which he is entitled. If stock which cost $100 per share sells in the market for any thing more than that amount, it is said to be above par; that is, above the sum equal to the first cost-the term par signifying equality. If it sells for less than that amount, it is below par; and the amount above or below par is spoken of as so much per cent. If it sells for 86 on the $100 in advance, it is 6 per cent. above par. If it sells for so much less, it is so much below par. EXAMPLES. 1. What is the value of $600 of stock, at 6 per cent. above par? Arns. 8636. 2. What is the value of $2000 of railroad stock, at 8St per cent.? Ans. $1750. 3. What is the value of $1500 of bank stock, at 108 per cent.? at 107 per cent.? at 115 per cent.? at 105 per cent.? BARTER. Art. 218.-BARTER is the exchanging of one commodity for another, according to prices or values agreed upon by the parties. RULE. Divide the value of that article whose quantity is giren, by the price of the article wlwse quantity is required; or, the question may be solved by the Rule of Three. EXAMPLES. 1. How many pounds of coffee, at 13~ cents per pound, must be given in barter for 1200 lbs. of sugar, at 8 cents per pound? Operation. By cancelling. 13=4-: 8:: 1200 How many lbs. coffee?11200 lbs. sug. 1200 cts. 401h$ 2 410)90010 13 1 4 j -240 1 lb. coffee. 3 [720 lbs. Ans. 720 lbs. Ans. EXERCISES IN BARTER. 215 2. How much tea, at 64 cents per pound, must be given in barter for 2 cwt. of chocolate, at 32 cents per pound? Ans. 112 lbs. 3. How many pounds of lead, at 9 cents per pound, must be given for 783 lbs. of iron, at 6 cents per pound? Ans. 522 lbs. 4. A. has broadcloth, at 16s. 6d. per yard. B. has linen, at Is. 4d. per yard. How many yards of broadcloth must be given in exchange for 660 yards of linen? Ans. 53- yds. 5. A. bartered 53- yards of broadcloth, at 16s. 6d. per yard, for 660 yards of linen. What was the price of the linen? Ans. Is. 4d. 6. How much sugar, at 8 cents per pound, must be given in barter for 13 cwt. of cinnamon, at 547 cents per pound? Ans. 12 cwt. 7. A. barters 1 cwt. of cinnamon, at 546 cents per lb., for 12 cwt. of sugar. What was the value of the sugar per pound? Ans. 8 cents. 8. A. has lined, worth 20d. per ell English, ready money, 5ult in barter he will have 2s. B. has broadcloth, worth 14s. 6d. per yard, ready money. What ought to be the price of the broadcloth, in barter? Ans. 17s. 4Ad. 9. B. has coffee which he barters with C. at o1d. per lb. more than it cost him, for tea which cost 10s.; but in barter C. puts it at 12s. 6d. What was the first cost of the coffee? Ans. 3s. 4d. 10. A. has 5 tons of butter, at $425 per ton, and 10-1 tons of tallow, ~33 15s. per ton, which he barters with B. for 316 barrels of beef, at 21 s. per barrel, and the remainder in cash. How much money does he receive? Ans. $2200.25. 11. C. and D. barter. C. has corn, at 75 cents, ready money, but in barter he will have $1. B. h as rye, at 50 cents, ready money. What ought he to have for his rye, in barter? Ans. 66- cents. 12. A. has rye at $1.44 per bushel, ready money, but in barter he will have $1.56 per bushel. D. has cotton, at 18 cents per pound, ready money. WVhat price must the cotton be in barter, and how many pounds of cotton must be bartered for 100 bushels of rye? Ans. Cotton, 19' cents per pound. 800 lbs. for 100 bushels rye. 216 SUPi'LEMENT TO INTELREI;31' DISCOUNT, ETC. 13. B.gives C. 250 vards of dlru"ret, at 18. per yard, foi 308k lbs. of pepper., What doe tahe pepper cost C. per lb.. Ans. Iod. 14. A. and B. barte;. A. has 4 I cwt. of hops, at 87.20 per cwt., for which B. gives him $90G il money, and the rest in prunes, at 10 cents per lb. What quantity of prunes does A. receive? lAns. 17 cwt. 3 qrs. 4 lbs. 15. How many acres of l.nd, worth ~40 10s. per acre, must be given for 600 acres, worth $8.50 per acr? Ans. 37-. 16. A. has 7- cwt. of sugar, at Sd. per pound, for which B. gives him 12- cwt. of flour. How much per pound was the flour? Ans. 44d. 17. A. has corn, at $1.25, ready money, but in barter he values it at $1.50 per bushel. B. has cotton at 20 cents per pound, ready money. What should be the price of the cotton, in barter, and how many pounds must be given for 100 bushels of corn? Answer to the last, 625 lbs. 18. A. has cloth, ylaled at $4 per yard, ready money, but in barter he will have $4.50. B. has cloth at 2 pounds per yard, ready money; at at hat pice ought B. to rate his cloth in barter, and how many yards must be given A. in exchange for 540 yards of cloth? Answer to the last, 324 yards. 19. D. has ribbon, at 2s. per yard, ready money, but in barter he will have 2s. 3d. E. has broadcloth, for which he will have in barter 3as. Cti. 3qr.s. What ought to be the cash price of E.'s cloth, and how many yards of ribbon ought D. to give him for 488 yards of broadcloth? E.'s cloth, 32s. Gd. Ans. 7939 yards ribbon. SUPPLEMENT TO INTEREST, DISCOUNT, BARTER, AND LOSS AND GAIN. Art. 219. —1. What is the interest of $365.25 for 1 year, 3 months, and 2 days? Ans. $27.515. 2. What will $1002.153 amount to in 4 years, 1 month, and 15 days, at simple interest? Ans. $1250.185. 3. What is the interest of $125000 for 1 day? Ans. ^20.833, SUPPLEMENT TO INTEREST, IDISCOUNT, ETC. 217 4. How much will ~300 amount to in 5- years, at 3- per cent.? Ans. ~356 Is. 3d. 5. What is the amount of ~10 15s. 6d., for 16 years and 10 months? Ans. ~21 13s. ld. 3qrs. 6. How much will $185.26 amount to in 2 years, 3 months, and 11 days, at 7 per cent.? Ans. $216.944. 7. How much will $298.59 amount to, from May 19th, 1797, to Aug. 11, 1798, at 8 per cent.? Ans. $327.913. 8. What is the interest of $658, from Jan. 9th to the 9th of Oct. following, at ~ per cent.? $2.467. 9. What principal will amount to $1319.90, in 5 years and 8 months? Ans. $985. 10. Took up a note, April 29, 1799, which amounted to $205.86, dated June 14, 1798, on interest at 53 per cent. What was the sum borrowed? Ans. $196. 11. A note of 6 years' standing amounted to ~3810; the principal was ~3000. What was the rate per cent.? Ans. 4. 12. At what rate per cent. will $420 amount to $520.80 in 8 years? Ans. 3 per cent. 13. At what rate per cent. will ~413 12s. 6d. amount to ~~46 3s. 8d. in 43 years? Ans. 6-. 14. In what time will $500 amount to $725, at 5 per cent.? Ans. 9 years. 15. In what time will a note of ~420 amount to ~520 16s., at 3 per cent.? Ans. 8 years. 16. What will be the amount of $597.75, in 20 years, at 6 per cent., compound interest? Ans. $1917.077. 17. Gave f Inote for ~450, payalble in 3 years, at 5 per cent., compound interest. To what did it amount? Ans. ~520 18s. 7Vd. 18. What is the amount of ~217, for 2-1 years, at 5 per cent., interest payable quarterly? Ans. ~242 13s. 4?d. 19. Bought a quantity of goods, to the amount of ~250, ready money, and sold them for ~300, payable in 9 months. What was the gain in ready money, discounting at 6 per cent.? Ans. ~37 Is. 7d. lqr. 20. What is the present worth of.1 000, payable one-half in 4 months, the other half in 8 months, discounting at the rate of 5 per cent.? Ans. 1975.345. 19 218 EUtA'rION OF PAYMENI'S. 21. How much tea, at 9s. 6d. per pound, must be given in barter for 156 gallons of wine, at 12s. 3~d. per gallon? Ans. 201 lbs. 1327 oz. 22. A. has 240 bushels of rye, at 90 cents per bushel, ready money, which he barters with B., at 95 cents, for wheat which cost 99 cents per bushel. How many bushels of wheat must he receive for his rye, and at what price? Ans. 218 T bushels, at $1.041 per bushel. 23. A. and B. barter. A. has cloth which cost him 28d., B.'s cost him 22d. B. puts his cloth at 25d., in barter. How high must A. rate his cloth, to gain 10 per cent. in the trade? Ans. 35d. 24. Bought 100 yards of cloth, at $2 per yard. How must I sell it per yard, to gain $50? Ans. $2.50. 25. Bought cloth at $1.50 per yard, which, not proving so good as I expected, I am willing to lose 171 per cent. How must I sell it per yard? Ans. $1.23'7+. 26. Bought 50 gallons of wine, at 4s. per gallon. By accident, 10 gallons leaked out. How must I sell the remainder per gallon, to gain 10 per cent. upon the whole cost? Ans. 5s. 6d. 27. A man sells a quantity of corn at $1 per bushel, and gains 20 per cent. Some time after, he sold of the same to the amount of $37.50, and gained 50 per cent. How many bushels were there itr the last parcel, and at what rate did he sell it per bushel? Ans. 30 bushels, at $1.25 per bushel. EQUATION OF PAYMENTS. Art 220 —- EQUATION OF PAYMENTS is the method of finding the mean time for the payment of several debts due at different times. 1. If a man owes me $10, to be paid in 4 months, and $5, to be paid in 7 months, and he wishes to pay the whole at once, in what time should the whole be paid? It is evident that the use of $10 four months is the same as QvursIeNr.-1. What is Equation of Payments? 2. Rule? EQUATION OF PAYMENTS. 219 the use of $1 forty months; and the use of $5 seven months is the same as the use of $1 for thirty-five months. Then, $10+$5=$15, and 40+35=75 months. Thus it appears, that the use of $10 for four, and $5 for seven months, is the same as the use of $1 for seventy-five months: $15, therefore, may be used -L as long as $1. That is, I of seventy-five months, 75- 15=-5 months, the answer. Hence the RULE. lMultiply each payment by the time when it becomes due, and divide the sum of the products by the sum of the payments, and the quotient will be the time required. 2. A merchant has owing him $420, to be paid as follows: $100 in 8 months, $100 in 2 months, and $220 in 5 months. In what time ought the whole to be paid at once? Ans. 5 months. 3. A. owes B. $800, to be paid as follows: $200 in 3 months, $150 in 4 months, and the remainder in 8 months. What is the equated time for the payment of the whole? Ans. 6 months. 4. A. owes B. $380, to be paid as follows: $100 in 6 months, $120 in 7 months, and $160 in 10 months. What is the equated time for the payment of the whole? Ans. 8 months. 5. A merchant has owing him $698, of which $181 is to be paid at the present time, $199 in 3 months, and $318 in 8 months. What is the equated time for the payment of the whole? Ans. 4~ months. 6. A. owes B. $500, of which 1 is to be paid in 3 months,. in 8 months, and the remainder in 2 months. What is the equated time for the payment of the whole? Ans. 3 months, 21 days. 7. A. has owing him $924, of which I is to be paid in 3 months, and a in 2 years. In what time ought the whole to be paid? Ans. 13 months. 8. I have three notes against a man: one of $400, due in 5 months; one of $500, due in 6 months; and the other of $350, due in 9 months; and he wishes to pay the whole at once. In what time ought he to pay it? Ans. 6.52 months. 9. A. owes B. $960, of which 1 is due in 3 months, I in 1months, in 9 months. What is the equated time for the payment of the whole? Ans. 3 months, 15 days. 220 FELLOWSHI P. 10. A merchant bought goods to the amount of $3000, and agreed to pay $500 ready money, $600 in 4 months, and the remainder in 9 months; but they agree to make one payment of the whole. What is the equated time? Ans. 6 months, 15 days. FE LL OWShI P. Irt. 221.-FELLOWSHIP is a rule by which merchants and others, trading in company, may ascertain their respective gain or loss, in proportion to each man's share in the joint stock. The money, or value of property vested in trade, is called the Capital, or Stock. The gain or loss to be shared by the company is called the Dividend. When the several stocks are employed without regard to time, it is called Single Fellowship. 1. Two men, A. and B., bought a horse for $60, of which sum A. paid $40, and B. paid $20. They sold the horse for $90. What was each man's share of the gain? It is evident, that each man's share of the gain should bear the same ratio to the whole gain, that his share of the stock bears to the whole stock. Now, the whole stock was $60, of which A. paid $40; then A. paid 4 =- of the whole stock, and B. paid $20== 2 0 of the whole stock. As the whole gain was $30, A.'s share is - of 30=$20, and B.'s share is ~ of 30=$10. Hence the RULE. As the whole stock is to each man's stock, so is the whole gain or loss to each man's share of the gain or loss. Or the question may be expressed thus: What gains each individual stock, if the whole stock give gain? Whole stock $60-A.'s stock- $40. Whole gain $30-B.'s stock $20. QUESTIONS.-1. What is Fellowship? 2. What is capital, or stock? 3. What is the dividend? 4. What is Single Fellowship? 5. What is the rule? 6. What is the method of pre of? EXERCISES IN FELLOWSHIP. 221 Operation. Operation. How much A.'s gain. 40$ A.'s stock. B.s gain. 20$ B.'s stock. Whole stock $60 30$ whole gain. 30$ $60 $ 20 Ans. A.'s gain. $110 Ans. B.'s gain Proof —Add together the respective gains, and if the work be right, their sum will equal the whole gain. 2. A., B., and C. trade in company. A.'s stock is $240, B.'s $360, and C.'s $600. They gain $325. What is each man's share of the gain? A.s gain, $65.00. Ans. B.'s gain, 97.50. C.'s gain, 162.50. 3. A. and B. bought a lot of land for $1280, of which B. paid $400, and A. the remainder. They sold it so as to gain $200. What was each man's share of the gain? An. A.'s gain, $137.50.. B.'s gain, 62.50. 4. A. and B. owned a ship, valued at $72000-lost at sea; insurance $50000. What was each man's loss, supposing A. owned 3 times as much as B.? Ans A.'s loss, $16,500.. B.'s loss, 5,500. 5. A man dying, leaves property to the amount of $3000. A. has a note of $600 against the estate, B. has a note of $1800, and C. a note of $1600. How much must each-lose? A.'s loss, $150. Ans. B.'s loss, 450. (C.'s loss, 400. 6. Three partners, A., B., and C., shipped 216 horses for the south. A.'s share of the cost of the horses was $2880; B.'s, $5760; C.'s, $4320. During the voyage they were obliged to throw 90 overboard. How many horses did each partner lose? A. lost 20. Ans. B. lost 40. (. lost 30.'7. A. and B. trade in company. A.'s stock was 60 guineas, and his share of the gain was -a. What was B.'s stock? Ans. 36 guineas. 8. Three men gained in an adventure $96. A. put in a certain sum, B. put in twice as much as A., and C. as much as A. and B. both. What was each man's share of the gain? A.'s, $16. Ans. B.'s, 32. 19 / C.'s, 48. 222 ASSESSMENT OF TAKES. 9. Two men trade in company. Their joint stock is $800, bf which B. put in - of 3 of 3 of - of 4 times the whole. What is each man's stock? A A.'s, $200. Ans. -B.'s, 600. 10. A., B., and C. trade in company. A.'s stock is $250, B.'s $300, C.'s $550. They lose 5 per cent. by trading. What is each man's share of the loss? A.'s loss, $12.50. Jlns. B.'s loss, 15.00. C.'s loss, 27.50. 11. A man by his will left his estate to his children, as follows: to A. he gave $5000, to B. $4500, to C. $4500, and to D. $4000; but his whole estate amounted to but $12000. How much did each receive? A. received $3333.33-. A s B. " 3000.00. Cns. " 3000.00. D. " 2666.662. ASSESSMENT OF TAXES. Art. 222.-IN order to the assessment of taxes on a town, the following facts should be known: 1. The amount of tax assessed by the Legislature of the State. 2. The inventory of all the rateable property in the town. RULE. I. From the tar raised by the town, deduct the amount of poll taxes. II. Find Ihe tax on a dollar, and multiply each man's inventory by it, and to the product add hs poll tax. EXAMPLES. 1. A town inventoried at $160,000, raises a tax of $3400, There are 400 rateable polls, taxed 50 cents each. What is the tax on a dollar, and what is A.'s tax, whose real and personal estate is inventoried at $1683, and who pays for one poll? QuESTIONS.-1. What facts should be known, in order to the assessment of taxes? 2. What is the rule? ASSESSMENT OF TAXES. 223 First deduct the amount of poll tax for 400 polls, at 50 cents each, 400 X.50=-200, amount of poll tax; then $3400-200 -=$3200 to be assessed on the whole property. Secondly, find the tax on a dollar. Operation. What tax. 1 $160,000 3200 Ans..02 cts. on $1. Then to find A.'s tax, multiply the amount of his inventory by the tax on a dollar, and to the product add his poll taxthus, $1683 x.02=-33.66+.50 —34.16, A.'s tax. Or, having found the tax on a dollar, a table may be formed, containing the tax on 1, 2, 3, or to 20 dollars; then on 30, 40, &c., to 100 dollars; then on 110, 120, &c., to 1000 dollars. Then, having the inventory of the property of an individual, his tax may be readily made out. TABLE. Tax on $1 is $.02 Tax on $17 is $.34 Tax on $150 is $3.00 " 2".04 " 18".36 " 160" 3.20 " 3".06 " 19".38 " 170" 3.40 4 ".08 " 20'.40 " 180 " 3.60 5".10 " 30".60 " 190 " 3.80 " 6".12 " 40".80 " 200 " 4.00 q.14 " 50" 1.00 " 300 " 6.00 " 8 ".16 " 60 " 1.20 " 400 " 8.00 9 ".18 "'0' 1.40 " 500 " 10.00 " 10".20 " 80 " 1.60 " 600 " 12.00 " 11.22 " 90" 1.80 "'700 " 14.00 I 12.24 " 100" 2.00 " 800 " 16.00 " 13 ".26 " 110" 2.20 " 900 " 18.00 " 14 ".28 " 120 " 2.40 " 1000 " 20.00 " 15 ".30 " 130 " 2.60 " 16 ".32 " 140 " 2.80 We find by the table, the tax on $1000 to be $20.00 " "' " " 600 " 12.00 " " " " 80 " 1.60,, ";, " 3 ".06 A.'s inventory, $1683 $33.66 Poll tax; 50 Amount of A.'s tax, $34.16 224 DOUBLE FELLOWSHIP. 2. The inventory of reat and personal estate in the town of,for the year 1849, is $800,000. The amount assessed on each rateable poll is $1. The number of polls is 400. The amount of town tax voted to be raised for the year 1849 is $2000. The proportion of state tax for said town for tlkat year is $400; county tax $200; the amount of school tax is $800; the highway tax is $1200. How much is A.'s town, state, county, school, and highway tax, whose whole estate is inventoried at $5000, and who pays for one poll? F Town tax, $11.847. State tax, 2.369. Ans. - County tax, 1.189. Total, $27.25. School tax, 4.739. Highway tax, 7.108. DOUBLE FELLOWSHIP. Art. 223.-1. Two men, A. and B., hire a pasture for $36. A. put in 8 oxen 6 weeks, and B. 12 oxen 8 weeks. How much must each pay? It is evident that the pasturage of 8 oxen, 6 weeks, is the same as of 1 ox 48 weeks; and the pasturage of 12 oxen, 8 weeks, is the same as of 1 ox 96 weeks. The shares of A. and B. are the same as though A. had put in 1 ox 48 weeks, and B. 1 ox 96 weeks; 96+48=144 weeks. Then A.'s share of the rent will be -1j = of $36=$12, and B.'s share will be 9I-6 2 of $36=-24. Hence the RULE, Multiply each man's stock by the time it is continued in trade, and consider the product his share of the joint stock, and proceed as in Single Fellowship Operation. Operation. How many * 48 weeks. How many $ 96 weeks. Weeks. 144 36 $ Weeks. 144 36 $ $12, A.'s share. $24, B.'s share. 2. A. and B. trade in company. A. put in $3000, for 6 months; B. put in $4000, for 10 monthsi and C. put in INVOLUTION. 225 $2500 for 12 months. They gained $880. What is each man's share of the gain? A.'s share, $180. Ans. \B.'s share, 400. C.'s share, 300. 3. Three men trade in company. A. put in 84000 for 12 months, B. put in 83000 for 15 months, and C. pat in l5000 for 6 months. The whole gain was $G15. What was their respective shares? A.'s gain, 0240. Ans. B.'s gain, 225..C.'s gain, 150. 4. A., B., and C. masde a stock for 2 years. A. put in at first $1000. At the end of 6 months he put in,500 more. B. put in 51600, and after 8 months took out`400. C. put in?2000 for 20 months, and then took out $1500. They gain P1000. What is each man's share? 5. A., B., and C. lost in trade $263.90. A.'s stock was 8580, for 6 months; B.'s stock was S580, for 91- months; C.'s stock for 01 2 2onths; C's stock was 870O, for 8- months. What is each man's share of the loss? (A.'s loss, $59.15. Ans. B.'s loss, 80.45. (.'s loss, 118.30. 6. A. commenced business on the first of January, with a capital of $3800; on the first of May he took in B. as a partner, with a capital of $2700; on the first of August, they admit C. as a partner, with a capital of $4000: at the end of the year they dissolve partnership; each took his share of the stock and gain, the gain being $4360. How much did each take? A. took $6080. Ans. B. 3780. C. " 5000. INVOLUTION. Art. 22 —-INVOLUTION is multiplying a number into itself. The product is called a power; the number so multiplied is called a root, or the first power. The product of any number multiplied into itself is called the second power, or square. If the square be multiplied by the first power, the product is called the cube, or third power. QUESTIONS.-I. What is Involution? 2. What is a power? 226 INVOLUTION. The power is sometimes denoted by a small figure, called the index, or exponent, of the power, placed above the given number at the right hand-Thus, 32 denotes that the second power of 3 is required, or it shows how many times 3 is to be involved or multiplied. This may be illustrated by the following: 22=2 x 2=4, the second power of 2. 23-2 X 2 x 2= 8, the third power of 2. 24=2 X 2 x 2 2=16, the fourth power of 2. 25=2 X 2 X 2 X 2 x 2=32, the fifth power, or sursolid. 26=2 x 2 x 2 X 2 X 2 X 2=64, the sixth power, or square cubes. The product of any two powers is always that power whose index is the sum of the indices, or exponents, of the power multiplied, thus: 0 1 2 3 4 5 6 7 8 9 10 1 2 4 8 16 32 64 128 256 512 1024 If 16, which is the 4th power of 2, be multiplied into 64, the 6th power of 2, we shall have 1024, the power indicated by the multiplication of 24 X 26 —26 +4=20=-1024. EXAMPLES. 1. What is the square, or second power, of 25? Ans. 625. 2. What is the square, or second power, of 145? Ans. 21025. 3. What is the cube, or third power, of 23? Ans. 12167. 4. What is the cube, or third power, of 159? Ans. 4019679. 5. What is the biquadrate, or fourth power, of 29? Ans. 707281. 6. What is the fifth power of 134? 7. What is the square of 1? 8. What is the cube of 1? Ans. 1; 1. 9. What is the square of? 10. What is the cube of -? Ans. I; - 4' 8' 11. What is the cube of 1.5? 12. What is the cube of 2.25? 13. What is the square of 22? Ans. 24=5.76. QUESTIONS.-3. What is a root? 4. What is the second power?-the third power? 5, How is a power denoted? EVOLUTION. 227 OBS.-Mixed numbers may be reduced to improper fractions, before involving: Thus, 22==2; or they may be reduced to decimal: Thus, 2-=-2.4. The powers of the nine digits, frorn the first to the ninth, may be seen by the following TABLE. Roots..... 1 2 3 4 5 6 7 8 9 Squares.... 14 9 16 25 36 49 64 81 Cubes.... 1 8 27 64 125 216 343 512 7Q9 4th power. 1 16 81 256 625 1296 2401 4096 6561 5th power..1 321 243 1024 3125 7776 16807 32768 59049 6th power 1 64 729 4096 15625 46656 117649 262144 531441 7th power.l 128 218 16384 78125 279936 823543 2097152 4782969 8th power..1 256 6561 65536 390625 1679616 5764801 16777216 43046791 9th power. 11 512 19683 262144 1953125 10077696 40353607 134217728 387420489 EVOLUTION Art. 225.-EVOLUTION, the opposite of Involution, is the extracting of the root of any number, or the finding of such a number as, when multiplied into itself a certain number of times, will produce a given number. Thus, 3 is the square root of 9, because 3 x3=9; -also, 3 is the cube root of 27, because 3X 3 X 3 = 27. Any given power may be found by a continued multiplication of the number into itself; yet there are numbers whose precise root can never be found; but, by'the use of decimals, we can arrive sufficiently near for all practical purposes. A number whose precise root cannot be found, is called a surd, or irrational number, and its root a surd root. The square root may be denoted by this character, /, called the radical sign, placed before the power; and the other roots by the same sign, with the index of the root placed over it, or by the fractional indices placed on the right hand. Thus, the quare root of 9 is expressed, 5/9, or 90, and the cube root of 27 thus: 2/27, or 273. QUESTIONS.-1. What is Evolution? 2. How may any given power be found? I. Can the precise roots oflll powers be found? 4. flow can we approximate suffiziently near for practical purposes? 5. What is a number called whose precise root cannot be found? 6. What is the advantage of denoting roots by the fractional indices? 228 EXTRACTION OF TIIE SQUARE ROOT. The method 9f denoting roots by the fractional indices is rrefelttble, as, by it, not the root only is denoted, but the power. Thie numerator of the index denotesthe power, and the denominator the root of the number over which it is placed. If the power is expressed by several numbers, with the sign + or - between thenm, a line, or vinculum, is drawn from the top of the sign over all the numbers. Thus, the square root of 12+4 is V1'2+4=4, and the cube root of 357-14 is a357- 14-7. EXTRACTION OF THE S QUARE ROOT. Formation of the Square, and Extraction of the Square Root. Art. 22g.-It hlas been shown, that to obtain the square of any number, whether entire or fractional, we have only to multiply that number into itself. Therefore, To extract the square root, is to find a number, which, multiplied into itself once, will produce a given number. The principle applied in the extraction of the square root, will be better understood by attending, first, to the formation of the square. The square of any number expressed by a single figure, will contain no figure of a higher dinomination than tens. (See Table )of Powters.) Numbers which are produced by the multiplication of a number into itself, are called perfect squares. Th'ere are but nine perfect squares among all the numbers, which can be expressed by one or tw-o figures. The square roots of all other numbers, expressed by one or two figures, will be found between two whole numbers differing from each other by unity. Thus, 37, which is comprised between 36 and 49, las for its square root a number between 6 and 7; and 95, wvhich is comprised between 81 and 100, has for its square root a number between 9 and 10. What is the square of 32? tens. units. 32 =3 +2 3+2 60 4 9 - 12 - 4= 1024 EXTRACTION OF THE SQUARE ROOT. 229 Thus, it appears, that the square of a number made up of tens and units, contains the square of the tens, plus twice the products of the tens into the units, plus the square of the units. What is the square root of 1024? It is evident, that the root will contain more than one figure, since the number is composed of more than Ttwo places; and it will contain no more than two, for 1024 is less than 10,000, the square of 100. It will also be perceived, from the foregoing process, that the square of the tens, the first figure of the root, must be found in the two left-hand figures, which we will separate from the others by a point; thus, 1024. The two parts, of two figures cach, are called periods. The period 10 is comprised between the squares, 9 and 16, whose roots are 3 and 4; hence, 3 is the tens, or the first figure of the root sought. 1024(32 n Q We write 3, the first figure of the root, on the right of the given num_9-2= ___21 ber, and its square, 9, we subtract 3 X 2=6)12j4 from 10, the left-hand period, and to 62 X 2= 124 the remainder we bring dolwn the next period. Having subtracted the square of the tens from the given number, the remainder, 124, contains twice the product of the tens into the units, plus the square of the units; but since tens into units cannot give a product of less name than tens, it follows that the right-hand figure, 4, can form no part of the double product of the tens into tke units; therefore, if we divide 12, twice the product of tens into the units, by twice 3, the tens of the quotient, we shall obtain the unit figure of the root. We will now write this quotient figure on the right of the other, and multiply 62 by 2, the last quotient figure. W:.thus obtain, 1st, the square of the units; 2d, twice the product of the tens into the units; hence 32 is the required root. What is the square root of 572? Operation. In this example the remainder, 43, shows that 572 is not a perfect square; but 23 is 572(23 the greatest square contained in 57.2; that _4 iis, it is' the entire part of the root. This 72 may be shown, thus: The dclifereac- between - )12';9.tPthe squarets of two concsec utie numbnbers, is 129 7 -. equal to tuice the less number, plus 1. The difference between the squares of 8 and 9 is 20 230 EXTRACTION OF THE SQUARE ROOT. 17=8 2+1, and 23 x2+I=47, which is greater than 43, the remainder, which shows that 23 is the entire part of the root, The foregoing rule may now be applied to finding the length of one side of a square surface, whose area is expressed by the given number. EXAMPLES. Art. 227.-1. What is the length of one side of a square garden, containing 576 square rods, or what is the square root of 570? We first distinguish* the number whose root is to be found, into periods of two figures each, denoted by the index of the root. By the number of periods, we perceive that the root will consist of two figures, a unit and ten. As the second power of ten cannot be less than a hundred, we look for the square of tens in the second, or left-hand period, which is 5. We find the nearest square in 5 to be 4, and its root 2, or 2 tens, which we place in the quotient as the first figure of the root; and its square 4, or 400, under the 576(2 period, and subtracting it, we have a 4 remainder of 1, or 100, to which we 176 add 76, the next period. Had the garden contained but 400 square rods, Fig. 1. we should now have obtained the length of one side, 2 tens=20, and 20 X 20=400; consequently, 400 rods o 20 would be disposed of in the form of a 20 osquare. (See Fig. 1.) But we have a v 40 remainder of 176 rods, to be added to 400 the square, and in such a manner that its form shall not be altered. We must, therefore, make an equal addi20 rods. tion on two sides. Then 20+20=40, the length of the whole addition. To find the width of the addition, we place the double of the root * It is distinguished into periods of two figures each, because the second power can never have more than twice as many figures as its root, and never but one less than twice as many. The third power can never have more than three times as many figures as its root, and never but two less than three times as many. Distinguish, therefore, any number into periods of as many figures as are denoted by the index of the root. EXTRACTION OF THE SQUARE ROOT. 231 already found, on the left hand of the dividend, for a divisor. If we divide 176, the number of rods 576(24 to be added, by 40, the length of the 44)176 addition, (or 17 by 4, rejecting the 176 unit figure of the dividend and diviProof: 24X24=576 sor,)we have 4 rods, the width of the addition. Then 40 X 4=-160, the number of rods added on the two sides; still there is a remainder of 16 rods. As the additions made are no longer than the sides of the square, there will Fig. 2. be a deficiency in the corner, 20X4=80 16 (see Fig. 2,) of a square whose sides are equal to the width of o the addition, 4 x4=16 rods. fco* XO ~ We therefore place 4, the last 20 X 20=400 - quotient figure, on the right of v. 11' the divisor, because its square.Wm~~ ~o is necessary to supply this deficiency. The whole divisor now multiplied by the last quotient ~figure, equals 176, the number 24 rods. of rods which were to be added 24 rods. to the square. We have now obtained 24, the root of 576, or the length of one side of a square garden containing 576 square rods. Proof by Involution: 24x24=5-6. From the preceding example and illustration we derive the following RULE. I, Distinguish the given number into periods of two figures each, by putting a dot over the units, and another over the hundreds, and so on. The dots show the number of figures of which the root will consist. II. Find the root of the greatest square number in the lefthand period, and place it as a quotient in division. Place the square of the root found, under said period, and subtract it therefrom, and to the remainder bring down the next period, for a dividend. III. Double the root already found, for a divisor; see hote often the divisor is contained in the dividend, (excepting the right-hand figure,) and place the result for the next figure in the root, and also on the right hand of the divisor. 232 EXTRACTION OF THE SQUARE ROOr. IV. Mlzltiply the divisor by the figure in the root lastfound, and subtract theproduct from the dividend. To the remainder, bring down the next period, for a neuw dividend. Double the root now found, obr a new divisor, and proceed in the operation as before, until all the periods are brought down. OBS.-Doubling the right-hand figure of the last divisor, observing to add 1 to the place of tens, when the double of the unit figure is over ten, is the same as doubling the root, or quotient. EXAMPLES. Art. 228s-2. What is the square root of 119716? Operation. 119716(346 Ans. 9 64)297 256 6S6)4116 4116 3. What is the square root of 1444? Ans. 38. 4. What is the square root of 59536? Ans. 244. 5. What is the square root of 124896? lAns. 353.4+. Ocs. 1. —When there is a remain(ler, after all the figures are brought downl, ciphers may be annexed, and the operation continued to any as-;igned degree of exactness. 6. What is the square root of 67321? Ans. 259.46+. 7. What is the square root of 25289? Ans. 159.02+. 8. What is the square root of 21027? Ans. 145.006+. 9. What is the square root of 6842.723400? Ans. 82.7207+. OBs. 2.-When there are whole numbers and decimals in the given sum, point off both ways fiom the units' place; if the decimals be an odd number, annex ciphers, and make them even. 10. What is the square root of 10.4976? Ans. 3.24. 11. What is the square root of 336.234? Ans. 18.333+. 12. What is the square root of.108241? Ans..329. 13. What is the length of a square field containing 7744 square rods? Ans. 88 rods. EXTRACTION OF THE SQUARE ROOT. 223 14. What is the square root of -9? Ans. 3 OBs. 3.-The square root of a fraction may be foand by extracting the root of the numerator and denominator. 15. What is the square root of 1-a? Ans. 5. 2 75. 16. Whatis the square root of -4-? Ans. 2. 17. What is the square root of -4? Ans. 7-. 18. What is the square root of ~? Ans..707+. OBs. 4.-When the numerator and denominator are surd numbers, re duce the fraction to a decimal, and extract the root as above directed. 19. What is the square root of A? Ans..866+. 20. What is the square root of -? Ans..9355+. 21. How many rows on one side of a square cornfield, containing 15376 hills? Ans. 124. 22. An army of 242064 men are drawn up in a solid body, in the form of a square. What is the number of men in rank and file? Ans. 492. 23. A man has 841 peach-trees, which he wishes to plant in the form of a square. How many must be planted in each row? Ans. 29. 24. There is a circular pond, containing 110889 square rods. What will be the length of a square field containing the same number of rods? Ans. 333 rods. 25. A number of men gave ~22 Is. for a charitable purpose, each giving as many shillings as there were men. What was the number of men? Ans. 21. 26. What is the length of one side of a square acre of land? Ans. 12.64+. 27. The diameter of a circle is 6 inches. What is the diameter of a circle 4 times as large? Ans. 12. OBn. 5.-Circles are to one another as the squares of their diameter; therefore, to find the required diameter, square the given diameter, multiply the square by the given ratio, and the square root of the product will be the diameter required. 28. The diameter of a circle is 24 feet. What is the diameter of a circle one-fourth as large? Ans. 12 feet. 29. In the right-angled triangle ABC, the side AC is 9 feet, and the side BC 12 feet. What is the length of the side AB? 20* 234 EXTRACTION OF THE SQUARE ROOT. In every right-angled triangle, B the square of the hypotenuse is equal to the sum of the squares of the base and perpendicular; therefore, the square root of the CD sum of the squares of the base,. / and perpendicular, will be the / a hypotenuse, and the square root / Fig. 3. of the difference of the square of the hypotenuse, and either of the other sides, will be the remaining side. A Base. 0 AC2= 92= 81 BC2 =122=144 AB2 - 225 AB =-/225=15 feet, Ans. 30. What is the distance between the opposite corners of a room, 20 feet in length and 15 in width? Ans. 25 feet. 31. If the distance between the opposite corners of a room be 25 feet, and the width of the room be 15 feet, what is the length? Ans. 20 feet. 32. If a room be 20 feet in length, and 25 feet betveen the opposite corners, what is the width? Ans. 15 feet. 33. Two men owning a pasture 32 rods in-width, and 50 rods between the opposite corners, agreed to divide said pasture into two equal parts by a wall running through it lengthwise. Suppose they pay 50 cents a rod for building the wall, what does it cost them? Ans. $19.209. 34. Suppose a ladder 50 feet long, to be so placed as to reach a window 30 feet from the ground on one side of the street, and without moving it at the foot, will reach a window 20 feet high, on the other side; what is the width of the street? Ans. 85.825+ feet. 35. Two men travel from the same place-one due east, the other due north. One travels 40 miles the first day, the other 30. What is the nearest distance between them at night? Ans. 50 miles. 36. A. and B. set out together, and travel in the same direction on parallel courses, which are 20 miles apart. A. travels 45 miles, and B. 25. What is the distance between them at night? Ans. 28+ miles. EXTRACTION OF THE CUBE ROOT. 235 37. Suppose a pine-tree to stand 25 feet from the end of a house 40 feet in length, the foot of the tree being on a level with the foundation of the chimney, which stands in the centre of the house, and a line reaching from the foot of the tree to the top of the chimney, be 75 feet, what is the height of the chimney? and if the height of the tree be - of -, of - of 14 of the height of the chimney, what will be the length of a line reaching from the top of the chimney to the top of the tree? Ans. 60 feet, height of the chimney. 75 feet; length of the line. Art. 229.-To find a mean proportional between two num bers. RULE. Multiply the given numbers together, and the square root of their product is the mean proportion sought. 1. What is the mean proportional between 3 and 12? Operation. 3 x 1236, and x/36=6 Ans. It is evident, that the ratio of 3 to 6 is the same as the ratio of 6 to 12; for 3=1, and 2 —=. 2. What is the mean proportional between 12 and 48? Ans. 24 3. What is the mean proportional between 9 and 81? Ans. 27. 4. What is the mean proportional between 25 and 625? Ans. 125. EXTRACTION OF THE CUBE ROOT. Formation of the Cube, and Extraction of the Cube Root. Art. 230.-The third powver, or cube of any number, is the product of that number multiplied into its square; and the cube root is a number which, multiplied into its square, will produce the given number. Roots and powers are correlative terms; that is, if 3 is the cube root of 27, then 27 is the third, power, or cube, of 3. 236 EXTRACTION OF THE CUBE ROOT. There are but nine perfect cubes among numbers expressed by one, two, or three figures; each of the other numbers has for its cube root a whole number, plus a fraction. Thus 64 is the cube of 4, and 27 is the cube of 3; therefore, the cube root of each number between 27 and 64 must be 3 plus a frac tion. What is the cube of 24? tens. units. 24=2+ 4 2+ 4 8+16 4+ 8 4+16+16 2+ 4 16+64+ 64 8+32+32 8+48+96+64=13824 It will be perceived, from the above process, that the cube of a number composed of tens and units, is made up of four parts, viz: 1. The cube of the tens, (8 thousands.) 2. Three times the product of the square of the tens into the units, (48 hundreds.) 3. Three times the product of the tens into the square of the units, (96 tens.) 4. The cube of the units, (64 units.) To extract the cube root is to find a number which, multiplied into its square, will produce the given number. What is the cube root of 13824? Operation. As this number is greater than'*,i(24 *1000, which is the cube of 10, but 1824(4 less than 1,000,000, its root will 8 consist of two figures, tens and 22 X 3=12)58124 units; but the cube of tens cannot be less than thousands; therefore, the three figures, 824, on the right, cannot form a part of it. Hence we separate these fiom 13 by a point, and look for the cube of tens in 13, the left-hand period. The root of the greatest cube contained in 13 is 2, -hich is the tens in the required root; for the cube of 20, which is 8000, is less, and the cube of 30, which is 27000, is greater than the given number; EXTRACTION OF THE CUBE ROOT. 237 therefore, the required root is composed of 2 tens, plus a certain number of units less than ten. We now subtract 8, the cube of the tens, from 13, and bring down the next period, 824. We have now 5824, which contains the three remaining parts of the cube, viz: Three times the product of the square of the tens into the units, plus three times the product of the tens into the square of the units, plus the cube of the units. Now, as the square of tens gives hundreds, it follows, that three times the square of the tens into units must be contained in 58, which we separate from 24 by a line. If we now divide 58 by three times the square of the tens, we shall obtain the units of the required root. - We may ascertain whether the unit figure be right, by cubing the quotient, or by applying the following principle: The diference between the cubes of two consecutive numbers is equal to three times the square of the least number, plus three times this number, plus 1. Thus, the difference between the cube of 3 and the cube of 4, is equal to 9x3 +3x3 1-= 37, which is the difference between the cube of 3 and the cube of 4. Therefore, had we written 3 in the unit's place, the remainder would have been equal to 3 times the square of 23, plus three times 23, plus 1, which would show that the unit figure must be increased. Thus far the illustration has been general, —applied to numbers merely-numbers in the abstract. We may now apply it to solid bodies. Numbers which represent, or stand for things, are called concrete, as question first below. EXAMPLES. Art. 231 —1. What is the length of one side of a solid block containing 13824 solid inches, or what is the cube root of 13824? OBs.-The foregoing operation can be better understood by blocks prepared for the purpose. It is necessary to have one cubical block, of a convenient size, to represent the greatest cube in the left-hand period, and three other blocks, equal to the sides of the first block, but of indefinite thickness, to represent the additions upon the sides. Then three other blocks, equal in length to the sides of the cube, and their other dimensions equal to the thickness of the additions on the sides of the cube. Lastly: a small cubic block, of dimensions equal to the thickness of the additions, to fill the deficiency at the corner. By placing these blocks as above described, the several steps in the operation may be easily understood It may be observed, however, that this illustration would serve only for concrete numbers, as in the above question. 238 EXTRACTION OF THE CUBE ROOT. Having distinguished the given number into periods of three figures each, denoted by the index of the root, we perceive, by the number of periods, that the root will consist of two figures. As the cube of ten cannot be less than a thousand, 10 x 10 X 10=1000, we look for the cube of tens in the second, or lefthand period. We find, Operation. by trial, the greatest cube in 13, or 13000, to be 8, 13824(24 root. or 8000, and its root, 2 23=2X2X2=8 or 2 tens, (the length of 22X 300+60=1260) 5824 one side of the cube, Fig. 1200 x 4-=4800 4,) which we place in the 604 x4= 960 quotient, as the first fig4 4x4- 964 ure of the root, and its cube,20 x 20 x 20 = 8000, 5824 under that period; and, subtracting it, we have a remainder of 5, or 5000-to which we bring down the next period. Had the cube contained but 8000 solid inches, we should now Fig. 4. have found its root, or the length o0 of one side. But we have 5824 inches to be added to the cube, and in such a manner that its cubic form shall not be altered. It is obvious, that an equal addition must be made on three. sides. As each side is 20 inches square, we lhave 20X20 X3= 1200; or, which is the same thing, 20 multiply the square of the quotient by 300. 2 X 2 X300= 1200 inches surface, to which the additions are to be made. It will be seen (Fig. 5) that there Fig. 5. are three deficiencies along the 20 sides, a a a, where the additions a meet, 20 inches in length, 20 X 3 =60, or multiply the quotient by 30; 2 X 30=60. We have,then, 1200+60=1260, which may be considered the points where. the additions are to be made. Then 5824- 1260-=4 inches, the thlickness of the addition, or the I a 20 EXTRACTION OF THE CUBE ROOT. 239 second figure of the root. The area of the sides multiplied by the thickness, 1200 X 4-=4800 inches, the amount of the addition upon the sides. Then the Fig. 6. number of inches necessary to fill 20 4 the deficiencies where the additions on the sides meet, is 60 x 4 x 4=960 inches. Still there is.. ii a deficiency of a small cube in the corner, (Fig. 6,) whose dimensions are equal to the thick- + ness of the additions: 4 x 4 X 4= 64 inches. This supplied, and the cube is completed. (Fig. 7.) The sum of all the additions will be a subtrahend equal to the dividend; 4800~] 960+064=-5824. We have now found the length of one side of the cube to be 24 inches. Proof by Involution: 24x 24 X 24- 13824. Fig. 7. 24 Irt, 232.-Hence it appears, that a cube is a solid body, having six equal sides, and its cube Iiiiii root is the length of one of those sides. From the foregoing example and illustration we derive the following 24 RUL E. I. Distinguish the given number into periods of three figures each, beqinning at the right hand. II. Find the greatest cube in the left-hand period, and place its root as a quotient in division. III. Subtract the cube from said period, and to the remainder brin/q down the next period, for a dividend. IV. Multiply the square of the quotient by 300, calling it the triple square, and the quotient by 30, calling it the triple quotient, and' the sum of these call the divisor. OBs.-The triple quotient is not indispensable in forming the divisor. V. Seekf:how many times the divisor is contained in the dividend, ald place the result in the quotient, for the second figure of the root. 240 EXTRACTION OF THE CUBE ROOT. VI. fMultiply the triple square by the last quotient figure, and write the product under the dividend; multiply the triple quotient by the square of the last quotient figure, and place this product under the last; under these write the cube of the last quotient figure, and call their sum the subtrahend. ubtbtract the subtrahendfrom the dividend, and to the remainder bring down the next period, for a new dividend, and proceed as before, till the work is finished. EXAMPLES. 2. What is the cube root of 1906624? Operation. 1X1X300=-300 1906624(124 Ans. x 30= 30 1 Divisor, 330 906 dividend. 300 X 2 -600 30 X 22 120 23- 8 728 subtrahend. 122 X 300+12 X 30=43560)178624 43200 X4 =-172800 360x42= 5760 43= 64 Subtrahend, 178624 3. What is the cube root of 941192? Ans. 98. 4. What is the cube root of 6331625? Ans. 185. 5. What is the cube root of 11543176000? tAns. 2260. 6. What is the cube root of 34.328125? Ans. 3.25. 7. What is the cube root of.000729? Ans..09. 8. What is the cube root of.003375? Ans..15. 9. What is the cube root of 5? of 3? 10. What is the cube root of 8-? Ans. 2 11. What is the cube root of 2-1? lAs.. 12. What is the cube root of 1-728? Ans. 12. 219 7 1 3' 13. What is the cube root of -7? Ans. 1. 14. A certain hill contains 11.543176 cubical feet. What is the length of one side of a cubical mound, containing an equal number of feet' Ans. 226 feet. EaXTRACTION OF TlHE' CUBE ROOT. 241 15. The contents of an oblong cellar is 9201 cubical feet. What is the length of one side of a cubical cellar, of the same capacity? A^. 21 feet. 16. A mer:hait bought cloth to the amount of 8393.04, but forgets the number of pieces, and also the number of yards in each piece, and what the cloth cost per yard; but remembers that he paid as many cents per yard as there were yards in each piece, and that there were as many in each piece as there were pieces. What did he pay per yard? Ans. 34 cents. 17. What is the width of a cubical vessel, containing 75 wine gallons, each 231 cubic inches? 18. Required the side of a cubic box that shall contain a bushel? Ans. 12.9 +inclhes. Art. 233.-Solids of the same form are to one another as the cubes of their similar sides, or diameters. EXAMPLES. 1. If a bullet, weighing 72 lbs., be 8 inches in diameter, wihat is the diameter of a bullet weighing 9 lbs.? Ais. 4 inches. Statement. 8'=512 2: 9::512: 643. Or thus:! 1 0$ 6n 3 43 14 Ans. 2. A bullet, 2 inches in diameter, weighs 4 lbs. What is the weight of a bullet 5 inches in diameter? Ans. 62- lbs. 3. If a silver ball, 9 inches in diameter, be worth $1400, what is the worth of another ball, 12 inches in diamlleter?:A i is. e>t4 as. 4 8.:, Ans..948.148+. Art, 231,-To find two mean proportionals between two numbers. RULE..Divide the greater by the less, (and extract the cube root of the quotient: multiply the lesser number by this root, and the product will be the lesser mean; mzzt,7liply tlis mlzCan b'y the same root, and the product will be the greater mean.z. t^\ 242 EXTRACTION OF ROOTS IN GENERAL. EXAMPLES. 1. What are the two mean proportionals between 4 and 256? 256 4=64; thenV/64=4, and 4x4=16, the lesser, and 16 x4=664, the greater. Proof, 4: 16:: 64: 256. 2. What are the two mean proportionals between 5 and 625? Ans. 25 and 125. 3. What are the two mean proportionals between 7 and 2401? Ans. 49 and 343. EXTRACTION OF ROOTS IN GENERAL. RULE. Art. 235.-I. Point the given number into periods of as many figures as the index of the root directs. Thus, for the square root, two figures; cube root, three; fourth root, four, etc. II. Find, by trial, the greatest root in the left-hand period, and subtract its power from that period, and to the remainder bring down the first figure of the next period, for a dividend.!. Involve the root, already found, to the next inferior power to that wuhich is given, and multiply it by the number denoting the given power, for a divisor, by which find the second figure of the root. IV. Involve the whole root now found to the given power; subtract it from the given num.ber, as before, and bring down the first figure of the next period to the remainder, for a new dividend, and proceed as before, till the work is finished. OBs.-Tlie roots of most of the powers may be found by repeated extractions of the square and cube root —hus: For the 4th root, take the square root of the square root. For the 6th " take the square root of the cube root. QUESTIONS. —1. Rule for finding a mean proportional between two numbers? 2. What is a cube? 3. What is a cube root? 4. What is it to extract the cube root? 5. What is the rule? 6. Why do you distinguish the given number into periods of three figures each? 7. Why do you multiply the square of the quotient by 300? 8. Why the quotient by 30? 9. Vhy the triple square by the last quotient figure? 10. Why the triple quotient by the square of the last quotient figure? 11. Explain the process of illustrating this rule by blocks. 12. What proportion have solids to one another? 13. Rule for finding two mean proportionala between two numbers? 14. Rule for extracting roots in general? ARITHMETICAL PROGRESSION. 243 For the 8th root, take the square root of the 4th root. For the 9th " take the cube root of the cube root. For the 12th " take the cube root of the 4th root. EXAMPLES. 1. What is the square root of 7569? Operation.'7569(87 8x 8= 64 =square, or 2d power, of the quotient. 8x 2=16, 16)116=dividend. 87 x 87= 7569=square of the quotient. 0000 2. What is the fifth root of 4084101? Operation. 4084101(21 2x2x2x2x2= 32 = 5th power of the quotient. 2 x2x2x2X5=80)88 = 1st dividend. 4084101 21 x 21 X 21 21 x 21=4084101=5th power of the quotient. 3. What is the fourth root of 140283207936? Ans. 612. 4. What is the seventh root of 4586471424? Ans. 24. 5. What is the ninth root of 1352605460594688? Ans. 48. ARITHMETICAL PROGRESSION. Art. 236.-ARITHMETICAL PROGRESSION is when a series of numbers increases by a common excess, or decreases by a common difference. When numbers increase by a common excess, they form thb ascending series, as 2, 4, 6, 8, 10, 12, etc. QUESTION.-1. What is Arithmetical Progression? 244 ARITHI1METICAL PROGRESSION. When mnmbers decrease byT a common difference, they form the descending seiies, as 12, 10, 8, 6, 4, 2, etc. The numbers forming the series are called the terms; the first and last terms are called the extremes, and the other terms the m~eans. When any even number of terms differs by Arithmetical Progression, the sum of the two extremes will equal the sum of any two means equally distant from the extremes; as 2, 4, 6, 8, 10, 12. The two extremes, 2+12=6+8, the two means. When the number of terms is odd, the double of the mean will equal the sum of the two extremes, or the sum of any two numlbers equally distant from the extremes; as 1, 2, 3, 4, 5. The double of the mean 3 x2 5- -1= 6. In Arithmetical Progression, five things are to be considered, viz.: the first and last terms, the number, common difference, and sum of all the terms; any three of which being given, the other two may be found. 1. If I buy 4 books, giving 2 cents for the first, 4 for the second, and so on, with a common difference of 2, what do I pay for the last book? It is evident, that if we add 2 cents, the common difference, to the price of the first book, we shall have the price of the second, and so on to the last; thus, 2+2 —=4, 4+2==6, 6+2 -8 cents, the answer. It will be seen that 2, the common difference, is added to every term but the last. If, then, we multiply the numnber of terms, less 1, by the common difference, we have the difference between the cost of the first book and the last; thus, 3x2=6, and 6+2=8, as before. Therefore, Art. 237,-When the first term, the number of terms, and common difference are given, to find the last term: RULE. Mfultiply the number of terms, less 1, by the common difference, and to the product add the first term, and the sum will be the last term. 2. If the first term of a series be 5, the number of terms 35, and the common difference 3, what is the last term? Ans. 35-1 x 3 =102 +5-107. QUEST1ONS. —2. When is the series ascending? 3. When descending? 4. What ia meant by the terms? 5. What is meant by the extremes? 6. By the means? ARITHMETICAL PROGRESSION. 245 3. If I buy 80 yards of cloth, giving 6 cents for the first, 10 for the second, and so on, with a common difference of 4, what do I pay for the last yard? Ans. 322 cents. 4. Suppose a man purchase 40 sheep, paying 3 pence for the first, 10 for the second, and so on, with a common difference of 7, what does he pay for the last sheep? Ans. 276-pence. 5. If 96 acres of land be sold at the rate of 10 cents for the first acre, 19 for the second, and so on, with a common difference of 9, for how much is the last acre sold? Ans. 865 cents. 6. If I buy 4 books, the prices of which are, in Arithmetical Progression, giving 2 cents for the first, and 8 for the last, what is the common difference in the prices of the books? This question is the reverse of question 1st. 8-2=6, 6 -3 =2, the common difference. It is plain, that the difference between the price of the first and last book, is the whole addition made to the price of the first book; and as the addition is made equalolto the three books, it is equally plain that the whole addition, divided by the number of additions, will be the addition made to the price of each book. ThereforeArt. 238.-When the extremes and number of terms are given, to find the common difference, we have this RULE. Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common difference. 7. If the first term of a series be 3, the last term 276, and the number of terms 40, what is the common difference? Ans. 7. 8. A man on a journey travels the first day 2 miles, and increases his travel daily by an equal excess for 15 days, so that the last day he travels 72 miles. What was the daily increase? Ans. 5 miles. 9. Bought books, paying 2 cents for the first, and 8 cents for the last, with a common difference of 2. What number of books did I buy? As the difference of the extremes, divided by the number of the terms less 1, will give the common difference, it is evident that the difference, of the extremes divided by the common 21* 246 ARITHMETICAL PROGRESSION. difference, will give the number of the terms less 1. Then, 8-2=6, the difference of the extremes, and 6 -2=3, which is one less than the number of terms; then 3 +1 = 4, the number of books purchased. ThereforeArt. 239.-When the first and last terms, and the common difference are Rven, to find the number of termsRULE. Divide the diference of the extremes by the common difference, and the quotient will be 1 less than the number of terms. 10. If the first term of a series be 2, and the last term 72, the common difference 5, what is the number of terms? AAns. 15. 11. A man bought sheep, paying at the rate of 3 pence for the first, and 276 for the last, with a common difference of 7. What number did he buy? Ans. 40. 12. A man has a number of sons, the Denmon difference of whose ages is 4 years; the youngest is 8,7he eldest 40 years old. How many sons has he? Ans. 9. 13. If I buy 4 books, paying 2 cents for the first, and 8 cents for the last, how many cents do I pay in all? If the price of the first book is 2 cents, and the price of the last is 8 cents, it is evident that the average price of the books is half way between 2 cents, the price of the first, and 8 cents, the price of the last book: 2+-8 -2= 5. Then 5, the average price, multiplied by the number of books, will give the whole cost: 5x4=20 cents. The same may also be shown by writing the double series, thus: 2 +4+6+8 8+6 1-4+2 10 10 10 10 It will be seen. by this formula, that the sum of any two corresponding terms in the double series is equal to the sum of the two extremes in the simple series; if, therefore, we multiply the sum of the extremes by the number of terms, we shall obtain a sum twice too large. ThereforeArt. 210,-When the first and last terms, and the number of terms are given, to find the sum of the series GEOMETRICAL PROGRESSION. 247 RULE. Multiply half the sum2 of the extremes by the number of terms. The product will be the sum of the series. 14. A man has 9 sons; the youngest ix 8, the eldest 40 years old. What is the sum of their ages? Ans. 216 years. 15. How many times will a clock strike in a day, if constructed like the clocks of Venice, to run till 24 o'clock? Ans. 300. 16. If a triangular piece of land, 60 rods in length, be 1 rod wide at one end, and 60 at the other, what number of square rods does it contain? Ans. 1830. GEOMETRICAL PROGRESSION. Art. 211.-A GEOMETRICAL PROGRESSION is a series of terms, which increase by a uniform multiplier, or decrease by a uniform divisor; as 3, 6, 12, 24, etc., increasing by a uniform multiplier, 2; or 54, 18, 6, 2, *-, etc., decreasing by a uniform divisor, 3. Th'e multiplier, or divisor, which produces the series, is called the ratio. 1. A man bought 5 yards of cloth, paying 3 cents for the first yard, 6 for the second, and so on, doubling the price to the last. What was the price of the last yard? 3 X 2 X 2 X 2 X 2=48, the cost of the last yard. From the above operation it will be seen that the cost'of the second yard is the product of the ratio multiplied by the cost of the first yard; and that the cost of the third yard, is the product of the second power of the ratio multiplied by the cost of the first yard, or the first term; and finally, that the cost of the fifth yard, or the last term, is the product of the fourth power of the ratio, multiplied by the cost of the first yard. It appears, also, that any term in the series may be found by inQUESTIONS.-1. What is Geometrical Progression? 2. What is an ascending series? 3. What a descending? 4. What is the ratio? 5. When the first term and ratio are given, how do you,find the last term? 6. When the first and last terms, and the ratio are given, how do you find the sum of the sriles'? 248 GEOMETRICAL PROGRESSION. volving the ratio to a power less 1 than the number of terms, and multiplying that power by the first term. OBs.-The process of involving the ratio to a high power, may be shortened by multiplying together those lower powers whose indices added equal the infaex of the power sought. To find the fifth power of 3, we may multiply together the second and third powers, for the index of the second power of 3, and the index of the third power added, equal 5, 3 3 and 32 X 339 X 27=243, the 5th power of 3. Art. 242.-When the first term and ratio are given, to find the last termRULE. Involve the ratio to a power whose index is 1 less than the number of terms, and multiply this power by the first term. The product will be the answer, if the series is increasing; but if it is decreasing, divide the first term by the ratio. 2. If I hire a man for 12 months, and agree to pay him 1 dollar for the first month, 3 for the second, and so in a trip] proportion, what must I pay him for the last month? I 2 3 4 5 6 6 + 5 s 11 3 9 27 81 243 729, 729X243=177147 X 1=$177147 Ani, It will be seen that the sum of the indices of the fifth and sixth powers added, equal 11, which is 1 less than the number of terms; and the fifth and sixth powers multiplied together, equal the 11th power of the ratio, which multiplied by the first term, gives the answer, or the last term. 3. A man bought 20 cows, paying 2 farthings for the first, 10 for the second, and so on, in a five-fold ratio. What was the price of the last cow? Ans. ~39736429850 5s. 2d. 2qrs. 4. A man bought 5 yards of cloth, giving 2 cents for the first, and 32 for the last; the prices forming a geometrical series, the ratio of which was 2. What was the whole cost of the cloth? The price of the cloth would be the sum of the following numbers: 2+4+8+16+32=62, the whole cost. It will be seen, that the whole cost is the same as the difference between the two extremes divided by the ratio less 1 added to the greater extreme: Thus, 32-2 = 30, and 30 1 = 30, and 30+32=62. GEOMETRICAL PROGRESSION. 249 Again, if any Lerm of a corresponding series be multiplied by the ratio, the product will be the succeeding term. We will now form a new series, and write it one step farther to the right of that from which it is formed; if we now subtract the first series from the second, we find that all the terms but the first in the first series and the last in the second, disappear, thus: 2 4 8 16 32 4 8 16 32 64 -2 64=64- 2=62 OBS.-If the ratio were 3 we should have double the first series, if it were 4 we should have triple; hence we divide by the ratio less one. Art. 243,-Hence, when the first term, last term, and ratio, are given, to find the sum of the series, we have the following RULES. 1. Multiply tlhe last term by the ratio, and from the product subtract the first term, and divide the remainder by the ratio less 1; the quotient will be the answer. II. Divide the diference between the two extremes by the ratio less 1, and add the quotient to the greater term; their sum will be the answer. 5. The extremes of a Geometrical Progression are 3 and 18673, and the ratio 11. What is the sum of the series? Operation 1st. 18673 X 11-3' 10=20540 Ans. Operation 2d. 18673-3 +118673-=20540 Ans. 11-1 6. If I discharge a debt by paying 1 dollar the first month, 4 the second, and so on, in a four-fold ratio, the last payment being 65536 dollars; what was the whole debt? Ans. $87381. 7. The first term in a geometrical series is 2, the nmnber of terms 10, and the ratio 3. What is the sum of the series? Ans. 59048. 250 INTEREST BY PROGRESSION OBS.-The last term may be found by rule first, or the two processes of finding the last term and the sum of the series may be reduced to one, thus: Art. 244.-When the first term, the number of terms, and the ratio are given, to find the sum of the seriesRULE. Involve the ratio to a power whose index is equal to the number of terms,from which subtract 1; divide the remainder by the ratio less 1, and the quotient, multiplied by the first term, will be the answer. 8. A man sold 15 yards of cloth; the first yard for 1 shilling, the second for 2, the third for 4, and so on, doubling the price of each succeeding yard. For how much did he sell the whole? Operation. 215=32768, and 32768-1x 1 =32767s. 2-1 210)327617 ~1638- s. Ans. 9. A man bought 20 yards of cloth, agreeing to pay 3 pence for the first yard, 9 pence for the second, and so on in a triple proportion to the last. What did le pay for the whole? Ans. ~21792402 10s. 10. A gentleman bought a horse, agreeing to pay what his shoes would amount to, at 1 cent for the first nail, 2 for the second, 4 for the third, and so on, doubling the price of each succeeding nail to the last. The number of nails was 32; what was the price of the horse? Ans. $42949672.95. 11. A laborer wrought 20 days, and received for the first day's labor 4 grains of rye, for the second 12, for the third 36, &c. How much did his wages amount to, allowing 7680 grains to make a pint, and the whole to be disposed of at $1 per bushel? Ans. $14187+. COMPOUND INTEREST BY PROGRESSION. At. 245. —1. What is the amount of $6, for four years, at 6 per cent., compound interest? INTEREST BY PROGRESSION. 251 Operation. $6 1st term. 1.06 636 2d. 1.06 3816 636 6.7416 3d. 1.06 404496 67416 7.146096 4th. 1.06 42876576 7146096 7.57486176 5th. It will be seen, that this question may be solved by the rule after Example 1st in Progression. The principal is the first term, the amount of 81 for one year the ratio, and the number of years, 1 less than the number of terms. The question may be thus stated: —If the first term be 6, the number of terms 5, and the ratio 1.06, what is the last term? 1.06= 1.262 x 6 =7.572 dollars. The amount of ~1, or $1, at 5 or 6 per cent., may be found by the table for Compound Interest, (see Art. 211.) 2..What is the amount of $30, for 7 years, at 5 per cent., compound interest? Ans. $42.213. 3. What is the amount of $7, for 4 years, at 9 per cent., compound interest? Ans. $9.881. 4. If the amount of a certain sum for 6 years, at 6 per cent., compound interest,, be $56.74040, what is that sum, or principal? It will be seen that this question is the reverse of the preceding. If the amount be the product of that power of the ratio denoted by the number of years and the principal, then the amount divided by that power of the ratio will be the principal. 56.74040 1.066 $ 252 ANNUITIES AT COMPOUND INTEIREST. 5. If the amount of $49, for 0 years, compound interest, be $56.74040, vwhat is the rate per cent. 56.74040 404 — 1.4185 1 tile 6th power of the ratio; 40 then, by extracting the 6th root, we have 1.06 for the ratio. Ans. 6 per cent. 6. If the amount of $40, at 6 per cent., compound interest, be $56.74040, what is the time? 56.74040 — 40-=1.41851=1.06 raised to a power whose in3ex 40 is equal to the time; therefore, if we divide 1.41851 by 1.06 until there is no remainder, it is plain that the number of divisions will be the time required; or, having found the power of the ratio, we may look in the table under the given rate per cent., and against the power we shall find the number of years. Ans. 6 years. 7. In what time will $60 amount to 875.74820, at 6 per cent., compound interest? Ans. 4 years. ANNUITIES AT COMPOUND INTEREST. Art. 246,-An annuity is a sum of money payable yearly for a certain number of years, or forever. When annuities are not paid at the time they become due, they are said to be in arrears. The sum of all the annuities remaining unpaid, together with the interest on each, for the time they have remained due, is called the amount. EXAMPLES. 1. What is the amount of an annual pension of $100, which has remained unpaid 5 years, allowing 6 per cent., compound interest? The last year's pension will be $100, without interest, because it is paid as soon as due; the last but one will be $106, the amount of $100 for one year; the last but two, $112.36, thll amount of $100 for two years at compound interest, and so on, forming a geometrical pro'gression. The sum of these amounts will be the sum of the series, or the amount due. ANNUITIES AT COMPOUND INTEREST. 253 Art. 247,-Hence, when the annuity, time, and rate per cent. are given, to find the amount, we have the following RULE. Involve the ratio to a power denoted by the number of years; from this power subtract 1; divide the remainder by the ratio, less 1, and the quotient, multiplied by the annuity, will be th# amount. The above example may be stated thus: If the first term be 100, the number of terms 5, and the ratio 1.06, what is the sum of all the terms? 1- x100=O 563.7. Ans. $563.7. 2. What is the amount of an annuity of $70, to continue 5 years, allowing 6 per cent., compound interest? Ans. $394.59. 3. What is the amount of an annuity of $160, to continue 10 years, at 5 per cent., compound interest? Ans. 2012.448. 4. If a yearly rent of $75 be in arrears 4 years, to what does it amount, at 6 per cent., compound interest? Ans. 328.087. 5. A salary of $600 remains unpaid 5 years. To what does it amount, allowing 6 -per cent., compound interest? Ans. $3382.255. Art. 248. —The annuity, time, and rate being given, to find the present worth. RULE. Find the amount of the annuity in arrears for the whole time; this amount, divided by that power of tiie ratio denoted by the number of years, will give the present worth. 6. What is the present worth of an annual pension of $96, to continue 4 years, allowing 6 per cent., compound interest? Ans. $332.643. 7. What is the present worth of an annual salary of $400, to continue 5 years, allowing 5 per cent., compound interest? Ans. $1731.792. QUESTIONS.-1. What is an annuity? 2. When are annuities said to be in arrears? 3. What is the amount? 4. What is the rule, when the annuity, time, and rate per cent. are given, to fin the amomut? 5. Rule, when the amount, time, and rate are giren, to find the present worth? 22 254 ANNUITIES AT COMPOUND INTERESF. TABLE, Showing the present worth of $1 or ~1 annuity, at 5 and 6per cent., compound interest,for any number of years from I to 40. Years. 5 per cent. 6. Y 5 p c 6 per cent. 1 0.95238 0.94339 21 12.82115 11.76407 2 1.85941 1.83339 22 13.163 12.04158 3 2.72325 2.67301 23 13.48807 12.30338 4 3.54595 3.4651 24 13.79864 12.55035 5 4.32948 4.21236 25 14.09394 12.78335 6 5.07569 4.91732 26 14.37518 13.00316 7 5.78637 5.58238 27 14.64303 13.21053 8 6.46321 6.20979 28 14.89813 13.40616 9 7.10782 6.80169 29 15.14107 13.59072 10 7.72173 7.36008 30 15.37245 13.76483 11 8.30641 7.88687 31 15.59281 13.92908 12 8.86325 8.38384 32 15.80268 14.08398 13 9.39357 8.85268 33 16.00255 14.22917 14 9.89864 9.29498 34 16.1929 14.36613 15 10.37966 9.71225 35 16.37419 14.49825 16 10.83777 10.10589 36 16.54685 14.62098 17 11.27407 10.47726 37 16.71129. 14.73678 18 11.68958 10.8276 38 16.36789 14.84602 19 12.08532 11.15811 39 17.01704 14.94907 20 12.46221 -11.46992 40 17.15909 14.92640 OBs.-To find the present worth of any annuity, at 5 or 6 per cent., by the above table:-First find the present worth of 1 or ~1 annuity; then multiply it by th3 given annuity, and the product will be the present worth. 8. What is the present worth of an annuity of $300, to continue 25 years, at 6 per cent., compound interest? The present worth of $1 annuity, by the table, for 25 years, is 12.'78335. Then, 12.78335 X 300 — 3835.005 Ans. 9. What ready money will purchase an annuity of $250, to continue 40 years, at 6 per cent., compound interest? Ans. $3731.6. Annuities taken in reversion at compound interest. Art. 249 —Annuities taken in reversion are certain sums of money, payable yearly for a limited period, but not to commence until after the expiration of a certain time. QTr-UsrIO.s. —6. What are arpnities Vkeni in reversion? 7. R.'le? ANNUITIES AT COMPOUND INTEREST. 255 RULE. Find the present worth, to commence immediately, and this sum, divided by the power of the ratio denoted by the time in reversion, will give the answer. 10. What is the present worth of a reversion of a lease of $40 per annum, to continue 6 years, but not to commence until the end of three years, allowing 6 per cent. to the purchaser? Present worth,.. 196.692801 Third power of the ratio, 1.19101 Ans. $165.147. The same result may be obtained by finding the present worth of the annuity to commence immediately, and to continue the whole time. Thus, 3+6=9 years, and from the present worth for this time subtract the present worth of the annuity for the time of reversion, 3 years. Or, by the table, find the present worth of $1 for the whole time; from the sum subtract the present worth of $1 for the time of reversion, and multiply the difference by the given annuity. Thus, The whole time,... 6.80169 The time of reversion,... 2.67301 Difference,. 4.12868 40 $165.14720 Ans. 11. What is the present worth of $50, payable yearly for 4 years, but not to commence until 2 years, at 6 per cent.? Ans. $154.1965. 12. What is the present worth of the reversion of a lease of $70 per annum, to continue 20 years, but not to commence until the end of 8 years, allowing 6 per cent. to the purchaser? Ans. $503.7459. 13. What is the present worth of a lease of $200, to continue 30 years, but not to commence until the end of 10 years, allowing 6 per cent.? Ans. $1513.264. Art. 250.-To find the present worth of an annuity to continue forever. RULE. Divide the annuity by the rate per cent., and the quotient will be the present worth. 256 PERMUTATION. 14. What is the present worth of a freehold estate whose yearly rent is $60, allowing 6 per cent. to the purchaser? 60 It is evident that the estate is.06 worth as much money as, at the given rate per cent., would give interest equal to the rent. 15. What is $300 annuity worth, to continue forever, allowing 5 per cent. to the purchaser? Ans. $6000. Art. 251.-To find the present worth of a freehold estate, in reversion, at compound interest. RULE Find the value, as though it were to be entered on immediately, by the foregoing rule, and divide this value by that power of the ratio denoted by the time of reversion; and the quotient will be the present worth of the estate in reversion. 16. Suppose a freehold estate, of $48 per annum, to commence two years hence, be put on sak. What is the value, allowing 6 per cent to the purchaser? 48 800 800 06 1.062 1.136-$711.997 Ans 17. Which is the more valuable, a term of 16 years, in an estate of $100 per annum, or the reversion of such an estate forever after 16 years, computing at the rate of 5 per cent., compound interest? Ans. The term of 16 years, by $167.551 +. -4 — PERMUTATION. Art. 252.-PERMUTATION is the method of finding how many changes may be made upon thie order of any given number of things. 1. How many changes can be made of the first three letters of the alphabet? QUETIONS. —8. Rule for finding the present worth of an annuity to continue forever? 9. Rule bor finding the present worth of a freehold estate in reveraionat compound interest 1. What is permutation of quantities? POSITION. 257 The letter a can occupy but 1 position; a and b can change places, and occupy 2 positions, ab and ba, 1 x 2=2. The three letters, a, b, and c, can, any two of them, leaving out the third, have two positions, 1 x 2=2; consequently, when the third is taken in, there can be 1 x 2 3 = 6 positions, which may be expressed thus: abc, acb, bac, bca, cba, cab. The same may be shown of any number of things. Hence, to find the number of changes which can be made of any given number of different things — RULE. Multiply all the terms of the natural series of numbers, from 1 up to the given number, continually together, and the last product will be the answer required. 2. Christ Church, in Boston, has 8 bells. How many changes can be rung upon them? x2x3X4x5x6x X8=40320 Ans. 3. Six men met at a public house, and agreed to remain so long as they could occupy different situations at the dinnertable. How long did they remain, and what was the price of their board, at 25 cents for each dinner? Ans. j 720 days.. _ $1080 board. POSITION. Art. 253.-POSITION teaches to find the true number by the use of false, or supposed numbers. It is of two kinds, Single and Double. Art. 254.-SINGLE POSITION is so called, because the true number is obtained by the use of one supposed number. 1. A., B., and C. travelled. C. paid a certain part of the expense; B. paid double, and A. treble the sum which C. paid. The amount of their expenses was $60. What did each one pay? Suppose C.'s expense was 8; then, by the conditions of the question, B.'s expense was 8 X 2=$16; and A.'s 8 x 3=$24; QU8rSTIONS.-2. Rule for finding the number of permutations? 3. What is Position? 4. What is Single Position 22* 258' SINGLE POSITION. and the sum of their expenses $8+$16+$24 —$48. As the ratios, in the true and supposed, are the same, it follows, that the true sum of their expenses will have the same ratio to the true expense of each individual, that the sum of their supposed expenses has to the supposed expenses of each individual. Thus: 48: 8:: 60: 10, C.'s expense; 48: 16:: 60: 20, B.'s expense; and 48: 24:: 60: 30, A.'s expense. RULE. Suppose any number, and proceed in the operation as though it were the true; then, as the result of the operation, or sum of the errors, is to the supposed number, so is the given number to the true number required. EXAMPLES. 2. A person, after spending 3 and ~ of his income, had $30 left. What had he at first? Suppose $60 -=30 1-20 60-50=-10 income left: Then 10:60: 30: 180 Ans. Or by fractions: j==, and =2; then +=, the income spent, and remains=$30; then -6-30 X6=$180, as before. 3. A certain sum of money is to be divided between 5 men, in such a manner that A. shall have, B. 1, C. I, D. Il, and E. the remainder, which is $40. What is the sum? Ans. $100. 4. A schoolmaster being asked how many scholars he had, replied, if he had as many more, ~ and X as many more, he would have 11 less than 99. How many had he? Ans. 32. 5. A man bought a horse, chaise, and harness for $216. The horse cost twice as much as the harness, and the harness one third as much as the chaise. What was the cost of the chaise? Ans. $108. 6. What number is that whose A 1,,and make 12'7? Ans. 90. DOUBLE POSITION. 259 7. A man being asked his age, said, If you add to its double g, g, y, and TI of my age, it will be 122. What was his age? Ans. 45. 8. A certain sum of money is to be divided among 4 persons, in such a manner, that the first shall have 4 of l of 2; the second Ar of I~ of 2; the third -j5 of 45; the fourth has $110. What is the sum divided? Ans. $240. 9. A. and B. having found a purse of money, disputed who should have it. A. said that, O, and - of it amounted to $35, and if B. would tell him how much was in it, he should have the whole; otherwise he should have nothing. How much did the purse contain? Ans. $100. DOUBLE POSITION. Art. 255.-DOUBLE POSITION teaches to discover the true, by the use of two supposed numbers. RULE. I. Suppose two numbers, and proceed with each according to the conditions of the question, as in Single Position, noting the error. The difference between the result and the given sum is the error. II. Multiply the first supposition by the second error, and the second supposition by the first error. III. If the errors are alike-that is, both too great or both too small, divide the difference of the products by the difference of the errors. IV. If the errors are unlike-that is, one too large, and the other too small, divide the sum of the products by the sum of the errors. OB. —This rule is founded on the supposition that the first error is to the second as the difference between the true and first supposed number, is to the difference between the true and second supposed number. When this is not the case, the exact number cannot be obtained by this rule. QUxSTIONS.-5. What is Double Position? 6. On what supposition is this rule founded? 7. Verify the principle. 260 DOUBLE POSITION. EXAMPLES. 1. A man being asked what his carriage cost, replied, If it had cost twice as much as it did, and $20 more, it would have cost $370. What was the cost of the carriage? Suppose, first, $120 Having supposed 120, and pro2 ceeded with it according to the con240 ditions of the question, the result ob20 tained is 260; then 37o0-260=110, 260 -20 the first error. Suppose, secondly, $160 Then 370-340=30, the 2 second error. 320 20 340 First sup. 120 110 first error. \/ /\ Second sup. 160 30 second error. 110 120 17600 3600 3600 110-30=80)14000(175 Ans. 80 600 Proof, i':5 x2+20=3T70 560 Verification, 110: 30: 55: 15 400 110 55 11 11 400 ^40 30 3015 3 3 The foregoing question may be thus solved: Let x equal the cost of the carriage; then by the conditions of the question, 2x+20=370. Solution. xz=cost of carriage 2x+20==370 2x=370 —20=350 350 2x=350=; then x=- =175 Ans..., B., and built a house, which cost 228. B. paid 2. A., B., and 0. built a house, which cost $228. B. paid DOUBLE POSITION. 261 $30 more than A., and C paid as much as A. and B. What did each pay? A. paid $42. Ans. B. paid $72. C. paid $114. 3. A. and B. have the same income. A. saves - of his annually, but B., by spending $120 per annum more than A., at the end of 6 years, finds himself 8120 in debt. What is their income, and what does each spend annuilly? T'heir income $400. Ans. A. spends $300, and B. spends $420. 4. A man has two silver cups of unequal weight, having one cover to both, weighing 5 oz. When the cover is put on the less cup, it weighs double the greater; when put upon the greater cup, it weighs three times the less. What is the weight of each cup? A n The less, 3 oz. n The greater, 4 oz. 5. There is a fish whose head is 3 feet long, his tail is as long as his head and half the length of his body, and his body is as long as his head and tail. What is the length of the fish? Ans. 24 feet. 6. A man being asked, in the afternoon, what o'clock it was, answered, that the time passed from noon was equal to - of the time to midnight. Required the time. Ans. 20 minutes past 1 o'clock. 7. A gentleman has two horses, and one carriage which is worth $100. If the first horse be harnessed into the carriage, he and the carriage together will be worth three times as much as the second horse; but if the second be harnessed into the carriage, they will be worth seven times as much as the first horse. What is the value of each horse? Ans. $20 and $40. 8. A laborer was hired 60 days upon this condition, that for every day he wrought he should receive 3s. 4d., and for every day he was idle he should forfeit Is. 8d. At the expiration of the time he received ~3 15s. How many days did he work, and how many days was he idle? Ans. He was employed 35 days, and was idle 25, 262 ALLIGATION MEDIAL ALLIGATION. Art. 260.-ALLIGATION is the method of mixing two or more simples,.of different qualities, so that the composition may be of a mean, or middle quality. When the quantities and prices of the simples are given, to find the mean price of the mixture compounded of them, the process is called ALLIGATION MEDIAL. Art. 261.-1. If I mix 8 lbs. of sugar, worth 10 cents a pound, with 10 lbs., worth 15 cents a pound, what is I lb. of the mixture worth? Eight pounds, at 10 cents a pound, are worth 10 x 8=80 cents, and 10 pounds, at 15 cents, are worth 15 X 10=150 cents; then, 80+150=230 cents, the price of the whole mixture, and 8+10=18 pounds, the whole mixture; then $2.30 18 lbs.=12- cts., the worth of I pound of the mixture. Hence the RULE. Multiply each quantity by its price, and divide the sum of the products by the sum of the quantities. The quotient will be the rate of the compound required. EXAMPLES. 2. A grocer mixes sugar, 5 lbs. at 6 cts., 8 lbs. at 5 cls., and 7 lbs. at 10 cts. a lb. What is 1 lb. of the mixture worth? Ans. 7 cts. 3. A farmer mixes 12 bushels of wheat at $1.75 a bushel, 8 bushels of rye at $1, and 6 bushels of corn at 80 cts. a bushel. What is a bushel of the mixture worth? Ans. $1.30. 4. A goldsmith melted together 12 lbs. of gold, 21 carats fine, 8 lbs. 20 carats fine, 9 lbs. 22 carats fine, and 7 lbs. 18 carats fine. Of what fineness is the mixture? Ans. 204 carats fine. 5. A merchant mixed 8 gallons of wine, at 4s. 2d. pe- gal ALLIGATION ALTERNATE. 263 Ion, with 10 gallons at 6s. ad., and 12 gallons at 8s. 4d. per gallon. What is a gallon of the mixture worth? Ans. 6s. 7d. 6. If 4 lbs. of tea, at 6s. per lb., 8 lbs. at 5s., and 6 lbs. at 3s., be mixed together, what is 1 lb. of the mixture worth? Ans. 45 shillings. ALLIGATION ALTERNATE. Art. 262.-ALLIGATION ALTERNATE is when the prices of the simples to be mixed, and the mean rate, are given, to find what quantity of each is to be taken at a given rate. 1. I have corn at 50 cents a bushel, and oats at 30 cents a bushel, which I would mix, so that the mixture may be worth 40 cents a bushel. What quantity of each must be taken? It is evident that equal quantities of each must be taken, for the price of the corn exceeds the mean rate as much as the price of the oats falls short of it, which is 10 cents in each case. We find, also, that the whole mixture, which is 20 bushels, at the mean rate, 40 cents a bushel, equals the price of 10 bushels of oats at 30, and 10 bushels of corn at 50 cents a bushel. RULE. I. Reduce the rates of all the simples to the same denomination, and write them in a column under each other, and the mean rate on the left hand. II. Connect the rate of each simple, which is less than the rate cc the compound, with one that is greater, and each that is greater with one that is less. III. TWrite the difference between each rate, and that of the compound against the number with which it zs connected. Then, if only one difference stand against any rate, it will express the relative quantity to be taken of that rate; but if more than one, their sum will express that quantity. EXAMPLES. Art. 263.-2. A farmer has wheat at $1.50, rye at $1.00,. corn at 90, and oats at 40 cents a bushel, which he mixes so 264 ALLIGATION ALTERNATE. that the mixture is worth IQ cents a bushel. What quantity of each does he take? Operation. Bushels. Bushels. F 1.50 55" Or 1.50 1 1 9 1100- 5 Ans 95 1'0 1 155 Ans..90 —[ 5.90 — 55.40~ 5 J L.40 — 5j By linking the price of the different simples, as above, their quantities are mutually mixed, and the portion taken of each depends upon the manner of linking them. In the first operation, the price of the wheat, which is greater than the mean price, is linked with the price of the oats, which is less. The price of the wheat is found to be as much greater than the mean rate, as the price of the oats is less; therefore an equal quantity of each is taken. The same is true of the corn and oats. In the second operation, the price of the wheat is linked with the price of the corn. The difference between the price of the wheat and the mean rate, is 55, and the difference between the price of the corn and the mean rate, is 5. Hence, it appears that the less the difference between the price of a simple and the mean rate, the greater will be the quantity taken of that simple; and the greater the difference the less the quantity. 3. A merchant has teas at 72 cents, at 62 cents, and 57 cents a pound, which he would mix, so that the mixture may be worth 67 cents per lb. What quantity of each must be taken? Operation. lbs. ( 72 10+-515 ) 67^ 62- 5=.5 Ans. ( 57 5- 5 The correctness of the above operation may be ascertained thus: The cost of 15 lbs. at 72 cents, is $10.80, and the cost of 5 lbs. at 62 cents, is $3.10, and the cost of 5 lbs. at 57 cents, is $2.85. Then the whole cost is $10.80+$3.10+ $2.85= —16.75, which, divided by 25 lbs., gives the mean price, $16.75-25=-67 cents. Hence, it appeals that Alligation Alternate is the reverse of Alligation Medial, and may be proved by it. ALLIGATION ALTERNATE. 265 4. A grocer mixes wines at 29s., 24s., 22s., and 17s. a gallon, so that the mixture is worth 23s. per gallon. How much of each sort does he take? F I gal. at 29s. 6 gal. at 29s. st Ans. 6 gal at 24s. d A " at 24s. 6 gal. at 22s. I " at 22s. 1 gal. at 17s. 6 " at 17s. r 7 gal. at 29s. 6 " at 24s. 3d Ans. 76 " at 22s. 7 " at 17s. As many different answers may be obtained to questions in ihis rule as there are modes of linking the prices of the simples. Let the number of simples be what it may, and with how many soever each one is linked, since the price of one that is less than the mean rate, is always linked with one that is greater, there will always be an equal balance of loss and gain between the two, and consequently an equal balance on the whole. 5. It is required to mix brandy at 12s., wine at 9s., cider at 2s., beer at ls., and water at Os., per gallon, so that the mixture may be worth 7s. per gallon. What quantity of each must be taken? 13 gals. brandy. 5 " wine. Ans. 2 "' cider. 5 " beer. h 5 " water. Art. 261.-When the composition is limited in quantity. RULE. Find the proportion.of each quantity as before; then say, As the sum of the quantities is to the given quantity, so is each of the differences to the required quantity. EXAMPLES. 6. Suppose a mass of pure gold, a mass of pure silver, and a mass which is a mixture of gold and silver, each weighing 9 oz.; by immersing them in water, it is found that the quantity of water displaced by the gold is 5; by the silver 8, and by 23 260 ALLIGATION ALTERNATE. the mixture 7. What part of the mixture is gold, and what part silver? 8 12 2 2: 6 silver. By a similar problem, Archimedes detected the fraud of the artist employed by Hiero, king of Syracuse, to make him a crown of pure gold. 7. A druggist has medicines at 6d., 3d., 9d., and 4d. per oz., and would form a compound of 15 oz., worth 5d. per oz. How much of each sort must he take? 1-7 oz. at 6d. Ans. " 3d. 33 " 9d. LI1 7 4d. 8. A goldsmith would melt together gold of 13, of 14, of 15, and of 21 carats fine, to form a composition of 35 oz. 18 carats fine. What proportion of each must he take? 5 of 13 5 14 Ans. ) 5, 14 carats fine. " 15 20 " 21 9. How many gallons of water, worth Os. per gal., must be nixed with wine worth 12s. per gal., so as to fill a cask of 20 gallons, and that a gallon of the mixture may be afforded at 9s. per gallon? Ans. 5 gal. water. ( 15 gal. wine. Art. 265.-When one of the simples is limited to a certain quantity. RULE. Find the proportional quantities, or differences, as before; then say, As the difference standing against the given quantity is to the given quantity, so are the other differences severally to the several quantities required. EXAMPLES. 10. A grocer mixes sugar at 9 cts., 12 cts., and 14 cts., with 16 lbs. at 15 cts. How much of each sort must he take, that the, mixture may be worth 13 cts. per lb.? DUODECIMALS. 26' 9 — 2 I14- I L15 ~ 4 against the given quantity. 2 8 lbs. at 9s) 4: 16:: I 4 " 12s. Ans. 1:4 " " 14s. ) 11. A grocer would mix flour, at $6, $5, $12 a barrel, with 10 barrels at $11 a barrel. How much of each kind must he take, that the mixture may be worth $10 a barrel? (4 at $16 Ans.. 2 at $5 per barrel. (8 at $12) 12. How much water must be mixed with 100 gallons of brandy, worth 7s. 6d. per gallon, to reduce it to 6s. 3d. per gallon? Ans. 20 gallons. 13. A farmer would mix barley at 50 cents, oats at 30, with 20 bushels of rye at 60 cents a bushel. How much of each sort must he take, that the provender may be worth 40 cents per bushel? Ans. 60 bushels of oats, and 20 bushels of barley. DUODECIMALS. Art. 266 — THIS rule is principally used in measuring surfaceg and solids. Calculations are generally made in feet, inches, or primes, seconds, thirds, fourths, and so on. These subdivisions of the foot are made by a common divisor, 12, an inch being I- of a foot, a second 1J of an inch, or T-~ of a foot, thus forming a descending series of a geometrical progression, whose first term is 1, and the ratio 12. Hence the term duodecimal. It is derived from the Latin word duodecim, which signifies twelve. Duodecimals, then, are fractions of a foot, as may be seen by the following: QueBTIOKN.-1. What are Duodecimals? 2, For what is the rule chiefly used. What are the divisions of the foot? 268 DUODECIMALS. TABLE I. 1' inch, or prime, is................ -2 of a foot. 1" second is A of.............. 1"' third is A of A- of......... 1"" fourth is A of A of I of 1 1 " The marks', ", "', "", placed over numbers, denote the denomination, and are called indices. In Multiplication of l)uodecimals, the denomination of the product is denoted by the sum of the indices. That is, if inches be multiplied by inches, the product will be seconds; thus, 2'X2'-4". If inches be multiplied by seconds, the product will be thirds; thus, 2' X 2=4"', etc. TABLE II. 12"" fourths make..........1"' third. 12"' thirds.............1" second. 12" seconds..............1' inch, or prime. 12' inches, or primes...... 1 foot. MULTIPLICATION OF DUODECIMALS. Art. 267.-1. How many square feet in a board, 11 feet 2 inches long, and 1 foot 4 inches wide>? Orai. Having written numbers of the same 1 2era tion. denomination under each other, as in 1 4' Multiplication of Compound Numbers, 4______ we commence with the feet in the mul11 2' tiplier, and say, once 2 is -2-, and 11 3 8' 8" feet multiplied by 1 is 11. Proceeding 14 10' 8" Ans. to the next figure in the multiplier, which is 4 inches, or -h-, we say, 4 times 2are 8, but 4 is 1, and 2 is 2; therefore, 4-2 = of a foot, or 8" seconds, which being less than 1 prime, we write it in the place of seconds, and proceed to the next figure in the multiplFcand, which is 11. Multiplying 11 feet by xT, we have 11 x 4_ = _2=3 feet 8' inches. Having written 8 in the place of inches, and 3 in the place of feet, we add the several partial products, and obtain 14 feet l0' 8", the answer. By examining the foregoing operation, it will be seen, that the first DUODECIMALS. 269 product, being the product of inches by feet, is inches. The second product is the product of feet, and consequently is feet. The third product is the product of inches by inches; the sum of the indices being two, it is 12ths of an inch. The fourth and last is the product of feet by inches, and is 12ths of a foot. Therefore, to multiply feet, inches, etc., by numbers of corresponding denominations, we have the following RULE. I. Write the several denominations of the multiplier under the corresponding denominations of the multiplicand. II. MAultiply first the lowest denomination in the multiplicand by the highest in the multiplier, observing to carry 1 for every 12 from a lower to a higher denomination. It is to be remembered that the denomination of the product of two numbers is denoted by the sum of the indices. EXAMPLES. Art. 268.-2. How many square feet in a marble slab, 5 feet 7 inches in length, and 4 feet 8 inches in breadth? Operation. 5 7' 4 8' 22 4' 3 8' 8" ft. 26 0' 8" Ans. Duodecimals may also be written as decimal fractions, observing to carry for 12 instead of 10. Thus, ft. 5 7'=5.7 4 8'=4.8 388 224 26.08=26 ft. 0' 8" 3. How many square feet in a room 15 feet 8 inches in length, and 14 feet 9 inches in breadth? Ans. 231 ft. 1 in. 4. What is the product of 15 ft. 2' 6" x 20 ft. 3' 7"? 5. How many solid feet in a block 4 ft. 8' long, 3 ft. 6' wide, and 2 ft. 9' thick? Ans. 44 ft. 11'. 23* 270 DUODECIMALS. OBS. 1.-The solid contents may be found by multiplying the length by the breadth, and that product by the thickness. 6. How much wood in a load 9 ft. 8' long, 8 ft. 7' wide, and 3 ft. high? Ans. 1 cord, 120 ft. II'. 7. How many square feet in a stock of 20 boards, 13 ft. 11' long, and 1 ft. 7' wide? Ans. 440 ft. 8' 4". 8. How many feet of flooring in a room 30 feet 6 inches long, and 19 feet 5 inches in width? Ans. 592 ft. 2' 6". 9. How much wood in a cubic pile, 12 feet 3 inches on each side, and what will it be worth at $4 25 per cord? Ans. to the last, $61.035. 10. How many square yards in the walls of a room, 14 feet 8 inches long, 11 ft. 6 inches wide, and 7 ft. 11 inches high? Ans. 46 yds. 0 ft. 0' 4" 10"' 8"". 11. How many cord-feet in a pile of wood 38 feet long, 7 feet 2' wide, and 4 feet 7' in height? Ans. 78 cord-feet, 0' 1" 9". 12. How many cord-feet in a load of wood 8 feet long, 3 feet 6 inches high, and 4 feet 5 inches wide? Ans. 7 cord-feet, 11 solid feet 8 in. 13. How much wood in a load 9 feet long, 3 feet 4 inches wide, and 2 feet 6 inches high? Ans. 4 cord:feet, 11 solid feet. OBs. 2.-Many mechanics and surveyors take dimensions in feet and decimal parts. This method is preferable, inasmuch as by it the calculations of the artificer are rendered more simple and easy. For such, it is convenient to have a rule, or scale, four feet long, divided into feet, and each foot into ten equal parts. One foot, on one end of the rule, should be divided into one hundred equal parts. The former division will be 10ths, and the latter 100ths of a foot. Dimensions taken. by this rule are calculated the same as other decimal fractions. 14. How many square feet in a board 20.5 in length, and 1.8 in width? Ans. 36-. 15. How much wood in a pile 40.5 in length, 5.4 in width, and 6.2 in height; and what will it be worth at $3.75 a cord? Ans. to the last, $39.724+. 16. What will be the cost of a marble slab, 13.9 feet in length, and 2.1 feet in width, at $1.18 per square foot? Ans. $34.444. MISCELLANEOUS RULES. MENSURATION. MENSURATION OF SURFACES. DEFINITIONS. 1. A point is a small dot; or, mathematically considered, is that which has no parts, being of itself indivisible. 2. A line has length, but no breadth. 3. A superficies, or surface, called also area, has length and breadth, but no thickness. 4. A solid has length, breadth, and thickness. 5. -A right line is the shortest that can be drawn between two points: as AB. 6. The inclination of two lines meeting one another, or the opening ~ between them, is called an angle: A as ABC; B the angular point. 7. If a right line fall upon another righ. line, so as to incline to neither side, but make the angles on each side equal, then those angles are called right angles, and the line is said to be perpendicular to the other line: as ABC, right angle. A 8. An obtuse angle is greater than L a right angle: as LBG. 9. An acute angle is less than a right angle: as ABL. A * A ~ B ~ " 272 MENSURATION OF SURFACES. 10. A circle is a round figure bounded by a single line, in every part equally distant from some A point, which is called the centre. 11. The circumference or peri- G phery of a circle, is the bounding line. 12. The radius of a circle (AO) _ is a line drawn from the centre to L the circumference. Therefore, all radii of the same circle are equal. B 13. The diameter of a circle (AC) is a right line drawn from one side of the circumference to the other, passing through the centre; and it divides the circle into two equal parts called semicircles. 14. The circumference of every circle is supposed to be divided into 860 equal parts, called degrees; and each degree into 60 equal parts, called minutes; and each minute into 60 equal parts, called seconds; and these into thirds, etc. Hence a semicircle contains 180 degrees, and a quadrant 90 degrees. 15. An arc of a circle (BCD) is any part of the circumference. 16. A chord (BD) is a right line drawn from one end of an arc to another, and is the measure of the arc. The chord of an arc of 60 degrees is equal in length to the radius of the circle of which the arc is a part. 17. The segment of a circle is a part of a circle cut off by a chord. 18. A sector of a circle (LOG) is a space contained between two radii and an arc less than a semicircle. 19. Parallel lines (LO and BE) are such as are equally distant from each other. 90. A triangle is a figure bounded C by three lines. 21. An eqilateral triangle (ABC) has its three sides equal in length to each other. (CE is the perpen- dicular height.) 22. An isosceles triangle has two of its sides equal. A E B 23. A scalene triangle has three unequal sides. 24. A right-angled triangle has one right angle. 25. An obtuse-angled triangle has one obtuse angle. 26. An acute-angled triangle has all its angles acute. DEFINITIONS. 273 2/. Acute and obtuse angled triangles, are called oblique-angled triangles, or simply oblique triangles; in which the lower side is called the base, and the other two, legs. 28. In a right-angled triangle the longest side is called the hypotenuse, and the other two, legs, or base and perpendicular. OB's.-The three angles of every triangle being added together, will amount to 180 degrees; consequently. the two acute angles of a right-angled triangle amount to 90 degrees, the right angle being also 90. 29. The perpendicular height of a triangle is a line drawn from one end of the angles perpendicular to its opposite side. C D 30. A square (ABCD) is a figure bounded by four equal sides, and containing four right angles. A B E F 31. A parallelogram (EFGH) is I a figure bounded by four sides, the i opposite ones being equal, and the i angles right. G iE 3 K 32. A rhombus (JKLM) is a figure bounded by four equal sides, but has its angles oblique. JN perpendicular height of a rhombus. L N E F 33. A rhomboid (EFGH) is a figure bounded by four sides, the opposite ones being equal, but the angles oblique. G ]H 274 MENS'RATION OF SURFACES. 34. The perpendicular height of a rhombus, or rhomboides, is a line drawn from one of the angles to its opposite side. 35. A trapezoid (ABCD) is a C D figure bounded by four sides, two of which are parallel, though of un- / -' \ equal lengths. A-' B 36. A trapeze, or trapezium, is a figure bounded by four unequal sides. 37. A diagonal is a line drawn between two opposite angles. 38. Figures which consist of more than four sides are called polygons; if the sides are equal to each other, they are called regular polygons, and are sometimes named from the number of their sides, as pentagon, or hexagon, a figure of five or six sides, etc. If the sides are unequal, they are called irregular polygons. An irregular plane figure. Irregular plane figure divided into triangles. Heptagon (The dotted lii es represent a division into triangles.) 39. The area of a figure is the space contained between the bounding ilnes of its surface, without regard to thickness. The area is reckoned so many square inches, square feet, square yards, or square rods, etc MENSURATION OF SURFACES. 275 Art. 269.-To find the area of a square, or parallelogram. RULE. Multiply the length by the breadth, or perpendicular height, and the product will be the area. 1. How many square rods in a field 28 rods on each side? 28 X 28784 rods, Ans. 2. What is the area of a square field, one side of which is 25.35 chains? Ans. 642.622 chains. 3. What is the area of a field 30.5 chains in length, and 24.5 in width? Ans. 747.25 chains. 4. How many square feet in a board 18.8 feet long, and 2.7 feet wide? Ans. 50.76. 5. How many acres in a rectangular piece of ground, 64 rods long and 24 rods wide? Ans. 93. Art. 370.-To find the area of a triangle. RULE. LMultipl! the perpendicular by the base, and one-half the product will be the area; or, multiply the base by half the-perpendicular height, and the product will be the area. 1. What is the area of a triangle whose base is 20 feet, and whose height is 18 feet? 18 x 20=360- 2= 180 feet, Ans. 2. What is the area of a triangle whose base is 55 rods, and its height 24.6 rods? Ans. 676.5. 3 How many feet of boards will it take to cover the gable end of a barn, 38 feet wide, the height from the beam to the top being 12.5 feet? Ans. 237.5. When the three sides of a triangle are known, the area may be found by the following RILE. Add together the three sides, and from half their sum subtract each side separately; multiply the half sum and the remnainders together continually, and the square root of the product will be the area. 4. What is the area of a triangle whose three sides are 14, 12, and 8 rods? 276 MENSURATION OF SURFACES. 14+12~+8-34 -2= 17, the half sum: then, 17 17 17 14 12 8 Rein. 3 x 5 X 9x17=2295: then /2295=47.9+ rds. 5. The three sides of a triangle are 6, 8, and 10 chains, What is the area? Ans. 24 chains. Art. 271.-To find the area of a trapezoid. RULE. Multiply half the sum of the two parallel sides by the perpendicular distance between them: the product will be the area. 1. What is the area of a piece of land that is 30 chains long, 20 chains wide at one end, and 18 chains at the other? 2o0 1 8 19, the half sum of the two sides: then 19 X 30=570 chains, Ans. 2. What is the content of a board, 10 feet long, 10 inches wide at one end, and 2 feet ten inches at the other? Ans. 181 feet. 3. What is the area of a hall, 40 feet long, and at one end 30 feet, and at the other 24 feet wide? Ans. 1080 feet. 4. How many acres in a farm 300 rods long, 80 rods wide at one end, and 60 at the other? Ans. 131 acres, 40 rods. Art. 272,-To measure any irregular plane figure. RULE. The whole may be divided into triangles, and measured separately. The sum of the area of the triangles will be the area of the whole. OF THE CIRCLE. The circumference of a circle is found by calculation to be about 3-i times the diameter; or more accurately, by decimals, as I is to 3.1416, or as 113 is to 355, so is the diameter to the circumference. Hence, if the diameter is given, to find the circumference, it may be found by multiplying the diameter by 31, or by 3.1416 or as 113 is to 355, so is the diameter to the circumference. MENSURATION OF SURFACES. 277 1. What is the circumference of a circle whose diameter is 42 feet? 42 X 3+=132 ft.; or 42 3.1416 =131.9472 feet; or 113: 355': 42: 131.946+ feet. By reversing the foregoing, the diameter may be found, the circumference being given. 2. If the circumference of a circle be 132 feet, what is the diameter? 132 —31 =42 feet, Ans. 3. Suppose the diameter of a circular pond to be 121 rods, what is the circumference? Ans. 380.28+ rods. 4. If the circumference of a circular field be 198 rods, what is the diameter? Ans. 63 rods. 5. What is the diameter of a tree, whose circumference is 9-1 feet? Ans. 3 feet. 6. If the circumference of the earth is 25000.8528 miles, what is the diameter? Ans. 7958 miles. 7. The diameter of the earth being 7958 miles, what is the circumference? Ans. 25000.8528. Art, 273.-To find the area of a circle. RULE. Multiply half the diameter by half the circumference; the product will be the areg. 1. What is the area of a circular grove, whose diameter is 147 rods, and circumference 462 rods? Ans. 462 -2X147-2=16978~ rods. 2. What is the area of a circle whose diameter is 28, and the circumference 88 rods? Ans. 616 rods. 3. How many square rods in a circle whose circumference is 63, and the diameter 20 rods? Ans. 315. Art. 274. —The diameter given, to find the area. RULE. Multiply the square of the diameter by.7854, and the proauct will be the area. 1. What is the area of a circle whose diameter is 28 rods? 28X28x.7854=615.7536 rods, Ans. 2, What is the area of a circle whose diameter is 59 rods? Ans. 2733.9774 rods. 24 278 MENSURAfION OF SURFACES. Art. 275.-The circumference given, to fia. the area. RULE. Multiply the square of the circumference by.07958, and the product will be the area. 1. What is the area of a circle whose circumference is 46 rods? 46 x 46 X.07958=-168.39128 rods, Ans. 2. What is the area of a circle whose circumference is 44 rods? Ans. 154.06688 rods. Art. 276.-To find the area of an oval, or ellipsis. RULE. Multiply the longest and shortest diameters together, and the product by.7854. The last product will be the area. 1. What is the area of an oval, whose longest diameter is 7 ft., and the shortest 5 ft.? 7 x5.7854=27489 ft. Ans. 2. What is the area of an oval, whose longest diameter is 15, and the shortest 13 feet? Ans. 153.153 ft. Art. 277 —To find the area of a globe, or sphere. RULE lfultiply the circumference by the diameter. The product will be the area. 1. What is the area of a globe, whose circumference is 44 feet, and the diameter 14? 44 x 14=616 ft. 4Ans. 2. low many square inches in the surface of a btll, 1 inch in diameter? Ans. 3.1416. 3. Hw\ many square miles in the surface of the earth, allowing its circumference to be 25000 and its diameter 8000 miles? Ans. 200000000. Art. 278.-Given the chord of an arc, and its height, to find the diameter of a circle, of which the arc is a part. RULE. Divide the square of half the chord by the height, and the quotient, added to the height, will be the diameter required. MENSURATION OF SURFACES. 279 1. Given the chord, BD, 287, A and the height, CE, 78 feet, to find the diameter, AC. Operation. 287-2=143.5, and 143.52= 20592.25, and 20592.25 -78 = 264, and 264+78=342, the required diameter. u 2. Given the chord 178, and the height 257 yards, to find the diameter. Ans. 287.821 yards. 3. Given the chord 843, height 648 links. Ans. 922.17 links. 4. Given the chord 40, height 12 yards, to find the diameter. Ans. 455 yards. 5. Given the chord 560, height 45 links, to find the diameter. Ans. 1787 links. Art. 279.-Given the radius and number of degrees in an arc of a circle, to find the length of the are. RULE Multiply the radius by the number of degrees in the arc, and by.0174533. Or find the circumference, multiply it by the degrees, and divide by 360~. 1. Required the length of an \ arc, AC, of 57~, in a circle of which the radius, AB, is 38 feet. C D Diameter. Operation..0174533 x 57 X 38=37.8038478 feet, Ans. 2. What is the length of an arc of 19~ 37', the radius being 98 yards? Ans. 0174533 x 19~.617 x 98=33.553 yards. 3. What is the length of an arc of 83~ 24', radius 32 poles? Ans. 1 furlong, 6 poles. 3 yards, 6.72 inches. 4. What is the length of an arc of 150~, radius 19 ells? Ans. 49 ells, 27 inches, 280 MENSURATION OF SURFACES. 5. What is the length of an arc of 17~ 50', radius 178 miles? Ans. 55 miles, 3 furlongs, 8 poles, 4~ yards. Art. 280.-To find the area of a sector of a circle. RULE. I. If the length of the arc be known, multiply half the arc oy the radius, or the arc by half the radius. II. If the angle of the sector be given, find the length of the arc, and proceed as before. 1. What is the area of a sector, of which the arc is 79, and the radius of the circle 47 yards? Ans. 9= —39.5 and 39.5 X47=1856.5 square yards. 2. What is the area of a sector, of which the arc is 17 feet 5 inches, and the radius 22 feet? Ans. 191.583 square feet=21 yards, 2.583 feet. 3. What is the area of a sector, of which the angle is 1270 16', the radius 133 feet? Ans. 1 rood, 32 poles, 4.845 yard,, OBS.-The area of the circle is 55571.63245; and this multiplied by 12714, and divided by 360-19645.60+. 4. What is the area of a sector, of which the angle is 2' degrees, the radius 97 miles? Ans. 2216.95 miles. 5. What is the area of a sector, of which the angle is 1370 20', the radius 456 links? Ans. 2 acres, 1 rood, 38 poles, 21.9 yards. 6. What is the area of a sector, of which the arc is 156 feet, the radius 478 feet? Ans. 3 roods, 16 poles, 28 yards, 6 feet. Art. 281 —To find the area of a ring contained by two concentric circles. OBs.-Concentric means having the same centre. RULE. Multiply the sum of the diameters by their difference, and that product by.7854. MENSURATION OF SURFACES. 281 1. Required the area of the ring ABC —DEF, of which the diameters - "... are 10 and 6, or OC, 5, and OF, 3 /, — \A feet. / \ \ Ans. 10+6=16, and 10-6=4, F p \ then 16x4x.7854=50.2656 feet. \ \ O 2. What is the area of a ring, of \ \,' which the diameters are 72 and 48 ft.? \ x Ans. 2261.952 square ft.=8 rods, " E.'B 9I yards. 3. Required the area of a ring, of which the diameters are 314 and 256 yards. Ans. 5 acres, 1 rood, 18 poles, 10 yards, 7.42 feet. 4. What is the area of a ring, of which the diameters are 246 and 228 inches? Ans. 46 feet, 77 inches. Art, 282.-To find the area of a space bounded on one side by a curve line. RULE. Let perpendiculars be erected upon the base, so numerous that the part of the curve between any two nearest to one another shall di;fer but little from a straiyht line. Then add the perpendiculars at the extremities of the base, if there are any, and to half the sum add the remaining perpendiculars. AMultiply the sum by the base, and-divide the product by the number of parts into which the base is divided by the perpendiculars: the quotient will be the area, nearly. C 1. Suppose the perpendiculars at b C the extremities of the base to be 10 BJ, and 16, and the others to be 11,: ~ 14, 16, and the base to be 20 feet..', Operation. A t r D 2 —+-=13, and 13+11+14+16=54; and 54 X 20-4=270 square feet, the area. 2. A. curve-lined space meets the base at one of its extremi. ties, and the perpendicular at the other extremity is 96; the other perpendiculars are 83, 70, 64, 51, 38, 25, and the base 325 links. What is the area? Ans. 175964 square links. 3. Perpendiculars were raised from the base to a curve; those at the ends were 364 and 578, the others were 396, 418, 453, 512, 554 links, and the base 1260 links. What is the area? Ans. 5 acres, 3 roods, 22 poles, 4 yards, 3.2 feet. 282 MENSURATION OF SOLIDS. 4. A curve meets the base at one extremity; the base is 2364; the perpendicular, at the other extremity, 758, and the others are 642, 587, 524, 432, 417, and 335 links. What is the area? Ans. 1119860- links-11 acres, 31 poles, 23.5 yards. MENSURATION OF SOLIDS. Art. 283.-The MENSURATION OF SOLIDS includes the men. suration of all bodies which have length, breadth, and thickness. DEFINITIONS. 1. Solids are figures, having length, breadth, and thickness. 2. A prism is a solid, whose ends are any plane figures, which are equal and similar, and its sides are parallelograms. OBs.-A prism is called a triangular prism, when its ends are triangles; a square prism, when its ends are squares; a pentagonal prism, when its ends are pentagons; and so on. 3. A cube is a square prism, having six sides, which are all squares. 4. A parallelopiped is a solid, having six rectangular sides, every opposite pair of which are equal, and parallel. 5. A cylinder is a round prism, having circles for its ends. 6. A pyramid is a solid, having any plane figure for a base, and its sides are triangles, whose vertices meet in a point at the top, called the vertex of the pyramid. 7. A cone is a round pyramid, having a circular base. 8. A sphere is a solid, bounded by one continued convex surface, every point of which is equally distant from a point within, called the centre. The sphere may be conceived to be formed by the revolution of a semicircle about its diameter, which remains fixed. A hemisphere is half a sphere. 9. The segment of a pyramid, sphere, or any other solid, is a part cut off the top by a plane, parallel to the base of that figure. 10. A frustum is the part that remains at the bottom after the segment is cut off. 11. The sector of a sphere is composed of a segment less than a hemisphere, and of a cone, having the same base with the segment, and its vertex in the centre of the sphere. 12. The axis of a solid is a line drawn from the middle of one end to the middle of the opposite end; as between the opposite ends of a prism. The axis of a sphere is the same as a diameter, or a line passing through the centre, and terminating at the surface on both sides. 13. The height, or altitude of a solid, is a line drawn from its vertex, or top, perpendicular to its base. MENSURATION OF;LOLIUS. 283 Art. 28. —To find the solidity of a cube. RULE. Multiply the length, breadth, and th;ckness together, and the product will be the area. 1. If the length of one side of a cubical block be 14 inches, what is its solidity? 14 x 14 x 14 =2744 inches, Ans. 2. How many cubical feet in a mound, each side of which is 25.5 feet? Ans. 16581.375 feet. Art. 285.-To find the solidity of a prism, or cylinder. RULE. Find ihe area of the end, and multiply it by the length. The product will be the area. What is the solidity of a prism, the area of whose end is 2.6 feet, and whose length is 16 feet? 2.6 X16=41.6 feet, Ans. Art. 286.-To find the side of the largest stick of timber that can be hewn from a round log The circle, PEON, represents A EB the end of a round stick of timber; ABCD, a circumscribed square, and PEON, an inscribed square. It will be perceived that the square ABCD is double the square P... PEON. But the square ABCD is equal to the square of PO, the diameter of the circle; but PO is equal to Pa+aO=Pa+aE. Now Pa+aE2=PE2, the side of I C the largest inscribed square. Hence the RULE. Extract the square root of double the square of half the diameter at the smallest end of the stick,for the side of the stick when squared. 1. What will be the side of the largest stick of square timber which can be hewn from a round log, 18 inches in diameter at the smallest end? /9 x 9x 2 =12.727 + inches, Ans. 284 GAUGING. 2. The diameter of a log at the smallest end is 24 inches. What will be the side of the largest stick of timber that can be hewn from it? Ans. 16.97 + inches Art. 287. —To find the solidity of a pyramid, (r cone. RULE. Multip7y the area of the base by one third of the he ght, and the product will be the area. 1. What is the contents of a cone, whose height is 21 feet, and the diameter of the base 9.5 feet? 9.5 x 9.5 x.7854 x 218-3=496.176 feet, Ans. 2. How many solid feet in a cone, whose height is 48 feet, and whose diameter at the base is 13 feet? Ans. 2123.7216 feet. Art. 288 —To find the solid contents of a globe, or sphere. RULE. Multiply the cube of the diameter by.5236, or multiply the square of the diameter by one sixth of the circumference. 1. What is the solidity of a ball, 9 inches in diameter? 9x99 x.5236=381.7044, Ans. 2. What is the solidity of a globe, whose diameter is 13 inches? - Ans. 1150.3492 inches. Art. 289.-To find the solid contents of the segment of a sphere, the height and base of the segment being given. RULE. To three times the square of the ratius of the base of the seg ment, add the square of the height, and multiply this sum by the height of the segment, and this product by.5236. How many cubic feet are there in a coal-pit, the diameter of whose base is 103 feet, and whose height is 9 feet? Ans. 37877.0931. G A UG I NG. Art. 290 — GATTGING is the art of measuring all kinis of vessels, such as pipes, hogsheads, barrels, etc. MEASURING GRAIN, ETC. 285 RULE. Add the square of the head diameter to the square of the bing diameter; multiply the sum by the length, and the product by.0014 for ale gallons, or by.0017 for wine gallons. 1. What is the contents of a cask, whose diameters are 18 and 26 inches, and its length 38 inches? 26 x 26- 18 x 18 x 38==38000; then 38000 X.0017=64.6 wine measure. 38000 X.0014=53.2 gallons, beer measure. 2. How many wine gallons will fill a cask 50 inches in length, bung diameter 38, head diameter 30 inches? Ans. 199.24 gallons. MEASURING GRAIN, ETC. Art. 291.-WHEN the grain is heaped in the form of a cone. RULE. Measure the perpendicular height of the heap, and also the slanting height, from the top to the floor, in inches; then multiply the difference of the squares of those two heights by the perpendicular height, and this product. by.0005. The last product will be the contents in bushels. 1. How many bushels in a parcel of wheat heaped in the form of a cone; the perpendicular height being 40 inches, and the slanting height 90 inches? The square of 90=8100 The square of 40= 1600 6500 difference of squares. 40 perpendicular height. 260000.0005 130.0000 bushels, Ans. 2. What number of bushels in a conical heap of rye, the perpendicular height being 35 inches and the slanting height 65 inches? Ans. 52.5 bushels. Art. 292.- When grain is heaped against the side of the barn. 286 TONNAGE OF VESSELS. RULE. Multiply the dfference of the squares of the heights by one half of the perpendicular height, and this product by.0005. The result will be the contents in bushels. 1. How many bushels of oats are in a heap, the perpendicular height being 30 inches, and the slanting bleight 60 inches? Ans. 20.25 bushels. 2. How many bushels of beans are in a heap, the perpendicular height being 25 inches, and the slanting height 50 inches? Art. 293.-When grain is heaped in the corner of the barn. RULE. IMutiply the dfference of the squares of the heights by one fourth of the perpendicular height, and this product by.0005. The result will be the contents in bushels. 1. Required the number of bushels of grain heaped in the corner of the barn; the perpendicular height being 40 ine.hes, and the slanting height 70. inches? 1Ans. 16.5. 2. How many bushels of barley in the corner of a box, tlo perpendicular height being 24 inches, and the slanting height 36 inches? Ans. 2.16 bushels. TONNAGE OF VESSELS. CARPENTERS' RULE. Art. 294.-For single-decked vessels, multiply the length and breadth at the main beam, and depth in the hold, together, and divide the product by 95, and the quotient is the tons. But for a double-decked vessel, take half of the breadth of the main beamfor the depth of the hold, and proceed as before. 1. What is the tonnage of a single-decked vessel, whoso length is 67 feet, breadth 24 feet, and depth 12 feet? Ans. 203 - tons. 2. What is the tonnage of a double-decked vessel, whose length is 80 feet, and breadth 30 feet? Ans. 3781 - tons. GOVERNMENT RULE. "If the vessel be double decked, take the lelgth thereoffrom MECHANICAL POWERS. 287 the fore part of the main stem to the after part of the stern-post, above the upper deck; the breadth thereof at the broadest part above the main wales, half of which breadth shall be accounted the depth of such vessel, and then deductfrom the length three fifths of the breadth; multiply the remainder by the breadth, and the product by the depth, and divide this last product by 95, the quotient whereof shall be deemed the true contents, or tonnage of such ship or vessel; and if such ship or vessel be single decked, take the length and breadth, as above directed, deduct from the length three fifths of the breadth, and take the depth from the under s/de of the deck plank to the ceiling in the hold, and then multiply and divide as aforesaid, and the quotient shall be deemed the tonnage." 1. What is the government tonnage of a single-decked vessel, whose length is 90 feet, breadth 40 feet, and depth in the hold 12 feet? Ans. 333, tons. MECHANICAL POWERS. Art. 295.-TIIAT body which communicates motion to another, is called a power: the body which receives the motion is called the weight. The mechanical powers are six: the Lever, the Wheel and Axle, the Pulley, the Screw, the Inclined Plane, and the Wedge OF THE LEVER.Art, 29. —The lever is a bar, moveable about a fixed point, called its fulcrum, or prop. It is, in theory, conside:ed an inflexible line, without weight. There are several kinds of lever used in mechanics. Tho more common kind is that which is here shown. It is a principle in mechanics, that the power is to the weight as the velocity of the weight is to the velocity of the power. 288 OF THE WHEEL AND AXLE. Art. 297.-To find what weight may be balanced by a given power. RULE As the distance between the body to be raised, or balanced, and the fulcrum, or prop, is to the distance between the prop and the point where the power is applied, so is the power to the weight which it will balance. 1. If a man, weighing 160 lbs., rest on a lever 10 feet long, what weight will he balance on the other end, supposing the prop to be 1 foot from the weight? 1: 9:: 160: 1440 lbs. Ans. 2. If a weight of 1440 lbs. were to be raised by a lever 10 feet long, the prop being 1 foot from the weight, what power must be applied to the other end, to balance the weight? Ans. 160 lbs. 3. At what distance from the prop must a power of 160 lbs. be applied, to balance 1440 lbs., 1 foot from the prop? Ans. 9 feet. 4. At what distance from a weight of 1440 lbs. must a prop be placed, so that a power of 160 lbs., applied 9 feet from the prop, may balance it? Ans. 1 foot. OF THE WHEEL AND AXLE. Art. 298.-The wheel and axle are here represented with the weight attached to the circumference of the axle, and the power - applied to the circumference of the wheel. The principle of the lever is employed in the wheel and axle. RULE. As the diameter of the axle is 7 to the diameter of the wheel, so, is the power applied to the wheel, to J the weight suspended on the axle. [ llllll ii llltli 1. If the diameter of the axle be 6 inches, and that of the THE PULLEY. 289 wheel be 60 incnes, what weight applied to the wheel will balance 10 lbs. on the axle? 60: 6:: 10: 1 lb. Ans. 2. If the diameter of the wheel be 60 inches, what must be the diameter of the axle, so that 1 lb. on the wheel may balance 10 lbs. on the axle? Ans. 6 inches.'3. If the diameter of the axle be 6 inches, what must be the diameter of the wheel, so that 10 lbs. on the axle may balance 1 lb. on the wheel? Ans. 60 inches. THE PULLEY. Art. 299:-The pulley is a small wheel, moveable about its axis by means of a cord, which ). passes over it. When the axis of a pulley is fixed, the pulley only changes the direction of the power; if moveable pulleys are used, an equilibrium is produced, when the power is to the weight as one to the number of ropes applied to them. If each moveable pulley has its own rope, each pulley will be double the power. Art. 300.-The number of moveable pulleys and the power given, to find what weight may be raised. RULE. As 1 is to twice the number of moveable pulleys, so is thi power to the weight. OBS.-Reverse the rule, to find the power. 1. What weight would balance a power of 45 lbs., applied to a cord that runs over 3 moveable pulleys? 2. If a cord, which runs over 2 moveable pulleys, be attached to an axle 3 inches in diameter, the wheel of the axle being 28 inches in diameter, and a power of 10 lbs. be exerted 25 290 THE SCREW. at the circumference of the wheel, what weight would be raised under the pulleys? Thus, 3:28:: 10 X 3 2:560 lbs. Ans. OF THE SCREW. Art. 301.-The screw is a spiral thread, or groove, cut round a cylinder, and everywhere making the same angle with the length of the cylinder. The power is to the weight which is to be raised, as the dis- tance between two contiguous threads of the screw is to the circumference of a circle, described i by the power applied at the end of the lever. RULE. Multiply twice the length of the lever by 3.1416, which will give the circumference of the circle; then say, as the circumference is to the distance between the threads of the screw, so is the weight to be raised to the power which will raise it. 1. The threads of a screw are 1 inch asunder; the lever, by which it is turned, is 30 inches long, and the weight to be raised is 1 ton=2240 lbs. What power must be applied to turn the screw? 30x2=60, and 60x3.1416-188.496 inches, the circumference. Then 188.496: 1: 2240: 11.88 lbs. Ans. 2. If the lever be 30 inches, the circumference of the circle described by the power 188.496, the threads of the screw 1 inch asunder, and the power 11.88 Ibs., what weight will be raised? Ans. 2240 lbs. 3. If the weight be 2240 lbs., the power 11.88 lbs., and the lever 30 inches in length, what is the distance between the threads of the screw? Ans. 1 inch. 4. If the power be 11.88 lbs., the weight 2240 lbs., and the threads 1 inch asunder, what is the length of the lever? Ans. 30 inches, nearly. THE WEDGE. 2S1 INCLINED PLANE. Art. 302.-An inclined plane is a plane which makes an acute angle with the horizon. To find the power that will draw a weight up an inclined plane. RULE. As the length of the plane is to the perpendicular height of the plane, so is the weight to the power. 1. An inclined plane is 40 feet in length, and 8 feet in perpendicular height. What power is sufficient to balance a weight of 2000 pounds? Ans. 400 lbs. 2. A certain railroad, 200 rods in length, has a perpendicular elevation of 20 feet. What power is sufficient to sustain a train of cars weighing 100,000 pounds 2 Ans. 606.23. THE WEDGk — Art. 303.-The wedge is composed of two inclined planes, whose bases are joined. When the resisting forces, and the power which acts on the wedge, are in equilibrium, the weight will be to the power as the height of the wedge to a line drawn from the middle of the base to one side, and parallel to the direction in which the resisting force acts on that side. Art. 304 —To find the frce of the wedgo RULE. As the breadth, or thickness, of the head of t9 wedge, is to one of its slanting sides, so is the power whick acts against its head, to the force produced at its side. Suppose 100 lbs. to be applied to the head of a wedge, 2 inches broad, and 20 inches long, what force would be effected on each side? Ans. 1000 lbs. 292 MATHEMATICAL PROBLEMS. MATHEMATICAL PROBLEMS. Art. 305 —PROB. I. The sum and difference of two numbers given, to find those numbers. RULE. Subtract the difference from the sum, and divide the remainder by 2. The quotient will be the smaller number. Thern add the given difference to the smaller number, and this sum will be the larger number. EXAMPLE. An assembly of 344 persons is conver.ed in two rooms, one of which has 142 persons more in it than the other. How many are in each? Operation. 344-142=202; then 202.2=101 persons, in one room; then 101+142=243, in the other. Art. 306.-PROB. II. The sum of two numbers, and the difference of their squares given, to find those numbers. RULE. Divide the difference of their squares by the sum of the numbers, and the quotient will be their difference. We then have their sum and difference, to find each number, by Prob. I. EXAMPLE. A. and B. played at marbles, having at first 14 each; but after playing several games, B. having lost some of his, would not play any longer, and it was found that the difference of the squares of what each then had, was 336. How many did B. lose? Thus 336-14+14=12 difference; 14=half sum, and 12 2=6=half difference. Then I-+6=20, A. retired with; and 14-6=8, B. retired with: then 14-8=6, B. lost. Art. 307.-PRo B. III. The difference of two numbers, and the difference of their squares given, to find those numbers. RI LE. Divide the difference of their squares by the difference of their numbers, and the quotient will be their sum; then proceed by Prob. I. MATHEMATICAL PROBLEMS. 293 EXAMPLE. Said William to John, Father gave me $12 more than he gave Charles; and the difference of the squares of our separate parcels is 288. How much did he give each? Thus, 288-12=24, the sum: then 24-12-~2=6; then 12-+6 $18, William's share; and 12-6=$6, Charles had given him. Art, 308. —PROB. IV. The sum of two numbers and their quotient given, to find those numbers. RULE. Add 1 to the quotient, and by this sum divide the sum of the two numbers: this will give the less number. Subtract the less numberfrom the sum, and you will obtain the greater number. EXAMPLES. 1. Divide 100 into two such parts, that if the greater be divided by the less, the quotient will be just 30. Operation. Thus 100- 30+1=- 3-1, the less part; then 100 —39623T, greater. 2. The sum of A. and B.'s ages is 45, and if you divide A.'s by B.'s the quotient will be 4. What is the age of each? Ans. A.'s 36 years; B.'s 9 years. Art. 309.-PROB. V. The difference of two numbers, and the quotient given, to find those numbers. RULE. The difference of the two numbers divided by the quotient less 1, will be the less number. Add the less number to the diference, and you will have the greater number. A greyhound, in pursuit of a hare, ran three times as fast as the hare, as when he overtook the hare he had run 30 rods more tha she. How many rods did each run? Operation. 30 3- 1=15 rods, the hare ran; then 15+30=45 rods, the greyhound ran. Art. 310.-PROB. VI. To find the true weight of any quantity when weighed. in each scale of a balance, whose beam is unequally divided. 25* 294 MATHEMATICAL PROBLEMS. RULE. Take the square root of the product of the different weights, for the true weight. A parcel of sugar weighs in one scale 25 lbs.; in the other 30 lbs. What was its true weight? /25 x 3o=2..856. Art. 311.-PRo B. VII. The base and perpendicular given, to find the hypotenuse. RULE. The square root of the sum of the squares of the base and perpendicular will be the hypotenuse. This rule is illustrated by the following figure. If the base of a right-angled triangle be 9 feet, and the perpendicular 12, what is the hypotenuse? / ^ / t~12 contents, s L_~~~~~/! Icontents. Art. 312.-PROB. VIII. Given the base and sum of the perpendicular and hypotenuse of a right-angled triangle, to find the perpendicular. RULE. From the square of the sum subtract the square of the base, and divide the remainder by twice the sum, and the quotient will be the perpendicular. MATHEMATICAL PROBLEMS. 295 A tree, 100 feet in height, is broken off-the top of the tree reaches the ground 30 feet from the bottom, while the part broken off rests on the stump. How high from the ground was it broken off? Ans. 45' feet. Art. 313 —PROB. IX. Given the base and the difference of the hypotenuse and perpendicular of a right-angled triangle, to find the perpendicular. RULE. From the square of the base subtract the square of the given difference, and divide the remainder by twice the difference. EXAMPLE. If the base of a right-angled triangle be 30 feet, and the difference of the other two sides 6 feet, what is the length of the perpendicular? Ans. 72 feet. Art. 314. —PRO. X. To find the diameter of the earth, from the known height of a distant mountain, whose summit is just visible in the horizon. RU LE. From the square of the distance, divided by the height, subtract the height. The highest point of the Andes is about 4 miles above the bed of the ocean. If a straight line from this touch the surface of the water at the distance of 1781 miles, what is the diameter of the earth? Ans. 7940. Art. 315. —PROB. XI. To find the greatest distance at which a given object can be seen on the surface of the earth. RULE. To the product of the height of the object into the diameter of the earth, add the square of the height; and extract the square root of the sum. 1. If the diameter of the earth be 7940 miles, and Mount Etna 2 miles high, how far can it be seen at sea? Ans. 126+ miles. -/7940 X 2+22=126. OBs.-The actual distance at which an object can be seen is increased by the refraction of the air. 296 MATHEMATICAL PROBLEMS. 2. A man standing on a level with the ocean, has his eye raised 5~ feet above the water. To what distance can he see the surface? Ans. 2- miles. Art. 316.-PROB. XII. To find the height of an object at sea, or on the surface of the earth, having only the distance given. RULE From the given distance, take the distance which the elevation of the eye above the surface will give, found by the last problem; then divide the square of the remainder by the diameter of the earth, and the quotient will be the height required. Art. 317.-PROB. XIII. To find the contents of squared timber. RULE. Multiply the mean breadth by the mean thickness: the product, multiplied by the length, will give the contents. Required the contents of a log, the length 24 feet 6 inches, mean breadth 1 foot I inch, and mean thickness 1 foot 1 inch. Ans. 28 feet 9 inches 6"' Art. 318.-PRoB. XIV. To find the contents of round timber. COMMON RULE. Take one fourth of the mean girt, and square it, and multiply it by the length, for the contents. OBs.-1. Tapering timber should be divided into pieces of eight or ten feet long, and these parts should be computed separately, and added. 2. In order to reduce the tree to such a circumference as it would have without its bark, a deduction is generally made of a or J. of an inch for every foot of quarter-girt for young oak, ash, beech, etc.; but 1, or even 1 J inch, must be allowed for old oak, for every foot of quarter-girt. 3. The common rule gives the contents too small, by 3 feet on every 11 feet of contents; yet it is universally used in practice, being originally introduced in order to compensate the purchaser of round timber for the waste occasioned by squaring it. RULE II.- Take one fifth of the girt, and square it, and multiply by twice the length, for the contents. 1. Required the contents of a tree 24 feet long, and its girts at the ends 14 and 2 feet? Ans. 96 feet, by the common rule; the true content is 122.88 feet. MATHEMATICAL PROBLEMS. 297 2. How much timber in a tree 18 feet long, and its mean girt 5 feet 8 inches? Ans. Common rule, 36 feet 1~ inch; trua.content, 46 ft. 2 inches 10" 6"'. LEVELLING. Art. 319.,PROB. I. To find the difference in the height of twvo places, by levelling rods. RULE. Set up the levelling rods perpendicular to the horizon, and at equal distances from the spirit level; observe the points where the line of level strikes the rods before and behind, and measure the heights of these points above the ground; level in the same manner, from the second station to the third, from the third to the fourth, etc. The diference between the sum of the heights at the back stations, and at the forward stations, will be the difference between the height of the first station and the last. If the stations are numerous, it will be expedient to place the back and forward heights in separate columns in a table, as in the following example. Back heights. Fore heights. Feet. Inches. Feet. Inches. 1st observation,.. 5 2 7 5 2d. 2 8 6 3 3d.. 3 6 5 9 4th " 4 5 3 2 5th ".. 8 7 1 7 24 4 24 2 24 2 Difference,.. 0 2 If the sum of the forward heights is less than the sum of the back heights, it is evident that the last station must be higher than the first. Art. 320.-PROB. II. To find the difference between the true and apparent level, for any given distance. OBS.-1. The true level is a curve, which either coincides with, or is parallel to, the surface of water at rest. 298 PHILOSOPHICAL PROBLEMS. 2. The apparent level is a straight line, which is a tangent to the true level, at the point where the observation is made. 3. The difference between the true and the apparent level is nearly equal to the square of the distance, divided by the diameter of the earth. 1. What is the difference between the true and apparent level, for a distance of one English mile, supposing the earth to be 7940 miles in diameter? Ans. 7.98 inches, or 8 inches, nearly. 2. A tangent to a certain point on the ocean strikes the top of a mountain 23 miles distant. What is the height of the mountain? Ans. 352 feet. PHILOSOPHICAL PROBLEMS. Art. 321.-PROB. I. To find the time in which pendulums of different lengths would vibrate, that which vibrates seconds being 39.2 inches. The time of the vibrations of pendulums are to each other, as the square roots of their lengths; or, their lengths are as the squares of their times of vibrations. RULE. As the square of one second is to the square of the time in seconds in which a pendulum would vibrate, so is 39.2 inches to the length of the required pendulum. EXAMPLES. 1. Required the length of a pendulum that vibrates once in 8 seconds. 12: 82:: 39.2 in.: 2508.8 in.=209 9 ft. Ans. 2. How often will a pendulum vibrate, whose length is 100 feet? Ans. 5.53+ seconds. Art. 322. —PROB. II. By having the height of a tide on the earth given, to find the height of one at the moon. RULE. As the cube of the moon's diameter, multiplied by its density, is to the cube of the earth's diameter, multiplied by its density, so is the height of a tide on the earth, to the height of one at the roon. PHILOSOPHICAL PROBLEMS. 299 EXAMPLE. The moon's diameter is 2180 miles, and its density 494; the earth's diameter is 7964 miles, and its density 400. If, then, by the attraction of the moon, a tide of 6 feet is raised at the earth, what will be the height of a tide raised by the attraction of the earth at the moon? Ans. 236.8+ feet. I. If the diameters of two globes be equal, and their densities different, the weight of a body on their surfaces will be as their densities. II. If their densities be equal, and their diameters different, the weight of a body will be as ~ of their circumferences. III. If their diameters and densities be both different, the weight will be as 2 of their semidiameters multiplied by their densities. TABLE. Density. Diameter. Semidiameter. ~ semidiameter. Sun........100 883246 441623 294415 Jupiter.......94.5 89170 44585 29723 Saturn........67 79042 39521 26347 Earth........400 7964 3982 2654 Moon........494 2180 1090 726 Art. 323,-PROB. III. To find how far a heavy body will fall in a given time, near the surface of the earth. Oss.-Heavy bodies, near the surface of the earth, fall 16 feet in 1 second of time; and the velocities they acquire in falling are as the squares of the times; therefore, to find the distance any body will fall in a given time, we adopt the following RULE. As 1 second is to the square of the time in seconds that the bocy is falling, so is 16 feet to the distance infeet, that the body will fall in the given time. How far will a leaden bullet fall in 8 seconds? 12: 82: 16 ft.: 1024 ft.:= Ans. Art. 324. —PRO. IV. The velocity given, to find the space fallen through, to acquire that velocity. RULE. Divide the velocity by 8, and the square of the quotient will be the distance fallen through to acquire that velocity. 300 PHILOSOPHICAL PROBLEMS. 1. The velocity of a cannon-ball is 424 feet pel second. From what height must it fall to acquire that velocity? Ans. 2809 feet. 2. At what distance must a body have fallen to acquire the velocity of 1024 feet per second? Ans. 3 miles, 544 feet. Art. 325 —PROB. V. The velocity given per second, to find the time. RULE. Divide the velocity by 8, and a fourth part of the quotient will be the time in seconds. 1. How long must a body be falling to acquire a velocity of 304 feet per second? Ans. 9~ seconds. 2. How long must a body be falling to acquire a velocity of 864 feet per second? Ans. 27 seconds. Art. 326. —PRB. VI. The space through which a body has fallen, given, to find the time it has been falling. RULE. _Divide the square root of the space fallen through by 4, and the quotient will be the time in which it was falling. IIow long would a ball be falling from the top of a tower, 900 feet high, to the earth? Ans.'7 seconds. Art. 327.-PRon. VIT. The time given, to find the space fallen through. RULE. Miultiply the time by 4, and the square of the product will be the space fallen through in the given time. 1. What is the difference between the depth of two wells, into each of which, should a stone be dropped at the same instant, one would reach the bottom in 5 seconds, and the other in 3? 5 x4=-20, and 20 x 20400: then 3 X 4=12, and 12 X 12 - 144: then 400- 144-256 ft. Ans. 2. A ball was seen to fall half the way from the top of a tower in the last second of time. How long was it in descending, and what was its height before its descent? Ans. 186,486+ feet. PHILOSOPHICAL PROBLEMS. 301 Art. 328.-PROB. VIII. To find the velocity, per second, with which a heavy body will begin to descend, at any distance from the earth's surface. RULE. As the square of the earth's semidiameter is to 6 feet, so is the square of any other distance frnm the earth's centre, inversely, to the velocity with which it begins to descend per second. With what velocity per second will a ball begin to descend, if raised. 3000 miles above the earth's surface? As 4000 4000: 16:: 4000+3000 x 4000 x 3000: 5.22449 feet, Ans. And if the height is required, and the velocity given, thus, as 16: 4000 x 4000:: 5.22449: 49000000, and /490000004000= 3000 miles, Ans. Art. 329. —PROB. IX. The weight of a body, and the space fallen through, given, to find the force with which it will strike. RULE. Multiply the space fallen through by 64; then multiply the square root of this product by the weight, and the product is the momentum, orforce with, which it will strike. There is a monument 64 feet high. Supposing a stone, weighing 4 tons, should fall from its top to the earth, what would be its force, or momentum? Ans. 573440 lbs. That is, it would strike the earth with more force than the weight of two hundred and fifty tons. Art. 330,-PRon. X. To find the magnitude of any thing, when the weight is known. RULE. Divide the weight by the specific gravity found in the table, and the quotient will be the magnitude sought. What is the magnitude of several fragments of clear glass, whose weight is 13 ounces? 13-2600-=.005 of a cubic foot; and.005 x1728=8.640 cubic inches, Ans. 26 302 PHILOSOPHICAL PROBLEMS. TABLE, Showing the specific gravity of several solid and fluid bodies, in Avoirdupois Weight. The specific gravity of a body is its weight compared with pure, or distilled water. A Cubic foot of Ounce. A Cubic foot of Ounce. Pb tina, rendered malleable Brick......................... 2000 Pnd hammered............. 20170 Live Sulphur............... 2000 Ve-y fine Gold................ 19637 Nitre........................ 1900 Standard Gold............... 18888 Alabaster................... 1875 Moidore Gold.................. 11140 Dry Ivory............... 1885 Guinea Gold................... 17793 Brimstone................... 1800 Quicksilver.................. 13600 Solid subs. Gunpowder.. 1745 Lead............................ 11325 Alum......................... 1714 Fine Silver..................... 11087 Ebony........................ 171 Standard Silver.............. 10535 Human Blood............. 1054 Rose Copper................... 9000 Amber..................... 1030 Copper......................... 8843 Cows' Milk.................. 1030 Plate Brass................... 8000 Sea Water.................. 1030 Steel.................... 7852 Pure Water............... 1000 Cast Brass..................... 850 Red Wine.................. 993 Iron.............................. 7645 Oil of Amber.............. 98 Block Tin.................... 1321 Proof Spirits................ 925 Cast Iron...................... 135 Dry Oak.................... 925 Lead Ore..................... 6800 Olive Oil.............. 913 Copper Ore.................. 3775 Loose Gunpowder........ 872 Diamond................... 3400 Spirits of Turpentine...... 864 Crystal Glass.................. 3150 Alcohol, or pure spirits.. 850 White Marble............... 2707 Elm and Ash............... 800 Black "................ 2704 Oil of Turpentine......... 172 Rock Crystal................. 2658 Dry Crab-tree.............. 65 Green Glass................... 2620 Ether........................ 732 Clear Glass.................... 2600 White Pine................ 569 Flint................. 2582 Sassafras Wood.......... 482 Stone Paving.............. 2570 Cork......................... 240 Stone, Cornelia..., 2568 Common.. e "Cornelia............ 2568 Common Air................ 1 - -54 Free................. 2352 Inflammable Air........... 0'-'L Art. 331.-PRoB. XI. The bulk and weight of any body given, to find its specific gravity. RUL E. Divide the weight by the bulk, and the quotient is the specific gravity. Suppose a piece of marble contains 8 cubic feet, and weighs 13531 pounds, or 21656 ounces. WVhat is its specific gravity? 21656 8=22707, the specific gravity, as required by the table. ASTRONOM[CAL PROBLEMS. 303 ASTRONOMICAL PROBLEMS. Art. 332.-PROB. I. To find the dominical letter for any year in the present century, and also to find on what day of the week January will begin. RULE. To the given year add its fourth part, rejecting the fractions; divide this sum by 7; if nothing remains, the dominical letter is A; but, if there be a remainder, subtract it from 8, and the residue will show the dominical letter, reckoning 0=A, 2=B, 3=C, 4 —D, 5-E, 6=F, 7=G. These letters will also show on what day of the week January commences. For, when A is the dominical letter, January begins on Sunday; when B is the dominical letter, January begins on Saturday; C begins it on Friday; D begins it on Thursday; E on Wednesday; F on Tuesday; G on Monday. 1. Required the dominical letter for 1825. 4)1825 8-6 - 2==B, dominical letter. __456 As B is the dominical letter, January will 7)2281 begin on Saturday, and the second day will be - 315 —6 the Sabbath. 2. Required the dominical letter for 1842. Ans. B. 3. Required the dominical letter for 1837. Ans. A. 4. What is the dominical letter for 1801? Ans. D. 5. What is the dominical letter for 1845? Ans. E. Art. 333.-PROB. II. TO find on what day of the week any given day of the month will happen. RULE. Find by the last problem the dominical letter for the given year, and on what day in January will be the first Sabbath; and the corresponding days in the succeeding months will be as follows: Wednesday for February; Wednesday for March; Saturdayfor April; Monday for Mag; Thursdayfor June; Saturday for July; Tuesday for August; Friday for September, Sunday for October; Wednesday for November; Friday for December. Having found the day of the week for any day in the month, any other day may be easily obtained, as may be seen in the following example. 304 OF BALLS AND SHELLS. 1. Let it be required to ascertain on what day of the week will be the 25th day of September, 1842. The domilrical letter for 1842 is B; therefore, the 2d of January will be the Sabbath; and, by the above rule, the 2d of February will be Wednesday; the 2d of March will be Wednesday; the 2d of April will be Saturday; the 2d of May will be Monday; the 2d of June will be Thursday; the 2d of July will be Saturday; the 2d of August will be Tuesday; the 2d of September will be Friday. If the 2d be Friday, the 9th, 16th, and 23d will be Fridays. And if the 23d be Friday, the 24th will be Saturday,'he 25th will be the Sabbath, the day required. 2. On what day of the week will be December 8, 1849? Ans. Saturday. 3. On what day of the week will happen July 4, 1857? Ans. Saturday. 4. On what day of the week were you born? OF BALLS AND SHELLS. Art. 331. —An iron ball, 4 inches in diameter, weighs 9 lbs., nearly; and a leaden one, 4-, weighs about 17 lbs., and a pound of gunpowder measures about 30 cubic inches. Given the diameter of an iron ball, to find the weight, and the converse. RULE. Divide the cube of the diameter by 73; the quotient will be the weight in pounds. Multiply the weight by 7,-. The cube root of the product will be the diameter. 1. What is the weight of an iron ball, of which the diameter is 31 inches? Ans. 3.5 -7~=6.0293 lbs. 2. What is the diameter of an iron ball which weighs 24 lbs.? Ans. /24 x 71= /170.6 =5.547 inches diameter. 3. What does an iron ball weigh,_ whose diameter is 5.5 inches? Ans. 23.3965 lbs. 4. What is the diameter of an iron ball weighing 48 lbs? Ans. 6.988 inches. PILING OF BALLS. 305 5. What does an iron ball weigh whose diameter is 4.6 inches? Ans. 13.688 lbs. 6. What is the diameter of an iron ball which weighs 36 lbs? Ans. 6.349 inches. Art. 335.-Given the diameter of a leaden ball, to find its weight, and the converse. RULE. Divide the cube of the diameter by 4a: the quotient will be the weight in lbs. Multiply the weight by 4~: the / of the product will be the diameter in inches. 1. What is the weight of a leaden ball, whose diameter is 4.25 inches? Ans. 4.253 -41=17.059 lbs. 2. What is the diameter of a leaden ball which weighs 36 lbs.? Ans. 5.45 inches. 3. What is the weight of a leaden ball, of 4.6 inches i\ diameter? Ans. 21.63 lbs. 4 What is the diameter of a leaden ball weighing 48 lbs. I Ans. 6 inches. PILING OF BALLS. Art. 336,-BALLS and shells are piled up in horizontal courses, upon a base of the form of an equilateral triangle, or of a square, or of a rectangle. The number of balls in a row. diminishes, till, in the two first forms, it ends in a single ball, and in the last in a single row. The number of rows is equal to the number of balls in the lesser side of the under row. The number in the top row of a rectangular pile is one more than the difference between the length and breadth of the bottom row. Art. 337,-PROB. I. To find the number of balls in a triangular pile. RULE. Multiply the number of balls in a side of the bottom row by that number increased by 1, and again by that number increased by 2: theproduct, divided by 6, will be the number of balls in the pile. 26* 306 TO FIND THE WEIGHT OF CATTLE. Required the number of balls in a triangular pile, of which each side of the base contains 30 balls. Ans. 4960. Art. 338.-PROB. II. To find the number of balls in a square pile. RU I E. To twice the number of balls in a side of the bottom, add 1, and multiply the sum by the number in that row, and by that number increased by 1: the product, divided by 6, will give the number of balls in the pile. Let the side of the bottom row of a square pile contain 20 balls. How many balls are in the pile? Ans. 2870. Art. 339.-PROB. III. To find the number of balls in a rectangular pile. RULE. From 3 times the number in the length of the bottom row, increased by 1, subtract the number in the breadth, and multiply the remainder by the breadth, and by the breadth increased by 1: the product, divided by 6, will give the number of balls in the pile. Suppose the number of balls in the length of a rectangular pile to be 59, and in the breadth 20, what is the number in the pile? Ans. 11060. Art. 340,-PROB. IV. To find the number of balls in an incomplete pile. RULE. From the number of balls in the complete pile subtract the number in the pile that is wanting, both computed as before: the remainder is the number in the incomplete pile. Required the number of balls in a rectangular pile of 15 courses, the numbers in the bottom row being 60 and 25. Ans. 14590. TO FIND THE WEIGHT OF CATTLE. Art. 341.-TAKE the girt behind the shoulder, and the length from the fore part of the sho-alder-blade to the buttock, both in feet. Multiply the square of the girt by 4 times the length, and divide by 21. Multiply this quotient by 16, and it will give the weight of the four quarters, nearly. MISCELLANEOUS QUESTIONS. 307 OBs.-The four quarters are li tle moie than I the whole weight. The skin weighs nearly'W, and the tallow very nearly y1. What will the four quarters of an ox weigh, whose girt is 6 feet 6 inches, and length 5 feet 10 inches. Ans. 6.52 x 23 — 21 x 16=751 +lbs. MISCELLANEOUS QUESTIONS. Art. 342.-1. What is the product of 2s. 6d., multiplid by 2s. 6d.? Ans. ~-1. 2. Purchased a book for 15 cents, and sold it for 18. What did I gain per cent.? Ans. 20 per cent. 8. Sold a book for 18 cents, and gained 20 per cent. What did it cost me? 4. Purchased a book for 15 cents; sold it so as to gain 20 per cent. What did I get for it? 5. If E of 3 of -7 of - of 5- of a vessel be worth ~378, how many dollars is 3 of it worth? Ans. $9450. 6. A person owning 2 of a ship, sold - of his share for $3750. What was the whole ship worth? Ans. $15000. 7. What is the sum of the third and half third of 3s. 4d.? Ans. Is. 8d. 8. How many solid feet in a stick of timber 17 inches square, and 6 feet 5 inches long? Ans. 12 ft. 1517 in. 9. A man owning - of a farm, sold i of his share for $245. What was the value of the farm? Ans. $1225. 10. A. holds B.'s note for $2000, dated June 1st, 1825, ou which are the following endorsements, viz: Received Sept. 1st, 1825................$96. Dec. 10th, 1825................ 15. Apr. 20th, 1826................ 36. July 1st, 1826................200. Jan. 10th, 1827................ 20. Mar. 25th, 1827................ 90. How much remains due June 1st, 1827? Ans. $1767.228. 11. What principal, tt 5 per cent., will amount to $725 in 9 years? Ans. $500. 308 MISCELLANEOUS QUESTIONS. 12. What principal will gain $150, in 1 year, at 6 per cent.? Ans. $2500. 13. What is the present worth of $590, due 3 years hence, discounting at 6 per cent.? Ans. $500. 14. What is the present worth of $200; $100 payable in 2 months, $50 in 3 months, and $50 in 5 months, discounting at 4 per cent.? Ans. 8198.01. 15. A note of $500 amounted to $725 in 9 years. What was the rate per cent.? Ans. 5 per cent. 16. In what time will ~420 amount to ~520 16s. at 3 per cent.? Arns. 8 years. 17. What will $1350 amount to in 3 years, at 5 per cent., compound interest? Ans. $1562.793. 18. A. has 150 gallons of wine, which he will sell at 7s. 3d. per gallon, ready money, but in barter he will have 8s. per gallon. B. has linen at 3s. 6d. per yard, ready money. How must B. sell his linen per yard, in proportion to the bartering price of A.'s wine, and how many yards will be equal to A.'s wine? Ans. Bartering price, 3s. 101d., and 310 yards 2 qrs. 3+nails. 19. A merchant bought hats at 4s. each, and sold them at 4s. 9d. What was the gain in laying out ~100? Ans. ~18 15s. 20. A merchant bought 10 tons of iron for ~200. The freight and duties amounted to ~25, and his own charges to ~8 6s. 8d. For how much per pound must he sell it to gain 20 per cent.? Ans. 3d. per lb. 21. Sold a watch for ~50, and by so doing lost 17 per cent., whereas I ought to have cleared 20 per cent. How much was it sold under its real value? Ans. ~22 5s. 94d. ~2. Four men trade in company. A.'s stock was $560; B.'s $1040; C.'s $1200; their whole stock was $3200. They gained in two years a sum equal to twice their stock, and $160 more. What was D.'s stock, and what was each man's share of the gain? D.'s stock, $400. A.'s gain, $1148. Ans. B.'s " $2132. C.'s " $2460. D.'s " $820. 23. A., B., and C. trade in company. A. put in $20, B. $30, and C. a sum unknown. The gain was $36, of which $16 MISCELLANEOUS QUESTIONS. 309 was C.'s share. What was C.'s stock, and what was A.'s and B.'s gain? ( C.'s stock, $40. Ans. A.'s gain, $8. B.'s " $12. 24. What is the square root of 42? Ans. 64. 25. What is the mean proportional between 24 and 96?.ins. 48. 26. A certain general has an army of 5625. How many must he place in rank and file, to form them into a square? Ans. 75. 27. Arrange 10952 men in such a manner that the number in rank may be double the file. Ans. 74 in file, and 148 in rank. 28. There is a circle, whose diameter is 4 inches. What is the diameter of a circle 3 times as large? Ans. 6.928-+. 29. Two boats start on a river at the same time, from places 300 miles apart; the one proceeding up the stream is retarded by the current 2 miles per hour, while that moving down the stream is accelerated 2 miles per hour; both are propelled by by a steam engine, which would move them in still water 8 miles per hour. How far from each starting-place will the boats meet? Ans. 11 2~ miles from the lower place, and 187-~ miles from the upper place. 30. There are 3 circular ponds; the diameter of the less is 100 feet; the area of the greater is 3 times the area of the less. What is its diameter? Ans. 173.2+. 31. What is the superficial contents of one side of a cubical stone, containing 474552 solid inches? Ans. 6084 inches. 32. If a cube of silver, whose side is 4 inches, be worth ~50, what is the side of a cube of like silver, worth 4 times as much? Ans. 6.349 inches. 33. The height of a tree standing on the bank of a river is 75 feet; a line reaching from the opposite shore to the top of the tree is 256 feet long. What is the breadth of the river? Ans. 244.7+ feet. 34. If a pipe 6 inches in diameter will discharge a certain quantity of water in 4 hours, in what time will 3 pipes, each 4 inches in diameter, discharge double the quantity? Ans. 6 hours. 35. Two men start from the same place and trav.el, one south'6 leagues, the other east 45 leagues. How far will they be apart? Ans. 88.3 + leagues. 310 MISCELLANEOUS QUESTIONS. 36. What is the side of a cubical mound, containing 5832 solid feet? Ans. 18 feet. 37. If a ball, 6 inches in diameter, weigh 32 lbs., how much will a ball of the same metal weigh, whose diameter is 3 inches? Ans. 4 lbs. 38. The side of a cubical box is 2 feet. What is the side of a box which will contain three times as much? Ans. 2 feet 10 inches. 39. A refiner mixed 3 lbs. of gold, 22 carats fine, with 3 lbs. 20 carats fine. What was the fineness of the mixture? Ans. 21 carats. 40. How many gallons of water must be put to wine, at 3s. a gallon, to fill a vessel of 100 gallons, so that a gallon of the mixture may be afforded at 2s. 6d. per gallon? Ans. 162 gallons. 41. IT 12 bushels of oats, at Is. 6d. per bushel, be mixed with barley at 2s. 6d., rye at 3s., and wheat at 4s. per bushel, how much barley, rye, and wheat must be mixed with the 12 bushels of oats, that the mixture may be worth 2s. 9d. per bushel? Ans. 12 bushels of each sort. 42. A man travelled 6 miles the first day, 9 the second, increasing each day's journey 3 miles. He travelled 61 days. How many miles did he travel the last day? Ans. 186 miles. 43. If 100 pears be placed in a right line, 1 yard asunder, how many miles will a man travel, to pick them and carry them 1 at a time to a basket placed 1 yard from the first pear? Ans. 5 miles, 1300 yards. 44. Suppose 550 men are in a garrison, with provision sufficient to last them 9 months. How many must depart, that their provision may last them 11 months? Ans. 100. 45. If a man labor for me 16 days, when the price of labor is $1.25, how long must I labor for him, to requite the favor, when the price of labor is 75 cents per day? Ans. 261 days. 46. The third part of an army were killed, the fourth part were taken prisoners, and 1000 fled. How many were in this army? Ans. 2400. 47. An ignorant fop, wanting to purchase an elegant house, met with a facetious gentleman, who told him he had one which he would sell him on the following very reasonable terms, viz: that he should give him 1 penny for the first door, 2 for the second, 4 for the third, and so on, doubling the price MISCELLANEOUS QUESTIONS. 311 of each door, which were 36 in number. "It is a bargain," exclaimed the simpleton, "and here is a guinea to bind it." What did the house cost him? Ans. ~286331153 Is. 3d. 48. A., B., and C. would divide $100 between them, so that B. may have $3 more than A., and C. $4 more than B. What is each man's share? A's $30. Ans. B.'s;33. (.'s $37. 49. Divide ~340 among 3 men, in such a mannet that the first shall have 3 times as much as the second, and the third 4 times as much as the first? 1st, ~634. Ans. 2d, ~21}. 3d, ~255. 50. A person having a certain number of dollars, said, if a third, a fourth, and a sixth of them were added, their sum would be $45. How many dollars had he? Ans. $60. 51. What number is that which, being multiplied by A, the product will be 15-? Ans. 21. 52. What number is that from which, if you subtract 5 of itself, the remainder will be 12? Ans. 20. 53. What part of 25 is A- of a unit? Ans. -. 54. What number is that which, being multiplied by 2, the product will be -? Ans. 3. 55. If - of a farm be worth ~3740, what is the whole worth? Ans. ~9973 6s. 8d. 56. A father dying, left his son a fortune, -1 of which he spent in 6 months; 2 of the remainder lasted him 12 months longer, when he had ~348 left. What sum did he receive? Ans. ~1284 18s. 5f-7d. 57. A young man received $210, which was - of his elder brother's fortune, and 3 times the elder brother's fortune was _ the father's estate. What was the value of the estate? Ans. $1890. 58. A man has 80 shillings to divide among his laborers, consisting of an equal number of men, women, and boys. To every boy he gives 6d., to every woman 8d., to every man Is. 4d. How many were there of each? Ans. 32. 59. Suppose a man pay to his laborers, men, women, and boys, ~7 17s. 6d.; to every boy he gave 6d., to every woman 8d., and to every man 16d.; for every boy there weie 3 women, and for every woman there were 2 men. How many were there of each? Ans. 15 boys, 45 women, and 90 men. 312 MISCELLANEOUS QUESTIONS. 60. A gentleman bought a horse, a chaise, and harness, for ~60; the horse cost twice as much as the chaise, and the harness half as much as the horse. What was the cost of each? ( Horse, ~30. Ans. Chaise, ~15. Harness, ~15. 61. Divide $1000 among 3 men, in such a manner that as often as the first has $3, the second shall have $5, and the third $8. (1st, $187.50. Ans. 2 2d, $312.50. d3d, $500.00. 62. A. can do a piece of work in 10 days; D. can do the same in 13 days. In what time will both working together do the same work? Ans. 5135 days. 63. If 6 lbs. of pepper be worth 13 lbs. of ginger, and 19 lbs. of ginger be worth 41 lbs. of cloves, and 10 lbs. of cloves be worth 63 lbs. of sugar, at 10 cents per pound, what is the value of 1 cwt. of pepper? Ans. $38.22. 64. A tradesman increased his estate annually one third, less $960, which he spent in his family. At the end of 34 years be found that his estate amounted to $30284. What had he at first? Ans. $13551.75. 65. A person wants a cylindrical vessel, 3 feet deep, which shall hold twice as much as another 28 inches deep, and 46 inches in diameter. What must be the diameter of the required vessel? Ans. 57.373 inches. 66. How long must be the tether of a horse which will allow him to graze quite round an acre of ground? Ans. 3 9 yards. 67. What number is that which, being increased by its, 2 - 1 ----- - --- -~- "-:2' 4' and 5 more, the number will be doubled? Ans. 20. 68. A man, after having spent I and 3 of his money, had ~2612 left. How much had he at first? Ans. ~160. 69. A vessel has 3 pipes; the first will fill'it in I of an hour, the second in I of an hour, and the third in I of an hour. In what time will all running together fill it? Ans. 9 of an hour. 70. A. and B. employed equal sums in trade. A. gained a sam equal to - of his stock; B. lost $225; then A.'s money was double that of B's. What was each man's stock? Ans. $600. MISCELLANEOUS QUESTIONS. 313 71. There is a certain number, - of which exceeds 4 by 6. What is that number? Ans. 80. 72. If $$60 be divided between 4 men in such a proportion that the first shall have 1, the second -, the third., and the fourth -, what will each receive? 1st 2. Ans. d 15, 3id, 1212 4th, 109o. 73. A., B., C., and D. spent 35 shillings at a reckoning, and being a little dipped, agreed that A. should pay -, B. i, C. 3, and D.,. What does each pay, in this proportion? A. 13s. 4d. Ans. B 10Os. C. 6s. 8d. D. 5s. 74. The wheels of a chaise, each 4 feet high, in turning within a ring, moved so that the outer wheel made two turns while the inner made one, and their distance from one another was 5 feet. What were the circumferences of the tracks described by them? Ans Outer, 62.8318 feet. -s Inner, 31.4159 feet. 75. A., B., and C. traded in company, and gained ~350, of which A. took a certain sum; B. took 4 times as much as A., and C. 8 times as much as B. What were their respective shares of the gain? A.'s gain, ~9 9s. 2t(. 1-7-qr. Ans. B.'s gain, ~37 16s. 9d. 012qr. C.'s gain, ~302 14s. Od.' 2qrs. 76. A gentleman divided his fortune among his sons, giving A. ~9 as often as B. ~5, and C. ~3 as often as B. ~7. C.'s dividend was ~15375-. To what did the whole estate amount? Ans. ~11583 8s. 10d. 77. If 57 of 3 f of a ship be worth2 of of }2 of the cargo, valued at ~1000, what was the value of both ship and cargo? Ans. ~1837 12s. 17 —d. 78. Three men purchase a lot of land in company. A. paid I, B. 3 of the whole, and C. paid ~256. How much did A. and B. pay, and what part of the land had C.? ( A. paid ~597 6s. 8d. Ans. B. paid ~640. C.'s share, -I-; 79. A gay fellow soon got the better of 2 of his fort!ln,. He then gave ~1500 for a commission, and his profusion coa27 314 MISCELLANEOUS QUESTIONS. tinued until he had but ~450 left, which he found to be just 3 of his money, after purchasing his commission. What was his fortune at first? Ans. ~3780. 80. A. and B. are on opposite sides of a circular field, 268 rods in circumference. They start both at the same time to go round it, and go the same way. A. goes 22 rods in 2 minutes; B. goes 34 rods in 3 minutes. How many times will B. go round the field before he will )vertake A.? Ans. 17 times. 81. How high above the earth must a man be raised to see j of its surface?* Ans. One diameter hioh. 82. The girt of a vessel round the outside of the hoop is 22 inches, and-the hoop is 1 inch thick. What is the true girt of the vessel? Ans. 15-. 83. The hour and minute hand of a watch are exactly together at 12 o'clock. When are they next together? Ans. I h. 5 m. 27T3 s. 84. Three men trade in company till they gain ~120. The sums put in were in such proportion, that as often as A. had ~5 of the gain, B. had;7; and as often as B. had ~4, C. had ~6. What was each man's share of the gain? A.'s, ~26 13s. 4d. Ans. B.'s, ~37 6s. 8d. C.'s, ~56. 85. A. and B. cleared by an adventure at sea 45 guineas, which was ~35 per cent. upon money adventured. With this gain they agreed to purchase a genteel horse and carriage, which they were to use in proportion to their sums adventured, which was found to be 11 to A. as often as 8 to B. What money did each adventure? As. A. ~104 4s. 21-d. B. ~75 15s. 99. 86. A military oicer drew up his soldiers in rank and file, h'lvi;g the number in rank and file equal. On being reinforced with three times his fi:st number of men, he placed them all in the s tme form, and then the number in rank and in file was just double what it was at first. He was again reinforced with three times his whole number of men; and, after placing the whole in the same form as at first, his number in rank and in file was 40 men each. How many men had he at first? Ans. 100 men. * This question is for the student in Geometry. MISCELLANEOUS QUESTIONS. 315 87. A general disposing his army into a square battalion, found he had 231 over and above, but increasing each side with one soldier, he wanted 44 to fill up the square. Of how many men did his army consist? Ans. 19000. 88. There arc 3 horses belonging to different men, employed to draw a load of salt from Boston to Lowell for $9.50. A.'s and B.'s horses are supposed to do 7 of the work, A.'s and C.'s —, B.'s and C.'s -z. They are to be paid proportionally. What is each man's share of the gain? A.'s, 3.288-. Ans. B.'s, 4.384-. C.'s, 1.826 l. 89. A., B., and C. are to share ~100, in the proportion of 3,, and 5, respectively; but C. dying, it is required to divide the whole sum properly between the other two. An. i A.'s share, ~57 2s. 10od. B.'s share, ~42 17s. 15d. 90. There is an island 50 miles in circumference, and 3 men start together to travel the same way round it. A. travels 7 miles a day, B. 8, and C. 9. When will they all come together again, and how far will each travel? 91. A man died leaving $1000 to be divided between his two sons, one 14 and the other 18 years of age, in such a manner that the share of each being let, at 6 per cent. interest, should amount to the same sum when they should arrive at the age of 21. What did each receive? Ans The eldest, $546.153+. ns The youngest, $453.846+. 92. A hare starts 12 rods before a greyhound, but is not perceived by him till she has been up 45 seconds. She scuds away at the rate of 10 miles an hour, and the dog, on view, makes after, at the rate f 16 miles an hour. How long will the course hold, and what space will be run over from the spot where the dog started? Ans. 97 seconds. 2288 feet. 93. There is a circular field, surrounded by a rail-fence 10 rails high. each one rod in length, and the number in the fence equals the number of acres in the field? What is the area of the field? Ans. 201062.4 acres. 94. How much greater is the circle described by the head of a man 6 feet high, than by his feet, in the revolution of the earth on its axis? 95. In an orchard, ~ the trees bear apples, J pears, - plums, 316 MISCELLANEOUS QUESTIONS. and 50 of them produce nothing. How many trees are there in all? Ans. 600. 96. Sound moves at the rate of 1142 feet in a second. If the time between the lightning and the thunder be 30 seconds, what is the distance of the explosion? Ans. 6.488+ miles. 97. In a thunder-storm I observed by my watch that it was 6 seconds between the lightning and the thunder. At what distance was the explosion? Ans. 6852 feet. 98. Tubes may be made of gold, weighing not more than at the rate of T-6~ of a grain per foot. What would be the weight of such a tube, which would extend across the Atlantic, from Boston to London, the distance being 3000 miles? Ans. I lb. 8 oz. 6 pwt. 3-1 grs. 99. Suppose one of those meteors called fireballs, to move parallel to the earth's surface, and 50 miles above it, at the rate of 20 miles per second, in what time would it move round the earth? The earth's diameter being 7964 miles, the diameter of the orbit will be 7964+50 x28064, and 8064x3.1416 =25333.8624, its circumference. Then 25333.8624 201266.693120=21' 6" 41"' 35"" 13""' 55""", the Ans. 100. In giving directions for making a chaise, the length of the shafts between the axletree and back-band being settled at 9 feet, a dispute arose whereabout on the shafts the centre of the body should be fixed. The chaise-maker advised to place it 30 inches before the axletree; others supposed that 20 inches would be a sufficient incumbrance for the horse. Now, supposing 2 passengers to weigh 3 cwt., and the body of the chaise 3- cwt. more, what will the horse, in both these cases, bear in addition to his harness? A 1161-. 77101. A piece of square timber is 10 feet lon each side of 101. A piece of square timber is 10 feet long, each side of the greater base 9 inches, and each side of the less 6 inches. How much must be cut off from the less end to contain a solid foot? Ans. 3.392 feet. 102. A carpenter put a curb of oak round a well: the inner diameter of the curb was 3 feet, and its breadth 7- inches. What was the expense of it, at 8d. per square foot? Ans. 5s. 2~d. BOOK-KEEPING. 317 BOOK-KEEPING. ALL mercantile transactions consist in exchanging articles of trade, either for money or its equivalent. A systematic record of such transactions is called Book-keeping. Every person should possess sufficient knowledge of this science to keep such record of his business as will at any time exhibit a true state of his affairs. The person who purchases goods, or receives any thing of me, is debtor to me; and he who pays me money, or delivers any thing to me, is creditor. The following is a plain and simple method of keeping accounts without a Day-book, and will be found sufficient for the purposes of farmers and mechanics, where their business is such, that charges are made only at considerable intervals; but in all cases where some three or four, or more charges are made daily, a Day-book should be regularly kept. 1837. James Trueworthy, Dr. $ C 1837. James Trueworthy, Cr.? $ C Jan 1, To 4 cords of wood, at $3, 1200 Jan. 1, By 3 gals. molasses, at I Jan. 9, To 1 load of hay, at $20, - 200 40 cts. 120 Feb.'2, To'20 bush. oats, at 50 cts. 10(00 March 1, By 1 lb. starch, at 12 cts. 112 June 1, To 15 lbs. veal, at 5 cts.- 75 April 2, By 50 lbs. coffee, at 14 cts. 700 July 6, To pasturing horse 3 June 13, By 12 Ibs. sugar, at 9 cts. 1 08 weeks, at 50 cts. - 1 50 20, By 3 yds. broadcloth, at $5 15 00 Aug. 19, By cash to balance, - -19 85 4425 4425 BOOK-KEEPING BY SINGLE ENTRY. By Single Entry two principal books are required, the Day, or Waste Book, and Ledger. The Day-book should be ruled with two columns on the right hand, for dollars and cents, and one column on the left for noting the month, day, and year when the charges were made. All charges should be made at or near the time when they bear date, or when they purport to have been made. Where 27* 318 BOOK-KEEPING. an individual is charged with articles delivered to him, he is a debtor for the amount of them, and against his name on the Day-book, " DR." is written to designate him as debtor. When such individual delivers articles to be allowed him on his account, he is said to be a creditor, and " CR." is written against his name on the Day-book, to designate that he is credited with such articles. THE LEDGER. The Ledger is a book to which the accounts entered on the Day-book are transferred from time to time, in order that each man's whole account may appear by itself. All articles for which he is debtor are usually entered on the left-hand side of the page, and all articles for which he is creditor are entered on the right-hand side of the page. The date of the charge, and the amount in dollars and cents, are entered in the same manner as on the Day-book; but where there are a number of charges of different articles made under one date on the Day-book, they are usually entered upon the Ledger as " Sundries," which is a general term for a variety of articles; and the total amount of all such articles charged on any one day, is carried out against such entry in dollars and cents. When a charge is posted from the Day-book to the Ledger, a bracket, or other mark, is made on the left of the charge on the Day-book, to denote that it has been transferred to the Ledger. When the account is to be settled, the sum requisite to balance the same is ascertained, and cash, or a note, is given, which is entered upon the book, and the account is thus made equal. Traders and mechanics most usually give receipted bills of their accounts on settlement. In some instances, a settlement is written at the bottom of the account, as follows: "Settled and balanced all accounts to this date," which is signed by the parties. BOOK-KEEPING. 319 FORM OF THE DAY-BOOK. 1837. 1 1837. John Silvers, Dr. Jan. 1, John Newton, Dr. Jan. 6, To 4 yds. broadcloth, at To 4 gallons molasses, at $4.5) 18 00 40 cts. 1 60 Cr. By 8 bush. of rye, at 15 lbs. sugar, at 10 cts. 1 50 4s.6d. 6 00 ~ ~__ ___ ___ _~ ~ — - By 15 lbs. dried apples, Joel Stephenson, Dr. at 5 cts. 75 To 6 lbs. chocolate, at Is. 1 00 40 lbs. cheese, at 10 cts. 4 00 Jedediah Jones, Dr. " 7, John Newton, Dr. To 50 lbs. fish, at 4 cts.. 2 25 To 3 yds. flannel, at 3... 1 50 2 lbs. starch, at 14 cts.. 28__ 5 6 lbs. coffee, at 15 cts.. 90 Joel Stephenson, Dr. _ ~ 3 _~ ~ To I bush. corn, at 9s.... 1 50, John Silvers, Dr. 1 fur hat 4 00 To 38 lbs. rice, at 5 cts... 1 90 2 lbs. cocoa, at 28 cts.. 56 " 8, John Newton, Cr. By I ton of hay -......... 15 00 Joel Mason, Dr. Dr. To 1 bbl. flour...... 7 50 To 50 lbs. iron, at 7 cts. 3 50 o 50 lbs. suar..... 500 1 barrel of flour,..... 50 -.. _ ~ ~~__ I____ ~_ -Jedediah Jones, Dr. " 4. Timothy Styles, Dr. To 38 lbs. rice, at 5 cts... 1 90 To 6 gals. of oil, at 6s.... 6 00 ~ _ _ _ 50 lbs. fish, at 51 cts... 2 25 Joel Mason, Dr. To 1 lb. Y. H. tea, at 3s... 50 5 John Silvers, Dr. I bush salt...... 1 00 To 40 lbs. of raisins, at I lb. cocoa...... 28 14 cts. 5 60 -- _- ~ ~_ ___ _ _ ~- ~- John Silvers, Dr. John Newton, Dr. To 1 pr. of boots........3 50 To 1 gaLoil........... 00 I pr. shoes......... 2 00 8 lbs. coffee, at 15 cts. 1 20 Cr. By cash, 10s. 6d.... 1 75 Timothy Styles, Dr. To 100 lbs. nails, at 6 cts.. 6 00 Joel Mason, Dr. To 3 bush. of salt, at 6s.. 3 00 Joel Stephenson, Dr. -~ ~ -~__ _ ~~ To 40 lbs. butter, at 20 cts 8 00 John Newton, Dr. 1 firkin, 3s......... 50 To 2 lbs. saleratus, at _ II cts. 22 " 12. John Silvers, Cr. By 2 days' work........ 2 00 " 10, John Newton, Cr. 3 cords wood, at 15s.... 7 50 By 40 lbs. dried apples,.. 2 00 Dr. To 1 lb. flour...... 0 10 Dr. To 2 lbs. Y. H. tea, at _ 3s.6. 1 16 " 13, Joel Stephenson, Cr. To 2 lbs. raisins, at 2s... 34 By 2 Cdrds bark, at 12s... 4 00 " 11, Jedediah Jones, Dr. Timothy Styles, Cr. To 2g gal. oil, at 6s....... -2 50 By 2 firkins butter, 80 Ibs, at Is. 13 34 Joel Mason, Cr. Cash................. 91 By 5 bush. apples, 2s... 1 67 _. _ 1 cords wood, at 15s.. 3 75 " 14, John Newton, Dr. Cash to balance ac't,.. 10 44 To 100 Ibs. fish, at 5 cts.. 5 00 ~ ~__ _ - - - l Cr. By 2 cords of wood, John Silvers, Cr. at 15s. 5 00 By 30 bush. potatoes, at By cash, to balance ac't. 2 27 ls. 6d. 7 50 100 lbs. cheese,..... 10 00 John Silvers, Dr. ~ ~ 12, Jedediah -J o, To 100 lbs. iron at 7 cts.. 7 00 * 12, Jedediah Jones, Cr. By cash rec'd of 1. Hardy. 300 " 15. Joel Stephenson, Cr. 2 days' labor, at 4s. 6d. 1 50 By 10 lbs. lard, at s..... 1 67 Cash to balance ac't,.. 3 33 20 lbs. butter, at 20 cts. 4 00 320 BOOK-KEEPING. FORM OF THE LEDGER. DR. JOHN NEWTON, CR. 1837. 1837. Jan 1..To 4 gallons molasses, at Jan. 5. By Cash.............. 1 75 40 cts. 1 60 8. By 1 ton of hay.......... 15 00 "* 5. "sundries............. 2 42 " 14. By 2 cordsofwood, at 15s. 5 00 " 7. "3 yds. flannel, at 3s... 1 50 "" " Cash to balance accounts. 2 27 " 8 " 8sindriA.......... 12 50 --- " 10. " sundries.......... 1 50 $26 02 14. 100. 1bfsh, at 5 cts 5 00 $26 0211 ________________ DR. JOEL STEPHENSON, CR. 1837. 1837. 4 Jan. 1. To 6 lbs. chocolate at Is.. 1 00 Jan.13 By 2cordsof wood, a 12s. 4 i0 " 7.1 "sundries.... 5 1.1 50 DR. JEDEDIAH JONES, CR 1837. ]. Jan. 1. To sundries.............43 11 J By undes...... 50 1837. 1837. sr..... 4 50 DR. JOHN SILVERS, CR. 1837. I l 1837. 7i Jan. 3.iTo sundries-..... t 2 4 J6 n. 6. By sundries..... 10 1 76 DR. JOEL MASON, CR. I -~ _ i - DR. TIMOTHY STYLES, CR. — I I^T I I,............:..... ) The student should be required to complete posting the above accounts from the Day-book. To each Ledger there is an Alphabet, or Index of Names, consisting of a small book with the letters of the alphabet pasted upon the leaves, so that each man's name can be readily found by turning to the leaf marked by the first letter of his sirname. BOOK-KEEPING. 321 INDEX TO THE LEDGER. A PAGE I H PAGE 0 PAOT Allen James............ I Hoit William.......... 8 Odlin Woodbridge.... 16 B Horsley Henry........9 P Bacon Samuel..........2 J Pratt Thomas........17 Brown Leonard........ 3 Jenkins Abraham... 9 Putney Andrew........ 18 C Jones Jedediah....... 10 R Canning George........ 3 K Ramsay Amos......... 18 D Knowles Michael......11 Rollins Zenas.......... 20 Dalton Levi......... 4 L S E Leeds William........ 12 Stephenson Joel....... 21 Edgell Oliver........4 M Silvers John........ 22 F Mason Joel.......... 13 Styles Timothy........ 23 Fulton Curtis........... 5 M'Farland Asa....... 13 T Freeman Nathan........6 N Trueworthy James.....24 G Newton John........ 14 W GrosvenorJacob........ 7 NelsonJohn........... 15 Wyndham Augustus... 25 Every person engaged in business should take an inventory of his notes and accounts, his stock in trade, and other property, and of the debts owed by him, once or twice a year. By comparing this with former inventories, he will know his gain or loss, from time to time. Inventory of Stock in Trade, Notes, Accounts, and other Property, taken from the foregoing example, July 1, 1837. 200 lbs. coffee, at s.......... $33 34 Am' bt brought up........ $567 34 2 kegs tobacco, 100 lbs., at Note againstJames Allen, for$50 20 cts. 20 00 and int., dated Jan. 1,1836... 53 00 5 bales cotton, 1500 lbs., 10 cts. 15 00 Note against Thomas Pratt for 3 boxes sugar, 1200 lbs., 9 cts. 108 00 $300, dated July 1, 1836, and 4 casks nails, 1600 lbs., 6 cts. 96 00 interest.................... 309 00 40 bushels corn, at 6s......... 40 00 Bal. on ac't against John Silvers 5 81 6 tons hay, at $10........ 60 00 do. against Joel Stephenson 5 33 00 bs. iron. at 6 cts........... 24 00 1 horse.....75 00 90 gals. Molasses, at 40 cts..... 36 00. 1chaise............... 125 00 Homestead.................. 1000 00 Amt caried up......$567 34 - 2140 48 Inventory taken Jan. 1, 1837... 1950 75 Ret gain in 6 months... 189 73 832 RECEIPTS. NOTXES, REZZX:XI!S, ETC. NOTES. (1.) Orford, August 10, 1837. For value received, I promise to pay to John True, or order, seventy-five dollars fifty cents, on demand, with interest. JOSEPH DENMAN. Attest: JOEL TRUSTY. (2.) Concord, July 4, 1837. For value received, I promise to pay to James Doughty, or bearer, ten dollars thirty-four cents, six months after date. JOHN MORSE. (3.) [By two persons.a New Haven, Oct. 6, 1837. For value received, we jointly and severally promise to pay to S. T., or order, seventeen dollars and eighty-eight cents, on demand, with interest. ALONZO FONTAINELLE, JAMES WHITEHEAD. Attest: TIMOTHY TRUSTY. For Bank Notes, see "Discount," page 232. RECEIPTS. (1.) Norwich, June 7, 1837. Received of Mr. Nicholas Jewett, five dollars, in full of all accounts. HENRY SLOCUM. (2.) RECEIPT FOR AN ENDORSEMENT ON A NOTE. New York, Sept. 9, 1837. Received of Mr. John Hadlry, (by the hand of James True,) twenty-five dollars fifty cents, which is endorsed on his note of May 6, 1836. PETER TRUSTY. ORDERS. —-FORM OF A BOND. 323 (3.) RECEIPT FOR MONEY RECEIYED ON ACCOUNT. Mount Holly, Dec. 9, 1837. Received of Mr. John Van Dyke, twenty dollars on account. THOMAS BEAN. (4.) RECEIPT FOR INTEREST DUE ON A BOND. Received, this fourteenth day of May, of Mr. S. W., the sum of six dollars, in full of one year's interest of 100 dollars due to me on the 16th day of April last, on bond from the said S. W. By me, C. B. ORDERS. (1.) Mr. Joel M'Knight: SirFor value received, pay to O. S., ten dollars, and place the same to my account. SUEL RYNO. Hooksett, Sept. 7,1837. (2.) Lowell, Oct. 10, 1837. SIR: —For value received, pay S. O. twenty cents, and this, with your receipt, shall be your discharge from me. To Mr. Daniel Holden. JUSTUS PRUDES. FORM OF A BOND. Know all men by these Presents, That I, [we] A. B., of C., in the county of R., and state of New HIampshire, gentleman, [and C. D., of, &c.] am [are] held and firmly bound to E. F., of said C., yeoman, [and G. H1. of, &c.,] in the sum of one hundred dollars, to be paid to the said E. F. [and G. H., or either of them,] or his [their] certain attorney, executors, administrators, or assigns, to which payment, well and truly to be made, I [we jointly and severally] bind myself, [ourselves,] my [our] heirs, executors, and administrators, firmly by these presents. Sealed with my [our] seal [s], and dated the tenth day of May, A.-D. 1837. With a condition to pay money. The condition of this Obligation is such, that if the said A. B., [C. D., &c., or either of them,] his [their] heirs, executors, and administrators, do and shall well and truly pay, or cause to be paid, to 324 FORM OF A RELEASE. the said E. F., [G. H., or either of them,] his [their] executors, ad. ministrators, or assigns, the sum of fifty dollars and interest, on or before the tenth day of May next; then this obligation to be void, otherwise in force. [Signed and sealed as the preceding forms.] GENERAL FORM OF AN AGREEMENT. Articles of Agreement, indented, made, and concluded the tenth day of May, in the year of our Lord one thousand eight hundred and thirty-seven, (or A. D. 1837,) between A. B., of C., in the county of M., yeoman, of the one part, and C. D., of said C., husbandman, on the other part. The said A. B., for the consideration hereafter mentioned, doth hereby covenant and agree, that, etc. And the said C. D. doth hereby covenant and agree, etc. In testimony whereof they have hereunto interchangeably set their hands and seals, the day and year above [or first above] written. [Signed and sealed as the preceding forms.] 4FORM OF A WARRANT OF ATTORNEY. TO CONFESS JUDGMENT.'To A. B., Esquire, of, etc., an attorney of court of -, to be holden at ~, on the - day of —, or to any other attorney of said court. This is to authorize you, or any of you, to appear for me, E. F., of, etc., in the said court, or any other subsequent term, at the suit of G. H., of, etc., and by non sum informatus, nil dicit, or otherwise, confess judgment against me unto him, the said G. H., in an action of debt for one hundred dollars, and costs of suit; and for your, or either of your so doing, this shall be your warrant. In witness whereof, I have hereunto set my hand and seal, this tenth day of June. BRIEF FORM OF A RELEASE. May 20, 1837. I, A. B., do hereby release to E. E. all suits, romises, covenants, and demands, which I have or can claim against im. [Signed, sealed, and witnessed, as other instruments.] OBs.-By a release of all demands are barred all rights and titles to lands, warrants, debts, duties, actions, judgments, and executions, and all contracts except those which are to be performed on a future contingency. By a release of all covenants and promises, are released all such covenants and contracts as are not released by the woril de. mand, etc.