UC-NRLF *B Eh2 521 ifll v.. .Y r; JiL.^ IvIBRARY OF THE University of California. l6. &. MuAA.^iAj k^A. Received ~7vcan^ > -cSp-. ... /^"f^^ Accession No. ^ 9/^..X4'^- '^^^^ ^"^ • .:■#■■ ..%■ » M. ADAMS'S NEW ARITHMETIC. IN WHICH THE PRINCIPLES OF OPERATING BY NUMBERS AR^ AKAX.YTZOAZ.Z.Y ZSXPX.AXNSD, AND SYNTKETXCALZ.V APPI.IED; THUS COMBINING THE ADVANTAGES TO BE DERIVED BOTH FROM THE INDUCTIVE AND SYNTHETIC MODE OF INSTRUCTING: yHE WHOLE MADE FAMILIAR BY A GREAT VARIETY OF USEFUL AND INTER- ESTING EXAMPLES, CALCULATED AT ONCE TO ENGAGE THE PUPIL IN THE STUDY, AND TO GIVE HIM A -FJ^LL KNOWLEDGE OF FIGURES IN THEIR j!^l>^ICATION TO ALL THE PRAC- X TICAL PURPOSES OF LIFE. DESIGNED FOR THE USE OF SCHOOLS AND ACADEMIES IN THE UNITED STATES. BY DANIEL ADAMS, M AUTHOR OP THE SCHOLAR'S ARITHMETIC; SCHOOL GEQGRAPHT^ &C, ItEENE, PUBLISHED BY JOHN PRENTISS 1828 DISTRICT OF NEW-HAMPSHIRE. District Clerk's Office. Be it remembered, That on the eighteenth day of September, A. D. 1827, in the fifty-second year of the Independence of the United States of America, Daniel Adams, of said district, has deposited in this oflRce the title of a book, the right whereof he claims as author, in the words following, to wit : " Arithmetic, in which the Principles of operating by Numbers are analytically ex • plained, and synthetically applied : tlius combining the Advantages to be derived both from the inductive and synthetic Mode of instructing : the whole made familiar by a great Variety of useful and interesting Examples, calculated at once to engage the Pupil in the Study, and to give him a full Knowledge of Figures in their Application to all the practical Purposes of Life. Designed for tl>e Use of Schools and Academies in the United States. By Dakiel Adams, M. D. Author of the Scholar's Arithmetic, School Geog- raphy, &c." In conformity to the act of Congress of the United States, entitled, " An Act for tho encouragement of learning, by securing the copies of maps, charts, and books, to the authors and proprietors of such copies during the times therein mentioned ;" and also to an act, entitled, "An Act supplementary to an act for the encourage- ment of learning, by securing the copies of maps, charts, and books, to the authors and proprietors of such copies during the times therein rnentioned ; and extending the bene- fits thereof to the arts of designing, engraving and etching historical and other prints." CHARLES W. CUTTER, Clerk of the D istrict of JVevj-Hampshire. A true copy. Attest, C. W. CUTTER, Clerk. Stereotyped at the fitereotyped at the Boston Type and Stereotype Foundry, A3 There aro two methods of teaching, — the synthetic and the analytic. In the synthetic method, the pupil is first presented with a general view of the science he is studying, and afterwards with the particulars of which it consists. The analytic method reverses this order: the pupil is first presented with the particulars, from which he is led, by certain natural and easy gradations, to those views wliich are more general and comprehensive. The Scholar's Arithmetic, published in 1801, is synthetic. If that is a. fault of the work, it is a fault of the times in which it appeared. The analytic or inductive method of teaching, as now applied to ele- mentary mstruction, is among the improvements of later years. Its introduction is ascribed to Pestalozzt, a distinguished teacher in Switzerland. It has been applied to arithmetic, with great ingenuity, by Mr. Colburn, in our own country. The analytic is unquestionably the best method of acquiring know ledge ; the synthetic is the best metliod of recapitulating, or remeicing it. In a treatise designed for school education, both methods are use- ful. Such is the plan of the present undertaking, v/hich the author, occupied as he is with other objects and pursuits, would willingly have forborne, but that, the demand for the Scholar's Arithmetic still con- tinuing, an obligation, incurred by long-continued and extended pa- tronage, did not allow him to decline the labour of a revisal, which should adapt it to the present more enlightened views of. teaching this science in our schools. In doing this, however, it has been necessary to make it a new work. In the execution of this design, an analysis of each rule is first given, containing a familiar explanation of its various principles ; after- which follows 3, synthesis of these principles, with questions in form of a sup- plement. Nothing is taught dogmatically /'no technical term is used till it has first been defined, nor any principle inculcated without a pre- vious developement of its truth ; and the pupil is made to understand the reason of each process as he proceeds. The examples under each rule are mostly of a practical nature, be- ginning with those that are very easy, and gradually advancing to those more difficult, till one is introduced containing larger numbers, and which is not easily solved in the mind ; then, in a plain, familiar manner, the pupil is shown how the solution may be facilitated by figures. In this way he is made to see at once their use and their ap- plication. At the close of the fundamental rules, it has been thought advisable to collect into one clear view the distinguishing properties of those rules, and to give a number of examples involving one or more of them. These exercises will prepare the pupil more readily to understand tho 4 PREFACE. application of these to the succeeding rules ; and, besides, will serve to interest him in the science, since he will find himseJi* able, by tho application of a very few principles, to solve many curious questions. The arrangement of the subjects is that, which to the author has appeared most natural, and may be seen by the Index. Fractions have received all that consideration which their importance demands. The principles of a rule called Practice are exhibited, but its detail of cases is omitted, as unnecessary since the adoption and general use of federal money. The Rule of Three^ or Proportion, is retained, and the solu- tion of questions involving the principles of proportion, by analysis^ is distinctly shown. The articles Alligation, Arithmetical and GeorftBtrical Progression^ Annuities and Permutation, were prepared by Mr. Ira Young, a mem- ber of Dartmouth College, from whose knowledge of the subject, and experience in teaching, I have derived important aid in other parts of the work. The numerical paragraphs are chiefly for the purpose of reference : these references the pupil should not be allowed to neglect. His at- tention also ought to be particularly directed, by his instructer, to the illustration of each particular principle, from which general rules are deduced : for this purpose, recitations by classes ought to be instituted in every school where arithmetic is taught. The supplements to the rules, and the geometrical demonstrations of the extraction of the square and cube roots, are the only traits of the old work preserved in the new. DANIEL ADAMS. Mont Vernon, (N. H.) Sept. 29, 1827. i^®a^. SIMPLE NUMBERS. p^^^ Numeraliou and Notation, 7 Addition, 12 Subtraction, 19 Multiplication, 26 Division, 37 Fractions arise from Division, 42 Miscellaneous Questions, involving- the Principles of the preceding Rules, *" 62 COMPOUND NUMBERS; Different Denominations, 56 Federal Money, - 57 , to find the Value of Articles sold by the 100, or 1000, . . 64. , Bills* of Goods sold, ..." 68 Reduction, 69 TablesofMoney. Weight, Measure, &c 69—82 Addition of Compound NumlDcrs, 85 Subtraction, • • • 89 Multiplication and Division, 93 FRACTIONS. Common, or Vulgar. Their Notation, 101 Proper, Improper, &c 102 To change an Improper Fraction to a Whole or Mixed Number, . . . 103 a Mixed Number to an Improper Fraction, 104 To reduce a Fraction to its lowest Terms, . 105 Greatest common Divisor, how found, 106 To divide a Fraction by a Whole Number a||wo ways, l(M To multiply a Fraction by a Whole Numbe'r; two ways, 110 a Whole Number by a Fraction, 112 one Fraction by another, 113 General Rule for the Multiplication of Fractions, 114 To divide a Whole Number by a Fraction, 1 15 one Fraction by another, 117 General Rule for the Division of Fractions, 118 Addition and Subtraction of Fractions, 119 Common Denominator, how found, 120 Least Common Multiple, how found, 121 Rule for the Addition and Subtraction of Fractions, 124 Reduction of Fractions, 124 Decimal. Their Notation, 132 Addition and Subtraction of Decimal Fractions, 136 Multiplication of Decimal Fractions, , 137 Division of Decimal Fractions, . . . 139 To reduce Vulgar to Decimal Fractions, 14ffi Reduction of Decimal Fractions, 146 To reduce Shillings, &c^to the Decimal of a Pound, by Inspection, . . 146 the three first Decimals of a Pound to Shillings, &c., by Inspectipn, 137^ 6 INDEXr Reduction of Currencies; .161 To reduce Endish, &c. Currencies to Federal Money, ...>... 153 Federal Money to the Currencies of England, &c 154 one Currency to the Par of another Currency, 155 Interest, 156 Time, Rate per cent-, and Amount given, to find the Principal, .... 164 Time, Rate per cent., and Interest given, to find the Principal, .... 165 Principal, Interest, and Time given, to find the Rate per cent., .... 166 Principal, Rate per cent., and Interest given, to find the Time, .... 167 To find the Interest on Notes, Bonds, dtc, wlien partial Payments have been made, 168 Compound Interest, 169 by Progression, 229 Equation of Payments, 176 Ratio, or the Relation of Numbers, 177 Proportion, or Single Rule of Three, 179 Same Questions, solved by Analysis, U 65, ex. 1 — 20. Compound Proportion, or Double Rule of Three, 187 Fellowship, 192 Taxes, Method of assessing, 195 Alligation, . . . - 197 Duodecimals, 201 Scale for taking Dimensions in Feet and Decimals of a Foot, 204 Involution, 205 1 Evolution, 207 Extraction of the Square Root, 207 Application and Use of the Square Root, see Supplement, . . . 212 Extraction of the Cube Root, 215 Application and Use of the Cube Root, see Supplement, Arithmetical Progression, . . . 222 Annuities at Compound Interest, 231 Practice, H 29, ex. 10—19. H 43. Insurance, IT 82. Buying and Selling Stocks, ![ 82. Geometrical Progression, . . . 225 Permutation, 237 Commission, U 82 ; TI>85, ex. 5, 6. Loss and Gain, TI 82 j Tl 88, ex. 1—8. Discount, IT 85, ex. 6 — 11. MISCELLANEOUS EXAMPLES. Barter, ex. 21—32. 1 Position, ex. 89—108. To find the Area of a Square or P^llelogram, ex. 148—154. of a Triangle, ex^lPo — 159. Having the Diameter of a Circle, to find the Circuniference j or, having the Circumference, to find the Diameter, ex. 171 — 175. To find the Area of a Circle, ex. 176—179. of a Globe, ex. 180, 181. To find the Solid Contents of a Globe, ex. 182—184. of a Cylinder, ex. 185—187. of a Pyraniid, or Cone, ex. 188, 189. of any Irregular Body, ex. 202, 203. Gauging, ex. 190, 191. | Mechanical Powers, ex. 192—201. Forms of Notes, Bonds, Receipts, and Orders, 259 Book-Keeping, 263 Ams^mii^ii®< NUDXHRATION. IT 1. ^ A SINGLE or individual thing is called a wiit^ unity^ or one ;r one and one more are called two ; two and one more are called three ; three and one more are called four ; four and one more are called five ; five and one more are called six; six and one more are called seven; seven and one more are called eight ; eight and one more are called nine ; nine and one more are called ten^ &c. These terms, which are expressions for quantities, are called numbers, < There are two methods ^of expressing numbers shorter than writing them out in words ; one called the Roman method by letters,* and the other the Arabic method by figures. The latter is that in general use.^ In the Arabic method, the nine first numbers have each an appropriate character to represent them. Thus, * In the Roman method by letters,! represents one; W,Jive: X, ten; \^, fifty ; C, one hundred ; J), fire hundred ; and M, one thousand. As often as any letter is repeated, so many times its value is repeated, unless it be a letter representing a less number placed before one representing- a greater ; then the less number is taken from the greater 5 thus, IV represents /own^ iX, nine, , &.C., as will be seen in the following TABLE. One I. Ninety LXXXX. or XC. Two 11. One hundred C. Three III. Two hundred cc. Four nil. or IV. Three hundred ccc. Five V. Four hundred cccc. Six VI. Five hundred D. or 10 * Seven VII. Six hundred DC. Eight VIII. Seven hundred DCC. Nine vim. or IX. Eight hundred DCCC. Ten X. Nine huixlred DCCCC. Twenty XX. One thousand M. or ClO.t Thirty XXX. Five thousand lOO.orV.t Forty Fifty XXXX. or XL. Ten thousand CCIOO.orX. L. Fifty thousand 1003. Sixty LX. Hundred thousand CCCIOOO.orC: Seventy LXX. One million M. Eighty LXXX. Two million MM. * Jo is used instead of D to represent five hundred, and ijpr every additional au Dexed at the right hand, the number is increased ten times. t CIO is used to represent one thousand, and for every C and put at each end, tb« number is increased ten times. X A line over any number increases its value one thousand times. 8 NUMERATION. V 1. A unity unity y or one, is represented by this character, Two Three Four ......... Five Six . . . • . Seven ......... Eight Nin^e ......... Ten has no appropriate character to represent it ; but is considered as forming a unit of a second or higher order, consisting of tens, represented by the same character (1) as a unit of the first or lower order, but is written in the second place from the right hand, that is, on the left hand side of units ; and as, in this case, there are no units to be written with it, we write, in the place of units, a cipher, (0,) which of itself signifies nothing ; thus. Ten One ten and one unit are called One ten and two units are called One ten and three units are called One ten and four units are called One ten and ^ve units are called One ten and six units are called One ten and seven units are called One ten and eight units are called One ten and nine units are called Two tens are called Three tens are called Four tens are called Five tens are called Six tens are called Seven tens are called Eight tens are called Nine tens are called Eleven Twelve Thirteen Fourteen Fifteen Sixteen Seventeen Eighteen Nineteen Twenty Thirty Forty Fifty Sixty Seventy Eighty Ninety Ten tens are called a hundred^ which forms a unit of a still higher order, consisting of Atmrfrctfe, represented by the same character (1) as a unit of each of the foregoing orders, but is written one place further toward the left hand, that is, on the left hand side of tens ; thus, . . . . One hundred One hundred, one ten, and one unit, are called One hundred and eleven 1. 2. 3. 4. 6. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15 16, 17. 18. 19. 20. 30. 40. 50. 60. 70. 80. 90. 100. lit,. IT 2) 3. NUMERATION. 9 IT 2« There are three hundred sixty-five days in a year. In this number are contained all the orders now described, viz. units, tens, and hundreds. Let it be recollected, units occupy the first place on the right hand ; tens^ the second place from the right hand; hundreds^ the third place. This number may now be decomposed, that is, separated into parts^ exhibiting each order by itself, as follows : — The highest order, or hundreds, are three, represented by this character, 3 ; but, that it may be made to occupy the third place, count- ing from the right hand, it must be followed by two ciphers, thus, 300, (three hundred.) The next lower order, or tens, are six, (six tens are sixty,) represented by this character, 6 ; but, that.it may occupy the second place, which is the place of tens, it must be followed by one cipher, thus, 60, (sixty.) The lowest order, or units, are five^, represented by a single character, thus, 5, (five.) We may now combine all these parts together, first writing down the five units for the right hand figure, thus, 5 ; then the six tens (60) on the left hand of the units, thus, 65 ; then the three hundreds (300) on the left hand of the six tens, thus, 365, which number, so written, may be read three hundred, six tens, and five units ; or, as is more usual, three hundred and sixty-five. yi 3. Hence it appears, that figures have a different value according to the place they occwpry, counting from the right hand towards the left, in Eh ID Take for example the number 3 3 3, made by the same figure three times repeated. The 3 on the right hand, or in the first place, signifies 3 units ; the same figure, in the second place, signifies 3 tens, or thirty ; its value is now increased ten times. Again, the same figure, in the third place, signi- fies neither 3 units, nor 3 tens, but 3 hundreds, which is ten times the value of the same figure in the place immediately preceding, that is, in the place of tens ; and this is a funda- mental law in notation,Uhat a removal of one place towards the left increases the value of a figure ten times. ^ j^Ten hundred make a thousand, or a unit of the fourth order. Then follow tens and hundreds of thousands, in the fiame manner as tens and hundreds of units. To thousands 10 NUMERATION. IT S. succeed millions^ billions, &c., to each of which, as to units and to thousands, are appropriated three places,* as exhi- bited in the following examples : O . n3 "^ § 2 s § 2 S V ^ o O* H pq S H tii ^^ n3 T3 ^O '^ Qj q^ OJ ty (y .^ I §.^ I g.^ I g.5 ^ g.5 I g.^ Example 1st. 3 174592 837463 512 Example 2d. 3, 1 7 4, 5 9 2, 8 3 7, 4 6 3, 5 1 2, **^ rt *t? ^ "^ «*-•.**:; •TS O '^ . ^ p^ m^ 'TIS 'T3 . •a.2^ '^.2 « 'S.o ^ .2.0 g .2.0 § -2,0 Phoj'O &. The reading of any number set down in figures, is called Numeration. After being able to read correctly all the numbers in the foregoing table, the pupil may proceed to express the fol- lowing numbers by figures : 1. Seventy-six. 2. Eight hundred and seven. 3. Twelve hundred, (that is, one thousand and two hun- dred.) 12 ADDITION OF SIMPLE NUMBERS. IT 3, 4 4. Eighteen hundred. 5. Twenty-seven hundred and ninet^n. ,6. Forty-nine hundred and sixty. 7. Ninety-two thousand and forty-five. 8. One hundred thousand. 9. Two millions, eighty thousands, and seven hundreds. 10> One hundred millions, one hundred thousand, one hundred and one. 11. Fifty-two millions, six thousand, and twenty. 12. Six billions, seven millions, eight thousand, and nine hundred. 13. Ninety-four billions, eighteen thousand, one hundred and seventeen. 14. One hundred thirty-two billions, two hundred millions, and nine. 15. Five trillions, sixty billions, twelve millions, and ten thousand. 16. Seven hundred trillions, eighty-six billions, and seven millions. ADBXTXON OF SIMPLE NUMBERS. IT 4. 1. James had 5 peaches, his mother gave * "' . Z peaches more ; how many peaches had he then ? 2. John bought a slate for 25 cents, and a book for eight cents ; how many cents did he give for both ? 3. Peter bought a waggon for 36 cents, and sold it so as to gain 9 cents ; how many cents did he get for it ? 4. Frank gave 15 walnuts to one boy, 8 to another, and had 7 left ; how many walnuts had he at first ? 5. A man bought a chaise for 54 dollars ; he expended 8 dollars in repairs, and then sold it so as to gain 5 dollars ; how many dollars did he get for the chaise ? 6. A man bought 3 cows ; for the first he gave 9 dollars, for the second he gave 12 dollars, and for the other he gave 10 dollars ; how many dollars did he give for all the cows ? 7. Samuel bought an orange for 8 cents, a book for 17 cents, a knife for 20 cents, and some walnuts for 4 cents j how many cents did he spend ? IT 4, ADDITION OF aiMPLE NUMBERS. 13 8. A man had 3 calves worth 2 dollars each, 4 calves worth 3 dollars each, and 7 calves worth 5 dollars each ; -^liow many calves had he ? 9. A man sold a cow for 16 dollars, some com for 20 dol- lars, wheat for 25 dollars, and butter for 5 dollars ; how many dollars must he receive ? The putting together two or more numbers, (as iri the foregoing examples,) so as to make one whole number^ is called Addition^ and the whol^ number iscalled the smuj or amount,- 10. One man owes me 5 dollars, another owes me 6 dollars, another 8 dollars, another 14 dollars, and another 3 dollars ; what is the amount due to me? 11. What is the amount of 4, 3, 7, 2, 8, and 9 dollars? 12. In a certain school 9 study grammar, 15 study arith- metic, 20 attend to writing, and 12 study geography; what is the whole number of scholars ? Signs. (, A cross, -(-, one line horizontal and the other per- pendicular, is the sign of addition. It shows tliat numbers, with this sign between them, are to be added together. It is sometimes read plus^ which is a Latin word signifying morft.J \ Two parallel, horizontal lines, -n, are the sign of equality. ; It signifies that the number before it is equal to the number after it. Thus, 5 -j- 3 r= 8 is read 5 and 3 are 8 ; or, 5 plus (that is, more) 3 is equal to 8. In this manner let the pupil be instructed ta.commit the following ADDITION TABIiX;. 2 + = 2 2 + 1 = 3 24-2=: 4 2 + 3 = 6 2 + 4 = 6 2 + 5 = 7 2 + 6 = 8 2 + 7 = 9 2 + 8 = 10 2 + 9 = 11 B 3 + = 3 3 + 1 = 4 3 + 2 = 5 3 + 3 = a 3 + 4 = 7 3 + 5 = 8 3 + 6 = 9 3 + 7^ 10 3 + 8 = 11 3 + 9 = 12 4 + 0= 4 f 5- - 4+1= 5 5n - 4 + 2= 6 6- . 4 + 3= 7 5- « 4+4= 8 5- -. 4 + 5= 9 5- « 4 + 6 = 10 5- - 4 + 7=11 5- - 4 + 8=12 5- . 4 + 9 = 13 6- - 0= 6 1=6 2= 7 3= 8 4=9 5=10 6 = 11 7= Ilk 8=13 9 = 14 14 ADDITION OF SIMPIiE NUMBERS. ^4,6 ADDITION TABLE— CONTINUED. 6+0= 6 7 + 0z= 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 3 + 8=14 7 + 8=15 8 + S = 16 d + S^zirf 6 + 9 = 15 7 + 9=16 8 + 9== 17 9 + 9 = 18 5 + 9 z= how many ? 8 + 7 = how many ? 4 + 3 + 2 = how many ? 6+4 + 5 =: how many? 2 + + 4 + 6 zz: how many ? 7+1 + + 8 ~ how many ? 3 + + 9 + 5 = how many ? 9 + 2 + 6 + 4 + 5 rz how many ? 1+3 + 5 + 7 + 8 zzz how many ? 1+2 + 3+^ + 5 + 6 = how many ? 8 + 9 + 0-1-2 + 4 + 5 = how many ? 6 + 2 + 5 + + 8 + 3 = how many ? IT 5. When the numbers to be added aveismaU^ the addi- tion is readily performed in the mind; but it will frequently be more convenient, and even necessary, to write the num- bers down before adding them, 13. Harry had 43 cents, his father gave him 25 cents mare ; how many cents had he then ? One of these numbers contains 4 tens and 3 units. The other nimiber contains 2 tens and 5 units. To unite these two numbers together into one, write them down one under the other, placing the units of one number directly under umis of the other, and the tens of one number directly under tens of ihe o4her, thus : 43 cejits. Having written the numbers in this znan* 25 cents, ner, draw a line underneath. X 5» ADDITION OF SIMPLE NUMBERS. 15 43 cents. ^^ ^^^^ begin at the right hand^ and add 25 cents. ^^® ^ units of the lower number to the 3 — * units of the upper number, making 8 units, 8 which we set down in unit's place. ) ^ We then proceed to the next column, and 43 cents, add the 2 tens of the lower number to the 25 cents. 4 tens of the upper number, making 6 tens, . "" ^ or 60, which we set down in ten's place, ^715. 68 cents. ^^^ ^^^^ ^,^^^1, -^ ^^^^ It now appears t!.at Harry's whole number of cents is 6 tens and 8 units, or 68 cents ; that is, 43 -f- 25 == 68. 14. A farmer bought a chaise for 210 dollars, a horse for 70 dollars, and a saddle for 9 dollars; what was the whole amount ? Write the numbers as before directed, with units under units, tens under tens, &c. OPERATION. Chaise^ 210 dollars. Add as before. The units will Horsey 70 dollars. be 9, the tens 8, and the hundreds Saddle^ 9 dollars. 2 ; that is, 2 10 -j- '''0 + 9 = 289. Ajiswer, 289 dollars. After the same manner are performed the following ex- amples : 15. A man had 15 sheep in one pasture, 20 in another pasture, acd 143 in another ; how mauy sheep had he in the three pr,stures ? 15 -|- 20 + 143 z=z how many ? 16. A man has three farms, one containing 500 acres, another 213 acres, and another 76 acres ; how many acres in the three farms ? 500 -|- 213 + '''6 = how many? 17. Bought a farm for 2316 dollars, and afterwards sold it so as to gain 550 dollars; what did I sell the farm for? 2316 + 550 =1 how many ? Hitherto the amount of any one column, when added up, has not exceeded 9 ; consequently has been expressed by a single figure. But it will frequently happen that the amount of a single column will exceed 9, requiring /if?o or more figures to express it. 18. There are three bags of money. The first contains i76 dollars, the second, 653 dollars, the third, 524 dollarsit vhat is the amount contained in all the bags ? 16 ADDITION OF SIMPLE NUMBERS. % 6, OPERATION. Writing down tlie numbers as First bag, 876 already directed, we begin with the Second bag, 653 Yight hand, or unit column, and Third bag, ^524 fjj^^j the amount to be 13, that is, Amount, 2053 ^^ units and 1 ten. Setting down the 3 units, or right hand figure, in unit's place, directly under the column, wc reserve the 1 ten, or left hand figure, to be added with the other teits, in the next column, saying, 1, which we reserved, to 2 makes 3, and 5 are 8, and 7 are 15, whi^h is 5 units of its 0W71 order, and 1 unit of the next higher order, that is, 5 tens and 1 hundred. Setting down the 5 tens, or right hand Tjgure, directly under the column of tens, we reserve the left hand figure , or 1 hundred, to be added in the column of hun- dreds^ saying, 1 to 5 is 6, and 6 are 12, and 8 are 20, which behig the last column, we set down the whole number, writing the 0, or right hand figure, directly under the column, and carrying forward the 2, or left hand figure, to tlie next place, or place of thousands. Wherefore, we find the whole amount of money contained in the three bags to be 2053 doJlors, — the answer. Proof. We may reverse the order, and, beginning at the top, add the figures downward. If the two results are alike, the work is supposed to be right. From the examples and illustrations now given, we de- rive the following RUIiE. I. Write the numbers to be added, one under another^ placing units under units, tens undfer tens, &c., and draw a line underneath. II. Begin at the right hand or unit column, and add to- gether all the figures contained in that column : if the amount does not exceed 9, write it under the column ; but if the amount exceed 9, so that it shall require two or more figures to express it, write down the unit figure only under the column ; the figure or figures to the left hand of units, being tens, are so many units of the next higher order, which, being reserved, must be carried forward, and added to the first figure in the next column. III. Add each succeeding column in the same manner, aud set down the whole amount at the last column IT 5. ADDITION OF SlJViPLE NUMBERC. 17 EXAMPLES FOR PRACTICE. 19. A man bonght four loads of hay; one load weighed 1817 pounds, another weighed 1950 pounds, another 2156 pounds, and another 2210 pounds; what was the amount of hay purchased ? 20. A person owes A 100 dollars, B 2160 dollars, C 785 dollars JD 92 dollars ; w^hat is the amount of his debts ? 21. A farmer raised in one year 1200 bushels of wheat, 850 bushels of Indian corn, 1000 bushels of oats, 1086 bush- els of barley, and 74 bushels of pease ; what was the whole amount ? Ans, 4210. 22. St. Paul's Cathedral, in London, cost 800,000 pounds sterling ; the Royal Exchange 80,000 pounds ; the Mansion- House 40,000 pounds ; Black Friars Bridge 152,840 pounds; Westminster Bridge 389,000 pounds, and the Monument 13,000 pounds ; what is the amount of these sums ? Ans, 1,474,840 pounds. 23. At the census in 1820, the number of inhabitants in the New England States was as follows : — Maine, 298,335 ; New Hampshire, 244,161 ; Vermont, 235,764; Massachu- setts, 253,287 ; Rhode Island, 83,059 ; Connecticut, 275,248 ; what was the whole number of inhabitants, at thot time, in those States ? Ans. 1,389,854. 24. From the creation to the departure of the Israelites from Egypt was 2513 years ; to the siege of Troy, 307 years more; to the building of Solomon's Temple, 180 years; to the building of Rome, 251 years; to the expulsion of the kings from Rome, 244 years ; to the destruction oi" Carthage, 363 years ; to the death of Julius Ca^sar^ 102 years ; to the Christian era, 44 years ; required the time from the Crea- tion to the Christian era. Ans. 4004 years. 25. 26. 2S63705421061 43675830214(53 3107429 3 15638 17523497136 2 62530 3 4792 6081275306217 247135 5652174630128 8673 8703263472013 B^ 18 SUFFLEMiCNT TO NUMERATION ANI) ADDITION. tF 6. 27. 28. 6364207681023 90237646821S5 2812345672948 283496732670 8 6057042087094 9306342167 33 1 3162835906718 2365478024369 76 042 86 53 7892 805 06 70 80 900 29. What is the amount of 46723, 6742, and 986 dollars.' 30, A man has three orchards ; in the iirst there are 140 it<^^ that bear apples, and 64 trees that bear peaches j in ifie second, 234 trees bear apples, and 73 bear cherries ; in the tlurd, 47 trees bear plums, 36 bear pears, and 25 bear cherries ; how many trees in all the orchards ? rO NUMERATION AND ADDITION. QUESTIONS. 1. What is a single or individual thing called? 2. What is notation ^ 3. Wliat are the methods of notation now in use ? 4. llow many are the Arabic characters or figures r 5. What is numeration ? 6. What is a fundamental law in notation ? 7. What is addition ? 8. What is the rule for addition ? 9. What is the result, or number sought. called? 10. W^hat is the sign of addition? 11. of equality ? 12. How is addition proved ? EXERCISES. 1. Washington was born in the year of our Lord 1732; he was 67 years old when he died ; in what year of our Lord did he die ? 2. The invasion of Greece by Xerxes took place 481 years before Christ ; how long ago is that this current year 1827? 3. There are two r.umbers, the less number is 8671, the difference between the numbers is 597 j what is the greater number. f By 6* SUBTRACTION OF SIMPLE NCTMBERS. 1 9 4. A man borrowed a sum of money, and paid in part (i84 dollars ; the sum left unpaid was 876 dollars ; what was the sum borrowed ? 5. There are four numbers, the first 317, the second 812, the third 1350, and the fourth as much as the other three j what is the sum of them all ? 6. A gentleman left his daughter 16 thousand, 16 hun- dred and 16 dollars ; he left his son 1800 more than his daughter ; wha^ was his son's portion, and what was the wnount of the whole estate - a S Son's portion, 19,416. ^^*' I Whole estate, 37,032. 7. A man, at his death, left his estate to his four children, who, after paying debts to the amount of 1476 dollars, received 4768 dollars each; how much was the whole estate? Ans. 20548. 8. A man bought four hogs, each weighing 375 pounds ; how much did they all weigh? Am. 1500, 9. The fore quarters of an ox weigh one hundred and eight pounds each, the hind quarters w^eigh one hundred and twenty-four pounds each, the hide seventy-six pounds, and the tallow sixty pounds ; what is tlie whole weight of the ox ? Ans, 600 10. A man, being asked his age, said he was thirty-four years old when his eldest son was born, who was then fif* teen years cf age ; what vv^as the age of the father? 11. A man sold two cows fur sixteen dollars each, tweiv- ty bushels of corn for twelve dollars, and one hundred pounds of tallow for eight dollars ; what was his due ? OF SIMPLE NUMBERS. t 6. 1. Charles, having 18 cents, bought a book, for which he gave 6 cents ; how many cents had he left ? 2. John had 12 apples ; he gave 5 of them to his brother j how many had he left ? 3. Peter played at marbles ; he had 23 when he began, but when he had done he had only 12 ; how many did be lose? 20 SUBTRACTION OF SIMPLE NUMBERS. ^ 6. 4. A man bought a cow for 17 dollars, and sold her again for 22 dollars ; how many dollars did he gain ? ^ 5. Charles is 9 years old, and Andrew is 13 ; what is the difference in their ages ? 6. A man borrowed 50 dollars, and paid all but 18; how many dollars did he pay? that is, take 18 from 50, and how many would there be left ? 7. John bought a book and slate for 33 cents ; he gave 8 cents for the book ; what did the slate cost him ? 8. Peter bought a waggon for 36 cents, and sold it for 45 cents; how many cents did he gain by the bargain ? 9. Peter sold a waggon for 45 cents, which was 9 cents more than he gave for it ; how many cents did he give for the waggon ? 10. A boy, being asked how old he was, said th:it he wa* 25 years younger than his father, whose age was 33 years ; how old was the boy ? The taking of a less number from a greater (as in the fbregoing examples) is called Subtraction.^ The greater ^number is called the jninuend^ the less number the^ mibtrch hendj and what is left after subtraction is called the differ- ence, OT^ remainder. 11. If the minuend be 8, and the subtrahend 3, what is the difference or remainder ? Ans. 6. 12. If the subtrahend be 4, and the minuend 16, what is the remainder ? 13. Samuel bought a book for 20 cents ; he paid down 12 cents ; how many cents more must he pay ? Sign. ^ A short horizontal line,"^ — , is the sign of subtrac- tion. It is usually read minnSy which is a Latin word signi- fying less. It shows that the number after it is to be taken from the number befare it. Thus, 8 — 3 r= 5, is read 8 mW mis or less 3 is equal to 5 ; or, 3 from 8 leaves 5. The latter expression is to be used by the pupil in committing tbe following IT 6, 7. SUBTRACTION OF SIMPLE NU31BERS. 21 2 — 2 = 3 — 2 = 1 4 — 2 = 2 6 — 2 = 3 6 — 2 = 4 7 — 2 = 5 8 — 2 = 6 9 — 2 = 7 10 — 2 = 8 SUBTRACTION TABLE. 6—3 = 3 7 — 3 = 4 8 — 3 = 5 9 — 3 = 6 10 — 3 = 7 3 — 3 = 4 — 3=1 5 — 3 = 2 4 — 4 = 5 — 4=1 6 — 4 = 2 7 — 4 = 3 8 — 4 = 4 9 — 4 = 5 10 — 4 = 6 5 — 5 = 7 — 7 = 6 — 5=1 8 — 7 = 1 7 — 5 = 2 9 — 7 = 2 8 — 5 = 3 10 — 7 = 3 9—5 = 4 10 — 5 = 5 8 — S=iO 9 — 8 = 1 6 — 6 = 7 — 6 = 1 8 — 6 = 2 9 — 6 = 3 10 — 8 = 2 9 — 9 — 10 — 9 = 1 10—6 = 4 7 — 3 =: how many ? 8 — 5 =z how many ? 9 — 4tzz how many ? 12 — 3 =: how many ? 13 — 4 =: how many ? 18 — 7 zz: how many ? 28 — 7 =z how many ? 22 — 13 = how many ? 33 — 5 = how many ? 41 — 15 =: how many ? IT 7. When the numbers are small^ as in the foregoing examples, the taking of a less number from a greater is rea- dily done in the mind; but when the numbers are large^ the operation is most easily performed part at a time, and therefore it is necessary to write the numbers down before performing the g«- ^ration. 14. A farmer, having a flock of 237 sheep, lost 114 of them by disease ; how many had he left ? Here we have 4 units to be taken from 7 units, 1 ten i* be taken from 3 tens, and 1 hundred to be taken from 2 hundreds. It will therefore be most convenient to write the less number under the greater, observing, as in addition, to place units under units, tens under tens, &c. thus : We now begin with the OPERATION. From 237 the minuend^ Take 114 the subtrahend^ units, saying, 4 (units) from 7, (units,) and there remain 3, (units,) which we set down directly under the column in unit's place. Then, proceed 'ng to the next column, we say, 1 (ten) from 3, (tens,) and here remain 2, (tens,) which we set down in teii^s place* 123 the remainder. S2 SUBTRACTION OF SIMPLE NUMBERS. IH T Proceeding to the next column, we say, 1 (hundred) from 2, (hundreds,) and there remains 1, (hundred,) which we set do^vn in hundTeiTs place, and the work is done. It now ap* pears, that the number of sheep left was 123 ; that is, 237—114 = 123. After the same manner are performed the following ex- amples : 15. There are two farms; one is valued at 3750, and the other at 1500 dollars ; what is the difl'erence in the value of the two farms ? 16. A man's property is worth 8560 dollars, but he ha» debts to the amount of 3500 dollars ; what will remain aftei paying his debts ? 17. James, having 15 cents, bought a pen-knife, for which he gave 7 cents; how many cents had he left? OPERATION. 15 cents. A difficulty presents itself here ; for we 7 ceMs* cannot take 7 from 5 ; but we can take 7 — , ^ from 1 5, and there will remain 8. Scents left. 18. A man bought a horse for 85 dollars, and a cow for 27 dollars; what did the horse cost him more than the cow ? OPERATION. The same difficulty meets us here as in Horsey 85 the last example ; we cannot take 7 from CkfW, 27 5 ; but in the last example the larger num- — her consisted of 1 ten ap ^ 5 units, which Difference, 58 together make 15; we therefore took 7 from 15. Here we have 8 tens and 5 units. We can now, in the mind, suppose 1 ten taken from the 8 tens, which would leave 7 tens, and this 1 ten we can suppose joined to the 5 units, making 16. We can now take 7 from 15, as be- fore, and there will remain 8, which we set down. The taking of 1 ten out of 8 tens, and joining it with the 5 units, is called borrowing ten. Proceeding to the next higher o> der, or tens, we must consider the upper figure, 8, from which we borrowed, 1 less, calling it 7; then, taking 2 (tens) from 7, (tens,) there will remain 5, (tens,) which we set down, making the difference 58 dollars. Or, instead of making the upper figure 1 less, calling it 7, we may make the lower figure one jnore, calling it 3, and the result will be thesam&( Cor 3 from 8 leaves 5, the same as 2 from 7. t 7, 8. SUBTRACTION OF SIMPLE NUMBERS. 2S 19. A man borrowed 713 dollars, and paid 471 dollars; how many dollars did he then owe? 713 — 471= how many ? Ans. 242 dollars, 20. 1612-— 465 z= how many ? ^7W. 1147. 21. 43751 — 6782 zz: how many ? Ans. 36969. IT 8. CThe pupil will readily perceive, that subtraction is the reverse of addition.^ 22. A man bought 40 sheep, and sold 18 of them; how many had he left ? 40 — IS zz: how many ? Ans. 22 sheep. 23. A man sold 18 sheep, and had 22 left; how many had he at first ? 18 -{- 22 = how many ? Ans. 40. 24. A man bought a horse for 75 dollars, and a cow for 16 dollars ; what was the difference of the costs ? 75 — 16 iz: how many ? Reversed, 59 -f- 16 zz how many? 25. 114 — 103rz:howmany? Reversed, 11 + 103 zi: how many ? 26. 143 — 76 zn how many ? Reversed, 67 + 76 zz: how many ? Hence, subtraction may be proved by addition^ as in the foregoing examples, and-addition by subtraction, *To prove subtraction^ (;^ve may add the remainder to the mthtrahend^ and, if the work is right, the amount will be equal to the minuend. "^ ^Jb prove oMition^ we may subtract^ successively, from the amount, the several numbers which were added to pro- duce it, and, if the work is right, there will be no re- mainder) Thus 7 + 8 + 6 zz: 21 ; proof, 21 —- 6 izr 15, and 15 — 8 zz: 7, and 7 — 7 zz 0. From the remarks and illustrations now given, we deduce tlie following RULE. I. Write down the numbers, the less under the greater, E lacing units under units, tens under tens, &c. and draw a ne under them."\ II. Beginning with units, take successively each figure in the lower number from the figure over it, and write the re- mainder directly below. IllJ^JWhen the figure in the lower number exceeds the figure over it, suppose 10 to be added to the upper figure; but in this case we must add 1 to the lower figure in tha Ufizt column, before subtracting. This is called borrowing 10^^ 24 SUPPLEMENT TO SUBTRACTION. Tl 8, exampl.es for practice. 27. If a farm and the buildings on it be valued at, 10000, .nd the buildings alone be valued at 4567 dollars, what is the value of the land ? 28. The population of New England, at the census in 1800, was 1,232,454 ; in 1820 it was 1,659,854 ; what was the increase in 20 years ? 29. What is the diifereuce between 7,648,203 and 928,671 ? 30. How much must you add to 358,642 to make 1,487,945 ? 31. A man bought an estate for 13,682 dollars, and sold it again for 15,293 dollars j did he gain or lose by it? and how much ? 32. From 364,710,825,193 take 27,940,386,574. 33. From 831,025,403,270 take 651,308,604,782. 34. From 127,368,047,216,843 take 978,654,827,352. TO SUBTRACTION. ,' QUESTIONS. 1. What is subtractio7i? 2. What is the greater numher called ? 3. the less number ? 4. What is the resuk or answer called ? 5. What is the sign of subtraction ? 6. What is the rizZe ? 7. What is understood by horrowing ten ? 8. Of what is subtraction the reverse ? 9. How is subtraction proved ? 10. How is addition proved by sub- traction ? EXERCISES. 1. How long from the discovery of America by Colum- bus, in 1492, to the commencement of the Revolutionary war in 1775, which gained our Independence ? 2. Supposing a man to have been born in the year 1773, now old was he in 1827 ? 3. Supposing a man to have been 80 years old in the year 1826, in what year was he born ? 4. There are two numbers, whose difference is 8764 ; the greater number is 15687 j I demand the less ? f 8, SUPPLEMENT TO SUBTRACTION. 2b 6. What number is that which, taken from 3794, leaves 865 ? 6. What number is that to which if you add 789, it will become 6 J50 ? 7. In New York, by the census of 1S20, there were 123,706 inhabitants; in Boston, 43,940; liow many more inhabitants were then in New York than in Boston? 8. A man, possessing an estate of twelve thousand dollars, gave two thousand five hundred dollars to each of his two daughters, and the remainder to his son ; what was his son'» share ? 9. From seventeen million take fifty-six thousand, and what will remain ? 10. VVhat number, together with these three, viz. 1301, 2561, and 3120, will make ten thousand? 11. A man bought a horse for one hundred and fourteen dollars, and a chaise for one hundred and eighty-seven dol- lars ; how much more did he give for the chaise than for the horse ? 12. A man borrows 7 ten dollar bills and 3 one dollar bills, and pays at one time 4 ten dollar bills and 5 one dol- lar bills ; how many ten dollar bills and one dollar bills must he afterwards pay to cancel the debt ? Ans. 2 ten doll, bills and 8 one d .11. 13. The greater of two numbers is 24, and the less is IC; what is their difference ? 14. The greater of two numbers is 24, and their ditTeft- cnce 8 ; w hat is the less number ? 15. The sum of two numbers is 40, the less is 16 ; what is the greater ? 16. A tree, p8 feet high, was broken off by the wind ; the top part, which fell, was 49 feet long; how hio^b was th« stump which was left ? 17. Our pious ancestors landed at Plymputh, Massacliti^ •elts, in 1620 ; how many years since ? 18. A man carried his produce to market ; he sold hit pork for 45 dollars, his cheese for 38 dollars, and his butter for 29 dollars; he received, in pay, salt to ihe value of 17 dollars, 10 dollars worth of sugar, 5 dollars worth of men lasses, and the rest in money; how much mo!»ey did !i« receive? Jus. MO (!o liars. 19. A boy bougfit a sled for 28 cents, and gave 14 ceuti C 20 MULTIPLICATION OF SIMPLE NUMBERS. If 8, 9. to have it repaired ; he sold it for 40 cents ; did he g 4. Thnie men bought a horse ; each man paid 23 doIVirs for his share ; how many dollars did the horse cost them ? 5. A man has 4 farms worth 324 dollars each; how many dollars are they all worth ? 6. In one dollar there are one hundred cents ; how many cents in 5 dollars ? 7. How much will 4 pair of shoes cost at 2 dollars a pair? 8. How much will two pounds of tea cost at 43 cents a pound ? 9. There are 24 hours in one day ; how many hours in 2 days ? in 3 days ? in 4 days ? in 7 days ? 10. Six boys met a beggar, and gave him 15 cents each ; how many cents did the beggar receive ? When questions occur, (as in the above examples,) where the same number is to be added io itself se\eral times, the operation may be mucli lacililated by a rule, called Multi' plication^ in which the number to be repeated is called the multiplicand^ and the number which shows how many times the multiplicand is to be repeated is called the multiplier. The multiplicand and multiplier, when spoken of collectively, are called the<^ac/or5,t (producers,) and the answer is called the product, 11. There is an orchard in which there are 5 rowsof treps, and 27 trees in each row; how many trees in the orchard ? In this example, it is In the first row, 27 trees, evident that the whole second .... 27 nu'iiber of trees will be third .... 27 equal to the amount of fourth .... 27 fice 27's added together. fifth .... 27 In adding, we find _ _ - , , , that 7 taken five times In the whole orchard, 135 trees. amounts to 35. We write down the five units, and reserve the 3 tens; the amount of 2 taken five times is 10, and the 3, which we reserved, makes 13 which, written to the left of units, makes the whole number of trees 135. If we have learned that 7 taken 5 times amounts to 35, and that 2 taken 5 times amounts to 10, it is plain we need write the number 27 but once, and then, setting the multi- t6 MULTIPLICATION OF SIMPLE NUMBERS. IT 9, IQ. pli**r under it, we may say, 5 times 7 aro 35, writing down the 5, and reserving the 3 (tens) as in add-tion. Again, 5 times 2 (tens) are Multiplicand^')^ trees in each row* 10, (tens,) and 3, Multipliers'^ brows, (tens,) v»hich we n J . ~7^ A reserved, make 13, ProdiK^t, y 13o trees, Am. (t^,,,^) ^ before, T\ 10. 12. There are on a board 3 rows of spots, and 4 spots in each row ; how many spots on the board ? % % ^ ^ A slight inspection of the figure will ^ show, that the number of fpots may be % % % % found either by taking 4 thite times, (3 # * * # times 4 are 12,) or by taking ^ Jour times, (4 tunes 3 are 12 ;) for we may say there are 3 rows of 4 spots each, or 4 rows of 3 spots each*; there- fore, we may use either of the given numbers for a multi- plier, as best suits our convenience. We generally write the numbers as in subtraction, the larger uppermost, with units under units, tens under tens, &c. Thus, Multiplicand, 4 spots. Note, 4 and 3 are the factors^ Multiplier, 3 ro^.os. which produce the product 12. Product, 12 Ans, Hence,—^ Multiplication is a short way of performing manff additions ; in other words,—// is the method of repeating any fl/uitttLiir nny given number of times. SiGis. ^wo short lines, crossing each other in the form of the letter X,\are the sign of multiplication. Thus, 3X4 m 12, signifies that 3 times 4 are equal to 12, or 4 times 3 are 12. Note, Before any progress can be made in this rule, the following table must be committed perfectly to memory. V 10. MULTIPLICATION OF SIMPLE NUMBERS. 29 \ MULTIPLICATION TABLE. 2 times are 4 X 10 = 40 7X 7 = 49 |10X 4= 40 2X 1= 2 4 X 11 =44 7X 8 = 56 !10X 5= 50 2 X 2— 4 4 X t2 = 48 7X 9 = 63 10 X 6= 60 2 X 3= 6 5X 0= 5X 1= 5 5X 2 = 10 5X 3 = 15 7X 10 = 70 10 X 7= 70 2X 4z=: 8 7X11 = 77 10 X 8— 80 2X 5rzzl0 7X 12 = 84 10 X 9= 90 2X 6z=12 8 X = 10 X 10 = 100 2X 7=1=14 5 X 4 = 20 8 X 1 = 8 10 X 11 = 110 2X 8zz:16 5 X 5 = 25 8X 2 = 16 10 X 12 = 120 2 X 9 = 18 5 X 6 = 30 8X 3 = 24 11 X 0= 2 X 10 zz: 20 2 X 11=22 5X 7 = 35 5 X 8 = 40 8X 4 = 8X 5 = 32 40 11 X 1= 11 1 1 X 2 = 22 2 X 12=z24 5 X 9 = 45 8X 6 = 48 11 X 3= 33 3X 0:^ 5 X 10 = 50 8X 7 = 56 1 1 X 4 = 44 3X 1= 3 5 X 11 =55 8X 8 = 64 1 1 X 5 = 55 3X 2= 5 X 12 = 60 8X 9 = 72 11 X 6= 66 3X 3z= 9 6X0—0 8X10 = 80 11 X 7= 77 3X 4=12 6 X 1=6 8X 11 = 88 11 X 8= 88 3X 5=15 6X 2—12 3X 3 mis 8X 12 = 96 11 X 9= 99 3 X 6 = 18 9X = ~0 11 X 10=110 3X 7 = 21 6 X 4 zz= 24 0X^1 = 9 11 X 11 = 121 3 X 8 = 21 OX 5 = 30 QX 6zz: 36 9X *2 = 9X 3 = 18 27 11 X 12=132 3X 9 = 27 12 X 0= 12 X 1= 12 12 X 2= 24 12 X 3= 36 3 X 10 = 30 3X 11=33 3 X 12 = 36 6 X 7 = 42 6X 8 = 48 6 X 9 = 54 9X 4 = 9X 5 = 9X 6 = 36 45 54 4 X 0= 6 X 10 = 60 9X 7 = 63 12 X 4= 48 4 X 1=4 6 X 1 1 = 66 9X 8 = 72 12 X 5= 60 4 X 2= 8 6 X 12 = 72 9X 9 = 81 12 X 6 = 72 4X 3=12 7X 0= 9X 10 = 90 12 X 7= 84 4 X 4=10 7X1=7 9X 11 = 99 12 X 8= 96 4X 5 = 20 7X 2 = 14 7X 3 = 21 9X 12 = 108 12 X 9=108 4X 6 = 24 10 X = 12 X 10=120 4X 7z=:28 7X 4 = 28 10X1 = 10 12 X 11 = 132 4X 8:x-32 7X 5 = 35 lOX 2 = 20 12 X 12=144 4X 9 = 36 7X 6 = 42 10 X^ 3 = 30 so MULTIPLICATION OF SIMPLE NUMBERS. IT 10. 9 X 2 zz: how many ? 4 X 3 X 2 = 24. 4 X 6 in how mail}' ? 3x2X^1=: how many ? 8X 9=z how many ? 7x1X2 = how many ? 3 X 7 iz: how many ? 3x3x2 = how many ? 6 X 5 zz: how many ? 3x2X4X5zz how many ? 13. What will 84 barrels of Hour cost at 7 dollars a bar- rel ? Ans. 5S8 dollars. 14. A merchant bought 273 hats at 8 dollars each; what did they cost? Ans. 2184 dollars. 15. How many inches are there in 253 feet, every foot being 12 inches ? OiPERATlON. The product of 12, with each of the signifi- 253 cant ligures or digits, having been commit- 12 ted to memory from the multiplication table, A on^ '^ ^'^ J"^^ ^^ *^^^^ ^^ multiply by 1 J as by a Am. 6[)6b ^jj^gj^^ ^g^j^^^ rj-j^^g^ J 2 ^^^^^^ 2 ^^^ 3g^ ^^ 16. What will 476 barrels offish cost at 11 dollars a bar- rel ? Am. 5236 dollars. ^7. A piece of valuable land, containing 33 acres, was sold for 246 dollars an acre ; what did the whole come to ? As 12 is the largest number, the product of which, with the nine digits, is found in the multiplication table, therefore, when the multiplier exceeds r2, we multiply by each figure in the multiplier separately. Thus : OPERATION. The mullipli- 2^6 dollars^ the price oj I acre. ^j. consists of 3 ^ ''^'''^^'' ^/ «^'-^5- tens and 3 units. 738 dollars, the price of 3 acres. ^^'^^^^^ raultiply- 738 dollars, the price of 30 acres. ^"? ^Y . ^^^ ^ . units gives us Ans. 8118 dollars, the price of 33 acres. 738 dollars, the price of 3 acres. We then multiply by the 3 tens, writing the first figure of tlie product (8) in ten'^s place, that is, directly under the figure by which we irndtiply. It now appears, that the product by tlie 3 tens consists of the same figures as the product by the three units ; but there is this difference — the figures in the product by the 3 tens are all removed one place further to- ward the left hand, V which their value is increased ten- fold^ which is as it should be, because the price of 30 acrci IT 10. MULTIPLICATION OF SIMPLE NUMBERS. 31 is e\ndently ten times as much as the price of 3 acres, that is, 7880 dollars ; and it is plain, that these two products, added together, give the price of 33 acres. These examples will be sufficient to establish the fol» lowing RULE. I. Write down the multiplicand, under which write the multiplier, placing units under units, tens under tens, &c., and draw a line underneath. II. When the multiplier does not exceed 12, hegin at the right hand of the multiplicand, and multiply each figure con- tained in it by the multiplier, setting down, and carrying, as in addition. III. When the multiplier exceeds 12, multiply hy each figure of the multiplier j^eparately, tirst by the imitSy then by the tens^ &c., remembering always to place the first figure of each product directly under the figure by which you multi- ply. Having gone through in this manner with each figure in the multiplier, add their several products together, and the sum of them will be the product required. EXAMPLES FOR PRACTICE. 18. There are 320 rods in a mile ; how many rods are there in 57 miles ? 19. It is 436 miles from Boston to the city of Washing- ton ; how many rods is it? 20. What will 784 chests of tea cost, at 69 dollars a chest ? 21. If 1851 men receive 758 dollars apiece, how many dollars will they all receive ? Ans, 1403058 dollars. 22. There are 24 hours in a day; if a ship sail 7 miles in an hour, how many miles vvill she sail in 1 day, at that rate? how many miles in 36 days ? how many miles in 1 year, or 365 days ? Aiis. 6i320 miles in 1 year. 23. A merchant hought 13 pieces of cloth, each piece containing 28 yards, at 6 dollars a yard ; how many yards were there, and what was the whole cost ? Ans. Tke whole cost was 2184 dollars. 24. Multiply 37864 by '235. Product, 8.S98040. 25 29831 ... 952 28399112. 26 93956 ... 8704. 817793024. S3 CONTRACTIONS IN MULTIPLICATION. IT II. CONTRACTIONS IN MULTIPLICATION. I. WTien the multiplier is a composite number* ir 11. Any number, which may be produced by the mul- tiplication of two or more numbers, is called a composite number. Thus, 15, which arises from the multiplication of 5 and 3, (5 X 3 rrr 15,) is a composite number, and the num- bers 5 and 3, which, multiplied together, produce it, are called component parts ^ or factors of that number. So, also, 24 is a composite number ; its component parts or factors may be 2 and 12 (2 X 12 — 24 ;) or they may be 4 and 6 (4X6 = 24 ;) or they may be 2, 3, and' 4 (2 X 3 X 4 =^3 24.) 1. What will 15 yards of cloth cost, at 4 dollars a yard? 15 yards are equal to 5 X ^. yards. The cost of 6 4 yards would be 5 X 4 zz: 20 dollars ; and because 15 5 yards contain 3 times 5 yards, so the cost of 15 yards — will evidently be 3 times the cost of 5 yards, that is, "^" 20 dollars X 3 =: 60 dollars.. Ans, 60 dollars. o 60 Wherefore, If the multiplier be a composite number^ we may, if we please, multiply the multiplicand first by one of the cowr- ponent parts^ that product by the other ^ and so on^ if the com- ponent parts be more than two;, and, having in this way multiplied by each of the component parts, the last product will be the product required. 2. W^hat will 136 tons of potashes come to, at 96 dollan per ton ? 8 X 12 1= 96. It follows, therefore, that 8 and 12 are component parts or factors of 96. Hence, 136 dollars, the price of 1 ton. 8 one of the component parts, or factors. 1088 dollars, the price of 8 tons. 12 the other component part, or factor. Ans, 13056 dollars, the price of 96 tons. 3. Supposing 342 men to be employed in a certain piece of work, for which they are to receive 112 dollars each, how much will they all receive ? 8 X 7 X 2 r:= 112. Ans. 38304 dollaw. IT 12, 13. CONTRACTIONS IN MULTIPLICATION. B$ 4. Multiply 3G7 by 48. Product, 17616. 6 853 ... 56. 47768. 6 1086 ... 72. 78192. II. Wien the multiplier is 10, 100, 1000, ^c. IT 12. It will be recollected, (IT 3.) that any figure, on be- ing removed one place towards the left hand, has its value increased tenfold ; hence, to multiply any number by 10, it is only necessary to write a cipher on the right hand of it. Thus, 10 times 25 are 250 ; for the 5, which was units before, is now made teits^ and the 2, v/hich was tens before, is now made hundreds. So, also, if any figure be removed two places towards the left hand, its value is increased 100 times, &c. Hence, Wien the mtdtiplier is 10, 100, 1000, or 1 with any number of ciphers annexed^ annex as maiy ciphers to the multipli- cand as there are ciphers in the muitiplier, and the m»ilti- plicand, so increased, will be the product required. Thus, Multiply 46 by 10, the product is 460. *..83... 100, 83V.0. 95 ... 1000, 95000. 1 SAMPLES FOR PRACTICE. 1. What wiil 70 barrels of flour cost, at 10 dollars a barrel ? 2. If 100 men receive 126 dollars each, how many dol- lars vvv!l they all receive ? 3. What will 1000 pieces of broadcloth cost, estimating each piece at 312 d'^llars ? 4. Multiply 5682 by 10000. 6 82134 ... 100000. IT 13. On the principle suggested in the last IT, it follows, When there are ciphers on the right hand of the multipli- cand, multiplier, either or both, we may, at first, neglect these ciphers, multiplying by the significant fig^ircs only; after which we must annex as many ciphers to the product as there are ciphers on the right hand of the multiplicand Mid multiplier, counted together. S4 CONTRACTIONS IN MULTIPLICATION. IT 13. EXAMPLES FOR PRACTICE. 1. If 1300 men receive 460 dollars apiece, how many dollars will they all receive ? OPERATION. '^^^ ciphers in the multiplicand 460 ^^^ multiplier, cointed together, 1300 3.re three. Disregarding these, we . write the sirjnificant figures ot" the 1^8 multiplier under the significant fig- 46 ures of the multiplicand, and multi- Ans. 198000 dollars. ?/>;> ^^^^/ which we annex three ciphers to the Mght hand of the product, which gives the true answer. 2. The number of distinct buildings in New England, appropriated to the spinning, weaving, and printing of cot- ton goods, was estimated, in 1826, at 400, running, on an average, 700 sph.dles each ; what was the whole number ol spindles ? 3. Multiply 357 by 6300. 4 8600 .... 17. 5 9340 .... 460, 6 5200 .... 410. 7 378 .... 204. OPERATION. 378 204 1522 I^^ the operation it will be seen, that multi- 000 l^b'i^g by ciphers produces nothing, There- 756 ^^^^' 77112 III. Wheit there are ciphers between the significant figures of t4ie multiplier y we may omit the ciphers, niultjplying by the significant figures onhj^ placing the first figure of each pro- duct directly undei the tig'ire by which we multiply. EXAMPLES FOR PRACTICE. 8. Multiply 154326 by 3007. 13» SUPPLEMENT TO MULTIPLICATION. 35 OrERATION. 154326 3007 1080282 462978 Product, 464058282 9. Multiply 543 by 206. 10 1620 ... 2103. 11 36243 ... 32004. S^UPPZiSIMESNT TO MULTIPLICATION. QUESTIONS. 1. What is multiplication ? 2. What is the number /o be multiplied called ? 3. to multiply by called ? 4. What is the result or answer called ? 5. Taken collectively, what are the multiplicand ana multiplier called ? 6. What is the ngn of multiplication ? 7. What does it show ? 8. In what order must the given number be placed for multiplication ? 9. How do you proceed when the multiplier is less i\\diX\ 12? 10. When it exceeds 12, what is the method of procedure ? 11. What is a composite number? 12. What is to be under- stood by the component parts, ov factors, of any number? 13. How may you proceed when the multiplier is a compo- site number 1 14. To multiply by 10, 100, 1000, &c., what suffices? 15. Why? 16. When there are ciphers on the right hand of the multiplicand, multiplier, either or both, how may we proceed? 17. When there are ciphers be- tween the significant figures of the multiplier, hew are they to be treated ? EXERCISES. 1. An army of 10700 men, having plundered a city, took 80 much money, that, when it was shared among them, each man received 46 dollars ; what was the sum of money taken? S6 SUPPLEMENT TO SIULTIPLICATIOTV. IT 13. 2. Supposing the number of houses in a ceitain town to be 145, each house, on an average, containing two families, ana each family 6 members, what would be the number of inhabitaixts in that town ? Ans, 1740. 3. If 46 men can do a piece of work in 60 days, hovr many men will it take to do it in one day ? 4. Two men depart from the same place, and travel in opposite directions, one at the rate of 27 miles a day, the other 31 miles a day; hov/ far apart will they be at the end of 6 days ? Ans. 348 miles. 5. What number is that, the factors of which are 4, 7, 6, and 20 ? Ans, 3360. 6. If 18 men can do a piece of work in 90 days, how long will it take one man to do the same ? 7. Vv^hat sura of money must be divided between 27 men, so that each man may receive 115 dollars? S. There is a certain number, the factors of which are 89 and 265 ; what is that number ? 9. What is that number, of which 9, 12, and 14 are factors ? 10. If a carriage wheel turn round 346 times in running 1 mile, how many times v/ill it turn round in the distance from New Yo.k to Philadelphia, it being 95 miles. Ans, 32870. 11. In one minute are 60 seconds; how many seconds in 4 minutes ? in 5 minutes ? in 20 minutes ? in 40 minutes ? 12. In one hour are 60 minutes ; how many seconds in an hour ? in two hours ? how many seconds from nine o'clock in the morning till noon ? 13. In one dollar are 6 shillings; how many shillings in 3 dollars ? in 300 dollars ? in 467 dollars ? 14. Two men, A and B, start from the same place at the same time, and travel the same y^'ay; A travels 52 miles a day, and B 44 miles a day ; how far apart will they be at the end of 10 days? 15. If the interest of 100 cents, for one yrar^ be 6 cents, how many cents will be the uiterest for 2 years ? for 4 years ? for 10 years ? for 35 years ? for 84 years ? 16. If the interest of one dollar, for one y^^ar, be six cents, Wiat is the interest for 2 dollars the same time ? 5 dollars ? 7 dollars ? — 8 dollars ? 95 doilar»l ^ 13, 14. DIVISION OF SIMPLE NTTMBERS. 87 17. A farmer sold 468 pounds of pork, at 6 cents a pound, and 48 pounds of cheese, at 7 cents a pound ; kow many cents must he receive in pay ? 18. A boy bought 10 oranges ; he kept 7 of them, and sold ^le others for 5 cents apiece ; how many cents did he receive ? 19. The component parts of a certain number are 4, 5, 7, 6, 9, 8, and 3 ; what is the number ? 20. In 1 hogshead are 63 gallons; how many gallons in 8 hogsheads ? In 1 gallon are 4 quarts ; how many quarts in 8 hogsheads ? In 1 quart are 2 pints ; how many pints in 8 hogs- heads ? DIVISION OF SIMPLE NUMBERS. IT 14. 1, James divided 12 apples among 4 boys ; how many did he give each boy ? 2. James would divide 12 apples among 3 boys; how many must he give each boy ? 3. John had 15 ap})les, and gave them to his playmates, who received 3 apples each ; how many boys did he give them to ? 4. If you had 20 cents, how many cakes could you buy at 4 cents apiece ? 5. How many yards of cloth could you buy for 30 dollars, at 5 dollars a yard ? 6. If you pay 40 dollars for 10 yards of cloth, what is one yard worth ? 7. A man works ^ ^J^ for 42 shillings; how many shil- lings is that for one day ? 8. How many quarts in 4 pints? in 6 pints? in 10 pints ? , 9. How many times is 8 contained in 88 ? 10. If a man can travel 4 miles an hour, tow many houn would it take him to travel 24 miles ? 11. In an orchard tliere are 28 trees standing in rows, 9iid there ar^e 3 trees in a row ; how many rows are there ? ReiTwrk. When any on^s thing is divided into two equal parts, one of thos^ parts is called s^ half; if into 3 equal parts, one of those parts is called a third; if into four equal parts, one part is called a quarter or a fourth; if into liye, •ne partis called a ///A, and so on. 38 DIVISION OF SIMRLE NUMBERS. M 14, 15* 12. A boy had two apples, and gave one half an apple to each of his companions ; how many were his companions ? 13. A boy divided four apples among his companions, by giving thorn one third of an apple each ; among how many did he divide his apples? 14. Huw many quarters m 3 oranges ? 15. How many oranges would it take to give 12 boys one quarter of an orange each ? 16. How much is one half of 12 apples ? 17. How much is one third of 12 ? ~n18. How much is one fourth of 12 ? li). A man had 30 sheep, and sold one fifth of them ; how many of them did he sell ? 20. A man purcliased sheep for 7 dollars apiece, and paid for ther. ail 63 dollars; what was their numbei ? 21. How many oranges, at 3 cents each, may be bought for 12 cents ? It is plain, that as many times as 3 cents can be taken from 12 cents, so many oranges may be bought; the object, therefore, is to find how many times 3 is contained in 12. 12 cents. First orangey 3 cents. We see in this example, thai — " we may take 3 from 12 foui _, , ^ times, after which there is no re Second orangey cents. niainder; consequently, subtrac- Q lion alone is sufficient for the ope- Tklrd orange, 3 cents. ^^^^0" 5 ^^^^ ^^'^ "^^y ^^^^^ to the — same result by a process, in most 3 cases much shorter, called Di- Fourth orange y 3 cents. vkion, IT 15. It is plain, that the cost of one orange, (3 cents,) multiplied by the number of oranges, (4,) is equal to the cost of all the oranges, (12 cents ;) 12 is, therefore, a pro- duct, and 3 one of its factors; and to find how many times 3 is contained in 12, is to find the other factor, which, mul- tiplied into 3, will produce 12. This factor we find, by trial, to be 4, (4 X 3 zn 12;) consequently, 3 is contained in 12 4 times. Ans. 4 oranges. 22. A man would d\\ ide 12 oranges equally among 3 chil- dren; how many oranges Svould each child have.? Here the object is to divide the 12 oranges into 3 equal iri5. DIVISION OF SIMPLE NUMBERS. 89 parts, and to ascertain the number of oranges in each of those parts. The operation is evidently as in the last example, and consists in finding a number, which, multiplied by 3, will pro- duce 12. This number we have already found lo be 4. Ans. 4 oranges apiece. As, therefore, mullipllcation is a short way of performing many additions of the same number ; so, dwisioii is a short way of performing many subtractions o^ the same number; and may be defined. The method of finding how mmiy limes one number is cont aided in another ^ and also of dividing a num- ber into any number of equal parts. In all cases, the process of division consists in finding one of the factors of a given product, when tiie other factor is known. The number given to be divided is called the dividend^ and answers to the product in multiplication. The number given to divide by is called the divisor^ and answers to ont of the factors in multiplication. The result^ or answer sought, is called the quotient^ ffrnin tlie I^atin word quoties, how many?) and answers lo the other factor. Sign. The sign for division is a short honzontal line be- tween two dots, -H. It shows that the number before it iu to be divided by the number after it. Thus 27 ~ 9 = 3 is read, 27 divided by 9 is equal to 3 ; or, to shoridi the ex- pression, 27 by 9 is 3 ; or, 9 in 27 3 times. In plpce of the dotSj the dividend is often written over the line, and the di- visor under it, to express division ; thus, ^^ = 3, read as before. ^^z=:3 ^z=9 DM'ISIOJV TABLE .* f =1* t=l J=l *=1 S=i 1=2* 1=2 1=2 -y = 2 V = 2 f =3 f =3 J,^ = 3 -^ = 3 ¥ = 3 f =4 ¥=--4 JjS = 4 ¥ = 4 ¥ = 4 ^ = 5 ^--=5 ^- = 5 ¥ = 5 s^~5 -^ = 6 V- = 6 ^ = 6 ?Ji_6 5/. = 6 4^ = 7 ¥ = 7 ^ = 7 ^ = 7 ¥ = 7 4^ = 8 ¥ = 8 \'=:8 H'- — ^ *^ = 8 ■y:=9 ¥ = 9 V = 9 V = 9 ¥=9 • The reading used hv tho jujpil in conimiiiin^ ihe table may be^ 2 by 2 U I, 4 by 2 is 2, &c. 5 or, 2 in" 2 one lime, 2 In 4 iwo times, &c. 40 DIVISION OF SIMPLE NUMBERS. IT 16, 16. DIVISION TABLE— CONTINUED. t ^ f .8.— 1 = 1 iS=l ii=l •^=2 f*=2 e=2 ^=3 fS = 3 «-=3 ¥-=4- f5 = 4 ^ = 4 V=5 f8 = 5 «=5 ^^=6 fS = 6 «=6 ^=7 t3 = '7 H=7 ^=8 U = s ff=8 V=9 f8=9 f*=9 if = 1 ff =2 ff=3 If =4 ff =5 e=6 H=7 If =8 Y/=9 28 -h 7, or ^- T=: how many ? 42 -f- 6, or ^(f- zz: how many ? 54 -T- 9, or ^ = how many ? 82 -f- 8, or 2^- z=z how many ? 33 -7- 11, or ^^zz: how manj ? 49 -7- 7, or ^ z= how many f 32 -T- 4, or ^*i zz: how many ? 99 -r- 1 1, or f f zz: how many ? 84 -:- 12, or ^^ =z how many? 108 -f- 12,or\<^-zz:howmany? ^ 16« 23. How many yards of cloth, at 4 dollars a yard, can be bought for 856 dollars ? Here the number to be divided is 856, which therefore is the dimdend; 4 is the number to divide by^ and there- fore the divisor. It is not evident how manj' times 4 is con- tained in so large a number as 855. This difficulty will be readily overcome, if we decompose this number, thus : 856 :zz 800 + 40-1- 16. Beginning with the hundreds, we readily perceive that 4 is contained in 8 2 times; consequently, in 800 it is contained 200 times. Proceeding to the tens, 4 is contained in 4 1 time, and consequently in 40 it is contained 10 times. Lastly, in 16 it is contained 4 times. We now have 200 -j- 10 -[-4 ==214 for the quotient, or the number of times 4 is contained in 856. Ans, 214 yards. We may arrive to the same result without decomposing the dividend, except as it is done in the mind, taking it by parts, in the following manner : For the sake cf convenience, we write down the dividend with the divi- sor on the left, and draw a 'ine between them ; we also draw aline underneath. Then, beginning on the left hand,. Dividend, Divisor J 4 ) 856 Quotient^ 214 IT 16. mVlSlON OF SIMPLE NUMBERS. 41 we seek how often the divisor (4) is contained in 8, (hundreds,) the left hand ligure; finding it to be 2 times, we write 2 directly under the 8, which, falling in the place of hundreds, is in reality 200. Proceeding to tens, 4 is con- tained in 5 (tens) 1 time, which we set down in ten^s place, directly under the 5 (tens.) But, after taking 4 times ten out of the 5 tens, there is 1 ten left. This 1 ten we join to the 6 units, making 16. Then, 4 into 16 goes 4 times, which we set down, and the work is done. This mawner of performing the operation is called Short Division, The computation, it may be perceived, is carried on partly in the mind, which it is always easy to do when the divisor does not exceed 12. RUL.E. From the illustration of this example^ toe derive this general rule for dividing^ when the divisor does not exceed 12 : I. Find how many times the divisor is contained in the first figure, or figures, of the dividend, and, setting it direct- ly under the dividend, carry the remainder, if any, to the next figure as so many tens. II. Find how many times the divisor is contained in this dividend, and set it down as before, continuiag so to do till all the figures in the dividend are divided. Proof. We have seen, (IF 15,) that the divisor and quo- tient are factors, whose product is the dividend, and we have also seen, that dividing the dividend by one factor is merely a process for finding the other. Hence division and m-dilpUcation mutually prove each other. To prove division^ we may mulliphj the divisor by the quo- tient, and, if the work be right, the product will be the same as the dividend ; or we may divide the dividend by the quo^ tientj and, if the work is right, the result will be the same as tlie divisor. To prove innllipUcation^ we may divide the product by one factor^ and, if the work be right, the quotient will be the other factor, ' EXAMPLES FOR PRACTICE. 24. A man would divide 13,462,725 dollars among 5 men % how many dollars would each receive ? D* 42 DIVISION OF SIMPLE NUMBERS. If I&, 17, In this example, as we cannot DMend ^""^^ ^ '° ^^^ '-^''^ '^Sure, (1,) we Divisor, 5 ) 13,462,725 ^''i^f *^° Y'^'' ^"^ f,^-^' ^ '° ^^ ' ' ' ' \vi crn y? timps. and fhprp s\Tf^ !^ will go 2 times, and there are 3 Quotienty 2,692,545 over, which, joined to 4, the next figure, makes 34 ; and 5 in 34 will go 6 times, &c. Proof. In proof of this example, we mul- Quotient, tiply the quotient by the divisor, 2,092,545 and, as the product is the same as 5 divisor. the dividend, we conclude that the 1^469 72^ work is right. From a bare in- ' ' spection of the above example and ifjs proof, it is plain, as before stated, that division is the re- verse of multiplication, and that the two rules mutually prove each other. 25. How many yards of cloth can be bought for 4,354,560 dollars, at 2 dollars a yard ? at 3 dollars ? at 4 dollars ? at 5 dollars ? at 6 dollars ? at 7 ? at 8 ? at 9 ? at 10 ? Note. Let the pupil be required to prove the foregoing, and all following examples. 26. Divide 1005903360 by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. 27. If 2 pints make a quart, how many quarts in 8 pints : in 12 pints ? in 20 pints ? in 24 pints > in 248 pints ? in 3764 pints ? in 47632 pints ? 28. Four quarts make a gallon ; how many gallons in 8 quarts ? in 12 quarts ? in 20 quarts ? in 36 quarts? in 368 quarts ? in 4896 quarts ? in 5436144 quarts? 29. A man gave 86 apples to 5 boys ; how many apples would each boy receive ? t)ividend. Here, dividing the Divisor, 5 ) 86 number of the apples r\ *• * *TX , ». . , (86) by the number of Quotient, 17-1 Remainder. ^^ys, (6,) we find, that each boy's share would be 17 apples ; but there is one apple left. V17, 6)86 In order to divide all the apples equal- — - ly among the boys, it is plain, we must di- * 'T vide this one remaining apple into 5 eqwd If 17. DIVISION OF SIMPLE JVUMBERS. 43 parts^ and give one of these parts to each of the boys. Then each boy's share would be 17 apples, and one fifth part of another apple; which is written thus, 17^ apples. A71S. 17 J apples eachc The 17, expressing whole apples, are called integers^ (that is, whole numbers.) The ^ (one fifth) of an apple, express- ing part of a broken or divided apple, is called a fraction^ (that is, a broken number.) Fractions, as we here see, are written v/ith two numbers, one directly over the other, with a short line between them, showing that the vpper number is to be divided by the lower. The upper number, ox dividend^ is, in fractions, call- ed the numerator J and the lower number, or divisor^ is called the denominator. Note, A number like 17-J, composed of integers (17) iind a fraction, (^,) is called a mixed number. In the preceding example, the one apple, which was left after carrying the division as far as could be by ivhole num- bers, is called the remainder^ and is evidently a part of the dividend yet undivided. In order to complete the division, this remainder, as we before remarked, must be divided into 5 equal parts ; but the divisor itself expresses the number of parts. If, now, we examine the fraction, we shall see, that it consists of the remainder (1) for its numerator^ and the divisor (5) for its denominator. Therefore, if there be a remainder^ set it down at the right hand of the quotient for the numerator of a fraction, under which write the divisor for its denominator. Proof of the last example. In proving this example, we 17-i find it necessary to multiply 5 our fraction by 5 ; but this is '~~ easily done, if we consider, that' the fraction ^ expresses one part of an apple divided into 5 equal parts ; hence, 5 times ^ is 1^= 1, that is, one whole apple, which we reserve to be added to the unitSj saying, 5 times 7 are 35, and one we re- served makes 36, &c. 30. Eight men drew a prize of 453 dollars in a lottery f \kovr many dollars did each receive } 44 mViSIOJS OF SIMPLE NUMBERS. IT IS, 19. Dividend, Here, after carrying the division as Divisor^ 8 ) 453 far as possible by whole numbers, we . 77" have a remainder of 5 dollars, which, Quotient^ 56f written d& above directed, gives for the answer 56 dollars and | (live eighths) of another dollar, to each man. TT 18. Here we may notice, that the eighth part of 5 dol- lars is the same as 5 times the eighth part of 1 dollar, that is, the eighth part of 5 doilarh is § of a dollar. Hence, f expresses the quotient of 5 divided by 8. Proof, I is 5 parts, and 8 times 5 is 40, that is, ^- =: 5, 66|- which, reserved and added to the product of 8 time* 8 6, makes 53, &c. Hence, to multiply a fractioUj -"— we may multiply the mmeratOTy and divide the ^^ product by the denominator. Or, in proving division, we may multiply the whole num- ber in the quotient onhj^ and to the product add the remain- der ; and this, till the pupil shall be more particularly taught in fractions, will be more easy in practice. Thus, 56 X 8 1= 448, and 448 + 5, the remainder, z= 453, as before. 31. There are 7 days in a week ; how many weeks in 365 days ? Ans, 52| weeks. 32. When flour is worth 6 dollars a barrel, how many barrels mav be bought for 25 dollars /* hov/ many for 50 dol- lars ? -' for 487 dollars ? for 7631 dollars? 33. Divide 640 dollars among 4 men. 640 ^ 4, or ^1^ = 160 dollars, Ans. 34. G78 -^ 6, or ^9- rr=: how many ? Ans. 113. 35. ^^^ =z how many ? 36. z.^^-^ = ho\\ many? 37. 34^ — how many? Ans, 384f 38. 2_zR^-=: how many? 39. 4(i.|jajLr=r:how mauy ? 40. afijijy?-X2rrr bow many? IT 19- 41. Divide 4370 dollars equally among 21 men. When, as in this example, the divisor exceeds 12, it is evident that the co.nputation cannot be readily carried on in the mind, as in the foregoing examples. Wherefore, it is more convenient to write down the computation at lengthy in the following manner : IT 19. DIVISION OF SIMPLE NUMBERS. 45 OPERATION. We may write the divisor Dlmsor. DividemL Quotient. and dividend as in short di- 21 ) 4370 ( 20822^. vision, but, instead of writing 42 the quotient under the divi- -^ dend, it will be found more ^'^^ convenient to set it to the ^^ right hand. ' 2 Remainder. taking the dividend by parts, we seek how often we can have 21 in 43 (hundreds ;) rinding it to be 2 times, we set do^vn 2 on the right hand of the dividend for the high- est figure in the quotient. The 43 being hundreds, it fol- lows, that the 2 must also be hundreds. This, however, we need not regard, for it is to be followed by tens and units, obtained from ihe tens and units of the dividend, and will therefore, at the end of the operation, be in the place of hun- dreds, as it should be. It is plain that 2 (hundred) . times 21 dollars ought now to be taken out of the dividend ; therefore, we multiply the divisor (21) by the quotient figure 2 (hundred) now found, naking 42, (hundred,) which, written under the 43 in the dividend, we subtract, and to the remainder, 1, (hundred,) bring down the 7, (tens,) making 17 tens. We then seek how often the divisor is contained in 17, (tens ;) finding that it will not go, we write a cipher in the quotient, and bring down the next figure, making the whole 170. We then seek how often 21 can be contained in 170, and, finding it to be 8 times, we Avrite 8 in the quotient, and, multiplying the divisor by this number, we set the product, 168, under the 170 ; then, subtracting, Ave find the remain- der to be 2, which, written as a fraction on the right hand of the quotient, as already explained, gives 208^^ dollars, for the answer. This manner of performing the operation is called Ij)ng Division, It consists in writing down the whole computation. From the above example, we derive the following RULE. I. Place the divisor on the left of the dividend, separate them by a line, and draw another line on the right of the dividend to separate it from the quotient. II. Take as many figures, on the left of the dividend, as 46 DIYISIO SIMt'LE NUMBERS. H 19. contain the divisor once or more ; seek how many times they contain it, and place the answer on the right hand of the di^adend for the first figure in the quotient. III. Multiply the divisor by this quotient figure, and write the product under that part of the dividend taken. IV. Subtract the product iVom the figures above, and to the remainder bring down the next figure in the divitlend, and divide the number it makes up, as before. So continue to do, till all the figures in the dividend shall have been brought down and divided. Note 1. Having brought down a figure to the remainder, if the number it makes up be less than the divisor, write a cipher in the quotient, and bring down the next figure. Note 2. If the product of the divisor, by any quotient figure, be greater than the part of the dividend taken, it is an evidence that the quotient figure is too large^ and must be diminished. If the remainder at any time be greatn than the divisor, or equal to it, the quotient figure is too snially and must be increased. EXAMPLES FOR PRACTICE. 1. How many hogsheads of molasses, at 27 dollars a hogs- head, may be bought for 6318 dollars ? Ans, 234 hogsheads, 2. If a man's income be 1248 dollars a year, how much is that per week, there being 52 weeks in a year ? Ars. 24 dollars per week. 3. What will be the quotient of 153598, divided by 29 ? Ans. 5296^1- 4. How many times is 63 contained in 30131 ? Ans, 478^ J times ; that is, 478 tiroes, and JJ of another time. 5. What v/ill be the several quotients of 7652, divided by 16, 23, 34, 86, and 92? 6. If a farm, containing 256 acres, be worth 7168 dollars, what is that per acre ? 7. Wh.*t will be the quotient of 974932, divided by 365 ? Ans. 2671 ^Vs- 8. Divide 3228242 dollars equally among 563 men ; how many dollars must each man receive ? Ans. 5734 dollars. 9. If 57624 be divided into 216, 586, and 976 equal parts, what will be the magnitude of one of each of these equal parts'* IT 20, 21. CONTRACTIONS IN DIVISION. 47 Ans. The magnitude of one of the kst of these equal parts will be 59^\% 10. How many times does 1030603615 contain 3215? Alls. 320561 times. 11. The earth, in its annual revolution round the sun, is said to travel 596088000 miles ; what is that per hour, there being 8766 hours in a year ? 12. JL2.3i.i6 2.8aii ::^ how mauy ? 13. Axixoa^k r= how many? 14. 5^6_<^9^^L _ 1^0 w many ? CONTRACTIONS IN DIVISION. I. When the divisor is a composite number. IT 20. 1. Bought 15 yards of cloth for 60 dollars ; how much wa> that per yard ? 15 yai^s are 3 X 5 yards. If there had been but 5 yajds, the cost of one yard would be -^^- -:= 12 dollars ; but, as there are 3 times 5 yards, the cost of one yard will evidently be but one third part of 12 dollars ; that is, ^ = 4 dollars. Ans. Hence, when the divisor is a composite number, we may, if we please, divide the dividend by one of the component parts, and the quotient ^ arising from that division, by the other : the last quotient will be the answer. 2. If a man can travel 24 miles a day, how many days will it take him to travel 264 miles ? It will evidently take him as many days as 264 contains 24. OPERATION. 24 = 6X4. 6)264 24)264(11 days, iln«. — 24 11 days. 24 3. Divide 576 by 48= (8 X 6.) 4. Divide 1260 by 63= (7 X 9.) 5. Divide 2430 by 81. C. Divide 448 by 56. II. To divide by 10, 100, 1000, &c. IT ai. 1. A prize of 2478 dollars Is owned by 10 men, is each man's share ? 48 CONTRACTIONS IN DIVISION. Tr21,22. Each man's share will be equal to the number of tens con- tained in the whole sum, and, if one of the figures be cut off at the right hand, all the figures to the left may be consid- ered so manv tens; therefore, each man's share will be 247f^ dollars." It is evident, also, that if 2 figures had been cut off from the right, all the remaining figures would have been so ma- ny hundreds; if 3 figures, so many thousands, &c. Hence we derive tbis general Rule /or dividing by 10, 100, 1000, &c. : Cut off from the right of the dividend so many figures as there are ciphers in the divisor ; the figures to the left of the point Xvill express the quotient, and those to the right, the remainder. 2. In one dollar are 100 cents ; how many dollars in 42400 cents? 'Ans, 424 dollars. 494'00 Here the divisor is 100; we therefore cut off 2 ' figures on the right hand, and all the figures to the left (424) express the dollars. 3. How many dollars in 34567 cents ? Ans. 345^^ dollars. 4. How many dollars in 4567840 cents? 5. How many dollars in 345600 cents ? 6. How many dollars in 42604 cents ? Ans. 426^^^. 7. 1000 mills make one dollar ; how many dollars in 4000 mills ? in 25000 mills ? in 845000 ? 8. How many dollars in b487 mills? Ans. 6-/^^^ dollars. 9. Hov/ many dollars in 42863 mills ? in 368456 mills ? in 96842378 mills ? 10. In one cent are 10 mills; how many cents in 40 mills ? in 400 mills ? in 20 mills ? in 468 mills ? in 4784 mills ? in 34640 mills ? III. When there are ciphers wi the right hand of the divisor. ^ CtSi. 1. Divide 480 dollars among 40 men ? In this example, our divisor, ^InNdftlrf^'^^^^' (^^?) ^^ ^ composite number^ 4|u;^4S|U (10 X 4 =r 40 ;) we may, there- 12 dolls. Ans. ^or^j divide by one component part, (10,) and that quotient by the other, (4;) but to divide by 10, we have seen, is but ta cut off the right hand figure, leaving the figures to the left ^ IT 52. SUPPLEMENT TO BITISION. 49 of the point for the quotient, which we diyide hy 4, and the work is done. It is evident, that, if our diviwsor had been 400, we should have cut off 2 figures, and have divided in the same manner ; if 4000, 3 figures, &;c. Hence this gene- ral Rule : — When there are ciphers at the right hand of the di- tlsor^ cut them off, and also as many places in the dividend ; divide the remaining figures in the dividend by the remain- ing figures in the divisor; then annex tlie figures, cut off from the dividend, to the remainder. 2. Divide 748346 by 8000. Dividend* Dirwor, 8|000)748|346. Quotient^ 93. — 1346 Remainder. An$. 93|3«S8. ?. Divide 46720367 by 4200000. Dividend. 42[00000)467|20367(ll3^^ Quoiienf, "47 / ''""'' 42 f UNW' 520367 Rmainder. 4. How many yards of cloth can be bought for 340500 dollars, at 20 dollars per yard ? 5. Divide 76428400 by 900000. 6. Divide 345006000 by 84000. 7. Divide 4680000 by 20, 200, 2000, 20000, 300, 4000, 50, 600, 70000, and 80. QUESTIONS. 1. What is division ? 2. In what does tlyb prccfim ctf dl» vision consist ? 3. Division is the r€«>ef5^ of what ? 4. >Vhdt k the number to he divided called, and to what does it aw- Fwer in multiplication ? 5. What is the number to divide by called, and to what does it answer, &c. ? 6. What is the result or answer called, &c. ? 7. What is the siffn of divi •ion, and what does it show ? 8. What is the other way of expressing division ? 9. What is shiyrt dimim^ and how it E 50 SUPPLEMENT TO DIVISION. If 2% it performed? 10. How is division proved! 11. Howls Tnultiplication proved? 12. What are integers^ or v/bole numbers? 13. What are fractions, or broken numbers? 14. What is a mixed number ? 15. When there is any thing left after division, what is it called, and ho\/ is it to be written? 16. How are fractions written! 17. What is the upper number called? 18. the lower number? 19. How do you multiply a fraction ? 20. To what do the numerator and the denominator of a fraction answer in di- vision ? 21. What is /ow^r division ? 22. Rule? 23. When the divisor is a composite number, how may we proceed ? 24. When the divisor is 10, 100, 1000, &:c., how may the operation be contracted ? 25. When there are ciphers at the right hand of the divisor, how may we proceed ? EXERCISES. 1.. An army of 1500 men, having plundered a city, took 2625000 dollars ; what was each man's share? 2. A certain number of men were concerned in the pay*- ment of 18950 dollars, and each man paid 25 dollars ; what was the number of men ? 3. If 7412 eggs be packed in 34 baskets, how many in a basket ? 4. What number must I midtiply by 135 that the pro- duct may be 505710? 5. Light moves with such amazing rapidity, as to pass from the sun to the earth in about the space of 8 minutes^ Admitting the distance, as usually computed, to be 95,000,000 miles, at what rate per minute does it travel ? 6. If the product of two numbers be 704, and the multi- plier be 11, what is the multiplicand ? Aiis, 64. 7. If the product be 704, and the multiplicand 64, what is the multiplier ? Am. 11. 8. The divisor is 18, and the dividend 144; what is the quotient ? 9. The quotient of two numbers is 8, and the dividend 144 ; what is the divisor ? 10. A man wishes to travel 585 miles in 13 days ; how far must he travel each day? 11. If a man travels 45 miles a day, in how many day* tnll he travel 585 miles ? fr 22. SUPPLEMENT TO DIVISION. 51 12. A man sold 35 cows for 560 dollars ; how much was that for each cow ? 13. A man, selling his cows for IC dollars each, received for all 560 dollars ; how many did he sell ? 14. If 12 inches make a foot, how many feet are there in 364812 inches? 15. If 364812 inches are 30401 feet, how many inches make one foot ? 16. If you would divide 48750 dollars among 50 men, how many dollars would you give to each one ? 17. If you distribute 48750 dollars among a number of men, in such a manner as to give to each one 975 dollars, how many men receive a share ? 18. A man has 17484 pounds of tea in 186 chests; how many pounds in each che?t ? 19. A man would put up 17484 pounds of tea into chests containing 94 pounds each ; how many chests must he have? 20. In a certain town there are 1740 inhabitants, and 12 persons in each house ; how many houses are there ? in each house are 2 families ; how many persons in each family? 21. If 2760 men can dig a certain canal in one day, how many days would it take 46 men to do the same ? How many men would it take to do the work in 15 days ? in 5 days ? ■ in 20 days ? in 40 days ? in 120 days? 22. If a carriage wheel turns round 32870 times in run- ning from New York to Philadelphia, a distance of 95 miles, how many times does it turn in running 1 mile ? Ans. 346. 23. Sixty seconds make one minute ; how many minutes m 3600 seconds ? in 86400 secoiids r — — in 604800 seconds ? in 2419200 seconds ? 24. Sixty minutes make one hour; how many hours in 1440 minutes ? in 10080 minutes ? in 40320 minutes ? in 525960 minutes ? 25. Twenty-four hours make a day; how many days in 168 hours ? ' in 672 hours ? in 8766 hovrs ? 26. How many times can I subtract forty-eight from four hundred and eighty ? 27. How many timea 3478 is equal to 47854 ? 28. A bushel of grain is 32 quarts ; how many quarts must I dip out of a chest of grain to make one half (^) of a bushel ? for one fourth (J) of a bushel ? for one eighth (}) of a bushel ? Aits, to the last^ 4 quarts. 50 MISCEttANEOUS QUESTIONS* IT 22, 23v 29. How many is i of 20? J of 48 ? j-of 247 ? ^ of 345678 ? ^ of 204030648 ? ^7i5. «o ^ of 45 ? i of 300 ? ^ of 478 ? f of 3456320 ? Ans. to the last, 1 152106^. 31. Whatisiof4? 1 of 20 ? i of 320? J of 7843 ? Ans, to the last, 1960|. MXS0£3Z.Z.ANE0nS QUESTIONS, Involving the Principles of the preceding Rules. Note. The preceding rules, viz. Numeration, Addition, Subtraction, Multiplication, and Division, are called the Fun- damental Rules of Arithmetic, because th^y are the foun- dation of all other rules. 1. A man bought a chaise for 218 dollars, and a horse for 142 dollars; what did they both cost ? 2. If a horse and chaise cost 360 dollars, and the chaise cost 218 dollars, what is the cost of the horse ? If the horse cost 142 dollars, what is the cost of the chaise? 3. If the sum of 2 numbers be 487, and the greater num- ber be 348, what is the less number? If the less number be 139, what is the greater number? 4. If the minuend be 7842, and the subtrahend 3481, what is the remainder ? If the remainder be 4361, and the minuend be 7842, what is the subtrahend? ^ 23* When the minuend and the subtrahend are given, how do you find the remainder? When* the minuend and remainder are given, how do you find the subtrahend ? When the subtrahend and the remainder are given, how do you find the minuend ? When you have the sum of two numbers, and one of them given, hov/ do you find the other ? When you have the greater of two numbers, and their difference given, how do you find the less number ? When you have the less of two numbers, and their differ- eace given, how do you and the greater number ? If 23, 24. MISCEr^LANEOUS QUESTIONS. 53 6. The sum of two numbers is 48, and one of the numbers is 19 ; what is the other 1 6. The greater of two numbers is 29, and their difference 10 ; what is the less number? 7. The less of two numbers is 19, and their difference is 10 ; what is the greater ? 8. A man bought 5 pieces of cloth, at 44 dollars a piece; 974 pairs of shoes, at 3 dollars a pair ; 600 pieces of calico, at 6 dollars a piece ; what is the amount ? 9. A man sold six cows, worth fifteen dollars each, and a yoke of oxen, for 67 dollars ; in pay, he received a chaise, worth 124 dollars, and the rest in money; how much money did he receive ? 10. What will be the cost of 15 pounds of butter, at 13 cents per pound ? 11. How many bushels of wheat can you buy for 487 dollars, at 2 dollars per bushel ? IT 24. When the price of one pound, one bushel, &c. of any commodity is given, how do you find the cost of any number of pounds, or bushels, &c. of that commodity ? If the price of the 1 pound, &c. be in cents, in what will the whole cost be? If in dollars, what? if in shillings? if in pence ? &c. When the cost of any given number of pounds, or bushels, &c. is given, how do you find the price of one pound or bushel, &c. In what kind of money will the answer be ? When the cost of a number of pounds, &e. is given, and also the price of one pound, &c., how do you ijttd the num- ber of pounds, &c. 12. When rye is 84 cents per bushel, what will be the cost of 948 bushels ? how many dollars will it be ? 13. If 648 pounds of tea cost 284 dollars, (that is, 28400 cents,) what is the price of one pound ? When the factors are given, how do you find the product? When the product and one factor are given, how do you find the other factor ? When the divisor and quotient are given, how do you find the dividend ? When the dividend and quotient are given, how do yoti find the divisor ? 14. What is the product of 754 and 25 ? E* 5t MISCELLANEOUS QUESTIONS. 1( ^4, 25. 15. What numoer, multiplied by 25, will produce 18850? 16. What number, multiplied by 754, will produce 18650 ? 17. If a man save six cents a day, how many cents would he save in a year, (365 days,) ? hov/ many in 45 years ? how many dollars would it be ? how many cows could he buy with the money, at 12 dollars each ? Ans, to the last^ 82 cows, and 1 dollar 50 cents remainder. 18. A boy bought a number of apples; he gave away ten of them to his companions, and afterwards bought thirty-four more, and divided one half of what he then had among four companions, who received 8 apples each ; how many apples did the boy first buy > Let the pupil take tlie last number of apples, S, and re- verse the process. Ans, 40 apples. 19. There is a certain number, to which, if 4 be added, and 7 be subtracted, and the difference be multiplied by 8, and the product divided by 3, the quotient will be 64 ; what is that number ? Ans. 27. 20. A clicss board has 8 rows of 8 squares each ; how many squares on the board ? TT S5. 21. There is a spot of ground 5 rods long, and 3 rods wide ; how many square rods does it contain ? Note, A square rod is a square (like one of those in the annexed figure) meas- uring a rod on each side. By an inspection of the figure, it will be seen, that there are as many squares in a row as rods on one side, and that the number of rows is equal to the number of rods on the other side ; therefore, 5 X 3= 15, the number of squares. Ans, 15 square rods. A figure like A, B, C, D, having its opposite sides equal and parallel, is called a parallelogram or oblong, 22. There is an oblong field, 40 rods long, and 24 rods wide ; how many square rods does it contain ? 23. How many square inches in a board 12 inches long, and 12 inches broad ? Ans, 144. r> c ! I f I -" ■ i I ._L_^^ 1 _ J. J A B t 25. MISCELLANEOUS Q^UESTIONS. 55h 24. How many square feet in a board 14 teet long and 2 feet wide ? 25. A certain township is six miles square ; how many square miles does it contain ? Ans, 36. 26. A man bought a farm for 22464 dollars ; he sold one half of it for 12480 dollars, at the rate of 20 dollars per acre; how many acres did he buy ? and what did it cost him per acre ? 27. A boy bought a sled for 86 cents, and sold it again for 8 quarts of walnuts ; he sold one half of the nuts at 12 cents a quart, and gave the rest for a penknife, which he sold for 34 cenis ; how many cents did he lose by his bargains ? 28. In a certain school-house, there are 5 rows of desks ; m each row are six seats, and each seat will accommodate 2 pupils ; there are also 2 rows, of 3 seats each, of the same size as the others, and one long seat where S pupils may sit; how many scholars will this house accommo- date? ' Alts. 80. 29. How many square feet of boards will it take for the floor of a room 16 feet long, and 15 feet wide, if we allow 12 square feet for waste ? 30. There is a room 6 yards long and 5 yards wide ; how many yards of carpeting, a yard wide, will be sufficient to co\ er the floors, if the hearth and fireplace occupy 3 square yards ? 31. A board, 14 feet long, contains 28 square feet; what IS its breadth ? 32. How many pounds of pork, worth 6 cents a pound, can be bought for 144 cents ? 33. How many pounds of butter, at 15 cents per pound, must be paid for 25 pounds of tea, at 42 cents per pound ? 34. 4 + 5+6+1+8= how many ? 35. 4 + 3+10 — 2 — 4+6 — 7 = how many ? 36. A man divides 30 bushels of potatoes among 3 poor men ; how many bushels does each man receive ? What is J of thirty ? How many are § {two thirds) of 30 ? 37. How many are one third (^) of 3 ? • of 6 ? of 9 ? of 282 ? of 45674312 ? 38. How many are tivo thirds (f ) of 3 ? of 6 ? of 9? of 282? of 45674312? 39. How many are J of 40 ? f of 40 ? ■ ^ of 60? fof60? lofSO? of 124? of 246876 ? f of 246876 ? 40. How many is -i of 80? |of80? ^ of 100? 41. An inch is one twelfth part i-^^) of a foot; how many 56 COMPOUND NUMBERS. IT 25, 26 feet in 12 inches ? in 24 inches ? in 36 inches ? in 12243648 inches ? 42. If 4 pounds of tea cost 128 cents, what does 1 pound cost ? 2 pounds ? 3 pounds ? 5 pounds ? 100 pounds ? 43. When oranges are worth 4 cents apiece, how many can be bought for four pistareens, (or 20 cent pieces ?) 44. The earth, in moving round the sun, travels at the rate of 68000 miles an hour ; how many miles does it travel in one day, (24 hours ?) how many miles in one year, (365 days ?) and how many days would it take a man to travel this last distance, at the rate of 40 miles a day ? how many years ? Aiis. to the last, 40800 years. 45. How much can a man earn in 20 weeks, at 80 cents per day, Sundays excepted ? 46. A man married at the age of 23 ; he lived with his wife 14 years ; she then died, leaving him a daughter, 12 years of age ; 8 years after, the daughter was married to a man 5 years older than herself, who was 40 years of age when the father died ; how old was the father at his death ? Ano. 60 years 47. There is a field 20 rods longj and 8 rods mide ; how many square rods does it contain t Ans, 160 rods. 48. What is the width of a field, which is 20 rods long, and contains 160 square rods ? 49. What is the length of a field, 8 rods wide, and con- taining 160 square rods ? 50. What is the width of a piece of land, 25 rods long, and containing 400 square rods ? IT 26. VA number expressing things of the same kind is called a simple number ;' thus, 100 men, 56 years, 75 cents, are each of them simple numbers Y but when a number ex- presses things of difterent kinds,'4t is called a compound nuwr her ; thus, 43 dollajs 25 cents and 3 mills, is a compound number; so 4 years 6 months and 3 days, 46 pounds 7 shillings and 6 pence, are compound numbers. NoteS Different kinds, or names, are usually called 3\g(25 which divided by 3, the ^ quotient is 175 cents, Answer^ 175 cents^ = $ V75, which, reduced to dollars, is $ 1'75, the answer. 2. Bought 4 bushels of corn for $ 3 ; what was that a bushel ? 4 is not contained in 3 ; we may, however, reduce tU^ (S3 to cents, by annexing two ciphers, thus : OPERATION. 300 cents divided by 4, the quotient 4)3^ |g Y5 cents, the price of each bush, of Am, <75 cents, corn. 3. Bought 18 gallons of brandy, for $42^75 ; what did it cost a gallon ? OPERATION. 18)42'75(2375 mills, = $2^375, the ansiver. 36 . 67 $ 42^75 is 4275 cents. After bringing 54 down the last figure in the dividend, and dividing, there is a remainder of 9 cents, which, by annexing a cipher, is reduced to mills, (90,) in which the divisor is con- 90 tained 5 times, which is 5 mills, and there 90 is no remainder. Or, we might have re* ^ duced $ 42*75 to mills, before dividing, by • '* annexing a cipher, 42750 mills, which, divided by 18, would have given the same result, 2375 mills, >vhich, reduced to dollars, is $ 2*375, the answer. 135 126 66 DIVISION OF FEDERAL MOBfUt; If SL 4. Divide $ 59^387 by 8. OPERATION. 8)59'387 Quotient j 7'423f , that is, 7 dollars, 42 cents, 3 mills, and | of another mill. The f is the remainder, after the last di- vision, written over the divisor, and expresses such fractional part of another mill. For all purposes of business, it will be sufficiently exact to carry the quotient only to mills, as the parts of a mill are of so little value as to be disregarded. Sometimes the sign of addition (-J-) is annexed, to show that there is a remainder, thus, $ 7'423 -[-. RUL.E. From the foregoing examples, it appears, that division of federal money does not diifer from division of simple num- bers. The quotient will he the answer in the lowest denomina- tionin the green sur/ij which may then be reducedto dollars Note. If the sum to be divided contain only dollars, or dollars and cents, it may be reduced to mills, by annexing ciphers before dividing ; or, we may first divide, annexing ciphers to the remainder, if there shall be any, till it shall be reduced to mills, and the result will be the same. exampl.es for practice. 5. If I pay $ 468'7o for 750 pounds of wool, what is the value of 1 pound ? Ans, $ 0'625 ; or thus, $ 0^62^, 6. If a piece of elcth, measuring 125 yards, cost $ 181^25, v/hat is that a yard ? Ans. $ 1'45. 7. If 536 quintals of fish cost $ 1913^52, how much is that a quintal ? Am. $ 3^57. 8. Bought a farm, containing 84 acies, for $3213 ; what did it cost me per acre ? Ajis. $ 38^25. 9. At $ 954 for 3816 yards of flannel, what is that a vard ? Ans. $0'25. 10. Bought 72 pounds of raisins for $ 8 ; what was that a pound ? ^,- =: how much ? Ans. $0411^; or, $0411+. 11. Divide $ 12 into 200 equal parts; how much is 07?e of the parts ? ^^^^r ~ liow mnch ? Ans. $ 0*006. 12. Divide $ 30 by 750. y^ ~ ^^^^''^ much ? 13. Divide $60 by 1200. yf^Ty = how much ? 14. Divide $215 into 86 equal parts; how much will one of the parts be ? -^g^- == ^^^^"^ much ? \ ^ 3^1. SUPPLEMENT TO FEDERAL MONEY. 6t 15. Divide $ 176 equally among 250 men ; how much will each man receive ? ^|4 z=. how much ? SUPPLEMENT TO FEDERAL MONEY. QUESTIONS. 1. What is understood by simple numbers? 2. *' by compound numbers? 3. by different denomina- tions ? 4. What is federal money ? 5. What are the de- nominations used in federal money ? 6. How are dollars distinguished from cents ? 7. Why are two places assigned for cents, while only one place is assigned for mills ? 8. To what does the relative value of mills, cents, and dollars correspond ? 9. How are mills reduced to dollars ? 10. • to cents? 11. Why? 12. How are dollars reduced to cents? 13. to mills? 14. Why? 15. How is the additioD of federal money performed? 16. subtraction? 17. multiplication? 18. divi- sion? 19. Of what name is the product in multiplication, and the quotient in division ? 20. In case dollars only are given to be divided,* what is to be done ? 21. When is one number or quantity said to be an aliquot part of another ? 22. What are some of the aliquot parts of a dollar "^ 23. When the price is an aliquot part of a dollar, how may the cost be found ? 24. What is this manner of operating called ? 25. How do 3^ou find the cost of articles, sold by the 100 or 1000? EXERCISES. 1. Bought 23 firkins of butter, each containing 42 pounds, for 16^ cents a pound ; what would that be a firkin, and how much for the whole ? Ans. $ 159'39 for the whole. 2. A mon killed a h^^i^ which he sold as follows, viz. the hind quarters, weighing 129 pounds each, for 5 cents a pound ; the fore quarters, one weighing 123 pounds, and the other 125 pounds, for 4J cents a pound; the hide and tal- low, weighing 163 pounds, for 7 cents a pound; to what did the whole amount ? Ans, ,$ 35*47. 3. A farmer bought 25 pounds of clover seed at 11 cents a pound, 3 pecks of herds grass seed for $ 2^25, a barrel of flour for $6'50, 13 pounds of sugar at 12J cents a pound; for which he paid 3 cheeses, each vv^eighhig 27 pounds, at 8^ cents a pound, and 5 barrels of cider at $ V25 a barrel. The balance between tlie articles bought and sold is 1 cent is it /or, or a^jainst the farmer ? 6S SUPPLEMENT TO FEDERAL MONEY. If 32. 4. A man dies, leaving an estate of $ 71600 ; there are demands against the estate, amountirg to $ 39876*74 ; the residue is to be divided between 7 sons; what will each one receive ? 5. How much coffee, at 25 cents a pound, may be had for 100 bushels of rye, at 87 cents a bushel ? Ans, 348 pounds. 6. At 12J cents a pound, what must be paid for 3 boxes of sugar, each containing 126 pounds ? 7. If 650 men receive $86'75 each, what will they all receive ? 8. A merchant sold 275 pounds of iron, at 6J cents a pound, and took his pay in oats, at $ 0'50 a bushel ; how many bushels did he receive ? 9. How many yards of cloth, at $ 4*66 a yard, must be given for 18 barrels of flour, at $9*32 a barrel ? 10. What is the price of three pieces of cloth, the first containing 16 yards, at $ 3'75 a yard ; the second, 21 yards, at $ 4*50 a yard ; and the third, 35 yards, at $ 5*12^ a yard ? IT 32. It is usual, when goods are sold, for the seller to deliver to the buyer, with the goods, a bill of the articles and their prices, with the amount cast up. Such bills are sometimes called bills of parcels. Boston, January 6, 1827. Mr. Mel Atlas Bought ofBenj. Burdett 12^ yards figured Satin, at $ 2*50 a yard, $31*2I» 8 ...;... sprigged Tabby,... 1*25 10*00 Received payment, $41*25 Benj. Burdett. Salem, June 4, 1827. Mr. James Paywell Bought of Simeon Thrifty S hogsheads new Rum, 118 gal. each, at $0*31 a gal. 2 pipes French Brandy, 126 and 132 gal. .. 1*12 J 1 hogshead brown Sugar, Of cwt. .. 10*34 .. cwt- 3 casks of Rice, 269 lb. each, .. *05 ..lb. 5 bags Coffee, 75 lb. each, .- *. *23 1 chest hyson Tea, 861b. .. *92 Received payment, $706*62j For Simeon Thrifty, ^ Pkter Faithful, ^ 32, 33. REDUCTION. 69 Wilderness, February 8, 1827. Mr. Peter Carpenter (See ir 30.) Bought of Asa Fdliree 5682 feet Boards, at $ 6 per M. 2000 8'34 800 Thick Stuff, .. 12^64 1500 Lathing, .. 4' 650 Plank, .. 10' 879 Timber, .. 2'50 ..... C. 236 2^75 Received payment, $ 101^849 Asa Falltree. Note, M. stands for the Latin mille^ which signifies 1000, and C.for the Latin word centum^ which signifies 100. IT 33. We have seen, that, in the United States, money is reckoned in dollars, cents, and mills. In England, it is reckoned in pounds, shillings, pence, and farthings, called denominations of money. Time is reckoned in years, months, weeks, days, hours, minutes, and seconds, called denomina- tions of time. Distance is reckoned in miles, rods, feet, and inches, called denominations of measure, &:c. The relative value of these denominations is exhibited in tables, which the pupil miist commit to memory. ENGLISH MONEY. The denominations are pounds, shillings, pence, and far- things. TABLE. 4 farthings (qrs.) make 1 penny, marked d. 12 pence - - - - i shilling, - - s. 20 shillings - - - - 1 pound, - - £, Note, Farthings are often written as the fraction of a penny ; thus, 1 farthing is written J d., 2 farthings, ^ d., S farthings, f d. 70 REDUCTION. How many farthings in penny? in 2 nt IT 35. 1 in 2 pence ? — — in 3 pence ? in 6 pence ? in 8 pence ? in 9 pence ? in 12 pence ? •- in 1 shilling ? in 2 shillings ? How many pence in 2 shil- lings ? in 3 s. ? in 4 s. ? in 6 s. ? in 8 s. ? in 10 s. ? in 2 shillings and 2 pence ? - in 2 s. 4 d. ? in 3 s. 6 d. ? in 2 s. 3 d. ? in 2 s. 6 d. ? in 4 s. 3 d. ? How many shillings in 1 pound ? in 2ie . ? mZ£,? in 4£ . ? in 4iS . 6 s. ? in 6^ . 8 s. ? in 3£. 10 s.? in 2£. 15 s.? How many pence in 4 far- things ? in 8 farthings ? in 12 farthings ? in 24 farthings ? in 32 far- thing - in 36 farthings ; in 48 qrs. ? How many shillings in 48 qrs. ? in 96 qrs. ? How many shillings in 24 pence ? in 3C d. ? in48d. ? in72d.? in 96 d. ? in 120 d. ? in 20 d. ? in 27 d. ? in 28 d. ? in 30 d. ? in 42 d. ? in 51 d. ? How many pounds in 20 shil- lings ? in 40 s. ? in 60 s. ? ■ in 80 s. ? in 86 s. ? -in 128 s. ? in 70 s. ? in 55 s. ? It has already heen remarked, that the changing of ojie kind, or denomination, into another kind, or denomination, without altering their value, is called Reduction. (IF 27.) 'J'hus, when we change shillings into pounds, or pounds into shillings, we are said to reduce them. From the foregoing examples, it is evident, that, when we reduce a denomina- tion of greater value into a denomination of less value, th« reduction is performed by multiplication ; and it is then call ed Reduction Descending, But when we reduce a denomina- tion of less value into one of greater value, the reduction is performed by division ; it is then called Reduction Ascending, Thus, to reduce pounds to shillings, it is plain, we must multiply by 20. And again, to reduoe shillings to pounds, we must divide by 20. It follows, therefore, that reduction descending and ascending reciprocally prove each other. IT 33, 34. REDUCTION. 71 1. Inl7ie. 13 s. 6Jd. how many farthings ? OPERATION. £. s. d. qrs 17 13 6 3 20 5. 3535. in 17ie. 13 5. \2d. 4242 d. 4^. 16971 qrs. the Ans, In the above example, be- cause 20 shillings make 1 pound, therefore we multiply \1£. by 20, increasing the product by the addition of the given shillings, (13,) which, it is evident, must always be done in like cases ; then, be- cause 12 pence make 1 shil- ling, we multiply the shillings (353) by 12, adding in the given pence, (6.) Lastly, because 4 farthings make 1 penny, we multiply the pence <'4242) by 4, adding in the given farthings, (3.) We then find, that in 17£. 13 s. 6fd., are contained 16971 farthings. ^ 34. The process in the fully examined, will be found To reduce high deTiomina- tions to lower, — Multiply the highest denomination by that number which it takes of the next less to make 1 of this higher, (increasing the pro- duct by the number given, if any, of that less denomina- 2. In 16971 farthings, hoTT many pounds .? OPERATION. Farthings in a penny, 4)16971 3 qrs* Pence in a shilling, 1 2 )4242 6 d- Shillings in a pound, 2|0)35l3 13*. 17^. Ans. 17 £. 135. 6fJ. Farthings will be reduced to pence, if we divide them by 4, because every 4 far- things make 1 penny. There- fore, 16971 farthings divided by 4, the quotient is 4242 pence, and a remainder of 3, which is farthings, of the same name as the dividend. We then divide the pence (4242) by 12, reducing them to shillings ; and the shillings (353) by 20, reducing them to pounds. The last quotient, 17iS., with the several re- mainders, 13 s. 6 d. 3 qrs. con- stitute the answer. Note. In dividing 353 s. by 20, we cut oiF the cipher, &c., as taught IT 22. foregoing examples, if care- to be as follows, viz. To reduce low denominations to higher, — Divide the lowest denomination given by that number which it takes of the same to make 1 of the next higher. Proceed in the same manner with each succeeding denomination, until you hav« n HEDUCTIOlSr, IT 34. tion.) Proceed in the same Planner with each succeeding denomination, until you have brought it to the denomination required. brought it to the denomination required. EXAMPLES FOR PRACTICE. B. Reduce Z2£. 15 s. 8d. ^o farthings. 5. In 29 guineas, at 28 s. ^ach, how many farthings ? 7. Reduce $ 183, at 6 s. each, to pence ? 9. In 15 guineas, how many pounds ? 4. Reduce 31 4*^2 farthings to pounds. 6. In 38976 farthings, how many guineas ? \8. Reduce 11736 pence tc dollars. 10. Reduce 21^2. to guin- eas. Note. We cannot reduce guineas directly to pounds, but we may reduce the guineas to shillings^ and then the shil- lings to pounds. TROY WEIGHT. By Troy weight are weighed gold,* silver, jewels, and all iquors. The denominations are pounds, ounces, penny- weights, and grains. TABLE. 24 grains (grs.) make 1 pennyweight, marked pwt. 20 pennyweights - - 1 ounce, ----- oz. 12 ounces - 1 pound. lb. 11. Bought a silver tank- ard, weighing 3 lb. 5 oz., pay- ing at the rate of $ I'OS an ounce ; what did it cost ? 13. Reduce 2101b. Soz. 12 pwt. to pennyweights. 15. In 71b. 11 oz. 3 pwt. 9 grs. of silver, how many .grains ? 12. Paid $ 44^28 for a sil- ver tankard, at the rate of $ 1^08 an ounce; what did it weigh ? 14. In 50572 pwt. how many pounds ? 16. Reduce 456S1 grains to pounds. * The fineness of gold is tried by fire, and is recxctied m carats, by which is understood the 24th part of any quantity ; if it lose ncoiinff in the trial, it is said to be 24 carats finej if it lose 2 carats, it is then 22 c£,r"Etsiiiie, which is the stand- ard for gokl Silver wl*ch abides the fire without loss is said Ic c« 12 ounces filne. Tlie standard for silver coin is 11 oz. 2 pwts. of fine s.".^ei, and 18 pwts. of copper HKjlted tog^ether IT 34. REDUCTION. 73 APOTHECARIES' WEIGHT. Apothecaries' weight* is used hy apothecaries and physi- ciaus, in compounding medicines. The denominations ar© pounds, ounces, drams, scruples, and grains. TABLE. 20 grains, (grs.) make 1 scruple, marked 9. 3 scruples - - - 1 dram, - - - 3 . 8 drams - - - - 1 ounce, - - - g . 12 ounces - - - - 1 pound, - - - ib. 17. In9tb. 8§. 15. 29. 1 18. Reduce 65799 grs. to 19 grs., how many grains. [pounds. AVOIRDUPOIS WEIGHT.t By avoirdupois weight are weighed all things of a coarse and drossy nature, as tea, sugar, bread, flour, tallow, hay, leather, medicines, (in buying and selling,) and all kinds of metals, except goU and silver. The denominations are tons, hundreds, quarters, pounds, ounces, and drams. TABLE. 16 drams, (drs.) make 1 ounce, - marked - oz. 16 ounces ----- 1 pound, ----- lb. 28 pounds ----- 1 quarter, ----- qr. 4 quarters ----- 1 hundred v/eight, - - cwt. 20 hundred weight - - 1 ton, ------ T. Note 1. In this kind of weight, the words gross and net are used. Gross is the weight of the goods, together with the box, bale, bag, cask, &c., which contains them. Net weight is the weight of the goods only, after deducting the weight of the box, bale, bag, or cask, &c., and all other al- lowances. Note 2. A hundred weight, it will be perceived, is 1 12 lb. Merchants at the present time, in our principal sea-ports, buy and sell by the 100 pounds. * The pouiic! and ounce apcthecarics' vvciglit, and the pound arid ounce Troy, are the same, only differonlly divided, and suodiddcd. t 175 oz. Troy =2: 192 oz. avoirdupois, and 175 )b. Troy -= lit lb. avoirdu- DOJs. 1 lb. Troy =^ .5760 grdns, and 1 )b. avoirdupois =s 7000 grains Troy* G 74 REDUCTION. IF 34. 19. What willScwt. 3 qrs. 171b. of sugar come to, at 12J cents a pound. 21. A merchant would put 109 cwt. qrs. 12 lb. of raisins into boxes, containing 26 lb. each ; how many boxes will it require ? 23. In 12 tons, 15 cwt. 1 qr. 191b. 6 oz. 12 dr. how many drams ? 25. In 28 lb. avoirdupois, how many pounds Troy ? 20. How much sugar, at 12 J- cents a pound, may be bought for $ 82*625 ? 22. In 470 boxes of raisins, containing 26 lb. each, how many cwt. ? 24. In 7323500 drams, how many tons ? 26. In 34lb. oz. 6 pwt. 16 grs. Troy, how many pounds avoirdupois ? CLOTK MEASURE. Cloth measure is used in selling cloths and other goods, sold by the yard, or ell. The denominations are ells, yards, quarters, and nails. TABILE, 1 nails, (na.) or 9 inches, make 1 quarter, marked qr. 1 quarters, or 36 inches, - 1 yard, - - - - yd. 3 quarters, - - - - - - 1 ell Flemish, - - E. Fl. 5 quarters, ------ 1 ell English, - - E. E. 6 quarters, ------ 1 ell French, - - E.Fr. 27. In 573 yds. 1 qr. 1 na. now many nails? 29. In 151 ells Eng. how many yards ? Note. Consult ^ 34, ex. 3 . 28. In 9173 nails, how ma- ny yards ? 30. In 188f yards, how ma- ny ells English ? LONG MEASURE. Long measure is used in measuring distances, or other things, where length is considered without regard to breadth. The denominations are degrees, leagues, miles, furlongs, rods, yards, feet, inches, and barley-corns. IT 84; REDUCTION. 75 TABL.E. 3 barley-corns, (bar.) make 1 incb, - marked - 1 foot, - - - - - l2 inches, 3 feet, ------ 5 J yards, or 16^ feet, 40 rods, or 220 yards, - - 8 furlongs, or 320 rods, - 3 miles, - - - - - 60 geographical, or 69J statute miles, - - 360 degrees, - - - 1 yard, - - - - - 1 rod, perch, or pole, - 1 furlong, - - - - 1 mile, - - - - - 1 league, - - - - -\ 1 degree, - - a great circle, or ence of the earth. m. ft. yd. fur. M. L. or ^ deg. circumfer 31. How many barley-corns will reach round the globe, it being 3G0 degrees ? Note. To multiply by 2, is to take the multiplicand 2 times ; to multiply by 1, is to take the multiplicand 1 lime; to multiply by j-, is to take the multiplicand half a time, that is, the half of it. Therefore, to reduce 360 degrees to stat- ute miles, we multiply first by the whole number,^ 69, and to the product add half the multi- plicand. Thus : J) 360 69^ 3240 2160 180 half of the multiplicand. 25020 statute miles in 360 de- grees. 33. How many inches from Boston to the city of Wash- ington, it being 482 miles ? 35. How many times will a wheel, 16 feet and 6 inches in circumference, turn round in the distance from Boston to Providence, it being 40 miles } 32. In 4755801600 barley- corns, how many degrees ? Note. The barley-corns be- ing divided by 3, and that quotient by 12, we have 132105600 feet, which are to be reduced to rods. We can- not easily divide by 16J on account of the fraction ^ ; but 16^ feet = 33 half feet^ in 1 rod ; and 132105600 feet =z 264211200 half feety which, divided by 33, gives 800G400 rods. Hence, when the divisor is encumbered with a fraction, ^ or J, &c., we may reduce the divisor to halves, or fourths, &c., and reduce the dividend to the same; then the quo- tient will be the true answer. 34. In 30539520 inched, how many miles ? 36. If a wheel, 16 feet 6 inches in circumference, turn round 12800 times in going from Boston to Providence^ what is the distance ? 76 REDUCTION. it S6. LAND OR SQUARE MEASURE. Square measure is used in measuring land, and any other tiding, where length and breadth are considered. The de- nominations are miles, acres, roods, perches, yards, feet and inches. IT 35, 3 feet in length make a yard in long measure ; but it requires 3 feet in length and 3 feet in breadth to make a yard in square measure ; 3 feet in length and one. foot wide make 3 square feet ; 3 feet in length and 2 feet wide make 2 times 3, that is, 6 square feet ; 3 feet in length and 3 feet wide make 3 times 3, that is, 9 square feet. This will clearly appear from the annexed figure. 3 feet = 1 yard. It is plain, also, that a square foot, that is, a square 12 inches in length and 12 inches in breadth, must con- tain 12 X 12 := X44 square inches. 11 5 o 144 TABIiE. square inches nz 12 X 12 ; that is, ^ 12 inches in length and 12 inches > make 1 square foot, in breadth ------ N 9 square feet =; 3 X 3 ; that is, 3 feet in length and 3 feet in breadth 30^ square yards — 5^ X ^y or 272^ square feet = 16^ X 16^, 40 square rods, ------ 4 roods, or 160 square rods, 640 acres, ------ 1 square yard. ( 1 square rod, ' ( perch or pole. . 1 rood. • 1 acre. • 1 square mile. Note. Gunter's chain, used in measuring land, is 4 rods in length. It consists of 100 links, each link being 7-^ inches in length ; 25 links make 1 rod, long measure, and 625 square links make 1 square rod. !r 35, 36. REDUCTIOX. 77 37. In 17 acres 3 roods 12 rods, how many square feet ? Note, In reducing rods to feet, the multiplier will be 272 J. To multiply by J-, is to take a fourth part of the mul- tiplicand. The principle is the same as shown IT 34, ex. 31. 39. Reduce 64 square miles o square feet ? 41. There is a town 6 miles square ; how many square miles in that town ? how many acres? ^ 38. In 778457 square feet, how many acres ? Note. Here we have 776457 square feet to be divided by 2 72 J-. Reduce the divisor to fourths^ tliat is, to the lowest denomination contained in it ; then reduce the dividend to fourths^ that is, to the same denomination, as shown 1132, ex. 34. 40. In 1,784,217,600 square feet, how many square miles ? 42. Reduce 23040 acres to square miles. SOLID OR CUBIC MEASURE. Solid or cubic measure is used in measuring things that have length, breadth, and thickness; such as timber, w^ood, stone, bales of goods, &c. The denominations are cords, tons, yards, feet, and inches. IT 36. It has been shovv^n, that a square yard contains 3 X 3 =r, 9 square feet. A cubic yard is 3 feet long, 3 feet wide, and 3 feet thick. Were it 3 feet long, 3 {^fti wide, ; nd one foot thick, it would contain 9 cubic feet ; if 2 feet !iick, it would contain 2 X 9 = 18 cubic feet; and, as it is 3 feet thick, it does contain 3 X 9 i= 27 cubic feet. This will clearly appear from the annexed figure. It is plain, also, that a cubic foot, that is, a solid, 12 inches in length, 12 inches in breadth, and 12 inches in thickness, will contain 12 X 12 X 12 = 1728 solid or cubic inches. 78 REDUCTION. 1[ S6^ TABLE. '« 1728 solid inehes, z= 12 X 12 X 12, ^ that is, 12 inches in length, > make 1 solid foot 12 in breadth, 12 in thickness, ) 27 solid feet, :^ 3 X 3 X 3 - - - - 1 solid yard. 40 feet of round timber, or 50 feet > , ^ , , of hewn timber, I' ' 1 ton or load. 128 solid feet, — 8 X 4 X 4, that ) is, 8 feet in length, 4 feet in > - - 1 cord of wood, width, and 4 feet in height, ) Note. What is called a cord foot ^ in measuring wood, is 16 solid Uiei ; that is, 4 feet in length, 4 feet in width, and 1 foot in height; and 8 such feet, that is, 8 card feet make 1 cord. 44. In 622080 cubic inche5> how many tons of round tim- ber ? 46. In 592 solid feet of wood, liow many cord feet r 48. In 8 cards of wood, how many cord feet ? 50. 2048 solid feet of wood, how many cord feet? how I how many cord feet? how many solid feet? jmany cords? 43, Reduce 9 tons of round timber to cubic inches. 45. In 37 cord feet of wood, how many solid feet ? 47. Reduce 64 cord feet of wijcd to cords. 49. In IG cords of w^ood, WINE MEASURE. Wine measure is used in measuring all spirituous liquors, ale and beer excepted ; also vinegar and oil. The denomi- nations are tuns^ pipes, hogsheads, barrels, gallons, quarts, pints, and gills. TABLE. 4 gills (gi.) - make - - 1 pint, marked pt. 2 pints -------1 quart, - - - qt. 4 quarts ------ ^ gallon, - - - gal. 3H gallons ------ 1 barrel, - - - bar. 63" gallons ------ 1 hogshead, - - hhd. 2 hogsheads ----- i pipe, - - - P. 2 pipes, or 4 hogsheids - 1 tun, - - - - T. Note, A gallon, wine measure, contains 231 cubic inches. TF 36. REDUCTION. 79 51. Reduce 12 pipes of wine to pints. 53. In 9 P. 1 hhd. 22 gals. 3 qts. how many gills ? 55. In a tun of cider, how many gallons ? 52. In 12096 pints of wine, how many pipes ? 54. Reduce 39032 gills to pipes. 56. Reduce 252 gallons to tuns. ALE OR BEER MEASURE. Ale or beer measure is used in measuring ale, beer, and milk. The denominations are hogsheads, barrels, gallons, quarts, and pints. TABLE. 2 pints (pts.) - make - 1 quart, - marked qt. 4 quarts ----- i gallon, ----- gal, 36 gallons - - - - i barrel, ----- bar. 54 gallons ----- i hogshead, - - - - hhd Note, A gallon, beer measure, contains 282 cubic inches. 57. Reduce 47 bar. 18 gal. of ale to pints. 59. In 29 hhds. of beer, how many pints ? 58. In 13680 pints of alC; how many barrels ? 60. Reduce 12528 pints to hogsheads. DRY MEASURE. Dry measure is used in measuring all dry goods, such as grain, fruit, roots, "salt, coal, &:c. The denominations are chaldrons, bushels, pecks, quarts, and pints. TABL,E. 2 pints (pts.) make - 1 quart, - marked - qt. 8 quarts ----- l peck, ----- pk. 4 pecks ----- 1 bushel, ----- bu. 36 bushels ----- 1 chaldron, - - - - ch , Note. A gallon, dry measure, contains 268|- cubic inches. A Winchester bushel is 18^ inches in diameter, 8 inches deep, and contains 2150f cubic inches. 80 61. In 75 bushels of wheat, how many pints ? 63. Reduce 42 chaldrons of coals to pecks. REDUCTION. U 36, 37. 62. In 4800 pints, how ma- ny bushels ? 64. In 6048 pecks, how ma- ny chaldrons ? TIME. The denominations of time are years, months, weeks, days, hours, minutes, and seconds. TABLE. 60 seconds (s.) - make - 1 minute, marked m. 60 minutes ------ i hour, - - - - h. 24 hours -------i day, - - - - d. 7 days -------i week, - - - - w. 4 weeks ------ i month, - - - - mo 13 months, 1 day and 6 hours, ^ 1 common, or ) or 365 days and 6 hours, 5 Julian year, 5 " ^^' IT 07, The year is also divided into 12 calendar months, which, in the order of their succession, are numbered as fol- lows, viz, January, 1st month, has 31 days. February, 2d, - - - 28 March, 3d, - - - 31 Mayl' ■ tt] -" "- I 31 ^^^ ^^^'kT^-!^ June - 6th 30 can be divided by 4 with- July, - TO^ 31 outaremamder,itiscall. August, 8th - - - 31 ed leap year, m which September, 9th, 30 February has 29 days. October, 10th, - - - 31 November, 11th, - - - 30 December, 12th, - - - 31 The number of days in each month may be easily fixed in the mind by committing to memory the following lines : Thirty days hath September, April, June, and November, February twenty-eight alone ; All the rest have thirty-one. ttSl REDUCTION. 81 The first seven letters of the alphabet, A, B, C, D, E^fF, G, are used to mark the several days of the week, and they are disposed in such a manner, for every year, that the letter A shall stand for the 1st day of January, B for the 2d, &c. In pursuance of this order, the letter which shall stand for Sun" dai/j in any year, is called the Dominical letter for that year. " The Dominical letter being known, the day of the week on which each month comes in may be readily calculated from the following couplet : At Dover Dwells George Brown, Esquire, * Good Carlos Finch And David Fryer. These words correspond to the 12 months of the year, and the first letter in each word marks the day of the week on which each corresponding month comes in; whence any other day may be easily found. For example, let it be required to find on what day of the week the 4th day of July falls, in the year 1827, the Dominical letter for which year is G. Good answers to July ; consequently, July comes in on a Sunday ; wherefjore the 4th day of July falls on Wednesday. Note, There are two Dominical letters in leap years, one for January and February, and another for the rest of the year. 65. Supposing your age to 66. Reduce 475047465 se- be 15 y. 19 d. lib. 37 m. conds to years. 45 s., how many seconds old are you, allowing 365 days 6 hours to the year ? 67. How many minutes from the 1st day of January to the 14th day of August, inclusive- ly? 69. How many minutes from the commencement of the war between America and Eng- land, April 19th, 1775, to the settlement of a general peace, which took place Jan. 20th, 17S3? 68. Reduce 325440 minutes to days. 70. In 4079160 minutes, how many years ? 62 SUPPLEMENT TO REDUCTION. IT 37* CIRCULAR MEASURE, OR MOTION. Circular measure is used in reckoning latitude and longi- tude ; alsd in computing the revolution of the earth and other planets round the sun. The denominations are circles, signs, degrees, minutes, and seconds. TABL.E. 60 seconds (^') - make - 1 minute, - marked - ' 60 minutes ----- i degree, ----- o 30 degrees ----- i sign, ------ s. 12 signs, or 360 degrees, - 1 circle of the zodiac. Note, Every circle, whether great or sriiall, is divisible into 360 equal parts, called degrees. 71. Reduce 9 s. 13^ 25' to! 72. In 1020300", how many seconds. | degrees ? The following are denominations of things not included in the Tables :— 12 particular things - make - 1 dozen. 12 dozen -------- i gross. 12 gross, or 144 dozen, - - - - i great gross. Also, 20 particular things - make - 1 score. 6 points make 1 line, ( used in measuring the length of 12 lines - - 1 inch, ( the rods of clock pendulums. 4 inches - - 1 hand, \ ""^f ™ measuring the height of ' ( noises. 6 feet - - 1 fathom, used in measuring depths at sea. 112 pounds - - make - - 1 quintal of fish. 24 sheets of paper - make - 1 quire. 20 quires -------1 ream. SUPPX-EMENT TO KEDUCTZON. QUESTIONS. 1. What is reduction? 2. Of how many varieties is re- duction ? 3. What is understood by different denominations^ as of money, weight, measure, &c. ^ 4. How are high de- IT 37. SUPPLEMENT TO REDUCTION. 8S nominations brought into lower ? 5. How are low denomi- nations brought into higher ? 6. What are the denomina- tions of English money ? 7. What is the use of Troy weight, and what are the denominations ? 8. avoirdupois weight ? the denominations ? 9. What distinction do you make between gross and we/ weight? 10. What dis- tinctions do you make between long, square, and cubic •neasure ? 11. What are the denominations in longmea-, sure ? 12. — in square measure ? 13. — — in cubic mea- sure ? 14. How do you multiply by ^ ? 15. When the di- visor contains a fraction, how do you proceed ? 16. How is the superficial contents of a square figure found ? 17. How is the solid contents of any body found in cubic measure ? 18. How many solid or cubic feet of v/ood make a cord ? 19. What is understood by a cord foot? 20. How many such feet make a cord ? 21. What are the denominations of dry measure ? 22. of wine measure ? 23. of time ? 24. of circular measure ? 25. For what is cir- cular measure used ? 26. How many rods in length is Gun- ter^s chain ? of how many links does it consist ? how many links make a rod ? 27. How many rods in a mile ? 28. How many square rods in an acre ? 29. How many pounds make 1 cwt. ? EXERCISES. 1. In 46i£ . 4 s., how many dollars ? Ans. $ 154. 2. In 36 guineas, how many crowns, at 6 s. 7d. each ? Ans. 153 crowns, and 9d. 3. How many rings, each weighing 5 pwt. 7 grs., may be made of 31b. 5 oz. 16 pwt. 2 grs. of gold? Aiis, 158. 4. Suppose West Boston bridge to be 212 rods in length, how many times will a chaise wheel, 18 feet 6 inches in circumference, turn round in passing over it ? ** Ans, 1892^^2 times. 5. In 470 boxes of sugar, each 26 lb., how many cwt. ? 6. In 10 lb. of silver, how many spoons, each weighing . oz. 10 pwt. ? 7. How many shingles, each covering a space 4 inches one way and 6 inches the other, would it take to cover 1 square foot ?;> How many to cover a roof 40 feet long, and 24 feet wide ? (See If 25.) Ans. to the last^ 5760 shingles, 8. How many cords of wood in a pile 23 feet long, 4 feet wide, and 6 feet high ? Ans. 4 cords, and 7 cord feet. 84 SUPPLEMENT TO REDUCTION. IT 87. 9. There is a room 18 feet iu length, 16 feet in width, •iU>d 8 feet in height; how many rolls of paper, 2 feet wide, and containing 1 1 yards in each roll, v/ili it take to cover the v/alls? Am, 8^. 10. How many cord feet in a load of wood 6^ feet long, 2 feet wide, and 5 feet liigh ? Ans. 4y^g^ cord feet. 11. If a ship sail 7 miles an hour, how far will she sail. £t that rate, in 3 w. 4 d. 16 h. ? 12. A merchant sold 12 hhds. of brandy, at §2'75 a gal- lon ; how much did each hogshead come to, and to how much did the whole amount ? 13. How much cloth, at 7 s. a yard, may be bought for 29£. Is.? 14. A goldsmith sold a tankard for 10£. Ss. at the rate of 5 s. 4 d. per ounce ; how much did it weigh ? 15. An ingot of gold weighs 2 lb. 8 oz. 16 pwt. ; how much is it worth at 3 d. per pwt. ? 16. At $ 048 a pound, what will 1 T. 2 cwt. 3 qrs. 16 lb. of lead come to ? 17. Reduce 14445 ells Flemish to ells English. 18. There is a house, the roof of which is 44J feet in length, and 20 feet in width, on each of the two sides; if 3 shingles in width cover one foot in length, how many shingles will it take to lay one course on this roof? if 3 courses make one foot, Iiow many courses will there be on une side of the roof? how many shingles will it take to cover one side ? to cover both sides ? A.}is. 16020 shingles. 19. How many steps, of 30 inches each, must a man take in travelling 54.j~ miles ? 20. How many seconds of lime would a person redeem in 40 years, by rising each morning j hour earlier than he now does ? "21. If a man lay up 4 shillings each day, Sundays ex- cepted, how many dollars would he lay up in 45 years ? 22. If 9 candles are made from 1 pound of tallow^ how many dozen can be made from 24 pounds and 10 ounces ? 23. If one pound of wool make 60 knots of yarn, how many skeins, of ten knots each, may be spun from 4 pounds 6 ounces of wool ? IT 38. ADDITION OF COMPOUND NUMBERS. 85 ADSXTZOKT OF COMPOUND NUMBERS. M 38. 1. A boy bought a knife for 9 pence, and a comb for 3 pence ; how much did he give for both ? Ans. 1 shilling. 2. A boy gave 2 s. 6 d. for a slate, and 4 s. 6 d. for a book ; how much did he give for both ? 3. Bought one book for 1 s. 6 d., another for 2 s. 3 d., an- other for 7 d. ; how much did they all cost ? Ans, 4 s. 4 d. 4. How many gallons are 2 qts. + 3 qts. + 1 qt. ? 6. How many gallons are 3 qts. + 2 qts. + 1 qt. + 3 qts. + 2 qts. ? 6. How many shillings are 2 d. + 3 d. + 5 d. -|- 6 d. -[- 7 d. ? 7. How many pence are 1 qr. -j- 2 qrs. -j- 3 qrs. + 2 qrs. + 1 qr. ^ 8. How many pounds are 4 s. -)- 10 s. + 15 s. + 1 s. ? * 9. How many minutes are 30 sec. -|- 45 sec. -f" 20 sec. ? 10. How many hours are 40 min. -f- 25 min. -f- 6 min. ? 11. How many days are 4 h. + 8 h. + 10 h. -j- 20 h. ? 12. How many yards in length are I L -{- 2 L -{- I L? 13. How many feet are 4 in. + 8 in. -]- 10 in. -|- 2 in. + 1 in. ? 14. How much is the amount of 1 yd. 2 ft. 6 in. + 2 yds. 1ft. 8in.? 15. What is the amount of 2 s. 6 d. +4s. 3d. + 7s. 8d.? 16. A man has two bottles, which he wishes to fill witli wine ; one will contain 2 gal. 3 qts. 1 pt., and the other 3 qts. ; how much wine can he put in them ? 17. A man bought a horse for 15iB. 14 s. 6 d., a pair of oxen for 20iS . 2 s. 8 d., and a cow for 5£ . 6 s. 4 d. ; what did he pay for all ? When the numbers a^e large, it will be most convenient to write them down, placing those of the same kind, or de- nomination, directly under each other, and, beginning with those of the least value, to add up each kind separately. ^ DERATION. In this example, adding up the column of pence, we.find the amount to be 18 pence, which being = 1 s. 6 d., it is plain, that we may write dpwn the 6 d. under the column o^ Aas, 41 3 6 pence, and reserve the 1 s. to be add- ed in with the other shillings. £. s. d. 15 14 6 20 2 8 5 6 4 H 86 ADDITION OF COMPOUND NUMBERS. V 38. Next, adding up the column of shillings, together with the 1 s. which we reserved, we find the amount to be 23 s. zn l£. 3 s. Setting the 3 s. under its own column, we add the 1 £ . with the other pounds, and, finding the amount to be UiS ., we write it down, and the work is done. Am. 41iB. 3 s. 6 d. Note. It will be recollected, that, to reduce a lower into a higher denomination, we divide by the number which it takes of the lower to make one of the higher denomination. In addition, this is usually called carrying for that number : thus, between pence and shillings, we carry for 12, and be- tween shillings and pounds, for 20, &c. The above process may be given in the form of a general Rule for the Addition of Compound Numbers : I. Write the numbers to be added so that those of the ame denomination may stand directly under each other. II. Add together the numbers in the column of the lowest lenomination, and carry for that number which it takes of he same to make one of the next higher denomination, j^roceed in this manner with all the denominations, till you ome to the last, whose amount is written as in simple num- bers. Proof The same as in addition of simple numbers. EXAMPIiES FOR PRACTICE. £. s. d. gr. 46 11 3 2 16 7 4 538 19 7 1 £. s. d. 72 9 6J- 18 10^ 36 16 5f £. 183 8 s. 19 17 15 d. 4 10 4 if), oz. jjwt. gr 36 7 10 11 42 6 9 13 81 7 16 15 Troy WeicIht. oz. piot. gr. 6 14 9 8 6 16 3 11 10 oz. 3 pwt. 13 7 J 18 16 4 Bought a silver tankard, weighing 2 lb. 3 oz., a silver cup, weighing 3 oz. 10 pv/t., and a silver thimble, weighing 2 p^vt. 13 grs. ; what was the weight of the whole ? 38. ADDITION OF COMPOUND NUMBERS. St ) Ayoirdupois Weight. r. ciDt, qr, lb. oz. dr. ctot. qr. lb. oz. dr. 14 11 1 16 5 10 16 3 18 6 14 25 2 11 9 15 i- 2 16 8 12 7 18 25 11 9 22 11 10 A man bought 5 loads of hay, weighing as follows, viz. 23 cwt. ( = 1 T. 3 cwt.) 2 qrs. 171b. ; 21 cwt. 1 qr. 16 lb. ; 19cwt. qr. 24 lb. ; 24 cwt. 3 qrs. ; 11 cwt. Oqr. lib.; how many tons in the whole ? Cloth Measure. yds. qr. na. E. Fl. qr. na. E. En. qr. na. 36 1 2 41 1 2 75 4 2 41 2 3 18 2 3 31 1 ' 65 3 1 67 1 28 3 1 There are four pieces of cloth, which measure as follows, viz. 36 yds. 2 qrs. 1 na. ; 18 yds. 1 qr. 2 na. ; 46 yds. 3 qrs. 3 na. ; 12 yds. qr. 2 na. ; how many yards in the whole ? Long Measure. Deg. mi. fur. r. ft. in. bar. 59 46 6 29 15 10 2 216 39 1 36 14 6 1 678 53 7 24 9 8 1 Mi. fur .poL 3 7 8 6 27 Land or Square Measure. Pol. ft. in. A. rood pol. ft. in. 36 179 137 56 3 37 245 228 19 248 119 29 1 28 93 25 12 96 75 416 2 31 128 119 88 ADDITION OF COMPOUND NUMBERS. IT 38. There are 3 fields, which measure as follows, viz. 17 A. 3r. 16 p.; 28 A. 5 r. 18 p.; 11 A. Or. 25 p.; how much laud in the three fields ? Solid or Cubic Measure. Ton. ft. in. yds. ft. in. cords, ft. 29 36 1229 75 22 1412 37 119 12 19 64 9 26 195 9 110 8 11 917 3 19 1091 48 127 Wine Measure. Hlid. gal. qts. pts. Tun. hhd. gal. qts. 51 53 1 1 37 2 37 2 27 39 3 19 1 59 1 9 13 1 28 2 A merchant bought two casks of brandy, containing as follows, viz. 70 gal. 3 qts. ; 67 gal. 1 qt. ; how many hogs- heads, of 63 gal. each, in the whole ? Dry Measure. Bus. ;?. qt. pt. Ch. bus. p. qts. 36 2 5 1 48 27 3 5 19 3 7 6 29 1 7 Time. Y. mo. ID. d. h. 772. 5. Y. m. w. d. 57 11 3 6 23 55 11 40 3 1 5 84 9 2 16 42 18 16 7 4 32 6 5 5 18 5 27 5 2 H 39, SUBTRACTION OF COMPOUND NUMBERS. 89 SUBTRACTION OF COMPOUND NUMBERS. ^ 39. 1. A boy bought a knife for 9 cents, and sold it for 17 cents ; how much did he gain by the bargain ? 2. A boy bought a slate for 2 s. 6 d., and a book for 3 s. 6 d. ; how much more was the cost of the book than of the slate ? 3. A boy owed his playmate 2 s. ; he paid him 1 s. 6 d. ; how much did he then owe him ? 4. Bought two books ; the price of one was 4 s. 6 d., the price of the other 3 s. 9 d. ; what was the difference of their costs ? 5. A boy lent 5 s. 3d.; he received in payment 2 s. 6 d. ; how much was then due ? 6. A man has a bottle of wine containing 2 gallons and 3 quarts ; after turning out 3 quarts, how much remained ? 7. How much is 4 gal. less 3 gal. ? 4 gal. — (less) 2 qts. ? 4 gal. — 1 qt. ? 4 gal. — 1 gal. 1 qt. ? 4 gal. — 1 gal. 2 qts. ? 4 gal. — 1 gal. 3 qts. ? 4 gal. — 2 gal. 3 qts. .? 4 gal. 1 qt. — 1 gal. 3 qts. ? 8. How much is 1 ft. — (less) 6 in. ? 1 ft. — 8 in. ? 6 ft. 3in. — . 1ft. 6in. ? 7ft. 8 in. —4ft. 2 in. ? 7 ft. 8 in. —■ 5 ft. 10 in. ? 9. What is the difference between 4£, 6 s. and 1^. 8 s.? 10. How much is 3£. — (less) Is.? 3^. — 2 s. ? 3^^. — 3s.? 3£, — 15 s.? Z£. 4s. — 2ie. 6s.? 10i£. 4s.— 5^.8. s? 11. A man bought a horse for 30iS. 4 s. 8d., and a cow for 5£ . 14 s. 6 d. ; what is the difference of their costs ? OPERATION. Xs the two numbers are large. Minuend 30 4 R ^ ^^^^^ ^^ convenient to write Subtrahmd, 5 14 6 them down, the' less under the ' greater, pence under pence, shil- Ans. 24 10 2 lings under shillings, &c. We may now take 6 d. from 8 d., and there will remain 2 d. Proceeding to the shillings, we can- not take 14 s. from 4 s., but we may borrow, as in simple num- bers, 1 from the pounds, zz= 20 s., which joined to the 4 s. makes 24 s., from which taking 14 s. leaves 10 s., which we set down. We must now carry 1 to the 5^., making 6£.^ which taken from 3Q,£ . leaves 24iS., and the work is done. Note, The most convenient way In borrowing is, to sub* 90 SUBTRACTION OF COMPOUND NUMBERS. ft S9, tract the subtrahend from the figure borrowed, and add the d'ifFerence to the minuend. Thus, in the above example, 14 from 20 leaves 6, and 4 is 10. The process in the foregoing example may be presented in the form of a Rule for the Subtraction of Compound Num- bers : I. Write down the sums or quantities, the less under the greater, placing those numbers which are of the same de- nomination directly under each other. II. Beginning with the least denomination, take succes- sively the lower number in each denomination from the up- per, and write the remainder underneath, as in subtraction of simple numbers. III. If the lower number of any denomination be greater than the upper, borrow as many units as make one of the next higher denomination, subtract the lower number there- from, and to the remainder add the upper number, remem- bering always to add 1 to the next higher denomination for that which you borrowed. Proof Add the remainder and the subtrahend together, as in subtraction of simple numbers ; if the work be right, the amount will be equal to the minuend. exampl.es for practice. 1. a merchant sold goodsto the amount of 136iS . 7 s. 6^d., and received in payment 50^. 10 s. 4fd; how much re- mained due ? Ans. S5£, 17 s. If d. 2. A man bought a farm for 1256dB. 10 s., and, in selling it, lost 87iB . 10 s. 6 d. ; how much did he sell it for ? Ans. U6S£, 19 s. 6 d. 3. A. man bought a horse for 27 £ . and a pair of oxen for 19<£ . 12 s. 8^ d. ; how much was the horse valued more than the oxen ? 4. A merchant drew from a hogshead of molasses, at one time, 13 gal. 3 qts. ; at another time, 5 gal. 2 qts. 1 pt. ; what quantity was there left ? Ans, 43 gal. 2 qts. 1 pt. 5. A pipe of brandy, containing 118 gal. sprang a leak, when it was found only 97 gal. 3 qts. 1 pt. ren^ained in the cask; how much was the leakage ? 6. There was a silver tankard which weighed 3 lb. 4 oz. ; the lid alone weighed 5 oz. 7pwt. 13 ^rs.; how much did the tankard weigh without the lid .? if Sfd. SUBTRACTION OF COMPOUND NUMBERS. 91 7. From 15 lb. 2 oz, 5 pwt. take 9 oz. 8 pwt. 10 grs. 8. Bought a hogshead of sugar, weighing 9 cwt. 2 qrs. 17 lb. ; sold at three several times as follows, viz. 2 cwt. 1 qr. 11 lb. 5 oz. ; 2 qrs. 18 lb. 10 oz. ; 25 lb. 6 oz. ; what was the weight of sugar which remained unsold ? Alls. 6 cwt. 1 qr. 171b. 11 oz. 9. Bought a piece of black broadcloth, containing 36 yds. 2 qrs. ; two pieces of blue, one containing 10 yds. 3 qrs. 2 na., the other, 18 yds. 3 qrs. 3 na. j how much more was there of the black than of the blue ? 10. From 28 miles, 5 fur. 16 r. take 15 m. 6 fur. 26 r. 12 ft. 11. A farmer has two mowing fields; one containing 13 acres 6 roods; the other, 14 acres 3 roods: he has two pastures also; one containing 26 A. 2 r. 27 p. ; the other, 45 A. 5 r. 33 p. : how much more has he of pasture than of mowing ? 12. From 64 A. 2 r. 11 p. 29 ft. take 26 A. 5 r. 34 p. 132 ft. 13. From a pile of wood, containing 21 cords, was sold, at one time, 8 cords 76 cubic feet ; at another time, 5 cords 7 cord feet ; what was the quantity of wood left ? 14. How many da^'s, hours and minutes of any year will be future time on the 4th day of July, 20 minutes past 3 o'clock, P. M. ? Ans. 180 days, 8 hours. 40 minutes. 15. On the same day, hour and minute of July, given in the above example, what will be the difference between the past and future time of that month ? 16. A note, bearing date Dec. 28th, 1826, was paid Jan. 2d, 1827 ; how long was it at interest ? llie distance of time from one date to that of another may be found by subtracting the first date from the last, observing to number the months according to their order, (^f 37.) OPERATION. * jy ^1827. Istm. 2d day. Note, In casting in- • ( 1826. 12 28 terest, each month is Ans. ~0 . Td^ reckoned 30 days. 17. A note, bearing date Oct. 20th, 1823, was paid April 25th, 1825 ; how long was the note at interest ? 18. What is the difference of time from Sept. 29, 1816, to April 2d, 1819 ? Ans, 2 v. 6 m. 3 d. 19. London is 51° 32', and Boston 42° 23' N. latitude; what is the difference of latitude between the two places ? Am. 9° 9', 92 SUBTRACTION OF COMPOUND NUMBERS. IT 40. 20. Boston is 71° 3', and the city of Washington is 77^ 43' W. longitude ; what is the difference of longitude be- tween the two places ? Am. 6° 40'. 21. The island of Cuba lies between 74^ and 85° W. lon- gitude} how many degrees in longitude does it extend? IT 40. 1. When it is 12 o'clock at the most easterly ex- tremity of the island of Cuba, what will be the hour at the most westerly extremity, the difference in longitude be- ing 11°? Note. The circumference of the earth being 360°, and the earth performing one entire revolution in 24 hours, it follows, that the motion of the earth, on its surface, from west to east, is 15° of motion in 1 hour of time; consequently, 1° of motion in 4 minutes of time, and 1' of motion in 4 seconds of time. From these premises it follows, that, when there is a dif- ference in longitude between two places, there will be a corresponding difference in the hour, or time of the day. The difference in longitude being 15°, the difference in time Avill be 1 hour, the place easterly having the time of the day 1 hour earlier than the place ivesterly, which must be par- ticularly regarded. If the difference in longitude be 1°, the difference in time will be 4 minutes, &c. Hence, — If the difference in longitude, in degrees and minutes, between two places, be multiplied by 4, the pro- duct will be the difference in time, in minutes and seconds, which may be reduced to hours. We are now prepared to answer the above question. 11° Hence, when it is 12 o'clock at tiie 4 most easterly extremity of the island,, "" — . it will be 16 minutes past 11 o'clock 44 minutes. ^^ ^j^^ ^^^^^ western extremity. 2. Boston being 6° 40' E. longitude from the city of Washington, when it is 3 o'clock at the city of Washington, what is the hour at Boston ? Ans. 26 minutes 40 seconds past 3 o'clock. 3. Massachusetts being about 72°, and the Sandwich Islands about 155° W. longitude, when it is 28 minutes past 6 o'clock, A. M. at the Sandwich Islands, what will be the liour in Massachusetts ? Ans. 12 o'clock at noon* ^B" 41* MULTIPLICATION OF COMPOUND NUMBERS. 93 MUKTXPKICATION & DIVISZOZff OF COMPOUND NUMBERS. IT 41. 1. A man bought 2 yards of cloth, at 1 s. 6 d. per yard ; what was the cost ? 2. If 2 yards of cloth cost 3 shillings, what is that per yard ? 3. A man has three pieces of cloth, each measuring 10 yds. 3 qrs. ; how many yards in the whole ? 4. If 3 equal pieces of cloth contain 32 yds. 1 qr., how much does each piece contain ? 5. A man has five bottles, each containing 2 gal. 1 qt. 1 pt. ; how much wine do they all contain ? 6. A man has 11 gal. 3 qts. 1 pt. of wine, which he would divide equally into five bottles ; how much must he put into each bottle ? 7. How many shillings are 3 times 8 d. ? 3 X 9 d. ? > 3 X lOd.? . 4X ^d. ? 7X6d.? 10 X 9 d. ? 2X3 qrs. ? 5X2 qrs. ? 8. How much is one third of 2 shillings ? ^ of 2 s. 3 d. ? ^ of 2 s. 6 d. ? i of 2 s. 4 d. ? ^ of 3 s. 6d.? fjyof 7s. 6d.? Jof l^d.? ^of 2^d.r 9. At 1£, 5 s. 8|-d. per yard, what will 6 yards of cloth cost ? 10. If 6 yards of cloth cost 7£. 14 s. 4Jd., what is the price per yard ? Here, as the numbers are large, it will be most convenient to write them down before multiplying and dividing. OPERATION. £. s. d. qr. 6)7 14 4 2 cost of 6 yards, 3 price of 1 yard. OPERATION. £. s. d. qr. 1 5 8 S price of 1 yard. 6 number of yards. Ans. 7 14 4 2 cost of 6 yards. 6 times 3 qrs. are 18 qrs. = 4 d. and 2 qrs. over ; we set down the 2 qrs. ; then, 6 times 8d. are 48 d., and 4 to carry makes 52 d. :zz 4 s. and 4 d. over, which we write down ; again, 6 times 5 s. are 30 s. 1 5 8 Proceeding after the man- ner of short division, 6 is con- tained in 7iS . 1 time, and 1 £ . over ; we write down the quotient, and reduce the re- mainder (1^.) to shillings, (20 s.,) which, with the given shillings, (14 s.,) make 34 s. j, 94 MULTIPLICATION AND DIVISION ^41. and 4 to carry makes 34 s. =: l£, audl4s. over; 6 times 1^. are 6jS., and 1 to carry makes 7iB ., which we write down *, and it is plain, that the united products arising from the leveral denominations is the real product arising from the whole compound number. 11. Multiply 3 jB. 4 s. 6 d. by 7. 13. What will be the cost of 5 pairs of shoes at 10 s. 6 d. a pair ? 15. In 6 barrels of wheat, each containing 2 bu. 3 pks. 6 qts., how many bushels r 17. How many yards of cloth will be required for 9 coats, allowing 4 yds. 1 qr. 3 na. to each ? 19. In 7 bottles of wane, each containing 2 qts. 1 pt. 3 gills, how many gallons ? 21. What will be the weight of 8 silver cups, each weighing 5 oz. 12 pwt. 17 grs. ? 23. How much sugar in 12 hogsheads, each containing 9cwt. 3qrs. 21 lb. ? 25. In 15 loads of hay, each weighing 1 T. 3 cwt. 2 qrs., how many tons ? 6 in 34 s. goes 5 times, and 4 s. over ; 4 s. reduced to pence = 48 d., which, with the given pence, (4 d.,) make 52 d. ; 6 in 52 d. goes 8 times, and 4 d. over ; 4 d. =: 16 qrs., which, with the given qrs. (2) = 18 qrs. ; 6 in 18 qrs. goes 3 times ; and it is plain, that the united quotients arising from the several denomina- tions, is the real quotient aris-* ing from the whole compound number. 12. Divide 22 £. lis. 6d. by 7. 14. At2ie. 12 s. 6 d. for 5 pairs of shoes, what is that a pair ? 16. If 14 bu. 2 pks. 6 qts. of wheat be equally divided into 5 barrels, how many bushels will each contain ? 18. If 9 coats contain 39 yds. 3 qrs. 3na., what does 1 coat contain ? 20. If 5 gal. 1 gill of wine be divided equally into 7 bot- tles, how much will each con- tain ? 22. if 8 silver cups weigh 3 lb. 9 oz. 1 pwt. 16 grs., what is the weight of each ? 24. If 119 c\vt. 1 qr. of su- gar be divided into 12 hogs- heads, how much will each hogshead contain ? 26. If 15 teams be loaded withl7T. 12 cwt. 2 qrs. of hay, how much is that to each team ? ir4i. OF COMPOUND NUMBERS. 95 When the multiplier ^ or divisor^ exceeds 12, the operations of multiplying and dividing are not so easy, unless they be composite numbers ; in that case, we may make use of the component parts^ or factors^ as was done in simple numbers. Thus 15, in the example above, is a composite number produced by the multiplica- tion of 3 and 5, (3x5 = 15.) We may, therefore, multiply 1 T. 3 cwt. 2 qrs. by one of those component parts, or factors, and that product by the other, which will give the true answer, as has been al- ready taught, (IT 11.) OPERATION. T. cwt. qr. 13 2 3 one of the factors. 3 10 2 5 the other factor. 17 12 2 the answer, 27. What will 24 barrels of flour cost, at 2ie . 12 s. 4 d. a barrel ? 29. What will 112 lb. of sugar cost, at 7^ d. per lb. ? Note. 8, 7, and 2, are fac- tors of 112. 31. How much brandy in 84 pipes, each containing 112 gal. 2 qts. 1 pt. 3 g. .? 33. What will 139 yards of cloth cost, at d £, 6 s. 5 d. per yard ? 139 is not a composite num- ber. We may, however, de- compose this number thus, 139 = 100+ 30 -f 9. We may now multiply the 15 being a composite num- ber, and 3 and 5 its compo- nent parts, or factors, we may divide 17 T. 12 cwt. 2 qrs. by one of these component parts, or factors, and the quotient thence arising by the other, which will give the true answer, as already taught, (U20.) OPERATION. T. cict. qr. One factor^ 3)17 12 2 The other factor, 6 ) 5 17 2 Ans. 13 2 28. Bought 24 barrels of flour for 62^. 16 s. ; how much was that per barrel ? 30. If 1 cwt. of sugar cost 3 £, 7 s. 8 d., what is that per lb, ? 32. Bought 84 pipes of brandy, containing 9468 gal. 1 qt. 1 pt. ; how much in a pipe ? 34. Bought 139 yards of cloth for 461^. lis. lid.; what was that per yard ? When the divisor is such a number as cannot be produced by the multiplication of small numbers, the better way is to divide after the manner of 96 MULTIPLICATION ANB DIVISION IT 41, price of 1 yard liy 10, which will give the price of 10 yards, and this product again by 10, which will give the price of 100 yards. We may then multiply tlae price of 10 yards by 3, which will give the price of 30 yards, and the price of 1 yard by 9, which will give the price of 9 yards, and these three pro- ducts, added together, will evi- dently give the price of 139 yards ; thus : £. s. d. 3 6 5 price of 1 yd, 10 price of 10 yds, 8 price of 100 yds. price of 30 yds, price of 9 yds, 461 11 11 price of 139 yds. Note, In multiplying the price of 10 yards (33£. 4 s. 2d.) by 3, to get the price of 30 yards, and in multiplying the price of 1 yard (^£, 6 s. 5 d.) by 9, to get the price of 9 yards, the multipliers, 3 and 9, need not be written down. 33 4 2 10 332 1 8 99 12 6 29 17 9 but may mind. be carried in the long division, setting down the work of dividing and re- ducing in manner as fol- lows : 139)46i 11 U{Z£, 417 44 20 891 (6 s. 834 12 695 ( 5 d. 695 The divisor, 139, is con- tained in 461 £, 3 times, (3i£.,) and a remainder of 44£., which must now be reduced to shillings, multi- plying it by 20, and bringing in the given shillings, (lis.,) making 891 s., in which the divisor is contained 6 times, (6 s.,) and a remainder of 57 s., which must be reduced to pence, multiplying it by 12, and bringing in the given pence, (11 d.,) together mak- ing 695 d., in which the di- visor is contained 5 times, (5 d.,) and no remainder. The several quotients, S£ , 6 s., 5 d., evidently make the answer. V4l. OF COMPOUND IfUMBCRS. 97 The processes in the foregoing examples may now be pre- sented in the form of a Rule for the Multiplication oj Compound Numbers. I. When the multiplier does not exceed 12, multiply sue cessively the numbers of each denomination, beginning with the least, as in multiplication of simple numbers, and carry as in addition of compound numbers, setting down the whole product of the highest denomination. II. If the multiplier exceed 12, and be a composite num- ber, we may multiply first by one of the component parts, Ihat product by another, and 80 on^ if the component parts be more than two ; the last product will be the product re- quired. III. When the multiplier exceeds 12, and is not a com- posite, multiply first by 10, and this product by 10, whi^h will give the product for 100; and if the hundreds in the?mul- tiplier be more than one, ravi' tiply the product of 100 by the number of hundreds ; for the tens^ multiply the product of 10 by the number of tens ; for the units^ multiply the mulii- vlicand; and these several pro- ducts will be the product re- quired. 1 Rule for the Division of Cam* pound Numbers. I. When the divisor does not exceed 12, in the manner of short division, find how many times it is contained in the highest denomination, un- der which write the quotient, and, if there be a remainder, reduce it to the next less de- nomination, adding thereto the number given, if any, of that denomination, and divide as before ; so continue to do through all the denominations, and the several quotientsi^ will be the answer. II. If the divisor exceed 12, and be a composite^ we may di- vide first by one of the com- ponent parts, that quotient by another, and so on, if the com- ponent parts be more than two ; the last quotient will be the quotient required. III. When the divisor ex- ceeds 12, and is not a com- posite pumber, divide after the manBcr of long division, set- ting down the work of di- Tiding and reducing. MULTIPUCATION ANP DITISION, &C, IT EXAMPLES FOR PRACTICE. 1. What will 359 yards of cloth cost, at 4 s. 7^ d. per yard ? 3. In 241 barrels of flour, 6ach containing 1 (iwt. 3 qr. 9 lb. ; how many cvvt. ? 6. How many bushels of wheat in 135 bags, each con- taining 2 bu. 3 pks. ? 3X9X5 = 135. 7. What will 35 cwt. of to- bacco cost, at 3 s. 104- tl. per lb. ? 9. If 14 men build 12 rods 6 feet of wall in one day, how many rods will they build in 7^ days ? . 2. Bought 359 yards of cloth for 83 jg . s. 4i d. ; what was that a yard ? 4. If 441 cwt. 13 lb. of flour be contained in 241 barrels, how much in a barrel ? 6. If371bu. Ipk. of wheat be divided equally into 135 bags, how much will each bag contain ? 5. At 759 i£. 10 3. for 35 cwt. of tobacco, what is that per lb. ? 10. If 14 men build 92 rods 12 feet of stone wall in 7^ days, how much is that per day? IT 42. 1. At 10 s. per yard, what will 17849 yards of cloth cost ? Notc^ Operations in multiplication of pounds, shillings, pence, or of any compound numbers, may be facilitated by taking aliquot parts of a higher denomination^ as already ex- plained in ^^Practice^^ of Federal Money, 1129, ex.10. Thus, in this last example, the price 10 s. =:^ of a pound ; therefore, ^ of the number of yards will be the cost in pounds. -i-^|49. — 8924 £, 10s. Ans. 2. What cost 31648 yards of cloth, at 10 s. or JiS. per yard ? at 5 s. ^ ^£ . per yard ? • at 4 s. = -^^ , per yard f > at 3 &. 4 d. :== ^£ . per yard ? at 2 s. r= ^V ^ • per yard ? Am. to last^ 34g4 iS . 16 s. 3. What cost 7430 pounds of sugar, at 6 d. z=i J s. per lb ? at 4 d. = ^ s. per lU? at 3d. := i s. per lb.? at 2 d.=:^ s. per «>.? at l^'d.=:^s. per lb. ? J|k Ans, to the last^ i4^ff|. = 928 s. 9 d. = 46 iS . 8 s. 9 d. 4. At $18'75 per cwt., what will 2qrs. =^cwt. cost? what will 1 qr. = ^ cwt. cost ? . what will 16 lb, = f cwt. cost ? what will 14 lbs. == ^ cwt. cost ? what will 8 lbs. = -ifjf cwt. cost ? Ans. to the last ^$^^^9- t 42. SUPPLEMENT TO COMPOUND NUMBERS. 9& 6. What cost 340 yards of cloth, at 12 s. 6 d. per yard ? 12s.6d. = 10s. {z=i£.) and2s.6d. (zz^ig.); there- fore, i)i)S40 170 iB . = cost at 10 s. per yard. 42 iS . 10 s. = at 2 s. 6 d. per yard. Ans, 212 ig 10 s. = at 12 s. 6 d. peryard. Or, 10 s. z= Jig.) 340 S 8. 6 d. = i of 10 s.) 170 ig . at 10 s. per yard. 42 jB . 10 s. at 2 s. 6 d. per yard. Ans. 212 ie . 10 s. at 12 s. 6 d. per yard. SUPPZiZSMENT TO THS ARXTHHHSTXC OF COXMCPOUND NUMBERS. QUESTIONS. 1. What distinction do you make between simple and compound numbers ? (^ 26.) 2. What is the rule for addi- tion of compound numbers ? 3. for subtraction of, &c. ? 4. There are three conditions in the rule given for multi- plication of compound numbers; what are they, and the methods of procedure under each? 5. The same questions in respect to the division of compound numbers ? 6. When the multiplier or divisor is encumbered with a fraction, how do you proceed ? 7. How is the distance of time from one iate to another found ? 8. How many degrees does the earth revolve from west to east in 1 hour? 9. In what time dees it revolve 1°? Where is the time or hour of the day earlier — at the place most easterly or most westerly ? 10. The difference in longitude between two places being known, how is the difference in Jime calculated ? 11. How may operations, in the multiplication of compound num- bers, be facilitated ? 12. What are some of the aliquot parts ofil£.? ofls.? oflcwt? 13. Whatisthi« manner of operating usually called ? 100 SUPPLSMBNT TO COMPOUND NUMBERS. TT 40. £X£RCIS£S. 1. A gentleman is possessed of 1 J dozen of silver spoons, each weighing 3 oz. 5 put. ; 2 doz. of tea spoons, each weigh- ing 15 pwt. 14 gr. ; 3 silver cans, each 9 oz. 7 pwt. ; 2 silver tankards, each 21 oz. 15 pwt. ; and 6 silver porringers, each 11 oz. 18 pwt. ; w^hat is tlie weight of the whole ? Ans. 18 lb. 4 oz. 3 pwt- Note, Let the pupil be required to reverse and prove the following examples : 2. An English guinea sh-^uld weigh 5 pwt. 6 gr. ; a piece of gold weighs 3 pwt. IT gr. ; how much is that short of the weight of a guinea ? 3. What is the weight of 6vchests of tea, each weighing 3 cwt. 2 qrs. 9 lb. ? 4. In 35 pieces of cloth, each measuring 27 yards, how many j^ards ? 5. How much brandy in 9 casks, each containing 45 gal. 3 qts. 1 pt. ? 6. If 31 cwt. 2 qrs. 20 lb. of sugar be distributed equally into 4 casks, how much will each contain ? 7. At 4^ d. per lb., what costs 1 cwt of rice ? 2 cwt ^ 3 cwt ? Note. The pupil will recollect, that 8, 7 and 2 are fac- tors of 112, and may be used in place of that number. 8. If 800 cwt of cocoa cost 18 iS . 13 s. 4 d., what is that per cwt. ? what is it per lb. ? 9. What will 9^ cwt. of copper cost at 5 s. 9 d. per lb. ? 10. If 6^ cwt of chocolate cost 72 i£. 16 s., what is that per lb. ? 11. What cost 456 bushels of potatoes, at 2 s. 6d. per bushel ? Note. 2 s. 6 d. is ^ of 1 £ . (See IT 42.) 12. What cost 86 yards of broadcloth, at 15s. per yard? Note. Consult ^ 42, ex. 5. 13. What cost 7846 pounds of tea, at 7 s. 6 d. per lb. ? at 14 s. per lb. ? at 13 s. 4 d. ? 14. At $ 94'25 per cwt, what will be the cost of 2 qrs, of tea? of 3 qrs. ? of 14 lbs. ? of 21 lbs,? of 16 lbs. ? of 24 lbs. ? NoU. Consult IT 42, ex. 4 and 6. T42,43. FRACTIONS. 10! 15. What will be the cost of 2 pks. aud 4 qts. of wheat, at $ 1*50 per bushel ? 16. Supposing a meteor to appear so high in the heavens as to be visible at Boston, 71° 3', at the city of Washington, IT 43', and at the Sandwich Islands, 155° W. longitude, and that its appearance at the city of Washington be at 7 minutes past 9 o'clock in the evening; what will be the hour and minute of its appearance at Boston and at the Sandwich Islands ? FRACTION'S. ^ 43. We have seen, (IT 17,) that numbers expressing whole things are called integers, or whole numbers ; but that, in division, it is often necessary to divide or break a whole thing into parts, and that these parts are called f-actions^ or broken numbers. It will be recollected, (IT 14, ex. 11,) that when a thing or unit is divided into 3 parts, the parts or fractions are call- ed thirds ; when into four parts, fourths ; when into six parts, nxths ; that is, the fraction takes its name or denwnination from the number of parts, into which the unit is divided. Thus, if the unit be divided into 16 parts, the parts are called six- teenths, and 5 of these parts ^^ ould be 5 sixteenths, expressed thus, -3^. The number below the short line, (16,) as before taught, (IT 17,) is called the denominator, because it gives the name or denomination to the parts ; the number above tlie line is called the mimerator, because it numbers the parts. The denominator shows how many parts it takes to make ^unit or whole thing; the numerator shows how many of these parts are expressed by the fraction, 1. If an orange be cut into 5 equal parts, by what frac- tion is 1 part expressed ? 2 parts } 3 parts ? 4 parts ? 5 parts ? how many parts make unity or a whole orange ? 2. If a pie be cut into 8 equal pieces, and 2 of these pieces be given to Harry, what will be his fraction of the pie ? if 5 pieces be given to John, what will be his fraction } what fraction or part of the pie will be left ? It is important to bear in mind, that fractions arise from division^ (IF 17,) and that the nviaerator may be considered a I * J expresses the quotient, of which J 102 FRACTIONS. It 4 J. dividend^ and the denominator a divisor, and the tjo/ttc of the fraction is the quotient ; thus, 5- is the quotient of 1 (the numerator) divided by 2, (the denominator ;) J is the quo- tient arising from 1 divided by 4, and f is 3 times as much, that is, 3 divided by 4 ; thus, one fourth part of 3 is the same as 3 fourths of 1 . Hence, in all cases, a fraction is always expressed by the sign oj divhion, 3 is the dividend^ or numerator . 4 is the divisor f or denominator. g. If 4 oranges be equally divided among 6 boys, what part of an orang**. is each boy's share ? A sixth part of 1 orange is ^, and a sixth part of 4 oranges is 4 such pieces, = |. Ans. | of an orange. 4. If 3 apples be equally divided among 5 boys, what part of an apple is each boy's share ? if 4 apples, what ? if 2 apples, what? if 5 apples, what? 5. What is the quotient of 1 divided by 3? of 2 by 3? of 1 by 4? of2by4? of3by4? of5 by7? ofebyS? of 4 by 5 ? of2byl4? 6. What part of an orange is a third part of 2 oranges ? one fourth of 2 oranges ? J of 3 oranges ? I of 3 oranges? iof4? i of 2 ? |of6? I of 3 ? — — :J- of 2 ? ^4 Proper Fraction, Since the denominator shows the num- ber of parts necessary to make a whole thing, or 1, it is plain, tliat, when the numerator is less than the denominator, the fraction is less than a unit, or whole thing ; it is then called a proper fraction. Thus, ^, f , ^c. are proper fractions. An Improper Fraction, When the numerator equals or ez- ceeds the denominator, the fraction equals or exceeds unity, or 1, and is then called an improper fraction. Thus, f, f, f, -y^, are improper fractions. A Mixed Number, as already shown, is one composed of a whole number and a fraction. Thus, 14^, 13 J, &:c. are mix* ed numbers. 7. A father bought 4 oranges, and cut each orange into d equal parts ; he gave to Samuel 3 pieces, to James 5 pieces^ to Mary 7 pieces, and to Nancy 9 pieces ; what was each one's fraction ? Was Jameses fraction proper, or improper 1 Why ? Was Nancy's fraction proper, or improper ? Why ? IT 44. FRACTIONS- 103 To change an improper fraction to a whole or mixed number, IT 44. It is evident, that every improper fraction must contain one or more whole ones, or integers. 1. How many whole apples are there in 4 halves (J) of 8in apple ? in f ? in f ? in -^^ ? in ^? in-^? inJ^I^? in^? 2. How many yards in f of a yard ? in § of a yard ? inf? — ^ — inf? in-^-? in-y^? in A^P inJgL? in^ - of a gallon = 1371 quarts. To reduce a fraction to its lowest or most simple terms, ^ 46. The numerator and the denominator, taken to- gether, are called the terms of the fraction. If ^ of an apple be divided into 2 equal parts, it becomes f . The effect on the fraction is evidently the same as if we had multiplied both of its terms by 2. In either case, ike parts are made 2 times as many as they were before ; hut they are only HALF AS LARGE ; for it will take 2 times as many fourths to make a whole one as it will take halves ; and hence it is that f is the same in value or quantity as ^. f is 2 parts ; and if each of these parts be again divided into 2 equal parts, that is, if both terms of the fraction be multiplied by 2, it becomes -|. Hence, ^ = J z= |, f 5 "^e reverse of this is evidently true, that -| izz f nr J. ' ^^ It follows therefore, by multiplying or dividing both terms of the fraction by the same number^ we chc^'ge its terms without altering its value. Thus, if we reverse the above operauon, and divide both terms of the fraction | by 2, we obtain its equal, f ; dividing again by 2, we obtain ^, which is the most simple form of th t fraction, because the terms are the least possible by which the fraction can be expressed. The process of changing | into its equal J is called re- ducing the fraction to its lowest terms. It consists in dividing both terms of the fraction by any ''umber which will divide them both without a remainder ^ and the quotient thence arising in the same manner ^ and so on^ till it apptars that no number greater tfian 1 will again divide them, A number, which will divide two or more numbers with- out a remainder, is called a common divisor, 01 common meor Buxe of those numbers. The greatest numbei that will do this is called the greatest common divisor. 106 FRACTIONS. fr 46, 41 1. What part of an sere are 128 rods ? One rod is j^ of an acre, and 128 rods are -Jff of an acre. Let us reduce this fraction to its loivest terms. We find, hy trial, that 4 will exactly measure both 128 and 160, and, dividing, we change the fraction to its equal Jg. Again, we find that 8 is a divisor common to both terms, and, di- viding, we reduce the fraction to its equal |^, which is uoxt in its lowest terms, for no greater number than 1 will again measure them. The operation may be presented thus : ^>.128 ^2 4 . . 2. Reduce f^g, j^^, ||-g-, and ^|| to their lowest terms. ^^' h h h ^^^ i- Note, If any number ends with a cipher, it is evidently divisible by 10. If the two right hand figures are divisible by 4, the whole number is also. If it ends with an even number^it is divisible by 2 ; if with a 5 or 0, it is divisible ^^/Ihis.; "i^K uce ^J|, ^, ifl, and f ^ to tlieir lowest terms. 1^7. Any fraction may evidently be reduced to its lowest terms by a single division, if we use the greatest common divisor of the two terms. The greatest common measure of any two numbers may be found by a sort of trial easily made* Let the numbers be tSic two terms of the fraction -fff . The common divisor cannot esiceed the less number, for it must measure it. We ^vill try, therefore, if the less number, 128, which measures itself, will also divide or measure 160. 128^160('l ^^^ ^^ ^^^ ^^^^ ^ time, and 32 re- ■loa mam; 128, therefore, is not a divisor of 160. We will now try whether this re- 32) 128(4 mainder be not the divisor sought ; for if 128 32 be a divisor of 128, the former divi- sor, it must also be a divisor of 160, which consists of 128 -[- 32. 32 in 128 goes 4 times, with- out any remainder. Consequently, 32 is a divisor of 128 and 160. And it is evidently the greatest common divisor of these numbers ; for it must be contained at least once more in 160 than in 128, and no number greater than their difference, that is, greater than 32, can do it. IT 47, 48. FRACTIONS 107 Hence the nth for finding the greatest common dioisor of iwo numbers : — Divide the greater number by the less, and that divisor by the remainder, and so on, always dividing the last divisor by the last remainder, till nothing remain. The last divisor will be the greatest common divisor required. Note* It is evident, that, when w^e would find the greatest common divisor of inore than two numbers, we may first find the greatest common divisor of two numbers, and then of that common divisor and one of the other numbers, and so on to the last number. Then will the greatest common divisor last found be the answer. 4. Find the greatest common divisor of the terms of the fraction f ^, and, by it, reduce the fraction to its lowest terms. OPERATION. 21)35(1 21 14)21(1 14 Greatest divis. 7) 14(2. Then, 7)—=:^ An$, 14 ^35 5 6. Reduce ^^^^ to its lowest terms. Am. ^. Ao/e. Let these examples be wrought by both methods ; by several divisors, and also by finding the greatest common divisor. ^? • f 6. Reduce -ff^r to its lowest terms. Ans, -J, f 7. Reduce J^f to its lowest terms. ^ Ans. f, 8. Reduce -^t%- to its lowest terms. Ans, ^^ 9. Reduce J-f f | to its lowest terms. Ans, ^, To divide a fraction by a whole number, IT 48. 1. If 2 yards of cloth cost f of a dollar, what does 1 yard cost ? how much is f divided by 2 ? 2. If a cow consume f of a bushel of meal in 3 days, how much is that per day ? J -i- 3 z= how much ? 3. If a boy divide | of an orange among 2 boys, how much will he give each one ? | -f- 2 z= how much > 4. A boy bought 5 cakes for ^J of a dollar ; what did I cake cost ? -ff -r- 5 = how much ? 108 TRACTIONS. 48 t 5. If 2 bushels of apples cost f of a dollar, what is that per bushel ? 1 bushel is the half of 2 bushels j the half of f is ^. Ans. ^ dollar. 6. If 3 horses consume ^f of a ton of hay in a month, what will 1 horse consume in the same time ? ^f are 12 parts ; if 3 horses consume 12 such parts in a month, as many times as 3 are contained in 12, so many parts 1 horse will consume. A7is. ^ of a ton. 7. If f I of a barrel of flour be divided equally among 5 families, how much will each family receive ? f|- is 25 parts ; 5 into 25 goes 5 times. Ans. -^^ of a barreL The process in the foregoing examples is evidently di- viding a fraction by a whole number ; and consists, as may be seen, in dividing the numerator^ (when it can be done without a remainder,) and under the quotient writing the denominator. But it not unfrequently happens, that the nu- merator will not contain the whole number without a re- mainder. 8. A man divided J of a dollar equally among 2 persons ; what part of a dollar did he give to each ? ^ of a dollar divided into 2 equal parts will be 4ths. Ans. He gave J of a dollar to each. 9. A mother divided -J- a pie among 4 children ; what part of the pie did she give to each ? ^ -- 4 :::= how much ? 10. A boy divided -J of an orange equally among 3 of his companions; what was each one's share? -J- -f- 3 = how much ? 11. A man divided f of an apple equally between 2 chil- dren ; what part did he give to each ? f divided by 2 =: what part of a whole one ? J is 3 parts : if each of these parts be divided into 2 equal parts, they will make 6 parts. He may now give 3 parts to one, and 3 to the other : but 4ths divided inio 2 equal parts, become 8tlis. The parts are now twice so many^ but they are only half so large ; consequently, f is only half so much as f . Ansi. f of an apple. In these last examples, the fraction has been divided by multiplying the denominator^ without char.ging the numerator. The reason is obvious ; for, by multiplying the denominator bj any number, the parts are made so many times smaller^ amce it will take so many more of them to make a whole IT 49, FRACTIONS. 109 pne ; and if no more of these smaller parts be taken than were before taken of the larger, that is, if the numerator be not changed, the value of the fraction is evidently made so many times less. IT 49. Hence, vi^e have two ways to divide a fraction by a whole number : — I. Divide the numerator by the whole number, (if it will contain it without a remainder,) and under the quotient write the denominator. — Otherwise, II. Multiply the denominator by the whole number, and over the product write the numerator. EXAMPLES FOR PRACTICE. 1. If 7 pounds of coiFee cost f^ of a dollar, what is that per pound ? f-J- -f- 7 iz: how much ? Ans, ^ of a dollar. 2. If ^ of an acre produce 24 bushels, what part of an acre will produce 1 bushel ? ^§ ~ 24 =: how much ? 3. If 12 skeins of silk cost ^J of a dollar, what is that a «kein ? ^ -f. 12 = how much ? 4. Divide f by 16. Note, When the dtvisor is a composite number, the in- telligent pupil will perceive, that he can first divide by one component part, and the quotient thence arising by the other ; thus he may frequently shorten the operation. In the last example, 16 = 8x2, and f -^ 8 ir: ^, and ^ -r- 2 = tV- ^^^' tV 5. Divide -^ by 12. Divide /^ by 21. Divide ff by 24. 6. If 6 bushels of wheat cost $ ^J what is it per bushel ? Note. The mixed number may evidently be reduced to an improper fraction, and divided as before. Ans, Jl = it of a dollar, expressing the fraction in its lowest terms. (TT 46.) ^ 7. Divide $ 4^^ by 9. Quot. ^ oi ^ dollar. ^ 8. Divide 12f by 5. QuoU -^=:2f 9. Divide 14f by 8, QiMt. Ifj. 10. Divide 184^ by 7, Am. 26^^. Note. When the mixed number is Utrge, it vnii be most convenient, first, to divide the whole number, and then re^ duce the remainder to an improper fraction ; and, after di- viding, annex the quotient of the fraction to the quotient of K 110 FRACTIONS. IT 49, 50. the ^vhole number; thus, in the last example, dividing 184J "by 7, as in whole numbers, we obtain 26 integers, with 2J ■=. ^ remainder, which, divided by 7, gives i% and 26 -f- A == 26^^, Ans. 11. Divide 27S6| by 6. Ans, 464f. 12. How many times is 24 contained in 7646 J-^ ? Aiiii. 318ffJ. 13. How many times is 3 contained in 462 J ? Ans, 154^. To multiply afraclion by a whole number. ^ 50. i. If 1 yard of cloth cost -J of a dollar, what will 2 yards cust ? ^ X 2 zi= how much ? 2. If a cow consume j of a bushel of meal in 1 day, how much will she consume in 3 days ? ^ X 3 izi hov/ much ? .3. A boy bought 5 cakes, at f of a dollar each \ what did he give for the whole ? f X 5 zn how much? 4. How much is 2 times ^ ? 3 times ^ ? 2 times I? 5. Multiply f by 3. f by 2. -J by 7. 6. If a man spend f of a dollar per day, how much will he spend in 7 days ? f is 3 parts. If he spend 3 such parts in 1 day, he will evidently spend 7 times 3, that is, \^- z_- 2| in 7 days. Hence, we perceive, a fraction is muliiplied by multiplying the numerator^ ivilhout changing the denominator. But it has been made evident, (1[ 49,) that multiplying the denominator produces the same effect on the value of the frac- tion, as dhndlng the numerator: hence, also, dividing the de- nominator will produce the same effect on the value of the fraction, as midtiplyinn the numerator. In all cases, therefore, where one of the terms of the fraction is to be multiplied^ the same result will be effected by dividing the ofhrr ; and where one tprm is to be diirided^ the same result may be effected by multiplying the other. This principle, borne distinctly in mind, will frequently enable the pupil to shorten the operations of fractions. Thus, in the following example : At -^^ of a dollar for 1 pound of sugar, what will 11 pounds cost ? Multiplying the numerator by 11, we obtain for the pro- duct II = I of a dollar for the answer. If 61. FRACTIONS. Ill IT 51< But, by applying the above principle, and dividing tlie denominator, instead of multiplijing the numerator^ we at once come to the answer, |, in its lowest ternris. Hence, there are two ways to muUiply a fraction by a zvhole number :-^ I. Divide the denominator by the whole number, (when it can be done witfiout a remainder,) and over the quotient write the numerator. — Otherwise, II. Multiply the numerator by the whole number, and un- der the product write the denominator. If then it be an improper fraction, it may be reduced to a whole or mixed number. EXAMPLES FOR PRACTICE. 1. If 1 man consume -3^^^ of a barrel of liour in a month, how much will 18 men consume in the same time ? — 6 men ? 9 men ? Ans. to the last^ \\ barrels. 2. What is the product of fVj multiplied by 40 ? ^^J^ X 40 r= how much ? " Ans. 23|. 3. Multiply ^W by 12. by 18. by 21. by 36. by *48. by 60. Note, When the multiplier is a composite number, the pupil will recollect, (IT 11,) that he may first multiply by one component part, and that product by th*^ other. Thus, in the last example, the multiplier 60 is equal to 12 X 5; therefore, j^'^X 12 ziz -{4^ and J-J X 5 1= -f | =: 5^^-, Ans, \ 4. Multiply 5J by 7. An. 40J. Nnfp. It ia evident, thnt the mixed number may be re- duced to an improper fraction, and multiplied, as in the pre- ceding examples ; but the operation will usually be shorter, to multiply the fraction and whole number separately^ and add the results top^ether. Thus, in the last example, 7 times 5 are 35; and 7 times J are ^5^=:5|, which, addjd to 35, make 40,{^, Ans. Or, we may multiply the friction first, and, writing down the fraction, reserve the integers, to be carried to the product of the whole number. «. 6, What will 9^1} tons of hay come to at $ 17 per ton } Am, $ 1642\j. 6. If a man travel 2£jj miles in 1 hour, how far will he 112 FRACTIONS. IT 61, 52S. travel in 5 hours ? in 8 hours ? — -— in 12 hours ? in 3 days, supposing he travel 12 hours each day? An», to the last, 77f miles. Note. The fraction is here reduced to its lowest terms ; the same w^ill be done in all following examples. To multiply a whole number by a fraction. 11 52. 1. If 36 dollars be paid for a piece of cloth, what costs f of it ? 36 X |- =^ how much ? f of the quantity will cost | of the price ; f a time 36 dol- lars, that is, f of 36 dollars, implies that 36 be first divided into 4 equal parts, and then that 1 of these parts be taken 3 times; 4 into 36 goes 9 times, and 3 times 9 is 27. Ans. 27 dollars. From the above example, it plainly appears, that the ob^ ject in mulliplying by a fraction^ whatever may be the multipli- cand^ is, to take out of the midtiplicand a part^ denoted by the. multiplying fraction; and that this operation is composed of two others, namely, a division by the denominator of the DfHiltiplying fraction, and a multiplication of the quotient by the numerator. It is matter of indifference, as it respects the residtj which of these operations precedes the other, for 36 X 3 ~ 4 = 27, the same as 36 -r- 4 X 3 zn 27. Hence, — To multiply by a fraction^ whether the multiplicand be a whole number or a fraction^ — RULE. Divide the multiplicand by the denominator of the multi- plying fraction, and multiply the quotient by the numerator ; or, (which will often be found more convenient in practice,) first multiply by the numerator, and divide the product by the denominator. Multiplication, therefore, when applied to fractions, does not always imply augmentation or increase, as in w^hole numbers ; for, when the multiplier is less than unityy it will always require the product to be less than the multiplicandy to which it would be only equal if the multiplier were 1 . We have seen, (IT 10,) that, when two numbers are multi- plied together, either of them may be made the multiplier, w^ithout affecting the result. In the last example, therefore, instead of multiplying 16 by f , we may multiply J by 16 (TT 60,) and the result will be the same. ^62,53. FRACTIONS. 113 EXAMPLES FOR PRACTICE. ^2. What will 40 bushels of corn come to at f of a dollar per bushel ? 40 X f = how much ? 3- What will 24 yards of cloth cost at f of a dollar per yard ? 24 X § == how much ? 4. How much is J- of 90 ? f of 369 ? ^^. of 45 ? j 5. Multiply 45 by y^^. Multiply 20 by J-. To muttiplij one fraction by another. TT 53. 1. A man, owning f of a ticket, sold f of his share ; what part of the whole ticket did he sell ? § of ^ is how much ? We have just seen, (IT 52,) that, to multiply by a fraction, is to divide the multiplicand by the denominator^ and to rnulti- ply the quotient by the numerator. ^ divided by 3, the de- nominator of the multiplying fraction, ('^49,) is -^r--, which, multiplied by 2, the numerator, (11 51,) is -j-\, Ans. The process, if carefully considered, will be found to con- sist in multiplying together the two numerators for a new mir merator^ and the two denominators for a nevj denom'utulor. EXAMPLES FOR PRACTICE. 2. A man, havi.ng I o^ ^ dollar, gave f of it for a dinner ; what did the dinner cost him ? Ans. J- dollar. . 3. Multiply I by f . Multiply fj by f . Product^ ^^. I, 4. How much is | of f of J of J ? Note. Fractions like the above, connected by the word ofy are sometimes called compound fractions. The word OF unplies their continual multiplication into each other. Ans. iU = ^(T' When there are several fractions to be multiplied con- tinually together, as the several numerators are factors of the new numerator, and the several denominators are factors of the new denominator, the operation may be shortened h^ dropping those factors lohich are the same in both termSy on the principle explained in If 46. Thus, in the last example, ^j §, 1^, I, we lind a 4 and a 3 both among the numerators and among the denominators; therefore we drop them, multiply- ing together only the remaining numerators, 2 X '^ ^=^ l'^? ^^^ a new numerator, and the remaining denominators, 5X8 = 40, for a new denominator, making +J- = J^, Am, as before. 114 FRACTIONS. IT 63, 54. 5. f of I of f of f of Y^j of f of I = how much ? Am, ^5^. f 6. What is the continual product of 7, ^, ^ of f and 3^ ? Note, The integer 7 may be reduced to the form of an improper fraction by writing a unit under it for a denomina- tor, thus, -J. Ans. 2-f|. 7. At ^\ of a dollar a yard, what will J^ of a yard of cloth cost ? 8. At 6f dollars per barrel for flour, what will -f-^ of a bar- rel cost ? 6|- := \i- ; then V^ X t\ = ^^1=$ 2^^, ■^^^' 9. At |- of a dollar per yard, what cost 7f yards ? Ans, $6^i. ^ 10. At $ 2^ per yard, what cost Gf yards ? Am, $ 14f f . II. What is the continued product of 3, f , f of f , 2f , and i4off oft ? Ans, fff. TT 54. TAc Rule /or the multiplication of fractions may 71010 be presented at one view : — I. To multiply a fraction by a whole number^ or a whole number by a fraction^ — Divide the denominator by the whole liumber, when it can be done without a remainder; other- wise, midtiply the numerator by it, and under the product write the denominator, which may then be reduced to a whole or mixed number. XL To multiply a mixed number by a whole number^ — Multi- ply the fraction and integers, separately^ and add their pro- ducts together. III. To multiply one fraction by another^ — Multiply together the numerators for a new numerator, and the denominators for a new denominator. Note, If either or both are mixed numbers^ they may fir€t be reduced to improper fractions, exampi.es for practice. 1. At $f per yard, what cost 4 yards of cloth ? 5 yds. > 6 yds. ? 8 yds. } 20 yds. ? Ans. to the last, $ 15. 2. Multiply 148 by i, -—by f by ^. by ^. Last product, 44^. 3. If 2^ tons of hay keep 1 horse through the winter^ IT 54j 55. FRACTIONS. 116 how much will it take to keep 3 horses the same time? . 7 horses ? 13 horses ? Ans. to last^ '^7-^ tons. 4. What will S^^ barrels of cider come to, at $ 3 per barrel ? 5. At $ 14f per cwt., what will be the cost of 147 cwt.? / 6. A owned f of a ticket ; B owned -f-^ of the same ; the ticket was so lucky as to draw a prize of $ 1000 ; what was each one's share of the money ? 7. Multiply 1 of f by f of f . Product, -J. 8. Multiply 7i by 2^^. Product, 15^. 9. Multiply i by 2f . Product, 2|. 10. Multiply f of 6 by |. Product, 1. * 11. Multiply f of 2 by -^ of 4. Product, 3. 12. Multiply continually together ^ of 8, f of 7, f of 9, and I of 10. Product, 20. 13. Multiply 1000000 vy f . Product, 555555f. To divide a whole number by a fraction^ IT 65. We have already shown (IT 49) how to divide a fraction by a whole number ; we now proceed to show how to divide a whole number oy a fraction. 1. A man divided $9 among some poor people, giving them J of a dollar each ; how many were the persons who received the money ? 9 -^ f z=: how many ? 1 dollar is |-^ and 9 dollars is 9 times as many, that is, -^ ; then J is contained in -^/- as many times as 3 is contained in 36. Ans. 12 persons. That is, — Multiply the dividend by the denominator of the dividing fraction, (thereby reducing the dividend to parts of the same magnitude as the divisor,) and divide the product by the numerator, 2. How many times is | contained in 8 ? 8 -j- f = how many ? OPERATION. 8 Dividend. 5 Denominator^ Numerator, 3)40 Quotient, 13^ times, the Answer, To multiply by a fraction, we have seen, (!! 52,) implied two operations — a division and a multiplication; so, also, to divide by a fraction implies two operations — a inultiplication and a division. 116 FRACTIONS. IT 59. IT 56. Division is the reverse of muUiplication. To multiply by a fraction, whether the multiplicand be a whole number or a fraction, as has been already shown, (TT 52,) we divMe by the de- nominator of the multiplying fraction, and midiiply the quo- tlent by the numerator. Note, In either case, it is matter of indifference, as it respects the result, which of these operations precedes the other; but in practice it will frequently be more convenient, that the multiplication precede the division. To divide by a fraction, whether the dividend be a whole number or a fraction, we multiply by the denomina- tor of the dividing fraction, and divide the product by the numerator. 12 multiplied by |-, the pro- duct is 9. In multiplication, the mul- tiplier being less than unity, or 1, will require the product to be less than the multipli- cand, (If 52,) to which it is only equal when the multi- plier is 1, and greater when the multiplier is more than 1. 12 divided by J, the quo- tient is 16. In division, the divisor be- ing less than unity, or 1, will be contained a greater number of times; consequently will re- quire the quotient to be great- er than the dividend, to which it will be equal when the di- visor is 1, and less when the divisor is more than 1. exampl.es for practice* 1. How many times is ^ contained in 7 ? 7 -r- ^ = bow many ? 2. How many times can I draw J of a gallon of wine out of a cask containing 26 gallons ? 3. Divide 3 by J. 6 by |. 10 by f. 4. If a man drink y^g^ of a quart of rum a day, how long will 3 gallons last him ? 6. If 2J bushels of oats sow an acre, how many acres will 22 bushels sow ? 22 -7- 2f = how many times ? Note. Reduce the mixed number to an improper frac- tion, 2f = \K Ans, 8 acres. 6. At $4f a yard, how many yards of cloth may be bought for $ 37 ? Ans. S/g- yards. 7. How many times is -f^r^ contained in 84 ? Ans, 90J times. it 66, 67. FRACTIONS. lit / 8. How many times is ^ contained in 6 ? Ans. f of 1 time. 9. How many times is 8| contained in 53 : Ans, 6^ times. 10. At f of a dollar for building 1 rod of stone wall, how many rods may be built for $ 87 ? 87 ~ f r= how many times ? To divide one fraction by another. ^ 57. 1. At § of a dollar per bushel, how much rye may be bought for f of a dollar ? f is contained in f how many times ? Had the rye been 2 whole dollars per bushel, instead of f of a dollar, it is evident, that f of a dollar must have been divided by 2, and the quotient would have been -^^ ; but the divisor is 3ds, and 3ds will be contained 3 times where a like number of whole ones are contained 1 time; conse- quently the quotient fV is 3 times too smallj and must there- fore, in order to give the true answer, be multiplied by 3, that is, by the deiiominator of the divisor; 3 times t% =i i^^bush* Ans. The process is that already described, IT 55 and IT 56. If carefully considered, it will be perceived, that the numerator of the divisor is multiplied into the denominator of the divi- dend, and the denominator of the divisor into the numerator of the dividend ; wherefore, in practice, it will be more con- venient to invert the divisor ; thus, § inverted becomes i} ; then multiply together the two upper terms for a numerator, and the two lower term^ for a denominator ^ as in the multiplication of one fraction by another. Thus, in the above example, ^X3_ 9 2'X5-10'^^^^^^'^- EXAMPLES FOR PRACTICE. 2. At J of a dollar per bushel for apples, how many bush- els may be bought for |^ of a dollar ? How many times is ^ contained in J ? Ans. 3 J bushels. 3. If |- of a yard of cloth cost f of a dollar, what is that per yard? It will be recollected, (If 24,) that when the cost of any quantity is given to find the price of a unit, w^e divide the cost by the quantity. Thus, f (the cost) divided by J (the quantity) will give the price of I yard. Ans. If of a dollar per yard. 118 FRACTIONS. IT 57, 59; Proof. If the work be right, (TT 16, " Proof,") the pro- duct of the quotient into the divisor will be equal to the dividend ; thus, ft X 5 = f • This, it will be perceived, is multiplying the price of one yard (f|) by the quantity (J) to find the cost (f ;) and is, in fact, reversing the question, thus, If the price of 1 yard be f-| of a dollar, what will | of a yard cost ? Ans. f of a dollar. Note. Let the pupil be required to reverse and prove the succeeding examples in the same manner. 4. How many bushels of apples, at ^^ of a dollar per bushel, may be bought for ^ of a dollar ? Ans. 4| bushels. J 5. If 4f pounds of butter serve a family 1 week, how many weeks will 36 1 pounds serve them ? The mixed numbers, it will be recollected, may be re- duced to improper fractions. Aiis. Syg^- weeks. 6. Divide^ by ^. Qnot.l. Divide^ by ^. Quot. 2. 7. Divide f by i. Quot. 3. / Divide | by j%. Quot. f |» 8. Divide 2^ by U. Divide 10| by 2f Quot. IJ. Quot. 4|^. 9. How many times is j\j contained in f ? Ans. 4 times. 10. IIow many times is f contained in 4§ ? Alts. 11§ times. 11. Divide g of J by J of -f . Quot. 4. ^ 68. The Rule /t-r dicision effractions may now he pr&* seated at one view : — I. To divide a fraction by a whole number, — Divide the numerator by the whole number, when it can be done with- out a remainder, and under the quotient write the denomi- nator ; otherwise, multiphj the denominator by it, and over the product write the numerator. II. To divide a ivhole number by a fraction, — Multiply the dividend by the denominator of the fraction, and divide the product by the numerator.^ III. To divide one fraction by another, — Invert ihe divisor^ and multiply together the two upper terms for a numerator, and the two lower terms for a denominator. Note. If either or both are mixed numbers, they may be reduced to improper u actions. IT 59. ADDITION AND SUBTRACTION OF FRACTIONS. 119 EXAMPLES FOR PRACTICE. 1. If 7 lb. of sugar cost -^^^ of a dollar, what is it pe? pound ? T^jy -f- 7 = how much ? f of y^^y is how much ? 2. At $ 1^ for f of a barrel of cider, what is that per bar- rel? 3. If 4 pounds of tobacco cost J of a dollar, what does 1 pound cost? 4. If |- of a yard cost $ 4, what is the price per yard ? 5. If 14f yards cost $ 75, what is the price per yard ? Ans. 5^3. 6. At 31 J dollars for IQj- barrels of cider, what is that per barrel ? Ans. $ 3. 7. How many times is | contained in 746 ? Ans. 1989^. 8. Divide ^ of f by f . Divide J by f of f . QuoL |. Quot. 3|f . 9. Divide J of ^ by ^ of f . Quot. ^. 10. Divide -J of 4 by ^^. Quot. 3. 11. Divide 4f by | of 4. Quot. 2^. 12. Divide | of 4 by 4|-. Quot. f ^ ADDITION AND SUBTRACTION OF FRACTIONS. tr 59. 1. A boy gave to one of his companions f of an orange, to another f, to another -J; what part of an orange did he give to all ? 1 + 1- + i — l^o^v much ? Ans. f . 2. A cow consumes, in 1 month, y\ of a ton of hay ; a horse, in the same time, consumes y\- of a ton ; and a pair of oxen, -^ ; how much do they all consume ? how much piore does the horse consume than the cow ? the oxen than the horse ? -f-^ + y\ + -j\ =: how much ? -^ — A = how much ? -^ — ^ = ^^^^ much ? 3- i + f +i = lio^vmuch ? f — ;i- = howmuch? 4. 2V+/^ + ^a + M + 2^(T = h0WmUch? il — T^ff = how much ? 5. A boy, having J of an apple, gave ^ of it to his sister; what part of the apple had he left ? f — i=: how much ? 120 ADDITION AND SUBTRACTION OF FRACTIONS. U 60. When the denominators of two or more fractions are alikey (as in the foregoing examples,) they are said to have a common denominator. The parts are then in the same denomina- tion, and, consequently, of the same magnitude or value. It is evident, therefore, that they may be added or subtracted, by adding or subtracting their numerators, that is, the num- ber of their parts, care being taken to write under the re- sult their proper denominator. Thus, ^-f-ir^^ifj i — f 6. A boy, having an orange, gave -J- of it to his sister, and ^ of it to his brother ; what part of the orange did he give away ? 4ths and 8ths, being parts of different magnitudes, or value, cannot be added together. We must therefore first reduce them to parts of the same magnitude, that is, to a common de- nmninaior. f are 3 parts. If each of these parts be divided into 2 equal parts, that is, if we multiply both terms of the fraction f by 2, (IT 46,) it will be changed to f ; then f and J are J. Ans, |^ of an orange, 7. A man had f of a hogshead of molasses in one cask, and f of a hogshead in another; how much mt>ie in one cask than in the other ? Here, 3ds cannot be so divided as to become 5ths, nor can 5ths be so divided as to become 3ds ; but if the 3ds be each divided into 5 equal parts, and the 5ths each into 3 equal parts, they will all become 15ths. The f will become ^^, and the |- will become -^ ; then, -f-^ taken from -J-J leaves IT 60. From the very process of dividing each of the parts, that is, of increasing the denominators by multiplying them, it follows, that each denominator must be di factor of the common denominator ; now, multiplying all the denomina- tors together will evidently produce such a number. Hence, To reduce fractions of different denominators to equivalent fractions, having a common denominator, — ^RuLE : Multiply together all the denominators for a common denotni' nator, and, as by this process each denominator is multiplied by all the others, so, to retain the value of each fraction, multiply each numerator by all the denominators, except it» own, for a new numerator, and xind'cr it write the common denominator. V 00. ADDITIOKT ANU SUBTRACTION OF FRACnOLiS. 121 EXAMPLES FOR PRACTICE, 1. Reduce §, J and ^ to fractions of equal value, having a common denominator. 3 X 4 X 5 n= 60, the common denominator. 2 X 4 X 5 = 40, the new numerator for the iirst fraction. 3X3X5 = 45, the new numerator for the second fraction. 3X4X4 = 48, the new numerator for the third fraction. The new fractious, therefore, are |{;, fj, and ||. By an inspection of the operation, the pupil will perceive, that the numerator and denominator of each fraction have been mul- tiplied by the same numbers; consequently, (IT 46,) that their value has not been altered. ^ ^ ^ ' 2. Reduce ^, f, ^ and f to equivalent fi€cfioni^, having' .. common denominator. Ans. jf g-, -}f J, J|3-, ^^g, 3. Reduce to equivalent fractions of a common denomi- nator, and add together, ^, f , and ^. ^ns. 1-5 + If + i^ = f J = Hh Amount, 74. Add together J and f . Amount^ 1 J|. \ 5. What is the amount of ^ -f- ^ -f 1 + -^ ? Arts, m=}AV 6. What are the fractions of a common denominator equivalent to J and f ? Ans. ^f and f £, or -fir and |J. We have already seen, (^ 59, ex. 7,) "that the common denominator may be any number, of which each given de- nominator is a factor, that is, any number which may be di- vided by each of them without a remainder. Such a number is called a common multiple of all its common divisors, and the least number that \vi\\ do this is called their least com- mon multiple; therefore, the least common denominator o^ imy fractions is the least common multiple of all their denominators. Though the rule already given will always find a common multiple of the given denominators, yet it will not always find their least comrao'/ multiple. In the last example, 24 is evidently a common multiple of 4 and 6, for it will exactly measure both of them; but 12 will do the same, and as 12 is the least number that will do this, it is the least common multiple of 4 and 6. It will theiefore be convenient to have A rule for finding this least common multiple. Let the num- bers be 4 and 6. It is evident, that on^ number is a multiple of another, when the former contains all the factors of the latter. The L 122 ADDITIOIV AlfD iSUBTKACTIOrf OF FRACTIONS. IT 61, factors of 4 are 2 and 2, (2 X 2 = 4.) The factors of 6 are 2 and 3, (2 X 3 z= 6.) Consequently, 2 X 2 X 3 = 12 contains tlie factors of 4, that is, 2 X 2 ; and also contains the factors of 6, thl PRACTICE 3. What is the value of f of a shilling ? OPERATION. Numer, 3 12 Den(m,S)Ze(4 d. 2 q. 32 4 4 Am, 9. |- of a month is howma- hy days, hours, and minutes ? 11. Reduce f of a mile its proper quantity; to ^' 13. Reduce -{^ of an acre A6 its proper quantity. ^X 15. What is the value of •fj of a dollar in shillings, pence, &c. ? 17. What is the value of ^j- of a yard ? 19. What is the value of y% of a ton ? EXAMPLES FOR PRACTICE. 4. Reduce 4 d. 2 q. to the fraction of a shilling. OPERATION. 4 d. 2 q. Is. 4 12 18 Numcr. 12 4 48 Denotth if = I Ans. 16(2 q. 16 5. What is the value of f ' 6. Reduce 7 oz. 4 pwt. to of a pound Troy ? the fraction of a pound Troy. / 7. What is the value off 8. Reduce 8 oz. 14f dr. to of a pound avoirdupois .'' the fraction of a pound avoir- dupois. Note. Both the numerator and the denominator must be reduced to 9ths of a dr. 10. 3 weeks, 1 d. 9 h. 36 m. is what fraction of a month ? 12. Reduce 4 fur. 125 yds. 2 ft. 1 in. 2f bar. to the frac- tion of a mile. 14. Reduce 1 rood 30 poles to the fraction of an acre. 16. Reduce 5 s. 7^ d. to the fraction of a dollar. 18. Reduce 2 ft 8 in. 1| b. to the fraction of a yard. 20. Reduce 4 cwt. 2 qr. 12 lb. 14 oz. 12y\- dr. to the frac- j tion of a ton* Note, Let the pupil be required to rex^erse and prove the following examples : 21. What is the value of -^^ of a guinea? 128 SUPPLEMENT TO FRACTIONS. IT 64, 22. Reduce 3 roods 17^^ poles to the fraction of an acre. '23. A man bought 27 gal. 3 qts. 1 pt. of molasses; what part is that of a hogshead ? / 24. A man purchased -f*^ of 7 cwt. of sugar; how much isugar did he purchase ? 25. 13 h. 42 m. 51f s. is what part or fraction of a day ? SUPPIiEMENT TO FRACTIONS. i QUESTIONS. 1. What 2iXQ fractions 1 2. Whence is it that the parts into which any thing or any number may be divided, take their name ? 3. How are fractions represented by figures ? 4. What is the number above the line called ? — Why is it so called ? 5. What is the number below the line called ?• — Why is it so called ? — Wliat does it show? 6. What is it which determines the magnitude cf the parts? — Why? 7. What is a simple or proper fraction? an improper fraction ? a mixed number ? 8. How is an improper fraction reduced to a whole or mixed number? 9. How is a mixed number reduced to an improper fraction ? a whole number? 10. What is understood by the terms of the fraction? 11. How is a fraction reduced to its most simple or lowest terms ? 12. W^hat is understood by a common di- visor? by the greatest common divisor? 13. How is it found ? 14. How many ways are there to multiply a frac- tion by a whole number? 15. How does it appear, that di- viding the denominator multiplies the fraction ? 16. How is a mixed wwmhex multiplied? 17. What is implied in multi- plying by a fraction ? 18. Of how many operations does it consist? — What are they? 19. When the multiplier is less than a unit, what is the product compared with the multi- plicand ? 20. How do you multiply a whole number by a fraction? 21. How do you multiply one fraction by ano- ther? 22. How do you multiply a mixed number by a mixed number ? 23. How does it appear, that in multiply- ing both terms of the fraction by the same number the value of the fraction is not altered ? 24. How many ways are there to divide a fraction by a whole number ? — What are' they? 25. How does it appear that diffraction is divided by multiplying its denwninatorl 26. How does dividing by ft 1r 64, 65. SUPPLEMENT TO FRACTIONS. J 2d fraction differ from multiplying by a fraction ? 27. When the divisor is less Iban a unit, what is the quotient compared with the dividend ? 28. What is understood by a common denominator ? the least common denominator ? 29. How does it appear, that each given denominator mi:st be a factor of the common denominator ? 30. How is the com- mon denominator to two or more fractions found? 3L What is understood by a multiple ? by a common miilti" plel by the least common multiple ? — What is the pro- cess of finding it ? 32. How are fractions added and sub- tracted ? 33. How is a fraction of a greater denomination reduced to one of a less? of a less to a greater? 34. How are fractions of a greater denomination reduced to in- tegers of a less ? integers of a less denomination to the fraction of a greater ? EXERCISES, 1. What is the amount of | and | ? of ^ and f ? of 12^, 3§ and 4 J ? Ans. to the last, 20-^- 2. To |- of a pound add f of a shill_^ng. Amount, 18^ s. Note, First reduce both to the same denomination. 3. |- of a day added to J of an hour make how many hours ? what part of a day ? Ans, to the last, f f d- 4. Add ^ lb. Troy to f^- of an oz. Amount, 6 oz. 11 pwt. 16 ^. • 5. How much is i less ^? t% — i-^ t^ — /*j^ ^H — 4^? 6-~4t? if? — ioff off? Am. to the last, ^4?. 6. From J shilling take J of a penny. Rem, 5^ d. 7. From ^ of an ounce take J of a pwt. Rem, 11 pwt. 3 grs, 8. From 4 days 7^ hours take 1 d. Q^V h. Rem, 2 d. 22 h. 20m. 9. At $^ per yard, what costs J of a yard of cloth ? IT 65. The price of unity, or 1, being given, to find the cost of any quantity, either less or more than uuity, multiply the price by the quantity. On the other hand, the cost of any quantity, either less or more than unity, being given, to find the price of unity, or 1, divide the cost by the quantity, Ans. $ ^ It l30 SUPPLEMENT TO FRACTIONS. IT 65. 1. If -fj lb. of sugar cost -^-^ of a shilling, what will Jf of a pound cost ?* This example will require two operations : first, as above, to find the price of 1 lb. ; secondly, having found the price of 1 lb., to find the cost of f f of a pound, t^ s. -7- -}^ (-ff of f^s. IT 57) nr y9^V s- the price of 1 lb. Then, ^^^ s. X . B (If of tU s. II 53) ^ m-i s. = 4 d. 3f?H q., the Answer, Or we rnay reason thus : first to find the price of 1 lb. : \^ lb. costs y^3- s. If we knew what -^ lb. would cost, we might repeat this 13 times, and the result would be the price of 1 lb. \^ is 11 parts. If ^^ lb. costs ^ s., it is ce- dent jV 1^' will cost ^j- of f"^ r= yj-j- s., and if lb. will cost 13 times as much, that is, -f^-^ s. = the price of 1 lb. Then, H of AV s. = 'iUl s., the cost of e of a pound, f ^f s, rr: 4 d. 3f fl^^ q., as before. This process is called solving the question by analysis. After the same manner let the pupil solve the following questions : 2. If 7 lb. of sugar co?;t f of a dollar, what is that a pound ? 4 of f = how much ? What is it for 4 lb. ? ^ of f =^ how much ? What for 12 pounds ? y of f zr how much ? Ans, to the last, $ If. 3. If 6 J yds. of cloth cost $ 3, what cost 9|- yards ? Ans. $4'269. 4. If 2 oz. of silver cost $ 2^24, what costs f oz. ? f. Ans. $*84. * 5. If ^ oz. costs $\i, what costs 1 oz. ? Ans. $ 1'283. i 6. If 1^ lb. less by ^ costs 13-J d., what costs 14 lb. less by h of 2 lb. ? . A71S. 4£ . 9 s. 9^ d. 7. Iff yd. cost $1, what will 40J yds. cost? Am. $59^062+. 8. If y^g- of a ship costs $ 251, what is -^ of her worth ? . Ans. $53^785+. \ 9. At 3|£. per cwt., what will 9f lb. cost? Ans. 6 s. 3^d. 10. A merchant, owning ^ of a vessel, sold § of his share for $957 ; what was the vessel worth ? Ans. $ 1794*375. 11. Iff yds. cost ^£-, what will -f^ of an ell Eng. cost? Ans. 17 s. 1 d. 2f q. • This and the followins^ are examples usually referred to theTuIe Proportion, or Rule of Three, See 1[ 95 ex. 35, tr 65. SUPPLEMENT TO FRACTIONS. 131 ^ 12. A merchant bought a number of bales of velvet, each containing 129^ yards, at the rate of $7 for 5 yards, and sold them out at the rate of $ 11 for 7 yards, and gained $ 200 by the bargain ; how many bales were there ? First find for what he sold 5 yards ; then what he gained on 5 yards — what he gained on 1 yard. Then, as many times as the sum gained on 1 yd. is contained in $ 200, so many yards there must have been. Having found the number of yards, reduce them to bales. Am. 9 bales. ^ 13. If a staff, 5f ft. in length, cast a shadow of 6 feet, how high is that steeple whose shadow measures 153 feet? Aris. 144-J- feet. , 14. If 16 men finish a piece of work in 28 J days, how 'long will it take 12 men to do the same work ? First find how long it would take 1 man to doit ; then 12 men will do it in ^ of that time. Ans. 37|- days. 15. How many pieces of merchandise, at 20^- s. apiece, must be given for 240 pieces, at 12^1- s. apiece ? Ans, 149^^^ 16. How many yards of bocking that is 1|- yd. wide will be sufficient to line 20 yds. of camlet that is f of a yard wide ? First find the contents of the camlet in square measure ; then it will be easy to find how many yards in length of bocking that is 1 J yd. wide it Avill take to make the same quantity. A71S, 12 yards of camlet. "t" 17. If 1|- yd. in breadth require 20^ yds. in length to make a cloak, what in length that is J yd. wide will be re- quired to make the same ? Ans, 34^ yds. 18. If 7 horses consume 2f tons of hay in 6 weeks, how many tons will 12 horses consume in 8 weeks ? If we knew how much 1 horse consumed in 1 week, it would be easy to find how much 12 horses would consume in 8 weeks. 2|. = -Lt tons- If 7 horses consume ^ tons in 6 weeks, 1 horse will consume | of Jji^ = 4^ of a ton in 6 weeks ; and if a horse consume ^ J of a ton in 6 weeks, he will consume ^ of a z= ^g. of a ton in 1 week. 12 horses will consume 12 timos yV^ =z -fff in 1 week, and in 8 weeks they will consume 8 times -iff =:: -^-f- z=i 6f tons, Ans, \ 19. A man with his family, which in ail were 5 persons, did usually drink 7^ gallons of cid'^r in 1 week; how much will they drink in 22^ weeks when 3 persons more are added to the family ? Ans. 280| gallons. 13^ DECIMAL FHAi^riONS. IT 66, 67* ^^ 20. If 9 students spend lOJiB. iu 18 days, liow much will 20 students spend in 30 days ? Aiis. 39^2 . 18 s. 4ff d. If S6, We have seen, that an individual thing or number may be divided into any number of equal parts, and that these parts will be called halves, thirds, fourths, fifths, sixths, &c., according to the number of parts into which the thing or number may be divided ; and that each of these parts may be again divided into any other number of equal parts, and so on. Such are called common^ or vulgar fractions. Their denom- inators are not uniform, but vary with every varying division of a unit. It is this circumstance which occasions the chief difficulty in the operations to be performed on them ; for when numbers are divided into different kinds or parts, they cannot be so easily compared. This difficulty led to the in- vention of decimal fractions, in which an individual thing, or number, is supposed to be divided first into te7i equal parts, which will be tenths ; and each of these parts to be again di- vided into ten other equal parts, which will be hundredths ; and each of these parts to be still further divided into ten other equal parts, which will be thousandths ; and so on. Such are called decimal fractions, (from the Latin worddeceniy which signifies ten,) because they increase and decrease, in a tenfold proportion, iu the same manner as whole numbers. IT 67. In this way of dividing a unit, it is evident, that the denominator to a decimal fraction will always be 10, 100, 1000, or i with a number of ciphers annexed; conse- quently, the denominator to a decimal fraction need not be expressed, for the numerator only, written with a point be- fore it (') called the sepa.ratrix, is sufficient of itself to ex- press the true value. Thus, -^^ are written ^6. tVtt '27. ^i^ '685. TTie denominator to a decimal fraction, although not ex- pressed, is always understood, and is 1 with as many ci- phers annexed as there are places in the numerator. Thus, *3765 is a decimal consisting of four places ; consequently, 1 with four ciphers annexed (10000) is its proper denomina- tor. Any decimal may be expressed in tlie form of a conH IT 67 DECIMAL FRACTIONS. 183 mon fraction by writing under it its proper denominator. Thus, ^3765 expressed in the form of a common fraction, When whole numbers and decimals are expressed to- gether, in the same number, it is called a mixed number. Thus, 25*63 is a mixed number, 25*, or all the figures on the left hand of the decimal point, being whole numbers, and '63, or all the figures on the right hand of the decimal point, being decimals. The names of the places to ten-millionths, and, generally, how to read or write decimal fractions, may be seen from the following TABIiE. CO 2 CD 3d place. 2d place. 1st place. " 1st place. 2d place. 3d place. 4th place. 5th place. 6th place. 7th place. » II II ii II II II 11 rf^ Ox *:| O Hundreds. Tens. Units. ooQOOoiOOCn o oa o Oi oo en oi o o o o en o o Q W o fc^ Tenths Hundredths. Thousandths. Ten-Thousandths Hundred-Thousandths. Millionths. Ten-Millionthfl. M O* bO •Jl 00 gp en "7 o> g^ S: S* Bon p. o Oi Ox OX CJt CO H c g 5 S S » p GO n> • § 134 DECIMAL FRACTIONS. IT 68, 69. From the table it appears, that the first figure on the right hand of the decimal point signifies so many tenth parts of a unit; the second figure, so many hundredth parts of a unit; the third figure, so many thousandth parts of a unit, &c. It takes 10 thousandths to make 1 hundredth, 10 hundredths to make 1 tentli, and 10 tenths to make 1 unit, in the same manner as it takes 10 units to make 1 ten, 10 tens to make 1 hundred, &c. Consequently, we may regard unity as a starting point, from whence whole numbers proceed, con- tinually increasing in a tenfold proportion towards the left hand, and decimals continually decreasing^ in the same pro- portion, towards the right hand. But as decimals decrease towards the right hand, it follows of course, that they in- crease towards the left hand, in the same manner as whole numbers. IT 68. The value of every figure is determined by its place from units. Consequently, ciphers placed at the right hand of decimals do not alter their value, since every signifi- cant figure continues to possess the same place from unity. Thus, ^5, '50, '500 are all of the same value, each being equal to ^^g-, or J-. But every cipher, placed at the left hand of decimal frac- tions, diminishes them tenfold, by removing the significant figures further from unity, and consequently making each part ten times as small. Thus, '5, '05, '005, are of different value, '5 being equal to y%-, or J ; '05 being equal to yj^, or 2V ; and '005 being equal to py%-^-, or ^-^-^y. Decimal fractions, having different denominatorSy are readily reduced to a common denominator^ by annexing ciphers until they are equal in number of places. Thus, '5, '06, '234 may be reduced to '500, '060, '234, each of which has 1000 for a common denominator. ^ 69. Decimals are read in the same manner as whole numbers, giving the name of the lowest denomination, or right hand figure, to the whole. Thus, '6S53 (the lowest denomination, or right hand figure, being ten-thousandths) is read, 6853 ten-thousandths. Any whole number may evidently be reduced to decimal parts, that is, to tenths, hundredths, thousandths, &c. by an- nexing ciphers. Thus, 25 is 250 tenths, 2500 hundredths^ 25000 thousandths, &c. Consequently, any mixed number IT 70. ADDITION AND SUBTRACTION OF DECIMALS. 136 may be read together, giving it the name of the lowest de- nomination or right hand figure. Thus, 25'63 may be read 2563 hundredths, and the whole may be expressed in the form of a common fraction, thus, ^^Vc^* The denominatfons in federal money are made to corre- spond to the decimal divisions of a unit now described, dol- lars being units or whole numbers, dimes tenths, cents hun- dredths, and mills thousandths of a dollar ; consequently the expression of any sum in dollars, cents, and mills, is simply the expression of a mixed number in decimal fractions. Forty-six and seven tenths =. 46^(j = 46 '7. Write the following numbers in the same manner : / Eighteen and thirty-four hundredths. Fifty-two and six hundredths. Nineteen and four hundred eighty-seven thousandths. Twenty and forty-two thousandths. One and five thousandths. 135 and 3784 ten-thousandths. 9000 and 342 ten-thousandths. 1 UISTI V 10000 and 15 ten-thousandths. X^P/^ pa, «p^ ' 974 and 102 millionths. ^''^4s^4k!£i^^ 320 and 3 tenths, 4 hundredths and 2 thousandths. 500 and 5 hundred-thousandths. 47 millionths. Four hundred and twenty-three thousandths. ADDITION AND SUBTRACTION OF DECIMAL FRACTIONS. ir 70- As the value of the parts in decimal fractions in- creases in the same proportion as units, tens, hundreds, &:c., and may be read together^ in the same manner as ^vhole numbers, so, it is evident that all the operations on decimal fractions may be performed in the same manner as on whole numbers. The only difficulty, if any, that can arise, must be in finding ichere to place the decimal point in the result. This, in addition and subtraction, is determined by the same rule ; consequently, they may be exhibited together. 1. A man bought a barrel of flour for $ 8, a firkin of but- 1S6 ADDITION AND SUBTRACTION OF DECIMALS. IT tO. ter for $ 3^50, 7 pounds of sugar for Sd^ cents, an ounce of pepper for 6 cents ; what did he give for the whole ? OPERATION. $ 8' = 8000 mills, or lOOOths of a dollar. * 3*50 = 3500 mills, or lOOOths. ^835 = 835 mills, or lOOOths. *06 = 60 mills, or lOOOths. Ans. $ 12'395 = 12395 mills, or lOOOths. As the denominations of federal money correspond with the parts of decimal fractions, so the rules for adding and subtracting decimals are exactly the same as for the same operations in federal money. (See IT 28.) 2. A man, owing $375, paid $175*75; how much did he then owe ? OPERATION. $ 375' z= 37500 cents, or lOOths of a dollar. 175*75 = 17575 cents, or lOOths of a dollar. $ 199*25 =: 19925 cents, or lOOths. The operation is evidently the same as in subtraction of federal money. Wherefore, — In th€ addition and subtrac- tion of decimal fractions, — Rule : Write the numbers under each other, tenths under tenths, hundredths under hun- dredths, according to the value of their places, and point off in the results as many places for decimals as are equal to the greatest number of decimal places m any of the given num- bers. EXAMPLES FOR PRACTICE. ^ 3. A man sold wheat at several times as follows, viz. 13*25 bushels; 8*4 bushels; 23*051 bushels, 6 bushels, and *75 of a bushel; how much did he sell in the whole ? Ans, 51*451 bushels. 4. What is the amount of 429, 21^^^, 355y^^, l^hr and 1-,^ ? Ans, 808^^^, or 808*143. ■^ 5. What is the amount of 2 tenths, 80 hundredths, 89 thousandths, 6 thousandths, 9 tenths, and 5 thousandths ? Ans, 2. 6« What is the amount of three hundred twenty-nine, and seven tenths ; thirty-seven and one hundred sixty-two Uiou- sandths, and sixteen hundredths ? IF to, 71. mul'TiplicAtiox of decimals. 137 7. A man, owing $43165 paid $376^865; how much did be then owe? Am. $3939436. 8. From thirty-five thousand take thirty-five thousandths. / Ans, 34999*966. 9. From 5'83 take 4*2793. Ans. 1*5507. 10. From 480 take 245*0075. Ans. 234*9925. ^ 11. What is the diff^erence between 1793*13 and 817* 05693? Alls. 976*07307. 12. From 4Tfg- take 2^^. Remainder^ '^-twh^ ^^^ l'^®- 13. What is the amount of 29^^, 374^^^^^^^, 97-^^^%, 315t(jW) 27, and 100^% ? Ans. 942*957009. MULTIPLICATION OF DECIMAL FRACTIONS. IT 71. 1. How much hay in 7 loads, each containing 23*571 cwt. ? OPERATION. 23*.571 cwt. =z 23571 lOOOths of a cwt. 7 7 Ans. 164*997 cwt. =1 164997 lOOOths of a cwt. We may here (*!T 69) consider the muUiplicand so many thousandths of a cwt., and then the product will evidently be thousandths^ and will be reduced to a mixed or whole num- ber by pointing off 3 figures, that is, the same number as are in the multiplicand ; and as either factor may be made the multiplier, so, if the decimals had been in the multiplier^ the same number of places must have been pointed off for deci- mals. Hence it follows, we must always i^oint ojf in the pro- duct as many places for decimals as there are decimal places in both factors. 2. Multiply *75 by *25. OPERATION. In this ex^ample, we have 4 de- J^^ cimal places in both factors ; we ^^ must therefore point off 4 places 3»75 for decimals in the product. The 250 reason of pointing off this num- ^ her may appear still more plain, U875 Product. if we consider the two factors as M* i$d MULTIPLICATION OF DJElCIMALS* IT 71^ common or vulgar fractions. Thus, '75 is -^^j and *25 is ^^s : now,f^^^ X xVo" = loVo^j == '1875, Ans. same as b^ fore. 3. Multiply '125 by '03. i|25 rlere, as the number of significant ¥uT) = '00375, the same as before. These examples v/ill be sufficient to establish the following RULE. In the multiplication of decimal fractions^ multiply as in whole numbers J and from the product point off so many figures for deci- mals as there are decimal places in the mxdtiplicand and midt'^ plier counted together^ and^ if there are not so many figures in the . product f supply the deficiency by jyrefixing ciphers, EXAMPLES FOR PRACTICE. 4. At $ 5'47 per yard, what cost 8'3 yards of cloth ? Ans, $45'40I. f 5. At $ '07 per pound, what cost 2G'5 pounds of rice ? Ans, $1'855. 6. If a barrel contain 1'75 cwt. of flour, what will be the weight of '63 of a barrel ? Ans, 1'1025 cwt. 7. If a melon be worth $ '09, what is '7 of a melon wortli ? Ans. 6^^ cents. 8. Multiply five hundredths by seven thousandths. Product, '00035. 9. What is '3 of 116 ? Ans. 34'8. 10. What is '85 of 3672 > Ans. 3121'2. 11. What is '37 of '0563? Ans. '020831. 12. Multiply 572 by '58. Product, 331 '76. i 13. Multiply eighty-six by four hundredths. Producty 3'44. 14. Multiply '0062 by '0008. 15. Multiply forty-seven tenths by one thousand eighty- six hundredths. ^71^72. DIVISION OF DECIMALS. 139 16. Multiply two hundredths hy eleven thousandths. 17. What will be the cost of tliirieeii huiiJiedths of a ton of hay, at $ 11 a ton ? 18. What will be the cost of three hundred seventy-five thousandths of a cord of wood, at $ 2 a cord ? J^ 19. If a man's wages be seventy-tive hundredths of a dol- lar a day, how much will he earn in 4 weeks, Sundays ex- cepted ? DIVISION OF DECIMAL FRACTIONS. IT 72. Multiplication is proved by division. We have seen, in multiplication, that the decimal places in the product must always be equal to the number of decimal places in the multiplicand, and multiplier counted together. The muki- plicand and multiplier, in proving multiplication, become the divisor and quotient in division. It follows of course, in di- vision, that the number of decimal places hi the divisor and quotient^ counted toijether^ must always be equal to the number of decimal places in the dividend. This will still further appear from the examples and illustrations which follow : 1. If 6 barrels of Hour cost $44'71S, what is that a bar- rel ? By taking away the decimal point, $44^718 nz 44718 mills, or lOOOths, which, divided by 6, the quotient is 7453 mills, zn $ 7^453, the Answer, Or, retaining the decimal point, divide as in whole num- bers. OPERATION. As the decimal places in the di- 6 )44 718 visor and quotient, counted toge- Ans. 7'453 ther, must be equal to the number of decimal places in the dividend, there being no decimals in the divisor^ — therefore point off three figares for decimals in the quotientj equal to the number of decimals in the dividend, which brings us to the same re- sult as before. 2. At $ 4^7^ a barrel for cider, how many barrels may be bought for $31 ? In this example, there are decimals in the divisor, and none in the dividend. $4'75=i:475 cents, and $31, by winexing two ciphers, = 3100 cents ; that is, reduce the di 140 DIVISION OF DECIMALS. H 72. vidend to parts of the same denomination as the divisor. Then, it is plain, as many times as 475 cents are contained in 3100 cents, so many barrels may be bought. 475) 3100 (6f|^ barrels, the Answer; that is, 6 barrels and 2850 Iff of another barrel. -- -— But the remainder, 250, instead of be- ing expressed in the form of a common fraction, may be reduced to lOths by annexing a cipher, which, in effect, is multiplying it by 10, and the division con- tinued, placing the decimal point after the 6, or whole ones already obtained, to distinguish it from the decimals which are to follow. The points may be withdrawn or not from the divisor and dividend. OPERATION. 4'75)31'00(6^526 + barrels, the Answer; that is, 6 bat- 2850 rels and 626 tliousandths of another •— -- barrel. ^^^^ By annexing a cipher to the first remainder, thereby reducing it to 1250 lOths, and continuing the division, 950 ^^ obtain from it '5, and a still fur- ther remainder of 125, which, by an- 3000 nexing another cipher, is reduced to 2850 lOOths, and so on. ■~T77 The last remainder, 150, is ^§ of a thousandth part of a barrel, whicli is of so trifling a value, as not to merit notice. If now we count the decimals in the dividend, (for every cipher annexed to the remainder is evidently to be counted a decimal of the dividend,) we shall find them to be five^ which corresponds with the number of decimal places in the divisor and quotient counted together. 3. Under IT 71, ex. 3, it was required to multiply 425 by '03 ; the product was '00375. Taking this product for a dividend, let it be required to divide '00375 by '125. One operation will prove the other. Knowing that the number of decimal places in the quotient and divisor, counted to- gether, will be equal to the decimal places in the dividend, we may divide as in whole numbers, being careful to retain the decimal points in their proper places. Thus, V 72, 73. DIVISION OF DECIMALS. 141 OPERATION. The divisor, 125, in 375 goes 3 425) *00375 (*03 times, and no remainder. We have ^^^ only to place the decimal point in QQQ the quotient, and the work is done. There are five decimal places in the dividend ; consequently there must be five in the divisor and quotient counted together ; and, as there are three in the di- visor, there must be two in the quotient ; and, since we have but one figure in the quotient, the deficiency must be sup- plied by prefixing a cipher. The operation by vulgar fractions will bring us to the same result Thus, 425 is ^2^%, and *00375 is yiAs^ • now, ^'^cu -^ A% = yf K-&S8ir = yfu = '^3, the same as before. IT 73. The foregoing examples and remarks are suffi- cient to establish the following RUIiE. In the division of decimal fractions^ divide as in whole mmi" beiSj and from the right hand of the qiiotient point off as many figures for decimals^ as the decimal figures in the dividend eX" ceed those in the divisor^ and if there are not so many figures in the quotient J supply the deficiency by prefixing ciphers. If at any time there is a remainder, or if the decimal figures in the divisor exceed those in the dividend, ciphers may be annexed to the dividend or the remainder, and the quotient carried to any necessary degree of exactness ; but the ciphers annexed must be counted so many decimals of the dividend. EXAMPLES FOR PRACTICE. >4. If $ 472^875 be divided equally between 13 men, how much will each one receive ? A?is, $ 36 '375. 5. At $ '75 per bushel, how many bushels of rye can be bought for $ 141 ? Ans, 188 bushels. f 6. At 12J cents per lb., how many pounds of butter may be bought for $ 37 ? Ans. 296 lb. 7. At 6^ cents apiece, how many oranges may be bought for $8? Ans. 128 oranges. 8. If '6 of a barrel of flour cost $ 5, what is that per bar- rel? Ans. $8'333-f^ 9. Divide 2 by 534. Quoi. ^037+. 142 REDUCTION OF COMMON OR Hi 73, 74r 10. Divide ^012 by *005. QiioL 2*4. II 1. Divide three thousandths by four hundredths. ^ QuoL '075, ' 12. Divide eighty-six tenths by ninet} four thousandths. 13. How many times is 47 contained in 8 ? REDUCTION OF COMMON OR VULGAR FRAC- TIONS TO DECIMALS. U 74. 1. A man has | of a barrel of flour; what is that expressed in decimal paits? As many times as the denominator of a fraction is con- tained in the numerator, so many whole ones are contained in the fraction. We can obtain no whole ones in |^, because the denominator is not contained in the numorator. We may, however, reduce the numerator to lentlis^ (IT 72, ex. 2,) by annexing a cipher to it, (wliich, in effect, is multiplying it by 10,) making 40 tenths, or 4^0. Then, as many times as the denominator, 5, is contained in 40, so many tenths are contained in the fraction. 5 into 40 goes 8 times, and no remainder. Ans, '8 of a bushel. 2. Express f of a dollar in decimal parts. The numerator, 3, reduced to tenths, is -Jg, 3^0, which, divided by the denominator, 4, the quotient is 7 tenths, and a remainder of 2. This remainder must now be reduced to hundredths by annexing another cipher, making 20 hun- dredths. Then, as many times as the denominator, 4, is con- tained in 20, so many hundredths also may be obtained. 4 into 20 goes 5 times, and no remainder, f of a dollar, there- fore, reduced to decimals, is 7 tenths and 5 hundredths, that is, '75 of a dollar. The operatic u may be presented in form as follows : — Num, Denom, 4 ) 3'0 (^75 of a dollar, the Answer. 28 20 20 If Y4. VULGAR FRACTIONS TO DECIMALS, 143 8. Reduce ^g^ lo a decimal fraction. The numerator must be reduced to hundredths, by annex* ing two ciphers, before the division can begin. 66 ) 4*00 ( '0606 +, the Ajiswer. 396 400 As there can be no tenths, a cipher must 396 be placed in the quotient, in tenth's place. 4 Note. ^ cannot be reduced exactly; for, however long the division be continued, there will still be a remainder.* It is sufficiently exact for most purposes, if the decimal be ex- tended to three or four places. From the foregoing examples we may deduce the follow- ing general Rule : — To reduce a common to a decimal froA> * Decimal figures, which continually repeat, like *06, in this exam- ple, are called Rcpetends., or Circulating Decimals. If only one figure repeats, as *3333 or '•1111 , &c.j it is called a single rcpeteud. If tzco ot more figures circulate alternately, as *06060(i, '234234234, &c., it is called a compound rcpeteud.. If other figures arise before those which circulate, as '743333, '143010101, &c., the decimal is called a mixed repetend. Ji single repetend is denoted by writing only the circulating figura with a point over it: thus, '3, signifies that the 3 is to be continually repeated, forming an infinite or never-ending series of 3's. Jl compound repetend is denoted by a point over the first and last re- peating figure : thus, '234 signifies tliat 234 is to be continually re- peated. It may not be amiss, hero to show how the value of any repetend m^j bo found, or, in other words, how it may be reduced to its equivalent vulgar fraction. If we attempt to reduce ^ to a decimal^ we obtain a continual repe- tition of tlie figure 1 : thus, 'lllll, that is, the repetejid '1. The value of the repetend '1, then, is -^ ; the value of '222, &c., the repetend *2, will evidently bo twice as much, that is, f-. In the samo manner, 3=» 6", and '4 = |, and 'f» == |-, and so on to 9, which = f = 1. 1. What is the value of '8 .? Ms. f . 2. What is the value of '6 ? Ans. f = f . Wliat is the value of 'G ? of '7? of '4.? of'o.^ of '9.? of*i.? If -^ be reduced to a decimal, it produces 'OlOJOl.or the repetend OJ. The repetend '02, being 2 times as much, must be ^g- and '03 = -jp^p aiv] *48, being 48 times a.s much, must be f f . and '74 === Jf , (&c. 144 REDUCT, OF VULG. FRACTIONS TO DECIMALS. IT 74, tion, — Annex one or more ciphers^ as may be necessary, to the numerator J and divide it by the denominator. If then there he a remainder, annex another cipher, and divide as before, and so continue to do so long as there shall continue to be a remainder, or until the fraction shall be reduced to any necessary degree of exactness. The quotient will be the decimal reqviired, which must consist of as many decimal places as there are ciphers annexed to the numerator ; and, if there are not so many figures in the quotient, the deficiency must be sup- plied by prefixing ciphers. exampl.es for practice. 4, Reduce .V, J, -^^-^, and y^^ to decimals. Ans, '5 ; '25 ; *025 ; '00797 +, /^5. Reduce fj, ^-^ft^^, yy\tj ^^^ ^uV^-^ to decimals. ' Ans. '692 + ; '003; '0028 +; '000183 +, 6. Reduce |^, ^^-q%, ^^^ to decimals. 7. Reduce |, ^V, ^^, h h tt? A? wi^ to decimals. 8. Reduce i, f , |, i, f , f , f, ^, ^\, '^% to decimals. If wh 1>6 reduced to a decimal, it produces '001 ; consequently, '002 = ef 9 , and '037 = /^V? »ind 425 = ff f, &c. As this principle will apply to any number of places, we have this general Rule for re- ducing a circulating decimal to a vulgar fraction^ — Make the given repetend the numerator, and the denominator will be as many 9s as there are repeating figures. 3. What is the vulgar fraction equivalent to '704 ? Ans. i%i* ^ 4. What is the value of '003 ? '014 ? '324 ? 'OlOSi ? "2463 ? '002103 ? ,ans. to last, TsWinr- 5. What is the value of '43 ? In this fraction, the repetend begins in the second place, or place of hundredths. The first figure, 4, is X(j, and the repetend, 3, is f- of -^j that is, "9^; these two parts must be added together, t^^ -f- ^ = f^ = -Jtj. Ans. Hence, to find the value of a mixed repetend^ — Find the value of the two parts, separately, and add them together. 6. What is the value of '153 ? i^^^T 4" F^TT = iU = t¥^7 >^ns. 7. What is the value of '0047 ? Ans. -^Mu* 8. What is the value of '138.? ^16.? '4123 .=» It is plain, that circulates may be added, subtracted, multiplied, and divided, by first reducing them to their equivalent vulgar fractions. IT 75. REDUCTIOIf OP DECIMAL FRACTIONS. 145 REDUCTION OF DECIMAL FRACTIONS. IT 76- Fractions, we have seen, (IT 63,) like integers^ are reduced from low to higher denominations by division^ and from high to lower denominations by multiplication. To reduce a compound num- ber to a decimal of the highest denomination, 1. Reduce 7 s. 6 d, to the decimal of a pound. 6 d. reduced to the decimal of a shilling, that is, divided by 12, is '5 s., which annexed to the 7 s. making 7'5 s., and divided by 20, is '375 £, the Ans. The process may he pre- sented in form of arw/e, thus : — Divide the lowest denomina- tion given, annexing to it one or more ciphers, as may be necessary, by that number ^^ilich it takes of the same to make one of the next higher denomination, and annex the quotient, as a decimal to that higher denomination; so-con- tinue to do, until the whole shall be reduced to the deci- mal required. EXAMPLES FOR PRACTICE. 3. Reduce 1 oz. 10 pwt to l)ie fraction of a poun^d. OPERATION. 20)10^0 pwt. 12)^5 oz. 425 lb, Ajis. N To reduce the decimal of a higher denomination to integers of lower denominations, 2. Reduce '375 £> , to in- tegers of lower denominations. ^375 £ . reduced to shillings, that is, multiplied by 20, is 7^50 s. ; then the fractimial part, *50 s., reduced to pence, that is, multiplied by 12, is 6 d. Ans, 7 s. 6 d. That is, — Multiply the given decimal by that number which it takes of the next lower de- nomination to make one of this higher, and from the right hand of the product point off as many figures for decimals as there are figures in the given decimal, and so con- tinue to do through all the de- nominations; the several num* bers at the left hand of the decimal points will be the value of the fraction in the proper denominations. EXAMPLES FOR PRACTICE. 4. Reduce 425 lbs. Troy to integers of lower denomina- tions. OPERATION. Ih. n25 _JL2 oz. 1*500 20 pwt. lO'OOO. Ans, loz.lOpwt. REDUCTION OP DECIMAI4 FRACTIONS. Tr7e. 6. What is the value of *2325 of a ton? ^8. hhd. What is the value of ^72 of heer ? t 10. What is the value of '375 of a yard ? k12. Wliat is the value of '713 of a day? 14. What is the value of '78125 of a guinea ? |16. AVhat is the value of '15334821 of a ton? 146 ; V 5. Reduce 4 cwt. 2| qrs. to the decimal of a ton. ,mte, 2{ — 2'6. .J^^7. Reduce 38 gals. 3'52 qts. of beer, to the decimal of a hhd. ff 9. Reduce 1 qr. 2 n. to the decimal of a yard. 11. Reduce 17 h. 6 m. 43 sec. to the decimal of a day. y 13. Reduce 21 s. 10^ d. to the decimal of a guinea. 15. Reduce 3 cwt. qr. 7 lbs. 8 oz. to the decimal of a ton. Let the pupil be required to reverse and prove the follow- iufir examples : yt, R,educe 4 rods to the decimal of an acre, 18. What is the value of '7 of a lb. of silver ? 19. Reduce 18 hours. 15 m. 50'4 sec. to the decimal of a day. 20. W^hat is the value of '67 of a league ? 21. Reduce 10 s. 9J d. to the fraction of a pound. IT fi^S. There is a method of reducing shillings, pence and farthings to the decimal of a pound, by inspection^ more simple and concise than the foregoing. The reasoning in relation to it is as follows : yV of 20 s. is 2 s. ; therefore every 2 s. is yV? or 4 i8. Every shilling is ^^ = y§^, or '05 £ . Pence are readily- reduced to farthings. Every farthing is ^^ M . Had it so happened, that 1000 farthings, instead of 960, had made a pound, then every farthing would have been y^Vry? or '001 £, . But 960 increased by ^ part of itself is 1000; conse- quently, 24 farthings are exactly to ^tt) o^ ^^25 £> ., and 48 farthings are exactly yf Htj-, or '050 £ . Wherefore, if the number of farthings, iu the given pence and farthings, be more than 12, ^ part will be more than ^; therefore add 1 to them : if they be more than 36, ^ part will be more than 1^; therefore add 2 to them: then call them so many* thousandths, and the result will be correct within less than i of TtrW of a pound. Thus, 17 s. 6f d. is reduced to tha 1r77. REDUCTION OF DECIMAL FRACTIONS. 147 decimal of a pound as follows : 16 s. = *8 i6 . and 1 s. = '05 £> . Then, 5 J d. = 23 farthings, which^ increased by 1, (the number being more than 12, but not exceeding 36,) is <024 £ ., and the whole is S74 £ . the Am. Wherefore, to reduce shillings^ pence and farthings to the decimal of a pound by inspection^ — Call every two shillings one tenth of a pound ; every odd shilling^ five hundredths ; and the number of farthings^ in the given pence and farthings^ so ma?nf thousandths^ adding one^ if the number be more than twelve and not exceeding thirty-six^ and tivo, if the number be more than thirty-six, IT 77. Reasoning as above, the result, or the three first figures in any decimal of a pound, may readily be reduced back to shillings, pence and farthings, by inspection. Double the first figure, or tenths^ for shillings, and, if the second figure, or hundredths, be five, or more than five, reckon ano- ther shilling ; then, after the five is deducted, call the figures in the second and tlJid place so many farthingG, abating one when they are above twelve, and two when above thir- ty-six, and the result will be the answer, sufficiently exact for all practical purposes. Thus, to find the value of ^876 £ . by inspection : — *8 tenths of a pound - - - =z 16 shillings. *05 hundredths of a pound - - zz: 1 shilling. *026 thousandths, abating 1, =: 25 farthings zir s. G^d, '876 of a pound - - - - z= 17 s. 6^ d. Am, EXAMPLES FOR PRACTICE. 1. Find, by inspection, the decimal expressions of 9 s. 7 d., and 12 s. Of d. Am, '479^2., and '603^2. 2. Find, by inspection, the value of ^523^3., and '691^2. Ans. 10 s. 5^ d., and 13 s. 10^ d. 3. Reduce to decimals, by inspection, the following sums, and find their amonnt, viz. : 15 s. 3 d. ; 8 s. 11^ d. ; 10 s. 6^ d. ; 1 s. 8^ d. ; J d., and 2^ d. Atnouht, £ 1'833. 4. Find the value of '47 £, • Note, When the decimal has but two figures, after taking out the shillings, the remainder, to be reduced to thousandths^ will require a cipher to be annexed to the right hand, or supposed to be so. Ans, 9 s. 4| i?. 148 SUPPLEMENT TO DECIMAL FRACTIONS. IF 77. 5. Value the following decimals, by inspection, and find their amount, viz.; *'785i^.; '357 £.; '916 i^.; '74 JS.; '6£.; '26£.; '09 JB.; and '008 iS. ^?i6. 3JS. 12s. 11 d. SUFS»Z.EM£NT TO DECZMAX. FHACTZONS. QUESTIONS. 1. What are decimal fractions ? 2. Whence is the term derived ? 3. How do decimal differ from coriimon frac- tions ? 4. How are decimal fractions written ? 6. How can the proper denominator to a decimal fraction be known, if it be not expressed ? 6. Hew is the value of ©very figure determined? 7. What does the first figure on the right hand of the decimal point signify ? the second figure ? < third figure ? fourth figure ? 8. How do ciphers^ placed at the right hand of decimals, affect their value ? 9. Placed at the left hand, how do they affect their value ? 10. How are decimJs read? 11. How are decimal frac- tions, having different denominators, reduced to a common denominator? 12. What is a mixed number? 13. How may any iclwle number be reduced to decimal parts ? 14. How can any mixed number be read together, and the whole expressed in the form of a common fraction ? 15. What is observed respecting the denominations in federal money ? 16. What is the rule for addition and subtraction of decimals, particularly as respects placing the decimal point in the results ? multiplication ? division ? 17. How is a common or vulgar fraction reduced to a deci- mal ? 18. What is the rule for reducing a compound num- ber to a decimal of the highest denomination contained in it ? 19. What is the rule for finding the value of any given decimal of a higher denomination in terms of a lower f 20. What is the rule for reducing shillings, pence and far- things to the decimal of a pound, by inspectmi ? 21. What ^i^the reasoning in relation to this rule ? 22. How may the three first figures of any decimal of a pound be reduced to« shillings, pence and farthings, by insjiectwn 1 M UK. SUPPLEMENT TO BECIMAt. FRACTIONS. 149 EXERCISES. 1. A merchant had several remnants of cloth, measuring as follows, viz. : 7 1 yds. 6f 1 i 9f 8i 3jV How many yapds in the whole, and what would the whole come to at $ 3'67 per yard ? Note, Reduce the common fractions to deci- mals. Do the same where var they occur in the examples which follow. Ans, 36^475 yards. $ 133^863 +, cost. J^. 2. From a piece of cloth, containing 36| yds,, a merchant sold, at one time, 7^^^ yds., and, at another time, 12| yds. ; how much of the cloth had he left ? Ans, IG'7 yds. 3. A farmer bought 7 yards of broadcloth for Sy^ i3 ., a barrel of flour for 2y\ ^ ., a cask of Hme for 1| £ ., and 7 lbs. of rice for ^ jS. ; he paid 1 ton of hay at 3^^^ £,^ I cow at6f £.j and the balance in pork at -^^ £, per lb.; how many were the pounds of pork ? Note, In reduci»:ig the common fractions in this example, it will be sufficiently exact if the decimal be extended to three places. Ans. l()8f lb. ^^ 4. At 12^ cents per lb., what will 37|- lbs. of butter cost? -4715. $4'718}. 5. At $17*37 per ton for hay, what will llf tons cost? Ans. $201'92f. 6. TTie above example reversed. At $ 201 '92| for 1 1 1 tons of hay, what is that per ton ? Ans. $ 17*37. I 7. If *45 of a ton of hay cost $ 9, what is that per ton ? Co7isult 'W 65. Ans. $ 20. 8. At '4 of a dollar a gallon, what will *25 of a gallon of molasses cost ? Ans. $*1. 9. At $ 9 per cwt., what will 7 cwt. 3 qrs. IG lbs. of sugar cost? Note. Reduce the 3 qrs. 16 lbs. to the decimal of a cwt, extending the decimal in this, and the examples which fol- low, to four places. Ans. 71'035-(-, 10. At $69*875 for 5 cwt. 1 qr. 14 lbs. of raisins, what is that per cwt. ? Ans. $ 13. 11. What will 2300 lbs. of hay come to at 7 mills per )b. ? Ans, $ itinO. 12. What will 765^ lbs. of coffee come to, at 18 cents per lb.? Ans, $137*79 150 SUPPLEMENT TO DECIMAL FllACTIOlfSv f 77/ 13. What will 12 gals. 3 qts. 1 pt of gin cost, at 28 cents per quart? Note. Reduce the whole quantity to quarts and the deci- mal of a quart. Am. $ J,4'42. 114. Bought 16 yds. 2 qrs. 3 na, of broadcloth for $ 100*125 ; what was tliat per yard ? Am. $ 6. 15. At $ 1*92 per bushel, how much wheat may be bought for $ '72 ? Ans. 1 peck 4 quarts- fl6. At $ 92'72 per ton, how much iron may be pur- chased for .$60'268r Ans. 13 cwt. 17. Bought a load of hay for $947, paying at the rate of $ 16 per ton j what was the weight of the hay ? A71S. 11 cwt. 1 qr.23lbs. /18. At $302*4 per tun, what will 1 hhd. 15 gals. 3 qts. of wine cost ? Am. $ 94*50. 19. The above reversed. At $94*50 for 1 hhd. 15 gals. 3 qts. of wine, what is that per tun ? Am. $ 302*4. Note. The following examples reciprocally prove each other, excepting when there are some fractional losses, as ex- plained above, and even then the results will be sufficiently exact for all practical purposes. If, however, greater exact- ness be required, the decimals must be extended to a greater number of places. 20. At $1*80 for 3^ qts. of wine, what is that per gal. ? 22. If f of a ton of pot- ashes cost $60*45, what is that per ton ? 21. At $2*215 per gal., what cost 3-j[- qts. ? 23. At $96*72 per ton for pot-ashes, what will f of a ton cost ? 24- If*Sofayard of cloth cost $2, what is that per > ard > 27. If 14 cwt. of pot-ashes cost 19 JS . 5 8., what is that per ton ? 25. If a yard of cloth cost $ 2*5, what will *8 of a yard cost ? 28. If a ton of pot-ashes cost 27 iB . 10 s., what will 14 cwt. cost ? 26. At $2*5 per yard, how much cloth may be pur- chased for $ 2 ? 29. At 27 ie. 10s. a ton for pot-ashes, what quantity may be bought for 19 iS. 5 s.? Note, After the same manner let the pupil reverse and prove the following examples : ir77j78. REDUCTION OF CURRENCIES. 151 30. At $ 18^50 per ton, how much hay may be bought for $ 12^025 ? 31. What will 3 qrs. 2 na. of broadcloth cost, at $6 pet yard ? ^'32. At $ 2240 for transportation of 65 cwt. 46 miles, what is that per ton ? 33. Bought a silver cup, weighing 9 6z. 4 pwt. 16 grs. foi* 3 iS . 2 s. 3 d. 3| q. ; what was that per ounce ? 34. Bought 9 chests of tea, each weighing 3 cwt. 2 qrs. 21 lbs. at 4 dB . 9 s. per cwt. ; what came they to ? 35. If 5 acres 1 rood produce 26 quarters 2 bushels of wheat, how many acres will be required to produce 47 quarters 4 bushels ? A quarter is 8 bushels. Note. The above example will require two operations, for which consult H 65, ex. 1. // 36. A lady purchased a gold ring, giving at the rate of *$ 20 per ounce ; she paid for the ring $1^25; how much did it weigh ? RliDUGTZO^I OF CUHRENCIBS. IT 78- Previous to the act of Congress in 1786 establish-' ing federal money, all calculations in money, throughout the United States, were made in pounds, shillings, pence and farthings, the same as in PJngland. But these denominations, although the same in name, were different in value in dii^ ferent countries. Thus, 1 dollar is reckoned in England, 4 s. 6 d., called English^ or sterling money. Nova'^Sc^^fa \ ^ ^' ^^^^^ Canada currency. ^ 6 s., called New England curreacf ^ The New Eng-^ land States, Virginia, Kentucky, and Tennessee, New York, ) Ohio, and /8s., called New Yhrk currency. N. Carolina, ) 153 REDUCTION OF CURRENCIES. IT 78 1 dollar is reckoned in 7 s. 6 d., called Pennsylvania currency New Jersey, Pennsylvania, Delaware, and Maryland, GeorS"'' ^""^ 1 ^ '• ^ ^'' '"'■"^'^ ^"^^'" currency. 1. Reduce 6£, 11 s. 6^-d. to federal money. Note, To reduce pounds, shillings, pence and farthings, in either of the above-named currencies, to federal money, — First, reduce the shillings, pence and farthings (if any be contained in the given sum) to the decimal of a pound by in- spection^ as already taught, IF 76. 6£. lis. 6id. — £6'576. English money. — Now, supposing the above sum to be bJnglish money, — 1£, is 20 s. =: 240 pence, in all the above currencies. 1 dollar, in English money, is reckoned 4 s. 6 d. =: 54 pence, that is, ^^/^j = ^u of -t pound. Now, as many times as -^tj^ the fraction which 1 dollar is of 1 pound, Eng- lish money, is contained in J2 6 '576, so many dollars, it is evident, there must be ; that is, — 7b reduce English to federal money ^ — Divide the given sum by £^^ the quotient will be federal money. je 6 W6 English money. ^ote. It will be 40 recollected, to di- vide by a fraction, 9) 263'040 we multiply by the l^^f federal money, Answer. tidf ihfp'roduc't by the numerator. Canada currency. — Supposing the above sum to be Cana- da currency, — 1 dollar, in this currency, is 5 s. := 60 pence, that is, -^-^^ z= ;|- of 1 pound. Therefore, — To reduce Canor da currency to federal money ^ — Divide the given sum by J, and the quotient will be federal money \ or, which is the same thing, — Multiply the given sum by 4. £6*576 Canada currency. 4 $ 27'304 federal money, Answer. V Tf8. REDUCTION OF CURRENCIES. 16S New England currency. — 1 dollar, in this currency, is 6 s. n: 72 pence, that is, -^^jj =: -f^j or '3 of a pound. There- fore, — To reduce New England currency to federal money j — Di- vide the given sum by '3. '3) £ . 6*576 New England currency. $21*92 federal money, Answer. New York currency. — 1 dollar, in this currency, is 8 s. = 96 pence, that is, -r?^^ = •^, or '4 of a pound. Therefore, — To reduce New York currency to federal money y — Divide the given sum by '4. *4) £ . 6*576 New York currency. $ 16*44 federal money. Answer, Pennsylvania crRRENcv. — 1 dollar, in this currency, is 75v e d. = 90 pence, that is, ^^^ ■=. f of a pound. Therefore, — To reduce Pennsylvania currency to federal money ^ — Divide by §, that is, multiply the given sum by 8, and divide the pro- duct by 3. £ . 6*576 Pennsylvania currency. 8 3)52*608 $ 17*536 federal money, Answer, Georgia currency. — 1 dollar, Georgia currency, is 4 »* 8 d. := 56 pence, that is, ^y^y z=z ^^ of a pound. Therefore, — To reduce Georgia currency to federal money, — Divide by -^j that is, multiply the given sum by 30, and divide the pro- duct by 7. £ . 6*576 Georgia currency. 30 7)197*280 $28*182f- federal money. Answer, From the forec^oing examples, we derive the following general Rule : — To reduce English money, and the currencies of Canada and the several States, to federal money, — First, re- duce the shillings, &c., if any in the given sum, to the deci- mal of a pound; this being done, divide the given sum by such fractional part as 1 dollar, in the given currency, it a fractional part of 1 pound. 152 REDUCTION OF CURRENCIES. IT 78 1 dollar is reckoned in New Jersey, ^ Kwi'erand f ^ '* ^ ^'^ ^^"^^ Pennsylvania currency Maryland, J S. Carolina and > . ^ , n j /-r Georo-ia, j 4 s. 8 d., called Georgia currency. 1. Reduce 6iB. 11 s. 6;^d. to federal money. Note, To reduce pounds, shillings, pence and farthings, in either of the above-named currencies, to federal money, — First, reduce the shillings, pence and farthings (if any be contained in the given sum) to the deciinal of a pound by i»- spection^ as already taught, U 76. 6£, lis. 6^d. ~ £6W6. English money. — Now, supposing the above sum to be bJnglish money, — 1^ . is 20 s. =z 240 pence, in all the above currencies. 1 dollar, in English money, is reckoned 4 s. 6 d. =z 54 pence, that is, ^^/tj = /^y of 1 pound. Now, as many times as ;j^75-, the fraction which 1 dollar is of 1 pound, Eng- lish money, is contained in i26'576, so many dollars, it is evident, there must be ; that is, — To reduce English to federal money ^ — Divide the given sum by ^^^ the quotient will be federal money. je 6 W6 English money. ^ote. It will be 40 recollected, to di- vide by a fraction, 9) 263^040 we multiply by the "io^f federal money, ^n.s«.^. denominator, and ** •" divide the product by the numerator. Canada currency. — Supposing the above sum to be Cana- da currency, — 1 dollar, in this currency, is 5 s. m 60 pence, that is, -^^ = ;|- of 1 pound. Therefore, — To reduce Canor da currency to federal money ^ — Divide the given sum by ^, and the quotient will be federal money ; or, which is the same thing, — Multiply the given sum by 4. i£6'576 Canada currency. 4 $ 27'304 federal money, Answer. U T^8. REDUCTION OF CURRENCIES. 15^ New England currency. — 1 dollar, in this currency, is 6 s. r= 72 pence, that is, -^-^jj =. -^^^ or '3 of a pound. There- fore, — To reduce New England currency to federal money j — Di^ vide the given sum by *3. *3) £ . 6*576 New England currency. $21 '92 federal money. Answer. New York currency. — 1 dollar, in this currency, is 8 s. = 96 pence, that is, ^^^ =z -^, or '4 of a pound. Therefore, —To reduce New York currency to federal money ^ — Divide the given sum by '4. '4) £ . 6 '576 New York currency. $ 16'44 federal money. Answer. Pennsylvania currency. — 1 dollar, in this currency, is 75. 6 d. z= 90 pence, that is, ^\% i= f of a pound. Therefore, — To reduce Pennsylvania currency to federal money, — Divide by §, that is, multiply the given sum by 8, and divide the pro- duct by 3. £ . 6 '576 Pennsylvania currency. 8 3)52'608 $ 17'536 federal money, Answer. Georgia currency. — 1 dollar, Georgia currency, is 4 »* 8 d. 1= 56 pence, that is, ^\^j z=z -^^ of a pound. Therefore, — To reduce Georgia currency to federal money, — Divide by -^j that is, multiply the given sum by 30, and divide the pro- duct by 7. £ . 6'576 Georgia currency. 30 7)197'280 $28'182f- federal money, Answer. From the foresjoing examples, we derive the following general Rule : — To reduce English money, and the currencies of Canada and the several States, to federal money, — First, re- duce the shillings, &c., if any in the given sum, to the deci- mal of a pound; this being done, divide the given sum by such fractional part as 1 dollar, in the given currency, Jg a fractional part of 1 pound. 166 INTEREST. IT 80, 81. Rates at which the following foreign coins are estimated at the Custom Houses of the United States, Livre of France, ---------- $ qg^ Franc do. $ 4Sf, Silver Rouble of Russia, -------- ^ '75. Florin or Guilder of the United Netherlands, - $ *40. Mark Banco of Hamburg, «------ ^ ^33^ Real of Plate of Spain, -------- ^ 40. Real of Vellon of do. $ <05. Milrea of Portugal, --------- $1*24. Tale of China, $1^48. Pagoda of India, ---------- ^ 1'84. Rupee of Bengal, ---------- ^ '50, 2. Reduce 8764 livres to federal money. 3. Reduce 10,000 francs to federal money. 4. Reduce 250,000 florins to federal money. 6. In $ 1000, how many francs ? IlffTERBST. T[ 81. In-terest is an allowance made by a debtor to a creditor for the use of money. It is computed at a certain number of dollars for the use of each hundred dollars, or so many pounds for each hundred pounds, &c. one year, and in the same proportion for a greater or less sum, or for a longer or shorter time. The number of dollars so paid for the u.^e of a hundred dollars, one year, is called the rate per cent, or per centum ; the words per cent, or per centum signifying by the hundred. The highest rate allowed by law in the New England States, is 6 per certt.* that is, 6 dollars for a 100 dollars, 6 cents for a 100 cents, 6 pounds for a 100, &c. ; in other words, Y§^ of the sum lent or due is paid for the use of it one year. This is called legal interest^ and will here be under- utood when no other rate is mentioned. ♦ In the State of New York; 7 per cent is the le^^al interest j in Enjlend th« twjiu iirteresi is 5 per cent. IT 8 1 . INTEREST. ] 57 Let us suppose the sum lent, or due^ to be $ 1. The 100th part of $ 1, or ^^-^ of a dollar, is 1 cent, and yf ^y of a dollar, the legal interest, is 6 cents, which, written as a de- cimal fraction, is expressed thus, ------ <06. So of any other rate per cent. 1 per cent., expressed as a common fraction, is y^; decimally, ----------- *oi. -j per cent, is a half of 1 per cent., that is, - - ^005. ^ per cent, is a fourth of 1 per cent., that is, - - '0025. J per cent, is 3 times J per cent., that is, - - - '0075. Note, The rate per cent, is a decimal carried to two placeSy that is, to hundredths ; all decimal expressions lower than hundredths are parts of 1 pei* cent. | per cent., for in- stance, is *625 of 1 per cent., that is, '00625. Ans. '025. Write 2J per cent, as a decimal fraction. 2 per cent, is '02, and ^ per cent, is '005. Ans. *U25. Write 4 per cent, as a decimal fraction. 4J per cent. 4|- per cent. 5 per cent. 7^ per cent. 8 per cent. 8| per cent. 9 per cent. 9 J per cent. 10 per cent. (10 per cent. is -x^; decimally, '10.) 10^ per cent. 11 per cent. 12^ per cent. 15 per cent. 1. If the interest on $ 1, for 1 year, be 6 cents, what will be the interest on $ 17 for the same time ? It will be 17 times 6 cents, or 6 times 17, which is the same thing : — $17 '06 1'02 Answer ; that is, 1 dollar and 2 cents. To fiad the interest on any sum for 1 year, it is evidei|t we need only to multiply it by the rate per cent, written as a decimal fraction. The product, observing to place the point as directed in multiplication of decimal fractions, will be the interest required. NotfiiA Principal is the money diiCj for which interest i« paid. Amount is the principal and interest added together. O 158 INTEREST. 1181,82 2. What will be the interest of $ 3245, 1 year, at 4^ per cent. ? $'32'16 principal rj.^^^^ ^^^^^ five de- __^045 rate per cent. ^-^^^^ pl^^^g jn the mul- jg()Y5 tiplicand and multiplier, 22860 ^^^'^ iigiires must be pointed olf for decimals Afis. $ 1 '446 75 from the product, which gives the answer, — 1 dollar, 44 cents, G mills, and -^^^^ of a mill. Parts of a mill are not generally regarded; hence, $ 1^446 is sufficiently exact for the answer. ,; 3. What will be the interest of $ 11^04 for 1 year, at 3 per cent. ? . at 5^ per cent. ? at 6 per cent. ? at 7} per cent. ? at 8^- per cent. ? at 9£ per cent, ? at 10 per cent. ? at 10^ per cent. ? at 11 percent.? at llf per cent.? at 12 per cent.: at 12J per cent. ? f 4. A tax on a certain town is $ 1627^18, on which the collector is to receive 2^ per cent, for collecting; what will he receive for collecting the whole tax at that rate ? Alls. $40^679, JVoie. In the same way are calculated commission, irv surance, buying and selling stocks, loss and gain, or any tfii])g else rated at so much per cent, without respect to time. { 6. What must a man, paying $ 0'374 on a dollar, pay on a debt of $ 132'25 ? " Ans. $ 49'593. 6. A merchant, having purchased goods to the amount ol $ 580, sold them so as to gain 12^ per cent., that is, 12^ cents on each 100 cents, and in the same proportion for a greater or less sum ; what was his whole gain, and what was the whole amount for which he sold the goods ? Ans. His whole gain was $ 72^50 ; whole amount ^, 652^50. 7. A merchant bought a quantity of goods for $ 763*37^ ; liow much must he sell them for to gain 15 per cent. .? Ans. $ 877^881. ir 82. Commission is an allowance of so much per cent- to. a person called a correspondenty factor, or broker, for as- Bksting merchants and others in purchasing and selling goods. IT 82. INTERESl. 159 ^ 8. My correspondent sends me word that he has pur- chased goods to the vahie of $ 1286, on my account; what will his commission come to at 2^ per cent. ? Ans. $32'15. ^9. What must I allow my correspondent for selling goods to the amount of $ 2317'46, at a commission of 3^ per cent. ? Ans. $75^317. Insurance is an exemption from hazard, obtained by tlie 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 insured 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. 10. What will be tha premium for insuring a ship and cargo from Boston to Amsterdam, valued at $ 37800, at 4^ per cent. ? c Ans. $ 170f. '11. What will be the annual premium for insurance on a house against loss from fire, valued at $ 3500, at f per cent. ? By removing the separatrix 2 figures towards the left, it is evident, the sum itself may be made to express the premium at 1 per cent., of which the given rate parts may be taken ; thus, 1 per cent, on $ 3500 is $ 35^00, and f of $ 35*00 is $ 26*25, Answer. / 12. What will be the premium for insurance on a ship and cargo valued at $25156*86, at ^ per cent. ? at f per cent. ? ^at f per cent. ? at f per cent ? at f per cent. ? Ans. At |- per cent, the premium is $ 157*23. Stock is a general name for the capital of any trading company or corporation, or of a fund established by govern- ment. The value of stock is variable. When 100 dollars of stock sells for 100 dollars in money ^ the stock is said to be at par, which is a Latin word signifying equal ; when for 7nore^ it is said to be aboMt par ; when for less^ it is said to be 6e- low par. ' 13. What is the value of $7564 of stock, at 112^ per cent. : that is, when 1 dollar of stock sells for 1 dollar 12^ 160 INTEREST. ^^ !I 82, 83 cents in tnoney^ which is 12^ per cent, above par, or 12^ per ceat. advance^ as it is sometimes called. Am, $ 8509*50. ^4. What is the value of $3700 of bank stock, at 95^ per cent., that is, 4J per cent, helow par ? Am. $ 35a3^50. 15. What is the value of $ 120 of stock, at 92 J per cent. ? at 86 J per cent. ? at 67f per cent. ? at 104^ per cent. ? at lOS^ per cent. ? at 115 per cent ? at 37^- per cent, advance ? / Loss AND Gain. 16. Bought a hogshead of molasses for $ 60 ; for how much must I sell it to gain 20 per cent. ? Am, $ 72. ^17. Bought broadcloth at $2'50per yard; but, it being damaged, I am willing to sell it so as to lose 12 per cent ; how much will it be per yard ? Am, $ 2*20. ' 18. Bought calico at 20 cents per yard ; how must I sell it to gain 5 per cent. ? 10 per cent. ? 15 per cent ? to lose 20 per cent. ? Aits, to the lasty 16 cents per yard. ^ 83. We have seen how interest is cast on any sum of money, when the time is one year ; but it is frequently ne- cessary to cast interest for months and days. Now, the interest on $ 1 for I year, at 6 per cent., being *06, is *01 cent for 2 months, *005 mills (or ^ a cent) for 1 month of 30 days, (for so we reckon a month in casting interest,) and *001 mill for every 6 days ; 6 being contained 5 times in 30. Hence, it is very easy to find by insj^ection, that is, to cast in the mind, the interest on 1 dollar, at 6 per cent for any given iime. The cents^ it is evident, will be equal to /ia(/*the greatest even number of the months ; the mills will be 5 for the odd month, if there be one, and 1 for every time 6 is contained in the given number of the days. Suppose the interest of $ 1, at 6 per cent, be required for 9 months, and 18 days. The greatest even number of the months is 8 half of which will be the cents, *04 ; the mills, reckoning 5 for the odd month, and 3 for the 18 (3 times 6 = IS\ days, will be *008, which, united with the cents, (*048,) give 4 cents 8 mills for the interest of $ 1 for 9 months and 18 days. IT 83. INTEREST. ^ 161 ^ 1. What will be the interest on $ 1 for 5 months 6 days ? — 6 months 12 days ? 7 months ? 8 months 24 days? 9 months 12 days ? 10 months? 11 months 6 days? 12 months 18 days? 15 months 6 days ? * 16 months ? -«^^ Odd days.' 2. What is the interest of $ 1 for 13 months 16 days ? The cents will be 6, and the mills 5, for the odd month, and 2 for 2 times 6 rz: 12 days, and there is a remainder of 4 days, the interest for which will be such part of 1 mill as 4 days is part of 6 days, that is, | = f of a mill. Ans. '06 7§. Si 3, What will be the interest of $ 1 for 1 month 8 days? 2 months 7 days? 3 months 15 days? 4 months 22 days ? — — 5 months 11 days ? 6 months 17 days? 7 months 3 days ? 8 months 11 days? 9 months 2 days ? 10 months 15 days ? (11 months 4 days ? 1 12 months 3 days ? Note, If there is no odd months and the number of days he less than 6 y so that there arc no 7nills^ it is evident, a cipher must be put in the place of mills ; thus, in the last example, — 12 months 3 days, — the cents will be '06, the mills 0, the 3 days ^ a mill. Ans. '060^. 4. What will be the interest of $ 1 for 2 months 1 da^ ? 4 months 2 days ? 6 months 3 days ? 8 months 4 days ? 10 months 5 days ? for 3 days ? for 1 day ? for 2 days ? for 4 days ? for 5 days ? 5. What is the interest of $5643 for S months 5 days ? The interest of $ 1, for the given time, is '040|- ; therefore, ^) and^) $56'13 principal. '040f interest of $ 1 for the given time, 2245*20 interest for 8 months. 2806 interest for 3 days. 1871 interest for 2 days. 2'29197, Ans. $2'291. '''■^■' 5 days = 3 days -j- 2 days. As the multiplicand is talcen Otice for every 6 days, for 3 days take h for 2 days take |, 163 INTfiRES'f. it 83. of the multiplicand. ^ -|- 4^ z= |. So also, if the odd days be 4 =: 2 days -{- 2 days, take ^- of the multiplicand tioice ; for 1 day, take -J. Note, If the sum on which interest is to be cast be less than $ 10, the interest, for any number of days less than 6, will be less than 1 cent ; consequently, in busmess, if the sum be less than $ 10, such days need not be regarded. From the illustrations now given, it is evident, — To find the interest of any sum in federal inonei/j at 6 per cent,, it is only necessary to multiply the principal by the interest of $ 1 for the given time, found as above directed, and written as a decimal fraction, rememhering to point otF as many pl-aces for decimals in the product as there are decimal places in both the factors counted together. EXABIPLES FOR PRACTICE. 6. What is the interest of $ 8749 for 1 year 3 months ? Ans, $6*533. » 7. Interest of $116,08 for 11 mo. 19 days? $6*751, 8 of $ 200 for 8 mo. 4 days ? $ 8432. ^ 9 of $ 0*85 for 19 mo. ? $ '08. 10 of §8'50 for 1 vear 9 mo. 12 days ? $'909. f il of $675 for 1 mo. 21 days r $ 5'737. i2 of $8673 for 10 days? $14*455. 13 of $0*73 for 10 mo.'? $ *036. 14./. of $ 93 for 3 days ? ) Note. The inte- ♦ 15 of $ 73*50 for 2 days ? ^ rest of $ 1 for 6 days 16 of $180*75 for 5 days? f being 1 mill, the dol- 17 of $15000 for 1 day? ) lars themselves ex- press the interest in mills for six days, of which we may take parts. Thus, 6 ) 15000 mills, 2*500, that is, $2^50, Arts, to the last. When the interest is required for a large number of years, it will be more convenient to fmd the interest for one year, and multiply it by the number of years ; after which find the interest for the months and days, if any, as usual. 18. What is the interest of $ 1000 for 120 years ? Ans. $7200 19. What is the interest of $520*04 for 30 years and 6 months? * Ans. $951*673. 1r 83, 84. INTEREST. i63 20. What is the interest on $400 for 10 years 3 months and 6 days ? Aiis, $ 246*40. 21. What is the interest of $220 for 5 years? for 12 years ? 50 years ? A^is. to last, $660. f 22. What is the amount of $ 86, at interest 7 years ? Alls. $12242. 23. What is the interest of 36 i3 . 9 s. 6|- d. for 1 year ? Reduce the shillings, pence, &c. to the decimal of a pound, by inspection, (*[T76;) then proceed in all respects as in federal money. Having found the interest, reverse the ope- ration, and reduce the three first decimals to shillings, &c., by inspection. (IF 77. ) Ans. 2 dC . 3 s. 9 d. I 24. Interest of 36 i2 . 10 s. for lS*mo. 20 days ? Ans, 3 £ . 8 s. U d. Interest of 95 <£ . for 9 mo. ? Aiis, 4 iE . 5 s. 6 d.- 25. What is the amount of 18iS. 12 s. at interest 10 months 3 days ? Ans. Id £ . 10 s. 9j d. ,' 26. What is the amount of 100 £ . for 8 years ? Ans. 148 ^. 27. What is the amount of 400 £. 10 s. for 18 months? A71S. 4^6 £. 10 s. 10 d. 3 q* 28. What is the amount of 640 i3 . 8 s. at interest for 1 year ? for 2 years 6 months ? for 10 years ? Ans. to last, 1024 i2. 12 s. 9^ d. Tf 84. 1. What is the interest of 36 dollars for 8 months, at 4^ per cent. ? Note. When the rate is any other than six per cent., first find the interest at 6 per cent., tlten divide the interest so found by such part as the interest, at the rate required, ex- ceeds or falls short of the interest at 6 per cent., and the quotient added to, or subtracted from the interest at 6 per cent., as the case may be, will give the interest at the rate required. 4^ per cent, is |- of 6 per cent. ; therefore, from the interest at 6 per cent, subtract j^ ; the remainder will be the interest at 4^ per cent. VQ8 Ans. 2. Interest of $64<81 for 18 mo., at 5 per ct.? A7is. $441. 3 of $500 for 9 mo. 9 days, at 8 per ct.? $31*00 4 of ^6242 fori mo. 20 days, at 4 per ct.? $*345 164 INTEREST. IT 84 j 85 6, Interest of $ 85 for 10 mo. 15 days, at 12 J percent. ? Ans, $9'295, I 6. What is the amount of $ 53 at 10 per ct. for 7 mo. ? Ans. $56^091 The timey rate per cent, and amount given^ to find the principaL •57fr 85. 1. What sum of money, pT\t at interest at 6 per cent, will amount to $61^02, in 1 yedi 4 months? The amount of $ 1, at the given rate and time, is $ 1*08 ; hence, $61^02 -- $ I'OS = 56'50, the principal required; that is, — Find the amount of $1 at the given rate and tbne^ by which divide the given amount j the quotient will he the priKci- pal required. ' Ans. $ 56*50. 2. What principal, ot 8 per cent., in 1 year 6 months, will amount to $ 8542 ? Ans. $ 76. 3. What principal, at 6 per cent., in 11 months 9 vlays, will amount to $ 99*311 ? Note. The interest of $ 1, for the given time, is *056^ ; but, in these cases, when there are odd days, instead of writing the parts of a mill as a common fraction, it will be more convenient to write them as a decimal, thus, *0565 ; that is, extend the decimal to four places. Ans. $ 94. 4. A factor receives $ 988 to lay out after deducting his commission of 4 per cent. ; how much will remain to be laid out ? It is evident, he ought not to receive commission on his oum money. This question, therefore, in principle, does not differ from the preceding. "* Note. In questions like this, where no respect is had to time, (1\ 81, ex. 4, note,) add the rate to $ 1. Ajis. $ 950. * 5. A factor receives $ 1008 to lay out after deducting his commission of 5 per cent. ; w^hat does his commission amount to ? A7is. $ 48. Discount. 6. Suppose I owe a man j$ 397*50, to be paid in 1 year, without interest, and I wish to pay him now ; how much ought I to pay him w^hen the usual rate is 6 per cent ? I ought to pay him such a sum as, put at interest, would, in 1 year, amount to $ 397*50. The question, therefore, does not differ from the preceding. Ans. $ 376. Note. An allowance made for the payment of any sum V 85, 86. INTEREST. 166 of money before it becomes due, as in the last example, i« called Discount. The sum which, put at interest, would, in the time and at the rate per cent, for which discount is to be made, amouiit to the given sura, or debt, is called the jnesent worth. 7. What is the present worth of $, S34, payable in 1 year 7 months and 6 d-ays, discounting at the rate of 7 per cent, r Ans. §750. 8. What is the discount on $321^63, due 4 years hence, discountiog at the rate of 6 per cent. ? Ans. $ 62^26. 9. How much ready money must be paid for a note of $ 18, due 15 months hence, discounting at tbe rate of 6 per cent. ? Ans. $ 16'744. 10. Sold goods for $ 650, payable one half in 4 months, and the other half in 8 months ; what must be discounted for present payment ? Ans. $lS'";fX A 11. What is the present worth of § 56^20, jiay able in 1 year 8 months, discounting at 6 per cent. ? at 4J per cent. ? at 5 per cent. ? at 7 per cent. ? at 7^ per cent ? at 9 per cent. ? Am. to the last, $ 48*869. The time^ rate per cent.^ and interest being g'wen^ to find th^ principal. •f 86, 1. What sum of money, put at interest IG months, will gain § 10*50, at 6 per cent. ? $1, at the given rate and lime, will gain *08; hence, $10*50-7- $*08= $131*25, the principal rec^jired ; that is, — Find the interest of $ 1, al the given rate and twie^ by which divide the given gain, or interest ; the quotient will be the principal required. Ans. $ 131*25. 2. A man paid $ 4*52 interest, at the rate of 6 per cent, ttt the end of 1 year 4 months ; what was the principal ? Ans. $56*50. 3. A man received, for interest on a certain note, at the end of 1 year, $ 20 ; what was the principal, allowing the rate to hiave been 6 per cent, ? -4??^. $ 333*333|. 166 , INTEREST. IT 87, 88* The principal^ interest^ and time being given^ to find the rate per cent. ir 87. 1. If I pay $3'78 interest, for the use of $36 for 1 year and 6 months, ^vhat is that per cent. ? The interest on $ 36, dXoneper cent, the given time, is $ *o4 ; hence, $ 3'78 -f- $ '54 = '07, the rate required ; that is, — Find the interest on the given sum, at 1 per cent. Jar the given time, by which divide the given interest ; the quotient mill be the rate at which interest was paid, Ans, 7 per cent. ' f 2. If I pay $ 2'34 for the use of $ 468, 1 month, what is 4he rate per cent. ? Ans, 6 per cent 3. At $46'80 for the use of $520, 2 years, what is that per cent. ? Ans, 4^ per cent The prices at which goods are bought and sold being given^ to find the rate per cent, of gain or loss. ^ 88. i. If I purchase wheat at $ I'lO per bushel, and sell it at $ 1'37^ per bushel, w]^at do I gain per cent. ? This question does not differ essentially from those in the foregoing paragraph. Subtracting the cost from the price at sale, it is evident I gain 27^ cents on a bushel, that is, ^ of the first cost '^ = '25 per cent., the Answer, That is, — Make a common fraction^ writing the gain or loss for the nnmcra- tor J and the price at which the article was bought for the de- nominator ; then reduce it to a decimal, 2. A merchant purchases goods to the amount of $ 550 ; what per cent, profit must he make to gain $ 66 ? Ans, 12 per cent 3. What per cent, profit must he make on the same purchase to gain $ 38*50 ? to gain $ 24'75 ? to gain $2'75? Note, The last gain gives for a quotient '005, which is ^ per cent. The rate per cent, it must be recollected, (If 81, note,) is a decimal carried to two places, or hundredths ; all decimal expressions lower than hundredths are parts of 1 per cent " \4. Bought a hogshead of rum, containing 114 gallons, at 96 cents per gallon, and sold it again at $ 1'0032 per gal- lon ; what was the whole gain, and v/hat was the gain per cent? 4 J $4'924, whole gain. ) 44- gain per cent. IT 88, 89. INTEREST. 167 5. A merchant bought a quantity of tea for $ 365, which, proving to have been damaged, he sold for $33245; what did he lose per cent. ? Ans. 9 per cent. 6. If I buy cloth at $2 per yard, and sell it for $2'50 per yard, what should I gain in laying out $ 100 ? Am, $25. 7. Bought indigo at $ 1^20 per lb., and sold the same at 90 cents per lb. ; what was lost per cent. ? Ans, 25 per cent. . 8. Bought 30 hogsheads of molasses, at $ 600 ; paid in duties $ 20-66 ; for freight, $ 40'78 ; for porterage, $ 6^05, and for insurance, $ 30*84 : if I sell them at $ 26 per hogs- head, how much shall 1 gain per cent. ? ^725.11^695per cent. The principal, rate per cent.^ and interest being given, to find the time, ^ 89. 1. The interest on a note of $ 36, at 7 per cent., v/as $ 3'78 ; what was the time ? The interest on $ 36, 1 year, at 7 per cent, is $ 2'52 ; hence, $ 3'78 -7- $ 2'52 = 1^5 years, the time required ; that is, — Find the interest for 1 year on the principal given, at the given rate, by which divide the given interest ; the quotient will be the time required, in years and decimal parts of a year ; the latter may then be reduced to months and days. Ans, 1 year 6 months, (2, If $31^71 interest be paid on a note of $226'50, what was the time, the rate being 6 per cent. ? Ans, 2 '33-^ := 2 years 4 months, 3. On a note of $ 600, paid interest $ 20, at 8 per cent. ; what was the time ? Ans. *416 -]- z=. 5 months so nearly as to be called 5, and would be exactly 5 but for the fraction lost. ' 4.- The interest on a note of $ 217^25, at 4 per cent, was $ 28'242 ; what was the time ? Ans, 3 years 3 months. Note, When the rate is 6 per cent., we may divide the interest by ^ the principal, removing the separatrix two places to the left, and tlie quotient will be tlie answer in months* 168 INTEREST. IT 90. To find the interest due an notes^ S^c. when partial payments have been iuade. ^ 90- In Massachusetts the law provides, that payments shall he applied to keep down the interest, and that neither interest nor payment shall ever draw interest. Hence, if the payment at any time exceed the interest computed to the same time, that excess is taken from the principal ; but if the payment be less than the interest, llie principal remains unaltered. Wherefore, we have this Rule : — Compute the interest to the first time when a payment w^as made, which, either alone, or together with the preceding payments, if any, exceeds the interest t/ien due; add that interest to the principal, and from the sum subtract the payment, or the $um of the payments, made within the time for which the interest was computed, and the remainder will be a new principal, with which proceed as with the first. 1. For value receioedj I promise to pay James Conant, or order^ one hundred sixteen dollars sixty-six cents and six mills^ with interest. May 1, 1822. g 116,686. Samuel Rood. On this note were the folio win cr endorsements : Dec. 25, 1822, received $ l&^QQQ ^ July 10, 1823, $ 1-666 Sept. 1, 1824, $ 5'000 June 14, 1825, ^33*333 April 15, 1826, $62^000 What was due August 3, 1827 r Ans. $ 23^775, Note, In finding the times for computing the interest, consult IT 40. The first principal on interest from May 1, 1822, $ 116*666 Interest to Dec. 25, 1822, time of the first pay- ment, (7 m.onths .24 days,) - - - 4*549 Amount, $ 121*215 Payment, Dec. 25, exceeding interest then due, 16*666 Remainder for a new principal, - - - 104*549 Interest from Dec. 25, 1S22, to June 14, 1825, (29 months 19 days,) - . - . 15*490 Amount carried^ forward, $ 120*039 IT 95, 91. COMPOUND INTEREST* 169 Amount broujrht fonrard, $120'039 Payment, July 19, 1823, Icj-s than interest thendwe, - - - - $ 1*666 Payment, Sept. 1, 1S24, less than interest then due, ... - 5*000 Payment, June 14, 1825, exceeding in- terest then due, - . - 33*333 $39^09 Remainder for a new principal, (June 14, 1825,) 80^040 Interest from June 14, 1825, to April 15, 1823, (10 months 1 day,) . * . . 4^015 Amount, $ fa4'055 Pa}Tnent, April 15, 1825, exceeding intereht then due, 62*000 Ren^ainder for a new principal, (April 15, 1826,) $ 22'055 Interest due Aug. 3, 1827, from April 15, 1826, (15 months 18 days,) . « - . 1<720 Balance due Aug 3, 1827, - - $ 23^775 2. For value received^ T promise to pay Jameq Lowell, qt ctdeTy eight hundred sixty-set eii dollars aiid thirty-three cents, with interest. Jan, 6. 1820, $867*33. Hiram Simson;* On this note were the follmving endorsements, vU» April 16, 1823, received $136*44. April 16, 1825, received $319. Jan. 1, 1826, received $518*68. What remained due July 11, 1827 ? Am* $215*li),i COMPOUND INTEREST. IT n» A promises to pay B $256 in 3 yean, ^^th in- terest annually; but at the end of 1 y^not fiacing it con- veaient to pay the interest, he consents to pny interest on the interest from that time, the same as on ths prin :ipAl. Note, Simple interest is that which is allowed for the principal only ; compoimd interest is that wlvich is allowed 170 COMPOUND INTEREST. IT 91, for both principal and interest^ when the latter is not paid at the time it becomes due. Compound interest is calculated by adding the interest to the principal at the end of each year, and making the cunoimt the principal for the next succeeding year. 1. What is the compound interest of $256 for 3 years, at 6 per cent. ? $ 256 given sum, or first principal. '06 15 256 •'36 interest, > . i u j ^ ^^v I'OO principal, 5 *<'''^^'''^*^"^*^g^*« 271'36 amount, or principal for 2d year. '06 10'2816 compound interest, 2d year, > addCvl to- 271 '36 principal, do. ) g<-'ther. 287'6416 amount, or principal for 3d year. '06 17'25846 compound interest, 3d year, ) added to- 28T'641 principal, do. ] gether. 304'899 amount. 256 first principal subtracted. Ans, $4S'899 compound interest for 3 years. / 2. At 6 per cent., what will be the compound interest, and wiiat the amount, of $ 1 for 2 year? ? what the amount for 3 years ? for 4 years ? for 5 years ? for 6 years ? for 7 years ? for S years ? Ans. to the last, $ 1'593-f-. It is plain that the amount of $ 2, for any given time, will be 2 times as much as the amount of $ 1 ; the amount of $3 will be 3 times as much, &c. Hence, we may form the amounts of $ 1, for several years, into a table of multipliers for finding the amount of any sum tor the same time. fr 91. COMPOUND INTEREST. 171 TABLE, Showing the amount of $ 1, or l£., &c. for any years, not exceeding 24j at the rates of 5 and compound interest. Tears. 1 2 3 4 6 6 7 8 9 10 11 12 5 per cent. 1'05 14025 1457G2 + 1'21550 4- 1*27628 + 1*34009 + 1*40710 + 1*47745 + 1*55132 + 1*62889 + 1*71033 + 1*79585 + 6 per cent. 1*06 1*1236 1*19101 + 1*26247 + 1*33822 + 1*41851 + 1*50363 + 1*59384 + 1*68947 + 1^79084 + 1*89829 + 2*01219 + Years. 13 14 15 16 17 18 19 20 21 22 23 24 5 per cent. 1*88564 + 1*97993 + 2*07892 + 2*18287 + 2*29201 + 2*40661 + 2*52695 2*65329 + 2*78596 + 2*92526 + 3*07152 + 3*22509 + number ot 6 per cent 6 per cent. 2*13292 + 2*26090 + 2*39655 + 2*54035 + 2*69277 + 2*85433 + 3*02559 + 3*20713 + 3*39956 + 3*60353 + 3^81974 + 4*04893 + Note 1. Four decimals in the above numbers will be suf- ficiently accurate for most operations. Note 2. When there are months and days, you may first find the amount for the years^ and on that amount cast the interest for the months and days ; this, added to the amount^ will give the answer. 3. What is the amount of $600*50 for 20 years, at 5 per cent, compound interest .'' at 6 per cent. ? $ 1 at 5 per cent., by the table, is $ 2*65329 ; therefore, 2*65329 X 600*50 =z $ 1593*30 + Arts, at 5 per cent. ; and 3*20713 X 600*50 z= $ 1925*881 + Aiis. at 6 per cent. |4. What is the amount of $ 40*20 at 6 per cent, com- pound interest, for 4 years .^ for 10 years ? for 18 years ? for 12 years ? for 3 years and 4 months ^ for 24 years, 6 months, and 18 days ? Am. tolasi^ $ 168*137. Note. Any sum at compound interest will double itself in 11 years, 10 months, and 22 days. From what has now been advanced we deduce the fol- lowing general RULE. I. To find the interest when the time is 1 year, or, to find /A# rate per cent, on any sum of mwiey^ without respect to timCy a» \U2, COMPOUND INTEREST. ^91. the premium for insurance^ commission^ &c., — Multiply tLe principal, or given sum, by the rate per ceut., written as a decinial fraction ; the product, remembering to point otf a?- many places for decimals as there are decimals in both the factors, will be the interest, &c. required. II. IVlten there are months and (lays in the given iime^ to find th2. I 8. What is the value of $ 800 United States Bank stock, at 112^ per cent. ? Am, $ 900. 9. What is the value of $ 560^75 of stock, at 93 per cent. ? Ans, $521 '497 10. What principal at 7 per cent, will, in 9 months 18 days, amount to $ 422'40 ? Ans, $ 400^ 11. What is the present worth of $ 426, payable in 4 years and 12 days, discounting ai the rate of 5 per cent. ? In large sums, to bring out the cents correctly, it w*ill sometimes be necessary to extend the decimal in the divisof to ^ve places. Aiis, $ 354'o06. 12. A merchant purchased goods for $250 ready money, and sold them again ior $300, payable in 9 months; what did he gain, discounting at 6 per cent. ? A?is. $37^081. f 13. Sold goods for $3120, to be paid, one half in 3 months, and the other half in 6 months ; what must be dis- counted for present payment ? Ans, 68'492. 14. The interest on a certain note, for 1 year 9 months, was $ 49^875 ; what was the principal ? Ans. $ 475. 15. What principal, at 5 per cent., in 16 months 24 days, will gain $ 35 ? Ans. $ 500. 16. If I pay $15*^50 interest for the use of $500,9 months and 9 days, what is the rate p'3r cent. ? f 17. If I buy candles at $467 per lb., and sell tliem at 20 cents, what shall I gain in laying out $ 100 ? ^71*. $19<76. 18. Bought hats at 4 s. apiece, and sold them again at 4 8i 9 d. ; what is the profit in laying out 100 ^ . ? Ans, 18 ig. 15 s* 19. Bought .37 gallons of brandy, at $140 per gallon, and sold it for $ 40 ; what was gained or lost per cent. ? 20. At 4 s. 6 d. profit on 1 £,, how much is gained in laying out 100 £ ., that is, bow much per cent. ? Ans, 22 iS . 10 s* • 21. Bought cloth at $4'48 per yard ; how must I sell il to gain 12;J per cent. ? ^ Ans. $ 5*04 176 EQUATION OF PAYMENTS. IT 91, 92 22. Bought a barrel of powder for 4 iC . ; for how much must it be sold to lose 10 per cent. ? Ans, 3 iS . 12 9. 23. Bought cloth at 15 s. per yard, which not proving so good as I expected, I am content to lose 47^ per cent. ; hoMr must I sell it per yard ? Ans, 12 s. 4Jd { 24. Bought 50 gallons of brandy, at 92 cents per gallon, fcut by accident 10 gallons leaked out ; at what rate must I sell the remainder per gallon to gain upon the whole cost al the rate of 10 per cent. ? Am, $ 1*265 per gallon. 25. A merchant bought 10 tons of iron for $ 950 ; the freight and duties came to $ 145, and his own charges to $ 25 ; how must he sell it per lb. to gain 20 per cent, by it f Ans. 6 cents per lb. EQUATXOZf OF PAimXENTS. IT 92. Equation of payments is the method of finding the mean time for the payment of several debts, due at different times. 1. In how many months will $ 1 gain as much as 5 dol- lars will gain in 6 months ? 2. In how many months will $ 1 gain as much as $ 40 will gain in 15 months ? Ans. 600. 3. In how many months will the use of $ 5 be worth as much as the use of $ 1 for 40 months ? 4. Bc-rrowed of a friend $ 1 for 20 months ; afterwards lent my friend $ 4 ; how long ought he to keep it to becomij indemnified for the use of the $1? 5. I have three notes against a man ; one of $ 12, due in 3 months ; one of $ 9, due in 5 months ,• and the other of $ 6, due in 10 months ; the man wishes to pay the whole at once; in what time ought he to pay it ? $ 12 for 3 months is the same as $ 1 for 36 months, and $ 9 for 5 months is the same as $ 1 for 45 months, and $ 6 for 10 months is the same as $ 1 for 60 months. 27 141 He might, therefore, have $1 141 months, and he may keep 27 dollars ^y part as long ; that is, ^^ = 5 monllii 6 -}- days, Answer, IT 93. ratio; or the relation of numbers. 177 Hence, To find the mean time for several payments^ — Hule; — Multiply each sum by its time of payment, and divide the sum of the products by the sum of the paijinents^ and tlie quotient will be the answer. Note, This rule is founded on the supposition, that what 13 gained by keeping a debt a certain time after it is due, is the same as what is lost by paying it an equal time before it is due ; but, in the first case, the gain is evidently equal to the interest on the debt for the given time, while, in the second case, the loss is only equal to the discount of the debt for that time, which is always less than the interest; therefore, the rule is not exactly true. The error, however, is so trifling, in most questions that occur in business, as scarce to merit notice. 6. A merchant has owing him $300, to be paid as fol- lows : $50 in 2 months, $ 100 in 5 months, and the rest in 8 months ; and it is agreed to make one paymeiit of the whole : in what time ought that payment to be ? Ans, 6 months. 7. A owes B $ 136, to be paid in 10 months ; $ 96, to be paid in 7 months; and $260, to be paid in 4 months: what IS the equated time for the payment of the whole ? Ans. 6 months, 7 days -{-. 8. A owes C $600, of whicli $ 200 is to be paid at the present time, 200 in 4 months, and 200 in 8 months ; what IS the equated time for the payment of the v/hole ? Ans. 4 months- 9. A. owes B $ 300, to be paid as follows : ^ in 3 months, ^ in 4 months, and the rest in 6 months : what is tlie equated time ? Ans, 4^ months. X'v ^^'^^^'^^ /^ OF THK \\ RATIO J i^^^^^^^^ OR ^X^r/.,.. TT 93. 1. What part of 1 gallon is 3 quarts? 1 gallon is 4 quart«, and 3 quarts is J of 4 quarts. An^, | of a gallon- 2. What part of 3 quarts is 1 gallon ? 1 gallon, being 4 quarts, is ^ of 3 quarts ; that is, 4 quarts is 1 time 3 (piarta and ^ of another time. Ans. J zz= 1 ^ nS RATIO ; OR THE RELATION OF NUMBERS. M 9^^ 3. What part of 5 bushels is 12 bushels ? Fii.ding what part one number is of another is the same as finding what is called the ratio^ or relation of one number to another; thus, the question, What part of 5 bushels is 12 bushels ? is the same as What is the ratio of 5 bushels to 12 bushels ? The Ansxoer is J^ zz= 2f . Ratioj therefore, may be defined^the number of times one number is contained in another; or, the number of times one quantity is contained in another quantity of the same kind. 4. What part of 8 yards is 13 yards ? or, What is the ratio of 8 yards to 13 yards? 13 yards is ^^- of 8 yards, expressing the ^\\\doi\ fractimall^ If now we perf:)rm the division, w^e have for the ratio \^ ; that is, 13 yards is 1 time 8 yards, and | of another time. We have seen, (IT Ib^sicjn^) that division maybe expressed fractionally. So also the ratio of one number to another, or the part one number is of another, may be expressed frao- tionfll}, to do which, make the number which is called the partj w^hether it be the larger or the smaller number, the nu- merator of a fraction, under which write the other number for a denominator. When the question is. What is the ratio, &c. } the number last named is the part ; consequently it must be made the numerator of the fraction, and the number first named the denominator. 5. What part of 12 dollars is 11 dollars ? or, 11 dollars is what part of 12 dollars ? 1 1 is the number which expresses i\iQ part. To put this question in the other form, viz. AVhat is the ratio^ &.c. r let that number, which expresses the part^ be the number last named ; thus, What is the ratio of 12 dol- lars to 1 1 dollars ? Ans. \^, 6. What part of 1 jg . is 2 s. G d. ? or,^Tiat is the ratio of 1 ie . to 2 s. 6 d. ? \ £, z=z 240 pence, and 2 s. 6 d. z=. 30 pence ; hence, ^^ =z ^, is the Answer, 7. What part of 13 s. 6 d. is 1 iB . 10 s. ? or, What is the ra- tio of 13s. 6d. to 1 £, 10s.? Ans. ^<^ 8. What is the ratio of 3 to 5 ? of 5 to 3 ? of 7 to 19 ? of 19 to 7? of 15 to 90 ? of 90 to 15 ? of 84 to 160 ? of 160 to 84 ? of 615 to 1107 ? of 1107 to 615 ? Ans, to the last, f \ V 94. KULE OF THREE. 179 PROPORTZON; OR * THE RUZS OF THKSE. IT 94. 1. If a piece of cloth, 4 yards long, cost 12 dollars, what will be the cost of a piece of the same cloth 7 yards long ? Had this piece contained twice the number of yards of the first piece, it is evident the price would have been twice as much ; had it contained 3 times the number of yards, the price would have been 3 times as much ; or had it contained only half the number of yards, the price would have been only half as much ; that is, the cost of 7 yards wnl be such part of 12 dollars as 7 yards is part of 4 yards. 7 yards is J of 4 yards ; consequently, the price of 7 yards must be J of Qie price of 4 yards, or J of 12 dollars. {- of 12 dollars, that is, 12 X J = -^j^- =r 21 dollars. Answer. 2. If a horse travel 30 miles in 6 hours, how many miles will he travel in 11 hours, at that rate ? 11 hours is -y- of 6 hours, that is, 11 hours is 1 time 6 hours, and f of another time ; consequently, he will travel, in 11 hours, 1 time 30 miles, and f of another time, that is, the ratio between the distances will be equal to the ratio be- tween the times. -y. of 30 miles, that is, 30 X V" ~ H^ = ^^ miles. If, then, no error has been committed, 55 miles must be -y- of 30 miles. This is actually the case ; for -|3 = -\K >V Ans, 55 miles. Quantities which have the same ratio between them are said to be praportional. Thus, these four quantities, hours, hours, miles, miles. 6, 11, 30, 55, written in this order, being such, that the second contains the first as many times as the fourth contains the third, that is, the ratio between the third and fourth being equal to the r^tio between the first and second, form what is called a pro- portion. ^ It follows, therefore, that proportion is a combinatian of two i^ual ratios. Ralin exists between two numbers ; but prO' porhioii requires at least three. tSO nULE OF THREE. IT 94, 95. To denote that tliere is a proportion between the numbers 6, 11, 30, and 55, they are written thus : — G ! 11 : : 30 : 55 which is read, 6 is to 1 1 as 30 is to 55 ; that is, 6 is the same* part of 11, thai 30 is of 55; or, 6 is contained in 11 as many times as 30 is contained in 55 ; or, lastly, the ratio or relation of 11 to 6 is the same as that of 55 to 30. ^ 95- Tlie first term of a ratio, or relation, is called the men per- form a piece of wor^ in 7 daj'«. how long will 3 men be in pedrnmug 'he fi».rno work ? Her« Us- requires viore. . i'ov the nuniher tf men, \yQiug less, wjJi re. 8 d. ' 20. If 1333 <£ . 6 s. 8 d. Massachusetts, be equal to 1000£ . sterling, how much sterling is equal to 4 «£. Massachusetts? Ans, 3 £ , 21. If 1000 £ . sterling be equal to 1333 i3 . 6 s. 8 d. Mas- sachusetts, how much Massachusetts is equal to 3 iB. ster- ling ? Ans, 4 £ , 22. If 3 cO . sterling be equal to 4 £ . Massachusetts, how- much sterling is equal to 1333 £ . 6 s. 8 d. Massachusetts ? Ans, 1000 ie. * 23. Suppose 2000 soldiers had been supplied with bread sufficient to last: them 12 weeks, allowing each man 14 ounces a day; but, on examination, they find 105 barrels, containing 200 lbs. each, wholly spoiled ; what must the al- lowance be to each man, that the remainder may last them the same time ? Ans, 12 oz. a day. IT 95. RULE OF THREE. 185 ' 24. Suppose 2000 soldiers were put to an allowance of 12 oz. of bread per day for 12 weeks, having a seventh part of their bread spoiled ; what was the whole weight of their bread, good and bad, and how much was spoiled ? . J The whole weight, 147000 lbs. ^^^' ^ Spoiled, - - 21000 fe. t25. 2000 soldiers, having lost 105 barrels of bread, weighing 200 lbs. each, were obliged to subsist on 12 oz. a day for 12 weeks ; had none been lost, they might have had 14 oz. a day ; what was the whole weight, including what was lost, and how much had they to subsist on ? . ^ Whole weight, 147000 lbs. ^"^* I Left, to subsist on, 126000 lbs. ' 26. 2000 soldiers, after losing one seventh part of their bread, had each 12 oz. a day for 12 weeks; what was the v/hole weight of their bread, including that lost, and how much might they ha\e had per day, each man, if none had been lost ? C Whole weight, 147000 lbs. Ans. < Loss, - - 21000 lbs. 1 ( 14 oz. per day, had none been lost. ^ 27. There was a certain building raised in 8 months by 120 workmen; but, the same being demolished, it is required to be built in 2 months; I demand how many men must be employed about it. Ans, 480 men. 28. There is a cistern having a pipe which will empty it in 10 hours ; how many pipes of the same capacity will empty it in 24 minutes ? Ans. 25 pipes. 29. A garrison of 1200 men has provisions for 9 months, at the rate of 14 oz. per day ; how long will the provisions last, at the same allow^ance, if the garrison be reinforced by 400 men ? Ans. 6 J months. 30. If a piece of land, 40 rods in length and 4 in breadth, make an acre, how wide must it be when it is but 25 rods long ? Ans. 6f rods. 31. If a man perform a journey in 15 days when the days are 12 hours long, in how many will he do it when the days are but 10 hours long ? Ans. 18 days. 32. If a field will feed 6 cows 91 days, how long will it feed 21 cows? A7is. 2G days. 33. Lent a friend 292 dollars for 6 months ; some time after, he lent me 806 dollars j how long may I keep it to balance' the favour ? Aiis. 2 months 6 + days. 186 KULE OF THREE. t &5. 34. If 30 men can perform apiece of work in 11 days, how many men will accomplish another piece of work, 4 times as big, in a fifth part of the time ? Am, 600 men. 35. If -J^ lb. of sugar cost f ^ of a shilling, what will ^f of a lb. cost ? Am, 4 d. 3^g J-^ q. Note, See IT 65, ex. 1, where the above question is solved by analysis. The eleven following are the next suc- ceeding examples in the same IT. I ?6. If 7 lbs. of sugar cost f of a dollar, what cost 12 lbs. ? Am, $lf ^ 37. If 64 yds. of cloth cost $3, what cost 9^ yds. ? Am, $4*269. 's 38. If 2 oz. of silver cost $ 2*24, what costs f oz. ? Am. $0*84. 39. Iff oz. cost $|J-, what costs 1 oz. ? Am, $ 1*283. 40. If 4 lb. less by ^ lb. cost 13^ d., what cost 14 lbs. less by -J of 2 lbs. ? Ans. 4 £ . 9 s. 9^^ d. 41. If f yd. cost $ ^, what will 40^ yds. cost ? Am, $59*062. f 42* If /^ of a ship cost $ 251, what is ^^g- ^^ ^^^^ worth r ^ Am. $53*785. 43. At 3| £ , per cwt., what will 9f lbs. cost ? Am, 6 s. 3-5^^ d. 44. A merchant, OAvning ^ of a vessel, sold f of his share for $ 957 ; what was the vessel worth ? Aiis. $ 1794*375. 45. If I yd. cost f ^ ., what will y\ of an ell English cost ? A71S, 17 s. 1 d. 2f q. 46. A merchant bought a number of bales of velvet, each containing 129^-f y ., what must be paid for the freight of 50 tierces, each weighing 2^ cwt., 100 leagues ? Ans. 92 je. 11 s. lOf d. 8. If 56 lbs. of bread be sufficient for 7 men 14 days, how much bread will serve 21 men 3 days ? Ans. 30 lbs. The same by analysis. If 7 men consume 56 lbs. of bread, 1 man, in the same time, would consume 4- of 56 lbs. =: ^- lbs. ; and if he consume -^^ lbs. in 14 days, he would consume y^ o( Af- =: || lb. in 1 day. 21 men would con- sume 21 times so much as 1 man; that is, 21 times f|| 1= J-|-P- lbs. in 1 day, and in 3 days they would consume 3 times as much ; that is, ^ |-^ = 36 lbs., as before. Ans. 36 lbs. Note. Having wrought the following examples by the rule of proportion, let the pupil be required to do the same by analysis. 4. If 4 reapers receive $11*04 for 3 days' work, how many men may be hired 16 days for $ 103'04 ? Am. 7 men. 'tr 97. SUPPLEMENT TO THE SINGLE RULE OF THREE. 191 5. If 7 oz. 5 pwt. of bread be bought for 4| d. when com is 4 s. 2 d. per bushel, what weight of it may be bought for 1 s. 2 d. when the price per bushel is 5 s. 6 d. ? Ans, 1 lb. 4 oz. 3|Jf pwts- 6. If $100 gain $6 in 1 year, what will $400 gain in 9 months ? Note. This and the three following examples reciprocally prove each other. 7. If $ 100 gain $6 in 1 year, in what time will $400 gain $ 18 ? 8. If $ 400 gain $ 18 in 9 months, what is the rate per cent, per annum ? \9. What principal, at 6 percent, per. ann., will gain $ 18 in 9 months ? - "i 10. A usurer put out $75 at interest, and, at the end of 8 ^ months, received, for principal and interest, $ 79 ; I demand ftt what rate per cent, he received interest. ! ' / Ans. 8 per cent. 11. If 3 mei^^rficeive fe^^ jB . for 19 J- days' work, how much must 20 men receive for 100^ days' ? A71S. 305 je. Os. 8d, SXTPPIJGIiyCSNT TO THE SZnC^Z.£S HUXiB OP TBUXIS. QUESTIONS. 1. What is proportion? 2. How many numbers are re- quired to form a ratio ? 3. How many to form a proporticfn ? 4. What is the first term of a ratio called ? 5. the second term ? 6 . Which is taken for the numerator, and which for the denominator of the fraction expressing the ratio ? 7. How may it be known when four numbers are in proportion ? 8. Having three terms in a proportion given, how may the fourth term be found ? 9. What is the operation, by which the fourth term is found, called ? 10. How does a ratio be- come inverted ? 11. What is the rule in proportion? 12. In what denomination will the fourth term, or answer, be found? 13. If the first and second terms contain different denominations, what is to be done ? 14. What is cqmpound proportion, or double rule of three ? 15. Rule ? 193 FELLOWSHIP. IT 97, 98, EXERCISES. 1. If I buy 76 yds. of cloth for $11347, what does it cost per ell English ? Ans. $ 1^861, 2. Bought 4 ]iieces of Holland, each containing ^4 eils English, for $ 96 ; how much was that per yard ? Ans. $0*80. 3. A garrison had provision for 8 months, at the rate ol 15 ounces to each person per day ; how much must be al lowed per day, in order that the provision may last 9^ Months? Ans. 12ff oz.^ 4. How much land, at $ 2'50 per acre, must be given in exchange for 360 acres, at $ 3'75 per acre ? Am, 540 acres, 5. Borrowed 185 quarters of corn when the price was 19 s. ; how much must I pay when the price is 17 s. 4 d. ? Ans. 202f4-. 6. A person, owning -g of a coal mine, sells f of his share for 171^. ; what is the whole mine worth ? Ans. 380^2. 7. If |- of a gallon cost f of a dollar, what costs | of a tun? Ans. $140. 8. At 1^ iS . per cwt, what cost 3^ lbs. ? Ans. lOJ- d. 9. If 4-j- cwt. can be carried 36 miles for 35 shillings, how many pounds can be carried 20 miles for the same money ? Ans. 907^ Ibg. 10. If the sun appears to move from east to west 360 de- crees in 24 hours, hov/ much is that in each hour ? in each minute ? in each second ? A'/is. to last, 15" of a deg. 11. If a family of 9 persons spend $ 450 in 5 months, how much would be sufficient to maintain them 8 months if 5 persons more were added to the family? Ans. $ 1120. Note. Exercises 14th, 15th, 16th, 17th, 18th, 19th, and *?Oth, " Supplement to Fractions,^^ afford additional example* in single and double proportion, should more examples be thought necessary. FSZiKOWSlIIP. If 98. 1. Two men own a ticket; the first owns ^, and the second owns f of it ; the ticket draws a prize of 40 del- ars ; what is each man's share of the money ? K 98. FELLOWSHIP ISS 2» Two men purchase a ticket for 4 dollars, of which one pays 1 dollar, and the other 3 dollars ; the ticket draws 4^ rofits, because he managed the concerns ; B was to have but 2 per cent. ; what was each one's gain ? and how much did A receive for his trouble ? Ans. A's gain was $117'142f, and B's $46,857|, and A received $ 70'285f for his trouble. \ 12. A cotton factory, valued at $ 12000, is divided into 100 shares ; if the profits amount to 15 per cent, yearly, what will be the profit accruing to 1 share ? to 2 shares ? to 5 shares ? to 25 shares ? Ans. to the Just. $ 450. 13. In the above-mentioned factory, repairs are to be made which will cost $ 340 ; what will be the tax, on each share, necessary to raise the sum ? on 2 shares ? on 3 shares ? on 10 shares ? Ans. to the last^ $34. i 14. If a town raise a tax of $ 1850, and the whole town //be valued at $37000, what will that be on $1? What will be the tax of a man whose property is valued at $ 1780 ? Ans. $*05 on a dollar, and $89 on $1780. ir99- FELLOWSHIP. 195 IT M. In assessing taxes, it is necessary to have an In- tentory of the property, both real and personal, of the whole town, and also of the whole number of polls ; and, as the polls are rated at so much each, we must first take out from the whole tax what the polls amount to, and the remainder is to be assessed on the property. We may then find the tax upon 1 dollar, and make a table containing the taxes on 1,2, 3, &c., to 10 dollars ; then on 20, 30, &.C., to 100 dollars ; and then on 100, 200, &:c., to 1000 dollars. Then, knowing the inventory of any individual, it is easy to find the tax upon his property. 15. A certain town, valued at $64,530, raises a tax of $2259*90; there are 540 polls, which are taxed $*60 each ; what is the tax on a dollar, and what will be A's tax, xvhose real estate is valued at $ 1340, his personal property at $ 874, and who pays for 2 polls ? 540 X *60 z= $324, amount of the poll taxes, and $2259*90 — $324 = 1935*90, to be assessed on property. $64530 : $1935*90 : : §1 : *03; or, x|f|4^=:*03,taxon $1. TABLE. dolls. dollR. Tax on 1 is *03 dollfl. dolls. dolls. Tax on 10 is *30 Tax on 100 2 .. *06 20 .. *60 200 3 .. *09 30 .. *90 300 4.. *12 40 .. 1*20 400 5 .. *15 50 .. 1*50 500 6 .. *18 60 .. 1*80 600 7 .. *21 70 .. 2*10 700 8 .. *24 80 .. 2*40 800 9 .. *27 90 .. 2*70 900 1000 Now, to find A's tax, his real estate being $ 1340, by the table, that The tax on - - - $1000 - - is - - j The tax on - - - 300 . - - • - The tax on - - - 40 - - - - - Tax on his real estate ----- - -- In like manner I find the tax on his personal > property to be ....>....^ 2 polls at ^0 each, are .....•• doDs. is 3* .. 6* .. 9^ .. 12* .. 15* .. 18* .. 21* .. 24* .. 27* .. 30* I find, ^30* 9* 1*20 Amoyntj $67*62 la^ FELLOWSMP. IT 99, 100, Vl6. What will B's tax amount to, whose inventory is 874 dollars reaL and 210 dollars perscmal property, and who pays % 3 polls ? ' Am, $ 34'32l 17. What will be the tax of a man, paying for 1 poll, whose property is valued at $ 3482 ? at $ 768 ? - at $ 940 ? at $ 4657 ? Am, to the last^ $ 140^31. 18. Two men paid 10 dollars for the use of a pasture 1 month ; A kept in 24 cows, and B 16 cows ; how much should each pay ? 19. Two men hired a pasture for $ 10 ; A put in 8 cows 3 months, and B put in 4 cows 4 months ; how much should each pay ? IT 10(^. The pasturage of 8 cows for 3 months is the same as of 24 cows for 1 month, and the pasturage of 4 cows for 4 months is the same as of 16 cows for 1 month. The shares of A and B, therefore, are 24 to 16, as in the former question. Hence, when time is regarded in fellowship, — Multiply each one'^s stock by the time he continues it in tradey and use the product for his share. This is called Double Fel-* lotvship, Ans. A 6 dollars, and B 4 dollars. n20. a and B enter into partnership ; A puts in $ 100 months, and then puts in $ 50 more ; B puts in $ 200 4 months, and then takes out $ 80 ^ at the close of the year, tliey find that they have gained $ 95 ; what is the profit of each ? . A ^ $ 43'711, A's share. ^"^* I $51^288, B's share. Ik21. A, with a capital of $500, began trade Jan. 1, 1826, and, meeting with success, took in B as a partnery with a_. capital of $ 600, on the first of March following ; four months after, they admit C as a partner, who brought $ 800 stock ; at the close of the year, they find the gain to be $ 700 ; how must it be divided among the partners ? C $. 250, A's share. Am. < $ 250, B's share. ( $ 200, C's share. QUESTIONS, 1. What is fellowship ? 2. What is the rule for operat- ing? 3. When time is regarded in fellowship, what is it called? 4. What is the method of operating in double fellowship ? 6. How are taxes assessed .? 6, How is feilowship proved ? fl 101, 102. ALLIGATION. 197 IT 101. Alligation is the method of mixing two or more gimples, 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, 1. A farmer mixed together 4 bushels of wheat, worth 150 cents per bushel, 3 bushels of rye, worth 70 cents per bushel, and 2 bushels of corn, worth 50 cents per bushel; what is a bushel of the mixture worth ? It is plain, that the cost of the whole^ divided by the ntm^ her of bushels, will give the price of one bushel, 4 bushels, at 150 cents, cost 600 cents. 3 at 70 210 2 at 50 100 ^^ =1 lOli cts, Ans, 9 bushels cost 910 cents. 2. A grocer mixed 5 lbs. of sugar, worth 10 cents per lb., 8 lbs. worth 12 cents, 20 lbs. worth 14 cents ; what is a pound of the mixture worth ? Am, 12^ J. 3. A goldsmith melted together 3 ounces of gold 20 carats fine, and 5 ounces 22 carats fine ; what is the fine- ness of the mixture ? Ans, 21|^. 4. A grocer puts 6 gallons of water into a cask containing 40 gallons of rum, worth 42 cents per gallon ; what is a gal- lon of the mixture woith ? Ans, 36 J§ cents. 5. On a certain day the mercury was observed to stand in the thermometer as follows : 6 hours of the day, it stood at 64 degrees ; 4 hours, at 70 degrees ; 2 hours, at 75 degrees, and 3 hours, at 73 degrees: what was the mean temperature for that day ? It is plain this question does not differ, in the mode of its operation, from the former. Ans, 69-^^ degrees. IT 102. When the mean price or rate, and the prices or rates of the several simples are given, to find the proportion$ or quantities of each simple, the process is called ALrgatim Alternate : alligation alternate is, therefore, the reverse of alligatiou medial, and may be proved by it 19S Ai.i.i«ATio]sr. t 102v 1. A man has oats worth 40 cents per bushel, which he wishes to mix with corn worth 50 cents per bushel, so that the mixture may be worth 42 cents per bushel ; w^hat pro- portions, or quantities of each, must he take ? Had the price of the mixture required exceeded the price of the oats, by just as much as it fell short of the price of the corn, it is plain, he must have taken equal quanlilies of oats and corn ; had the prince of the mixture exceeded the price of the oats by only J as much as it fell short of the price of the corn, the compound would have required 2 times as much oats as corn ; and in all cases, the less the difference between the price of the mixture and that of one of the simples, the greater must be the quantity of that sim- phj in proportion to the other ; that is, the quantities of the simples must be wverseli/ as the differences of their prices from the price of tlie mixture ; therefore, if these differen- ces be mutually excha7iged, they will, directly^ express the relative quantities of each simple necessary to form the com- pound required. In the above example, the price of the mixture is 42 cents, and the price of the oats is 40 cents ; consequently, the difference of their prices is 2 cents : the price of the corn is 50 cents, which dificrs from the price of the mixture by 8 cents. Therefore, by exchanging these differencee, we have 8 bushels of oats to 2 bushels of corn^ for the proportion required. Alts. 8 bushels of oats to 2 bushels of corn^ or in that proportion. The correctness of this result may now be ascertained by the last rule ; thus, the cost of 8 bushels of oats, at 40 cents, is 320 cents; and 2 bushels of corn, at 50 cents, is 100 cents ; then, 320 -f- 100 z=: 420, and 420, divided by the num- ber of bushels, (8 -}- 2,) = 10, gives 42 cents for the price of the mixture. 2. A merchant has several kinds of tea ; some at 8 shil- lings, some at 9 shillings, some at 11 shillings, and some at 12 shillings per pound ; what proportions of each must lie mix, that he may sell the compound at 10 shillings per pound ? Here we have 4 simples; but it is plain, that what has just been proved of two will apply to any number of pairs, if in each pair the price of owe simple is greater, and that of the otlier less, than the price of the mixture required. Hence we have this T103, ALLIQATIOI?. 1^9 RULE. The mean rate and the several prices being reduced to the same denomination, — connect with a contimied line each price that is less than the mean rate witK one or more that is GREATER, and each price greater than the mean rate with one or more that is less. Write the di^'erence between the mean m/e, or price j and the price of each simple opposite the price with which it is connected; (thus the difference of the two prices in each pair will be mutually exchanged;) then the sum of the differ^ ences, standing against any price^ will express the relative QUANTITY to he taken of that price. By attentively considering the rule, the pupil will per- ceive, that there may be as many different ways of mixing the simples, and consequently as many different answers, as there are different ways of linking the several prices. We will now apply the rule to solve the last question : — • OPERATIONS. Ihs. -2^ Or, —2) (12 2 ~2) / &s. — - U2s Here we set down the prices of the simples, one directly linder another, in order, from least to greatest, as this is most convenient, and write the mean rate, (10 s.) at the left hand. In the first way of linking, we find, that we may take in the proportion of 2 pounds of the teas at 8 and 12 s. to 1 pound at 9 and 11 s. In the second way, we find for the answer, 3 pounds at 8 and 11 s. to 1 pound at 9 and 12 s. 3. What proportions of sugar, at 8 cents, 10 cents, and 14 cents per pound, will compose a mixture worth 12 cent* per pound ? ^715. In the proportion of 2 lbs. at 8 and 10 cents to 6 lbs. at 14 cents. Note, As these quantities only express the proportions of each kind, it is plain, that a compound of the same mean price will be formed by taking 3 times, 4 times, one half, or any proportion, of each quantity. Hence, When the quantity of one simple is given, after findicg 200 ALLIGATION. IT 102. the proportional quantities, by the above rule, we may say, As the VROPORTiofi XL quantity : is to the given quantity:: so is each of the other proportional quantities : to the re- quired quantities .0/ each. 4. If a man wishes to mix 1 gallon of brandy worth 16 s. with rum at 9 s. per gallon, so that the mixture may be worth 11 s. per gallon, how much rum must he use? Taking the differences as above, we find the proportions to be 2 of brandy to 5 of rum; consequently, 1 gallon of brandy will require 2^ gallons of rum. Ans, 2^ gallons. 5. A grocer has sugars worth 7 cents, 9 cents, and 12 cents per pound, which ha would mix so as to form a com- pound worth 10 cents per pound ; what must be the pro- portions of each kind ? Am, 2 lbs. of the first and second to 4 lbs. of the third kind, 6. If he use 1 lb. of the first kind, how much must he take of the others ? if 4 lbs., what ? if 6 lbs., what ^ if 10 lbs., what? if 20 lbs., what? Ans, to the last^ 20 lbs. of the second, and 40 of the third. 7. A merchant has spices at 16 d. 20 d. and 32 d. per pound ; he would mix 5 pounds of the first sort with the others, so as to form a compound worth 24 d. per pound ; how much of each sort must he use ? Arts, 5 lbs. of the second, and 7J lbs. of the third. 8. How many gallons of water, of no value, must be mixed with 60 gallons of rum, worth 80 cents per gallon, to reduce its value to 70 cents per gallon ? Am, S^ gallons. 9. A man would mix 4 bushels of wheat, at $ 1*50 per bushel, rye at $146, corn at $'75, and barley at $ '50, so as to sell the mixture at $ '84 per bushel ; how much of each may he use ? 10. A goldsmith would mix gold 17 carats fine with some 19, 21, and 24 carats fine, so that the compound may be 22 carats fine ; what proportions of each must he use ? Ans, 2 of the 3 first sorts to 9 of the last 11. If he use 1 oz. of the first kind, how much must he use of the others ? What would be the quantity of the compound ? Am, to last^ 7^ ounces. 12. If he would have the whole compound consist of 16 oz., how much must he use of each kind ? if of 30 oz., how much of each kind ? if of 37^ oz., how much ? Ans, to the last^ 5 oz. of the 3 first, and 22 j^ oz. of the last If 102, 103. DUODECIMALS. 201 Hence, when the quantity of the compound is given, wis may say. As the sum of the proportional quarttities, found by the above rule, is to the quantity required, so is each PROPORTIONAL quantity^ found by the rule, to the required quantity of each. 13. A man would mix 100 pounds of sugar, some at 8 cents, some at 10 cents, and some at 14 cents per pound, so that the compound may be worth 12 cents per pound; how much of each kind must he use ? We find the proportions to be, 2, 2, and 6. Then, 2+2 + 6 = 10, and C 2 : 20 lbs. at 8 cts. ) 10 : 100 : : < 2 ! 20 lbs. at 10 cts. V Ans, (6 : 60 lbs. at 14 cts. ) 14. How many gallons of water, of no value, must be mixed with brandy at $ 1^20 per gallon, so as to fill a ves- sel of 75 gallons, which may be worth 92 cents per gallon? Ans. 17^ gallons of water to 57^ gallons of brandy. 15. A grocer has currants at 4 d., 6 d., 9d. and lid. per lb. ; and he would make a mixture of 240 bis., so that the mixture may be sold at 8 d. per lb. ; how many pounds of each sort may he take ? Ans, 72, 24, 48, and 96 lbs., or 48, 48, 72, 72, &c. Note, This question may have five different answers. QUESTIONS. 1. Wh?t is alligation? 2. medial? 3. — - the, rule for operating ? 4. Wha.t is alligation alternate ? 5. When the price of the mixture, and the price of the several simples, are given, how do you find the proportional quanti- ties of each simple ? 6. When the quantity of one simple is given, how do you find the others ? 7. When the quantity of the whole compound is given, how do you find the quan- tity of each simple ? DUODliCIBSiL£S. IT 103. Duodecimals are fractions of a foot. The word is derived from the Latin word duodecimy which signifies twelve, A foot, instead of being divided decimally into ten equal parts, is divided duodecimally into twelve equal parts, tali MULTIPLICATION OF DUODECIMALS. IT 103* called inches, or primes^ marked thus, ('). Again, each of these parts is conceived to be divided into twelve other equal parts, called seconds^ ("). In like manner, each second is conceived to be divided into twelve e»qual parts, called thirds, ('") ; each third irto twelve equal parts, called fourths^ {"") ; and so on to any extent. In this way of dividing a foot, it is obvious, that 1' inchy or prime^ is ----- • xV of a foot V^ second is -^ of y'j, - - - iz: y^ of a foot V third is -j^ of 1^ of 1^, - - =z yy^ of a foot V" fourth is y^ of -j^ of -j^ of y^, z= ^jj}-^ of a foot. r'^'' fifthisyVof T^of yVofiVof-iV, = ttVfs-IT of afoot, &C- Duodecimals are added and subtracted in the same man- ner as compound numbers, 12 of a less denomination making 1 of a greatery as in the following TABLB. 12''" fourths make V' third, 12''' thirds . - - 1'' second, 12" seconds - - 1' inch or prime, 12' inches, or primes, 1 foot. ^ Note. The marks, ', ", '", "", &c., which distinguish the different parts, are called the indices of the parts or denomi- nations. MULTIPLICATION OF DUODECIMALS. Duodecimals are chiefly used in measuring surfaces and solids. 1. How many square feet in aboard 16 feet 7 inches long, and 1 foot 3 inches wide ? Note. Length X breadth = superficial contents, (IT 25.) OPERATION. 7 inches, or primes, = ^^ of a Length 16 7' ^^^*' ^"^ ^ inches = y^ of a foot ; BrAdJh \ ^' consequently, the product of 7' X ' J_l__ 3' = y% of a foot, that is, 21" 4 1' 9" = 1' and 9" ; wherefore, we set 16 7' down the 9", and reserve the 1' A ^ — 1^1 ' ^ ^^ carried forward to its proper Ans. 20 8 9" place. To multiply 16 feet by 3' IT 103. MITLTIPLICATION OF DUODECIMALS. SOS is to take -^ of J/^ = ff , that is, 48' ; and the 1' which w« reserved makes 49', z= 4 feet 1' ; we therefore set down the 1', and carry forward the 4 feet to its proper place. Then, multiplying the multiplicand by the 1 foot in the mul- tiplier, and adding the two products together, we obtain the Answer, 20 feet, 8', and 9". The only difficulty that can arise in the multiplication of duodecimals is, in finding of what denomination is the pro- duct of any two denominations. This may be ascertained as above, and in all cases it will be found to hold true, that the product of any two denominations will always be of the denomi- nation denoted by the sum of their indices. Thus, in the above example, the sum of the indices of 7' X 3' is " ; con- sequently, the product is 21" ; and thus primes multiplied by primes will produce seconds ; primes multiplied by seconds produce thirds ; fourths multiplied hy fifths produce ninths, &c. It is generally most convenient, in practice, to multiply the multiplicand first by the feet of the multiplier, then by the inches, &c., thus :-^ A 16 ft. X 1 ft. = 16 ft., and 7' X 1 ft. = r. Then, 16 ft. X 3' = 48' = 4 ft., and 7' X 3' = 21^' = 1' 9". The two products, added together, give for the Ajiswer, 20 ft. 8' 9', as before. 20 8' 9" 2. How many solid feet in a block 15 ft. 8' long, 1 ft. C wide, and 1 ft. 4' thick ? OPERATION. ft- Length, 15 8' The length multiplied by the Breadth, I 5' breadth, and that product by the thickness, gives the solid con' tents, (IF 36.) 16 r 1 3' 16 7' 4 V 9" 15 8' If te 6 6' 4" 22 2' 4" Thickness , 1 4' 22 2i 4" 7 4' 9" 4'" Am. 29 7' 1" 4" jfdi MULTIPLICATION OF DUODECIMALS. IT 104* From these examples we derive the following Rule : — Write down the denominations as compound numbers, and in multiplying lemember, that the product of any two de- nominations will always be of that denomination denoted bj the sum of their indices, EXAMPLES FOR PRACTICE. 5. Hoyy many square feet in a &tpck of 15 boards, 12 ft. 8' in length, and 13' wide ? Ans, 205 ft. 10\ 4. What is the product of 371 ft. 2' 6" multiplied by 181 ft, V 9"? Ans. 67242 ft. 10' 1'' 4'" 6''". Note, Painting, plastering, paving, and some other kinds of work, are done by the square yard. If the contents in square feet be divided by 9, the quotient, it is ^evident, will be square yards. 5. A man painted the walls of a room 8 ft. 2' in height, and 72 ft. 4' in compass ; (that io, the measure of all its sides ;) how many square yards did he paint ? f Ans. 65 yds. 5 ft. 8' 8''. *6. There is a room plastered, the compass of which i$ 47 ft. 3', and the height 7 ft. 6' ; what are the contents ? Ans. 39 yds. 3 ft. 4' 6^'. \ 7. How many cord feet of wood in a load 8 feet long, 4 feet wide, and 3 feet 6 inches high ? Note, It will be recollected, that 16 solid feet make a cord foot, Ans. 7 cord feet. 8. In a pile of wood 176 It. in length, 3 ft. 9' wide, and 4 ft. :i width ? and what will be its cost, at 75 cents per square foot ? Ans, ITT feet; and it will cost $8'775. f 12. Hew many cords in a load of wood 7'o feet in length, 3^6 feet in width, and 4'8 feet in height ? Ans. 1 cord,ly%ft. ^13. How many cord feet in a load cf wood 10 feet long, 3^4 feet wide, and 3'5 feet high ? Ans, l-f^, QUESTIONS. 1. What are duodecimals? 2. From what is the word derived? 3. Into how many parts is a foot usually divided, and what tire the parts called ? 4. What are the other de- nominations ? 5. What is understood by the indices of the denominations? 6. In what are duodecimals chiefly used? 7. How are the contents of a mrface bounded by straight lines found? 8. How are the contents of a so/iVZ found? 9. How is it known of what denomination is the product of any two denominations ? 10. How may a scale or rule be formed for taking dimensions in feet and decimal parts of a foot ? ^ 105. Involution, or the raising of powers, is the mul- tiplying any given number into itself continually a certain number of times. The products tlnis produced arc cafled the powers of the given number. The number itself is cUled the first power, or root. If the first power be multiplied hy itselfy the product is called the second power or square ; if the square be multiplied by the first power, the product is called the third power, or cuhe^ &c. ; thus, 5 is the root, or 1st power, of 5. 5X5= 25 is the 2d power, or square, of 5, =5^ 5X5X5=125 is the 3d power, or cube, of 5, =5^ ''>X5X5X5=:G25 is the Ithpower, orbiquadrate,of 5, =5^ 506 INVOLUTION. H 105. The number denoting the power is called the indeXj or €Xp(r>icnt ; thus, 5* denotes that 5 is raised or involved to the 4th power. 1. What is the square, or 2d power, of 7 ? Ans. 49. 2. What is the square of 30 ? Ans, 900. 3. What is the square of 4000 ? Ans, 16000000. 4. What is the cube, or 3d power, of 4 ? Ans. 64. 5. What is the cube of 800 ? Ans. 512000000. 6. Vv^hat is the 4th power of 60 ? A.ns, 12960000. 7. What is the square of 1 ? of 2 ? of 3 ? of 4? A71S. 1, 4, 9, and 16. 8. What is the cube of 1 ? of 2 ? of 3 ? of 4 ? Ans. 1, 8, 27, and 64. 9. What is the square of f ? of |- ? of |- ? Ans. I, if, and ^. 10. What is the cube of |? of | ? of f ? ^ns. /y,f.\,andM. il. Wliat is the square of ^ ? the 5th povrer of i? A?is. ^, and -gV. 12. W^hat is the square of 1'5 ? the cube ? Ans. 2^25, and 3'375. 13. What is the 6th power of 1^2 ? Ans. 2'9859S4. 14. Involve 2^ to the 4th power. Note. A mixed number, like the above, may be reduced to an improper fraction before involving : thus, 2^ 1= |- ; or it may be reduced to a decimal ; thus, 2^ z=: 2'25. Ans. -^J>ry^ =1 2oiU. 15. What is the square of 4|- ? ' Ans. J |f J^ =: 23|-|-. 16. What is the value of 7^, that is, the 4th power of 7 ? Ans. 2401. 17. How much is 9^ ? 6^ ? 10^ ? Ans. 729, 7776, 10000 18. How much is 27 ? 3^ ? 4-'^? 5^ ? 65 ? 103 ? Ans. to last, 100000000. The powers of the nine digits, from the first power to the iifth, may be seen in the following TAB1.E. Roots - or 1st Powers 1 | 2 ] 3 | 4 5 6 1 7 1 8 9 Squares or 2d Powers 1 j 4 | 9 | IG 25 36 J 49 1 64 81 Cubes - .or 3d Powers 1 | 8 | 27 | 64 125 216 5 tm\ 612 72^1 Biquadrates or 4th Powers 1 |16 81 256 Sursolids or 5lh Powers 1 |32 243 1024 625! 3125 1296 1 ^401 4096 6661 7776 116807 32768 50ai9 If 107 EXTRACTION OF THE SQUARE ROOT. 207 X:VO£UTZO]N< IT 106. Evolution, or the extracting of roots, is the me- thod of finding the root of any power or number. The rootj as we have seen, is that number, which, by a continual multiplication into itself, produces the given power. The square root is a number which, being s<.|uared, will pro- duce the given number; and the cube, or third root, is a num- ber which, being cubed or involved to the 3d power, will produce the given nusiber: thus, the square root of 144 is 12, because 12^ =i 144; and the cxihe root of 343 is 7, be- cause 7 2, that is, 7 X 7 X 7, :ii= 343 ; and so of other num- bers. Although there is no number which will not produce a perfect power by involatisn, yet there are many numbers of which precise roots can never be obtained. But, by the help of decimals, we can approximate, or approach, towards the root to any assigned degree of exactness. Numbers, whose precise roots cannot be obtained, are called surd numbers, and those, whose roots can be exactly obtained, are called rational numbers. The square root is indicated by this character .\/ placed before the number ; the other roots by the same cliaracter, with the index of the root placed over it. Thus, the square root of 16 is expressed \/i6 ; and the cube root of 27 is expressed -v^27; and the 5th root of 7776,^^7776? When the power is expressed by several numbers, with the sign -f- or — between them, a line, or vinculum, is drawn from the top of the sign over all the parts of it ; thus, the square root of 21 — 5 is \/ 21 — 5, &c. EXTRiiCTZOnr OF TWSL SQirARE HOOT. IT 107. To extract the square root of any number is to find a number, which, being multiplied into itself, shall pro- duce the given number. 1. Suppesing a man has 625 yards of carpeting, a yard wide, what is the length of one side of a squaie room, the 208 EXIRACTION OF THE SQUARE ROOT. IF lOt. iioor of v/hicli the carpetiiiof will cover ? tliat is, what is one side of a square, which contains 625 square yards ? We have seen, (11 85,) that the contents of a square sur- face is found by multiplying the leiigth of one side into it- self, that is, by raising it to the second power; and hence, having the contents (625) given, we must extract its square root to find one side of the room. This we must do by a sort of trial : and, 1st. We will endeavour to ascertain how many figures t':ere will be in the root. This we can easily do, by point- ing off the number, from units, into periods of two figures each ; for the square of any root always contains just fwice as many, or one figure less than twice as many figures, as are in the root ; of which truth the pupil may easily satisfy him- self by trial. Pointiug off the number, we find, that the root will consist oHwo figures, OPERATION. C25(2 4 225 Fig. I. a ten i^nd a unit. 2d. We will now seek for the first figure, that is, for the tens of the root, and it is plain, that we must extract it from the left hand period 6, (hundreds.) The greatest square in 6 (hundreds) we find, by trial, to be 4, (hun- dreds,) the root of which is 2, (tens, 1= 20;) therefore, we set 2 (tens) in the root. The root, it will be recollected, is 07ie side of a square. Let us, then, form a square, (A, Fig. 1.) each side of which shall be supposed 2 ten -, = 20 yards, expressed by the root now obtained. The contents of this square are 20 X 20 z= 400 yards, now disposed of, and which, consequentl};, are to be deducted from the whole number of yards, (625,) leaving 225 yaids. This dedr.ction is most readily performed by subtracting the square number 4, (hundreds,) or the square of 2, (the figure in the root already found,) from the period 6, (hundreds,) and bring- ing down the next period by the side of the remainder, making 225, as before. IT 107. JEXTRACTION OF THE gqUAKE ROOT. 209 3d. The square A is now to be enlarged by the addition of the 225 remaining yards ; and, in order that llie figure may retain its square form^ it is evident, the addition must be made on tivo sides. Now, if the 225 yards be divided by the length of the two sides, (20 -f- 20 = 40,) the quotient will be the breadth of this new addition of 225 yards to the sides c d and b c of the square A. But our root already found, ^i^^tens, is the length of one side of the figure A ; we thergfor.e take double this root, iir 4 tens, for a divisor. OPERATION— CONTINUED. 625(25 "^ 4 45)225 225 Fig. SOyds. II. 5 yds. B 20 5 100 5 D J 25 d A 20 20 400 C c 20 5 100 a b 20 yds. 5 yds. , The divisor, 4, (tens,) is in reality 40, and we are to seek how many times 40 is contained in . 225, or, which is the same • thing, we may seek how many times 4 (tens) is contained in *22, (tens,) rejecting the right hand figure of the dividend, because we have rejected the cipher in the divisor. We find our quotient, that isj the breadth of the addition, to be 5 yards ; but, if we look at Fig. II., we shall perceive that this addition of 5 yards to the two sides does not com- plete the square ; for there is still wanting, in the corner D, a small square, eac»i side of which is equal to this last quotient, 5 ; we must, therefore, add this quotient, 5, to the divisor, 40, that is, place it at the right hand of the 4, (tens,) making it 45; and then the whole divisor, 45, multiplied by the quotient, 5, wiil give the con- tents of the whole addition around the sides of the figure A, which, in this case, being 225 yards, the same as our divi- dend, we have no remainder, and the work is done. Con- •equently, Fig. II. represents the floor of a square room, 25 210 EXTRACTION OF THE SC^UARE ROOT. iT lOt. yards on a side, which 625 square yards of carpeting will exactly cover. The proof may be seen by adding together the several parts cf the figure, thus : — The square A contains 400 yards. The figure B 100 Or we may prove it C 100 by involution, thus : — J^... 1> 25 25 X 25 = 625, as be- Proo/,~625 ^''''^* From this example and illustration we derive the follomng general RULE FOR THE EXTRACTION OF THE SQUARE ROOT. I. Point off the given number into periods of two figures each, by putting a dot over the units, another over the hun- dreds, and so on. These dots show the number of figures of which the root will consist. II. Find the greatest square number in the left hand pe- riod, and write its root as a quotient in division. Subtract the square number from the left hand period, and to the re- mainder bring down the next period for a dividend. III. Double the root already found for a divisor ; seek how many times the divisor is contained in the dividend, except- ing the right hand figure, and place the result in the root, and also at the right hand of the divisor; multiply the di- visor, thus augmented, by the last figure of the root, and subtract the product from the dividend ; to the remainder bring down the next period for a new dividend. IV. Double the root already found for a new divisor, and continue the operation as before, until all the periods are brought down. Note 1. If v/e double the right hand figure of the last divisor, we shall have the double of the root. Note 2. As the value of figures, whether integers or decimals, is determined by their distance from the place of units, so we must always begin at unit's place to point off the given number, and, if it be a mixed number, we must point it off both ways from units, and if there be a deficiency in any period of decimals, it may be supplied by a cipher. li is plain, the rcfn n urt ^Iv ; - 'consist of so many integer* ir 108. EXTRACTION OF THE SQUARE ROOT. 211 and decimals as there are periods beloDging to each in the given number. exa3ipl.es for practice. 2. What is the square root of 1034265G ? OPERATION. 10342656 ( 3216, Am, 9 62 ) 134 124 641 ) 1026 641 6426 ) 38556 38556 3. What is tlie square root of 43264 ? OPERATION. 43264 ( 208, Ans. 4 408 ) 3264 3264 4. What is the square root of 998001 ? Atuf. 999 5. What is the square root of 234'09.? Ans, 15^3. 6. What is tlie square root of 964*5192360241 ? Am. 31*05671. 7. What is the square root of '001296 > Ans, '036. 8. What is tlie square root of '2916 ? Ans. '54. 9. What is the square root of 36372961 ? Am. 6031. 10. What is the square root of 164 ? Ans, 12'8 +. ir 103. In this last example, as there was a remainder, after bringing down all the figures, we continued the opera- tion to decimals, by annexing two ciphers for a new period, and thus we may continue the operation to any assigned de- gree of exactness ; but the pupil w ill readily perceive, that he can never, in this manner, obtain the precise root ; for the last figure in each dividend will always be a cipher, and the 212 SUPPLEMENT TO THE SQUARE BOOT. IT 108. last figure in eacli divisor is the same as the last quotient figure ; but no one of the nine digits, multiplied into itself^ produces a number ending with a cipher ; therefore, what- ever be the quotient figure, there will still be a remainder. 11. What is the square root of 3 ? Ans, 1*73 ■ 12. What is the square root of 10 ? Ans, 346 ■ 13. What is the square root of 184'2 ? Ans. 13'57+. 14. What is the square root off? Note. We have seen, (U 105, ex. 9,) that fractions are squared by squaring hath the numerator and the denomina- tor. Hence it follows, that the square root of a fraction is found by extracting the root of the numerator and of the de- nominator. The root of 4 is 2, and the root of 9 is 3. Ans. J. 15. "\Yhat is the square root of /^? Ans. ^ 16. What is the square root of y\j%^? Ans. -^ 17. What is the square root of -f^^? Ans. ^^ = f, 18. What is the square root of 20 J ? Ans. 4^. When the numerator and denominator are not exact squares^ the fraction may be reduced to a decimal, and the approximate root found, as directed above. 19. What is the square root of f zn '75 ? Ans. '866 +. 20. What is the square root of f f ? Ans. '912 -|-. SUPPI^SMEH^T TO THE SQUARE ROOT. QUESTIOIVS. 1. W^hat is involution ? 2. What is understood by a power ? 3. the first, the second, the third, the fourth power ? 4. W^hat is the index, or exponent ? 5. How do you involve a number to any required power ? 6. What is evolution ? 7. What is a root? 8. Can the precise root of all numbers be found ? 9. What is a surd number? 10. a rational? 11. What is it to extract the square root of any number? 12. Why is the given sum pointed into periods of two figures each ? 13. Why do we double the root for a divisor ? 14. Why do we, in dividing, reject the right hand figure of the dividend ? 15. Why do we place the quotient figure to the right hand of the divisor ? 16. How may we fr 108. SUPPLEMENT TO THE SQUARE ROOT. 213 prove the work? 17. Why do we point off mixed numbers both ways from units? 18. When there is a remainder, how may we conthiue the op.eratiou ? 19. Why can we never obtain the precise root of surd numbers ? 20. How do we extract the square root of vulgar fractions ? EXERCISES. 1. A general has 4096 men ; how many must he place in rank and i>le to form them into a square ? Ans, 64. 2. If a square field contains 2025 square rods, how m?.ny rods does It measure on each side ? A7is. 45 rods. 3. How many trees in each row of a square orchard con- taining 5625 trees ? Ans. 75. 4. There is a circle, whose area, or superficial contents, is 5134 feet ; what will be the length of the side of a square of equal area ? \/5184 nz 72 feet, Ans. 5. A has two fields, one containing 40 acres, and the other containing 50 acres, for which B offers him a square field containing the same number cf acres as both of these ; how many rods must each side of this field measure ? Ans. 120 rods. 6. If a certain square field measure 20 rods on each side, how much will the sidii of a square field measure, contain- ing 4 times as much ? V'20 X 20 X 4 zn 40 rods, Ans. 7. If the side of a square be 5 feet^ what will be the side of one 4 times as large ? 9 times as large ? 16 times as large ? 25 times as large ? — — 36 times as large ? Answers, 10 ft. ; 15 ft. ; 20 ft. ; 25 ft. ; and 30 ft. 8. It is required to lay out 288 rods of land in the form of a parallelogram, which shall be twice as many rods in length as it is in width. Note. If the field be divided in the middle, it will form two equal squares. Ans. 24 rods long, and 12 rods wide. 9. I would set out. at equal distances, 784 apple trees, so that my orchard may be 4 times as long as it is broad ; how many rows of trees must I have, and how many trees in each row ? Ans. 14 rows, and 56 trees in each row. 10. There is an oblong piece of land, containing 192 square rods, of which the width is |- as much as the lengtli ; re- quired its dimensions. Ans. IC by 12. SI 4 SUl»PLJBJ^Nt' TO THE SQUARE ROOT. If 109» \ 11. There is a circle, whose diameter is 4 inches ; what is the diameter of a circle 9 times as large ? Note, The areas or contents of circles are in proportion to the squares of their diameters^ or of their circumferences. Therefore, to find the diameter required^ square the given diameter, multiply the square by the given ratio, and the square root of the product will be the diameter required. /\/4 X 4 X 9 = 12 inches, Ans. 12. There are two circular ponds in a gentleman's pleasure ground ; the diameter of the less is 100 feet, and the greater is 3 times as large ; what is its diameter ? Ans, 173'2-|- feet 13. If the diameter of a circle be 12 inches, what is the diameter of one -^ as large ? Ans, 6 inches. IT 109. 14. A carpenter has a large wooden square ; one part of it is 4 feet long, and ^he other part 3 feet long ; what is the length of a pole, which will just reach from one end to the otlier ? A Note, A figure of 3 sides is cilled a triangle, ^ and, if one of the corners " be a square corner^ or rigkt angle^ like the angle at B in the annexed figure, it is called a right-angled trian- gle^ of which the square of the longest side, A C, (called the hypotenuse,) is equal to the sum of the squares of the other two sides, A B and B C. 42 — 16, and 3^ — 9 ; then, \/9 + 16 1= 5 feet, Ans, 15. If, from the corner of a square room, 6 feet be mea- sured off one way, and 8 feet the oiher way, along the sides of the room, what will be the length of a pole reaching from point to point ? Ans, 10 feet. 16. A wall is 32 feet high, and a ditch before it is 24 feet wide ; what is the length of a ladder that will reach frcm the top of the wall to the opposite side of the ditch ? Ans, 40 feet 17. If the ladder be 40 feet, and the wall 32 feet, what is tlie width of ih^ ditch ? Ans, 24 feet 18. The ladder and dkch given, required the wall. Am, 32 feet / .^^ X--" / / / c Baso. B 9 T 110. EXTRACTION OF THE CUBE ROOT. 215 ^19. The distance between the lower ends of two equal rafters is 32 feet, and the height of the ridge, above the beam on which they stand, is 12 feet; required the length of each rafter. Ajis. 20 feet. 20. There is a building 30 feet in length and 22 feet in width, and the eaves project beyond the wall 1 foot on every side ; the roof terminates in a point at the centre of the building, and is there supported by a post, the top of which is 10 feet above the beams on which the rafters rest; what is the distance from the foot of the post to the corners of the eaves ? and what is the length of a rafter reaching to the middle of one side 1 a rafter reaching to the middle of one end ? and a rafter reaching to the comers of the eaves ? An^oers, in order, 20 ft. ; 15'62 + iX, ; 18^86 -f ft. ; and 22^36 + ft. •) 21. There is a field 800 rods long and 600 rods wide ; ^ what is the distance between two opposite corners ? Ahs, 1000 rods. 22. There is a square field containing 90 acres ; how many rods in length is each side of the field ? and how many rods apart are the opposite corners ? Answers, 120 rods ; and 169'7 -f- rods. 23. There is a square field containing 10 acres; what dis- tance is the centre from each co^^ier ? Am, 28'28 + rods. &ZTRJLCT£0»T OF THIS CUBE ROOT. IT 110. A solid body, having six equal sides, and each of the sides an exact square, is a cube, and the measure in length of one of its sides is the root of that cube ; for the length, brcfidth and thickness of such a body are all alike ; con- sequently, the length of one side, raised to the 3d power, ^ves the solid contents. (See IT 36.) Hence it follows, that extracting tne cube root of any num- ber of feet is finding the length of one side of a cubic bo- dy, of which the whole contents will be equal to the given number of feet. ^ 1. What are the solid contents of a cubic block, of which each side measures 2 feet ? Ans. 23-— 2X2X2r=8 feet 2. How many solid feet in a cubic block, n^easuring 6 feejt Oft each side? Aiv3, 6'^' z=z 125 feet. ^16 EXTRACTION OF THE CUBE ROOT. ffllO. 3. How many feet in length is eacli side of a cubic block, containing 125 solid feet? Ans. /v/125 = 5 feet. Note. The root may be found by trial. 4. What is the side of a cubic block, containing 64 solid feet? 27 solid feet?-- — 216 solid feet? 512 solid feet ? A7iswers, 4 ft. ; 3 ft. ; 6 ft. ; and 8 ft 5. Supposing a man has 13824 feet of timber, in separate blocks of 1 cubic foot each ; he wishes to pile them up in a cubic pile ; what will be the length of each side of such a pile ? It is evident, the answer is found by extracting the cube root of 13S24 ; but this number is so huge, that we cannot so easily fmd the root by trial as in the former examples ; — We will endeavour, however, to do it by a sort of trial ; and, 1st. We will try to ascertain the number of figures, of which the root will consist. This we may do by pointing the number off into periods o( three figures each (IT 107, ex. 1.) Pointing off, we see, the root will consist of two figures, a ten and a tmt. Let us, then, seek for the first figure, or tens of the root, which must be extracted from the left hand period, 13, (thousands.) The greatest cube in 13 (thousands) we find by trial, or by the table of powers, to be 8, (thousands,) the root of which is 2, (tens;) therefore, we place 2 (tens) in the root. The root, it will be recollect- ed, is one side of a cube. Let us, then, form a cube, (Fig. 1.) each side of which shall be supposed 20 feet, expressed by the root now obtained. The contents of this cube are Sim feet, Contents. 20 X 20 X 20 ~ 8000 solid feet, which are now disposed of, and which, consequently, are to be deducted from the whole number of feet, 13824. 8000 taken from 13824 leave 5824 feet. This deduction is most readily performed by subtracting the cubic number, 8, or the cube of 2, (the figure of the root already fouud,) from OPERATION. 13824(2 8 ■~5824 Fig. L IT 110. EXTRACTION or HB CUBE ROOT. 217 the period 13, (thousands,) and bringing down the next pe- riod by tlw side of the remainder, making 5824, as before. 2d. The cubic pile A D is now to be enlarged by the ad- dition of 5824 solid feet, and, in order to preserve the cu'bic form of the pile, the addition must be made on one half of its sides, that is, on 3 sides, a, 6, and c. Now, if the 5824 solid feet be divided by the square contents of these 3 equal sides, that is, by 3 times, (20 X 20 — 400) = 1200, the quo- tient will be the thickness of the addition made to each of tlie sides a, ^, c. But the root, 2, (tens,) already found, is tlie length of one of these sides ; we therefore square the root, 2, (tens,) zz: 20 X 20 i= 400, for the square contents of one side, and multiply the product by 3, the number of sides, 400 X 3 = 1200 ; or, which is the same in effect, and more convenient in practice, v/e may square the 2, (tens J and mul- tiply the product by 300, thus, 2 X 2 =z 4, and 4 X 300 == 1 200, for the divisor, as before. The divisor, 1200, is con- tained in the dividend 4 times ; consequently, 4 feet is the thickness of the addition made to each of the three sides, a, Dim^or, 1200) 5824 Dividend, Z>, c, and 4 X 1200 nz 4800, is OPERATION— CONTINUED. 13824(24 Root. 8 4800 960 64 5824 0000 Fig. II. the solid feet contained m these additions; but, if we look at Fig. II., w^e shall per- ceive, that this addition to the 3 sides does not complete the cube ; for there are deficiencies in the 3 corners w, tz, n. Now the length of each of these deficiencies is the same as the length of each side^ that is, 2 (tens) =: 20, and their ?vidth and thickness are each equal to the last quotient figure, (4) ; their contents, therefore, or the number of feet required to fill these deficiencies, will be found by multiplyin^the square of the last quotient figure, (4^) nz 16, by the length of all the deficiencies, that is, by 3 timei T 218 EXTRACnOir OF THE CUBE ROOT* IT 110. the length ©f each side, which is expressed by the former quotient figure, 2, (tens.) 3 times 2 (tens) are 6 (tens) = 60 ; or, what is the same in effect, and more convenient in practice, we may multiply the quotient figure, 2, (tens,) by 30, thus, 2 X 30 = 60, as before ; then, 60 X 16 = 960, con- tents of the thri.e deficiencies «, ??, ii. Looking at Fig. III., we perceive there is stiil a de- ficiency in the corner where the last blocks meet. This deficiency is a cube, each side of which is equal to the last quotient figure, 4. The cube of 4, therefore, (4 X 4 X 4 — 64,) will be the solid contents oi'this corner, which in Fig. IV. is seen filled. Now, the sum of thesft sev- eral additiolis, viz. 4800 + 960 + 64 — 5824, will make the subtrahend, wbich, sub- tracted from the dividend, leaves no remainder, and the work is done. Fig. IV. shows the pile which 13824 solid blocks of one foot each would make, when laid together, and the root, 24, sbows the length of one side of the pile. The correctness of the work may be ascertained by cubing the side now found, 24 », thus, 24 X 24 X 24 :z= 13S24, the 2i feet. given number; or it may be proved by adding together ^le contents of all the several parts, thus. Feet 8000 =: contents of Fig. I. 4800 = addition to the sides a, Z>, and c, Fig. I. 960 z= addition to fill the deficiencies w, ?i, w. Fig. II. ^ = addition to fill the corner €, e, c, Fig. IV. 13824 = contents of the whole pile, Fig. IV., 24 feet on «ch,sida i^ 1 10. KKTBACriON OF THE CUBE ROOT. 2TS From the foregoing example and Ulustraiion we derive ike following FOR EXTHACTING THE CUBE ROOT. I. Separate the given number into periods of three figures each, by putting a point over the unit figure, and every third fig«re beyond the place of units. II. Find the greatest cube in the left hand period, and pu* its root in the quotient. III. Subtract the Cube thus found from the said period, and to the remainder bring down the next period, and call this ihe dimdeiid, IV. Multiply the square of the quotient by 300, calling in the divisor. V. Seek how many times the divisor may be had in the dividend, and place the result in the root; then multiply the divisor by this quotient figure, and write the product under the dividend. VI. Multiply the square of this quotient figure by the former figure or figures of the root, and this product by 30, and place the product under the last ; under ail write the cube of this quotient figure, and call their amount the sub- trahend, ■; VII. Subtract the subtrahend from the dividend, and to tha remainder bring down the next period for a new dividend, with which proceed as before ; and so on, till the whole is finished. Note 1. if it happens that the divisor is not contained in the dividend, a cipher must be put in the root, and the next period brought down for a dividend. Note 2. The same rule must be observed for continuing the operation, and pointing oiF for decimals, as in the square root. Note 3. The pupil will perceive that the number which we call the dioisor, whea multiplied by the last quotient figure, does not produce so large a number as the real sub- tnJiend ; hence, the figure in the root must frequently be Mttaller than the quotient figure. £80 SUPPLEMENT TO THE CUBE ROOT. IT 110 exampl.es for practice. ^ 6. WTiat is the cube root of 1860867? OPERATION. 1860867(123 Ans. 1 1 2 X 300 = 300 ) 860 first Dividend, 600 22 X 1 X 30 HZ 120 23 1= 8 728 first Subtrahend. 122 X 300 = 43200 ) 132867 second Dividend 129600 32 X 12 X 30 = 3240 ^ 33 = 27 132867 second Svbiraheiid, 000000 ^7. What is the cube root of 373248 ? Ans. 72. 8. What is the cube root of 21024576 ? Ans, 276. 9. What is the cube root of 84^604519 ? Am. 4*39. 10. What is the cube root of ^000343 ? A)is. '07. 11. What is the cube root of 2 ? Am. 1'25 -}-. 12. What is the cube root of ^^ ■: Am. §. Note. See IT 105, ex. 10, aud IT 108, ex. 14. 13. What is the cube root of ^| ? Am. | 14. What is the cube root of ^V^g- ? ^R5. iV 15. Wliat is the cube root of -^^-^ ? Am. 425 +• 16. What is the cube root of y^^- ? Am. \. SXyPPliEMSNT TO THE CUBE HOOT. QUESTIONS. 1. What is a cube } 2. What is understood by the ciibe root ? 3. What is it to extract the cube root ? 4. Why is the square of the quotient multiplied by 300 for a divisor ? 5. Why, in finding the subtrahend, do we multiply the square of the last quctient figure by 30 times the former figure of the root.^ 6. Why do we cube the quotient figure ? 7. How do we prove the operation } H 111c SUPPLEMENT TO THE CtJBE ROOT. 2Z\ EXERCISES. i. What is the side of a cubical mound, equal to one 288 feet long, 216 feet broad, and 48 feet high ? Am, 144 feet. "^ 2. There is a cubic box, one side of which is 2 feet ; how many solid feet does it contain ? Ans. 8 feet. 3. How many cubic feet in one 8 times as large ? and what would be the length of one side ? Alls, 64 solid feet, and one side is 4 feet. 4. There is a cubical box, one side of which is 5 feet ; what would be the side of one containing 27 times as much ? 64 times as much ? 125 times as much ? Ans, 15, 20, and 25 feet. 5. There is a cubical box, measuring 1 foot on each side ; what is the side of a box 8 times as large ? 27 times ? 64 times ? Ans, 2, 3, and 4 feet. ^ 111, Hence we see, that the sides of cubes are as the cube roots of their solid contents^ and, consequently, their con- tents are as the cubes of their sides. The same proportion is true of the similar sides^ or of the diameters of all solid figures of similar forms. 6. If a ball, weighing 4 pounds, be 3 inches in diameter, what will be the diameter of a ball of the same metal, weigh- ing 32 pounds? 4 : 32 : : 33 : 6- , u4«5. 6 inches. \ 7. If a ball, 6 inches in diameter, weigh 32 pounds, what will be the weight of a ball 3 inches in diameter ? Ans. 4 lbs. 8. If a globe of silver, 1 inch in diameter, be worth $ 6, what is the value of a globe 1 for in diameter ? Ans, $110368. "0. There are two globes ; one of them is 1 foot in diame- ter, and the other 40 feet in diameter ; how many of the fmalier globes would it take to make 1 of the larger ? Ans, 64000. 10. If the diameter of the sun is 112 times as much as the diameter of the earth, how many globes like the earth would it take to make one as large as the siin ? Ana, 1404928. ^,. 11. If the planet Saturn is 1000 times as large as the earth, and the earth is 7900 miles in diameter, what is the diameter of Saturn ? Ans, 79000 miles. 12. There are two planets of equal density; the diameter of the less is to that of the larger as 2 to 9 ; what is the ra- tio of their solidities ? Ans. ^S-h- : or. as 8 to 729. Zit ARITHMETICAL PROGRESSION. IT 111,112^. Note* The roots of most powers may be found by the square and cube root only : thus, the biquadrate, or 4th root, is the square root of the square root ; the 6th root is the cube root of the square root ; the 8th root is the square root of the 4th root ; the 9th root is the cube root of the cube root, &c. Those roots, viz. the 5tb, 7th, 11th, &c., which are not resolvable by the square and cube roots, seldom oc- cur, and, when they do, the work is most easily performed by logarithms ; for, if the logarithm of any number be divided by the index of the root, the quotient will be the logarithm of the root itself. ARXTiiniXiTSCAXi PROaHESSIOSr. ^ 112. Any rank or series of numbers, more than two, increasing or decreasing by a constant difference, is called an Arithmetical Series^ or Progression, When the numbers are formed by a continual addition of the common difference, they form an ascending series ; but when they are formed by a continual subtraction of the com- mon difference, they form a descending series, Th « \ ^' ^) "^j ^) 11? ^^j 1-5) &:c. is an ascending series. ^^' ( 15, 13, 11, 9, 7, 5, 3, &c. is a descending series. The numbeis which form the series are called the terms of the series. The first and laM terms are the extremes^ and the other term.s are called the means. ^^ There are live things in arithmetical progression, any three of which being given, the other tioo may be found : — 1st. The/r5^ term. 2d. The last term. 3d. The number of terms. 4th. The common difference, 5th. The sum of all' the terms. 1. A man bought 100 yards of cloth, giving 4 cents for the first yard, 7 cents for the second, 10 cents for the third^ and so on, with a common difference of 3 cents ; what was the cost of the last yard ? As the common difference^ 3, is added to every yard except the last, it is plain the last yard must be 99 X 3, = 297 cents, more than the iirst vard. Ans. 301 cents. TT 112. ARITHMETICAL PROGRESSION. 22S Hence, wheii the first term^ the common difference^ and the number of termSj are given^ to find the last term^ — Multiply the number of terms, less 1, by the common difference, and add the first term to the product for the last term. 2. If the first term be 4, the common difference 3, and the number of terms 100, what is the last term ? Ans» 301. 3. There are, in a certain triangular field, 41 rows of com ; the first row, in 1 corner, is a single hill, the second contains 3 hills, and so on, with a common difference of 2 ; what is the number of hills in the last row.^ Ans. 81 hills. 4. A man puts out $1, at 6 per cent, simple interest, which, in 1 year, amounts to $ V06, in 2 j^ears to $ 142, and so on, in arithmetical progression, with a common dif- ference of $ ^06 ; what would be the amount in 40 years ? Ans, $3*40. Hence we see, that the yearly amounts of any sum, at simple interest, form an arithmetical series, of which the vrincipal is the first term, the last amount is the last term, the yearly interest is the cominon difference^ and the number of years is 1 less than the number of terms. 5. A man bought 100 yards of cloth in arithmetical pro- gression ; for the first yard he gave 4 cents, and for the last .301 cents ; what was the common increase of the price on each succeeding yard ? This question is the reverse of example 1 ; therefore, 301 — - 4 = 297, and 297 ~ 99 =i 3, common difference. j^jMence^ when the extremes and number of terms are given, ^fl^d the common difference, — Divide the difference of the extremes by tlie number of terms, less 1, and the quotient will be the common difference. 6. If the extremes be 5 and 605, and the number of terms 151, what is the common difference? Ans, 4. 7. If a man puts out $ 1, at simple interest, for 40 years, and receives, at the end of the time, $ 3'40, what is the rate ? If the extremes be 1 and 3^40, and the number of terms 41, what is the common difference ? Ans, *06. 8. A . man had 8 sons, whose ages differed alike ; the youngest was 10 years old, and the eldest 45 ; what was the common difference of their acres } Ans. 5 years. 224 ARITHMETIC AI^ PROGRESSION. IF 112, 9. A man bought 100 yards of cloth in arithmetical series; he gave 4 cents for the first yard, and 301 cents for the last yard ; what was the average price per yard, and what was the amount of the whole ? Since the price of each succeeding yard increases by a con- stant excess^ it is plain, the average price is as much less than the price of the last yard, as it is greater than the price of the first yard ; therefore, one half the sum of the first and last price is the average price. One half of 4 cts. -\- 301 cts. -. 152^ cts. = average ^ price ; and the price, 152^ cts. X lOO^^i 15250 cts.=: > Ans. $152*50, whole cost. ) Hence, when the extremes and the number of terms are given^ to find the sum of all the terms^ — Multiply J the sum of the ex- tremes by the number of terms, and the product will b« the answer. 10. If the extremes be 5 and 605, and the number of terms 151, what is the sum of the series ? Aiis. 46055. 11. What is the sum of the first 100 numbers, in their natural order, that is, 1, 2, 3, 4, &:c. ? Ans, 5050. 12. How many times does a common clock strike in 12 hours ? Ans. 78. 13. A man rents a house for $ 50, annually, to be paid at the close of each year; what will the rent amount to in 20 years, allowing 6 per cent., simple interest, for the use of the money ? The last year's rent will evidently be $ 50 without inteiest, the last but one will be the amount of $ 50 for 1 year , last but two the amount of $ 50 for 2 years, and so o: arithmetical series, to the first, which will be the amount $50 for 19 years ~ $107. If the first term be 50, the last term 107, and the number of terms 20, what is the sum of the series ? Ans. $ 1570, 14. What is the amount of an annual pension of $ 100, being in arrears, that is, remaining unpaid, for 40 years, allowing 5 per cent, simple interest ? Ans. $ 7900. 15. There are, in a certain triangular field, 41 rows of com ; the first row, being in 1 corner, is a single hill, and the last row, on the side opposite, contains 81 hills ; how many hills of corn in the field } Ans, 1681 hills» nt of II 112, 113. GEOMETRICAL PROGRESSION. 225 16. If a triangular piece of land, 30 rods in length, be 21 rods wide at one end, and come to a point at the other, wlic^ number of square rods does it contain ? Ans. 300, 17. A debt is to be discharged at 11 several payments, in arithmetical series, the first to be $5, and the last $75; what is the whole debt ? the common difference be- tween the several payments ? Ans. whole debt, $ 440 ; common difference, $ 7. 18. What is the sum of the series 1, 3, 5, 7, 9, &c., to 1001 ? Ans. 251001. Note, By the reverse of the rule under ex. 5, the differ- ence of the extremes 1000, divided by the common difference 2, gives a quotient, which, increased by 1, is the number of terras z=z 501. 19. What is the sum of the arithmetical series 2, 2J-, 3, 3^, 4, 4^, &c., to the 50th term inclusive ? Ans, 712^. 20. What is the sum of the decreasing series 30, 29f , 29^, 29, 28f , &c., down to ? Note, 30 -i- ^ 4- 1 = 91, number of terms, xins, 1365. QUESTIONS. 1. What is an arithmetical progression? 2. When, is the series called ascending 1 3. • ^vhen descending! 4. What are the numbers, forming the progression, called ? 5. What are the first and last terms called ? 6. What are the other terms called? 7. When the first ?er??i, common difference, aiid number of terms, are given, how d^^ou find the last term ? 8. How may arithmetical progression be applied to imple interest? 9. When the extremes and number of is are given, how do you find the common difference ? 9. how do you find the sum of all the terms ? ^. TF 113- Any series of numbers, continually increasing by a constant multiplier, or decreasing by a constant divisor, ig called a Geometrical Progression. Thus, 1, 2, 4, 8, 16, &c. IS an increasing geometrical series, and 8, 4, 2, 1, ^, jr, ^fcc* is a decreasing geometrical series. tt6 GEOMETRICAL PROGRESSION. IT 11$^ As in aritlimetical, so also in geometrical progression^ there are five things, any three of which being given, the other two may be found : — 1st The /r5/ term. 2d. The last term. 3d. The number of terms. 4th. The ratio. 5tii. The sum of all the terms. The ratio is the multiplier or divisoTy by which the series i# formed. 1. A man bought a piece of silk, measuring 17 yards, and, by agreement, was to give what the last yard would come to, reckoning 3 cents for the first yard, 6 cents for the second, and so on, doubling the price to the last; what did the piece of silk cost him ? 3X2X2X2X2X2X2X2X2X2X2X2X2 X2X2X2X2=: 196608 cents, = $ 1966*08, Answer, In examining the process by which the last term (196608) has been obtained, we see, that it is a product, of which the ratio (2) is sixteen times a factor, that is, one time less than the number of terms. The last term, then, is the sixteenth power of the ratio^ (2,) multiplied by ih^ first term (3.) Now, to raise 2 to the 16th power, we need not produce all the intermediate powers ; for2^i=i2X2X2X2z=:16, is a product of whi<,'h the ratio 2 is 4 times a factor ; now, if 16 be multiplied by 16, the product, 256, evidently con- tains the same factor (2) 4 times -(- 4 times, z=. 8 times; and 256 X 256 =: 65536, a product of which the ratio (2) is 8 times -{- 8 times, zn 16 times, factor ; it is, therefore, the 16th power of 2, and, multiplied by 3, the first term, gives 196608, the last term, as before. Hence, When the first term, ratio, and number of terms, are given, to find the last term, — I. Write down a few leading powers of the ratio with their indices over them. II. Add together the most convenient indices, to make an index less by one than the number of the term sought. III. Multiply together iha powers belonging to those t7>- dices, and their product, multiplied by the first term^ will b« the term sought f 113. GEOMETRICAL PROGRESSION. 227 2. If the first term be 6, and the ratio 3, what is the 8th term ? Powers of the ratio, with ^\ l 0^7 v ft i ~ sil^ v ^ fir«f their indices over them. ) ^' /' ^^' X, ^^ ~ 7^^ X ^ ^^^ ( term, ziz 10935, Answer, 3. A man plants 4 kernels of corn, which, at harvest, produce 32 kernels ; these he plants the second year ; now, supposing the annual increase to continue 8 fold, what would be the produce of the 16th year, allowing 1000 ker- nels to a pint ? Am, 2199023255'552 bushels. 4. Suppose a man had put out one cent at compound in- terest in 1620, what would have been the amount in 1824, allowing it to double once in 12 years ? 217 — 131072. ^715. $1310*72. 5. A man bought 4 yards of cloth, giving 2 cents for the first yard, 6 cents for the second, and so on, in b fold ra- tio ; whai did the whole cost him ? 2 + 6 + 18 + 54 = 80 cents. Ans. 80 cents. In a long series, the process of adding in this manncy would be tedious. Let us try, therefore, to devise some shorter method of coming to the same result. If all the terms, excepting the last^ viz. 2 -|- 6 + 18, be multiplied by the ratio, 3, the product will be the series 6 -f- 18-f-54 subtracting the former series from the latter^ we have, for the remainder, 54 — 2, that is, the last term, less the first term^ which is evidently as many times the first series (2 -f- 6 -j- 18) as is expressed by the ratio, less 1 : hence, if we dlmde the difference of the extremes (54 — 2) by the ratio, less 1, (3 — 1,) the quotient will be the sum of all the terms, ex" cepting the last^ and, adding the last term, we shall have the xohole amount. Thus, 54 — 2 z= 52, and 3 — 1 z= 2 ; then, 52 -i-2 :z=: 26, and 54 added, makes 80, Answer, as before. Hence, when the extremes and ratio are given, to fmd the gum of the series, — Divide the difference of the extremes by the? ratio, less 1, and the quotient, increased by the greater terviy will be the answer. 6. If the extremes be 4 and 131072, and the ratio 8, what is the whole amount of the series ? 131072 — 4 , 131072 = 149796 Anruf^r. 8—1 ' &28 GEOMETRICAL PROGRESSION. IT 113. 7. What is the sum of the descending series 3, 1, ^, J, 2^, &c., extended to infinity? It is evident the last term must become 0, or ind^^finitely near to nothing ; therefore, the extremes are 3 and 0, and the ratio 3. Ans, 4^. 8. What is the value of the infinite series 1 + i + iV + ^V, &c. ? Alls. 4. 9. What is the value of the infinite series, -J^y + tott + T7T0U + T^iTTU? ^^-y ^^3 what is the same, the decimal 41111, &c., continually repeated? Ans. -J. 10. What is the value of the infinite series, y-^-^y -)- y^jg^nj) &c., descending by the ratio 100, or, which is the same, the repeating decimal ^020202, &c. ? -4725. -^^, 11. A gentleman, whose daughter was married on a new year's day, gave her a dollar, promising to triple it on the first day of each month in the year ; to how much did her portion amount? Here, before finding the amount of the series, we must find the last term^ as directed in the rule after ex. 1. Ans. $265^720 The two processes of finding the last term^ and the omount, may, however, be conveniently reduced to o?ie, thus : — JVhen the first terrn^ the ratio, and the number of terms, are given, to find the sum or amount of the senes, — Raise 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. Applying this ^ule to the last example, 3^2 — 531441, .a»d 531441 — 1 ^ ^ __ 265720. Ajis. $ 265^720, as hdofi o 1 12. A man agrees to serve a farmer 40 years without any other reward than 1 kernel of corn for the first year, 10 for the second year, and so on, in 10 fold ratio, till the end of the time ; what will be the amount of his wages, allowing 1000 kernels to a pint, and supposing he sells his corn at 50 cents per bushel ? 10T IT, 4- 8^7846 _ = K^-,thatis, — - — 3-. 1^4641 z=: the 4th power of the ratio ; and then, by extracting the 4th root, we obtain 140 ^r the ratio. Aiis. 10 per cent. 8. In what time will $6 amount t<» $8'7846, at 10 per cent compound interest ? p^ = RT., that is, — g— = 1^4641 =r 140^- ; therefore, if we divide 1'4641 by 140, and tlien divide the quotient thence arising by 140, and so on, till we obtain a quotient that will not contain 140, the nziiider of tliese divisions will b« the number of years, Ans, 4 year*. *r 114, 115. GEOMETRICAL PROGRESSION. 2S1 9. At 5 per cent, compound interest, in what time will $ 40 amount to $ 68'40 ? Having found the power of the ratio 1*05, as heforc, which is 1*71, you may look for this number in the table^ under the given rate, 5 per cent., and against it you will find the number of years. Ans. 11 years. 10. At 6 per cent, compound interest, in what time will $ 4 amount to $ 5*352 ? Ans. 5 years. Annuities at Compound Interest, IT 115, It may not be amiss, in this place, briefly to show the application of compound interest, in computing the amount and present worth of annuities. An Annuity is a sum payable at regular periods^ of one year each, either for a certain number of years, or during the life of the pensioner, or forever. When annuities, rents, &c. are not paid at the time they become due, they are said to be in arrears. The sum of all the annuities, rents, &c. remaining un- paid, together with the interest on each, for the time they have remained due, is called the amount, 1. What is the amount of an annual pension of $ 160, which has remained unpaid 4 years, allowing 6 per cent, compound interest? The te/ yearns pension will be $ 100, without interest ; the last but one will be the amount of $ 100 for 1 year; the last but two the amount (compound interest) of $ 100 for 2 years, and so on ; and the sum of these several amounts will be the answer. We have then a series of amounts, (hat is, di geometrical series, (IT 114,) to find the sum of all the terms. If the first term be 100, the number of terms 4, and the ratio 1*06, what is the sum of all the terms ? Consult the rule, under IT 113, ex. 11. 1«06*— -1 —^ X 100 ~ 437*45. Ans. $437*45. Hence, when the annuity, the time, and rate per cent, are given, to find the amount, — Raise tlie ratio (the amount of 232 GEO:VI£TRfCAL PROGRESSION. IF 115,116. $ 1, &c. for 1 year) to a power denoted by the number of years; from this power subtract 1 ; then divide the remaiiH der by the ratio, less 1, and the quotient, multiplied by the annuity, will be the amount. Note, The powers of the amounts, at 5 and 6 per cent, up to the 24th, may be taken from the table^ under 1\ 91. 2. What is the amount of an annuity of $ 50, it being in arreais 20 years, allowing 5 per cent, compound interest? Ans. $]653<29. 3. If the annual rent of a house, which is $ 150, be in arrears 4 years, what is the amount, allowing 10 per cent. compound interest ? Am, $69645, 4. To how much would a salary of $600 per annum amount in 14 years, the money being improved at 6 per cent, compound interest? in 10 years? in 20 years ? in 22 years ? in 24 years ? Am, to the last, $ 25407'75. IT 116. If the annuity is paid in advance, or if it be bought at the beginning of the first year, the sum which ought to be given for it is called the present worth, 5. What is the present worth of an annual pension of $ 100, to continue 4 years, allowing 6 per cent, compeund interest ? The present worth is, evidently, a sum which, at 6 per cent, compound interest, would, in 4 years, produce an amount equal to the ainoimt of the annuity in arrears the same time. By the last rule^ we find the amount zz: $437'45, and by the directions under ^ 114, ex. 4, we find the present w^orth = $346'51. Am. $346^51. Hence, to find the present worth of any annuity, — ^First find its amount in arrears for the whole time ; this amount j divided by that power of the ratio denoted by the number of years, will give the present worth, 6. What is the present worth of an annual salary of $ 100 to continue 20 years, allowing 5 per cent ? Ans. $ 1246'22. If 116. GEOMETRICAL PROGRESSION. 233 The operations under this rule being somewhat tedious, we subjoin a TABL.E, Showing the present worth of $ 1, or 1 £, annuity, at 5 and 6 per cent, compound interest, for any number of years from 1 to 34. Yea«. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 5 per cent. 0'95238 1^85941 2^72325 3'54595 4^32948 5^07569 5'78637 6'46321 74 0782 7'72173 8'30641 8^86325 9^39357 9'89864 10'37966 10^83777 11^27407 6 per cent. | Yeans. 0'94339 18 1 '83339 19 2^67301 20 3'4651 21 4^21236 22 4'91732 23 5^58238 24 6^20979 25 6'80169 26 r36008 27 7*88687 28 8*38384 29 8'85268 30 9*29498 31 9*71225 32 10*10589 33 10*47726 34 5 per cent. 11*68958 12*08532 12*46221 12*82115 13*163 13*48807 13*79864 14*09394 14*37518 14*64303 14*89813 15*14107 15*37245 15*59281 15*80268 16*00255 16*1929 6 per cent. 10*8276 11*15811 11*46992 11*76407 12*04158 12*30338 12*55035 12*78335 13*00316 13*21053 13*40616 13*59072 13*76483 13*92908 14*08398 14*22917 14*36613 It is evident, that the present worth of $ 2 annuity in 2 times as much as that of ^ 1 ; the present worth of $ 3 will be 3 times as much, &c. Hence, to find the present worth of any ammity^ at 5 or 6 per cent.^ — Find, in this table, the present worth of $ 1 annuity, and multiply it by tlie given annuity^ and the product will be the present worth, 7. What ready money will purchase an annuity of $ 150, to continue 30 years, at 5 per cent, compound interest? The present worth of $ 1 annuity, by the table, for 30 years, is $ 15*37245 ; therefore, 15*37245 X 150 :=z $ 2305*367, Ans. 8. What is the present worth of a yearly pension of $ 40, at 6 per cent, compoaud interest? to continue 15 years ? — 20 34 years ? to contmue 10 years, at 5 per cent. ? — years ? 25 years ? Ans. to last, $ 647^716. 234 GEOMETRICAL PROGRESSrblS?. It 116. When annuities do not commence till a certain period of time has elapsed, or till some particular event has taken place, they are said to be in reversion, 9. What is the present worth of $ 100 annuity, to be continued 4 years, but not to commence till 2 years hence, allowing 6 per cent, compound interest ? The present worth is evidently a sum which, at 6 per cent, compound interest, would in 2 years produce an amount equal to the present worth of the annuity, were it to commence immediateli/. By the last rule, we find the present worth of tbe annuity, to commence immediately, to be $346 '51, and, by directions under IT 114, ex. 4, we find the present worth of $346'51 for 2 years, to be $30S'393. Am. $308'393. Hence, to find the present worth of any annuity taken in reversion, at compound interest, — First, 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 ready money will purchase the reversion of a ease of $ 60 per annum, to continue 6 years, but not to com- mence till tlie end of 3 years, allowing 6 percent, compound interest to the purchaser ? The present worth, to commence immediately, we find to be, $ 295^039, and -r,;r^ == 247^72. Am, $ 247*72. 1 06*^ It is plain, the same result will be obtained by finding the present v/orth of the annuity, to commence immediately, and lo continue to the end of the time, that is, 3 -|- 6 = 9 years, and then subtracting from this sum the present worth of the annuity, continuing for the time of reversion, 3 years. Or, wc may find the present worth of $ 1 for the two times by the table, and multiply their difference by the given an- nuity. Thus, by the table. The whole time, 9 years, m 6*80169 The time in reversion, 3 years, = 2*67301 Diff*erence, = 4*12868 60 $247*72080 Ans, 11. What is the present worth of a lease of $ 100 to con- tinue 20 years, but not to commence till the end of 4 years, 1r lit. GEOMETRICAL PROGRESSION. 235 allowing 5 per cent. ? what, if it be 6 years in rever- sion ? 8 years ? 10 years ? 14 years ? Am, to last, $ 629^426. TT 117. 12. What is tb^ worth of a freehold estate, of which the yearly rent is $ 60, allowing to the purchaser 6 per cent. ? In this case, the annuity continues forever, and the estate is evidently worth a sum, of which the yearly interest is equal to the yearly rent of the estate. The principal multiplied by the rate gives the interest; therefore, the interest dimded by the rate will give the principal ; 60 -r-«06 nz 1000. A?is. $1000. Hence, to find the present worth of an annuity, continuing forever, — Divide the annuity by the rate per cent., and the quotient will be the present worth. Note, The worth will be the same, whether we reckon simple or compound interest ; for, since a yearns interest of the mice is the annuity, the profits arising from that price can neither be more nor less than the profits arising from the an-- nuity, whether they be employed at simple or compound in- terest. 13. What is the worth of $ 100 annuity, to continue for- ever, allowing to the purchaser 4 per cent. ? allowing 6 percent.? 8 percent.? 10 percent. ? 15 per cent. ? 20 per cent. ? Ans, to last, $ 500. 14. Suppose a freehold estate of $ 60 per annum, to com- mence 2 years hence, be put on sale ; what is its value, al- lowing the purchaser 6 per cent. ? Its present worth is a sum which, at 6 per cent, compound interest, would, in 2 years, produce an amount equal to the worth of the estate if entered on immediately, — — z= $ 1000 == the worth, if entered on immediately, tfc 1000 and — jTQg^ = $ 889^996, the present worth. The same result may be obtained by subtracting from the worth of the estate, to commence immediately, the present worth of the annuity 60, /or 2 years, the time o/ reversion. Thus, by the table, the present worth of $ 1 for 2 years is 1^83339 X 60 == 110*0034 m present worth of $60 for 2 years, and $ 1000 -- $ 110'o634 =: $889*9966, Ans, as before. £36 GE03IETRICAL PROGRESSION. IT 117. 15. What is the present worth of a perpetual annuity of $ 100, to commence 6 years hence, allowing the purchaser 5 per cent, compound interest ? what, if 8 years in re- version ? 10 years ? 4 years ? 15 years ? 30 years ? Ans, to last^ $ 462'755. The foregoing examples, in compound interest, have been confined to yearly payments ; if the payments are half year- ly, we may take half the principal or annuity^ half the rate per cent. J and twice the number of years ^ and work as before, and so for any other part of a year, QUESTIONS. 1. What is a geometrical progression or series ? 2. What is the ratio ? 3. When the first term, the ratio, and the num- ber of terms, are given, how do you find ^^ last term! 4. When the extremes and ratio are given, how do you find the sum of all the terms ? 5. When the first term, the ratio, and the number of ternis, are given, how do you find the amount of the series ? 6. When the ratio is a fraction^ how do you proceed ? 7. What is compound interest ? 8. How does it appear that the amountSj arising by compound in- terest, form a geometrical series ? 9. What is the ratio^ in compound interest ? the number of teriJis ? the first term? the lust term? 10. When the rate, the time, and the principal, are given, how do you find the amount ? 11. "When A. R. and T. are given, how do you find P. ? 12. When A. P. and T. are given, how do you find R. ? 13. When A, P. and R. are given, how do you find T. ? 14. What is an aminity ? 15. When are annuities said to be in ar- rears ? 16. What is the amount ? 17. In a geometrical series, to what is the araoimt of an annuity equivalent ? 18. How do you find the amount of an annuity, at compound interest? 19. What is the present worth of an annuity ? how com- puted at compound interest ? how found by the table ? 20. What is understood by the term reversion ? 21. How do you find the present worth of an annuity, taken in rever^ sion ? by the table ? 22. How do you find the present worth of a freehold estate, or a perpetual annuity ? the same taken in reversion 1 by the table ? It 118, 119. MISCELLANEOUS EXAMPLES. SSt PJESRIKEUTATZON. IT 118. Permutation is the method of finding how many different ways the order of any number of things may he varied or changed. 1. Four gentlemen agreed to dine together so long as they could sit, every day, in a different order or position ; how many days did they dine together ? Had there been but two of them, a and 6, they could sit only in 2 times 1 (1 X 2z=:2) different positions, thus, a 6, and b a. Had there been three^ a, b, and c, they could sit in 1 X 2 X 3 zz: 6 different positions ; for, beginning the order with a, there will be 2 positions, viz. a b Cj and ac b ; next, beginning with by there will be 2 positions, b a c, and b c a; lastly, beginning with c, we have c a bj and c b a, that is, in all, 1 X 2 X 3 = 6 different positions. In the same manner, if there be fouvy the different positions will De 1 X 2 X 3 X 4 z=: 24. A?is, 24 days. Hence, to find the number of different changes or permu- tationSj of which any number of different things are capable^ — Multiply continually together all the terms of the natural series of numbers, from 1 up to the given number, and the last product will be the answer. 2. How many variations may there be in the position of the nine digits ? Ans, 362880. 3. A man bought 25 cows, agreeing to pay for them 1 cent for every different order in which they could all be placed ; how much did the cows cost him ? Ans. $ 155112100433309859840000. 4. Christ Church, in Boston, has 8 bells ; how many changes may be rung upon them ? Ans. 40320. TaiSCHI^I^AMlSOUS SXAmTIiES. IT 119. 1. 4 + 6 X 7 — 1 =z60. A line, or vinculum^ drawn over several numbers, signifies, that the numbers under it are to be taken jointly, or as one whole number. jS38 miscellaneous examples. If 119. 2. 9 — 8 + 4X8 + 4 — 6 = how many ? Ans. 30. 3. 7 + 4 — 2 + 3 + 40X5 = how many ? Ans. 230. 4. 3 + 6 — 2X4--2 __ ^^^ ^^^y P ^^ 3^ 2X2 6. There are two numbers ; the greater is 25 times 78, and their difference is 9 times 15 ; their sum and product are required. Ans, 3765 is their sum ; 3539250 their product 6. What is the difference between thrice five and thirty, and thrice thirty-five ? 35 X 3 — 5 X 3 + 30 = 60, Ans. 7. What is the diff*erence between six dozen dozen, and half a dozen dozen ? Ans. 792. 8. What number divided by 7 will make 6488 ? 9. What number multiplied by 6 will make 2058 ? 10. A gentleman went to sea at 17 years of age; 8 years after he had a son born, who died at the age of 35 ; after whom the father lived twice 20 years ; how old was the father at his death ? Ans. 100 years* 11. What number is that, which being multiplied by 15 the product will he ^? J -i- 15 = g^y, Ans. 12. What decimal is that, which keing multiplied by 15, the product will be '75 ? '75 -r- 15 = '05, Ans. 13. What is the decimal equivalent to ^ ? Ans. '0285714. 14. What fraction is that, to which if you add f , the sum will be f ? Ans. i^. 15. What number is that, from which if you take f, the remainder will be ^ ? Ans. f §. 16. What number is that, wiiich being divided by f, the quotient will be 21 ? Ans. loj. 17. What number is that, which multiplied by f pro- duces ^ ? Am. |. 18. What number is that, from which if you take | of itself, the remainder will be 12 ? Ans. 20. 19. What number is that, to which if you add f of J- of Itself, the whole will be 20 ? Ais. 12. 20.^ What number is that, of which 9 is the § part ? Ans. 13^. 21. A farmer carried a load af produce to maiket: he sold 780 lbs. of pork, at 6 cents per lb. ; 250 lbs. of cheese, at 8 cents per lb. ; 154 lbs. of butter, at 15 cents per lb. : V 119. ^ MISCELLANEOUS EXAMPLES. 239 in pay he received 60 lbs. of sugar, at 10 cents per lb. ; 15 gallons of molasses, at 42 cents per gallon ; ^ barrel of mack- erel, at $ 3'75 J 4 bushels of salt, at $ 1'25 per bushel ; and the balance in money : how much money did he receive ? Ans. $68'85. 22. A farmer carried his grain to market, and sold 75 bushels of wheat, at $ 1'45 per bushel. 64 rye, ... $ *95 142 com, ... $ '50 In exchange he received sundry articles : — 8 pieces of cloth, each containing 31 yds., at $ V75 per yd. 2 quintals of fish, ... $ 2^30 per quin. 8hhds. of salt, ... $4^30 per hhd. and the balance in money. How much money did he receive ? Ans, $38*80. 23. A man exchanges 760 gallons of molasses, at 37j^ cents per gallon, for 66^ cwt. of cheese, at $ 4 per cwt. ; how much will be the balance in his favour? ^t?^. $ 19. 24. Bought 84 yards of cloth, at $V25 perya^d; ho\% much did it come to ? How many bushels of wheat, a $ VoQ per bushel, will it take to pay for it? Am. to the lasty 70 bushels- 25. A man sold 342 pounds of beef, at 6 cents per pound, and received his pay in molasses, at 37J cents per gallon ; how many gallons did he receive ? Arts, 54'72 gallons. 26. A man exchanged 70 bushels of rye, at $ '92 per bushel, for 40 bushels of wheat, at $ 1^37 J- per bushel, and received the balance in oats, at $ '40 per bushel ; how many bushels of oats did he receive ? Ans. 23^. 27. How many bushels of potatoes, at 1 s. 6 d. per bushel, must be given for 32 bushels of barley, at 2 s. 6 d. per bushel ? Ans. 63^ bushels. 28. How much salt, at $ 1'50 per bushel, must be given in exchange for 15 bushels of oats, at 2 s. 3 d. per bushel ? Note. It will be recollected that, when the price and cost lire given, to find the quantity-, they must both be reduced to the same denomination before dividing. Ajls. 3 J bushels. 29. How much' wine, at $2^75 per gallon, must be givon in exchange for 40 yards ©f cloth, at 7 ». 6 d. per y^d ? Ans. 18^ gallon*. MO MISCELLA?fEOUS EXAMPLES. IT 119. 30. A had 41 cwt. of hops, at 30 s. per cvvt, for which B gave him 20 £ . in money, and the rest iu prunes, at 5 d. per lb. ; how many prunes did A receive ? Ans. 17 cwt. 3 qrs. 4 lbs. 81. A has linen cloth worth $'30 per yard; but, in bar- tering, he will have $ '35 per yard ; B has broadcloth worth $ 3"75 ready money ; at what price ought the broadcloth to be rated in bartering with A ? '30 : '35 : : 3'75 : $ 4'375, Aiis, Or, i|^ of 3'75 = $ 4'37^, Ans, The two operations will be seen to be ex- actly alike. 32. If cloth, worth 2 s. per yard, cash, be rated in barter at 2 s. 6 d., how should wheat, worth 8 s. cash, be rated in exchanging for the cloth ? Ans, 10 s., or $ 1'666§. 33. If 4 bushels of corn cost $ 2, what is it per bushel ? Ans. $'50. 34. If 9 bushels of wheat cost $ 13'50, what is that per bushel? Ans, $ 1'50. 35. If 40 sheep cost $ 100, what is that per head ? Ans, $2'50. 36. If 3 bushels of oats cost 7 s. 6 d., how much are they per bushel ? Ans, 2 s. 6 d., = $ '41f . 37. If 22 yards of broadcloth cost 21 £ . 9 s., what is the price per yard r An^, 19 s. 6 d., = $ 3'25; 38. At $ '50 per bushel, how much corn can be bought for $ 2'00 ? Ans, 4 bushels. 39. A man, having $ 100, would lay it out in sheep, at $ 2'50 apiece ; how many can he buy ? Ans, 40. 40. If 20 cows cost $ 300, what is the price of 1 cow ? of 2 cows ? of 5 cows ? of 15 cows ? Ans. to the last^ $ 225. 41. If 7 men consume 24 lbs. of meat in one week, how much would 1 man consume in the same time ? 2 men ? 5 men ? 10 men ? Ans, to the lasty 34^ lbs. Note, Let the pupil also perform these questions by the rule of proportion. 42. If I pay $ 6 for the use of $ 100, how much must I pay for the use of $ 75 ? Ans, $ 4'50. 4?. What premium must I pay for the insurance of my house against loss by fire, at the rate of ^ per cent., that is, ^ dollar on a hundred dollars, if my house be valued at $2475? ^715. $12'3W. IT 119. MISCELLANEOUS EXAMPLES, 241 44. What will be the insurance, per annum, of a store and contents, valued at $9S76'40, at 1^ per centum? Am, $148446. 45. What commission must I receive for selling $478 worth of books, at 8 per cent ? Ans. $ 38^24. 46. A merchant bought a quantity of goods for $734, and sold them so as to gain 21 per cent. ; how much did he gain ? and for how much did he sell his goods ? Ans, to the last^ $ 88844. 47. A merchant bought a quantity of goods at Boston, for $ 500, and paid $ 43 for their transportation ; he sold them so as to gain 24 per cent, on the whole cost ; for how much did he sell them ? Ans. $ 673^32. 48. Bought a quantity of books for $ 64, but ior cash a discount of 12 per cent, was made ; what did the books cost? Ans, $56^32. 49. Bought a book, the price of which was marked $ 4'50, but for cash the bookseller will sell it at 33^ per cent, discount ; what is the cash price ? Ans. $ 3^00. 50. A merchant bought a cask of molasses, containing 120 gallons, for $ 42 ; for how much must he sell it to gain 15 per cent. ? how much per gallon ? Aus. to last, $ '40^. 51. A merchant bought a cask of sugar, containing 740 pounds, for $ 59^20 ; how must he sell it per pound, to gain 25 per cent. ? Ans. $ 40. 52. What is the interest, at 6 per cent, of $ 71^02 for 17 months 12 days ? Ans, $ 6478 -|-. 53. What is the interest of $487*003 for 18 months ? Ans. $43'83+. 54. What is the interest of $ 8*50 for 7 months ? Ans. $<297J. 65. What is the interest of $ 1000 for 5 days ? Ans. $*833f 5d. What is the interest of $ *50 for 10 years ? Ans. $*30. 57. What is the interest of $84*25 for 15 montlis and 7 days, at 7 per eent. ? Ans. $ 7*486 -f-- 58. What is the interest of $ 154*01 for 2 years, 4 months and 3 days, at 5 piT cent. ? Am. $ 18*032. 59. What sum, put to interest at 6 per cent., will, in 2 years and 6 months, amount to $ 150 ? Note. See IT 85. Am. $ 130*134 -V-. 60. I owe a man $ 475*50, to be paid in 16 months witW- X 242 MISCELLANEOUS EXAMPLES. IT 11 9, out interest; what is the present worth of that debt, the use of the money being worth 6 per cent. ? Ans, $ 440'277 -|- 61. What is the present worth of $ 1000 payable in 4 years and 2 months, discounting at the rate of 6 per cent. ? Ans, $800. 62. A merchant bought articles to the amount of $ 500, and sold them for $ 575 ; how much did be gain ? What per cent, was his gain ? that is, How many dollars did he gain on each $100 which he laid out? If $500 gain $75, what dof^ $ 100 gain? Am. 15 percent. 63. A merchant bought cloth at $ 3^50 per yard, and sold it at $ 4'25 per yard ; how much did he gain per centum ? Ans, 21^ per cent. 64. A man bought a cask of wine, containing 126 gallons, for $ 283'50, and sold it out at the rate of $ 2^75 per gal- lon ; how much was his whole gain ? how much per gal- lon ? how much per cent. ? Am, His whole gain, $63'00; per gallon, $'50; which is 22f per centum. 65. If $ 100 gain $ in 12 months, in what time will it gain $ 4 ? $ 10 ? $ 14 ? Ans, to the last^ 28 months. 66. In what time will $ 54'50, at 6 per cent., gain $ 248 ? Ans. 8 months. 67. 20 men built a certain bridge in 60 days, but, it being carried away in a freshet, it is required how many men can rebuild it in 50 days. (lays. days. men. 50 : 60 : : 20 : 24 men, Am. 68. If a field will feed 7 horses 8 weeks, how long will it* feed 28 horses ? Am, 2 weeks. 69. If a field, 20 rods in length, must be 8 rods in width to contain an acre, how much in width must be a field, 16 rods in length, to contain the same? Ans, 10 rods. 70. If I purchase for a cloak 12 yards of plaid |- of a yard wide, how much hocking l^ yards wide must I have to line it ? Am, 5 yards. 71. If a man earn $75 in 5 months, how long must he work to earn $ 460 ? Ans. 30| months. *72. A owes B $ 540, but, A not being worth so much money, B agrees to take $ '75 on a dollar ; what sinn must B receive for the debt ? Ans, $ 406. 73. A cistern, whose capacity is 400 galloas, is supplied V 119. MISCELLANEOUS EXAMPLES. 243 by a pipe which lets in 7 gallons in 5 minutes ; but there is a leak in the bottom of the cistern which lets out 2 gallons in 6 minutes ; supposing the cistern empty, in what time would it be filled r In 1 minute ^ of a gallon is admitted, but in the same time f of a gallon leaks out. Ans, 6 hours, 15 minutes. 74. A ship has a leak which will fill it so as to make it sink in 10 hours ; it has also a pump which will clear it in 15 hours : no^v, if they begin to pump when it begins to leak, in what time will it sink ? In 1 hour tl>e ship would be -£^ filled by the leak, but in die same time it would be -^ emptied by the pump. Alls, 30 hours. 75. A cistern is supplied by a pipe which will fill it in 40 minutes ; how many pipes, of the same bigness, will fill it in 5 minutes ? Ans, 8. 76. Suppose I lend a friend $500 for 4 months, he promising to do me a like favour; some time afterward, I have need of $ 300 ; how long may I keep it to balance the former favour ? Ans, 6f months. 77. Suppose 800 r5oldiers were in a garrison with pro- visions sufficient for 2 months ; how many soldiers must de- part, that the provisions may serve them 5 months ? Ans, 480. 78. If my horse and saddle are worth $ 84, and my horse be worth 6 times as much as my saddle, pray what is the value of my horse ? Ans, $ 72. 79. Bought 45 barrels of beef, at $ 3^50 per barrel, among which are 16 barrels, whereof 4 are worth no more than 3 of the others; how much must t pay ? Ans, $ 143*50. 80. Bought 126 gallons of rum for $110; how much v;ater must be added to reduce the first cost to $ '75 per gallon ? Note, If $ '75 buy 1 gallon, how many gallons will $ 110 buy? Ans, 20§ gallons. 81. A thief, bavins: 24 miles start of the officer, holds his way at the rate of 6 miles an h^ur ; the officer pressing on after him at the rate of 8 miles an hour, how much does he gain in 1 hour ? how long before he will overtake the thief? Ans, 12 hours. 82. A hare starts 12 rods before a hound, but is not per- ceived by him till she has been up H minutes; she scuds away at the rate of 36 rods a minute, and the doir, on view. 244 MISCELLANEOUS EXAMPLES. IT iW^ makes after, at the rate of 40 rods a rriinutc ; how long will the course hold ? and what distance will the dog run ? Am, 14^ minutes, and he will run 570 roda. 83. The hour and minute hands of a watch are exactly together at 12 o'clock ; when are they next together ? In 1 hour the minute hand passes over 12 spaces, and the hour hand over 1 space ; that is, the minute hand gains upon the hour hand 1 1 spaces in 1 hour ; and it must gain 12 spaces to coincide with it. Ana. 1 h. 6 m. 2T^ s. 84. There is an island 20 miles in circumference, and three men start together to travel the same way about it ; A goes 2 miles per hour, B 4 miles per hour, and C 6 miles per hour j in what time will they come together again ? Ans. 10 hours. 85. There is an island 20 miles in circumference, and two men start together to travel around it; A travels 2 miles per hour, and B 6 miles per hour ; how long before they will again come together ? B gains 4 miles per hour, and must gain 20 miles to over- take A ; A and B will therefore be together once in eveiy 5 hours. 86. In a river, supposing two boats start at the same time from places 300 miles apart ; the one proceeding up stream is retarded by the current 2 miles per hour, while that mov- ing down stream is accelerated the same ; if both be pro- pelled by a steam engine, which would move them 8 miles per hour in still water, how far from each starting place will the boats meet ? Ans. 112| miles from the lower place, and 1 87 j miles from the upper place, 87. A man bought a pipe (126 gallons) of wine for $275; he Vv^ishes to fill 10 bottles, 4 of which contain 2 quarts, and 6 of them 3 pints each, and to sell the remainder so as to make 30 per cent, on the first cost ; at what rate per gallon must he sell it ? Ans. $ 2^936 -f-- 88. Thomas sold 150 pine apples at $'33^ apiece, and received as much i.iCney as Harry received for a certain number of watermelons at $ '25 apiece ; how much money did each receive, and how many melons had Harry ? Ans. $ 50, and 200 melons. 89. The third part of an army was killed, the fourth pari taken prisoners, and 1000 fled ; how many were in this army? This and the eighteen following questions are usually V 119. MISCELLANEOirS EXAMPLES. 245 wrought by a rule called Position, but they are more easily solved on general principles. Thus, ^ -j- i ^=^ t^j ^^ ^^ army; therefore, 1000 is ^^^ ^^ the whole number of men; and, if 5 twelfths be 1000, how much is 12 twelfths, or the whole ? -A^is. 2400 men. 90. A farmer, being asked how many sheep he had, an- swered, that he had them in 5 fields ; in the first were ^ of his flock, in the second J, in the third |, in the fourth y^, and in the fifth 450 ; how many had he ? Ans. 1200. 91. There is a pole, ^ of which stands in the mud, ^ in the water, and the rest of it out of the water; required the part out of the water. Ans. y^. 92. If a pole be -J in the mud, |- in the water, and 6 feet out of the water, what is the length of the pole ? Ans. 90 feet. 93. The amount of a certain school is as follows : ^^ of the pupils study grammar, f geography, y^j arithmetic,^ learn to write, and 9 learn to read : what is the number of each ? Ans, 5 in grammar, 30 in geography, 24 in arithmetic ; 12 learn to write, and 9 learn to read. 94. A man, driving his geese to market, was met by another, who said, " Good morrow, sir, with your hundred geese ;" says he, '^ I have not a hundred; but if I had, in ad' dition to my present number, one half as many as I now have, and 2^ geese more, I should have a hundred :" how many had he ? 100 — 2^ is what part of his present number ? Am. He had 65 geese. 95. In an orchard of fruit trees, ^- of them bear apples, ^ pears, ^ plums, 60 of them peaches, and 40, cherries ; how many trees does the orchard contain ? Ans, 1200. 96. In a certain village, ^ of the houses are painted white, J- red, ^ yellow, 3 are painted green, and 7 are unpainted; how many houses in the village ? Ans, 120. 97. Seven eighths of a certain number exceed four fifths of the same number by 6 ; required the number. ^ — ^=z ^\ ; consequently, 6 is ^ry of the required num- ber. Ans, 80. 98. What number is that^ to which if ^ of itself be added, (he sum will be 30 ? Ans, 2d. 99. What number is that, to which if its^ and I be added, the sum will be 84 ^ 84=l+^-f-^zz:J times the required number. Atis, 48. X* 246 MISCELLANEOUS EXAMPLES. If 11^ 100. What number is that, which, being increased by f and I of itself, and by 22 more, will be made three times &s much ? The number, being taken 1, f, and f times, will make 2^ times, and 22 is evidently what that wants of 3 times. Am, 30. 101. What number is that, which being increased by f, | and |- of itself, the sum will be 234f ? Ans. 90. 102. A, B, and C, talking of their ages, B said his age was once and a half the age of A, and C said his age was twice and one tenth the age of both, and that the sum of their ages was 93 ; what was the age of each ? Ans, A 12 years, B 18 years, C 63 years old. 103. A schoolmaster, being asked how many scholars he had, said, " U I had as many more as I now have, J as many, ^ as many, -^ and { as many, I should then have 435 ;" what was the number of his pupils ? Ans, 120l 104. A and B commenced trade v/ith equal sums of money ; A gained a sum equal to -^ of his stock, and B lost $ 200 ; then A's money was double that of B's ; what was the stock of each ? By the condition of the question, one half of f, that is, ^ of the stock, is equal to -I of the stock, less $ 200 ; conse- quently, $ 200 is f of the stock. Ans, $ 500. 105. A man was hired 50 days on these conditions, — that, for every day he worked, he should receive $ '75, and, for every day he was idle, he should forfeit $ '25 ; at the ex- piration of the time, he received $ 27'50 ; how many days did he work, and how many was he idle ? Had he worked every day, his wages would have been $ '75 X 50 131 $ 37'50, that is, $ 10 more tlian he received ; but everyday he wa.s idle lessened his wages $ '75 -j- $ ''^^ =:= $ 1 ; consequently he was idle 10 days. Ans, He wrought 40, and was idle 10 days. 106. A and B have the same income; A saves -J of his; but B, by spending $ 30 per annum more than A, at the end of 8 years finds himself $ 40 in debt ; what is their income, and what does each spend per annum ? Ans, Their income, $200pjr annum; A spends $175, and B $205 per unniim. 107. A man, Ijing at tlic p^nnt of death, left to his three ious his property ; to A J wanting $ 20, to B ^, and to C IT 119. MISCELLANEOUS EXAMPLES. 2i1f the rest, which was $ 10 less than the share of A ; what was each one's share ? Ans. $ SO, $ 50.and $ 70. 108. There is a fish, whose head is 4 feet long; ; his tail is as long as his head and ^ the length of his body, and his body is as long as his head and tail; what is tlie length of the fish ? The pupil will perceive, that the length of the body is ^ the length of the fish. Ans. 32 feet 109. A can do a certain piece of work in 4 days, and B can do the same work in 3 days ; in what time would both, working togethei, perform it ? Ans. 1^ days. 110. Three persons can perform a certain piece of work in the following manner : A and B can do it in 4 days, B and C in 6 days, and A and C in 5 days : in what time can they all do it together ? A7is. 3-^^ days. 111. A and B can do a piece of work in 5 days ; A can do it in 7 days ; in how many days can B do it ? Aiis. 17J^ days. 112. A man died, leaving $ 1000 to be divided between his two sons, one 14, and the other 18 years of age, in such proportion, that the share of each, being put to interest at 6 per cent., should amount to the same sum when they should arrive at the age of 21 ; wliat did each receive ? ^W5. The elder, $546453+ ; the younger, $453^846+. 113. A house being let upon a lease of 5 years, at $60 per annum, and the rent being in arrear for the whole time, what is the sum due at the end of the term, simple interest being allowed at 6 per cent. ? A7is. $ 336. 114. If 3 dozen pair of gloves be equal in value to 40 yards of calico, and 100 yards of calico to three pieces of satinet of 30 yards each, and the satinet be worth 50 cents per yard, how many pair of gloves can be bought for $ 4 ? Ans. 8 pair. 115. A, B, and C, would divide $ 100 between them, so as that B may have $3 more than A, and C$4 more than B ; how much must each man have ? Ans. A $ 30, B $ 33, and C $ 37. 116. A man has pint bottles, and half pint bottles; how much wine will it take to fill 1 of each sort ? how much to fill 2 of each sort ? how much to fill 6 of each sort? 117. A man would draw off 30 gallons of wine into 1 nint and 2 pint bottles, of each an equal number; how 248 MISCELLANEOUS EXAMPLES. IT 119. many bottles will it take, of each kind, to contaia the 30 gallons ? ^ Ans, 80 of each, 118. A merchant has canisters, some holdings pounds, some 7 pounds, and some 12 pounds ; how many, of each an equal number, can be filled out of 12 cwt. 3 qrs. 12 lbs. of tea ? Ans. 60. 119. If IS grains of silver make a thimble, and 12 pwts. make a teaspoon, how many, of each an equal number, can be made fromlSoz. 6 pwts. of siher? Ans. 24 of each. 120. Let 60 cents be divided among three boys, in such a manner that, as often as the first has 3 c^^nts, the second shall have 5 cents, and the third 7 cents ; how many cents will each receive ? Ans. 12, 20, and 28 cents. 121. A gentleman, having 50 shillings to pay among his labourers for a day's work, would give to every boy 6 d., to every woman 8 d., and to every man 16 d. ; the number of boys, women, and men, was the same ; I demand the number of each. Ans. 20. 122. A gentleman had 7 £. 17 s. 6 d. to pay among his labourers ; to every boy he gave 6 d., to every woman 8 d., and to every man 16 d. ; and there were for every boy three women, and for every woman two men ; I demand the num- ber of each. Ans. 15 boys, 45 women, and 90 men. 123. A farmer bought a sheep, a cow, and a yoke of oxen for $ 82'50 ; he gave for the cow 8 times as much as for the sheep, and for the oxen 3 times as much as for the cow ; how much did he give for each ? Ans. For the sheep $ 2'50, the cow $ 20, and the oxen $60. 124. There was a farm, of which A owned f, and B ^, the farm was sold for $ 1764; what was each one's share of the money ? Ans. A's $ 504, and B's $ 1260. 125. Four men traded together on a capital of $3000, of w^hich A put in ^, B ^, C ^, and D y^ ; at the end of 3 years they had gained $2364; what was each one's share of the gain? ( A's $1182. . ) B's $ 591. ^"^* J C's $ 394. ( D's $ 197. 126. Three merchants accompanied ; A furnished f of the capital, B f, and C the rest; tliey gain $1250; what i[ 119. MISCELLANEOUS EXAMPLES. 249 part of the capital did C furnish, and what is each One'rf share of the gain ? Ans. C furnished -/j^ of the capital ; and A's share of the gam was $500, B's $ 468'75, and C's $281'25. 127. A, B, and C, traded in company; A put in $ 500, B $f350, and C 120 yards of cloth ; they gained $332^50, of which C's share was $ 120 ; what was the value of C's cloth per yard, and what was A and B's shares of the gain ? Note, C's gain, heing $ 120, is f JJg^- = y^j of the whole gain : hence the gain of A and B is readily found ; also the price at w^hicli C's cloth was valued per yard. (C's cloth, per yard, $4. Am, < A's share of the gain, $ 125. f B's do. g 87^50. 128. Three gardeners, A, B, and C, having bought a piece of ground find the profits of it amount to i20iS. per annum. Now the sum of money which they laid down was in such proportion, that, as often as A paid 5£ ., B paid 7£ ., and as often as B paid 4ciC., C paid 6JS. I demand hovt^ much each man must have per aniium of the gain. Note. By the question, so often a-^ A paid 5 j£ ., C paid f of 7£. Ans, A2G£, 13 s. 4 d., B 37je. 6 s. 8 d., C b6£. 129. A gentleman divided his fortune among his so~#Sy giving A 9i£. as often as B b£,, and C 3j2. as often as B 7£,j C's dividend was 1537|J^.; to what did the whole estate amount? ^1//^. 11583ciK. S s. 10 d. 130. A and B undertake a piece of work for ^^54, on which A employed 3 hands 5 days, and I^ employed 7 hands 3 days ; what part of the work was done by A, Vv^hat part by B, and what was each one's share of the money ? Ans. A-j^, and B-/2-; A's money $22^50, B's $3V50. 131. A and B trade in company for one year only; on the tirst of January, A put in $ 1200, but B could not put any money into the stock until the first of April; what did he then put in, to have an equal share with A at the end of the year? Ans, $ 1600. 132. A, B, C, and D, spent 35 s. at a reckoning, and, be- ing a little dipped, agreed that A should pay f , B ^, C ^, and D ^ ; wliat did each pav in this proportion ? Ans, A 13 s. 4 d., B 10 s., C 6 s. 8 d., and D 5 s. 133. There are 3 horses, belonging to 3 men, employed to draw a load of plaster from Boston to Windsor for $26'45; 250 MISCELLANEOUS EXAMPLES. IT 119/ A and B's horses together are supposed to do | of th« work, A and C's j^^j, B and C's ^g ; they are to be paid proportionally ; what is each one's share of the money ? ( A's $ll'oO (=^^0 Ans, \ B's $ 5^75 {=z J^,) ( C's $ 9^20 {= ^V) Proof, $ 26'45. 134. A person, who was possessed of f of a vessel, sold 1^ of his share for 375 iS . ; what was the vessel worth ? Aiis, 1500£. 135. A gay fellow soon got the better of f of his for- tune ; he then gave 1500^ . for a commission, and his profu- sion continued till he had but 450c£. left, which he found to be ju"t f of his money, after he had purchased his commis- sion ; what was his fortune at lirsl? Ans, 3780^. 136. A younger brother received 1560iB ., which was just y^ of his elder brother's fortune, and of times the elder brother's fortune was | as much again as the father was ^vorth ; pray, what was the value of his estate ? Ans. 19165 jB . 14 s. 3^^ d. 137. A gentleman left his son a fortune, y^^ of which he spent in three months; f of -| of the remainder lasted him nine months longer, when he had only 537 £. left; what was the sum bequeathed him by his father ? Ans. 2082 i3. 18 s. 2-j\ d. 138. A cannon ball, at the first discharge, flies about a mile in eujht seconds ; at this rate, how long would a ball be in p.^ssing from the earth to the sun, it being 95173000 miles distant? Alls. 24 years, 46 days, 7 hours, 33 minutes, 20 seconds. 139. 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? Am. 19000. 140. A and B cleared, by an adventure at sea, 45 guineas, which was 35 iB. per cent, upon the money advanced, and with which they agreed to purchase a genteel horse and carriage, whereof they were to have the use in proportion to the sums adventured, which was found to be 11 to A, as often as 8 to B ; what money did each adventure ? Ans. A 104£. 4 s. 2-\^- d., B 75£. 15 s. 9-9^ d. 141. Tubes may be made of gold, weighing not more than at the rate of j-^^-^ of a grain per foot ? what would be the weight of such a tube, which would extend across the IT 119. MISCELLANEOUS EXAMPLES. S51 Atlantic from Boston to London, estimating the distance a| 3000 miles ? Aiis. 1 lb. 8 oz. 6 pwts. 3^^ gra 142. A military officer drew up his soldiers in rank and file, having the number in rank and file equal ; on being reinforced with three times his first number of men, he placed them all in the same form, and then the number in rank and file was just double what it was at first ; he was again reinforced with three times his whole number of men, and, after placing tbem all in the same form as at first, his number in rank and file was 40 men each ; hi>vv many men had he at first ? Ans, 100 men. 143. Supposing a man to stand SO feet from a steepje, and that a line reaching from the belfry to the man is just 100 feet in length ; the top of the spire is 3 times as high above the ground as the steeple isj what is the height of the spire ? and the length of a line reaching from the top of th« spire to the man? See IT 109. Ans. to last J 197 feet, nearly. 144. Two ships sail from the same port ; one sails direct- ly east, at the rate of 10 miles an hour, and the other direct- ly south, at the rate of 7^ miles an hour; how many miles apart will they be at the end of 1 hour ? 2 hours ? 24 hours ? 3 days ? Ans. to last^ 900 miles. 145. There is a square field, each side of which is 60 rods ; what is the distance between opposite corners ? Ans. 70*71 + rods. 146. What is the area of a square field, of which the op- posite corners are 70'71 rods apart ? and what is the length of each side ? Ans. to last^ 50 rods, nearly. 147. There is an"* oblong field, 20 rods wide, and the dis- tance of the opposite corners is 33^ rods ; what is the length of the field ? its area ? Ans. Length, 26§ rods; area, 3 acres, 1 rood, 13 J rods. 148. There is a room 18 feet square; how many yards of carpeting, 1 yard wide, will be required to cover the floor of it ? 182 _, 324 ft. zzz 36 yards, Ans. 149. If the floor of a square room contain 36 square yards, how many feet does it measure on each side ? Ans. 18 feet When one side of a square is given, how do you find its Ofca, or supptficial contents ? W?ien the tbtea^ or supetficial contents, of a square is g»V3n> lioiw do you find one side 1 tb2 MISCELLANEOUS EXAMPLES. TT 119, 150. If an oblong piece of ground be SO rods long and 20 rods wide, wliat is its area? A^ole, A Parallelogram, or Oblongy has its opposite sides equal and par^ allt'l, but the adjacent sides unequal. Thus ABC D is a parallelogram, £ A F 15 and also E F C D, and it is easy to see, that the contents of both are equal. Ans. 1600 rods, zz: 10 acres. 151. What is the lenr»h of an oblong, or parallelogram, whose area is 10 acres, and whose breadth is 20 rods? Ans. 80 rods. 152. If the area be 10 acres, and the length 80 rods, what is the other side ? When the lenr/th and breadth are given, how do you find the area of an oblong, or parallelogram ? When the area and one side are given, how do you find the other side ? 153. If a board be 18 inches wide at one end, and 10 inches wide ?.t the other, what is the mean or average width of the board ? Ans. 14 inches. When the greatest and least width are given, how do you find the mean width ? 154. How many square feet in a board 16 feet long, 1*8 feet wide at one end, and 1'3 at the other ? Mean width, J-L+ ^'^ ~ 1*55 ; and 1*55 X 16 = 24*8 feet, Ans. ^ 155. What is the number of square feet in a board 20 feet long, 2 feet wide at one end, and running to a point at the other ? Ans. 20 feet. How do you find the contents of a straight edged board, when one end is widei than the other ? If the le;igth be in feet, and the breadth in feet, in what denomination w^iil the product be ? If the length be feet, and the breadth htcheSy Vf)i2X parts of afoot will be the product? 156. There is an obloiig field, 40 rods long and 20 rods wide ; if a straight line be drawn from one corner to the op- posite corner, it will be divided intx) two equal right-asigled Uiangles ; what is the area of each ? Ans. 400 square rods = 2 acres, 2 roods IT 119. MISCELLANEOUS EXAMPLES. 25S 157. What is the area of a triangle, of which the base is 30 rods, and \\\(t perpendicular 10 rod.s? Ans. 150 rods. 158. If the area be 150 rods, and the base 30 rods, what is the perpendicular ? Ans, 10 rods. 159. If the perpendicular be 10 rods, and the area 150 rods, what is the base ? Am. 30 rods. When the legs (the base and perpendicular) of a right- angled triangle are given, how do you find its area ? When the area and one of the legs are given, how do you find the other leg ? Note, Any triangle may be divided into two right-angled triangles, by drawing a perpendicular from one corner to the opposite side, as may be seen by the annexed figure. Here A B G is a triangle, di- vided into two right-angled trian- gles, A d C, ard d B C; there- fore the whole base^ A B, multi- plied by one half the perpendicular d C, will givev the area of the whole. If A B = 60 feet, and d 1 inch ; therefore, the ratio will be found by dividiiig the circumference of a circle^ whose diaineler is twice the length of the levery by the distance FORMS OF XOTES. 259 between the threads of the screw, 120 X 3^ = 377^ cir- 3774. cumference, and ^ =. 377f , ratio, Ajis, 200. There is a screw, whose threads are ^ of an inch asunder; if it be turned by a lever 10 feet long, what weight will be balanced by 120 pounds power? Ans. 30171 pounds. 201. There is a machine, in which the power moves over 10 feet, wliile the weight is raised 1 inch ; what is the power of that machine, that is, what is the ratio of the weight to the power ? Ans, 120. 202. A man put 20 apples into a wine gallon measure, which was afterwards fdled by pouring in 1 quart of water; required the contents of the apples in cubic inches. Ans. 173|- inches. 203 A rough stone was put into a vessel, whose capaci- ty was 14 wine quarts, which was afterwards filled with 2^ quarts of water ; what was the cubic content of the stone ? Ans, 664^ inches. foruss of sroTsa bonds, re- CBXPTS, AMU ORSBRS. NOTES. No. I. Ovcrdean, Sept. 17, 1802. For value received, I promise to pay to Oliver Bountiful, or order, sixty-three dollars fifty-four cents, on demand, with interest after three months. William Trustyi Attest J Timothy' Testiiviony. No. II. Bllfort, Sept. 17, 1802. For value received, I promise to pay to 0. R., or bearer, " " dollars cents, three months after date. Peter Pencii?* 260 FORMS OF NOTES. No. III. By two Persons, Arian, Sept. 17, 1802. For value received, we, jointly and severally, promise to pay to C. D., or order, dollars cents, on demand, with interest. Alden Faithful. Attesty Constance Adley. Jajmes Fairface. Observations, 1. No note is negotiable unless the words " cr order^^^ other- wise " or bcarer^^^ be inserted iri it. 2. If the note be written to pay him " or order,^^ (No. I.) then Oliver Bountiful may endorse this note, that is, write his name on the backside, and sell it to A, B, C, or whom he pleases. Then A, who buys the note, calls on William Trusty for payment, and if he neglects, or is unable to pay, A may recover it of the endorser. 3. If a note be written to pay him " or bearerj^^ (No. II.) then any person, who holds the note, may sue and recover the same of Peter Pencil. 4. The rate of interest, established by law, being six per cent, per annum^ it becomes unnecessary, in writing notes, tO mention the fate of interest ; it is sufficient to write them for the payment of such a suni, with interest, for it will be understood legal interest, which is si^l per cent. 5. All notes are either payable oh demand, or at the ex- piration of a certain term of time agreed upon by the parties, and mentioned in the note, as three months, a year, &c. 6. If a bond or note mention no time of payment, it is always on demand, whether the Words "ow demand'''^ be expressed or not, 7. All notes, payable at a certain time, are on interest a» soon as they become due, though in such notes there be no mention made of interest. This rule is founded on the principle, that every man ought to receive his money when due, and that the non- payment of it at that time is an injury to him. The law, therefore, to do him justice, allows him interest from the time the money becomes due, as a cou.pensation for the injury. 8. Upon the same principle, a note, payao'*^ on demand, without any mention made of interest, is on inioicst after » f'OKMS OF BONDS. 261 demand of payment, for upon demand snch notes imme- diately become due. 9. If a note be given for a specific article, as rye, payable in one, two, or three months, or in any certain time, and the signer of such note sutfers the time to ehipse without de- livering such article, the holder of the note w411 not be obliged to take the article afterwards, but may demand and recover the value of it in money. BONDS. A Bond, with a Condition, from one to another. Know all men by these presents, that I, C. D. of, &c., in the county of, &c., am held and firmly bound to E. F., of, &c., in two hundred dollars, to be paid to the said E. F., or his certain attorney, his executors, administrators, or assigns, to which payment, well and truly to be made, I bind myself, my heirs, executors and administrators, firmly by these presents. Sealed with my seal. Dated the eleventh day of , in the year of our Lord one thousand eight hun- dred and two. The Condition of this obligation is such, that, if the above- bound C. D., his heirs, executors, or administrators, do and shall well and truly pay, or cause to be paid, unto the above- named E. F., his executors, administrators, or assigns, the full sum of two hundred dollars, with legal interest for the same, on or before the eleventh day of next en- suing the date hereof, — then this obligation to be void, or otherwise to remain in full force and virtue. Signed, &c. A Condition of a Counter Bond, or Bond of Indemnity, where one man becomes hound for another. The condition of this obligation is such, that whereas the above-named A. B., at the special instance and lequest, and for the only proper debt of the above-bound C. D., together with the said C. D., is, and by one bond or obligation bear- ing equal date with the obligation above-written, held and firmly bound unto E. F., of, &:c., in the penal sum of dollars, conditioned for the payment of the sum of, &c., with legal interest for the same, on the day of ■ ' ■ 862 J'ORMS OF RECEIPTS. next ensuing the date of the said in part recited obligation, as in and by the said in part recited bond, with the conditiop thereunder written, may more fully appear ; — if, therefore, the said C. D., his heirs, executors, or administrators, do and shall well and truly pay, or cause to be paid, unto the said E. F., his executors, admini.'strators, or assigns, the said sum of, 6ic., with legal interest of the same, on the said day of, &c., next ensuing the date of the said in part re- cited obligation, accordirg to the true intent and meaning, and in full discharge and satisfaction of the said in part re- cited bond or obligation, — then, &c. — otherwise, &c. Note, The principal difference between a note and a bond is, that the latter is an instrument of more solemnity, being given under seal. Also, a note may be controlled by a special agreement, different from the note, whereas, in case of a bond, no special agreement can in the least control what appears to have been the intention of the parties, as» expressed by the words in the condition of the bond. -iversity) receipts. ^/ - rr-^^,^ Sitgrieves, Sept. 19, 1802. Received from Mr. Durance Adley ten dollars in full of aH accounts. Orvand Constance. ^' Sit^ieves, Sept. 19, 1802. Received of Mr. Oiivand Constance five dollars in full of all accounts. Durance Adlev. Receipt for Money received on a Note. Sitgr-eves, Sept. 19, 1802. Received of Mr, Simpson Eastly (by the hand of Titus Trusty) sixteen dollars twenty-five cents, which is en- dorsed on his note of June 3, 1S02. Peter Cheerful. A Receipt for Money received on Acco'unt^ Sitgricves, Sept. 19, 1802. Received of Mr. Grand Landike fifty dollars on ac- count Ei PRO Slackley, FORMS OF RECEIPTS, &C. BOOK-KEEPING. 203 Receipt for Money received for another Person. Salem, Aug. 10, 1827. Received from P. C. oae hundred dxjilars for account of J. B. Eli Tuuman. Receipt for Interest due on a Note. Amherst, July 6, 1827. Received of I. S. thirty dollars, in full of one year's in- terest of $ 500, due to me on the day of • last, on note from the said I. S. Solomon Gray. Receipt for Money paid before it becomes due. Hillsborough, May 3, 1827. Received of T. Z. ninety dollars, advanced in full for one year's rent of my farm, leased to the said T. Z., ending the first day of April next, 1828. Honestus James. Note. There is a distinction between receipts given in full of all accounts^ and others in full of all demands. The former cut off accounts only ; the latter cut off not only ac- counts, but all obligations and right of action. ORDERS. Archdal*?, Sept. 9, 1802. Mr. Stephen Burgess. For value received, pay to A* B., or order, ten dollars, and place the same to my account Samuel Skinner« Pittsburgh, Sept. 9, 1821. Mr. James Rorottom. Please to deliver Mr. L. D. such goods as he may call for, not exceeding the sum of twenty- five dollars, and place the same to the account of your humble servant, Nicholas Reubens. It is neceesary that every man should have some regular, uniform method of keeping his accounts. What this method ghall be, the law does not prescribe ; but, in cases of dis- pute, it requires that the book, or that on which the charges were originally made, be produced in open court, when b^ will be required to answer to the following questiong :— m4 BOOK-KEEPING. 1 - >^':S -9 0) o O g- o O C P ,£5 §2 c r •S '^ '5 4) •ij ^1 ^ -E «* ^ ft) o o o s :s t; n P3 n ^00 OS Oi §■ lO O o « 1> iO o oi ■^ on fi cc CO -1 f-4 t>^ ^■^ i o s i-i «4M 1 ?2 •s •«*» (n -^ o s ird CO »-• 1 o o -a o 50 t 1 1 'c3 § CO ■^ ^ iii "^ o « fl ^ > ft •^ o c;> CD CO 1> lO S ^ 00 f'l Sr >» rH CTJ VH ;3 A H5 r*:i ^. YA 02387 ,'.-./• * * M \e ev : it ^