Library of Emory University 'rroofj J'UL 5 1934 THE SOUTHERN ARITHMETIC. . A PHILOSOPHICAL AND PRACTICAL TREATISE €JU THE - • SCIENCE OF NUMBERS. DESIGNED FOB THE USE OF SCHOOLS, BY MORGAN H. LOONEY. PRINCIPAL OF THE FAYETTEVILLE SEMINARY, , FIRST EDITION. NEWNAN, GA. PUBLISHED B.Y J. A. WELCH, 1858. Entered, according to the Act'of'Congress, in the year 1858r by Morgan H. Looney, in the Clerk's Office of the District Court of the United States, in and for the Northern District of Georgia. PREFACE. To educate is to develop; and all education is barren and worthless—and indeed: no education at all—that fails to develop; to strengthen; and to mature thereasoning powers. A system of education then. ..which but strengthens the memory by the retention of facts, notwithstanding that memory is indispensable and facts necessary, is inadequate to the great objects of education. That system only is sub¬ lime which, through its teachings and agencies, develops the reasoning powers of the human mind. The science of num¬ bers, and the art of calculation by them, have from the ear¬ liest ages till the present day formed an important branch in the pursuits- of the student, and the aspirer after knowh edge. ■. Because, the study of numbers, their relations, their dependencies, and their powers, of all others is best adapt¬ ed to fix thc^attention, concentrate the thoughts, and develop the mental resources. The generally taught rules of arith¬ metic arc but parts of the great whole of mathematical science —^fragments, if you please, of'the stupendous quarry, but out from the fagments, if. rightly put together, rises the noblest of all statu'es, the statue of human thought and hu¬ man reason. But the importance of the study of arithme¬ tic, as well as the best methods of teaching it, is too often misappreoiated and misunderstood. The memorising of i;ulcs and processes is with too many teachers the great de¬ sideratum ; and so, while facts are piled mountain high upon the memory, reason slumbers in inglorious ease, and in lethargic stupob. It has seemed to the author that a 6 PREFACE* work, combining the synthetic and analytic methods of reasoning has been a want of ourschools, a work adapted to our own southern latitude, a want of our people. He hps ventured therefore to put forth this hook, and has endea¬ vored, while carrying the student through the usual routine of rules and processes, to instil into his mind a love of in¬ quiry after the reason of things. In his "Solutions of Va¬ rious Problems" it has been his aim particularly to lead the mind of the student into a habit oH' reasoning—reasoning upon premises that are beautiful in their simple truth, ben. eficial in their- itse, and sublime in their effect. He claims not fa have invented, or created any thing new in this splen¬ did science. To be original in a work on arithmetic is a well know*n impossibility. The principles herein embodied are as old as the hills, and have been taught in one way or other for ages. It is only in their combination, arrange¬ ment, and adaptation to particular localities, that there can be predicated any claims to originality. To the schools and the people, then, of his native South he commends this work, hoping that it may receive from them the same gen¬ erous, and kind consideration, which he himself has met with in the few years of his labor in the great field of learn¬ ing and teaching. He can but promise in conclusion, that should this work become a test book in Southern Schools, he will be ever ready and willing, by giving due and re¬ spectful consideration to suggestions from" teachers and those interested, to criticise its defects, and will spare no pains or labor to make the second edition a vast improve¬ ment on the first. THE AUTHOR. FayetteVILLE, Ga., January 5, 1S5S. CONTENTS. PAGE. Numeration 11. Addition < 13. Subtraction 15. Multiplication ,..17. Division ; 21. Tables of Money, Weights, and Measures .'.23. Compound Addition 27. Compound Subtraction ,..30. Compound Multiplication - 32. Compound Division a 35. Recfuetion of Compound Numbers... 37. Fractions 1 41. Addition of Fractions 54. Subtraction of Fractions —...5/'- 'Multiplication of Fractions ..,...6(). Division of Fractions .*6-2. Decimal Fractions 661 Addition of Decimals v.i....6& Subtraction of Decimals. 1.60. Multiplication of Decimal^ 70_ Divisioli of Decimals ..71. CONTENTS. PAC Reduction of Decimals ." 74. Ratio '8- Proportion 70. .Single Puleof TLrce ; 7J. Compound Proportion, or Double Rule of Three— 01. Percentage -• Partnership . — Partnership on Time l-'5. Interest. - I'-'S. Promissory Notes 110. Compound Interest 125. Discount 128. Ranking 1 ,• 131. Tolling 135. Barter 137. Equation of Payments 140. Commission 142. Profit And Loss v-148. Involution 153. Evolution 155. Square Boot 156. Appplication of the Square Boot 159. Application .of Square Boot to Circles 164. Cube Boot 168. Application of Cube Boot 170. A General Ilule For Extracting Any Boot-- 172. Double Position 174. Arithmetical Progression 178. Geometrical Progression 182. Infinite Series 186. Permutations And Combinations--..; 187. Alligation Medial 190. Alligation- Alternate 192. CONTENT^. ' 0 PAlrE. Mensuration 195. Triangles 190, <..) ua< 1 r i I a t a r al s • 200- < 'ircli • 208. Mensuration of Solids 218. Prism* and Cylinders * 218. Pyramids and Cones-5 , 221. spheres 225. Spheroids •. 228. Cylindrical Kings ■••■280. Mensuration of Lumber 231. (iauging 234. Solutions of Problems by Analytical Methods 238. Miscellaneous Examples ---275. Value of oertain Foreign Coins and Currencies-'-' 283. Table showing the weight of a Cubic Foot of certain Substances "-284. ARITHMETICAL SIG^S. = Sign of equality ; 8=8, read eight equals eight, -f- Sign of addition ; 5-(-4=9, read five added to four equals nine, or five plus four equals 9. — Sign of subtraction; 5—4=1, read five minus four is equal to one, or five less four is equal to 1. X Sign of multiplication; 4x^=12, read four multiplied by three is equal to twelve. -7- Sign of division; 8-^-4=2, read eight divided by four is equal to two. The division of one number is also shown by placing one under the other thus J, which signifies that three is to be divided by four. : :: : Sign of proportion; 2:4 :: 8:16, read 2 is to 4 as 8 is to 10. 73 Signifies that 7 is to be raised to the second power, or multiplied into itself. 43 Signifies tliat 4 is to be raised to the third power, or multiplied into its square. -j 710 or 40- denotes that the square root of 16 is to be ex¬ tracted. f 27 or 27" denotes that the cube root of 27 is to be ex- * traded. Note.—Pressing engagements preventing the Author's supervis¬ ing the proof, a few typographical errors have occurred in the printing, the most important of which will be fouud corrected in an Errata at the end of the work. arithmetic. Arithmetic '(from the Greek arithmas, a ■■number,) is thh science of numbers and the'art of calculation.. Addition, Subtraction, Multiplication, and Division are the fundamental rules by. Which the operations of Arithme¬ tic are performed. x u mek atiox. numeration is the process of reading numbprs when expressed by figures.' ' . ■ ( , , ' > • There are ten characters, called humeral figures, used to express numbers in Arithmetic; viz. 1, 2, &, 4, 5, 6, 7, 8,, 9, 0,—which. are read one, two, three, four, five, six, seven, eight, nine, cipher or nought. (. ' The value, which these figures have when connected to¬ gether in a horizontal line, depends on the plaee which they occupy. This may be best illustrated bv a table, thus— 2 12 JftMEBATIOX. O rs B -j. 3 S S § 2= e -2 Q .2 . g «■ g.2 r=g § ^ g C^T: EH ,o ^ ^ pa ^SS 'So . oci-^=:«c<-ic?,+-'!:®P<:<-';:£?;cy2? 4:o-E±:03r3c>=-3 = cr50-2-3 . a: £^1®SsSssJS = SoS=-- § 5^5 S *C o ,c ^ o »- £2 EH C? ^ihhH^nwH^MhHHhf-' 1^2 3 4 5 6 7 8 91 234-5-6 7 89 The table is numerated from right to left, by merely railing the names as they ai-e put down. But it is read from left to right, and reads as follows: One hundred and twenty-three Quadrillions, four hundred and fifty-six Tril¬ lions, seven hundred and eighty-nine Billions, one hundred and twenty-three Millions, four hundred and fifty-six thou- sapd, seven hundred and eighty-nine. ' - • Here any figure occupying the unit's place-has only its simple value; but the figure standing in the second or tens' pi,ace denotes ten times its simple value, in tfie third <>r hundreds' place, a hundred times its simple value, and so on. . . The cipher when standing alone denotes npthing, but when, placed to the right of a significant figure, by moving that figure'one 'place to the left, increases its value iu a ten¬ fold proportion. Thus 5 is simply five, because it occupies the units' place, and has its simple value; but 50 is fifty because the cipher removes the 5 to the tens' place, and it then has ten times its unit value, and ten times 5 is fifty. 'If the aboye table were carried farther to the left by the annexing of other figures at the left, the names of the ADDITION. 13 groups of figures, taking three at' a time, would V as fol¬ lows, beginning where we left off m the table: Quintillions, Bextillions, Septillions, Octillions, Nonillionsy. Decilliohs, UndecillionS, Phodecillions, Tredecillions, Quartuordecil- lions, Quindecillions, &c. But, of course, the Teacher paust practice the student in all this. ' t , ADBITIOK Addition is the process by -which the shm of two or more'numbers is found. RULE. - Set down the given numbers, putting- units under units, tens under tens, &C., Then begin at "the tight hand and add the, figures together which constitute the units column ; if the suiti of the column be less 'than ten; set it down under tJie units column, and proceed to add the column of tens in the. same manner ; if the sum of the column be more than ten set down the right hand, figure, and carry the left hand figure to the next column, and so]on~till every column ha& ix-en added, setting down the whole result, of the last or left htmd lumn. 14 ADDITION. EXAMPLES. . 1.' A man ha$ his hogs in- three fields; in one he has 325, in another 578, and in the other 496/ How many hogs had he ? Ans. 1399.. OPERATION. 325 In this operation the column of units 578' being added makes 19 j we set down the 496 9 (units) and carry the 1 (tens) to the ' ' ■ tens column, and so on. 1399 2„' 3. 4. . 5. <5. . 324 4567 2345 4875 98765 846 3842 4370 .5898 3789G 478. 4789 9084 4321 45678 523 1243 7895 7000 58423 2171 14441 23694 22094 ^ 240762 ' - 1. r 8. 9. » 10. 89764 3284 987432 750321 3245 48568 48329 139 ' 486 724892 498 48 28 38419 7204 3497 • 842 4980 18- 78910 3979 25 4 - 483298 98339 820168 1043485 1316213 11. 12. " 13. 584329 "3456789 12345078 478942 '4829478 . - 23450789 584887 5392709 345.61891 796875 7243218 45078910 450789 0789870 .50789123 "894321 5328941 67891234 SUBTRACTION, 15 14.. Add the following numbers, 538, 4327, 58432, and 19. _ . Ans, 63316. 15. Add>28, 384, 4897,'58785 and 10000 together. • / ' ' Ans. 74094. »" 16. Add three hundred and ten, 4568, 36789, and one million together ■ * . ' . ■ Ans. 1041667. 17.- Add the following numbers, 18782,: 24, 328, 489, 5876, and 1008,, - • Ans. 26507. ■ 18,.Add the following numbers, 87, 18,182,3485, 756, 98428,-'.and 300000. . ' «. Ansi 402956. ' • ,v-19. I buy 55 bushels of Corn from one man, 320 bushels from another,. 1000 bushels from another, 78 bushels from another, and 247- bushels from another.' How Inany bush¬ els did I buy in all? ' Aps/1700. 20. A man made 75barrpls of CJorn on one field, 458 barrels on another field, and 3840 barrels on another field. How much did he make in "all? Ans., 4373. SUBTRACTION. Subtraction is the process of finding the difference be¬ tween two" numbers, by taking the legs from the'greater. The larger number is called the Minuend, the less is' ealled'the Subtrahend, and the result obtained by Subtract¬ ing is called the difference or remainder. ' ' • \ ' RULE. Place the less number under ther greater so that units mag be under .units, tens under tens, &c. Commencing at the right hand subtract edch figure of the subtrahend: from the figure above it. If'any figure of the subtrahend be larger than the figure above, it, add 10 to the upper figure 16 SUBTRACTION. before! subtracting, and then carry one or add 1 to the next fig we of the subtrahend, • EXAMPLES. From 845 In this question 845 is the minuend, and . Take .498 498 the subtrahend, .We proceed to subtract the 3 from the 5 and it leaves 2 Remainder 352 which we write down in the units place ; we then proceed to subtract 9 from 4, but this we can not do because 9 is tbe largest; we then add 10 to the 4 , making it 14, and subtracting the 9 from 14 leaves 5, which we' write down in the tens place; we then add 1 to the 4 making it 5, and subtract it from 8, leaving 8 which we write down in the hundreds place. From 3846 Take 2891 Rem. '955 From 8984 Take 4892 Rem. 4092 4. From 98784 Take 37856 Rem. 60928 5. From 4856 Take 3984 From 28571 Take 3845' From 98643 Take 89784 8. From 109 take 28. _ . Ans. 81. 9. From 3846 tak,e 2084. . Ans. 1762. ' 10. From- 38482 take. 19878. ' _ Ans. 18604. •11. From 1000 take 999., Ans. 1. 12. From 275846 take 37849. Ans. 237997. 13. From 78561 take 69845. Ans. 8716. ■ 14. From 300801 take 20008. Ans. 280793. V 15. If I have 90 bushels of Corn ,and* sell 65 bushels, how much will I have left? Ans. 25 bushels. 16. If from a piece of land containing 540 acres I sell 499 acres, how many acres shall I have left? Ans. 41. MULTIPLICATION. IT 17- A drover has 785 hogs, find sells 149 of them to A, n)d 284 of them to B. How many, will be* left in the Irove.? Ans, 352. 18.. A farmer has 94 shceip in one field, and 119 in kno¬ ttier; if he sells 157, how many will he have remaining? • * Ans. 19. What is the value, of 8975—7857? .Ans. 1118.. i 20. What is the value of 100'—1? " " Ans. 99'.r MULTIPLICATION Multiplication is the process of taking1 one number as many times as there are uilits in'another number. If 5 apples are multiplied by 4 it means simply that 5 apples are taken 4 times. ' • The number to be multiplied or taken is called Multi¬ plicand. i The number by which we-multiply is called the 'Multi¬ plier. , • •' The result obtained by the multiplication is called the Product. _ , The Multiplicand and Multiplier in reference to each other are called Factors, because they make ot produce the produ6t. * ' ' . Sincp the multiplication, of numbers however large de¬ pends upon the product of one digit by another we append a table of the products of each digit by any other, usually called The Multiplication Table, which the student should thoroughly commit to memory.. 18 MULTIPLICATION, TABliE OF PYTHAGORAS, 1- 2 3 4 5 ' 6 7 8 9 10 11 12; 13 14 15 Y Y ■~6 ~8 To Yb Yb Ye ■ 18 ~20 Yb YYjYY .28 Yib 1 ~6 19 12 15 18 21 "24 ~~27 YY "33 36 J 39 42 TE Y "8 1*2 16(20 *24 ~~28 ~32 "36 To YY Yt8,Yb YY 60 ~*5 10 15 20 25 "30 35 40 45 To -55 60 65 To 75 T 12 18 24 30 ' 36 Yb "48 "54 60 ,66 Yb Ys 841 90 I 14 21 28 35 "42, Yb i^6 Yl3 ~70 TiTl 98 105 ~8 16 24 32 40 ,48 56 64 72 "80 B8 96 104 112 120 ~9 18 27 36! 45 t>4 ~63 72 ~81 To Yb' 108 117 126135 To 20 3b 40 50 To "70 Yib Yib lob 110 120 130,71-40* 150 n 22 33 44J 55 ■66 "77 "88 799 Ho 121 132 143,154,165 12 24 36 48 60 Yb ~84 "96 Tob 12b 132 144,156 168.180 13 26 39 52 65 9! 104 117 130 lib 156 169 182 .195 14 28 42 56 7b "84 98 112 126 140,154 168 182196 210 1 i> 30 45 60 75 90 105 120 135 150 165 180.195 210 22? . The produptmust |?e of the same denomination t as tho multiplier, since taking a quantity any number of times doe4 not alter its nature : tli.us 4 an abstract number an abstract number:' so § yards X2=16 yards. ' The multiplier must always be considered abstract, for it simply means to take the multiplicand a certain number of times. Say you have bought 4 horses at 890 each) you find the post of all the horses by'tnu.ltiplyipg 890 by 4, It M.ULTIPJJJCATION; 19 would seem to you perhaps that you are here multiplying dollars by horses. But that is absurd. ■ You consider your 4 abstract. Ifi is abstract in the operation, for it is a num¬ ber showing how many times you will have to take $00 out of your pocket (j'f you please) to pay for the 4 horses. The product of two factors is the same whichever is the multiplier. The cost of two books at '3 dollars each, is just the same as the cost of 3 books at 2 dollars each. The ta* king apy thing a half a time js the same as taking half of that) thing one time. , , , j The product of any number ©f factors is the same; what¬ ever the order ip which they ar . "When! the dividend does not contain the divisor an ex¬ act number of times, the excess is called a remainder j fiut by putting the divisor under it with a short line between them it beconjes a fractional part of thte quotient. Division is the reverse of multiplication. The dividend answers to the product, and the divisor and quotient to the factors, which would make that product. . , To divide simple, numbers the following, is tho RULE. , - 4 ■ . Begin /ft' iJie left hand and 'f ind how many times the divisor is'contained in the fewest figures of the divi¬ dend, that will contain it, -far the first quotient figure. Multiply the divisor by this quotient figure, and subtract the product from the figures of the dividend used. With the remainder un ite the next figure of the dividend ; and find how many times the divisor is contained in the number thus formed for a new quotient figure, which write at the right of the former quotient figure. t And so on till all the figures of the dividend are divided. • * examples. 1. Divide 498-4 by 7 Ans. 712. '22 pivisiott. OPERATION. Divisor 7)4984 Dividend. 712 Quotient. . 2, Divide 2684 by 17' Ans. 21Q{|« OPERATION.1 . Divisor 17)3684(216 Quotient 34 28 17 114 102 ' )L2 Remainder. ' 8.< Divide 576 by 12. • 'Ans. 48. 4. Divide 8424 by 8. ' Ans. 1053. 5. Divide! 3687 'by 5. Ans. 737 A 6. Divide 1248 by4. Ans. 812. v7. Divide 78569 by.9. . t Ans. 8729|. . 8. Divide 24l2526270 by 10. , , , Ans. 24252627. 9. Divide 78565<3_by.11.. . .Ans. 71423^. 10,. Divide 3742448,by 12. • Ans. 311870^* 11. Divide 576 by 24. * , •> Ans. 24. 12. Divide 2160 by 18. , Ans. 120. Id, Divide 37840 by .20. Ans. 1892. 14. What is the value of 1728 -5-144 ? Ans. 12. 15.. What is the value of 78569 -f-.18 ? Ans. 4364-]?. 16. .What is the value of 437895 -5- 25 ? . Ans. 17ol5|9. 17. What is, the value of 756329 -5- 294? Ans. 2572i||. -MONEY. . 23 18. If $785(1 be divided amoung 60.met), bow many dollars will each man get ? . ' ( .. i. ■ Ans. ^157. 19. If 475894 acres of land, cqst 9321480 dollars, what Was the cost of one acre ? ,, , ^Ans,—^ . 20. If 15 dollars is to be. divided among 16 mfen, how much will each man get T r' •• Ans. If of a dollar. TABLES OF MONEY, WEIGHTS, AND MEASURES! tJNlTED STATES MONEY. 10 Mills mate ' • 1 Cent, marked . c. 100 Cents ( u 1 Dollar, " , $. 10 Dollars " 1 Eagle, " . E* ENGLISH MONEY. 4 Farthings make ' 1 Penny, marked d. 12 Pence ' u 1 Shilling, " S. 20 Shillings" " 1 Pound, " £, 21 Shillings sterling " ^ .1 Guinea' (e - G. Notb..—One Pound Stirling-is equal to $4.84" exchange value. 24 WEIGHTS AND MEASURES. TROY WEIGIHT. 24 Grains make 1 Pennyweight, marked dwt. -20 Pennyweights " 1 Ounce, . " 12 Ounces 1 Pound, oz. lb. Note.—Gold; Silver, and jewels are weighed by this weight. .20 Grains 3. Scruples '8 Drams 12 Ounces APOTHECARIES WEIGHT- , make 1 Scruple, marked sc. .1 Dranr, dr. " 1 Ounce, " -oz. " 1 Pound, , " lb. NotE.^—This is the weight 1>y which apotliacaries mix their me¬ dicines; they buy and sell by Avoirdupois. AyOIRDUPOIS WEIGHT. .16 Drams make 16 Ounces • ' 28 Pounds '' 4 Quarters " '20 IlundredWeitrht" 1 Ounce, marked 1. Pound,.' • 1 Quarter, " 1 HundredWeight" 1 Ton, " oz. >lh. qr. cwt. ton. Dy this weight are weighed almost every kind of goods, and all metals except gold and silver. LONG MEASURE. 3 Barl'ycorns or 12 Lines make I lech - marked in. 12 Inches • 1 Foot " ft. • 3 'Feet " 1 Yard,. " ' yd. . 5} Yards ' "1 Rod or pole, rd" or p. 40 Rods \ "1 Furlong " fur. 8 Furlongs " 1 Mile no 3 Miles " 1 League, " lea . r>V) A IMilcs nearly " 1 Degree, " Deg. 360 Degrees "1 Circle of the Earth. Not a. -A hand is 4 inches ; a fathom 6 fept; a Degree at.tho equator i3 aboilt,69^ miles, but usually considered (>9b MEASURES. 24 Inches 4 Nails. •. 4 Quarters .3 _ Quarters 5 Quarters CROTH MEASURE, make 1 Nail • marked 1 Quarter of a yard " " ,1 yard * • " 1 .Ell Flemish, " ' u 1 Ell English, *5 na. • qr- • yd. E. F. E! E» SQUARE MEASURE.. 144 Square Inches make 1 Square Foot,' marked sq. ft. 9 Square Feet '1 ^ 1 Square Yard H • sql yd. 1 Squarq Rod or Pole' sq. rd. 301 Square Yards <( 40 Square Rods " 4 Roods " 43560 sq. Feet • " 4840 sq. yds, u ■I60 sq. rd> 640 Acres (1 Rood, " 1 Acre, (( J u U q 'u it .'I'" u. v " 1 Square Mile, DRY MEASURE, 2 Pints 8 Quarts ■4 Pecks 36' Bushels 8 . Quarts 4 Pecks? <5 "Bushels make '( , .1 Quart, marked 1 Peck, " \1 Bushel, " 4 Chaldrofy u ■ ' . UORN MEASURE; make '" 1 Peck, " I Bushel 1 Barrel marked A. m. qt.. _pk. bti. ch pk. • bu. bar. Note,—A Winchester Bushel is 18£ inches in diarrieter, and 8' inches deep, apd contains 2150.4252 cubic inehes.. 2 Pints *4 Quarts 54 Gallons 2 Hogsheads 2 Butts ARE AND BEER MEASURE, make 4 Quart, • 1 Gallon, 1 Hogshead, marked Butt, Tun, qt- gal. hhd. butt tun. Note.—The ale gallon contains 282 cubic inches. 26 MEAStjfiE, ?IME, ANi)'MOTION, ^ COMMON UlQtftD MEASURE.. .4 Gills • ' % Pints 4 Quarts 42 Gallons - 63 Gallons 2 Tierces / • 2 Hogsheads 2 Pipes ni^ke Pint Qtiart Gallon Tierce IJogshead Puncheon . Pipe or Butt Tun ' zharked H pt. qt. gal. tier, hhd. pun. .pi- tun. Note.—The common wine gallon contains 231 cubic inches. GO Seconds 00 Minutes 24 Hours 7 Bays 4 Weeks 12 Months i • TIME., Uiake 1 Minute, . marked in. " ' 1 Hour,, " h. " 1 Day, " • d. u 1 Week, "• w. " 1 Month,' ' ' " ' mo. ft 1 Year, w " y- Note.—This table ig-in common use," but not correct in several particulars. In calculating .itllerest it is usual to count 30 days to the month,-and 12 months to the year^ t . The days in-each month really areas follows: January, March.. May, July, August, October, and December have each 31 days; April, ^une, September and November have pach 30 days ; Fel>» l-uary has 28 days, excepting Leap-year When ft has 23. . CIRCULAR MOTION. 60 Seconds (or 60") make 1 Prime minute, marked ' 60 Minutes " 1 Degree, 11 ° 30 Degrees " 1 Sigiq " s 12 Signs, or 360 Deg's "■ • 1 Circle. • > " c. COMPOUND ADDITION. 27 - - surveyors' measure.. 7jqq Inches , . make 1 Link, marked* 1- 25 Links ■ t u: 1 Pole,.' " ' V/ 100 Links, 4- Poles, or 66 Feet J Chain, , " , el?. 10 Chains' " 1 Furlong, . " fur. 8 Furlongs , " .1 Mile. surveyors' square measure. 10000 Square Links . make 1 Square. Chain. 10 Square Chains * '".-1 Acre. 0100 Square Chains , t _ _ " 1 'Square Mile.' . v ^ ^cubic or solid measure.' ' '' * ' .' 1728 Cubic Inches make 1 Cubic Foot, marked cu. ft. % 27 Cubic Feet 1 Cubic Yard, " .cu. yd. 128 Cubic Feet " 1 Cord of Wood. Note.—A pil-fe of wood 8 feet long, 4 feet wide^ and 4 feet liigh, is a Cord. ' COMPOUND ADDITION. When we have several numbers of different denomina¬ tions which We wish to add together, we call the process by which this is done Compound Addition,. Compound Addition, then, is the process of finding the amount of two or more compound numbers. , 3 28 COMPOUND ADDITION.' . 4 'RULE. ' IYritputt the given numbers so thai units of the same d<" 'nomination may stand in the same column. Add the right hand column or columns, v:hich is the lowest denomination, as in Simple Addition set the sum off and divide. it by as many as-it takes of that denomination to ninxke 1 of the next higher denomination. Write the remainder under the column added, and carry the quotient to the column of the next higher denomination. Proceed in the same manner till the la^t or left hand, column, which represents the high¬ est dehorQ-fnation, is added, which last sum vrrlte down as in Simple Addition., EXAMPLES. Adding the column of pence the sum is 27. yfa divide this 27 by 12 (because 12 pence make a shilling) and the quotient is 2, remainder 3. We set down the 3 and carry the 2. The sum of the two columns of shillings is 15 wliieh we divide by 20 (because 20 shillings make a pound) and the quotient is 2, and remainder 5. Set¬ ting down the remainder,, we carry the 2 and add the three columns of pounds as in Simple Addition. VotE.'— Farthings '.are frequently expressed by fractions, thus: one farthing, 1 ;. two farthings, i ; three farthings, f. In such cases consider every J- to be J, and add the upper figures of each fraction) ani\ divide the sum by 4. flic same rule applies to those fractions when they are united with cents in United Statt-4 juoney. . ' ^ £ ' •18 s. 12 fl. a 21 8 4 132 0 8 45 . 10 7 23 4 2 241 5 o O ^ COMPOUND ADDITION. 29 2. £ s. <3. Hhd. gats. qts. • pts P». 13 . 3 .27- 31 3 1 132 18. 1 132 9 2 1 3.20- •4 ,8 / 24 , 45 1 O 75 8 3 895 17 3; ' 1 9 11 10. v " 5 1 0 ' 4. , 5. ' A. R. * po. j(L Lea. m. • fur." p. yd. ft. 384 3 25 ■ 28- 75 2 '7 •'4 '-3, 2 275 '2 18 30 ■ 84 1 6 r29 2 1 942 1 •4. 8 28 2 5 18 4 2 728 3 39 10 134 1 4 32 ' 2 1 273 ^ 2 27 3 1845 2 '3 8 -5 2 6. 7. T. ■ ew.t. qr. . Ib- ©z. dr. fd. qr. na. 5 18 3 A4 12 1$. . 185 3 3 4 10 2 24 >£' 8 284 I 2 8 i 3 .9 5 13 ' 17 2 1 12 14 1 18 15 14 488 1 . 0 u 25 •8' 2 • 7' ' 8 •r 75 3 2 9 • 13 1" 27 >12 •' 3 ■ ■ 8 2' 3 8. Add together 45 Bar. 4 bu. 2 pe ; 18 Bar. 3,bu. 3 ' pe.; 10 Bar. 2 bu.~ 1 pe.^ 7 Bar. 4 bu. 3 ge.; 27 Bar. 3 pe.; 2 Bar. 3 bu.; 18 Bar. ,4 bu..3. pe. Ans, l.?l.Bar. 3 bu. 3 pe. 3* 30 compound subtraction. 9. Add together 1-6 lb. lloz. 19 dwt., 23 gr.; 31 lb. 10 oz. 1$ d'wt. 16 qr.; 63 lb. 9 oz, 12 dwt. 15 gr.; 17 lb. 8 oz. 13 dwt. 12 gr.; 61 lb. 1 oz.,12 dwt. 16 gr.; 1-7 lb. 6 oz. 17 dwt. 22 gr. Ans. 209 lb. 7 oz, 15 dwt. 8 gr. 10. Add together^27£. 8s. 4d.; 93£.'15s. 8d.; 29£. 18s. 9d.; 32£. 7s. lid.; 4£. 6s. 2d. Ans. 187£. 16s. lOd. 11. Add together 21d. 8h. 35m. 28s.; 18d. 20h. 38m. 29s.; 48d. 12h. 14m. 12s.; 30d. 25m. 13s. Ans. 118d. I7h. 53 m. 22s. 12. Add together 8cwt. 3qr. 71b. lOoz. 14dr. ;* 17c-wt. 2qr. 12dr. 271b. 15oz. 9dr.; 38cwt. 3qr.; 5cwt. lqr. 231b. lOdr.. ,, Ans, 70cwt. 3qr. ,21b. lloz. 13dr COMPOUND SUBTRACTION. Compound Subtraction is the process of finding the dilference between two compound numbers. RULE. Write the less number wider the greater so that units of the same denomination shall stand in the same column ; anif subtract as in. Simple Subtraction. Tf airy number in the subttahend is larger than the number above it add > 'COMPOUND SUBTRACTION. 31 to the upper number as many as malce< 1 of jthe next higher denomination/, Arid carry 1 to the next number of the.sub¬ trahend. EXAMPLES, ' . •l. £. s." a.' Froip '27 8 9'' We subtract the 8 pence from the 9 l"ake 18 . 9 8' pence;, and '-the- remainder is' 1, .which -we set dqwn. But we cannot subtract 8 19 1 the 9 shillings from the 8 because the 9 is .the larger, We therefore add 20 (be¬ cause 20 shillings make a pound) to -the 8, making it 28; then 9 from 28 leaves 19 which We write down. We car¬ ry 1 to the 18 making it'19,. and subtracting 19 from 27 we-have 8 remaining which we write down. 2. , 3. Yd. qr. na. A. ft, po! sq. yd. sq.ft. sq. in. 13 3 2 18 2 10 4 ' 1 8 8 2 3. 10 . 8 ., 24 2 2 . 7 5 ' 0 3 7 2 20 1 8- 1 i • . 4. From 28£. 18s. 8d. take 18£. 4s. lOd. " Ans. 10£. 13s. lOd. 5. From I857y. 3mo. 20dv take 1843y. 8ino. 24d'.. Ans. 13y. 6mo. 26d. ' 0. From 19cwt. 3qr. 181b., take 12cwt. lqr. 241b. Ans. 7cwt. lqr. 221b. 7. From 25hhd. l7gal. 3qt., take 62gah 2qt. Ans. 24hhd. 18gal. lqt. - 32 compound multiplication. 8.' From 38bu. 3pe. 5qt., take 20bu. 3pe. 7qt. k ' Ans. 17bu. 3pe. 6.qt. > 9. From 25fb. 8oz. 4dr., take 241b. 9oz 4dr. • 1 ' Ans. 15oz. 10. From 51b. 7oz. 10diH..18grv take 4lb. 8oz. 3dwt. And.'lloz. 7dwt. 18gr. 11. From 112A. 3R. 18pov take 20A. 2R. 28po. A'ns. 92A._ 30po. -12. From lSeii.yd. 20e.ii.ft. 1443cu. in., take 26cu.fL iOOOcu.in. Ans. 17c». yds; 21cu. ft. 443cu. in. • COMPOUND MULTIPLICATION. - Compound Multiplication is tbe process oF taking a compound number a proposed number of times. ) RULE. *■ *■ ' Multiply each, denomination of the,-compound number, as in multiplication of simple numbers^ beginning at the right hand, and carry as in Compound Addition. examples. 1. What will 7 bags of cotton cost at 13£. 9s. 8d. per bag? • Ans. 94£. 7s. 8d. compound - multiplication. .operation. , Having written the multiplier under £. s. d. : the lowest denomination of the multi- 13 9 3 ■ plicand,'we first multiply' 8d. by 7 7 ' .which gives 56d.; dividing this 56 by r^r ' • 12 (because 12"d, make a shilling) we 94 7 8 . -firjd it gods '4 times, and 8d. remainder. We set down the\8cf. and carry the 4 to the next product, &c. t , 2. Multiply 12A. „3R. 18po.; by 6. Ans. 77A. OR. 28|>o, 3. Multiply^ 17yd. 3qr. ,2na., by 8. . " , ' /' . , * Aris. 143yd, Oqr. Oua. * 4. Multiply 24bar. 3bu. 2pk,, by 9- . Ans. 222bar. ibu.2pki 5'. Multiply 8hhd. 31 gals. 3qt. Ipt., by 7. Ans. 59hhd.' 34gal,:0qt. Ipt. 6. Multiply 27yd, 2ftj Oin., by 12. .;v •' ^ ■ Ans,, 335yd, Oft, Oin, Multiply 9cwt. '3qt. -101b,, by 27. Ans. 265cwt. 2qr, 18ib, S - ) operation. • We first multiply 101b! by 27, which gives (!wt. qri 11). '270; we divide 270 by 28 ; it is contained 9 9' 3 10 times and 18 remainder; we write down tie 27' 18, and multiply' 3qv.' by 27, which gives '81, — :— and taking in the 9 we have 90 ; we divide 265 2 18 this 90 by 4; it is contained' 22 'times and 2 remainder; we- write down the 2, and multi¬ ply 9cwt. hy 27, taking in the 22, and we obtain 265, which we write down. 34' COMPOUND MULTIPLICATION". 8. Multiply 20gal. 3qtvlp4,-by 17. Ans. 354gal. 3qt. lpt. ^ 9. Multiply 5y. 8mo. 20d., by 3.1. ' Ans. 177y. 4mo. 20d: 10.. Multiply- 13b. 32m. 30sec.., by 50. ' . Ans. 67jh. 5m. Osec. 11. Multiply 14A. 311. lO.p. 20sq. yd., by 360. . ^ Ans. 5333A.. SR. 38p. 5s.cp yd. » 12.. Multiply 18k. 2pk. 6qt.,.by 92. Ans. 1719bu. lpk. Oqt. • 13. Bought 18Jiogslieads of sugar eacb weighing 12cwt. 2qr. 141b,what was the weight of the Vhole ? • ' Ans. 227cwt.lqr.01b. • 14.. Bought 18 bags of cotton 'gt 4£. ts. 8d.'per bag; what was'the cost of the whole.? Ans. 78£. 18s. Od. 15. What will 36yds. of cloth cost at 10s. Sd. per yard ? ?' Ans. 19£, 4§. Od. 16. What wilb21 paiv of boots be worth at 9s. Tid.-per pair? , • - Ans.-10£. 2s. 11. 17*1 If 1 Spoon weighs 3oz, 5pwt. 15gr., what will 91 spoons weigh ? Ans. 241b. lOoz. llpwt. 21gr. !8t ' If in a day a traveler goes 121ea, 2mi. 7fur., how- far would he travel in a'month. Ans. 388Iea. 2mi. 2fur. 19. 'If my daily.expenditure be 2£. 17s. 9d-, what Will I spend in a year ? ' ' Ans. 1053£. l8s? 9d. 20. If with. 1 dollar Lean bwy 1A. 2R. 18po. of land, how much can I,buy'with 897? Ans. 156A. IB. 26po, COMPOUND DLVfclON.. 35 COMPOUND DIVISION. Compound Division is the !process of dividing com¬ pound numbers into any. proposed' number of' equal fwtrts. kuiA Begin at tlieflefthand and divide each denomination in order, as in Simple Division. If there be a remainder re¬ duce it to the next denomination, adding in the number already contained in the dividend of thjs denomination, and divide as before examples'. 1 1. Divide 9jL£. 7s. 8d.,.by 7. Ans. 13& 9s. 8d, operation- £. s. d. We find that the dlVisor is 7)94 7 8 contained in the 94£ 13 times 13 D 8 and 3 remainder. We write the 13 ulider the 94, and the remainder 3 we multiply by 20 to reduce it to shillings; 3 X20==60, and taking in the 7 shillings we have 67 as the next dividend. Dividing 67 by 7 we obtain 9 quotient arid 4 remaindei*. We write the 9 under the shillings, and multiply the remainder 4 by 12 to reduce it to pence; 4xl2==48, an-d adding in,the 8 wo have 56, and 56-t-7.=8d. 36 COMPOUND' DIVISION. 2. Divide 77A. OR. 28p. by "6. Ans, 12A. 3R. 18p. 3. Divide 143yds. Oqr. Ona. .by 9, Axis. 17yd. 3qr-. 2na. 4. Divide 222bar. lbu. 2pk. by 9. •Aus. 24bar. 3bu. 2pk. 5. Divide 59hlicL 34gal. Oqt.-lpt. by 7. Ans. *hhd. 31 gal. 3qt. 1ft. (*. Divide 335yd. Oft. Gin. by 12. 1 * ' Ans. 27yd. 2ft. 9in. 7. Divide 265 cwt. 2 qr. ,181b. by 2"t. Ans. 9ewt. -3qr. 101b. OPERATION. "cwt. ' qr. ' lb. •27)265 2 18(9cwt. 243 22 4 90(3qr. 81 9 28 . • 252 . 18 ' 270(101b. ■ 270 • , 8. Divide 354gal, 3qt. Ipt. by >17. • Ans: 20gal. 3qt. Ipt. COMPOUND' DIVISION. 37 ■ 9. Divide 177y. 4mo. 20da. by 31. Ans. 5y. 8mo. 20da. 10. Divide 677h. 5min: Ose'c. by 50. Ans. 13K. §2mi'n- 30sec. 1.1. Divide 5333A, 3R. 38p: Isq. yd. by 360. Ans. 14A. 3R. l'Op. 20sq. yd? 12. Divide |.719b.u. lpk, .Oqb- by, 92.' - • <•, 4-ns,-18b:a. 2pk. 6qt. • 13. Bought 18 hogsheads of sugar weighing 227c\vt. lqr. 01b.what itvas the weight of each ? h Ans. 12cwt. 2qr. 141b. 14. Bought 18 bags of cotton for 784. 18s. 0d.; what did each bag cost? ' . * ' Ans. 44.:7s. 8d. 15. Bought 36 yards of ,clpth ,fo,r 194. 4s.-0d.; what was Jth©' prie.e per yard ? . . Ans. 10s. 8d. 16. Bought 21 pair of boots for 104. 2s. l.|d.; what was the price'of each pair ? ' Ans. 9s. 7id. 17. The weight of 9l.spoons is241b. lOoz. llpwt. 21gr. • what is the weight Of 1 spoon ?• Ans/ $pz. 5pwt. 15g. 18. In a month a man travels 8881ea. ,2mi. 2fur.; how far was his journey per day. Alis- ^121'ea. 2mL 7fur. REDUCTION OF COMPOUND tfUMBEK^. Reduction is the process of finding the equivalent of one denomination-in another higher,or lower denomination. Reduction'is of -two kinds u>xcndirig and descending. When a lower denomination is reduced to a higher it is called ascending ; when a higher to a lower it is called de¬ fending. Ascending reduction. is performed by division ; descending by multiplication. REDUCTION". DESCENDING REDUCTION. , . RULE. •" Multiply the highest denomination by whatever of the nestf lower denomination it takes to make 1 of the denom- inajtioji multiplied, and 'add in the given number pf said lower denqminatiott; and so on until it is reduced to the den om in at ion req u ired. examples. • 1. In 10£:'8s. 9d. how rdany pence ? Ans. 25053. operation. ' £. s. • d'. " 10 ' 8 9 20. 208 12 2505 An?. f t 2 In 35bu. 3pk. 7qt.; how many quarts? r • • ■ Ans. 1151. 3. In 27 milesv3fur.'20po. j how mapy poles? « Ans. 8780>. 4. In 2y.H 9ipo. 21da.; how many, days? Ans. 1011. 5. RcduCe 82A. SR. 18po. to yards. Ans. 1590541. REDUCTION. 39 REDUCTION ASCENDING. RULE. Jli indc the. given denomination by ■whatever • it takes - of that denomination to make 1 of the next higher denomina¬ tion. Divide the resulting quotient-by whatever it takes to make 1 of the the next higher denomination, and so on unj tij it is brought to 'the denomination required. The last quotient) ibith the remainders resulting all'along, (if any) will be the answer. ■ EXAMPLES. 1. Reduce 2505 pence to pounds, sterling. Ars. ,10£. 8s. 9cl. OPERATION- 12,)2505d. • pr ' 12)2505(208 ' ' — 24 20)208-f9d. 105 10£+8s _ 9,6 9,d. Ans, 20)208(10£. Ans. ' 20 8s. Ans. , Therefore, Ans. 1Q£. 8s. 9d. 2. Reduce 1151qts. to bushels. Ans. 35bu. 3pk. 7qt« 3. Reduce 8780 poles to miles. ' Ans. 27mi. 3fur. 20po. 40." KEDftCTfON. 4. Reduce 1011 days to years. Ans. 2y. 9mo. 21da. 5. Reduce -1590542 sq. yds. to Acres. Ans. 32A. BR. 18po. Note.—To divide by 301, multiply"both divisor andr dividend by 4, and perform the division, and let the remainder (if any) be di¬ vided by 4. Thus 122-5-30^ is the same as 488-5-121 ; and 488-5- 121—4, quotient, and 4-5-4, or 1, remainder. miscellaneous examples in reduc: •tion. . < • 1. In 95£ 18s. lOd. how marny pence ? . \ Ans* 23026d. 2. In 9480 inches how many yards ? Ans. 263yd. 1ft. ■ 3. In 20 cu, ft. 142 cu. in. how many cubic inclies ? ^ Ans. 84702 cu.'in. 4. In 1000 pints'how many gallons ? Ans. 125'gal. 5>. Reduce 30 days to seconds. Ans. 2592000see. 6. In a million of seconds how many hours ? k Ans. 277h. 46m. 40see. I. Reduce 13cwt 3qr. 201b to pounds. Ans. 15601b. 8. Reduce 597 drams (Apothecaries') to pounds. Ans. 61b. 2oz. 5dr 9. How many tons are there in 32-10856 drams (Avoir- ddpois) f Ans. 5 tons, 13owt. Oqr. 31b. 9oz. 8dr. 10. How many hhd. are there in 720 gallons? Ans. llhhd, 27gals. II. Reduce 2011) (Troy) to grains. , Ans. 115200gr. FRACTION^ 41 i . 12-. In 86,0'barrels and 2 bushels of-corn hdw.many pecks ? . Ans. 72.03pks. 13-. Reduce 185.7^ pence to-pounds, sterling-. .Ans. 7£ 14s. 9Id. 14. In 2 miles*how.many inches?. . . . Ans. 126720 inches. 15-. In 14hhd. 29gal. 3qts-. how many pints ? * •. . ' . ,Ans. 7294pt. 16. Reduce .1250 half pints tQ- gallons. , , Ans. 78gal. lpt. FRACTIONS* In Arithmetic or Algebra a Fraction (from'the Latin /radio, from franc/0, to brfeak,) signifies^ a pc{rt of an inte¬ gral. f .Any division of a unit into parts expressed' by the denominator, and any number of those parts being taken, expressed by the numerator—as | or £—is called a Vulgar Fraction. In the fraction £ 4 is the denominator and shows that the unit is divided into brokenparts each of which is a fourth; 3 is the numerator, and shows how many of those parts are taken. It is i important, also, that the student re¬ member that division is denoted by a' fraction, - and that. £ means 3 divided by 4. .'Suppose you wanted, to'divide 47 by 9;—the number 9 is not'contained any 'even number of times -in ,47.- It is contained 5'times ih45; but there, are 2 not yet divided by the 9, and.that .division, is "expressed by putting tbq 9 under the 2. Hence 47,-rAJ==5£. Enough 42 ERACTIO&S.- has perhaps been said to render the nature of a fraction understood. There are, in general and necessary use, four kinds of vulgar fractions. Proper, Improper, Mixed, and Compound. A proper Fraction .is one whose nuinerottor is leSs than the denominator, ae % an-dmproper is one whose numerator is larger than the denominator, as £ :■ a Mixect Fraction is a O ^ | B ' whole number-owirf a fraction^ as • -a Compound Fraction is a fraction of a fraetidil, as i of . CASE I. To reduce a Proper Fraction to its lowest terms. ■A Proper Fraction is reduced to its lowest terms by di¬ viding both numerator and denominator by any figure that will divide them both without a remainder. To illustrate we'll take the fraction -|g. Now there is no particular ne¬ cessity, for reducing this fraction, or any other proper.frac¬ tion,—for is as good as J; but there is sometimes acoa- venirnce in it, and an answer is generally given in the low. est terms. Take the Example -Jf. " You see that the nu¬ merator and denominator may b'oth be divided by 2, 4, 6, or 12. It'is"bestto divide -by 12 to prevent" a second divi¬ sion. Tlpis 12)| §)( J which can not be again divided and is therefore the answer for the lowest term's. EXAMPLES. 1. Reduce || to its lowest terms. 2. Reduce to its lowest terms. 3.;Reduce to its lowest terms. 4. 'Reduce to its lowest ternfis. 5. Reduce to its Ipwestterms. 6. Reduce to its lowest terms > -7. Reduce to its lowest terms.' Ans.-A Ans. |. Ans.|-. Ans. J (y A ns -J. 3 i Ans. f'gf Ans. go FRACTIONS. 8. Reduce }$fg to its lowest terms. ' » Ans. |f. ' U. Reduce to its lowest terms. Ans. 10. Reduce -}|§£ to its lowest terms. ; .Ans* CASE II. To reduce an improper fraction to an equivalent whole or mixed number. RULE. Divide th'e numerator 1>jj the denominator. EXAMPLE. 1. Reduce to a mixed number. OPERATION. 7)48 61 Ans. EXAMPLE!*. 2. Reduce 1g8 to a whole or mixed number.- Ans. 0. 3. Reduce J|5 Ans. I5f. ■4. Reduce ff An&. 2r^. 5. Reduce- 3T8g4 Ans. 29T7g. 41 FRACTIONS, 0. Reduce 12 85 T9 Ans-.: 67 j|. 7. Reduce 5840 2 1 Ans. 278/-,-. 8, Reduce 360 1 4 -Ans. 25-^. 9. Reduce 1840 "3 6 Ans. pl-J. 10. Reduce 1 72S T 7 2 tt Ans. 1. CASE III. ' To reduce a mixed to an improper fraction. RULE. Multiply the whole number by the denominator of the fractional part and add' in the numerator; place the de¬ nominator under the result. ' EXAMPLE; 1. Reduce 8| to an improper fraction. OPERATION 8 % 4, */ Ans. 3/, 4 4 • EXAMPLES. 2, Reduce 18|.to ad improper fraction, o. Reduce 37 £ 4, Reduce 28 f 5, Reduce 07 T% Ans. Y • Ans. y. Ans. 'I9. Ans, 9y. 25 19 2 0 314 1 '1 3 FRACTIONS. 45 6. Reduce 10 j| ' Ans. 5yV 7. Reduce 125A9. .Ans.- 8. Reduce 287-}{j. Ans. 9. Reduce 1247|$. Ans. 6%3090 10. Reduce 975* Ans. 4Yv92 As the reduction of compound fractions is precisely the same as multiplication of fractions, it "will be ommitted here. CASE IV. To reduce fractions to^a common denominator. There are two methods i-n geheral use. I'll illustrate both * these methods. I shall give the simplest first. It is the simplest, but not the one use,d by good arithmeticians, because of its being more tedious than the other. First Method. RULE. Multiply each numerator by all the denominators ex¬ cept'its own, then multiply all the denominators togeth¬ er for a common denominator, which, placedMmder the pro¬ ducts arising from multiplying the numerators by the denom¬ inators gives the answer. EXAMPLE. , 1.' Reduce atid f to a common denominator. 0 4* FRACTIONS* operation.- 4 3. . 2. a' ,4? ;s, 4 _ 3 2 denominator*- 4 5,4 1G 15 8 4" 48 45 40 20 60 Then, is, la, ±o \nf, co' 60' 60 examples. Reduce f, and A to a common denominator, Ans ,1'OP ISO ^Ci " 4 00' 4 0 0-' 401 Reduce 7°5, |, and g. Ana />;£$ 640 6 SO 12])> ^o1 7 30 Reduce -JE, Jand &. . . Ans AO 04 ,4 0 3-2. 4 0 0 6 2 1 fi 4 4 3 68' 4 3€b> 4 3 65' 4 3 65 Reduce Eg, J, and 5$. . Ans -_fio j 3.8 0 a1,s' 24 0' 24it' 24 0 2E ACTIONS. 'm -Second Method. ' HULK 4 Kind some number • (called a, conrmoYi multiple) that «'ach denominator of the given.sum will divide , even into. Divide each denominator into said- number, and multiply fche quotient into each numerator, putting spid common multiple under the product., and you have the answer. EXAMl'tE. I. It educe f, -A- and to a common denominator. OPERATION ILLUSTRATED. l>y looking at the denominators 8r 12'and 10 'we set; At onqcJ that they will each divide even into 48. So they would into 96, 192, &c.;' but 48 is the least commpii multiple, and so the least com. Denominator, and therefore best. Now take the fractions up in Order: -§- first; divi¬ ding 48 by 8 gives 0; 0 multiplied into 8 gives 18/ then JA one of the answers ;— again, A^ next; as before, divi¬ ding 48 by 12 gives 4, and 4 into' 5 gives 29; therefore is the second answer; lastly, dividing 48 by 10 uives & which multiplied into 11 gives. 33; and || last answer. ■ 48 FRACTIONS. . Therefore, 3g, -4g, -4j Now, rri mOst cases," this is^ decidedly a better method than the other, for its brevity if nothing else. But there are other good reasons. By this method you can- always get the least common denomintor ^without subsequently re¬ ducing, which in the other you c-annotr always do. But the best reason is that the study required sometimes to find the common multiple, by merely looking at the denomina¬ tors, is an excellent exercise for the young mind.- in cages where you cannot readily discover 3 common multiple, of eourse have recourse to the other luethod. But -to work. EXAMPLES. 2. Bcduee Ts3, ?, and J to a common denominator. * Ans 3 3 3 i~Z 12 12" 3. Beduce T90, and fi. 4. Keduce and §|. 5. Beduce T* , and oj. 6. Keduce T9B, and g. A ns. £? r'~ 7 0 7 0 7 0- Ans et! «« is 6j 04 V u4 j 04' 04' Ans £4 i>5 309 , 58' bS' S b " Ans 64 Q 5 36 72 0' 72 0' 720" I would advise the student in all the foregoing cases to extend his practice by making examples for himself. FRACTIONS, 49 CASE V. To reduce a denominate fraction from one denomination io another. , RULE. •If the reduction is descending, reduce the numerator, (as an the Rule of Reduction) and' put the denominator under the result. 1EXAivtrr/E. Reduce | of u Found to the fraction of a penny. OPERATION. 3x20xl2=7§° Ansv Note.—X denotes multiplication ~ . 4- denotes division; -f- denotcs addition ; — denotes subtraction. EXAMPLES. 2. Reduce f of a league to the fraction of a mile. Ans. | of a mile. 3. Reduce T9a of an acre to the fraction of a yard. Ans. 4 3y5a6 0 of a yard. '4. Reduce | of a yard to the fraction of an inch. Ans. 1|° of an inch, h. Reduce j-£ of a barrel to the fraction of a peck. Ans. ^°y° of a peck. 50 FRACTIONS. 0. 11 educe-/f of a mile to the fraction of a foot.- Ans. 15t\40 of a foot. If the reduction is ascending. Multiply the denomina¬ tor, of the given fraction by whatever it require? of the givfen denomination to make 1 of -the next higher de¬ nomination, till you re^eh the denomination requi/ed^ iu squared =;3Gsq. inches. And 108sq,' iii.X36sq. in.=144sq. in., or 1 footy Ans. FRACTIONS. 57 MISCELLANEOUS EXAMPLES IN ADDITION OP FRACTIONS.. 1.. Add t35, ^and Ag,together. • Ans. |f. 2. Add and 41. Ans. 5X^. 8. Add -f of a yard to I of-a quarter. ' Ans. 2qr. 4. Add J, j%, if, and 4. ■ Ans. 6f 5. Add x7g of a sq. foot to I of a foot squared. Ans. I a foot, I 0. Add \ of a shilling to 4 of a penny. Ans. 6Id. 7. Add 7§, 8|, and 164. . Ans. 27. 8. To 22^- square yards, add I a rod square. Ans. 1 sq. rod. SUBTRACTION OF FRACTIONS. To subtract one proper fraction from another. RULE. Reduce the fractions to a common denominator (Case 4th,) and subtract one new numerator from the other, put¬ ting the.com, denominator nnder the remainder. EXAMPLE; L From |f tate ||. OPERATIQN. [7 reduced to a common denominator ^4} 824j—323--1. Putting denominator under, ayc have sh AnS' 5$ FRACTIONS,; EXAMPLES,' 2. From take Ans. T"'g. 3. From take ?. Ans. 4. 4. From take |. Ans. T*g5R. 5. From r8f take ?. Ans. Rg- G. From j--| take Ans. 4- 7. From |;-j take -[.f. Ans. 3^. Where mixed fractions arg given. RULE. Reduce the mixed to improper' fractions (Case second.) and proceed as in the last examples. EXAMPLE. 1. From 7f take Gf. OPERATION. 7f by case third, — 275; 6| by ease third = ^7. Then from ~33 take 27, and the result is 41, which is the answer. EXAMPLES. 2. From 84 take 5-4. Ans. 3. 3. From 91 take 8R Ans. l(k. 4. From 18-J take 124. Ans. Gf. 5. From 27fi take 26'R Ans. 7'A To T<> x' 4 (1# Where denominate fractions are given. RULE. Reduce one of the given numbers to the same denomiR- atioh of the other, then proceed as before. EXAMPLE. 1. From | of an Acre take :j of a rod. OPERATION. I of an Acre (by ease 5th, 5 X160) 800 • ,, 3 ■ . .or 100 rods 100 rods—of a rod=99£ rods Ans. fcRACTlONS.' 59 EXAMPLES. 2. From § of a league take of a, mile. Ans. lip-j. 2f., 16p. 3. From 1 of. a barrel take |ofabu£]tel.. , Ans. fbtj.,'0T3ype. 4,. From T7T of an hour tak§ f qf a iRinuie. Ans. 87min.? 2o-}Ssec. 5. From f of a square foot take \ a foot squared. Ans. \ a square foot. ' MISCELLANEOUS EXAMPLES IN SUB¬ TRACTION OF FRACTIONS, 1. From | take 4* 2. From7 take f. 3. From 81 take 7§. 4. From f of a gallon takp f of a quart, Ans. 4|<- Ans. 61, ■Ans. Ans. 2qt.; GT9ept. GO TRACTION^.v MULTIPLXCATIQX O.F FRACTIONS. To multi-ply two or more fractional' expressions together. RULE. After reducing tlie mixed numbers, if any, to improper fractions, (Case 3rd.) multiply all the given numerators together, and all the given denominators. EXAMPLE. L Multiply £, t7t, and, 3y together. OPERATION. • *First reduce 3 J; it becomes iv; then | Xy7pX5—fUj which reduced (Case 2d.)=iyl!,g Ans. In ijiany cases cancelling may be introduced, which very materially yhorteirs the work. EXAMPLE. 2. Multiply £, 1;, -J, •*, & together. OPERATION T>Y CANCELLING. ^vd,2'.iwl 1 . LX-X-X-X-=- Ans. G £ ^ ^ G The student' will notice that the same figure occurring above and below the line cancels out. The meaning is that FRACTION'S.* 61 5 is contained in 5 1 time; but as the 1 would not increase other factors by multiplication, I always omit it, except in cases wliere^a?£ the figures above the line are marked out,— thcit I pubd6wn 1 as the numerator to the answer. EXATVIELE., V). Multiply 42, /j, and | together. * ORATION. ly±x[=l Ans. p$f>. 0 3 £ 3 The student after'taking down his numbers beside each other, (in the above example) will perceive that 9 is .con¬ tained 5 times in 45; he therefore marks out 9 and 45 putting 5 under the 45; next 2 is contained in 4 two times, he therefore marks out 4 and'2, putting 2 above the 4, and so on fill he has Cancelled all he can. He finds, when through, that all the figures tire marked out above, and that :i is ieft he.luvf. Then assuming, as I before said, 1 for the numerator J is the answer.) Where several figures are re-- maining above and below tlie'linc after the cancelling, he lias but to multiply those'remainiiig figures together, g's in example 1st., atid he will have the answer. , A comoouiid fraction, 4 of |\is nothing more thaii a sum in multiplication, and is reduced to a simple fraction thus 1^1 £X.6T5' T'n addition or subtraction these .compound, fractions-must always be reduced before attempting to add or subtract, ■ . 62 REACTIONS* EXAMPLES.' 4. Multiply f, T9T, and |I together. Ans. -^V 5. Reduce § of f off to a simple fraction. Ans. f. 6. Multiply 8| by 5f. Ans. 45. 7. Multiply 8 J, #, and | together. Ans. 1J. 8. Multiply |, and 5 together. Ans. 3f. 9. What will 18|'yds. of cloth come to atTfcts. per yd.! Ans. 81.43|cts. 10. Find the worth of 6 Fibs. of butter at 6] c-ts. perlh. Ans. SQJg cts. 11. '124 lbs. of coffee at 7f cts. per lb. Ans, 95 cts. 12. Find the worth of 24bushels of corn at 62 Jets, pCrlm Ans. 815.00., 13. Multiply |, and 9 together. • Ans. 5J. ' * % Note.—"Where £ whole numbsris given 1 is the denominator. DIVISION OF FRACTIONS. RULE. Invert the- divisor and proceed precisely as in inultipli oation of fractions.' EXAMPLE. 1. Divide f by ^ OPERATION. It is expressed thus, * |-f-T9c; performed thus, |XV ==bt Ans'. FRACTIONS* 68 EXAMPLE. 2. Divide 18| byo^. .STATEMENT. ' OPERATION. 75 25 3.-2 — jtfr $ 2 ,3 4 8 -X-;-X"=0 Afts). 4 1 1 " Inverting the divisor " says the young student, " puz¬ zles iie;—I can not understand it!" Thfen, I'll explain it. Suppose you wanted to divide 8 by 4. ..The afiswer is- %— Hut let's do it by fractions. 8 is 4 is f. Now divide' j by 1'-, X 4==f QT 3. lou sea the result-; and if it holds good in integrals, if holds good in fractions.-*—\ The argument I'll give you anon. EXAMPLES. 3 Divide i by • . Ans. 4. Divide 7£ by 8f. Ans. lg, 5. Divide 12 by f, _ Ans. l(i. 0. Divide 4| by ■§ ©f 4. Ans. 2-^. 7. Diyidc £ of 7 by f of 8! Ans. i||- 8. Divide ! of.f by £. Ans. f, I). Divide G8£ by 34$. Ans.. 2. - Before I conclude "fractions," I will introduce another kind called complex, .They are such a,s have a fraction for numerator or denominator, or both, As 41 7_ 8| !> U 92. To resolve, such from their complexity, -the student has 64 FRACTIONS., but to Remember tbat I have told him that a fraction de¬ notes a division—the numerator being the dividend, the denominator the divisor. .4\ then is simply 4 4-^9, or £ ~9 °r | X| *or i. Likewise 7 is-simpty 7'-t-8 ■}, \>r y/Pr \Xyf> if- ' 84 . ' 82 «n tnn J 2 S_^_3 9 2J5 y _4 104 iTO, TOO, —O 3 . J4 ^ . 4 A 315 Yl7* Ans. 42. Ans. 6. Ans. 12. Ans. Ans. 2. Ans. 12f. Note.—The student -would do" well to mate himself some u syms " in addition, subtraction, Ac., with this kind of fractions. I am often asked " why is it that multiplying by a frac¬ tion diminishes ?v as also why docs dividing by a fraction increase.. Before I attempt'to answer this, a few remarks concerning multiplication in general may be' necessary.— When any two numbers are multiplied together one at least of the numbers m ust be abstract To try to multiply two concrete numbers together, as 4 apples by 5-apples is ab¬ surd; it ii^ equally absurd to talk about multiplying. 25 •J I EXAMPLES. 91 1. Reduce— tb a sample fraction. ' * 41 ' ■2; Reduce _ to .a simple fraction. i 8 , 3. Reduce dLto a single fraction. . i o Q2 4. Reduce —1 to a simple fraction. 8f t 5. Reduce I to a simple fraction. 6. Reduce to'a simple fraction. "FRACTIONS. €5 cents by 25 cents. " Every multiplier must be considered purely a number," (Day's Algebra, Art. 91.) The opera¬ tion of multiplying, then, rnealis nothing more nor less than to take a certain numberL or oi\e or mo're, things a certa in number of times. Now just remember that when you mul¬ tiply you take a certain number, or n'jimber of things so many times. -You- go to a barrel of apples 5 -times, and each time you take 8 apples* You have illustrated mtilti- '• plication; you haactually multiplied. 8 by 5,. and you have 40 apples. , But how would you go about taking?; apples to /apples times.—it is simply ridiculous!. But " to the question." Multiply 12 by £ lyad you have 9,— You take 12 of a, time, i.e., you'take i' of 12 - one time, and you have' ft. Vipe versa, you divide 9 by and you have 12. Divide it, and thereby increase it'— "Well, look at it one minute; E CIM AL- FRACTIONS: • .Decimal. Fractions, (from the Lathi 'decern, ten) are fractions whoso.. denominators are each some-power of 10. A decimal fraction originates from dividing a unit (1) into ten equal parts, and then-bach of these into ten other equal parts, &c., so that its fractional units arefenths, hundredths, thousandths, Ac., &C. ... * Decimal fractions are commonly expressed by writing the numerator only, with the decimal point (.) at the left of it. There is no difference in the nature of a- common (or vulgar,) and a decimal fraction., /()- may he written -.7, and is read seven-tenths, iffo " " u '" twenty-five hundredths. U ' '* " U ■" 1Jlilie thousandths. In the last you will notice ciphers are put-in those places not requiring a digit. : , Mixed numbers, where the fractional part has a denom¬ inator a power of 10, are also expressed decimally. 8 A may be written A5 and read three and five tenths. 87T-j}<) " " " 87.09 " <* thirty seyen and nine hundredths, Ciphers placed oh the right of decimal figures do not af¬ fect their vqluC) because, as in common fractions fL and .fA the same in value, so'in decimals .9 and .90 are the same. But ciphers on the left of decimal figures, if between them and the decimal point do affect their value, because each cipher removes the decimal figure one place to the right, and thus dimii{idics its value tenfold. .3 .is three DE-CIAIAL FRACTION. 67 tenths, but .08 is three hundredths, fen, tidies less, or (in better English perhaps) only one teu'th as much. In our money, as a hundred, cents make a dollar, and every cent is therefore the -hundredth of a dollar, dollars and cents may be considered as a whole number and deci¬ mal, and in our calculations ivc always consider theig so. ,'The .Notation and Numeration of Decimals,yan be best illustrated,by a. TABLE. 3 "S3 HC riil gt ^ rfl ,.3 ' 'a ""3 "S ^ o , •-3 05 13 o 2 C'gia.2 3 C = 33! c g ;£< Wt2^^WEuWHt®Ha h H h1 SH'h-i fcH b-t i—( fc-n i—t P-t ; t a a i 9 « a r» n t s o Whole numbers arc numerated from right to loft, deci¬ mals from left to right, as in the table. Thc-above is read— Three hundred and forty-five, and One hundred and twen ty three millian'four-hundred and fifty-six thousand seven hundred and eighty nine Billionths. Decimals figures are read then just like whole numbers, but the whole called by the name of the last decimal figure.. Now in numerating .the above -table from left to right I •merely did it to find the name (if you please) of the last decimal' figure. I found itto be Billionths. I then went back, as usual from right to left till I reached the decimal point, and read the 1 £8.4 5 C 7» S 9 just as I would read a wliplc number, but called it Bittianths. 08 DECIMAL FRACTIONS, Addition of decimals Is very feasy. It is only necessary in setting down the num¬ bers, "to hotiee that', as in whole numbers uAi'ts should be directly under units, so in decimals,tenths should.be direct¬ ly under tenths, hundredths under hundredth^, See., the decimal points makihg an exact perpendicular column.— Begin at the entire right, and add just as in Simple Addi¬ tion, and put the decimal point in the answer exactly under those above. EXAMPLE 1st. OPERATION. 84.5 127.25 • 3142.184 ' 17.0009 4.405 3375.3399 2. Add 35.4, 184.05, "27.2S5, 10.9 together. ■ Ans. 257.635. 3. Add 75.018, 312.'109, 2-4008, and.369.8 together. Ans. 78083278. 4. Add .4.2, .85, 325., and .175 together. • Ans. 330.225. ' 5. Add .8, .49, .132, and .4328. together. .Aus. 1.8548. 6./ Add 24o, .32, 324, .1082 together: s ' ' Ans. 569.4282. " 7. Add 27,4," 45.05, 12.006, 31.0007, -iand 4.00008 to gether. . ' Ans. 119.45678. DECIMAL FRACTIONS. 61) SUBTRACTION OF DECIMALS Is also very easy. Same,care in regard to the point; then commence at thp right and proceed just as in Simple Sub¬ traction. If any of the decimal figuveg in the subtrahend have no figure above them, irriagisof a cipher above each, and go ahead. EXxVMPLE. 1. From 245.18 take-182.324. OPERATION. . 241)JS 182.324 '62.850 Ans. EXAMPLES. "2. From 2142.184 take 142.18.' Ans. 2000.004. 3. From 85.192 take 72.4321. Ans. 12.7599. 4. From 17.8 take 17.08. ( Ans. .72. .5. From 324.847 take 18.9. Ans. 305.947. G. From 508.75 take 03.7. Ans. 500.05. 70 DECIMAL FRACTIONS. MULTIPLICATION OF DECIMALS. HULL.' Set the numbers clown without reference to the point, and multiply, us in Simple Multiplication. Point oT in the product as many figures for decimals as are contained in the multiplicand and multiplier. If the product should not have as many figures in it as there are decimals in the multiplicand and multiplier, supply the deficiency by placing ciphers between the product and the decimal point. example. 1. Multiply 18.45 by 4.3. operation. 18-.45 There arc two dec-i- "4.3 mals iii'tbe muTtipli- —. cand, and one in the multiplier; therefore three in the product. ;>.>•> o 7380 .79.335.' example. 2. Multiply .123 by .013. operation. .123 .013 369, 123 .001599. There were three deci¬ mals iu each, therefore the product m ust have six. But the operation produced but four. I therefore prefix two, and then the decimal point. Decimal fractions. 71* EXAMPLES. 3. Multiply 18.45 by 80.051. Ans. 72P.49095' 4. Multiply Twenty nine and'four hundred and thirty five thousandths by seven and eight hundredths'. ' Ans. 208.3998. 5. Multiply 17-85 by 8.4. Ans. 149.94. 0. Multiply 8575 by 4.8. Ans. 411.6. 7. Multiply nineteen hundredths by eighty five thou¬ sandths. •Ans. .01615. 8. Multiply • One hundred and forty-five by eight ten- thousandths. Ans. .116. DIVISION" OF DECIMALS. The quotient arising from dividing a decimal fraction by a decimal fraction with an equal number of decimal figures i» a whole number. 8 tenths, -r- 4 tenths = not 2 tenths but 2, i. e., .4 is contained in .8 two tfmes. Hence, for di¬ viding decimals the following. •" • DECIMAL FRACTIONS. RULE. Annex ciphers to the divisor or dividend (if necessary) till the number of decimal figures is the same in both, then divide, as in Simple Division, and the resulting quotient "will be a whole number. If then there be a remainder, annex ciphers to sdicl remainder and continue the division , and for every cipher thus annexed you^will obtain a deci¬ mal figure m the quotient. EXAMPLE. 1. Divide 87.IS by 2,13. . OPERATION. . 2.13)87*48(36 Ans. As there are two de- 729 cimal figures in each of the given numbers, it is , 1458 nut necessary to annex 1458 ciphers to either. EXAMPLE. Divide 11.97 by 5.70. OPERATION. 5.70)11.97(2.1 Ams., 1140 570 annexed 570 ti<;nt 2# and a remainder 57 the remainder, and proceed, and we .obtain .1. In this example the dividend has two decimal figures,, ,the divisor only one. We therefore an¬ nex a cipher to the divi¬ sor (which <}oes* nut ef¬ fect its value) and .after dividing we have a"qu*> We then annex a cipher to DECIMAL FRACTION'S. n ' EXAMPLE. 3. Divide 210.5 by .05. OPERATION. .05)240.50(370 Ans. 195 455 455 0000 EXAMPLE. 4. Divide 23.04 by 90. The dividend having two decimals and the di¬ visor none we annex two to the divisor. It then goes 0-times, and tljc re¬ mainder is 2304. Then annexing ciphers to get decimannd -figures we pro¬ ceed, and the result is ' M. EXAMPLES. r>. Divide 148.05 by 4.23. Ans. 35. 0. Divide 707.2 by 2.72. Ans. 200. 7. Divide 308 by 43.8. *Ans. 8.4018+ 8. Divide 20,575 by 7.9. Ans. 3.363+ 9. Divide 0.3 by 35. Ans. .18. 10. Divide 4.725 by 8.542. Ans. .55314+ OPERATION. 9C-.00)23.04(0.24 ' 0000 • 230'40 annexed, 19200 38400 annexed, 38400 DECIMAL FRACTIONS. REDUCTION OF DECIMALS. .To reduce a decimal to a common or vulgar fraction. RULE. Erase the decimal point and write under «the numerator its decimal denominator, and reduce to its lowest terms. • example. 1. Reduce .36 to a common fraction. Ans. operation. .36= r3((t_ which reduced is .examples. 2. .Reduce .875 to a common fraction. Ans. |. 3. Reduce .9375.to a conjmon fraction. Ans. 4. Reduce 5.45 to a common fraction. An?. 5^. - 5. Reduce .125 to a common fraction. Ans. i. 6. Reduce .0018 to common fraction. Ans. -f,9^. To reduoe ,a. common fraction to a decimal. RULE. Consider-the-numerator a dividend and the denominator a divisor, and' proceed as in division of decimals. example. 1. Reduce f to a decimal. Ans. .625. operation. . 8)5(0.625 • 0 50 48 20 16 40 40 - tol.MAL FRACTIONS- example. 12. Reduce 13:} to t, decimal. . Ans. 18.75. operation. Let tlie 18.be as jt is; 'reduce'the $ thm 4)310.75 cf . -.M . 28 20 20. thcii placathe decimal .75 beside the interval, tlmS 18.7b. ' examples.' » " 3. Reduce T5;i to a decimal. Aiis.'v07fJ92. 4. Reduce 194 to a decimal. Ans. 19.125. 5. Reduce 315? to M decimal. Aus; 315.875. 0. Reduce 3 to a'decimal. Aus. .09318. > • To find the value of a decimal it* 'whole numbers'of lower denominations. RULE. . Multiply the decimal (or given nfumber) by that number which will reduce it to the jiext low^r denomination,, and point off a"s in Multiplication of decimals.. Then multiply the decimal part by the next lower, and point off as before, Ac.* EXAMPLE. 1. \\Tbat is the value of 4321) ofhn Acre.' ' Ans. 2 Roods, 2Upo. 6 Id •DECIMAL FRACTIONS/ OPERATION. .025 The whole numbers left 4 after pointing off, cousfci- ' tute the answer. It, 2,500 . 40 W 20.000. LX-JAirLES. •3. What is the value o£.47£. ? ' • ' ' ' • Ans. Os. 44.1 ?>. What is the value of 4.57 yds. ? Ans. 4yds. lft. It/Uin. 4.- W hat is the vaiue'of .725 of a mdlon ? v- Ans. 2qts. 1.8pt, 5. What'is the value of .G05hhd. ? • Ans. HSgai. Sgills-f- To reduce a simple or compound number to a decimal of a higher denomination. RULE. Reduce the given'number to a common fraction Of flie higher'denomination (by Case 7th. common fractions.) and thon-Aeduce that to a decimal. EXAMPLE 1. Reduce 2 roods, and 20po. to the decimal of an Acre* Ans. .6*25 A. DECIMAL. FRACTIOUS. operation. By case 7 th. com. frac. }—2r. 2{)po reduced is lOOpo.; X acre_, reduced to .poles is 1G0. ' The fommCm fraction then is i JXg or f, -jj reduced to a decimal ia .025 Ans. 2. Reduce .8 shillings to the decimal of a pound.^ > > ~ • Ans. .4. 3. Reduce lqr: 3na. to the-decimal of a yard-. c Ans. .4375. 4. Reduce 5fur. 85rds. 2yds. Oin. to the decimal of a milq. > ' Ans. .736032^f- 5. Reduce X8gal. 3qts. lpt. ta the decimal of a hhd.- . Ans. .299.6-h HATIO. . RATIO: Uatio is flic relation that one number Lours to anot:mr. • , 1 * * Ti ft is cxp*e4s«etl I>yhvo acts (:) between t:ic xcrm.-. t -mr I,wo numbers forming a ratio are called to-fit* ol the ratio. The first term is called fin*' rrrjnit, and the last. eoii*''_ii'<'nt. •The two terin.s takmi together are calkAa couplet. The index or exponent of a rotio is. the quotient »rbin<_ where one of the-terms is' divided Ly tue other, into' ratm of C> to 3 (expressed. 0 : 3) is 2: A ratio is either, direct or utm-sc;—a direct ratio h tin- quotient of the ont-'ccda\t by the' cv„:--q.v:,,t; and inn.<<-qu':y>f by the unt^ctd, <4 to 7 ? • A^s. 2. Where the numbers are' concrete they must -be of the same denomination. - A Wliat is the ratio of Gd. to 2s.? Ans. f. It. What is the ratio of Sqts. lpt. to 2gal. ? • Ans. R. 10. What is the inverse ratio of Gin, t« 5 ft. ? Ans. 10. PROPORTION OR''SINGLE.RLXE Otf THREE. 79 .FRO,PORTION. Proportion is simply equality of ratios.* 'Any four nrrfh- lu-rs 'arp proportfontil wheiqthe ratid of tile first to tile sec¬ ond is the same as the ratio of the third to the foitrth. Iii. the proportion 8 *: 4 : : 20 : 1Q the ratio intthe first couplet is 2; it is the sapie'in the second; thbrefore the numbers are "in proportion.1 ' ' ' ' , "The terms ofoa proportion aye the four numbers which ' j'o'rnt {hUproportion. The first and1 third are antecedents, the second itmT fourth are consequents.* The first antl lisf ate tilledfictctntcK, theasCdond anil third, mdart's. And in any proportionttiiC product of the,e.i fiemcs lilust equal the product of the.?n,<1U=8or 112=l'l2. Tleiree, ■3 f the extremes and one of 4jie mean!? ai'e given, the oth¬ er moan may he found. by dividing the 'product* of the ex¬ tremes by the givfen meap. Or, ' j If th.e means.and one' of the extremes are given the other -extreme may be found by dividing the product of the means by the given extreme. On this principle is founded,- ■ TTtE SINGLE RULE OF -THREE, *in which three terms-*of a proportion are given to find a fourth. RULE.. •M ■ v ■ j'ut the given number which (he answer'is- required to belike in the th ird place. Put the term of demand (jot. the term w'hich has " what" to it in the enunciation) in the isrovd' place * and the remaining term in the first place. BO PROPOHTION im BISUILE iuLE OF THAEE. * Multiply the second and third terms together, and divide the product by the first, 'or di'vide the second o f third by the first^and'multiply thq quotient on to the remaining^ terip. —If the first and second terms are pf different denomina¬ tions, they must be brought'to the same by Reducing, one or both. - » 1 " ' ■ T. * example. 1. ff 45 Acres of land cost .345-0, "what will 78 A. cost.? , ' ' ' . t Ans. 3780/ operation. We?, see that the answer is requited to Be in dollars, therefore we must put 3450 in the -3rd. place; we see also that 78 acres is the term of demand arn^ we put it;m the^nd. &e. Thus, • » 45A. : 78A.: : 3450- 78 3000 3150 15)35103(780 dollars-Ans. dlt) ' 860 360 0 . Or same statement thus, #A. : 78A. : : 8#0 10 10 3780. The answer like tht third term. PROPORTION OR SINGLE RULE QP. THREE. 81 EXAMPLE. .2. If 5ft's, and 7oz, of tobaccopost $3.90 what will 121bs, cost? 'Ans. $8.60-f * OPERATION. - 5ft. 7oz. : 12ft. : i $3.90 16 16 ' 870?. *' 192oa. : '$£.90 Then 390x192=74880; .74880.n-37=i=i8!(JQ-f . EXAMPLES. ' 5. What js brandy per gallon when a qrtart and a pint are worth 27 cents ? An's'. ,72cts. 4. If 13 acres and 3 roods of land produce 170 bushels of gr^in, what will 5 Acres produce ? . . Ans» 61bu. 3T8Tpo7 h. If 14 gallons.of brandy cost $8.96,'how much ca-n I "buy for $6.72 ? • ' ' Ac's. lO^lgalv ; / . ® ■ 6. If .a pole 13 feet high rtake^a- shadow 39 feet, 'what will be'the. length of a-shadow made-by a. tree 126 feet high i? ' t Abg, 360ft. ' 7. The shadow of a tree 63ft, high is 72 feet; what is. ihe height of a tree whose shadow is 92 feet and 4 -inches? - ' ' "Ans. 76ft fljin. 8> The circumference of the earth being 360 degree, and revolving on its axis Otiee in 24 hours, how far arc the in^ habitants at the equator carried in 10 minutes? • Ans. 17'3f lpilcs. •9. If 6 and 9 were 20, what, on the -same- -supposition, would 3 and 6 be ? ' Abs.-- 12. 8j2 SINGLE ,RJJLE OF THREE. j JO. What must^be paid for 21A. 3B. 20po.-ef'land, if 80A. ofl. cost $ 1200 ? ' w*Ans. $750. T J J . A fox has 90yds. flie start of adotg; the dog pursues him. running 8y'dk; while ' the fox runs Sr. How'many yards must the dog run ,to overtake the fox ? Ans. 240yds. flii this problem subtract 5 from 8 which' will show you Jiow many yds. the dp£ gains iiwevery 8: • Then the state¬ ment 8 : 8 : : 90 to'the answer.) 12. A rabbit starts 145yds. a.hfad ^o'f. a dog; .the plou runs '21yds. while the rabbit is l'unning 19. How many yards mhst thfc dog run th cut oh'it. ' . . Ans. 15224yds. / 1 • i. A in a journey has 17 miles the start of !>, A trav¬ els 5 Yhile's an boftr ; B travels THuiles an hour". How far must B travel to overtake A, and liow knur will it reouire V . . „ - V- ■ ' V .' , ^ = f miles, t '"18 hours and 20 minutes. ' 14'. B lias beofi g;om? from a certain place 9 hqurs when A starts in ffilrsuit of hi in." B travels at "the 'rate of J 9 miles per horny; A at the rate of 13 miles per hour. |L w fat add how long must A travel to overtake him ? » ( 890 miles. " ' ( 30 hours. WllVAiF TJIHTHHIPOHTIOX OP PATIO-IS JHVKHSK.' ' EXAMPLE. • I... It' 1 (linen pan accomplish a job in 8 'days how long wilj. J 0 wien he in doing it ? • ,. SIXOLE RULE'OF'THREE' • §8 statement anT> operatiox. / It is evident that 10m. 15m. : : Hda. the moi-e men to work* 10 the less time required. — The ratio, then, ie in- ' 15)80(51 days versej 'and we make » 75 the second the divi- — » ding term instead of '5 . the first. • . EXAMPLE."y. ■ 2. If 12 'men can do ft piece of work in. 20 cjays, how long- will it take 0 men to,do it ? ' . ' 1 • • Aha 40da'. 8. tf 7 men can accomplish a thing in 8 daj'sj in u;ha time can#40 men do it ? ' • ' Ams. G hours', ISmin. "4. If when flour is §5.GO per hundred the baker makes his. five-cent loaf to weigh Hoz. what should it' weigh when flour is §8.20 ?.' ' - • , AllS. 1-107.. 5. -IVheii. corn is fifty cents per bushel the distiller sells ' his whiskey at G5 .cents per gallon, Ah at then would he sell one gallon for when corn' i's eighty 'cents a'bushel ? .• , Ansa 40jj cents. WTTERE CGMMOX FRAFTTOXS ARE IX- TRODITClvD. 11ULE. Fropai'e the fraction by whatever process most pertinent. State as'in tine preceding examples. Invert the finst term, and just conshW the whole as a.'■< sunr" in multiplication of fractions. 84" SINGLE RULE 'OF THREE. examples. 1. If 44yds. of calico cost GSJ -cents, wliat will SjLyds, cost? . - A"ns'. cents. operation. 4 syds. : o jUyds.': : G8f Preparing fractions and inverting first term it becomes jl v $ J- y 5 —'1 r>-" ?5 cfr 74i-9 14 4 , ~ Z<£4 ui /example. % What will 18 Jibs, of sugar cost, when |lb. costs \ of 12|-centSf ' . , Ans. £1.423. « < v operation.' |lb. : 18-}Ibs: : : 4 of 12gets, * . - - • When prepared | r 5^5 : : M39 Inverting first i X V ^ V whieli will admit of cancelling itfius, i p. 15 0 H 19 — X—X^—=:345 or 1.424 Ans, ji }4 ' ■ examples.- 3. If £ of a yard of tape cost T73 of a cent, what will b$ • the price of J jl of a yard- ? ^ Aris, i of a Cent; 4. If J of 7 he 2, what will 1 of 9 be ? • . • > / , . . Ans. l jj <- 5. If I of a bushel cost 49cts., what will 4 of 5 bushel? e>, st? " Ans. 934 cents, •single* ru£b.of three. 85 0* If 7j\ gallons of molasses cost 85.281, what will i a •gallon cost ? . '' ' Ans. 35-||cts. 7. A fox- starts up 68£ yards 'ahead of the hounds; the -hoCmds run 8f yards while' the fox is running 7f yards. How piany yards must the. hounds run to overtake him ? Ans; 65.0. 8* A fox starts at 4 o'clock in the morning aipd ". slopes " at'the rate of 13 miles per hour ' At id minutes before 5 o'clock, the same morning, a grev-hound starts from the same point, and .pursues the fox'at the rate of 15 J miles per hour. At what, tjme in the day will the grey-hound catch the fox, and how far will he have run ? . . » a I ^ minutes after 8 o'clock1 A. M». i • Ans. Tlic distance run 6059&, WHERE DECIMAL FRACTIONS ARE USED. ihjLe. * I 'J Rtdte as before. See that the first and second, terms are of the same denomination, if nof .make them so. Then.'if cither the first or second-terms have mbrc 'decimal figures than the othersupply tbcdcficicntone by annexing'ciphers ; and proceed without rcfercnoc to the decimal point, and the answer w.ill be like tbe third term. example. 1If 5.4 yards of muslin cost 83.24, what wi'd be tbe price of 8.75 yards? ^ Ans. 85.25. Sft SJN«LE KV^E OF iHHEE. OPERATION. 5.4yds. : 8.75yds.' : ! .83.24 Annexing' cipher to the first, 5.40 : 8.75 •: S3.24 8-75 102,0 2268- 2592 . ■540428330055,25 •2700 • 1350 1080 •2700 . 2700 As,the multiplication and division in the above operation is performed without reference to decimal promts some might think We should be at!a loss to know what our52p A. 13ut not so. It must he just like the third term ; and. as the third term is dollars and cents so 525 will be dol¬ lars'and cents; therefore, Ans.. 85.25. EXAMPLES. 2. If 7.21 lbs. cost; 82.45, what will .0 lb. cost? \ \ 80.30o8— . • ' *• ' HS' ( or 30 cents —. 3. If .15 bushels cost 2.5s., what will Sbu. 2pe. come to? Ans. 2£. 18s. 4d. 4. What time will be' rcquhied.'for 4 mUn to do a piece of work which it took 20-men 7.8 hours to do? • • Ans. 39 hours. SINGLE RULB-OP THREE. 87 . • '5. A ship passes a certain point on the river when a boat is "18.75*miles below.' The.ship descends 8:43 miles while the boat descends 7.18 miles. At what distance* below th«'. point will they be together ? Ans. 12dm. 8fur, Alp. lie fore We give the miscellaneous examples we would re¬ mark thht operations in Proportion may be much abridged often by /dividing the diciduitj termand either of the others by any* number- that will divide both without a remainder. Take.the/proportion 320 : 330 : :'2T. Dividing first coup¬ let by 40 it stands 8:0: :-21 .which will mhke .the work much shorter. M [SCELL*A2sEQFS EXAMPLES IX PRO POTi- TIOX. ."■1. At 05- cents pqr bushel, what will 3 barrels 4bu. ipul 2pe. of corn' Qo*tne to ? . Ans. 812.G7p 2. AVJmt is the, cost of 17# yards of muslin at G8| ctmts per yard 7 Ans/812.Pit p.- 3. In what time can 1^ boys do a piece of work, which 8 men Could do in 17 days, each boy doing f as much work in a day as a man? . ' ♦ Ans, 17 days. ■ 4. A pol-q 25 feet high casts a shadow on the level ground to the distance qf 33 feet 10 inches1.- A tower 250'feet high standi just o» the' bank of h ri-ver and casts its "shadow 18 feet'and-6 inches beyond the opposite bank. What is»the width of the 'river ? ' ' ' Ans." 819ft. lOin 88 single rule of three. 5. If the penny loaf weighs- 8oz- when dour is 7 dollars phi* barrel, what should it weigh when flour is 85.00 per barrel ? Ans. lOozx 6. If by working 14 hours per day a farmer can ^laht half of a held'in 0 day?, in what time will li'e ul'.'nt the re¬ mainder-working 10 hours per^day ? j Ans. 121 da. -7. A sets out on a journey, and travels £7 miles a day; B sets out and travels'the same roa.1 40 miles a day. • Hut A has 9 days the start. In-how many days will B, over¬ take A ? Ans. 111. 8. A sets oat on a j.oagney and travels-£0 miles a day ; B sets out--On the tame road'and travels 3G inilgg a day, and in 30 days he overtakes A. How many days had A the utart. Ans. G. 9. If 18 2lbs. of tobacco cost -| of a dollar, what will be the price of 51 ribs. ? ' ' * Ans. $1,712. 10. If the eld lady's "ginger-Cake " weighs 2.1.7 ounces when molasses costs G*2.5 cents per galloii, what should it weigh'when molass.es is ,only 43:73 cis. per gallon j k ' Ans 3.1 ounces. 11. The circumference of a wagon-wheel is 12 -"feet, 6^ inches ; how many times will it tnrn over in going a mile ^ Ans 420. ' > T. ' * ' \ *- • 12. The ,earth, is 300 degrees in circumferencev and re¬ volves cue?-in 21 bouts. A degree of longitude at -the ■equator is G9l miles; but in the latitude of 40 degrees a degregof Jpngitilde is only 53 lyiles. How many miles are SINGLE rule of three. lh,e "inhabitants 'at the equator and a$ 40 degrets carried by this motihn in one minute ? ■ ■ . t f'At the eqdator 17g miles, < ills' ( At 40 dc-<^. 131 miles. Note.—Degrees of longitude are to each other in length as the 'cosines of fhgir latitudes. For every 10 degrees of latitude'they are as follows. Equator, 69 J- Miles (generally considered 09 J); 10 ' dcg. 68.1-milcs; 20 dog. 65-miles; 30 deg. 511.9 miles; do deg.'53 miles; 50'deg. 44.5 mil'es; 6'0 deg. 34.6 miles; 10deg/23.'?miles ; K0 deg. 12.5 miles ; 90 deg. 0. 13. A detachment of 2000 soldiers were supplied with bycad sufficient to last 12 weeks, allowing each mart 14 oun¬ ces a-day; huton,examinat*ion thereore 105barrels..each con¬ taining 200 pounds, wholly Spoiled : how much per day may each man eat. that the' remainder may supply .them 12 weeks ? , Ans. 12 ounces. U. If 360'men be plabed in a garrison, and huve'pro'vis- ions for G months, how many men muA he sent away at the end of.4' months that the remaining provision's may last those remaining 8 months. Ans. 270. 15. If 4 oxen or 5 cowS consume a certain amount- of provender in 9f days,.in what'-time Will 1 oxen'and 3 cows consume tho same quantity ? . . An,s'. 4yE\ days. Q % - 71' 16. If A of a bushel of corji cost, what will be the' 9 ' 10 < • 111 worth of a bag .of cotn containing just —: i bu. ? - 22 g ' ' • Ans. 724^ cents.- . 17'. A starts from a certain place on Monday morning at precisely 10 minutes before 9 o'clock, and travels' at the 00* SINGLE.RULE,OF THREE* uniform rate qf 3f miles per liour ? B starts from the same, place at. 12 minutes after 10 o'clock, tlie same morning, and travels at tlie rate pf 3 t miles, per hour. Now.-suppose they continue travelling' at the above rates, not stopping for night or any thing else, at what time will they be.together ? A f Ou the succeeding Friday* t at 42 minutes after -1 P. M. 1*. At SI.SO per Acre, .what is a f-inail lot of land, it be¬ ing an oblong square 3d0 feet in length and 280 in width, worth ? ■ . Ans. Sll'.KOU 10. If 181b. oo£. of"cheese cost 83.75, what will I of 1*.* ounce come to ? ' , Ans. til i| mills.. 20, If 35cwt, 3ipr. 181hs. cost §1n54, what iri.u.-*t.I pav for 161bs.? j Ans. 87.37-j- COMPOUND PRGPO&Tlofr. COMPOTJ^p .PROPORTIOK OR .DOITBLR RULE OF THREE. 'A problem'in compound Proportion is one tliat requires two statements. I mean that a firoldeni in the Rouble Rule «>f Three may be solved by two statements in the single Rule of Three* .For illustration take the following problem. ex-ample. 1. Iff) men in men : 9 men ! C40 which will get bTsh' This 72 dollar^ is what 9 men caircarn inT days. Rut the 9 men were to work 10 days, henectanother statement is required,•—in- H days the 9 men can earn 72 dollars, what will they- cam' in 10 days?—thus o clays : 10 dftvs r: £72 which <>"ets 90' » J ■ »■' o dollars Arcs*. EX.ViU PI.E. 2. If 7 oxen in '4 days cat 14 bushoB of corn, how lung vrill f» oxen be in eating SO bushels? ■ . . . ' -Ans.- 12 clays. ■ 7 COMPOUND PROPORTION. ' Argument.—If 7 oxen in 4 days eat 14 bushels, how long will 5 oxen be in eating thos§.14 bushels. Then har- , ing found the number of days (5f)- which it requires for the 5 oxen toeuf 14 bushels, we can easily find how long it requires for the same 5 oxen to'eat 80 bushels. f Fir ft statement 7 oxen": 5 oxen : : -f "days j which by inverse ratio gets foii4th term b-?d. ■ DELATION, -j yej4 ,^ta.temes;t 14bu. r-80b.u. 5y days . 1 which gets the answer 12 days. Tl^e ahove', I hope, is sufficient to explain 'the nature of qyuipouiuf proportion;, and in that way every problem can he-solved, but to render the*optratkgi easy we have a dul by'which even the ve'ry young pupil can solve questions in the 'Double Rule of Three without difficulty. RULE. "Without reference to the marks of proportion, consider fh-refe placesou the slate or board. Put die lii'ing or moving term in i\iQjirst place; that which dan nates time or dit- tnneI under thoso/ffi-c tlieni. When having thus stated, • notice where \}ic-hh>idv is. If under the third ternq multiply the first and second for a "divisor, and the oilier thrge together for a dividend. If the Miink is not under 'the third term, multiply the third and fuvrth /or a divisor and the other fur* a dividend. The answer will be like the term under which the hhnnh isi . Note.—Problems not pertinent to the above rule may be solved by two statements, as already explained. COMPOUND PROPORTION."' 9$ EXAMPLE. 3. If 12 workmen in 1'0 days can reap 500 ' Ac-res of wheat how many acres can 5 men reap in 7 .days ? .OPERATION. ^ ) First place. ' Second place. Third place. 12 lntj). 10 days. . . ,500 Acre*.' Oiuen.' ( 7,duys. ' ' , The hlan'h- is.under (kosthird. term, therefor® we multiply the first and second terms together for a divisor, 12x10= 120 divisor; and the other'three together for a' dividend, 500 X7 X5=17500 dividend ; Then ' 17500=120=Ans. 1151 Acres. EXAMPLE. 4. If 2401bs. of tohqcco will Iagt'3 men '5'yearn. How long Would" it require for 7 men to consufne 320-lbf?. ? ' Arts. 21 years. OPERATION. . Second place. Third place. " 5 years . 2401b. ' 3201b. # • The blank in the above statement is mo?'under the third term, so wC multiply the third '(2401b.) and fourth (,7 men) ' together ft>V a divisor. 240X1—1080, divisor'-; 'the othti" • three terms 3X0X020^4800, dividend. 4800"-t-1G80= 22 y«ars, Ans. ' ' ?* First place. 3 men 7 men 94 COMPOUND PROPORTION. EXAMPLES. 5. If 13'mechtinies liave 8240-^or 10' day's work. what will "be the wages of 31 mechanics for 8 di}"s ? ^ Ans. $457.84^3. (A If-8 oken consume 208'bus-hels of corn in 52 day?, • how many bushels.would 12 cows consume in 30 days, each cow eating jt as much'as an ox ? j ' ' ' . Ans. 150 bushels. . 7. If 3-mcn 4§ <4ft. X$ft. 5L.X324. 80ft. x 5ft. X^'h *. First .place. Spcd-nd place., Third place. ' ILeia,cing terms, 7 men • 80h. .SOO-.ctibicgfeet. x ' 160k. §2.00 cubic feet. Ttaii 800x 100=1-28000, divisor ; /3S00X80.X7=. 171V2000 dividend, l792000-f-T280&0'=14 rgen-, ;Ans.- Tliojibove work mpy be vastly abridged by. dividing tiid terms unddr'tlu/second arid third places.byi gtimc -number that will divide' them withpuf a -remainder. In the 2nd. place, both terms may be divided by 80, in, the 3rd.,, both may be divided by 800. The. sjbitemeiit .will then'stand thus— , . «/ ■ * / 7 men lb. . ,Tcu.,ft'« 21i. 4 cu. ft. Then 1 v 3 =2, divisor • 7 Xd=28, dividend•' 28-=-2t—' " 14 men, Ans. • • 12. If 5 men, working 7 hours a day for 3 days, dig '& ■ditch 12 feet long,,' 4 feet wide,•§ feet deep, how many men must be? employed* to dig a ditch CO fcpt iofjg, 8 filet .wide, att.d 0 feet deep, by working 0 hour's'a day for'14 clays, ' ' . -Ans. 23 rnOn. • 13. Haw many men could-dig-a pit 15ft. long, Oft wide, 6 ft. deep, by working 2 hours a day for 44'days, wlieh it takes ten m'en Y hours a day for 8 days to dig erte 18fh long, 15ft. wide, 12ft: deep ? ' " , Ans. '-3. >06 COMPOUND PROPORTJOJj", 14. IF 17 men., in 9'-days 8 hours a day, can' build a wall 14 ft.- long, 4 ft- thick, and 12ft. high, how many hours a day must 84 men work for 12 days to build a wall 28ft. loQg, 24ft. high, and 4ft. tj^ick ? ^ ' Ans. 12 hours a day. OPERATION. ■ ( Where there is & partial blank, as in the above, the term constituting the partial'blank is a factor of the- divisor. ■First place.,. Sec5Udpla.ee. Third place. , ' 1-7 men > 8h-X%h •• . 14ft.><4ft.xl^ft. , . 34 .men - 12h. 2bft/x24ft. ;<4ft. In the-above 'statement the partial Hunt- being! untie* .the second term, the third and fourth, as before, constitute the divisor together with the 12h. and it may be solved as the preceding examples; but it is perhaps hc.it -performed by cancdlinrp,pfocess I .will now illustrate.' For working any problem in compound proportion by cancel¬ ling we bave the. following, ItULE. „ State the proposition according to the rule bread)- given. Then police'the terms w^ose product would constitute the dividend, as also tl}o^e whose product would make the di-' visor j then express the whole in.the form- of a fraction, making the factors of the dividend (.with the sign of multi¬ plication between them), the numerator* and tliQse of the divisor, denominator. Ifp, partial blank;'occur, p>ut thgt term as a -factor of the denominator or divisor. Take the statement above: COMPOUND PROPORTION. OPERATION BY CANCELLING. ) X 8 -X 9 X ^ —j —c :—1—I—■—=7y •> [of cent.'is mean 5 out oPa hundred, or, 5 hundredths, and is Est espresso- d'edimalljr, thus .05 ; so six per cent, means six Jtundridih and .should he expressed .00, fee. The mAper cent, is the amount allowed for every huu d'red. v - • To find the percentage oh any given lYumber at a giv*e: rate per cen£ RPLE. Multiply the given number by .tire given rate espresso- d>. What percentage of 36-lb. is i0 per .cent, ? " . . . • . ' ^ Ans. 3.61b. 4. My agertt cbllects for 7 per cent., what is due Jiim for collecting $240 ? . - ' Ans-.. $16.80. 5\ What'pereentage of $57.20 is 6 per cent. ? • . • . Ai?s. $3,432. 6. What .is 1 per cent'-of 8100 ? -• Ans. 25 cent';. 7. What id 1 p^ l-xcent, of $3280. , ' ; % ' ' " ' ' • '$16'.40. ' Jo put on'per cent, on any given number. ItULE. Multiply tlic'given number by 100 -f- the rateper cent., and' point wfi" two.figurc.-4.f0r decimals.' , '' i EXAMPLES'.. v * • ■ ' ' 1. I buy corn at SO1 cents- per bushel. X want tQ put cu 20 per cent; - what will be my-pulce for i\ bushel / ' . ' ' ' Ahs. 60cts. OPERATION.. __ 50 COtttS. 420 •- ■ 60.00 100/ PER'CEXTAGE. 2, To gjain 30 per cfenF, what must I sell a yard o£ mus¬ lin ^br that oast 00 cents ?. . ' ' , Ans. 78cts. "3. A gentleman/worth §20000, by a fortunate trade in¬ creased-his money 50 per. cent.'; what is he then.worth ? Ans. $30000. 4. Cloth which cost $4.00 must be sold'for what to gain •25 per cent. ' ■ • Ans. 85.75. 5. Bought a h®rse for 81.50 ; what, will'1 get for him. if I gain o31 per cent.' • » ' Ans. 8200,. To take off per cent. r ' •' . . RULE. Multiply the given, number by 103 — the per cent and point off as before. EXAMPLES. 1. Having in Bank 83230 I wish- to withdraw -15 per cent.. How much will'remain in the bank? *v Ans. 82738.00. ' 'operation. 100—15==S5; SiSO'y'* .85—2788.00, 2. A hoi'sc worth $120 loses One eye, which, depreciates him 20. per cent.,' what is he now worth?" . , _ , Ans. 800.00. 3.. From 00 bushels of rye deduct 30 pev cent.; how much remains'? 1 Ans. Gvlbu. PERCENTAGE'. 101 4. A man buys sugar at 8f cents per lb., and sells it at a loss of 20 per cent., what does he get for it per lb. ? • . ' Ans; T cents. 5. I gave1 62} cents for a knife*; sold it at a loss of 5 .per eent.; what did I .get for it ? ■ Ans. oO-jj-cts. To hike off percentage that has been put on. RULE. - Annex two ciphers to the given number and divide by '100 + the a-ate per cent.' examples. \ K ' J- • • 1. Sold a# article.for'7,5 cents, and gained 25 per eent. ; what did it first cost ? , Ans. 60 cents. operation. 100+25—125; annexing ciphers to 1 75=750(,);. 7500,-^l2o=60'Ans. 2". 25 pe-r cent. has been/p$£ on a certain •dumberand the amount is 30, what is the number? •» ... t Ans. 24- 3. What number is that Vhieh increased by 40 per eent. becomes 65. ' -J Ans. 45.<- . 4. By selling ^ horse for $132 I gain 10; per cent. ; what did he cost fiie ? ' ' Ans. '$12(1 s 5. A number increased by 20 per cent, is 6; what is the n'uiuber ? ' , Ans. o. ' To put on a per cent, that hits Jbeen tak'ert off; 10,2 yBUCENTIA-GEy RULE. Annex two' ciphers and divide % 100 minus the rate per cent. EXAMPLES. 1. 'Sold an. article for 03 cents, and lost 10 per c,ent.; what did it cost,. ' Ans. 70 cents. (jpekation. 'Go with ciphers«anHexed becomes G300 ; 100—10—90 • 0300-^90=70 Ans. 3. A boy losing 12 per cent, of his apjples had 44 ye- malnlng; how. many Igvl he at first ? ' 1 Ans. 50. • 3. After taking 15 per cent, out of my purse .there re¬ mained £102 : hojv much did it contain at first Ans. £120. • * 4. "What was the length of a pole, which having 10 per cent, byokeii oil is still 21.0 feet long ? . Afts.. 24 feetl To find whapper cent, one number is of amother. x RULE. Ap-nAc two ciphers to the number denoting the' percent¬ age, ami divide by the number on which the percentage is reckoned. EXAMrp.ES. t 1What rufe per cent.- of 10 is' 2 ? ' Ans. 121. PARTNERSHIP. , operation. 2"P0^-r6r=121 Ans. ' 2\ "^Vhat rate per cent. of. 24 is G*? 'Ans. 25. .. > 3. What rate per pent, of 110 is 11 ? 'Alls., 10. 4. A boy.had. 12 marbles,—lost 2 what per .cent. is. that? • . ^ns. 10 H. ' 5. A pole 13 feet long, had ,3ft. broken off;-what per c-ent'.?, • •' Ans. .28-J-. Nor/.—As this thhig of percentage will he farther investigated hereafter, I d'eern the above sufficient for the present. v I'AUTin ERSII IP.' / ' Partnership is an association of two or move pel-sons in business, in which the profits and losses ni;e to-be shared in proportion'to the amount of stopk each 'contributes tej tlie capital. ... ■ EXAMPLES. I 1. In a partnership business A contributes-£40, B 850, C 800. Tiiey gain $30, What is each ipan's share of the gain"? C A. 88:00. Ans. ] B. 810.00. ( C. 812.00. 104 PARTNERSHIP'. OPERATION. A'c. stock 840 BV , " ' 8o0 <"s. ; 800 <1apital 8150, of which A contributes 1f'»j or pk ; IX vVo 01 ■--! (-1, tVo'01' %' ^ ig evidently entitled, to pk <4 830 ; B, to X pf it ; C, to § of it. . To i^et A's share tlien multiply 830 by pk=88 for A. To yet B's ilf . '*• 8o0 by -1=810 /'or B. To yet C's " " 830 by 3=812 for (_'. . The correctness of the argument in the above is evident: therefore the following IlTTLE. I. Multiply the whole gain or loss by Cac-h numbs part of the capital expresses! fractionally. Or, As the capital is -to the gain or loss so is each man's stock to his share. » 2. Two merchants A $ind\B, trade fn partnership, with a ettpita'1 of 812000 ; of which A furnishes 87000, B 85000. They gain 82400.' ■ What is the share of each ? ' ' . ( A's 81*00.. ... A"f' IB'sSlki. 3. "Two merchants M and N trade in partnership. M contributes 885.50, N, 874,50. They lose 10 per cent, on their capital. What is each one's .share of the loss? » „ ) M's 88.55'. Ans' [ N's «7.«. PAEIXEBSHIP., ' 10& 4. Three men A, 13, C, trade with a capital,of §1G50. A put in $440"; ITput in, 25.per cent/more than A ; C.pnt in 20 per cent, more than 33/ T-hey gain 40, per cept. of tlieir capital. What, is each man's shape of the gain ? f A'aSl-70. 'Ans. -] B\§220. (_ C's §204. •"). A "merchant failing, owes the following debts, to If §4241, to G .§4502, to \t §5000. His effects anjount t<> §7000,' which lie delivers up tp them. How much -will each receive towards' liis demand ? . ' C F §1772.00-'- AnsJ Cf $2494.25 — (_ n§2783.74-|- 0. .James, John, -'and IJenry buy a. horse for §150.— Jatncs paying $00,. Johh §50, and Henry the'balance. They sell- the horse for §1 to. -What doeS each.map 'make "'eieajf^ by the transaction ? * ' - { James §10.00. • Ans. - .Jphn * f Henry §0,007' lAVHTXERSIIIP OX TIME. ' - To find each partner's share of profit or loss when the stock is employed for different periods 'of time. RULE. ' , <• % Multiply each -partner's stock by its time p thin as the su,m of these products is to each product so is the whole gain or lo§s to each partner's share of the gain or loa*. PARTNERSHIP. EXAMPLES. 1. A put in $50 •for*-1 months ; B put in $60 .for .') months j-Mmd they gain $t5. What is the shave of each '( Ans.. A. $$>; B. $0. .fil'ERATKtX. 50 X4=^200' 500; 200 : $15 : 0 Aris. for A. c (>0x5—B00 50'0 : BOO : 15 : 0 Aris. for 15. , 500 . The principle op which the aby>ve is clone is -simply the reducing of eaehstockto a common time.. ^8 50 for .4. months is the saine as $200 for 1 month ; $00'for 5 months is the same as $300 for 1 month. Then A- lias in. $200 ;for 1 montji, and B ha^.in $300 for the same time ; and the cap¬ ital is $500. llcnce the Bnle, as we gave it. > * f '2. A commenced business'on the 1st of January with a capital of $1,200 ; on the 1st of April he took.in T> .with ;w 'capital of'$1000 ; on the 1st of September he took, in (' dith a capital of $000. They gained by t]ie end of the year $10$0. What is each man's share of the gain ? i A's $4$0. Ans..- B's $480. (WA812U. j 5. Tlirec men Lire a pasture for $17.00. ~ A.pubs,in S oxen for 10 weeks; B puts in 12 oxen fori weeks ; C puts in 0 oxcu lbr 15 weeks. How much ought each to pay of the money ? ' ' A, 85.54rVV' B, $5 82Tk'^ C, $6.23TtV. PARTNERSHIP. IOt 4. A lures a carriage for 87.20 to go 00 miles-; Graving jgoue 30 miles lie takes in B;. and having gone 80 miles farther lie takes in C> How much must each man pay ? f A must pay 84.40. - B " « 2.00. (C « « ".-80. .'Solution of the above Probhtn- V alopc goes 80 miles A k B together go .80 miles A, B k C together go 30 miles 00 • • ■ 00 \ 80*: : 87.20-: 240 A pays. 001 -80 ; : 87.20 : 240 A & B pay. ' 00 : 80 ; : 87.20 ; 240 A? B k c'pay. Xow A owes 2.4Q-]-'1.20'-f .80^44.40 Ans. for A. H B owes 1.20-f"^0- -42.00 Ans. for Bv 2 C owes ,80,y.80 Ans. for (1. Jfhe student will notice.ip distributing the prices that A not only' has to pay'h.U 82.40 for riding 80 miles alone but nlsoJie must pay hatyof the 82.40 for the 30 miles'' ride that ho was accompanied by B, and also a third of .the 4240 fop the ride in which 4m was accompanied by B & C. And B-hut only owes half of the 82.40 which accrues while accompanied by A,Jmt aJsd n third of that Which Wats duo for the laet 80 miles. 108 INTEREST. 5. TVo men A and B in Boston lure a jearriage for to go to Concord; N. II. the distance being 72 miles (there, and back 144 miles). Having gone 20 miles t-hey take in C ; at Concord they take in X>; and when within oO h>ile.> of Boston on their return they take in E. Tv hat must each pay V f A must pay 87.00,0 — IB" " r.00,0 — Ans.'' C " " f».S7,3*— | J) " ■< 2.80,4-}- | E " " 1.04?,1-'- INTEREST. Ihtetest is the Compcnsatiorr which .the borrower of money makes to the lender, or which the debtor makes to the creditor when the moi^ey remains in his hands after k is doe; and E generally reckoned at& certain rate per cciut. usually for one year. ' * . " The Principal is the sum due on which interest is com-' puted. . - • Tliekl»?OM>k is the interest and principal added together. Simple Interest is that reckoned 6n the principal only and is .what is meant when the term intei\\i" Is used 'alone. Legal interest is the rate per cent.'established by law: IXhEIiEST. 100 It* a rate per cent., higher than is allowed by law is takSh it "is called usury. ' ' The rate per cent, established hy law-is not the same in1 all the States. _ > In California it is 10 per cent. ■ In Ajabama and Texas 8 per cent. t , » In Nqw Tbrk, Michigan, Wisconsin, Minnesota, Geor¬ gia and Strath Carolina, it is 7 per cent. » ' In Lousiana it is 5 p'cr-cent. ' . In the other States of the,Union it is 0 pc^ cent." In Kansas and Nebraska Territories'it is G per cent. In the District of Columbia, and on debts in favor of .the United States it is G per cfent. The rate' per cent as we have" before said, is so much on a 'hundred, or so many hundredths. 5 per cent, then is is sipiply 105 or of the principal To find the'interest tin any Sura for-any number of years at aijy ratb.per'cent, ' RULfi!.. < Jfydtiply the principal hi/ the rate*+pcr aunt., and that product by the years, and point off tiro decimals. Examples. . 1,, IVhat is the interest of $75 for 2 years at 7 percent? Ans. $10.50. . 'operation". Yfe first multiply $75 by 7 per Cent, whicli gives $5.25. Multiplying s by .the per cent, and pointing off two fig- tir*s always gets the: interest 'for. one ye^r. $5.25 is tpe in¬ terest of <175 for 1 year- at 7 per cent. Then multiplying ly the 2 gives it for 2 years' Thus $5.25/ 2 ' $10.50ctd < - 2. What k the interest of $17 60 for 8 years at 7 prr cent? 83.OO.OOti?. hoTK--~lf decimals are given iti the principal, there trrr'si It twp t.{tra decimals ill the product. JTlnrrcfore. if rents are givrp them wiit bd four decimals in the .product) the first tv.0 of.which Re .may consider cents, provided thcrt are as many a? four lV-ur^s in the product- Yfb.cre dollar's alone are given the product Mill' he doiki-s and cents'provided there arc as many as three figures hi tim product. 8. Wligt-is the itvhlrest of $12 for T yehf at 7 per fifth '■ AitSc- 8-1 r-olivs of ■f.'W. . 4. \V hut is Urn ifttoR-st of $51.00 for 4 Jeturs at x per dibit <1 Aiisi ?2 per fulitr vrhdl. is the interest of $37 for 2 yit;rs f -' • - Atis, $8.70. 0). At G per eeht. for © years what k the interest of $uo y • Atts..$t(h£0, INTEREST. Ill' 7/ What is the interest of $75.20 for .8 years at X2§ per' - iiion would then be thus— 720 . 7 50.40 . 7 30)35280 11.76 Ans. 2. What is the interest of $100Q for-180 days at 7 per cent. 'I . Ans. $35.00. .3, The interest of $17,20 for 60 days at 8 per cent. ? Ans. 0.2299 or 22 cents 9 inills. .114 INTEREST. 4. The interest of §47.65 for 18 days at iQper cent £ Ans. 17 cents, 2 mills. 5. The interest of §3600 for 3 dqys at 7 per cent. ?' "Ans. llO. " •Note.—The process iu the above examples considers SO days to the month and 12 months to the year, which has the sanction t>f general usage and the decision of the Courts though not entirely accurate. Where perfect accuracy is required 355 must' be'usci instead of 35th Where the interest is required for months and days. RULE. Reduce the g luen time to days and proceed as in the I a examples. •' '* EXAMPLES'. 4 *1. What is the interest of §75 for 9 months asd 20 days at 7 per cent. ? Ans. 4.22{L The interest of §3.00 at 6 per cent, for 11 months . ana 10 days ? Ans. 204'eents- . 3. The interest of §790 fo-r 4 months and 2 days'at 7 per cent. ? Ans. §18.744- 4. The interest of §49.81 for 1 month and 28 days at 8 per cent. ? «-v Ans. 64 cercts-i- 5." The interost of §7300 for 3 months and 15 days r®fc 7 per cent ? . * Ans. §149.04 INTEREST '115 ■ l'u dud thp interest W-liere years, inon'ths, and days are -Wen.' BULTfi. Multiply the principal by the per apt. and that product by, the years,. and take the-aliquot parts for the months a nd day a, addi-nf/ said parts to the prqdtictyoblained bj> rnulti- jdi/ivf by the years. * ' , " I , - _ EXAMPLES. ' 1. What "is the interest, of $240 fur .5 years, 4-'months and 15 days, at 7 per cent. •?». * Ans. aoo-.ho. OPERATION. • *:&mo 4 months arc 1 15 days of 4\no. arc I lb*.St) interest-for 1 year. 84.00 x, 'a 5. years, sA o •'.70 $90.30 An». 2. The interest of $720 for 2 years, 1 -month,' and- 10 "days iit. 7 per cerit. ? * 4 •' •• ■ Ans. $100.40: 3. "The interest of $8'4.09 for 3'years 7 months and 12 days at 8 jier centA Ans, $2447711. 116 INTEREST. 4. Jhe interest pf $19.60 for 1 year,, 1 month and IS days, at 6 per cent-? • • ' . ^ , Ans.^Sl.33-28. 5. The interest of 96 cents fur,4 years and 20 days ut 7 per cent. ? ^■ Ajis..25cts.^- . To 'find th&-Principal when tipe interest, the time, and and the rate per cent, are given. BULE. IHciflc the given interesfb;?/ the ' interest' of SI - far -the given time a.t the green rate pefr cent.' examples. t , * 1. A certain sum is lent out at 5 per cent, for 4 years, and the interest is $4.80.; u'liaf is the-sum lent ? • Ans. $24. • operation. $4.80-^.20=4624 Ans. The interest of 81 ut 5- per cent' for 4 yours is 20 cents or 8 .20. There, fore (according to division of detinmls^ 4.S0-f-.20= a v.'hole munher 24, -which is dollars. ; 2. I borrowed a sum qf jnynoy, ajid t]ac interest for 2 years was $44* the rate per cent, being 10 j what was the siim borrowed ' ' Ans. 8400. 3. "What prijiciptvl at interest 5 years at 7 pet cent, will gain 845.96 ? •. > ' Ans. $4.5,00. INTEREST, it 17 \\ Lent a rfina.ll -sum, of money ^Ahe interest at-8 per cent, was 84,08 for S hears, and .8 -months,. IVhat was, the sunn lent'( ' . ' • ' ,-v . , • .. AnsA $18. 5. Of a certain principal 840- is $he .interest* at 8 per cent. for.50 'days ; what is the principal ? ,• * , v, ' i . Ans-. 83000. . t , 1 *1 ' ,0. A sum* of'money wag on interest. 7. .monflm ; the in¬ terest at -5r per cent, was 810.00 what was the sum ? ^ ( ' • ; * » $-®7t- • *J. What principal at interest for 3 years and 5 months will gain 85&.Q8L at 5 per cent. ? • • ; - v ■ . .w ««. 8. What principal at interest for 3 j'ears. ajtd 7 «pjonfiiis; the rate per cent, being 7§ will gain §G$.68 pE, ■ ' Ans. 8250. T6 find the rat'c'jicr en\t\ 'the prihcipal; ihc'interest. and the time Ufeing given'.- " 4 '' • "il'L'LK Dibule the given/interest by 'the irtferest of (lit gie/n principal 'at-1 per,cent., for the given time. < v EXAMPLES. * l' 1. The interest gf 8100 for 2 year,sA is 822.40 ; what is the rate per cent. ? * Ans. 7. • operation. 8160X-01X2 years =83.20;• '22.40-f-v 3,20=7 Ans. 2. The interest of 8240. for 3 years is 843.20 ; what is the rate per cent.'!' • • . Ans. 13 per cent. 118 INTEREST. . 8. The interest of 024.60 for 2-years and" 5 feontlA id $4,756 • what is the rate per cent-. f ( Aps. 8, 4. The interest of $12480 for 3 years and 1 month was $1924; what was the rate per cfent.? . ' ' *' , Anl 5. 5/ ITent a'man $39.60* lie kept it! 10 months and 18 days, and paid $2.i>2^- interest; what'Was the rate'per dent.2 ' ' ' - Ans.- 7-}. <1. The amount of 0360 for 2 years h> $410.40; wlmt its the rate per cent. ? ' . Ans. 7. To find lhe Time, the principal, interest, and rate per cent, being'given, RULE: i Divide fee pi Sen interest by, the interest of the pu'eit principal for one year at the tjiveh rate per 'cent, EXAMPLE?. 1. IIow long must 01250 be on inlerqst to gain S21S.75 at 7 per cent.• Ans. 2 years, 6 mo. OPERATION. 12.50' .07 ' 87.50)218.75(2y. ' 175 00 ' 43.75 12 ( 52500( Oiuo. 52500 " INTEREST' W . 2, MaW long must $05.40 be on interest at 0 per cent, to gain 61Q.501. , Am. ly. lOmo.-p 8. ■Haw'kmg must 6l2GObe on, interest at jf per cept,. to gain -6 si 7.50 ? ' Ans. ]y 4inov 20(L 4. I lent 8oV2 at S per .cent.,,.and received for interest _ $17.40; how long "Was it out? : Ahs. 7 months •5. l)6w k»i0 must 67^0 be on interest at»5 per cent, to gain 61.00? ' Ans. 10 days, G. Itow long must 625v00 he rfn interest a£ 12 h,per e'en i. to gain.670.SO? -Am. 24 years. 'I'lhOM LSS-ORNT K"OT;ES, • ' 1 A promh-'sary n'qte or notp of hand is ail engagement in Writing to pay a specified sjuni,-either to a person named in the note or to his order, or fq the hearer of the note.' ■ 1 . A note should contain the words "Value i&eciYcd/' and the sum promised should he expressed in written words, 'the usual way is to express tlio dollars in words, and /lie fotds, as hundredths of a dollar, in the form of a common fraction, /fo find tho am on vi t of A promissory note on whreh there has been- ho payment'made, lib,LIE tSuhtftici the xldtc bit which the note i$.'ihiC ft'Otii the toilnkjehivh the interest is to'be reckoned The renta lntUe will he the time the note.hat been bit interest^ Theft <'.re<] aebwdititf to rides 'already yieen. • I'liO I^TEftKSf". EXAMPLE'S '1. $360. ' ' < Augusta', GTa., July 18th, 1*54.' One tlay after date I promise to pa/. J$mes Smith bearer, Three H'tindrod and Si^-ty Dollars for vahie re¬ ceived. . • * T i motliy 'Doofi tt-1 e• What was the note worth Sept. 28th, 1856? _ X 'Ans.8415.a0. OPERATION. 1856,' 9,. 28 1854, 7, • -18 t i. ^ . | 2y., 2m@., lOdays, tune far which it draws interest. •Then,' 83 .fit) 007' i ^mo. 10da. \! 25.20 in. for 1 wear. t> . 50.40 u {i 2 years. •4 20 .70 55 30 in. for the whole-time., Adding principal, 360 841'5.30 Ans. 2,836.60 1 ' ,Atlanta, jQa., April lOtli, 1855. One day after date 1 promise to pay Ilenry Lang or bearerHlurty^six &' TtJ°o'<1o11oa*s for value received. '' * " ' v' ' John Dbe. I^TERlilSTi • i2i N,Wh"at Was ilrrc August' 2 nth; 1857 ? Ahfe/$42.08475. ' *• •» 3. $392.18. Savannah, Ga., Oct. 19th, 1S5G. . Rgr vahic received I -promise, to pay .Richard Roe or order, o'n- demands, thrco hundred and ninety two dollars tyid'.eighteencerits, .with interest, y. . • ^ John-Doc". > * What wa? due July lOtl). 1§57 ? » ' Ajis. 8412.08 + * « 4; 824.48. ' Charleston, S: C., July.3rd, 185-3. ' Six months aftei' date I promise to pay. Gideoh Miles or bearer, twenty-1'our dolhlrs .and forty eight cent^fot value received. - " - "Srmon Craft." ' Wlk't was due December 25th 1857 ? .*'' »•« Ans.-834.30 nfeaA^y'. 5. 81280 ,'Mobilq;,'Ak., Jami-ary lOt^i, 1853. . . _ 15y the. 25th, of Debetnbei: next, I pronyse to pay Saniucl Sini]+ins iQ-i^jbCaref one thousand tvpK) fiun d-rexl.an+eit+ty dollars -for value yeceived. ' ' , Arthur-.Ileltoh,, W'hathvas due May412tji,. 1856 ? . . Ans. $1523.76,-8. 6. $1500-. A/ . . Macon, pa, March.4th.) 1856. X+ie day after date •£ premise., to pay-SanGSliok, 04.'-beater' Sftebn 'hunxlred' dollars for value? receive A > ■. » \ 4 i , ' . Major Jones. . What is due Dec. 25th) 1857 ?■ ' ' ' Ans. '$1689.87, * . , \ , "Note.—pit isr-cuStonuirj+vliero the author lives, to .cbrfeider 30 days to the niontfl W'ilhoat' 2h?'ti Action, Scf have -we done'in 'tin? above examples.. ■ ' . » 122. , YnterestP Partial payments on" notes, .bonds, or <>fcher oLlig?4 Compute th'(\ interest to the time pf thc_ Jhrst pa.pnunl, ivfyichj vkmc or xqitjo precedinggagriuente, exceeds f./sc in~ terest then due* • ■ fAddrthat interest, to the. prineiptd, rindd>01n lip. w-ovrd $ubt'r.&ct ithc pftymeiit or payments thus far ma dd. ' The rdnumde# wilt form et new principal^ 'on K&rhputf the? interest;proceeding a» hrforc. •• As* the'.''above rtd^ia'the saftie sh that ■sanctioned by'btw and c'omipoit usage in hur .State (Noorgia} I shall So' Vo«n-v pute in the following, EXAMJPLES. ( 1,. 83'20. . . • (frifftn, Uco„ June 20, ls54. ' One- day after date ' I promise to pay Timothy .Ronsii- liert'd eft bearer,the sum of three hltndrcd -and twentv'Vkrl- for value rtech'ed.,- Pefer Snooks, On, tliis note are the the following indorsements, , August I0r 1E55, received $75. Oetoihm; 15,*185d, receiv¬ ed 820,0. . k f ' * ^ ' • I , Row Iftuhh remmns due 185?. , Amu jlOO.G2pU INTEREST. •123 OPERATION, t P/^ncipal carrying interest from June, 20, 1854. . $320.00,0 interest from Jutie 20,1854to Al\g. 10', 1855(ly. Ini, 20'd.') 25.51,0 Amount,-' 345.hi,0 First payment Aug. 10, 1855, 75.00.O Balance- for new principal, 270.51,0 Interest from Aug. 10, 1855, to Oct. 15,1856 (ly. 2in. 5pay T. B. Frost or bear¬ er, thirty-seven dollars and.ninety cents, for value received. Pat D'onnovan. Payments': May S,l§57, fifteen dollafs and twenty cents; August 8, 1857, ten dollars and eighteen cents; October 10,. 1857, fivp. dollars. What is due Dec. 25, 1857 ? Ans. §0.17.,6. 1 J ' .4. $1800,. Augusta, Ga., Sepd. 9, .1855. •One day* after date I promise to pay A. Griggs or beayery the sum of eighteen hundred dollars, fbr value received. FL IV. Flatfoot. •ludorsements: November 10, 18513, seven hundred dol¬ lars-.; . April 20,1857, one thousand dollars. What is the note worth Jan. 1,.185S ? Ans. 8308 024-. 5, $>3580. " Columbus, Ga., Jan. 3, 1854. • . hour months after uaie I promise to pay 8. W. Lyiln or Lcaher three thousand five'hundrpdyand eighty dollars for value received. T. B. Littlesense.' Payments Apfil 18, 1S55, twelve hundred dollars; January 28, 185 Principal for 3d year, : 561.8000 " ' 1 f. - .06 » " • v - Interest for 3d yeefr,., 33.708000 ■ ' ' * 561.80- Amount for 3- years,.T.'. '. : .v. .*.8595.508000 -i - < . s. 2. Wliat is the. compound interest of $360 for 2 years- and 6 months, at 7 per c^nt. ? , . Ans. $66.58974. COMPOUND INliBBST. ©PEHATION. Principal for 1st year,.. * $360 * * t Interest for 1st'year, v l, .. 25.20 360. Principal for 2fl year, • 385.20 .07 Interest for 2d year,... 26.9640 385.20 Prici'pal for.£ the 3d year, .412.1640 .07 2)28.851480 Interest for J the 3d yeaj,.....n..,.^.?» .'.......14.425740 412.164 'Amount for 2fi years, A .* 426.589740 . * . • ' < . ,360'. *' ' , ■ ■ ' j Compound interest for 2-4 years^. .>.$66.589740 3. What is thp amount of, $1000 Jor 4 years ,at 8 per •cent .compound. interest? - • Ans.. $1360A8896. , 4. What is th<§ amount of $280 for 3 years and 3 months at 5 per cent. ? , • ' Ans'. $328x18cts/and46 milis—(—, 5'.; What is thd com'porlnd interest of ''$300 for 4 ■«&»•« and' 6 months at 7 per cent. ? 'i « Ans. lOT.00,-2- 128 DISCOUNT. 6. What is the compound'interest of $120 for 3 years 1 month and 1.5 days at 6 per cent-'? ,/ Afis. 28,09—}—- 7. What is the amount of $500 for 7 years at 7 per cent, compound interest? i ■ ' . Ans. $802.90 nearly. '8. What is the compound interest of $20 for 20 years ? Ans. $57.89-p- DISCOUNT. Discount is a deduction made &S a compensation for pay¬ ing money before it is due. t Present worth is the amount of ready money that will •satisfy a debt before it is due. The two kinds of discount in general useT shall distin- gush by the terms true and nominal. The interest or pcrccntag-e of any sum cannot properly be taken as the true discount;' for the interest;,as wb'have al¬ ready seen, is the fractional part of the sum at interest, with" the late pp"r feent, for a muuerator and 100 for denominator. Whereas1 the discount of a sum for one year is the Frac¬ tional part of said sum with the rate percent, for nutoierator, and 100 plus the rate per cent, for denominator. To illus¬ trate the difference, the interest of 100 dollars for a year at 5 per cent, "is or of the sum; but the discount of a DISCOUNT. 12d smim that would produce, 100 .dollars present ftorth; or °f ^1C sum discounted- ' * v w - . Business men., however, often'take off the interest of any sum to be discounted, by which they do not obtain the true but the nominal discount. Thfs method'necessarily obtains where a due note is to be discounted, for the discount is mado'without reference to fime. > .• The true present-worth of$100 due'a-year hence* i§ sim-< ply a sum of money thatwduld amount to $100 a your hence, if'put out at interest.now. • ^ ' ' To find the true present worth of any £um for afty time at any rtite per cent. RULE. Divide the given sum by the, amount of%\ jbr the gu-e?[ time at the given: rate per cmt., and the quotient will he the true present wortfj.. _ . From the given sum subtract the present worth, and the remainder will be the discount. EXAMPLES. iVhat is the present -worth of $802.b0'*at 7 per -cent., for 1 year ? ' Ans. $750. , r OPERATION. $1 1,07)802.50($750 Ans. .OS 74!) Interest for 1 year, .07 535 ' 1.00 535 Answer for 1 *year, 1.07 0 130 D1SCOITUT. <> ,2* h-hold A's np'te for 893.60,.due 2 years hence; what is it worth if he pay it now, discounted at 7 per cent. ? , , . Abs. 3. What'is the discount of $117.60 due one year hence si'l2 per-ccnt.? •• <■ • Ans. 812.60 > 4. A note for 818.60 is due 1-year and 6 months hence, wlvtfc is-the discount on it if paid now, at 6 percent. ? . ' • - Ans. $1.58-j^3g.• , 5. What is the present worth'of 86-10.80 due in 1 year and 4 months, at 6 per cent. ? Ans. S5Q3.331. 6. What is the discount of 8263-01 due in 2 years and 0 months at 8 per cent. ? . • * • Ai>s. 848.51. 7. What is the present worth of 642.60 due in'7 months and 18 days, at 12 per"cent, discount ? . ■ ' ' . ' 'Ans. 809,77i§i. 8. What is the present worth of 816.011 for 2 'years, S months and 15 days, at 10 per cent, discount? •, ' Ans. 81*2.60.' 0. X bought of Wm. Kay "in Atlanta a hill of books a- mountjng 'to 878.50 ; for ready money he agreed to deduct 20 per cent/ What, amount did I pay in cash for the books ? Ans. 862.80. 1U. A bill of goods bought on time amounts to 8060; but for cash down the merchant deducts 10 per ceut. What must bo paid down for the goods ? Aus. BS^l- BANKING. BAKKING Is a general business done at banlgs. A ba'nk is a joint-stock company established to -receive d6posIis; fend money/ deal in exchange, or isetie bank bills, as a circulating medium, redeemable 111 specie at its plafec of business. ' '' The capital of a bank is the money paid in by thevstock- holders, as a basi's of business. The affairs of a bank are usually managedvby a bmrd_ qf directors. The chief officers pf a bank are a presided, ■cashier, and •teller. The president and cashier sign th'O bills issued ; the cashier superihteuds tlie bank accounts; and the teller attends to tile receiving and paying out of money. Bank discQunt is nothing more than the simple interest of a note, &e. deducted from it in advance. The interest is computed for three days more than the jspyeified time, which three are called1 " days of grace." The legal-ratfe of discount is generally the same as the legal rate.-of interest. > 1 A notQ is said to be discounted a,t a bank when it is re¬ ceived. as security .for the money paid for it, after deducting the interest for the»time it has to run. The time a note has to run is counted in days frPm the day on which'it is discounted to the day of its being due plus the three days o-f grace. To find the bank discount of any note for any time at any rate per cent. 132 BACKING. RULE. Find the interest on the note, or sum discounted, for the time it has to run, .(which is the givpn time'pins three daps) AYha,t is.the bank discount on a note for $7800, Guy in 90 days1, at G per cent.? Ans." $'120.90. ( OPERATION. ' '' * ' 7800 .06 468.00 140400 421200 300-)4352400( 120.90 Ans. 800 732, 720 8240 8240 We add, the three days of grace to the 90; and then get the interest for '93 days, Which is the bank discount. This work may be abridged by taking thy parts for 93 days thus— 7800 .06 • 90 days 46800 » 117.00 3.90 120.90 AnK BANKING'. 133 2. Wl^at is. the bank discount on a note for §1090, having 63 days to run, at 7 per cent. ? Ans. §12.25. What Is the bank discount on a note of §1200, pay¬ able in GO days, at 7 per cent.,? Ans. §l4.70, 4. A note for $*78G0 dije.in & months was discounted* at a bank, at G per cent • W,hat did the holder of the note £ot for it ? • . ' ' . •' • Ans. $76?j&.27. '•). llQiat is the present wor.th'of'a note for $72 due in 8 months, discduiifed at a bank at 7 per cent: ? Ans. §68.594. . '' G.' §3G0. . Atlanta, Gra., Dec. 3, 185G. '• Hix months after date, f6'r value received, I promise to pay J. W. Dun or order thn*ee hundred and sixty dollars' at the Bank of Fultop. '• \ ' S.S.'Toddlewell. If the above pote was 'discounted April 3; 1857, how' long had it to-run', and* what was received for it? . f Had to G3 days.. •ns. I procce(]s $355.50. OPERATION. ' , year ■ m d Time-note was given • 1856, 12* .3 6, 0 Time note Vas due 1857, ■ 6, 3 Time it was discounted 1857, > 4, 3 Time till note is due ' 2, 0 Adding days of grace 3 I 1 Time it hijs to xun 2, 3 or G3 days Ans. 134 BANKING. $p60 principal for 63 davs. ,4)7 60 3 25.20 interest for 1 year. 4.20 21 $360- $4.41 bank discount. $4.41=$355.59 Ans.' * 7. $1440. Augusta, Cra., Jan _8th, 1857t Bight' months after date I promise tg pay,, for value re¬ ceived, tt> Scranton, Smith & Co. fourteen hundred and forty dollars, at the Bank of Augusta. Tom. Tompkins. , [f the above note was discounted April 23, 1857, how long had it to run and what was received for it? ,• » ( To run 138 days. Il0' [ Prese'nt worth $1401.36. 8. $1250. ' Colurnhps, Ga., July' 18th, 1856. Six mouths after date, for value recefved I premise to pay Julius Snow,, or order, twelve hundred and fifty, dol¬ lars, at the Bank of Savannah. Sam'. Pluspigly. If the above note was discounted at the bank on the 25th of December, 1856, liow long had it to run, and what was fcho bank discount ? Vns f26days' w-m- TOLLING. IS© TOLLllSTGr. Toll is a eerfcdo' amount bf grain takfefi at the mill as a compensation for grinding, of of cotton, taken at the gin as a compensation for ginping. 1*011 (from the Latin tollo, \o take away), is taken out of tli'e. produce to he ground or gihned. If a milled grinds for "the eighthit means that, if you. fca-vry eight bushels of grain to his- mill to be ground, he will take out 1 bushel'as toll, and! grind 7 bushels for you. tfo, considered dereference to the whole it is the eighth, ' +hafc he grinds for; but considered in reference to what is left after (he tolling, or what he really grinds for his cus¬ tomer, it "is the seventh: " • To find thb toll of any quantity-of grain .or cotton, when- the toll is paid- out olthe grain or cotton carried, ' •' RULE. Multiply the xchole quantity by the part taken as folk expressed- fractionally. <' EXAMPLES. I. If I carry two bushels of wheat to a mill where the loll is an eighth, what is t(ie toll ? Ans. 1 peek. OPERATION.. - 2 bu-X g^b or i bu;==;l peck. vm T.OLLING. 2. What toll, the toll being an eighth, must be taken fvevp 18'bu. 2 pe of corn!* Ans. 2 bu., 11 pe., 3c What quantity of cotton must be taken from 1300 pounds of seed cotton at a gin- where they gin for the thir¬ teenth ? ' Ans. 100 lbs. 4. A carries to B'.s gin 3600,lb. of seed cottqn; if B gins for the 12th/ how niauy pounds will he take ? ' Ans. 30Q." ' 0. A carries to B's mill 2Q bushels of graiij ; if B grinds for the 8th how .much will bo, the toll.? • • . . An,s. 2} bu. ,-To find the tpllv where the toll is not taken out at the tigie of grinding or ginning, or threshing, RULE. Multiply tfie amount ground, ginned, or threshed hy {he- part representing .the toll\ expressed fewHonally- qs hr/ifr-,:, hut* ivith the denominator diminished' by 1. EXAMPLES. 1.^ A gins for the twelfth. B carries to A's gin 1800 lbs. of seed cotton,'saying that he wants It all ginned'and put ■ into one bag, and that he will bring the toll to-morrow.— how many pounds must he bring ? • • . Ans. 16,8T7T lb. .OPERATION. . 1,800 lb. X11=163^ lb. Ans. 2. A'grinds for the sixteenth/ B earries 16 bu. of grain, and has -it all ground; how much does! he owe for toll? ■ >' Ans. lbu. Op. 2 p.- qts. BARTER. 137 3. I thresh wheat for my .neighbors for the tenth. An my absence, my overseer sent away 20Q bushels, and re¬ tained, 20 bushels for• toll*1 how'much did he lose mc^by that operation ? „ Ans. 2 bushels. 4. I carry'to a g"m; where they take tke't2th for toll, 2400 lbs." of, cotton. If the toll is taken out as ushal, i| will be 200 lbs.; if it is to be paid to-morrow what will it be? > . Ans. 218yd lb.- 5. I carry tb a mill where they take the sixteenth'for toll, 32 bushels of wheat. 1'say to '.the miller, "firirid it and I'll bring you your"2 bushels1, for toll to'-rijorrow."— He agrees'to it. IIow much does he lose ? ' Ans.-yv ofh bushel. BARTER. Barter is the exchange of one kind -of merchandise for another Without loss to either party. ' To find what quantity of anything at a given pfce is equal hi value to some'other quantity Whose price is given,- RULE. • Divide the value of1 the given quantity at its' price tfy the price of the quajitity which you wi$h to ascertain <{he quotient will te the quantity. To find at whaf price a commodity has been disposed of 138 barter*. when bartered for another commodity whosp price is given, RULE. f)iride the value of the commodity yihos'e price ■isi given by the quantity of that whose price you wish to ascertain ■ the quotient will be the price. EXAMPLES. 1.. -How many, pounds of sugar at 8 cte. per 'lb. must be given for 20 Jbs. of'cojtee at 1£> c^nts per lb. Ans. 2b lbs. ' OPERATION. ^ . 20' lb.x.10=82.00; 82.00-~.08=25'lbs. Ans. 2. Bought 18 IB. of tobacco at 00 cts. per ft>. and paid for it-81 lbs of butter; what was the butter per lb. ? Ans. 20 cents. OPERATION. 18' lb.X.00=816.20; 610.20^8l-lb.=20 cents.-Am. 3. How much sugar at 13 cents per pound must be givcfi in barter for b80. lbs of raising at 0 cts.' a pound?' ' ' ( . , . r Ans. 401y\ 11a 4.' A has sngap.Worth 121 cents a pound, for which B wishes to barter coffee worth 11J cents a pound; hoW much coffee must B give'■A for 150 lb., of his sugar? Ans. 105^5 I],. 5.-B has wheat Worth 75-cents a bush eh for.40.bushels of which- A gives him' 50 bushels of corn; what was the ecn'u worth per bushel ? s • , • Ans. 60 cents. 'barter. 189 9. 1 go to a store and buy 81 yards of osnaburgs at 13 cents per yard,, 15, lbs. of tobacco at 60 cents a pound, and a coat worth &-12-, Irpay 820 in cash, anil ant to pay the rest ifi butter at«l8f cents a-porind) how many pounds of butter will it take-? Ans. 26 Fb.., 13 qz.-^p 7. I have whiskey wdcth.68£ cents a gallpn; A barters me I7f gallons of brandy for 19 \ gallons of.my whiskey; whatis the price of gallon of-his brandy ' Ans-. 7.5|||'.ceptsj 8. A shoemaker*'buys of' a merchant; 14 'yds.'of " home¬ spun" at 20 ots. .a''yahl;/I8|- lb. ©f cofled at 81 cents a pound, and 18$ Ife. of sugar at 10"§ .cents a pound; he pays towards the amount-8 bushels of corn at cents a bushel, aiid.is to pay'the-'balance in shops at 83.00 a pair. What number of shoes. wTiH> the merchant get? t . . Ang. 1 shoe. 9. I bought 3 dozen apples at. 10-cents a dozen ; paid for them 4 dozen eggs; at what price per dozen were the eggs? and what was the difference in the pripes of 1 apple and 1 0X2^ 100 £100x5— 500 £150x8=1200 300 1800 ■ 300)1800(0 months Ans. ' 1800 2. A owes B £00 to. be paid in 8 months; £100 to bo paid equation oe payments.' 141 in 7 month's; and $120- to be paid in 6 months.- - At what Jfchn'e mtist it'air bo paid ?,> . Aps 6moJ 27da.'+: 3. ,1 owe'A $1QQQ;' of which ""$400 is .duc in? 5 months ; $35Q' im6 months; and the remainder in 11 months. • What is the equated time to pay it all'? . kJ ' •• • • ' , . Ans. 7 months, 3 days. 4-. A merchant has $144 due him,' p& be paid in 7 months; but the debtor agrees to pay oqe half ready mon¬ ey, and one third in 4 months ;-r-what, time should be al¬ lowed him to pay the rest? • Ans. 34 months. 5. I owe A $'000 dun in 10 months; I agree to pay one third of it'doWn, and one fourth in 5 months;—What time should be allowed me to pay the rest? Ans. 1 year, 9 months. 0. I have pm'ehasOd ejopds.of A, at different 'times, and on different ; 121=4187.50; $187.50X.02=$3.75 Ans. rjtOFIT' ANDTLOSS. 14'3 ' - • 4 2. A sells for B -the following1; 320 bushehf of 'com 'at 75 cents a bushel, 8200JK ,of bacbn at 11 ^ cents - per "lb.,, and 4 bags-df cotton, averaging 6001b. each, at 131 centls a pound. What is. his commission at If per cent. ? 4ms. %L6.%4, • 8. A collects for me a note for $32.80, which had been on interest at 7 per cent, for 2' years, 7 months, add- 3' days. What shall I'receive, his commission being 4f per cent.? ' Ans. $87\ 4. A.commission merchant buys for me 17cwt. 3qr. and 181b. of sugar at^lO. cents abound; Ms commission being 2 i per cent., and tile freight $9.60; what does the sugar cost me per lb.f f v, '4,ns- 1^4" cents nearly. 5. A buys 300001b. of cotton for B; what is his commis¬ sion at i per cent:,fire #cottbn\being at 11-dollars 374.cents per hundred pounds? • , .Ans. $$.53i. . . 6. 'A. commission merchant buys on my -account 98fib. of coffee at 11 \ cents a pOuhd, and 113|lb. of sugar a't 14' cent* per lb.; what is his commission at 2f per cent. ?' " "Ans. 72/^ cents. 'IiOFITAAiSrD.LOSa By this rule merchants estimate '.their profit' and loss in, buying and Celling goods. One of four things is generally required by this rule:— 1. To find what per cent.-is gained or.lost in a transaction. 2. To find what price to sell an article-at to gain or lose a certain per cent. 144 PROFIT AND 'LOSS*- 3. To -ascertain the' prime 'cost' of an article when we know what per cent, has been gained or lost. 4. - If a certain per cent, has been gained'or lost by sell¬ ing at one price, to ascertain what that per cent, would be by selling at another priee. To ascertain what per cent, has been gained or lost. .RULE. Take the difference between th-e cost and sale, to that dif¬ ference annex two ciphers and divide by the cost; the To find'.the cost when a certain"pe'r cent has been gained, i. e. to talx off a' per centage that has been put on, or more mathematically -speaking, to .deduct a certain p.er centage f-Vom^the sum of the basis, and per centaue. ' EtJLEf Divide the given price 'icitK tito ciphers annexedI, by 10(4 plus the per cent.; the■ quotient willdje, the basis of per rentage "or prime cost. EXAMPLES. •1. I s&ld it hat fur.$5, and gained 25 per cent".;" what did the hat cost me at first? Ans.. $4. OP5RATlt)tSr. •"100+25=125; '5.004-1.25 =$4 Ahs. / • i •' Some students' may perhaps think it strange that wfe fake' ojf -this per centage in a different way from that pursued in the preceding examples. They, jnay at first view be disposed tp. argue thus : "If you sold it at a gain of 25 per cent., you addefi on to it .1; then why not deduct i'td'o-et the cost?" To answer .this—The hat cost $4, O then every dollar was a fourth of the. cost, and using the cost as a basis of per centage, you can put on the 25 per¬ cent. by adding,a -fourth of the four dollars to the four dol¬ lar^. It has now become 5 dollars, and 1 dollar is no longer the fouifth of it but-the fifth, and to-take it off yon 148 PROFIT AND LOSS,; Must deduct a fifth instead of a fourth. The simple reason is, that the basis of per centage is not the same. • 2. I sold a yard of muslin for 72 cents, and.gained ^0 per cent.; what did it first cost ? , Ans. 60 cents. i,. What was the cost of an article which .being sold for $1.20 gained 50 per cent.? Ans; 80 cents. 4. A horse was sold by A to B,. at 8280, and the gain per cent, was 40; what did the horse c.ost A ? , . Vs- Ans. -8300. 5- A yard of cloth sold for 211 cents, gained 16# per cent.) what was its cost £ Ans. 18# cts. To find the. first cost when a certain per cent. Jias been, lost by ,the sale. RULE. Divide {he given sum with two ciphers annexed bg 100 less theper cent. Tie quotient tciU be the cost or basis of per -centage. EXAMPLES. .1. In selling a pair of shoes for ^1.20 Hose 20 percent; what was the first cost of the shoes ? "> . . ' Ans. $1.50 operation. 100—20—80; 1.20,00-4-80=81.50. PROFIT AND LO^S. 149 2. By selling an article for 66 cents I lose 40 per cent. What did'it cost ? ' ' ■ AnS. $1,10: 3. By selling 40; bushels of salt for $27.60 I lose 80 per cdn't.' What did the salt eost me per bushel? ' Ans. 98|,ct&:> 4. Selling a horse for $87.50 loses me 12' joer" cent. • what'did I give 'for the horse ? ' . ' / Ans. $99.43t\. b. By selling sugar 71b. to the dollar I lose 10 per cent, on thg original cost,: what was the cost per lb; ? Ans. 151| cents. 6. I, sell baggirig'at 17§ cents per yard, and lose 15i per cent, by it: what was the original cost of the bagging? ' Ans. 2P0-i'H! cents. When an article sold at one price gains' or loses a certain per cent, to ascertain the gain or loss per cent, if sold at some other price. RULE. I . Find the prime, >cost 'according to one \f the preceding rales, and proceed to fiiid the'per cent by the first rule given pi profit and loss. " EXAMPLES. • .1. By selling a yard of calico for 15 cents my gain per cent is 25 ; if I had sold it for 16 cents what would have been my gain per cept. ? Ans. 33J per cent. OPERATION. p5Q0-f-125=12 c^nts prime cost;. 16—12^=4; • ■ 400^l2=83i Ans. 150; PROFIT AND LOSS. 2,. When i .spll-a bushel of corn at 85 cents I lose .15 per cent.; if I sell it for $1.20, do I gain or lose, and what;per cent-/? . . Ansr Gain 20 per cenh. 5. .If by selling .my horse for $100 I wbuld lose 33J .per cent., what would be my loss per cent, if I were to sell him for $120? « i Ans^ 20 per' cent. 4., If by selling coffee 81bv to the dollar a merchant gains 25'per cent,, what would be his gain per cent. if.he. soli 1 lib. to th6 dollar ? i ,. • . .•* .. , ' , _, v _ Ans. - He would lose 9^ per cent.' 5. .By selling a yard of cloth for $8.00 I lose ,20 per eent.'i by selling it at $5.00', woubl I gain", or lose? .and what per cent. ?, • "Ans. Gain 11| per cent.' MISCELLAXEOU3 EXAMPLES. 1. .1 bought apiece, of .land,containing 240 acres for $2500; sold it £t $15 per Acre';—what is t/ie gain, and gain per cent. ? Ans. \ Sain 51™' ( per cent 44. 2. I bought a box of tobacco for $10.20 ; sold it at 80 cents per lb. and gained 25 per cent.;—liow many pounds were there.in the box?- Aits'. 15{|lE 8. -Sold 40 bushels* of potatoes for $9.20, and gained 8*0 per cent.;—what did the potatoes cost me per bushel? Ans. 17t93 cent?. 4. I 'bou'ght 65 bushels bf wheat at 80 cents per bushel, and sold 20 bushels at'75 cents per bushel, and § of the PROFIT -ANU. LOS$. 1®1 Remainder at 70 cents per hush el;—n'ow what must I' sell the test at per blisliel that I may make 5 per cent/ on the whole? Ahs. $1.10,5. • h. A' merchant -has cotton goods which cost him '36 cents a yard: he wants Jto make 33 J per cerit on it;■—what must hoiW^qskincj pj;ice so that he. can fall- 10 per cent.,' and still make 38 j «per cent. ? ' , Aijs. 531 cents. I).; A f'urmcy has a hag of cotton whieh weighs 6501b., and for which" he is offered 13 cents a pound, hut .will pot fiell, _ lie Jvepps.it avyear;/jt loses meanwhile f percent, in weighty -and..he. gets J4 cen'ts for it—what has- he gained 'by " holding on.? to his cotton, ©ur .interest being . 7 per .cent.? . ■ , , ^ Ans. h'e has l'gst' a dime lacking \ of accent. ^ 7. A ,ha£ Corpyrorth 70 cents per bushel; but in barter ■he is willing to .put^i| -at .65 "cepts,-., provided he can have wheat .worth £0 cents, per bushel, for ,74 cents., Will«A guin or Jose, .a?4. what per Cent.. , , _'(i Ans/gains // per cent. 8. td'old' horse for $75 and lost '25 per cent by it; where¬ as 1 ought to. haVe gained 30' per cent. How much under his real value did 1 sell the horse? 'Ans! .$56. '• *9. '-Bought a hogshead of molasses fOr|56.40, but*a'rmnu ber of gallons hav& leaked out,'I Sell the remainder ht 90 oents'a-gallon, and lose 101 per cen,t. on the cost; how xnuch'h;ld le'alled'out ? ' 'Ans. 13'-gallons.' 10. S. .Jenkins has 73T9^ bushels 'of -corn which is wprth 7s-'i>er -bushel, but in barter 'he is willing to put* it at 6s. 8d.) provided he can'have wheat-worth 7s. 6d. per bushel for 7s.-3d. Will he.g^iupr lose, and how'mUeh percent?. ■ • Ans. lose 1 ® per cent. 152 PROFIT AND LOSS. 11. When 100 boxes- of prunes co'st $2.10. each, and by selling them at 83-.50 per. cwt. the gain is twenty-five pet cent., what is the Weight of each box ? r Ans. 84 lb. 12. Sold a watch which cost me $90 for l20 on a credit of 7 months ; thfe ■ rate per cent, of ' interest in this State being 7, how much did I make by ihe trade, and what per cent. ? , f Made $25.29tVA. nS" | Gain per cefit. 28^2^1 13.' A,sells a horse to B for 8100,'and gains 25 per cent. B sells him, to C at a loss of 10 per cent.; C sells him to I) and' clears 331 per Cent;—what p'er cent, would D have to deduct fr'om what 'he gave for him in order to sell him to A at what A first gave for him t ' - Ans. 33|. 14': My commission merchant irf -Charleston writes me that -cotton is worth' 14 cents a pound; .now what' can I afford to give per lb. for five bags, averaging 500 lb. each, the freight from the place where I buy tQ Charleston being 75 cents a bag, and his .commission for selling beings 2- per cent., that I may clear 10 per cent by the trade ? r Ans. 12-j3t cents. 15, Bought a hoi'se for 8130; I wish to make 20 per ' cent on him;' what must I ask for him so that.'I may fall 20 per £ent afid still make 20 per pent; ? ' Ans. 8195. 16. My asking price for an article is 7-5 cents, butM can fall 10 per cent, and make 3D per cent still; what did the article cost me ? Ans. at! 4 ~ ets» 47 Bought clotli at 24 cents U yard-; my asking price is 40, cents; what per centanust I fall, to clear -25 per cent. ? • Ans. 25 percent. INVOLUTION. 153 IYVOUUTlOY. In;volution (iuvolvo,"Latin) irs the process, of •invoicing a quantity tp a higher power, or what is usually called the raising of powers. ■ „ A power of a number is the product of that number ltiul ■ tiplied into itself one or more times. The ; number' so in¬ volved is called the root. , ' The first power of arty number is the'number itself. The second power is.tlie product of ,the number multi¬ plied by itself once, or used twice as a factor. The third pow-er is the root multiplied by itself twice, jjw used three times as a factor. . • • . The index or exponent is a small figure written at. the right, and at the top of the IrOot,' to indicate the power to which it is to be involved, or the' number times it is to be taken as a factor. Thus the 2nd power (or squre) of 5 is -expressed 52; the third power of 6'is written 63} &c. Where the power of a fraction is to be indicated,rthe> frac¬ tion is included in a parenthesis, thus, (I)4 by which the fourth power of f is indicated- v '' To raise a' numbpr to any required power. , RULE. Multiply the nuvibqr by itself as many times# fetching one, as there aire units in .the index of the power ta which ■you wish to invoice or 'raise it, Or, ' ( Set the number down horizontal1y as many times as the index of the required poicer has units, using the sign" of multiplication between the factors, and get the product. involution. examples. ' »l * 1. Wiiat is the second power (or square) of 2 ? Ans. .4. operation.' 2x2=4. * % 2. What is ,the third power (or cube of 5) ? Ans. 125. operation. 5x5.x5=i2p. 3. What is thesfourth power of 6 ? . Ans. 1290., ' • •operation-." 6x6x^x6—1296. ■* * - ' ' ^Note-.l.=The' fetufient may sometimes save himself tfdious multiplication by multiplying two of the .higher powers to'getbet. For instance^'in raising 3 to the 7th power, "when he gets it tp the fourth pqwer, h(j can complete.it "by multiplying the third power on to (he fourth, for the product of any two powers of a number is a power denoted by the sum of their exponents. •Note 2.—A fraction is raised'by Involving both, its nutherato? hnd denominator. 4. What is thfe-5th power of 8 ? Ans. 327(38. * 5. What is the 4th power of 7 ? 4 . .Ans. 240J7 6. What is the 7th power of 2 ? ' Ans. 128. 7. What is the square of)14? < ^ Ans. 196. 8. What is the cube of 9 ? ' Ans. 729. 9. What is. the 6th power of 7? < . Ans-. 117649. 10. What is the 4th power of f,' . Ans. 11. What is the value of 8 s . Ans. 512. 12- What is the value of (T90)'2 I _ , Aire. Tqv. ■ 13.. What is the value (2f)3 ? ' ' .Ans. 157T2(A . 14. What is the tlrird power of .8 ? N . , Ans .512. "EVOLUTION ■ 155 f* I 15. What is the 4th po;\ver of .045? Aqs. 000004100625: 16. What is the indicated prdduct of 2 3 by 23 ? ' '' ' , Ans. 25:. ■ 17. What is the value of 83X^ '? Ans. 32768, 18. What is the 10th power of 7? ' Ans. 282475240.' EVOLUTION. Evolution, the' revchse of involution, it> the process of extracting; the roots of jpian^itics. The root of any. number is such a ffictor as multiplied by itself a certain number ox times will, produce tliat.Vfuastity o/ number. • •' ( ■ The root takes the name, of, the power from which it is evolved. If a ftuniberds second -power, its root is' called second or square root; if it is, third power, the root is called third or cube root) if it is fourth power, the root is fourth root, if fifth power., fifth root, &c. We may find' exactly any power of any ro<5t; but it does not therefore, necessarily follow that wo may find exactly the root of any number, for some numbers really haw, po exact root, vand we can by the aid of'Vccimals obtain only a proxitnu>t& root, near enough powever for all. practical purposes. , ■ . * Roots, wtiich can be found exactly, are Called rational- 11 ,1'5»3 EV©HrTl03Sr: -SQUAHE ROOT. , Routs, wliibh only* approximate, are called^wrt/. . fete-usually denoted by ■uniting the. radical si"ea y, before the power, w-ith th.e index of tha root> over it, .ex¬ cept in *ease of" sgndre roof, where the 'in-dex.is omitted.— Thus the'square root "of S6 may. bes expressed.^/3{J j, the cube r&ot t>f 27 may be .denqted by--^27 ;,.the t4th-row^ of 30 by 4|/1'0, and so On. Hoots are also denoted' by a fractional index of which he numerator indicates, the power and the denominator the roof. ' Thus the second root of J5 may be dcifptedby 10" ; ' • 1 ' HrT " the third root of 27, by 27 > and the square of the cute rqot of 8, or tlie ctibe root of the square of 8 may be deno-_ ted by 8 . ' *KXTT*ACTION OF THE SQUARE EOOT, •. .v • > # i * . * ' . . The extraction of a square root of a number is the prebesp of finding what root squared of multiplied Bv itself wjll make that number. * '■ ltULJT. - • * • -• Separate' trie given vtwnbcY into' as nuy beginning at the right, mid iisiiig .a p»int ietw-ceh the periods rit the top. ** ■ " > ' 'Find'the mot of the greatest square in the left hand pet tod, and write it to thy right rf'tiie.given'huhibcv' as you would a quotient 'in division,'and subtract its pqjtare from the I> ft hand period. ' * • '. EVOLUTIONSQUARE ROOT. 157' Bring down the .next period to the right of the remain¬ der for fit dividend, aiid double the -root, already found for the tens.2ar is always so much larger than y the units, we' may'consider '2.xy(y as nmrjy equal to -all the remainder. . Then faking 2x (which* id5 double''the root already obtained)'for a (rial divisor and' going back to our 2+/++- we find it is contained, times fn+lie 2xy, exclusive of the right hand term y ~ y then by annexing the y to the trial' divisor 2x.we obtain the true' divfeor 2x-j-,y, which multiplied by y, the last-obtained root, produces 2xy+y*., ■» From the, above it appears that a numbar consisting of ■tens and upits'when squared isimqua! to the square of the .ten's plus twice the product of the tens and the units plus the square of the'finite." The number 86 squared equals 1296. 38 consists of* 3 tens (or 30) and 1 unit, 6. 30 squared is 9DO, -and 2(89x6).—369, and squared equals 88 ; 900 + 860+36 = 1296. .Then making two terms of 36, thus 30+8 and squaring we obtain 900+360+36. Now,_ to extract the square" foot of this 9(?0+86Q+o8, wo EVOLUTION : SQUARE .ROOT. find that. 80 is the rational square root of 900', which squared and subtracted there remains 360-{-36 j but 360-}- ■36 is the same as (00-|-6),X'6 out of which wo musVobtain 6 the second term of-the root. But, as 60 twice the tens is so much larger than 6 the unite, we may consider 60 X0 as nearly equal to the whole remainder. Taking .the 60 (which is twice the 30 already obtained) for a trial divisor, and trying it into the 800-(-36 we' obtain 6 as the second term of the root. Anuexing this to our trial divisor we have 60-J-6 for the true divisor, which'multiplied by the last obtained root 6 produces 360-J-36, and Subtracting nothing remains. Now we hate obtained 30-j-'6 as the square root of 000—}^3G0—f-3G, Or 36 as the square root of 1293.' ' ( The- principle applies equally to glf numbers' whether there are two figures or more than two hi the root. Examples. . . 2. What is th,e Square root.of 576 ? • ' -Ans.,.24. $.■ What iS tlip-square root "of-4225 ? Ans. 65.' 4. What is the square root of 18769 ? Ails. 137. 5. What is the square root of 15G25 ? Ans. 1.25. 6. What ifj tfic square-root'of *6056521.? Ans. 2461. 7. What is the square root of {14 ? ■< Ans. jq. 8. What is the square root of 10.24 ? " Ans. 8.2. \ 9. What is the square root of .000729 ? • Ans. .027. 10. What is the square root of. 17.3,056 ?' Ans. 4.16. 11. What is the square root of § ? r Ans. .79056--}-. 12. What is the square root of 384? Ans. 6.22896 . APPLICATION OP THE SQUARE ROOT.. A circle is a figure hounded by a line equally distant From the centre, A triangle is a figure of three sides. . • . r An equilateral triangle is a figure'whose-three sides are all equal. ' An isosceles triangle is one which has two sides equal. A se dene triangle is one which has its three sides unequal. 160 Evolution : square root. A right angled triangle is a triangle with one right angle. See Fia.l. fig-1- The longest side in any right angled triangle is called the. ' hypolhenuse, the horizontal side' the base, the other, perpendi¬ cular. Ease. In any right angled triangle^ the square of the longest side is equal' to the suin-of the squares of the other two sides. See demonstration by Fig. 2. In.all Similar triangles the sides'about the equal an-dca are in direct proportion to each othe'r. ° Fiff. 8 Fig, 4., Let ABC and D EF be two Similar triangles. • Then BC will be to EF iis AB toDE; and AC will be. to DF as EC to EF; &c. F ,\ . V «SQUARE ROOT. 161 Circles are to each other as the'squares of their diameters, semi diameters, or circumferences. If'the diameter of a circle he multiplied by 3.141592 the product will he the 'Circumference, * If the square of tfie diameter of a circle he multiplied by .7854 the product will he its area. • To find the hypothenuse of a right angled triangle when the base and perpendicular are- given. lltJLE. Square the 5u.se and jieryendtcular, add the squares io- y ether, and extract the square.root of (he-Sum, EXAMPLES. • 17. The hase of a right angled triangle hejng 12ft., the perpendicular lGft,; what will Be tho hypothenuse ? . . • Arts. 20ft. OPERATION. 12*^=144; 10^=250; 144q-25G=4(K); 1/49O=r20 Ans- 18. What is the longest side of a right-angled triangle, the'othet two being, the bug'18, the other 80? Ans. 82." 19. «A tree, 144 feet high,'stands just on the bapk of-a river, 60 feet wide"; what must-he the length of a ling to reach from the opposite hank to the top of the tree?' Ans.-156 feet. • 20. A ladder; planted exactly in the,,middle, of a street SO feet wide,, just reaches a window 20, feet high at one eiiie of the street;, what is the length of the ladder? Ang 25ft.'4 21. A starts from a certain point and travels exactly 162 SQUARE,HOOT., north 201.6 miles; Bstarts from tlie same point and travels exactly west 7>8.8 miles;—how far. apart are they in a straight .line ? Ans. 21.0 miles. "\Vhen the hypothenuse and either the hase or perpendi¬ cular are given to find the other' side. RULE. Square each of the given stales, take their difference. and extract the square root of that difference. EXAMPLES. 22. The longest side of a right angled, triangle is '26^ the base 10; what is the perpendicular? .Ans, 24. OPERATION. 26x26=67^1 10)<10=100; 670—100=576; 1 '576= 24. Ans. 23. Er>m the ton of a tree 100 feet high, a line 125 feet long just reaches to, the opposite bank of a river atthe edge of whose wat-cr tare tree-stands ;—what is the width of the river ? Ans. 75 feet.x 24. Ir. a circular island 90ft. in diameter there stands a phle right in tlifc ventre, from the top of which a lino -50 feet l>.ig will just reach to. the edge of the water;—what is the height of the pole? . ' \ A use 14 feet. 25., The lrypothenusc being 40t the perpendicular 36, what is the proximate base ? Ans. 17.435-4-. SQUARE ROOT. 1G3 26. The base being 15, the hypothenuse 17, what is the • perpendicular? Ans. 8. , - To find the diagonal of a cube: RULE. £j:t,racf the square root of the product of the square of One of "the sides multiplied by 3«, EXAMPLES. 27. "What' is the diagonal of a block-six inches every way'? OPERATION. 6x6=36; 36x3=108; 1/108=l6.39 in. Ans 28. I have a,cubical box 13 inches on .each side; what the distance'from one of the corners diagonally to the op¬ posite corner If . . . • . •. " ' Ans. 20.78 rn. 29. A certain robm is 20 feet long;'16. feet/wide,1 and. 12 feet high ; how 1'ong must a line be to ■extend- fiom one of tlie lower corners.to an opposite upper corner-? i Ans. .28.284 feet. OPERATION. 202- 400 ; 16-=256 ; 12'J=14i-:."l00-r2h0-14i=s SOU ; 1/899=28.28-1 Ans„ ' ' 30. A certain -room is 30 feet long, 20 feet wide, and 13 . leet high;- what must be the length of aline that will1 reach-... 164 . , SQUARE IlOOT. "from' one bf the lower corners diagonal!^ td the opposite "Uf- per one ? • y - - . . • * Ans. 38.32.feet. 31. A rooin y? 18ft. 6iri. long, lift.*4in. witje,' land 10ft. 8ii>. higlij what i3 the distance* diagonally fropi an upper to a lower corner ? L Ani 24.48+ feet. APPLIED TO CIRCLES. » - 0 EXAMPLES. % 32. If foP 83.60 I buy land in a circle whose diameter is . 40 yards, what pho'uld he the diametq? of a circle that I could buy. for 814.40 ? . • * ' i * * ,#*T • y ' ' * Ans. 80 yards. \ • ' * BTAREMEXT PY Proportion. $3.60 r^814.40 :; 1500yds. :i6400; v; 6400^80 Ans. !*• < \ Circles being to each other as the squares of their diam¬ eters, the given "diameter 40 must be squared, and then the root of the •resulting fourth term-- be extracted, which getls the answer required! • , . 33. If for 820 I can buy a circular garden 90 yards in di¬ ameter; what rnu^tbe the diameter of a garden that I could ' buy for 830 ? ' 1 * , ' •« ^ Ans!-110.22+yard?. 81. If a lead pipe | of an inoh/in diameter fill a cistern in 3 hours, what musrt be its diameter to fill it in 2 hours? . * , • Ans. .918+'of an inch. If a circular plate oF sheet-iron 10 inches in diame- SQUARE ROOT. 165 tcr Weigh.' 8oz, what must be the diameter of a plate of the same thickness to weigh 13oz, , ( ... .. Ans. 12.74-f inches. 1 ,36, If a grindstone, 14 inches in diameter weighs 301b., what would be the weight of one, of the same thickness, 28 inches iji diameter? . 'Ans. 120. lb. . OPERATION. 1 1^196; 28*^78 i; then 1.93 ; 734 :: 391b : 12C»lb. Ans. The student will observe that the square root of the quo* lUiU/is not to be extract0-1, unless tho third term of the proportion has boon squared. • ' 37 If a circular piece of land 30 rods in diameter ,be bought for f7 6J, what would anoiher piece of the sapie land be worth if 45.rods in diameter? • . Ans. £17.10. 38. If within a circle of 111 yards in diameter there were contained '2 acres of land, how much would be • contained •within & circle 200 yards in diameter ? ' Ans. 6.49 acres. 39. What must be the length of a rope in yards, that, tied to a horse's oeck, would allow him to feed Over an a me of ground ?' • * Ans. 39.25 t-yds. OPERATION. / ' 4810-r-.7851 ^6132.43 f; 1/d"ISin = 78.5-f ; 78.C>~-2 • — 39.25 Ans. ' • In an acre there are 4310 square yards. Now it is evl- • dent that the horse will" feed within a circle whose area is 108 SQUARE ROOT. > , 4840yds.; and as-the area of a circle is obtained by squar» ing its diameter, tmd multiplying b»y .the decimal '.7854,—*» so dividing the area of any circle by .7854, and ex¬ tracting the square .root of the''quotient will .give' the dikhieter. But as the rope is fastened to something in the centre ,of the circle, on every.sideof which it extends equal¬ ly, it is only half the diameter; and therefore ve .divide ' ' ■ ■ • ' ' ' . • 40. What must be the length of a rope in yards that a hoyse may feed-over 2 acres ?• V Ahg. 55.51yds. t 41. What must be"' the length of the,, rope in yafds that he jnay feed over § of an acre ? . • . - Ans. 34 yards ver-g nearly. '• 42. What must be the length of the rope in rods that the horse ftiay feed oyer ajracre and "a half? • * w' • '• Ans. 8.734- rods'- - • 43. Three mem, A,. B ' and 0, •boitglxt a grindstone 32 inches in diameter, A paying 70 cents, B 50- cents, C 40 cents, They agree that A slialj "grind off his--share first, then B; and then C. Ilbw much will each grind off? ' > Ans. A-4in.; B 4in , 0 Sinches. ' ■ SORRTIOX. We have, a eiretb 82 inches in diameter, and A is entitled to t7& of its. area because he paid T7g of the money; B is entitled to of its area, and C to First we find the area by squaring .the diameter.32, and multiplying by the decimal .7854, 1024X-7854=804.2490 square inches; ' 11 o\v deducting ^ of this area for A, we -leave 452.2904 square inches for B and C; ytJ of 804.3496 is.351.8592; 804.2496—351.8592=452.3904, which is the number of square inches left for B an pi Q pftcr A grinds his off. To get the diameter of this 452.3904- we divide by .7854, and SQUARE ROOT, 167 extract the square root of the .quotient/.452.3904-=.7854 s=576; 1/57612=24,1'"thc'diamfeter of thb grindstoho after' A grinds his Qff. To get A's, subtract this 24 from 32, and it le'aves_8, which'divided by 2 gives A's 4 inches. The ■griqdstone now goes to 13, who; as we Said before, is entitled to of it/—not /r of it as-if now is, bjit y/ of it as it was at first, because -he paid T5(J of', the money. Thesquare inches on its side now are ,452.3004, gnd T5^ Of the .whole 804.2496 is- 251.3280; this'subtracted from 452.3904; leaves 201.0624 square inches for C. The diameter of,this 201.0624,,found as we fouigl the.other, is 16 inches, which is just the diameter of the grindstone when it goes to'C,— Thi£ 16 Subtracted from .24 (which was the'diameter left for B' and C)lcaves 8 fur' B, - which divided by 2 gives 4 inches B's;-and 10-^2=8 C's, . Jt\ the .above solution we/iavp, for the sake of illustration, been intentionally tedious. The work, however may be grOatly abridged, by remembering' that circles are to each other as,their seniidiameters, and working with the serni- (Iiamgtorof the grindstone,,-and" rejecting the'use of the' decimal .7854 entirely. > Thus— _ • The semidiauicter" 16'2=256; /g of 256=112; 256— 112=144; • 1/1-14^=12 seipidhunetef for - B' and C ; 16—12=t4 (inchcs Ans.- for A- Then 144-—80,—-64, (that 80 is -j5jT, pf 356 to which B is entitled ;)'^ i/64=8 in., An,4. fof C. ' Then 12—8=^4 ihelms, Ana. fpr B. Note.—Where there is pothing said ahout what..each pays, we veryreasonably suppose- they all pay the"same, It' there are three men, each is entitled td j of the grindstone,'if four, each to j, &e, 44. A B and C buy a grindstone 5 feet and 4 inches in diameter/or 83.20, ofwjiich A pays 81.40, B 61.00, and-C the rest;, how./af down on the radius must each grind to, to get hi's shave ?, ' ' T A,' 8 j riches. Ans. -< B, SAuc-hes. ^ • f C, 16 inches, w 16.8 vCUfffr ROOT. . ".45. A, B ami -C,buy a grindstone 38 inches .in diameter} A pays 50 cents,. B pays 6(> cents; C 80Cents.% llyw much, mind -each grind off? *' . . . ! . ... (A, 2.Q9 in'ches, ,, 1 ' •' • ". Ans. -< B, 3.98 inches. ' " " . ' 0,12-33-' inches. * 45: Foar meri, A, B, 0 and 1) bought a- grindstone whose diameter was 4 feet. They agreed that A should grind off his share'first, and that each mariNshould have1 it alternate¬ ly, unt'4 he had ground off his share! Hoiv mu&h did jfach'grTndyOff? ' * ' ' ' , * ' * * | Ans, A) 3.22'+B/3.81+; C, 4.Q7+; D, 12 h. •JfiATU^CTIO^ OF .TIIE .CURE ROOT. . » • * The extraotion of the cqb'e root off a number is a process of finding what number or .root raised' to- the third power Will produce that nupiber". ' 1 * The principles ' on which the fcule is founded may be explained in a in ah nor similar to that by wliich-we explained the principle's in the extraction pf the square root. _ : y * I1ULF. iSepafate 'the given number finto 'as:Tnary, period* as poF.si >Te of three, figures each, beginning at.the units' jpfac. • Fiiidlhr. rout of the greatest cube in the It ft hand period. and write 'it to the right of the gi;cen number (t£ ike first CUEE-ROOj. 16D jijuri of t}i& required root. » Subtract its 'cute fropi the left hanrtpe/io f nnd to the remainder bring down ihe newt period, fur' a.dividenrf. • , Multiply the .square of the-root already found by ^ ami to the product' aungx two ciphers for a 'trial divisor, -and ste hojt) oflcpp this divisor f contained in ihe dividend, und write the result as the newtfiguic of the root. • / '' , > ' • '« • ' ' • * (>toTK.—fn this iri.tl-for U>e soeoorl fi^;r;e o.f ihe root the divisor will not1k)(? contained by oiicytnvT>rtnd s;>:aetruds thfee Units as many times As'it apneas tliyt-it.will.) • - * Add to tie trial divisor three tidl"-s thk- product erf the rout-'before dbt.al)\ J 4X60=240 ' . 42x180=71)00 4 Ca= ' 86 244 7593 170 CUBE ROOT. " 2. What is the cube root of 12812004;? ' .Ans. 234*. 3t Whafc'is the'cube root of 83621568? Ans. 432- 4. What is the cube root of 3 a 12.136 ? Ans. 14.6. 5. Tlie cube root df 12S95.213G25? Ans. '23.45. '"6. The cube rooVof 5294751*29 ? Ans.'800. ' 7. What'is the cube lVjot of+~_ ? . - Ans.'.585-}-. 8. What is the yaltle of (-|)3 f " Ans. .854-^ 9. What is. the value of Ans. 10. What is the difference between the expressions 8 3- and 27^ ? ■ Ans. 24. APPLICATION OF CUBE BOOT. Spheres are to each other as th6 cubes of their diameters*. If the cube of the diameter of a sphere be multiplied by the decimal .5"230, the product will bo-its solidity. Cubes and all solid bodies, whose correspondin g parts are similar and proportional to each other, are to.each other as the cubes ot their diameters, or of their homologous sides. EXAMPLES. 1. If a ball of lead-4- inches in diameter weighs 201b , what will be the weight of one. 3 inches in diameter? ' . -Airs. 1601b. 2. If a cannon-ball 8 inches in diameter weigh 401b., ■ what will be the 'diameter pf one whose weight is 201b. ? Ans, 6.34+ in. 3. How many balls 2 inches in diameter can be run out of. one 8 inches in diameter? Ans. 04. CUBE EOQT. 171 4. If a stack of hay 9 feet high weighs 2001b.. what would be the weight of a similar stack. 27 feet ,high.? Ans. 54001b. 5. If a globe of gold 3 J inches in diameter be worth $5145, what Will be the diameter of a globe of gold worth $120? ' ' Ans. 1 inch. 6. If a bell 10 inches high? 5 inches, wide,, and 1 inch thick weigh 3 pounds, what must be the height) width and thickness of a bell of similar shape to weigh 811b. ? ( 2 feet Gi.n. high, Ans. - 1 foot 3in, wide, (_ 3 inches thick. 7. If an ordinary man 6 feet in stature weighs .180 pounds, and 180000 pounds be taken as the weight of Polyphemus, the fabled Sicilian shepherd who carried the butt end of a pine tree for a walking stick, how high was he ? < Ans. 60 feeti 8. Three women bought a ball of butter fgr 90 cents; the first paying 40 cents, the second 30 cents, and the third 20'cents. " Now the diameter of the ball being 41 inches, how deep into it shfill each take for her share ? C First takes .40—f—fn. Ans. < Second '.49-f-m- ( Third 1.36-{-in. The above problem is solved precisely as the grindstone problems except that where you there squared and extrac¬ ted square root, you here cube and extract cube rpot: J9. If a horse that girths 7 feet, weighs" 8001b., what is the girth of a horse of similar proportions- whose weight is 15001b. ? Ans. 8.6-j-ft. 10. A stack of hay 20 feet high and 10 feet diameter at its base has been bought by A, Band C. Its shape bein^ that of a regular cone how far down from the top 12 172 CUBE ROOT- inust A cowne for his share, Bow far from thence must B come for his, and what will be the height of the frustrum left for C ? f A 13.86 feet. 1 B 3.61 feet. 2.53. feet,. ' A fieneraWlule for Extracting any Boot. BULE. ]yv}{avc the given number, by beginning at the ■units', place, and pointing off info periods as the required, root diverts, (i. e. if the. 1th root is required there will be four figures in a period f if the Oth, five, yd. of lining; 15yd. at 20cfs.ayard =3.00 According to the supposition the whole would be worth- $16,50 But the whole should have been worth 14.40 So We get as an error of excess , * $2.10 Again suppose 10 yards of silk; which at 90cts. comes to $9.00 Thdn there must have been 20'y d. of lining ; which at 20c.ts. comes to ~ 4.Q0 Making the whole cost to be • $13.00 When it should have been . 14.40 So,we get, as, an error of defect, $1.40 First supposed number multiplied by last .error, 15X1-40=21.00 .Last supposed numbeT multiplied by first 'error, 10X2-10=2,1.00- Adding the products (because the errors are unlike) gets, $42.00 Adding errors 2,iO-j-l'.4(J=3.50, 17 G DpUBLE POSITICW.- Dividing sum of products by gum of errors, 42. QO-v-3.50=E?5ilk 12y(L Subtracting the silk from the whole, 30yd.—l?yd.=Lining 18yd. 2/Bought cloth for a cloak at' 85-per yard, and lining at SI per yard; the whole number of yards was- 22, the <;ost Was $02. * How many yards were there of each ? . ( The Cloth 10 yards. ds. ^22 yards.' 3. A and B invested equal sums in trade. A gained a sum equal to i of his capital, and B lost $225; then- A's nloney -waS double B's. , "What did each 'invest ? Ans 8GOO 4. I have two watches, and a chain worth 840, if I put the.ehain on the first watch, it will-be worth twice as xnucji as- the second; if- I put the chain on the second watch, it will be worth, two and a half times the -first- What are the watches Aorth I ■ . - f First 830. "( Second- 835. 5. A says to B give me $9 of your money, and I'll have four times -as niifch as you'll have left; B says to A give me $9 of y-ours, and I'll have the same that you'll have ' left. IIow much had each? , f A had 839. A|-\r, « p. . G." I have three lines of different lengths : The first Is 1* feet long; the second is as long as the first and } of the third; the third is f of the lepgth of the first and second. What are the lengths of the second ayid third ? , f Second 30 feef. Ans- \ Third 36 feet. 7. I employed a workman for 60 days, on condition that for every day-he. worked 1- should give him §3, and for every day he idled ho should give me1 two dollars be¬ sides forfeiting his wages,' At the end of the time tie re- V V • ' , i DOUBLE POSITION. ITT ceivccl $55. IIow many days did he work and how many did he idle ? . { Worked 35 days. Ans' j Idled 25 days. 3". I employed a workman 30 days, and agreed to pay< him $40 and a coat; he worked 15 dgys, and I payed hiui $10 and the coatwhat was the coat wofth? • Ans. $20. . 9. A father ;divides-his property between hhs-threesons :■ to A he gives I of the whole lacking $10.0; to Jlj .} of the whole and.$100; .to C, the resf, which is $100 more than IPs. What is the whole sum, and the share of each ? . ( The sufn $1200. . j A's share $300. Aus") B's 0 $400. ■ -t'O's' •" '$-500. ■ •1*0.' There is a street 70/eet wide, andn house on one side of the street has a window 40 feet frorh the ground, and a house on the other side has a window, exactly opposite the, other window, 3,0 feet "high. At what point m the street in a direct line between the two windows must'a ladder be planted, to reach both windows ? . 30 feet from the house with Ahs. window 40 feet high. Note 1.—The most of the above problems dre very easily solved by analysis, as willbe seen in Our." Solution of Yarious Problems."' Xoxn 2.—I have intentionally ommitted Single Position, because I thiiik1 every problem applicable to it can be done by a better proeesk, 178 ARITHMETICAL' PRQaKESSION. ARITHMETICAL ^PROGRESSION,. ■ Arithmetical Progression is 5 series that ind-rease^ or de- creases by &'common difference. The'series is galled ascending when each 'term.after the first exceeds* the one before it ; it is called descending when each term after the first is less than tjie one before it.-— Thus, 2, 4, 6, B, Ac., is an ascending series, hawing 2 for the common difference between the terms; while 14,11, K, 5, &c , is a descending series having 3 for -a common difference. The first and last forms are called extremes. To find.the common difference; tl>e extreme^ and number of terms being given. ^ ' RTPE. Divide flic difference of the extremes by the number of terms less 1. , • "V EXAMPLE.-?. " ' * x ' ' * ' « 1. T11 an arithmetical series tlip extremes .are h and rib. and the number of terms 16; what is the common difference'/ Aug. 2. OPERATION. 35—fc=?»0; 1G—1==15; 30^-15=2 Ans. •2- a certain school the number of, students is 12; AR+THMETItfAL PROGRESSION. 17$ thy oldest 89 years old, the youngest (r. Ivow if trheir ages are in arithmetical progression, what is the common difference?" Ans. 3 years. 3. Bought 9 yards of clotli'the prices per yard being in arithmetical progression ; for the first yard I paid 10 cents; for the last 50 cents;' what \fas' the commoit difference ? Ans. 5 cents. 4-. A man travels a certain distaneein 7 days: heroes S miles the first day, and increases equally each day so that he goes 50 miles the last day.. What is his dqjly increase ? Ans. 7 miles. ,To find the number of -terms, the first term, last term, and common difference being given. RUBE. Divide the difference of the extremes hy the common difference, and add 1 to the 'quotient* EXAMPLES. 1. If the extremes of an arithmetical series are 5 and 25, and the common difference 2, what is the number of terms? . Ans. 11. OPERATION. 2*5—5=20;'20=2=10; 10+1=11 Ans. • 2., Bought a "number of yawls of cloth, the prices'per yard being in arithmetical progression, the first yard cost 8 180. ARITHMETICAL PROGvRESSjON. cents, the last 28, .the, common. difference? 4 cents,. what was the number of yards ? ,• - ; Ans. 0. 3. I'buy cloth paying 4 cents for the "first .yard, 7 cents for the second, 10 cents for the third, &c., and for the last yard I pay 28 cents. How pi any yards were there? -Ans. 0-» 4. A^man travels 20 miles the first, day, 30 miles the second,. 4Q •miles the third, &c.? and-the^ last day he goes 70 ,miles. IIow many days did he Jtravhl. Ans. O To find the sum of the terms, the extremes and num¬ ber of terms being given. RULE. JluLtifiy half the, sum of the extremes fry the number e,f terms, 'or the sum of the extremes by half the( number of terms. EXAMPLES. 1. The extremes of an arithmetical series arc 3 and 45, the number of terms 9 ; what is the. sum of the series ? Ans. 210. ' OPERATION. 45+3=48; 48--2=24{ 24x9—21G Ans. 2. Bought 9 yards of clotliydhe prices per yard being in arithmetical progression ; for the first yard J paid 4 cents' for the last 28 cents. What did the cloth all come to? ' Ansl 81.44. f>. A man traveled from Atlanta to Macon in 5 days he goes 10 miles, the first .day$ and .3fr miles t}m last day. ARITHMETICAL PROGRESSION. 181 What is1 the distance between the two places ? Ans. 100 mi. 4. The extremes are 4 and 48, the number of termes (>; what is the sum of all the terms ?•• '. , • Ans. 150. To find the sum of the terms, the extremes and' common dilference being given. RULE. I DijjuJe the, difference of J]'te .extremes hy the common diffeperiQC,-;and add 1 to the quotient; and multiply thin fun} by half tl\e sum of the 6%tremc§. EXAMPLES: 1. If the two extremes .of an arithmetical series .afe 3 and 10, and the common difference % what is the sum of the scries? t ■ Ans: 00. OPERATION. 19—3=10: 10^2^8; 8^1=9.: 9XV=99 to- 1 s ~ 2.. The extrem.es .are 4- and 04, and the 'common differ¬ ence 5, xyhat is the.sum,of the series ? Ans. 442. 3. "Bought cloth, paying 8 cents for the first, yard,'10 cts. for the second yard, &c., and'fpr the last X paid 81 24: what did the cloth come to? Ans. 838.94. 4. Bought whisky, paying 4 cents for the first gallon, 7 cents for the second,. 10 cents for the third, &c., till the last gallon cost me 31 cents; what did the whisky cost? 1 7 . 'Ans. 81.75 To find One of the extreme^, when the other extreme and the number of the terms, and the sum of the scries-are given-. . GEOMETRICAL PROGRESSION* RULE/ Divide twice the sum of the seines by the number of terms,; front the quotient take the given extreme, and the remain¬ der will be-the extreme required. EXAMPLES. 1. If 3 be the1 first term of an arithmetical series, 9 the number of terms, and 99 the' sum of the Series,' what is the last'term? •" Ans. 19. OPERATION. 99x2—198; 198-f-9^=22—8=49 Ans. ' 2. Paid $1.75 for ten gallons of whisky,1 if the first gallon cost four gents, and it increases by arithmetical seties, what did the last gallon cost ? * Ans. 31 cents. 3. Paid $38.94 for 59 yards of cloth, and the last yard cost $1.24; the prices per yard being an ai'ithmetical series what did I pay for the fii'st yard' ? , ' x Ans. 8 cents. 4. I paid $600 for five horses, and the lagt horse cost me $140 ; what did I giveTot the first' one? the difference being., common ? ' Ans. $100. . • geometrical progression. V tieoinot'rical Progression,-op progression by quotients,'is q series' of numbers that incr'ease by *a common multiplier or decrease by a common divisor. The common multiplier or GEOMETRICAL PROGRES£lfof. 188 divisor, is ,ealled tlie radio. - The series is ascending when eacli term after the first increases By- a constant ratio; de- .srcn6?, what is the last term ? Ana. l5. A. If I have my horse shod, and agree to pay Ucent fcfr the first nail, 2 eents for the second, 4 cents for the'third, &c., and the number of nails being'32 what will the smith receiVe for the last nail ? : • Ans. $21474836.48. 6. Bought five yards of cloth, the prices per yard 'being in Geometrical progression; for the'first yard-I paid 27 eents; now the ratio being 1 what-did the last yard cost ? Ans. i of a cent. To find the sum of the series, the first term, the ratio, and the number of terms being given. RULE, ; Find the last term as 7/1* the preceding examples, and multiply it by the ratio f from that prodyqt subtract the frst term, antf dicidc the remainder by the ratio less 1. Note 1.—In a descending series (in which the ratio is consid¬ ered as a fraction) the product of -the last term multiplied by the ratio must be subtracted from the first term, and the remainder divided by 1 lg#s tjic ratio. No jr. —;When a descending se.ries.is carri^d'to infinity it be¬ comes what is called da Infinite iSeries, and it^ last term is- Regarded as 0; EXAMINES, 1. The first term in a Geometrical Series is 3, the ratio 2, aud the number of termfe 6; what is the sun* of the scries ?, Ans. 189. geometrical progression. 185 operation. * V J .25 =32 ; 82X3=% last term; 96X2=192—3=^189; 189^2^=1^=189 Ans. 2. The first term is 96, the ratio 1, and tine number of terms 6 ; what is the sum -of the'series ? Ans." 189. operation. (*).'=& 5 35X96=3 last topn; 8Xi=li; .96—11=941; 1—1=1 ; 94i-i-2=189 Ans. 8.v The first term' is 50/the ratio 1.06, "the#number of terms 4; what is the sum of the series ? ■ " Ans. 218.7308. 4. I buy 7 apples, giving for the first 3 cents, for the secogd 1 J, cents, for the third f of a cent, &c.; what will they cost inc ?. , Ans. 5gt -cts.'. 5v A gentleman at the Birth of his son. deposited in a chest 1 cent towards his'portion, promising to double it at the return of .every birth day until he should he 21 years old: wliat was'4 his' portion ? . ' • *. Ans. $20971.51. (J. A loafer bought 6 gallons of brandy", promising to pay 371 cents for the first gallon, 18f cents for the second, and so on, decreasing 'by the ratio of 1; what did the whole cost him ? ■ Ans. 73|| cents. To -find the suiii of an infinite series. 186 GEOMETRICAL 'PROGRESSION. BULE. Divide the first term by 1 decreased by the fractional ratio, and the quotient will be the sum required. • . ' ■ • ' ♦, 7. If the series 3, 1}, f, f, &e., be, carried to infinity ■what will be tbe sum of the infinite series ? Ans. 6. OPERATION^ ' 1—1=1] 3-=-2=GAns. 8. If the series f, 4, I, ,1,, &c., be carried to infinity what will be its sum ? Ans. 1A 9. What is the sum of G, 2, g, &c., carried to infinity'' xYns. 9. 10. If a body bo put in motion by a force, which carries it 20 miles the first hour, 19 miles -the next, and so on in the ratio of Aj for ever, what number of miles would it pass over ? Ans. 400. To find the ratio, the extremes and number of terms being given. BULE.. Divide the last extreme by the first, and extract that root of thc^quotiihit, whose index is 1 Ass than the number of terms tjiven. EXAMPLES. vl. If the first term of a geometrical series be 5, the last 32Q, and the number of terms 7, tvhat is the ratio? Ans. 2. PERMUTATIONS AND COMBINATIONS. 1-87 OPERATION. 320-4-5-64; 64^=2 Arts. 2. If the first term be 384, the-last term 3., and the number of terms 8, what is the ratio.? Ans. 2. . OPERATION. 3-381=5|j or Th;.{3f)"*=i «y%=!bAm. . 3. In an. ascending series thd first term is 4^ the.last term 324, and the number of terms ,5 ; what is the ratio ? •" •' Ana. 3. 4. In a descending series of 4 terms, the first term is 135 and the last term 5,-what i»the ratio? " • Ans. p 5. In a geometrical series the first term is 5, the last term, 500000, and the number of terms 6, wha'tis the ratio? Ans. 10. 6. In a descending series 375 is the first term, and 3 the last term; - the number of terms being 4 what is the ratio ? * Ans." 4. PERMUTATIONS AND COMBINATIONS." Permutation is the process of finding how many changt catf-bc made in the arrangement of any given number o things. " 13 188 ' TEPsMyTATHONS .AND €§51 BINATI0N3. Combination is a process of finding hot? often a less •number of tilings cati be taken out of a greater without respect to their order. To find tfie number of changes that can be made with an j giVen number of things all taken together. RULE. 'Multiply together all the terms of - the natural "series of numbers from 1 vp to the given number inclusive, and the product 'mill be the number required. EXAMPLES. i. II. >w many changes may be rupg on seven bells ? Ans 5040. OPERATION, IX^XlxlXoXOXl—1040 A?s- • 9 * A Ifow%iuany changes of order do the first 5 letters of the alphabet'admit of? ' * ' ' ' • Ans. 120. 8. For how many days can S persons be seated in a diff¬ erent position at dinner ? ' Aus. 40320. » 4. IFow many ^hauges can be made with the 9 digits?. Ans. 362880. To find how many changes may be made by takiug each time any number of things lesa than all. RULE. , Take a Series ojf numbers beginning with the number of things given, and decreasing by 1 until the number of PERMUTATIONS' AND COMBINATIONS. 1 i8D terms fquals the number of things to be taken at once, and the product of all the terms will lie the answer required. EXAMPLES. 1. How many change can be fling with 4 bells out of 6? - , Ans. 360. OPERATION. ' ^ 6X5X4X3—360. Ans. 2. IIow many'sets of 3 loiters bach maybe formed -out of 8 different letters? • ' _Ans, 336. 3. How many different numbers can. be made with 4 figures out ©f the number 123547 , . . Ans. 120. To find the number of Combinations that can be made from a given nuiftber of different things, a given number takfen at d time. PtUJLE- Talce the natural series 1, 2, 3, 4 &q., up to the-less hum* her of things, aiul find the product of the ferms, Tahe'fl- so tfc series of numbers, beginning wit\the greater number if things, and decreasing by 1 until the number ojf terms equals the less number of things, and find tfye product of the terms'. Tfhe tatter product divided by the jformw mill give the number required. ** EXAMPLES. ' 1. How many combinations can be made of.7 letters out S all t . 1-3* of 10 the letters all being different? • •' P ' KM. 120. 190 iXLiGATioar.. OPERATION. 5 3 WxPxsxtxvwkt : —120' i x?x*x4xtxitxt 2. Iiow many combinations oan be made of -4 letters out of 5? -Ans- 5- 3. X boy wishing to buy some apples, asked tbe vender liow much be would charge for 10. Tbe vender replied 25 •cents. Tbe tjo-y saying this wiys too much, tbe vender told hira be would sell bim 5 if be would give him 1 eant fcir every different choice of 5 in 10. To which the boy agreed. How much did lie pay for the b apples '! Ans. S2.52. * -AJL-LllxATJLU-N. »*All'gatidli is it process by which we are enabled to.solve * questions .reUitiiig to the mixing of articles of different ' qualities or Values. * t- . * r It is of two kinds ; Alligation Medial, and Alligation Alternate. ALLIGATION MEDIAL r - * Is the ^process of finding the average ratet of ^ mixture composed of articles uf different qualities iv valued,■ tjie' quantity and rate-of each* being given." * ALLIGATION.- 191 .Tq fehis average or medium, value. RULE. . Find the' valuta of each of thefiven articles at tlicifj~/icen rates, and divide the sum of their values by the sum of the articles. EXAMPLES t 1. If a_mQrchant mixes 301b. of coffee wmfth. 10 yents a pound, 401b. worth £ ceniga pojund, mid ,50 pounds,, Worth 7 cent?} a pound, what is .the mixture worth per poundj Ans. 8-,Lcts. _ operation. 3Olb.Xl0c.l=§T0O 401b. X 8c.= 3.20 • 50.1b. X 7c.==' 3.50 1:201b. ■ $9.70 • ■ 120)970 ^ - 2. If I mix 10 bushels of wheat at 80 eents a bushtd, with ^0 bushels of rye at 1)0 cts. a buhel, 30 bushels of bar¬ ley at GO cts. a bushel, what would a bushe-l of the-.mixture be worth ? ; , ' Ans. 7371 ■cents. ."3. If 2. pounds of gold'18 carats fine be mixed with 3 pounds 20 carats fine, -and 4 pounds 22 catats fine, what is the.fineness of the mixture? - - . -*Ans. 20.4. carat?. 4. If 20 gallons of whisky .worth 40 cents, per gallon. 192 ALLIGATION. 0 be jfnixed with 25 gallons of rum worth 50 cents per gallon what would a gallon of the Resulting M stuff" be worth J ^.ns. 45|k 5. A grocef* mixes 101b. of sugar worth 7 cents a pound with 101b. "tforth 8 eents a pound, and.lQlb. worth 9 cents a pound. What is U pound of the mixture worth? Ans. "8 eents. ALLIGATION ALTERNATE Is a proces?of finding proportional quantitiesone to anoth¬ er, in which articles of different "Quality qr value'must be takep to 'compose a mixture of' -a given average or m^an rate of value or quality. Toi find which proportional quantities. RULE. Place the given, prices Hn'Igr each other in- the order of their value. Link the price of cadi injrr.d ient} which is less than the required, average, with one that'is greater.— Take the di.ffth-encs between the average or required mean aiyl each price'and place it opposite the prict with which it is competed'or linked. EXAtiirLps.-' l.'IIow much whisky at 40 cents a gallon, runj fet 50 cents atgallqn, byandy at 70 cents a gallon, and wine at 80 cents a gallon, must be mixed that the compound may be worth GO cents a gallon ? ' . * -' Ans. 20 gallons o'f whiskey ; 10 of .rum 5 10 of brand t; 20 of win®. A£ueATl0N4 • f 40~ r|=20afc40cte. t mr^O^ 1 1=10" 50cts. ■Mean rate 60 1 ^ J { ^ .» (80— 20 " 80fcts. 2. It. is re'qijired to mix several sorts of wine at (>0 cents, 80 (fcpts, and $1.20 with water, that the mixture may ho worth 75 cents per gallon; how much of'each must he ta¬ ken ? Ans. 45gals) of water; 5ga!s. at GO; 15guls. at 80; 75gal§. at J-20.v ■ 3. How muell pdieui at 80 cents, rye at'60' cehts, and barley at 50 cents must be tahen.to make a mixture worth 70 cents ? ■» • • 'Ans/OOhu. of wheat-; 10 of-rye; 10 of barley. •_, .4. A goldsmith would mix gold o£ 19 caVals fipe with sortie of 15, 23, and 24-carats, that the .Qompound may he 20 carats fine, what quantity of each must he take ? ' . * Ans. 4ozr of 15 carats; 3oz. of 19; loz. of 23*; 5oz. of 24. " ' • Note.—Cajat'is a weight that expresses rlhe fineness of -cold. ThI^c an ounce, pound, or any quantity of,gold and divide ft into 24.equal parts ; each of those parts is a-carat. If jhe gold is fwe it is 2-t carats fine. If alloyed by some-ba'ser metal—say for in¬ stance 23 of those parts are geld and 1 of them copper,—then the goldls 23 cafaits fine,—if there.are 20- part? gold, and 4 some father metal, then It is 20, carats fine, &c\ TVlien'one of the ir>gredierrts is limited to a certain quau- fcity. RULlA Take the difference between each price, and the. mean rate as before ; then sap, as the diffrmee of that simple, \eh.o>e ALJLIOfATlOft. quantity is given, is 'to the rest of the differences severalty, so is th^ quantity yfven to th$ several quantities required, EXAMPLES. 1. \Tfiat quantity .of sugar a^t. 10 cents, and 12 cents must be mixed with 181b. at 6 cents to make a mixture Vrwth 0 cents per lb. ? ( ' Ans, 13Jib. at 10 cents'; 13£11) at 12 cents. OPERATION. o f i £l |~5S+1==4 Then 4:3: :'l8 : .13} afelfr ' 13 ' 4 :3 : ; 1& :. 13} at»l2. V 2. How much tea at 86 cents, at 94 c&nts> and at 81.05 \>er lb. ough4 t® be mixed with 61b. at 75 cents for ^mixture to* be worth 92 cents per Jb. -? * - - Ans. TSlb. &t $1.05*5- 5Hb. at 94 cents'? 391b. at 80 cents. 3. IIow mfith gold at 15, *11, and 22 carats* fine*must be ihlxcd wifh*5oz. of 18-'carats fine, So that tire comp,®sition may be 20 carats fine ? .\ns. 5oz..ef 15. carats.) 5 of 17; and 25oz. of 22 carats. When tjie sum qnd quality of the ingredient^ are given. " RULE, Pi * Find an answer .as.befqYk by linking; then $ag,dsfhes*im of the,quantities oi' differences thus determined? is to the giv¬ en gallops of water must be mixed Avith wine at "$1.50 per gallon, so as to fill a vessel containing* 100 gallons, that it may he sold at §1.20 pei*gallon? " 1 v * ' Ans. 20giils. outvoter; 4* 80g'ate.kof winb. 195 OPERATION. \r» .n 190 f =30 Then 1&9g.: 100g.:: 30 :-20 Wa/ " 1-0 X50 l=X2Q " 150:100 :': 120.: 80, wi. 150 2. A merchant has. sugar at Sets., at 10. cents, 12 cents, and 20 cents a pound; witli these he would fill a hogshead that would contain 2001b., how much of each kind must ho take, that he may sell the*, mixture at 15 cents per pound ? Ans. 33flbr at 8, 10, and 12* cents; and'lOOlta. at 20 cents. 8. How much brandy at 40 cents,, 50 cents; and 60 cents« a gallon must be mixed with water, and how much water, to makel mixture that will fill a cask holding 40gals. to sell, at 45 cents'a gallon? Ans Segals, of water; 2fgaE. ,of brandy at 40, and 50 ; and 251 gals, at 60. • MEIsTSUEATTOA. Mensuration is the processubf determining the areas of surfaces, and the solidity "of solufs. - A surfaccis that which has length and breadth without thickness; . A solid is a body having length, breadth and thickness. The aura'of a surface.is'its quantity of surface, expressed in square inches, square feet, square, yards, &c. , • ' The solidity of a solid body is. its vqlumc expressedyr? cubic inches, cubic feet, cubic yards. &c. ' MfERSURATIOX OF SURFACES. A jdane figure is .an enclosed plane surface; if bounded by straight lines it is called a rectilinear figure; or polygon. 1% 'The perimeter of a figura' ia - Che measure around it, or it? boundary. ' • ^ - \ ' - • A polygon, of .three sides is cajled n triangle;' Of"fyuf sides-a quadrilateral ; of fivq sides -a'pentagon, &e. -TRIANGLES. An equilateral triafigle is one'whose gides are alll equal. 'Ah isosceles triangle js one which has two of its sides equal. ' , ' ' A scalene triangle is one whose sides are unequal: it may have one right qugle, of may not. ( llic altitude of a triangle is a line drawn from its vertex perpendicular to it s base, as AB. • • ' ' it To dad thy altitude orf^n equjUiteral triangle. ME3TSU,RAXI0N% 19T • RULE. jFrom the square of one side subtract the square of half «f one side, and extract the square root.- - EXAMINES. 1". What is the altitude of an equilateral triangle whose sides are each 30 inches ? Ans. 25.98-(-in. « „ . OPERATION. 3Q2==9a0j lo2—223; 90b—225=675; y675=25.98' 2. What Is the altitude of a triangle whose sides are each 40 feet ? Ans. 34.641+ft. 3* Whaf is the altitude'.of a triangle whose sides Ure each 35 feet? » Ans. 30.31-fffc. . 4. What is the altitude of a .triangle' 'whose sides are each 12| inches ? . ; Ans>. 1a).825 finches.. To find the area of anj triangle, the base and altitmfe being given. RULE. Multiply the base by half the altitude, or the altitude by half the base ; and the product will be the area. , EXAMPLES. _ 1. What is the area of a triaUgle whose base is 20 feet and altitude 1G feet 1 Ans. 1.60. feet. 198 Mensuration. * OPERATION,. 16a-2-=8>; 20x8=160 square feet, Ans. * <-* 2. The base of a triangle being 98 inches, its altitude 24 inches, what is its area? - Ans. 8sq. feet, 24sq. inches. 3. A piece of land is triangular in shape, base 90 rods, altitude 67 rods,—'what does it' -contain ? ' Ans. ISA., 135 rods. 4. A triangle has a base 12 feet, 4 inches, and altitude 8 feet, 5 inches; what isifs area'? , ' * Ans. 5sq, yds., Gsq,. ft., lo.Osq. inches. 0 « % .To find the area of a triangle when the three sides are given. KCLE. Add thelhree sides together, and tatce half their sum • From thishatf subtract each side separately. Then rnnfiphj together the half sum and each of the three rcmrtindcrs, and extract the square root of tht product ■which will be the area. EXAMPLES,' 1. Of a scalene triangle the sides are 30 feet, 40 fe.et, and 50 feet; what is its area? Ans. GOOsq. feet. OPERATION. 00 ' CO * 00' 30 . 40 50 30--lsL rem. 20 2d. rem. 10 3d. rem. . CO half sum. . 30 40 50 2)120 mensuration. 19D Then to obtain the "product we hayd, ^0 X30 X^0.xl0-3GOOOO j. «. ' . ! ^ t ' • ' . 1 - ■ « From which we find .the area, fhus p/360000^600 Ans. I » 2. I have 9, triangular garden,t whose sides/in rods, are 25^ 39 and '401 what does it contain ? • -Ans. 2A., T48 rods. ' 3. What Is the area of a small •scalene triangle, whose s'det; are respectively 17, 25 and 28finches? Ap§. 210sq. incites. 4. 'What is* the number of square feet on the surface of a'figure whose "three sides are.15 feet, 41yfect and 82 feet? '• Ans. 234. Of iW many-Acres.are there in a .piece of land' 0? tri» angular shape, .the sides being respectively 610 feet, 91*0 feet, and 1000 ? , ' Ans. 6A-, IK., 2po., 22-j-yArc}s.. . 6. The perimeter of a certain field'In the form of an equiktoral triahgle is 336 yards ; wbht is the area of tlie. field? x *- " Ans. lA;, l9-psq. rods. Note.—The area of an equilateral trigon may be fouild by squar¬ ing the.side and multiplying bjr .433. ( ' • To find the altitude of a'sdaLen'e triangle, tire three sides ' ^ being given. EULE. Find the area, as above, and divide it by half the side . you consider the base. EXAMPLES. 1. What is the altitude of the triangle, whose sides, p^'e 40, 30, and £>0, if 50 be taken as the base ? i . > ■ t ..Ans'. 24. MENS Fit ATION. 2. A soalone'triaflgle has its, sides as follows: <25 feet, 39 feet, and 40 feet, 39 being the- base, what -is ,t!ie-alti¬ tude ? • Ans 24 feet. . 2. What is the altitude ef a triafngle whose'sides are 50 inches, 50 inches, and 34 inches, taking 50 as the base ? • ' Ans. 30 inches. 4. *A triangle has ."one side 121 foot, another 184 feet, ^ ' 1 ° ° and the other 20 fcp't;'iiqw 184 being'tlie base, what is tjie altitude ? t , . Ass, 12 feet. 5. In the last triangle, 20 being the bgse, what will be the altitude ? t Ans. IQff ft. QU AI>RIL ATJAd ALS". A .parallelogram is any quadrilateral whose opposite sides are parallel. • A square is a parallelogram whase sides [ are equal, and whose angles are all right ,j. angles A rectangle is "a .parallelogram, whose apposite sides are equal and parallel, and its angles all right an- gAs. A rhombus -is a parallelogram, •whose sides'are equal, hnd'its angles hat right angles. MEN6UR4,XI0Jf. 201 A rhomboid is a parallelo- grarq whosQ opposite sides are equal and.parallel, and* its , angles tiot rialit armies. A' trapezoid is a ' quadrilateral Vliieli has two of.its sides mtralkl.- A trapezium is a quadri¬ lateral which lift3 no Uvq sides parallel. To find tlio area of a square. .. RULE. Square the nunxber*dei\oting one side, and the result u iti t Ic"the area. EXAMPLES. N 1. What is the area of a square piece of land, whose sido is 70 yards ? • Ans. 49Q0sq, yards, OPERATION. 70X?O=^4900 Ana. , MENSURATION. 2. I have a square piece of land whose side is 80 tods; how many acrep are in it ? r' * Ans., 8^(5 Acres. 8. .What is the area of a board 71 inches square? •Ans. 501 square inches. 4. Whht is tlie area of apiece of land square, whose side is 120 yards and 2 feet ? Ans. 3A., OR., Ipo., 10-v yards. 5. How many square yards of carpeting will cover the floor of a room 18 feet square? Aps. 36sq. yank."1 Re litid one side of a square, the area being given. RULE. Extract the square root of the area. jdxAMPLES. * G.. A square board contains 570 square inches On its sur¬ face ; what is the length of one side ? Ans. 2 feet. OPERATION, j 570—24 inches oe 2ft. Ans. » ' * i 7. I have a square pi (ice of land containing 10 Acres ; what number of yards on ope side of it ? Ans. 220 yards. 8. A lot of land is 2021- Acres; what is the length of orig side in rods ? ' Ans~ '180 rods. 0. If it takes 30 square yards of carpeting to cover a floor what is the size of the room, i't being square ? Ans. 18 feet square. ' MENSURATION*. "203 10. If the area of a square be I of a square foot, what is the length of one side? ' '' Ans. £ a foot. To find the arda of a square, its diagonal being given. RULE SqiCare the number denoting the diagonal, and tahe half of it for the area. EXAMPLES. 11. From one corner of a room- to the opposite corner is. 2Q'feet .;vhow many square feet on the floor ? Ans. 200. OPERATION. . 4 202=400; ^00H-2=20Qsq. ft. Ans, 12. Across a squ'are piece of land diagonally from corney' to corner is 320 yards; what is its area ? Ans. 51200sq. yards, 13. The distance across -a square piece of land diago. Rally is 254 rods; what is its area ? Ans. 201A., 21^., 18po. To find the area of a re'ctangle, or oblong square. RULE.' ' / , , 'V , Multiply its length by its width or its-base ty its altitude. EXAMPLES. 1. .What is the area of a rectangular bo^rd, whose length is 20 feet, and its breadth 9 inches ? - Ans. 15sq. feet.' 14 204 MENSURATION. OPERATION- 9in=fft.; 20ft. Xf=15sq. feet Ans. 2. A lot of land of rectangular sliape is 180 j-ards long, and 210 feet wide ; what is its area? Ana. 2 A.c 2920sq. yardg. • 0. How many square yards of carpeting will it require to cover the floor of a room 18 feet long, and lGft. wide ? Ans. 32sq. yards. 4. Wllat is the area of a billiard table 12 feet and 4 inches long and 6 feet 3 inches wide ? Ans. TTjfsq: fee't. r>. What is the. area of a rectangular piece of land, a •mile long and 90 yards \fide ? '' Ans. 32A., 2R., 36pov 11yds. To find the other side of a reetangle, the area and one si d e hei n g gi yen. RULE. Divide the area by the given 'side ; the quotient will be thq other, sulf. Notk.—If\he given numbers are not of the same denomination, make them so by reduction. EXAMPLES. G. The side fronting the street is 40 feet, how far back must the lot reach to contain £ an acre ? Ans. 544£ feet. OPERATION. ^■^317S-0sq.ft.; 217804-40=544^ feet Ans. MENSURATION. 205 7. The*are3 of a table, 4ft. long, is 14s

0.75; i(48xl8)~57G; then 5rG-j-GO.75=630.75 Aps.» . MENSURATION. 213 2. Require! the area of a segment whose height is> 15 rods and chord 24 rods. - . An?'. 1A. 3R. 30/gpo. 8. Required, the area of a segment of >a eirc-le, whose height is 4 and chord 1C, Ans. 44.666+. 4. What is th« area of the segment of a circle, whoso height is 10, and chord 24. Ans. 180.833+ To find the height of a segment, when its .chord and the diameter of the .circle are given. RULE. s Square the radius of the circle, and sqnq,rc half the chord; taf&flkrir different, and extract its square root. This root subtracted from the radius will leave the. height of the segment. EXAMPLES. 1. The diameter of k circle- is 20, what is the height of a segment whose chord is 16? Ans. 4. OPERATION. 102=10082^84;'100—64=30; +36=6; 10—6= 4 Ans. • 2. The diameter of a circle being 82 yards, what is the height of a segment whose chord is 80 yards ? Ans. 32 yards. 3. The diameter of a circle being 34 feet, what is the height of a segment whose chord is 30 fpet ? Ans. 9 feet. 214 mensuration! ' 4. The diameter of-a circle being 26 miles,- yhat is the height of a segment whose chord is 10 miles ? Ans. 1 mile. • To find the chord of a segment, when the diameter of the circle, and height of the segment are given. RULE. Subtract the height of the segment from the rad'us of the* circle,; then frogn the square of the radius subtract the square of this remainder, and extract'the square root of the difference. The result multiplied• by two ' will'give,'the chord. EXAMPLES! . 1., The diameter of a circle being 82, what is the chord dfia segment whose height is 32 ? Ans. 80. OPERATION. 41—32=9; 413=1681; 93=81; 1681—81=1600; 1690-=;40; 40x2=80 Ans. / 2. The diameter of of a circle being 34 feet, and ihe height of a segment 9 feet,—what is its chord ? Ans. 30 feet. 3. .What is the chord of a Segment 50 feet high, the . , -of circle being 122 feet. Ans. 120 feet. To find the side of a square which shall have an area equal to a given circle. • RULE. Find the area of the circle and extract its square root. Or, Multijily the diameter of the circle by .8862. MEN SITUATION*. 213 EXAMPLES. 1. The diameter erf a circle Is 20 fbet: what is the side of a square of .equal area ? • Ans. 17.7244-ft. FIRST OPERATION. 202=400; 400X.7854=314.1600; 3,14.1600-Wl7.724-rft. > ■ SECOND OPERATION. 20X-8'862=17.724ft. This process is the shortest, when the diameter is given. 2. I have a circular meadow 60 rods in diameter, and will exchange it for a square one; what must be the side of the square one to equal the other in area? . Ans. 53.172 rods. 3. What must be the' side of a square table to equal in area a round one whose diameter is 3 feet and 3 inches ? . Ans. 2 feet 10} inches. 4. I have a circular field containing 4 acres*; what must be the side of a square to contain the same ? Answer in yards. Ans. 13.9.14 yards+.. To find the side of the targest square that jean be inscribed in a circle of q given diameter. RULE. Multiply the diameter hy .7071. EXAMPLES. 1. What is the side of the Jafgest square that may be in¬ scribed in a circle of 30 inches diameter ? Ans. 21.213in. 21ft mensuration. operation. 30x.70tl=21.2l3' 2. I have a piece of round timber 50 Indies in diarfteter> how large a square sill can be hewn from it ? All's. 35.355in. square. 3. I have a circular island 100 yards in diameter ; I wisly to build a square garden as large as possible in it—what will be the length of one side of the garden ? Ans. 70yds., 2ft., ljin. To find the side of the largest,equilateral triangle that can be inscribed in a circle of given diameter. RULE. Multiply the given diameter Ly .866. •examples. »1, What' will be the side of thy largest Equilateral triangle that Can be inscribed in a circle 20 inches is diameter? Ans. 17.32in. OPERATION, ' " 20X.866=17.32 Ans. 2. The circle's diameter being -43 yards, what will be the side of the triangle? 0 Ans. 37vd$., 8? inches. 3. The diameter of a circle is 18| rods ; what is the side of an inscribed triangle ? Ans. 16.23| rods; To find the diameter of one of the three largest equal circles that can be inscribed in a given circle. RULE. Divide the diameter of the given circle by 2.155. MENSURATION. EXAMPLES. 1. Required the diameter of each of three circles inscrib¬ ed in a circle 100 inches in diameter. Ans. 4-6.4-f-in.che3. OPERATION. lOO-f-2.155—46.4-j-Ans. 2. I have a circ'ular field 1077 j yards in diameter, and I wish to have three circles as large as cm be made, all equal, vrithin the periphery of the field —icouired the diameter of each of the smaller circles. Ans. 500 yards. 0. I have a circle, marked on a board, 21J-^ inches in diameter, and wish to describe the boundaries of three equal circles that shall just touch each, other externally and touch'the larger circle internally. Required how far apart must the legs of the compasses be placed to describe the three circles. Arts. 5 inches: To find the area of an ellipse, RULE. Multiply the two diameters' together^ nnd (heir product by ijt* decimal .7854. EXAMPLES. 1. What is the area of an ellipse whose transverse axis 6*0 feet, and whose conjugate axis is 40'feet? 21$ MENSURATION OPERATION. 00X-10 X.7854=1884.96 Ans. 2. ,'What is the area of ah ellipse whose major, aiis'is 70 feet, and its minor axis 50 ? v . . " Aris. 2748.9sq. ft. "3. What is the area of an ellipse whose diameters-are 24 and 18 ? Ans. 339.2928. 4. The diameters of an "ellipse are 15| and yards; what is its area ? Ans.i50.5488sq:y^rds. • MENSURATION OF SOLIDS. PRISMS. A pr.ism is a solid whose sides .are parallelograms, .and whose ends or bases.are equal polygons parallel to each'dther. A triangular prism is one whose Jiase is a tri¬ angle. A square prism, is one 'whose, base is a square. A paralldopipedan is one whose ends or bases are parallelograms as well as-its sides. CYLINDERS. ' v A-cylinder-is around solid of .uniform diameter, and-whose bases are circular and parallel to each other. The perimeter of a prishi or cylinder is the ■whole line that bounds its end or base. mensuration. 219 The height,' or attitude', is tlie perpendicular di'stafie'e from 'one of the bas.es to the other. To find the Surface of a prism or cylinder. RULE.. ' • * x . • Multiply the perimeter of its base by thq height fur the ■ convert surface, to which add the arias the ttioo ends, ■ where the wltble surface is required. .EXAMPLES. * 1. What is the entire Surface of a. square prism 10 Met long, and' its base {2 feet squarS ? "An?. 88jsq. ft. *' OPERATION. ' 2X^=8 perimeter of the hase ; 8x16=80, convex sur¬ face; 2~=4, area of one base; 4 X2—8, area of both bases; tlEen 80-}-8=88 Ans. 2. What is the entire surface of a cylinder 49 inches long, anil 10 inciids in diameter ? ' ' , % Ans; 1413.72sq.'inches. OPERATION. 10 Xo.1416=31.410, p>qrimetor of base;' 31.410X40=12 56.64 'copv'ex surfaoe1Q2 X .7854=78.54 area of one base; ' 78.54X2=157.08, area of both bases;' .then l25'6.64-j-157 .08=1413.72' .Ans. . * • • ' / * * * 3.. What is the surface of a cube, the length pf each side being ^0 inches ? ■ ' Ans. 2400sq. inches. 4. What is the 'entire surface cf a triangular prism 20 15 , ,2:20 MENSURATION. feet/long,' and' its.base aa equilateral triangle having «e&eh of its sjdes 18 inches ? ' ^ « » * Ans. 9H949,sq. ft. 5. How many square feet of surface has a cylinder which it 8 feet and 4 iriehes long, and the circumference of its base 15.708 inches ?. « • • ' / , ^ns. ll.18.sq. ft. • 0. What is'tpe entire Surface of a parallelopipedon 8 feet long 4 inches wide, and 3 inches thick ?, ' 'Ans. loGSsq. inches. 7. A triangular prism is 10 feet lo&g, audits base in the . form of a scalene-triangle haying its sides' 30, 40 and 50 iyehes. What is its entire surface?, Ans". 10£:q. ft., 48sq. in. To find the solid contents of a prism or- cylinder. EULE: ' Mu{tjq>ly the area of one'end lUnd its slant height pO.feof. . t ' A»s. 4398 A4sq. ft. " 3.. Required the surface of a triangular pyramid, 'the . perimeter of its base being'120 inches, and its slant height ' 8"feet and 4 inches. Ans. 6392.S2sq. hi. To find the surface of afrustruin of a pyramid, or of a cone. ' RULE. Multiply the sum of the perimeters of the tiro ends ly MENSURATION*, half the slant heiyht, and to the product add the the two ends. EXAMPLES. 1. What is the entire surface of the frustrumof a square pyramid whfese'slant height is 10 inches, and each side of the lowercase 5 inches, gnd.each side of the upper base.$ inches. ' ' Aug. 194sa..4n. OPERATION. 20+1^=32, 'sum of perimeters; 32 ><5=1GQ, cpnvex* sur- ' face; 25-]-9=34, sum'of areas; lG0,-f-34=194; Aris. 2. W*hat is the' surface of a frus{rum of a cone whose., slant?'height is 2® feet, and the diameters of tlie ends It) feet and 3 feet ? * Ans. 49}.01t>Gsq. ft. 3. What is the surface-of the frustrum of a triangular , pyramid, whose slant height is 4 -feet and 2 inches and one of "the equal sides of the lower basS *12 inchcg,' and one of .the upper base-G inches ?.•>.*. ■ , ' Ans. 1427 935sq. in. v * % * "To find the solid contents ®f,a pyramid os cone. ntiLfc Multiply the area of'the hase hy %'of the a&tfflde. EXAMPLES,' , 1. The. height 'of a regular, cone being GO feet, and the ' diarnqter of its base 40 feet, what is its solidity ?., , _ Ans. 25132.8ca.'ft . 223 areas of operation. 1600'x.7854=1266.64, £,rea the base; .1256.64x20^ • 25132.8 AnSi' 2r A pyramid is 90 feet high, and its base*40 feet square ; hW-*many> solid feet are therein it? ,"i " ; , Ans. 48000.** 3. *A. triafigular pyramid is^50 fee thigh, and each side,of Its Base is 10 inches: how many cubic inches are there m it? • Ans. 8600.2cu. in! *4.'If there is a piece of lead'in the form of a* regular cqne_, the height of which! is '10 inches, and the diameter of it's base Q inches,* what is. its weight, the, weight of A cubic fpo't of lead being 709 £flb».^* •" • '* _ ' "'An's. SSib. ito'z.i * 5.^ I ha%e a conical wine.glass 5 inches high and *4. inh ched/Vfidq at the tbp ;4 how mych wine will it hold; a gallon of wine occupying 231 cubit? inches'? * ' •* . ' x - * * Ans. 2t% gills. 4 ^ f , t • t 9 ~. * * • ♦ To find .the solidity of a frustrum of a '.pyrniyid'or. of a cone'. - ' ' htft'e. Multiply thn areas*of thjf tied-ends together find' extract i\e square rootnf,the product.to this root add-,'the two areas, g,nd multiply^ t-he s.uni by J qf ihe altitude v examples*. 1. If tlie length of a frustrum of a Sqjiare pyramid be 18 feie't, -and the side of its^grSater ba&e bti 4" feet, ^nd of. the less 3 feet, wh5,t i§ its^sdlidk^? : ' .* ' . * . k V + Ans. 222cu, feet.. ' M-J5N SUK ATIpJT".' 'OPERATION. 16x9=144 j 144^=12; l2+4lC+9=37; 37x6-222. 2. If the frustrum of a cone be 20 inchfes long, and 10 inches in diameter at one end and G inches in diameter at the other end, what is its-volume ? ' « . '♦ *Ans. 1026.256cu. in. . . o\ Tf the frustrum of a triangular'pyramid, he 30 feet Mng/and one end has-its side 2 leet and 8,'inche's, the other 2 feet and G inches,, jvhdt number of cubic "feet does it com t^in ? . - , , "Ans. 'SGcu.#ft.4123G-d-cu. in, 4,-Suppose a water bucket be 12 inches dcipf and th'6 diameter#of one end 11 inches, of ,tha.Qth'er end 1(1 inches j how many gallons'wijl it hold? . • Ans. 4£-jtgallt)ns/ SPriKEES.- A* sphere is a Solid .bounded by a curved surface evei^-part of which* is equally distant from a,point-within called the centre.^ #' The diameter 6r 'axis of a sphere is ' a lihe passing {hrough'the centre, and terminating at th£ surface,. ' , * ' *.'*>» * The radius of a sphere is .a line drawrf from the centre - • to the surface^ it is half the diameter. « • A segment of a sphere as a part of it f cut dff bj a plane. ^ The pique is •called the 'lajse. of the segment; and the"- 220' . , ' MENSURATION ' .perpendicular distance from its-vertex to the centre' of the base is its height. . • To, find^the surface of a sphere. rjji/E'. Multiplythb diame{cn by the circimfevense^ EXAMPLES. 1. Required the convex surface of a globe vthose di&mc-.' ter'is 30 inches., Ans. ,2827.44sa. in/ • , , * 1 2. Y/hat istliQ convex surface of"a sphere whose diame¬ ter is 2 J iftchcs? ' • ' f • • Ans. 10.03 jsq. in. • 3. What number of square miles on the surface of the earth its diaipeter being 70b§ miles? ' •: - Ans. lOS9507.$G-j-sq. miles. To find the solidity of a sphere. -RULE. . I Multiply ike cube of [he clutmc'fcr by .5230. • /• • EXAMPLES. » 'i., flow many solid inches in a Spherfi of 20 inches libvaicter? " • ( An$.,4188.Sou. in, » * V > OPERATION. • 20*48000 ; 8C00X-O236==41S8.S000 Ans. 2.'Ir the diameter of the earth' he taken as 8000 miles, .how many solid miles of earth are there in it? » '»■»/ - . • Aps,* 268O^82O0OQO,Qiy mi^s'. MENSUEATIOlSr. • 227 3. I have a hcjjlow sphere the diameter of. Which Is 31 inches ; now, the shell being half an-inch thick, how many gallons of water does i£ contain ? , ' AnS Clgals., § quart. 4. What is the weight of a small silver globe 3| inches in diameter, a cubic fodt of silver wciehiid? G54i pounds? Ans.AO.SB+lbs. ■ % To find the convey surface of a spherical segment. RULE. A" Multiply the height of the segment by the circumference of the sphere. EXAMPLES. 1.* If the circumference of a sphere be 2(Mnches, •jvhat is the convex, suffacfrof a segment qi it 3 inches high ? . Ans'. GOsq.'id. OPERATION. * . 20 X 3=6 0 Ans. • 2. The pirclimference«of a sphere bcidg 30 fbet and 6 inel\es, what is the. convex .surface of a segment, the height ' of which is 2 feet*ahd 4 Inches ? - ' • , Ans. 7-Hsq. ft. To find the solidity of t^he' segment of a sphere. » RULE. Po three times 'the square of the radius op the base, add the square of the 'height', multiply this-sum by-the'height, and the product by :5236>" Tint remit icitt, be the solidity of the segments • . * 228 MENSURATION, • N(Vrk.—The radius of the base is found, when the diameterr-gof Ihq sphere is givpu, precisely as half the chord of the segment-sof' a.'circle is l'ouad. ' EXAMPLES. 1. If the diataetcr of a.sphcre be# 20 inches, and a seg¬ ment,' the heigh^of'which is 4 inches, he cut oil,,what is the solidity of the segment ? Ahs. 485+cu. in. OPERATION. To hnd the radius -10—4=0- 10-^=100; 82=2A;r 100-^00=04; C4-=f8, the radium of Hie'segment's base. Then 8-X3=122; 192-f-4-=20S; 208X-A=-'o2 .: S&f gg • 52oO^=A.o'j.ooo2i .xiiS. . 2. If 'the diameter' of a sphere be'82 feet, and ar segment 1 foot high be cut a oh", what will tactile solidity, of the aegthcut'{ ■ ■ , Anc 127.7584eu. ft. 3. What is the ■solidify of useguxnt cut off from a spherp ■34 inches in-diameter-, the height of .the segment being(9 ' inches? • Ans. 85-j2.5744eu. in." 4. From a sphere which is 50 'inches in diameter. I cut a segment 18 inches high, how nhmy cuhic inches iiVit? • Ans. 191*39.6896cu. inches. SPHEROIDS. spheroid is a figure rosemhr C linb' a sphere ; ^about as much' like • , _^fifpB§|jp|5^ a spherp as an ellipse' is Jike a oiiw ele.. It may;be formetlby-the revo¬ lution of an ellipse about one of its' axis.' 'MENSURATION* ' -229' ' t If the ellipse revolves about its transverse or longer axis (AB) tha solid "describeebis called & prolate spheroid ; if it revolves about its conjugate or shorter axis, the solid, de¬ scribed by the revolution is called oblate spheroid. ' ' . . * r ' " *1 The earth is an cbhtfe spheroid because the ax;3 or 'diam¬ eter about which it revolves ps 30 odd miles shorter than the diameter perpendicular t,o it. ' ^ lip find the solidity of a spheroid. EULE. * * % * Multiply the square of fhe revolving axis by the fixed axisj ahd that product by .5330. ' . EXAMPLES. " ' * • 1. In "a prolate spheroid the transverse' or fixed axis is 59. inches, and the revolving axis £0 inches; 'wbift is the solidity? , Ans. 235G2cu. inches. ORE-RATION. : 302=90Q; 900)^50X15230=23502.0000 Ans. * * * ,• * , * * * ' « • 2. What is the solidity of p, pr6lat£ spheroid whose fixed axis is >200 feet and revolving axis 10 feetc? ' . 5* ' Ans. l'0472cu. in. ;3. What' is, the solidity ofa'n oblate spheroid whpsfe fixed axis is 60 and revolving axis 100 ? • " , -Ans. 314160eu.'irr. 4. What 'is the solidity, of a prolate spheroid whose trans¬ verse -axis is 50, and conjugate axis-40 ^ : ... . Ans.-41888cu. in. t 51 "What is the solidify-.of an* o.blate spheroid whose diam¬ eters Ve 20, and 10 ? '' ' '■ Ans. 2094..4cu. in. 230 MENSURATION. CYLITOPJCAL -RINGS. v • « / If a cylinder Be bent uniformly till the two, ends meet, tlte figure forme'd will be a cylindrical ring.. To fijid the convex Surface of a c-yl- •indrical ling. RULE. Add the thickness of the ring to the inner diameter; multiply this svm la/ the piic7cness,and'thatgroductht/D .87 • , To fifid the solidity of a cylindrical ring. RULE.. Multiply the sun\of.ihe thickness'and inner diameter by' the'squarc if the thickness,.and that j>rod act Vy 2-4674. S. . ' • EXAMPLES. 1. The thickness of a cylindrical ring is 4 iriefiesgand "the inner diameter 18 inches; what is tire convex surface and solidity ? N \ ( '868.50sq. in. liS' i.8GS.524jBcu. ih'. 0 ' 2. Th'a.th'ickncss of a cylindrical ring is 2 inches and the outer diaincter 24 inches; what is thf csnvex surface and. solidity ? * * . s " i { Surface 434.2Ssq. in. n-"|'Solidity217.1ol2cu. in. 3. The thickness of a cylindrical' ring is half an*inch,v and the- outer diameter 5 inches; ybat'is its solidity? .' . f Surface 22.2-frsq# inches. ns' [ Solidity 2.775-j-vCu. -inches. MENSintXTfON. "231 4.- Find the surface and solidity of a big'ring, the inner diameter of which, is 10 feetr and its'thickness 10 fn clips. a f Surface 8!)sq ft», 15sq. inches. 1 '•( Solidity 18du. ft., 372.2cu. in. Note.—The area of any regular polygon, may be found by squar¬ ing the side of the polygon, ant} multiplying the square so-found by the number representing^ the area of a similar polygon* whose sidgis 1, , ' We give the numbers for eight regular polygons : * 0 _ Triangle,....?. .*..*. 43tf, I Heptagon, '. 3.634, Square, ,...v.\ 1. ' pOctagon, v* 4.8284, Pentagon,..: !....'. ......1.72, I'Nonagon, 6.1816, Hexagon, 2.508, } Decagon,- 7.694-2.' MENSURATION OE LUMBER. . Board's (in this portion of the country generally called planks) are measured by the square foot.* Scantling, Joists, sleepers, sills,'. &cj, are' measured by* hoard pleasure, the board bgfng considered an inch thick. iVhen-tjiey say that a plqrrk contains so many feet of lum- 'ber,' tlicy mcaiftli.at there are in it so many squares,12 inches each way and one inch thick. So, too,, when they say a piece of scantling oontains 8 foot of lumber they mean that it contains 8 square# 12" inches each way and 1 inch thick. / To find the number of feet in a board qr'irich-plank. RULE. Multiply the length in feet by the width in inches, and #*232 • -LlftlBEE,. the product divided bp 12 v:til be the answer, -(dr multi- * pip tile length in feet bp the ■width in feet, dnd the product will be the amic^r. • " ■> , * .* * * * Note.—If the board t ipers add -the width of the two ends to¬ gether and take half of it.for the mean width. EXAMPLES. ' 1. What are the contents of a board 21 feet long and 8 inches wide ? s % • • . An9. 10 feet. 2.. What*are the contents of a board 1 Allot long, and 1G inches wide ? ./ ' An?.. 24 feet. 3.- What are the contents of a tanerirjg board 13ft. long; I A . and-14 inches wide at one end- and G inches wide at the 'other? ' A::*-. 10 feA. .'4..»XIow many feet are /here in.10 boards 21 feet.long a$d 11 inches wide ? ' Ans. edd feet. 4 % 'To fiild th'c number cf feet, beard nicasurc, in scantling, joists, silts/ &c. >t ■ EULE. Multiply the uddlh in inches, bp thd thiclenccs in inchc£,* and this product bp the lengthinfeet'; andjhp lc\stpjwt^uct, divided bp 12 will give the answer :nVci7. . -» ^EXAMPLES. * • \ . *. 5. How many feet of lumber are tlierein 20 i.cnsfs wliich ore 18 feet lcjng', 5 inches wide, and 3 inches ihipk ? . Ans. 450ft. G. How many feet are there iivlS -sleepers 15 feet long, ■ 10 inches wide and three inches thick ? . Ans. G75 feet. LUMBER: •2.3.3 7. IIow many feet are there in . a sill 11 inches square and 21 feet long ? < ' ' Ans. 2111 fcet. if the scantling, joist, or s/leeper; or whatever it may be, tapers both'-in' wicfch aitd in- thickness, .thq cqmmdri rule is to add the areas of the two .ends together^;tgko J of the sum, and njuftiply by. the length in feet and divide by 12. 8. A piece of scantlipg 20 feet long is'lQ'inbhe^ square at- one, end, and 6- inches square at the other. How many feet does it contain ? ^Ans. fl3£.. 9.' How mhn>y feet'Jboarcj measure, are there* in>a rafter 80#fcct lon^, 4 inches by 3" .at -one end, and 0 inches by 2 at tl;c otli'er ? * , . Ans. '22• 10... J low many feet, board measure, in',10 pieces of scant¬ ling 20 feet long each' and 5 by 4 inches, at one end, and f> by 2. at the other ? ' Ans,'210 if. Tn those'last examples 'the exact- answer hi ay be get# by- considering the -pieces, frustra of pyramids', and working accordingly ; but. we have givfen the, rule which we think is common among saw-mill men. . KMUGIlfG. GA'XIGIITG-. • , • # ' ... Gatfgirrgis the process of finding, thq capacities fifties', boxes, cask's, $'c. •* *' / • " * To find how many barrels of corn, in the .ear, a crib of ^given dimension^ will contain. RULE.' Multiply ike length, hrehillh, and»ctrpth of. the cvib, ta¬ ken in'feet, togetherf and that*prdrlnet by .08; iRe% resist will he the-contents 'in harrelijnegrly. ' • . \ . . * ' * Note.—Th^ fule.is founded on the supposition that.ei'er^ hun¬ dred cubic feet cotftaim'8 barrels, Wh'yh is,« very nearly correct-— The truth is there cau be n&F VARIOUS PROBLEMS. ■ been drawjuofE, the Ji^upr* fe onjy 1,0. vfehes de^p in'the cgsk* How much is there in it? Ans. 28.68gals.. The contents in gallons of .any vessel -©f any shape may be found by dividing the cubic inches of space within it by 231 which is the number of cubic inches in a common wine gallon. 7. flow many, gallons "frill a cubical »hov 3ft. long, 12 inches, wide, and 10 inches deep hold T • . , • Ans* IS.7 gallons^ 8. Ht)W many gallons will a cylindrical, vessel 2 feet high and 10 inches in. diameter* contain ? Ans..S.10gaIs. 91. How many gallons of water will a* cistern contain that -•is 6 feet long, 5 feet deep, and 4 feet wide ? . Ans. 897.Gga!s. .SOLUTIONS OF VARIOUS PROBLEMS. Ef ANALYTIC METIIOPS. * 1. A gentleman divides his property among^his three sons A, B and 0, giving B twice as much as A, and C twice as much as B; and tlie whole apmunt.divided is .$7000. How much'.is the share of each ? Ans. A, $1000 j B, $2000; C, $4000. SOLjjTib^s op YARtous >rob£!emS; : 239 OPERATION. A "l v 1x1000=1000 A's." B 2 Then 2x1000=2000 B's. C 4 •: 4x1000=4000 C's, 7)7000(1000. • 7000 • . 2. Divide 9856 Between A, B and (Dp giving B tlir.ee times as mu.ch as A, and C as much as both of them. Ans. A, 8107; 33, 8321; C, 8428. 3. I have three Yards of cloth which cost me $1.20'; the second cost twice as much as I the first, and the third twice"' as much as both the others. What were their prices?' , *4 Ans* 13Jets.; 2GjCts.; 80 cfcnts. 4> A tlrdvcr has a lot of horses, mules afid oYch ; 'there are twice as many mules aS horses, -and as manj oxen as mules and horses. He sells his horses at 8100 a head, his mules at 880, and his oxim at 820; and got for the whole $2560. Ilowhiiany 6f each ? • ' . - Ans. 8 horses; 16 niylcs; 24 oxen. OPERATION. H. 1XS100=$100 1x8--= 8) M. 2X 880=8160 2X8=16 } Ans. ^ ox. ?X 820= .86a 3x8=24} 8320)82560(8' 2580 240. SOLUT^ON^ OF VARIOUS' PROjBRpMfe. 5. A far are? sold a cart load of water melons, musk mel- t I ons, and gourds, for 821; .the water melons were sold at 30 cents a piece, the musk melons at 20., and the gourds at 10. Now, if the number of musk melons were double the number of .water melon?, and the number of gourds equal to the number of water melons and musk melons together, how many had he oi each ? . Ans.. 21w.; 42m.; 63g. 0. I have some apples at } a cent a piece> twice as ma¬ ny peaches at -} of a c-cnt o piece; and four times as many jjcars at 1 of a cent a piece;—the whole woith 20 cents. How many of each? ■ . Ans. 12 apples; 24 peaches; -fS pears. 7. Bought some calico at 18 cents a yard, and three times as-much muslin at 40 cents a yard; sold tlie calico at a gain of 334 per cent, and the mruslin at a gain of 2~i per cent.; and received for the whole 812.18 How many yards were there of each. Ans. 7y"ds. of calico; 21 of muslin. OPERATION*. cal. lx.is=:i$;'..18xl8&j= .24 1x7=7). mus. 3X.40=1.20; 1.20x125=1.50- 3x7=21 \ Ans" 81.74)12.1S'(7 lllS "Where percentage, which is an ev0n p'art of 100, as 331 or 25, is to be put 6n, it'is perhaps best done by expressing it fractionally, for instance, if an article cost 12 Cents, and you want to put on 331, it will l>e done thus 12xf==10. SOLUTION'S OPVXilOUS PRO'BLfiMP. 241, The rtcken ufKjiis process is evident. ' Th6 cost' 12 cents va -%) 33 d ppr feerit. is another Ihird which maHes f( 2=36") Second f=-jf Tlien 16 1C )\ 2=32'Ans. Third * 15 15>h2—36 ) 49108(2 9,s ;j We. simply,invert tji-o given fractions, and reduce tlicm io a common denominator p; then take the new numerators and proceed as ItDthelast examples. The argument' is the same as in tb^ lust} 1. e. the first will get'three halves of a ' doHahy when the second gets four thirds of a dollar, &c. SOiUlloM OF V.AHIOUS PROBLEMS: 243 % 14, Let $74 be divided between1 A and B so that f* of Ms may equal 4 of Bin' , *; • . * • * . An% 'A,-$82; B, 842. 45. .Bet the number 25 be so- divide'd into three parts- "that \ of the first may be equal to -jj- of the second;'-aneL»f of tho third. . • • * , *16.vA gentleman being as-Iced his age replied "ohe half and three ci-gths of my age 111 ike 35 years." ( What wag his ago ? 1 ; Ans. 40 years. operation. J-_j,n=7. 85-^=40 Ans. It is evidenfthat ^ of his age is*85 years; it is.alsp evi¬ dent that, if o5 is of any thing, * of 35 is all of that -thing } Hence, we simply add the fractional parts togeth¬ er and divide the given number.by the result. . ■ - 17>• | add t of nry "money is'894: How inuelf have I ? • Ans. $il'2. 18. A gets J ©f an estate, B, if of it, and C, of it, and they all receive 82500; What is/the.cstqtc ? ' ■ Ans. $3191.48|4. 19. I'have three purses;, in the fh-st 1 have § of my money, in the second § of my money,' and in the third 4 d@!larg. IIow much-money have . • » Aps. $90 OPERATION. ' •* * 1+Wli i—ii-pft; 4-^=90, Afts.,' i ' T'his example is, not like the last three. In them, tha- purii of th OPEIM-Tiad..-, : • ' 5 000^7=1215, .Ans. , ^ What the Qifternl-contajrjs'^a nothing to do with -it? . It„ may contain onfe gallon Pe a million^-the result is the sante.', It i$ cvideh't that the aperture, which empties fhc distern in ^6I^UTJON3 X)J? ^ARIOU^J^ROBLEMS. '^7 ' ^0 lninrrt?%, would empty 3^ -of it in one piinute ; so the sec¬ ond \vQuld eyipty -£e of it in a minute, and the third ' 5*0. Adding^thebe three fractions together we obtgin^what'part ;of it ad tbflea would eimpty in,a mkulte, wliicli is •What¬ ever number multiplied infto'47.Y'H make it 600 is#t]ie num¬ ber of Required to ejnpty the vdiole ci,stern.. That number yiay be obtained by dividing 009 by 47. 29. A 12=1=12,-Ans. ,After working 4 clays there is certainly just.J tliQ work to do yet, for they could have completed it la. S days; -and it takps B 12 days to finish h ah' the work;, of caursc if 5 can* do half of it in 12 days he can do all of it in 24 clays. It'then becomes a problem just like the wiristy problem •Which. we haddast. So'1—hj equals what A Can dd in a day. • 41. A and B engage to do a job in 12 days; hut, after working at it S days, B quits, and it thon takes A 10 days to finish it. Required how lohg B alone would be in doing the'job. . ' • ' Ans. '20 days. ' 42," A and B engaged to reap a field in 4 da}Ts'; they both wotk at it one day and B quits; it then takes A 9 SOLUTIONS OF VARIOUS PROBLEMS. 251- % days to finish it. Required the time e'afih .would- |j£ in reaping the field. . ' , v f A, 12 Jays., A"S'.{B, 6 Jays.. 43. Several men can do a piece of, work, in 6'days'; if 5 men more work with them they can do it in 4 days. Re¬ quired the number of men: 'Ans'. 10 meif." . OPERATION. 1^5=^; or 1(^-1=1.0AOS, % v * Subtracting } from 1 we find what the 5 men could do' • in a day, which is + of the work. If 54 men can do +- one man can do V»f it in,a day; in G'daySone maA would" do ^ or -+ of the work; And, if in." 6 day,8; one. man does -+ of the work, it will take 10 men to- do alRthe w'ork. * 44. Several men can do a job in 10 days; if there were 3 men in or e^ they could do it in 8 days. How many, men are there ? Ans. 12. 45. A market woman bought some.eggs at •} a, cent a piece, and as many more at. R of a cent each, and sold them id! at | of a -cent a -piece; and lost 4 cents-, by.the' trade;— Required the number of eggs'she bought. •' Ans.- 240: , • OPERATION. t i 1 5*. 5- •' 9 * • 5 —J-• 4—J —'HA \ n'<< l + f —J4' Tg yo.' •* ■ ao _Yni>. If she bought some at a a cent, and the same number + k 252 SOLUTIONS OE VARIOUS PROBLEMS, •if' a •ce&fc th3n adding \ and -j together and dividing by 2 we get -fj as the average price she paid for each egg4 Then subtracting | from y~ shows what ihe lost on each egg, which is-g1-^ of a cent. But; as she lost 4 cents in all, then dividing 4 by ^ -we obtain the number of eggs, 240. 46. A market woman buys some eggs at J of a cent a piece arnicas many more at 4 of a cent apiece, and sells them allaf -fa of'tx cent a piece, ami makes Scents by the trade. How many eggs did she h'uy in all ? ' , ■ Ans. 40. 4T. A market woman buys a-certain quantity of eggs at of a cent each and as many others at % of a. cent each and sells them all at | of'a cent each, and makes 12 . cents by tire trade . How-many eggs did she buy in all ? Ansr 00. 43. A boy buys some apples at f of a cent a piec-e^ and twice' 43 mai-iy'others'at f of a cent a piece, and sells"ikem all at | of a cent a piece and loses 5 cents. How many apples did the "bpy buy ? ■ • Ans.' 36. OPERAXlOJr. H-iNftl f§-f==,V §-^=36, Ans. We cannot get the average price of one apple by adding dim 4 and-f togother and dividing; by 2 as we did in .the preceding examples; but, as there were iicice as many ap¬ ples bought at I as-at 5, we just doubleTthe I whk'h gets the price of two apples of the second lot: then adding f and -J together we hdvc tire price o((hrce apples'; and dividing by 3 gives us the cost per apple. Then subtracting n from qf. gives us the loss on each apple. And, as he lost.5 cents in all, dividing 5 by ~/g gets 36 apples, which is the answer required t SOLUTIONS 01? VARIOUS PROBLEMS. 253 49. Bought some apples at $ of a cent a piece, twice as' tnany others at £ a cent; sold them all-at | a.cent, and, ©lost 4 cents;—how many d,id t buy ?. - ■ ' Ang. 90-. 50. Bought eggs at f- of a cent, and three times as many othdfs at f of a'cent: sold them all at | of a .cent and gain¬ ed 9 cents. How many did 1 buy ? Ans. 80. 51. I have a room 18 feet long and 16 feet Vide;• bow many yards of carpet' i of a yard wide will it re'quir.e to cover the floor ? - Ans. 40y4s. OPERATION. '18X16=288 288-^-9=3?; 32-~i=40-J Ans. Multiplying 18 by 1'6 we obtain the'afca of the floor- in square feet; dividing by 9. we obtain the Area in square 4 •I • yards; but Us,the c'arpet is.ouly | ©f a yard wide it will take | of a yard of it to cover a square yard. ? that is, it wifl take 5 yards of the carpet to cover four square ybrds.-r- Ilence multiplying the area, 82-, by f produces* the-ansWef required. , - 52. How many yards of carpet f of a yard wide will be required to cover a floor ID feet square ? Ans! 32yds. 53. There Is. a room whoae dimensions' are 20 fe'et by 18 • the carpet with which we wish to cover it js -85 inches wide. How many yards must we take? ' " AnWdl-t. 54. The area of a certain floor is 288 square' feet, and the side and end are to each oth'er as 9 to 8. Required the length .of the side-and end. Ans. 18 and 16 feet. 17* 254 SOLUTIONS -OF VARIOUS PROBLEM®. OPERATION. ' SX9=?2'; 288-72=4; q/4-2; I j'xCiC Ans. , Wherc'tlie product'df two numbers and their ratio is giv^n 'j&e jjmply multiply the two numbers denoting the ratio together, and-by this product divide the given pro" duct 5 then extract the square root of the quotient, and mul¬ tiply, the "root into the numbers denoting the ratio. As njultiplying the terms of a ratio by the same number docs not alter the ratio.it is evident that some number mul¬ tiplied! into 9 and S owll give the true dimensions of 'the ifoom, Jf that number were known, and multiplied into 9 and 8-, itus evident tha/ th5 prpduc-t of the. two products woukl be equal to 288, ghe given product. And, as in that ■inultiplieat-ion the, common factor (which is- unknown) •would be squared, it is evident that the product' of 9. and 8 lacks, the squarQ of that factor (itseljfto be taken as a fac- -tor) to make the 72 equal to the 2-88. Then it is evident that 72- is contained in 288 a humbei; of times equal to the square of that factor. Therefore by extracting the square root we obtain the factor which multiplied into 9" and • S will raise them to their proper value. 55. 'I-wish to set out an orchard of 2400 apple trees "so /that the lqngt^ shall be to the breadth as 8 to ^ 2. IIow .many trees must there be-in length, and in breadthh i ) 00 in'length. "" j 40 in breadth. SOLUTIONS "OF VARIOUS'PROflLFMF.- '255 ■ 06. I have a room the side and ehd'of which ivt> to each' other as 3 to 2, and if I b,uy carpet ^ of a'yard wide it-will ttike 108 yards to cover the floor. Kequfred the length pf the side and end. . • ' . Ans. 36 and 24 feet." • 57. The difference of two .numbers is 3,-and the ^dif¬ ference of their squares is 61. Whht arfe the numbers? Ans. ,10 and 7.' OPERATION. 51—3=17 ; 17-j-^—20; • 20-1-2=10, Ans. for the larger number. • ' ' Where, the difference bf the squares .of two nmqbers -are, given together with their difference, dividing the differ¬ ence of their squares by their difference gives the sum o* the numbers. Adding the 'sum, and difference of two numbers.together and dividing by 2 gives the larger number- Where $ip .sum pf the numbers an/1 diff'eiVnce-. of their squares, are giveitf-dividing the -difference of .squarps.hy the sum gives the difference .of the number^ ' .58. The'diffcrence of t]ie squares erf two numbers' * is 16; tlie sum- of the nuipb'ers 8. -What are the numbers ' , .. ^ ' Ans. 5 atid 3. 50. if he sum of'-two number^'is" 8, and'their" product' 15*.' What are the nurpbers ? ' Ans. 5 arid- 3/ • 9 ■ * v * * QPERA'TION.* 84464; 15^4=^60; 64^60^4; 1?4=2 pS+^tO'; '10 « . • ' v' ; . f -*-2=5, Ans. v r The sum of any two n itpibcrs and their product b.eirig given to fipdthe numbers^' Squhre tli'e sum and' from' the1 square subtract four '.time^ the1" product tfie remainder is' the 256 somjtzons'of various problems. sqtiare -of the difference of the numbers, and extracting the square r06t.gives the «difference. Adding the sum and difference together and'chviding by 2 gives the larger of the tw® numbers, dO. The sum of two numbers is. 9, andtheir . product 8 ; what ^re tine numbers ? , Ans. 8 and 1. Gl. The perimeter of a rOom is 72 feet, and the area of the floor is 820. square feet. Required the dimensions of the room. . * Ans. 20 feet long, 16 feet wide. G2. Iq a right-ahgled triangle the sum of the hypothe¬ nuse and perpendicular is 90'feet? and th£ base 30 feet.— Required''the perpendicular. ' Ans. 40 'feet. OPERATION. t 9 O2=8100'; 302=900.; 8100—900=7200; 7200->TS0 =40, Ans. Where 'the sum of the hypothenuse and perpendicular is given'with the bags to find^ th'e peqpenditpilar. SquaVa the sum of the hypothenuse a>nd perpendicular, and square the base, and diyide their difference by _twice 'the puhi of the hypothenuse and perpendicular. 03, A tree 32 feet high stands on the bank of a stream 24 feet wide. Req-uired tlie'height of the stump to throw the top of the. tree to the opposite hank, thq tree- still cling¬ ing to its stump. . . ' Ans. 7 feet. By Prob. 62, we -simply divide the difference of the squares of 32 qnd 24 by twice.the^ height of -the tree.— (Those whp prefer it Rait solve it hj proportion t of similar SOLUTIONS OF VARIOUS PROBLEMS. 257 triangles'. See Greenleaf s Key to his National Arithmetic, .Example 39 in Square*Itoot.) [ 64. A tree 72 feet high stands on level .ground; 'how high must the stump be to throw the top of the - tred to »a point 60 feet from the bottom of the'tree, the end 'where it is cut off still resting on the stump ? Ans. 11 feet. 65. A tree standing on level ground is 50 feet high, what must be the height of the stump to throw the top of the tree 30 feet from the bottom of the tre§ ? Arts.- 16ft. 66. I have a vessel in the shgpe of-a conital wine .glass whose height is 1q inches, and width at the top 16.inches, I wish to-know the diameter of a sphere-tKat.will just sink into it till the vertex of the sphere Is even With the top' of the-vessel. Ans. Oi?.inches, . To solve this let? the student draw a diagram 'represent¬ ing the vessel,,and within that describe' a circle just ev,en horizontally with-the top of the vessel; let the 1-ine repre¬ senting the height of the vessel .be drawn right through the centre of thfe oh'cle and mark that line 15in., and mark half the line representing the horizontal width pf the vessel at the lop 8in. lie now has a right angled' triangle whose perpendicular-js 1'5 and base 8, with which he can.find the plant "height of the vessel. He- finds that slant height'fo be 17 niches. 'Thep let him draw from the centre @f the cir¬ cle. a radius perpendicular, to the side of the vessel where it is tangent to the* circle* Tpat line will * e'fit off from the line representing the side of the vessel a poffcion at the top equal to. 8 inches.I. h'ot .thpte will be-a regular figure form¬ ed, of four sides, two of its side's being radii of the circle, and the third side '8 inches, 00d of course the fourth ship *258 solution's of -various problems. will hi 8 inches too. Cutting off' 8 inches at the • top will leave 9 inches below. He now has> two similar triangles, the "perpendicular and base of one being 15 'and 8, and the perppndicularCof the other being 9. Then by propor¬ tion of triangles (see square root) he can make- the state¬ ment thus, 15 : <8 : : 9 : 4*] which is half the diameter re¬ quired. ' 07. What must be the diameter of a sphere which will sink even with the .top into a copical vegsel 24 inches .high and 14 inches wide at the top. Ans. 10 fin. jfS. • A conical vessel is 12 inches high and 10 inches w-ide at the top.' Required the diameter'of a sphere that will go down into it till the vertex- of the sphere is just even with the top of.the vessel. • . * . Ans. 0# inches. 09. .Taking the same vessel we. last had,, required tin" diameter of a bailor sphere that will go just half of itself iilto it ■ I mean a Sphere.that wiil go far enough down into it to leave exactly half itlie,sphere -above the horizontal plane of the top of the vessel. < 1 j ■ . . -. ^ . jAns. 9133 ipches. 1. 70. I employ a map k to. work-for me 30 days', oh,. eondi-> tioii that for every day'he works. I -am-tp pay him 83, and % everyday lie.iclgvs he is to pay me 2 dollars besides 'for¬ feiting ,his'wages ; /it the end of the time lie received* S40. How many days clid lie i'dle ? . Ans. l0 days. OFERATJoX, o'o'.X 3---90; 90-^-40=50 • 50-=-5t=10, Ans. • - , e-; ' . if'h'e liad worked all the time' at S3 a day he would hav# SO^UTI-OXS OF 'VARIOUS PROBLEMS. 253 got 830; but he only gets 810 ; he has therefore lost 850. Now tliip 850 divided by his loss per day will certainly show how many daj^s he was idle ; and his loss per day is 85, for lie losses his wages 83 and pays a forfeit of 82 be¬ sides. This kind of problem is generally s'olved byDouble Position. " 71. I employ a man for 90 days, on condition that for every day he frorks he is to have 82, and'for every day he idles he is to forfeit* his wages and 81 besides. At the end of the time- he received 8135. Required the number of days he worked, and sthe number he idled. 1 Worked 75 days. Ans' \ Idled ' 15 " f T 72. I employ a (man for,12 days; for every day lie works he is to have 81, and for every day he idles he is to forfeit his wages, and pay me 82.'. How long doeS lie idle if at the.end of the time he receives $115 ? » * 1 Ans. i a day. 73. A and B invested equal sums -in 'trade'; A gained a sum equal to -} of his stock, and B lost 8225; A,s money was then doVble B's. What did each invest ? ' .,Ans. $000. OPERATION. , 5^=1; 1-1^:225^3=1500,^. * J t -is pyident", if A gaing that his stock is f ; but he has twfce as.mnoh.as B„ 'and .B's, is therefore tie .half '7 • which is. § ; and.'if B potf lias | he has. spent, § y but. he spdnt$225; 8225, therefore is of his capital, and. | of 8225 i/.the capital. 260 SOLUTIONS OF -VARIOUS PROBLEMS. t 74. A and B invested equal sums in trade; A gained a sum equal to ^ of his capital, and B lost 8500; A's money was then 4 times as much as B's. Reqhired the capital in¬ vested by each. Ans. 8700. 75. A' and B invest equal sums intrude; B gains, 20 per cent, on his capital, and A loses 8l00; A's money is now f of B's. Required the capital of each. •Ans. 8500. 7(5. I employed a workman for 30 days, and agreed to pay him $40 and a coat; he worked 15 days, and I paid him 810 and the coat. What was the coat worth ? Ans. 820. OPERATION. 840 and the coat—810 and •coat=830 : 830-f-15— ♦ 1 82; 30X2=60; 60—40=820, Ans. We reason thus: If he was to have £40 and--coat, and only gets 810 and coat, he has'lost 830 by not working 15 days ; his wages per. diem then were evidently 82, which for 30 days would be '860; and. as we wei'o to give him 840 and the coat, the coat must be 820 to raise 840 to 860. 77. I employed a,man for 120,days, and agreed to pay -him 825Q and a horse; but he only worked 70 days an,d I paid him 8100 and"the horse.- What was the. horse worth t ; . * • ' • Ahs. 8110. 78. A man agr.ee S to work 14'days for ^20 ami a pair of boots ; he works 5 -days'and gets'8^ and the hoots. What' are the boots worth ?' Ahs'. 88V SOLUTIONS OF VARIOUS PROBLEMS. 261 79. A and B receive the same salary; A saves one third of his evdry year/but B by spending $250 more than A per year, fitids himself at the expiration of 7 years $350'in debt. Required the salary of A and B. Ans. $600. To solve the above we reason thus: If in 7 years B gets to be' $350 in debt, he gets to be $50 in, debt a year, that is, he spends $50 more than his income each year; and as# he spends $250 more than A does; and only goes $50 be¬ yond his income1, A must save $200 of his income; and; as he.saves 1 of his income, then $200 is J of his income,.and his income is'$000. 80. A and B get the same salary; A saves I of his every year, but B by spending-$150 more than A per year, finds himself $250 in debt in "5 years, Required their salary. Ans. $400. 81r A father divides his property among his three sons; to A he gives half of the whole wanting $40, to *B-£ of -the whole and $12, and to C the rest which is $8Q less than B's share. Required the whole amount divided. i Ans. $576. FOUyt OF STATEMENT,. A B V $—40 £+12 £—68. OS H-"*+*=S ■ 40 1=4 108 12 ' 96-r-£=576. 96 As A gets 40 less than half, 'setting the 40 off and giv- 262 SOLUTIONS OP VARIOUS PROBLEMS. ing him J, the 4 is evidently 40 too large for his share,; and as B gets | 'and 12, setting the 12 'aside, and leaving . him with I, it is evident that he lacks 12 of having his share; hut C was to have $80 less than B, and as B gets $12 Over a third C will get $68 under a third, and setting this $68 aside and giving him a third it is evid'ent that his 4 is $68 too large. Now we have given A \ which is $40 too much, and we have given C 4 which is $68 too much ; A's i, and C's i then have in them $108 too much; hut B's 4 lacks $12 of having in it enough, which reduces the $108 to $96. - Now it stands thus : A has 4 ofit, B i of it, 0 a of it, but in having these fractions they have $96 too much. Adding -5, 3,.and 4 together we obtain | which is ■?r more than all ofit; and as they had $96 too much, $96 ! must be i of the whole. Therefor^ the answer is $576. 82. A father distributes a sum of money to his three sons ; to A- he gives I of the whole lacking $100, to.B' l of the whole and $100, and to C the "remainder wjiich is 8100 more than B's. Bequired the sum distributed. Ans. $1200.- 8o.. I wish .to divide $85 among 5 bo}*s and 7 girls so that each boy shall have twice as much as a girl. Bequir-' ed the shane of each bov and each girl. . f Each boy 10 dollars. ns' { Each girl 5 dollars. OPERATION. 5X2=10 v .2X5=101 . 7Xl-J_ 1x5= 5 ) » 17)85(5 : ,85 SOLUTIONS OF VARIOUS PROBLEMS. 268 84. Divide $159 among 7 men,, 8 women, and 9 chil¬ dren, giving to each child some, and to each woman twice as much, and to each man twice as much as a woman. f $12 to each man. Ans. 3 $6 to each woman. , ($3 to each child. 85. A gentleman dying,, his wife being enceinte at the time, directed in,his will that if she'had a son she should have } of the estate and the s6n f (but if she should have a daughter, the daughter shoiild have 1 and the mother f. Now it sd happened that she had tioins, a son and a daughter. IIow must the estate be divided, it being $7000 ? f Daughter 1000. Ans. ] Mother 2000. (Son 40,0.0'. • It is evident from the, will that he wants the mother to have twice as much as the daughter, and the son twice as much as the mother. The operation is therefore easy. See Prob. 1. 86. Another just such case as that happening where the estate was worth $49.70, how should'it be divided '! ( Daughter $710. Ans. < Mother 1420. (_ Son 2840. 87. What will be the side of the largest cube that can be cut out of a sphere 20 inches in diameter? Ans. 11.547in. operation. 20x20^400; 400-^-3=133.333333 •, p 133.338333=11 .547 Ans. ' Wc Simply square the diameter, divide by 3 and, extract, the square.yoot. (See Square Root, Prob,■ 27.) 284 SOLUTIONS OP VARIOUS PROBLEMS. 88. What will he the side of a cube cut from a sphere 30 inches in diameter ? Ans. 17.32in. 89. What will be the side of the largest cube that ca'n be cut out from a globe 60 inches in diameter ? Ans. 34.64in. 90. I have a stove pipe 4 inches in diameter; how long a piece must I cut off to hold just a gallon ? Ans. 18.38in OPERATION. 4x4=16; 16x'o 854=12.5664; 23l-12.56Q4=LS.3S. Since a gallon contains 231 inches; and the contents of a piece of the pipe, that would contain that much, would be found by multiplying the area of the base by the height ; therefore the contents divided by the area of the base must i y J give the height. 91. What must be the height of a cylindrical vessel 10 . • inches in diameter, to contain 3 gallons ? Ans. 8.823-fin. 92. Suppose a tin pipe an inch and a half in diameter; how much of it must be cut off to hold a'quart ? Ans. 32.7-pinehes. . 93. A gentleman being asked the time replied that J of the time that it was past noon was equal to A of the time that it Wanted to midnight. What ,■was the time ? ' Ans. 3 oVlock P, M. OPERATION.' From noon to midnight is 12 hours. The proposition is then -to divide 12 hours into two such parts, that'A of the first may be equal to A of the second. • fee Prob. 10. SCHLUMOXS OF VARIOUS PROBLEMS. 265 04. When -I of the time' from noon is equal to yV of the time that it Wants till midnight^ what is the time ? Ans. 30min. after 4, P. M. 05. If f of the time from midnight is equal to of the time that it wants to noon, what is the time ? Ans.. minutes after 3, A. M. 96. What time in the day is it .when | of the time froni sunrise is equal to f of the time till sunset, the sun rising at 5, and setting at 7 '/ Ails.- 46min. 27y?rsec. after 11, A'. M. 97. Out of a cask of wine*, which had leaked away a third pari} 21 gallohs were affprwards drawy ) and the' cask watf then half full how much did.it hold ? Ans., 12Giral, t OPERATION. g-^==i; 21T'.ro—126. Ans. If a third*.had leaked away there were f left in the cask. Drawing off 21 gallons more rcduocd the $ to I; therefore the difference between f and shoWs what part of .the whole 21 gallons are.. We find that difference to be. i j 21 gallops, then, are l of the whole, and hence the whole 126 gallons. » # # ' 98. Out of a cask of wiitd, which had leaked away a fourth part, 39 gallons Werd afterwards drawn ; and the cask was then' | full : .how many gallons .did it contain when full ? Ans. 80. 99. Out of a cask of brandy l had leaked away, and then 8 gallons being drawn out, it was left full > how much did it hold ? , • Ans. 60 gall.ons. 100. Ajmle is divided into three parts. No. 1 .is 15 feet long ; No. ,3 is .as long as N'o> 1- and half of 'No. 2 5 266 SOLUTIONS OF VARIOUS PROBLEMS. Nd. 2 is as long as No. 1 and No. £> together. Required the length of No. 2 and No. 3. . f No. 2, 60 feet. . { No. 3, 45 feet. To solve this problem we represent the whole length oi No. 2 by 1', then No. 3 will be £+15; but No. 1, and No. 3 added .together will get No. 2, which we find to be £+30 ; No. 2, then, is J of itself and 30 feet long. Its whole length then must be GO. 101. A fish wa^ caught Whose tail weighed 0 pounds, his head weighed as much as his tail and half his body, and his body weighed as much as his liead apd tail togeth¬ er. What was the. weight of the fish ? Ans. 721bs. 102. Ih4ve three purses : the first contains 8 dollars, the second as much as the firsthand -J of the third, and the third as much as the first and second together. Required the con¬ tents of the second and third purses. . f Second 16 dollars. Ans* {Third 21 dollars, 103. Two travellers, A and B have stopped on the road¬ side to eat their dinner ; A has 5 loaves, B has 4. Anoth¬ er traveller coming up eats with them, all eating equally of the bread. The third traveller pays 18 cents for what, he has eaten. How much of it goes to A, and how much to B ? Ans. 12 cents to A; 6 cents to B ? OPERATION. 5+4=9;<6-e-3^=t3^ lS-e-3^=6 ; 5—3=2 ; '4—3=1: 2x6=12, A's; 1x6=6, B's. Adding the. five and four together-'we* obtain,!' loaves ; dividing 9 by 3,men wc find that each man eats 3 loaves; SOLUTIONS OF VARIOUS* PROBLEMS. « 2'GT •dividing 18 cents by 3. we get the pi-fee of one loaf to be t centd. As A liad 5 loaves, the third traveller must have bat 2 of A's loaves, and 1 of B's. Therefore multiplying 2 by 6 .we get 12 centa for A Ac. 104. A has 3 loaves, C 2; C cats with them and pays .15 cents: how. much does A get-? and B>? ; . A 12. cents. Ans- " B 3 cents. _ r lOth If in a pair of s;cales a b0(Ty yreiglis- 901b. in one scal^g.and only 401b. in the other; requhed^the true weight., and the, proportion-of tfyc lengths of the ffwo arms', of th^ balance-beam on.c^ch side of the point\of suspension. 4. Ans. True weight 601b." Pdoporti&ri 3 to *2. IVc dimply ,find'a geometrical mean between 90, and 40 by extractingkthe square root of their product, which gives 00lb. as the true weight. • Then the • difference between this 60- and 90 is"80 ; -and the difference J) etwee q 60 and 40 iff 20. And 801b. : 2Qlb. : : 31b. : 21b. or 3ft.': 2ft. 106.* In one scale ar. body weighs 121bs., in the other on¬ ly 3lb,; what is t]ie B-ue weight and the proportion of the arm§"? Ans 1 TruQ CJ>- { Proportion 2 to 1. 107. In one scale suppose a body to weigh 51b., in the other 801b.; what is'its true weight, and what are the fro-, portionate lengths of the ar'hs of scale beam ? • / ' Ans.'I W*20:fb. •' •' i Porportioii 4 to 1. 108. A person in play lost 1 of his money, and then won 3 shillings; after which he lost -J of what he then had, and had but 12 ' shillings left." How much-had he at first ? ■ ,. . Ans. 20 shillings. 18 ' 288 SOLUTIONS -OF VARIOUS- PROBLEMS.' OPERATION. , J 12=3=18 j 18—3=15; 15-=-f—20, Ans. ) If at the last he Ipst'J and iiad 12 shillings left, 12 shill¬ ings must he t.of what-he -had before the last loss, ^ and therefore 18 shillings,is what he had-after,winning the 3 shillings; and 18-^3=15 shillings, what he had before he won the threp'shillings, arid* just after his first loss.— His first loss was -} of the whole, and so 15 shillings must be of the. wliole ; itherefore 15=1=20, Ans. 100. A gambler lost f of his money and then won 5 dollars; .he tit£1^ lost f of what he had, and had but 10 dol¬ lars.remaining, IIow-much had he at first ? Ans. 870. 110. In playing three games a gambler loses as follows: In the first game' lib loses g of his capitalin' the second game he'Joses S06 dollars ; in the third game he loses* | of what he*had left after the second game; he now has but 200 dollars remaining. How much was his capital ? ' Ans. 81600. 111. A and 13 departed fronf different places and traveled towards paeli other. On meeting "it appeared that A had travelled 18 nriles more than B ; and that A could haVe gurie B's journey in I04 days, -but B would have been- 28 days L;i performing A's'jouruoy. Now supposing them to have started at thy euriie tinje, how fhr did .each travel? . , ( A, 72 miles.*. . ,1,1!, 54 . - SOLUTIONS QF' VARIOUS Plto^LEMS. 269 , OPERATION. f-r=ii i8~i=54; Ans. for B; 54-j-18==72, Ans. for A. Jo understand the atiqve operation we'll take the answers 5^ and 72, and exhibit their relations according to the conditions. First,'54 divided by 15f is evidently A'a journey per day j, and 72 divided by is B'js journey per. day.' Now B's daily journey must be to A's daily journey, as B's whole journey is to A's whole journey. Hence we have, the proportion, ; ' • 7 54:72::J?;^i •28 ; 151 542 722- * and • — ; diyiding denominators by 15.| we' have ' 15| 28 0 J 54 722' . m 54 —^=...7 ; extracting the square r6ot we havq =3 1 V 1 ,. " • 1' TA-, that is 4 of 54 wpuld be equal to 72; but 72 is l'8. 4 * •. 3 ' larger than 54, and therefore l'& must be i of 54. "So di¬ viding 18 by J" we get the answer for B. „ • From the above we* deduce the following r,ftle for solving such problems. " 1 . 'divide the larger number representing th.e time in which ojie eovld hdve gone the otlieds journey, by the smaller.;, and .extract the square root of the quotientt Subtract -one from 18* • ' • . 270' SOLUTIONS OF VARIOUS PROBLEMS, the foot so found-and divide the given difference ■in 'dis¬ tance by the remainder and you have the answer for the shorter distance. . . OPERATION BY ALGEBRA. * Let cc=B's journey, And'xc+18=A's journey. Then"A?L_ will represent A's daily journey • 15J ■ * ' ' ' - * And will represent B's daily journey! . Then x : ce+18.: : : JL* \ . 28 , 151 ^ And , 152 .2^- • .* Dividing denominators by 15| we have ^==^ZtL^L_ ° J 1 1J5 W * '' ^ /p_I ¥ Extracting square root we have -—— ; ' ^ .if * • "Or; 4.r=A3x-j-;54: *" * And, a;r=54, Ans. 112. A and B departed from different" points atthesame time, and traveled towards-each other. On meeting it ap¬ peared that A had travelled 11 miles farther than B; and A could have gone B's journcy'in S(j hours, "but B wofald have 12' hours in going AA * Required how fan each tra¬ velled, a-fid their distance per hour. Ans ^ ,1-1i1[OS • enul G miles per hour. 1 " | B 50 miles; and 5uniles pci'houf. 11S. A and B started, at the- same moment from different points and came together. When met, it appeared that A had gone 9 miles' farther than B -y and A could have gone SOLUTIONS.OF VARIOUS PROBLEMS, 271, Bh journey-it} 7| hours, but B would have been 10^ hours in' gnihg A's j.ourney. Required; as m the last. • » *( A travelled 72 miles; B, 68. (•A wdnt'8 milejs-.per hour; B, 7. 114. 'A and B engage to bhild a wall for $36 ; and, as A could do it it 6 days-,, tlrcy. agree to do. it' in 3 days.— But after'working-2 (lays.-they are obliged >to call in C, a superior workman to assist them the last day. In conse^- quence of-which B received 3 dollars less than he would lyive done otherwise. Is. what time could B build the veall ? and in what time could G ? Ans f R in-V2 da-vs- - AM' { C in 4 days. . OPERATION. They were to have $36 fof building'the wstll,- and as they agreed to finish it in 3-dayg, and A himgelf could' dolt in 6 days, he is evidently entitled to § of the $-38, whieh-is $18. There are then $18 left' for 4 B and- C. It. appears 7>y * the question that B loses $3 by employing C. Jt doc's not ap¬ pear by the question how much A loses by employing C, and that is the first thing we have- to find iii the operation. "What B receives is to what A receives, as what B loses is to What A loses by employing C. We have here a question in proportion, wh.ere wfe have, only the second and third terms given to find the first an,d" fourth- We may make the state¬ ment thus, supplying Jji<3 place df the first and.fourth terms with,an asterisk : . * : $18 'V: $3 : * Now it is. evidgqit "that the product of the two unknown terms-in' the above statement.is 54, for.the product of the extremes in any porportion'i^ e'qual to the product.of the 272 SOLUTIONS OF- VARIOUS PROBLEMS. % means. It is also evident that the sum of. those, two terms is 15, for $3 has already beenjtaken out'as what B jays to . C,. and represented, in.our third term, leaving $15 out of- which B is to receive his, wages, arid A is to pay his quan¬ tum of .loss to C.' Now, where the sum and product of two numbers are given, .to find the numbers, we have seen (Prob. 59) that by squaring the sum and subtracting four times the product we have the square of the difference, ex¬ tracting the sqtfare root of which gives us the difference of the required numbers : And by adding the sum' and diff¬ erence together and dividing by 2 we have the larger of the two numbers. ■ Squaring the sum 15, then, gives us 225 ; multiplying the product 54 by 4 gives us 21G ; and 225— 216=9; the square root of 9 Is 3, the difference of the numbers; adding 15 and 3 together and dividing by 2 we have $9, the-shm which B "receives, and 15—9=6, the sum-which C gets from. A. B, and Q then, each, received $9; for.'C receivfed §6 from A and $3 from B." Then to find'the number of days»required .by B and C seperately to do t£ie" work is easy. Thus : $9 : $36 :3 days : 12 days, Ans. for B. I 89 : $36 : : 1 day i 4 days, Ahs. for C. 115.' Two reapers, A and B, engaged to reap a field for $24; and as'A 'could do it in 0 days they ngree'to do it in 5 days. But they are compelled to fall in, 0 In assist them the last-day; whereby B receives $1 less than he would oilier- wise have done. In bow many dayS would B and 0 sepa¬ rately reap the field ?. f B in 15 days. nS-|'C in . 9 days. il<5. A and \B.engaged, to reapXfield for 90 shillings. ■ A could reap it, in 9 days and .tliey promised to complete-It • SOLUTIONS 'OF VARIOUS' PRdBLEMS* 27Sf im 5'days'. They, found, however, that they were-obliged to cull in C, an inferior workman, ,to' assist them the last' two days. In'consequence of which B received 3s. 9d. 16ss than he otherwise would have done. In what time could B and C eachT-eap the field ? . f B in '15 days. ! " ' " 2 Ds'. [ C in 18 days- h' 117. I have two horses, and' a saddle worth $20. If T put thesaddle.on the first horse it will make liis value of the second horse ; If I put the saddle on the sdcond it will make his value | of the first. Required the value of each horse. Ans., first horse $50 ; second, $-10. OPERATION. Assuming 1 as the value of the first •horse, the second: horse must be worth | .of the value of'the first horse ami saddle together | of X—{—20 is 4 —)—; the .value; of second horse. • Putting the saddle on the second horse, it stands 4-Jr3l5; now the first horse is, worth |" of this, which is Jyo_[l5pJq We now see that the value of the first horse ks Afl of his value and-4/y0 dollars; then 5^T° dollars must be ij-of h'is value. Hence his value-is $50. Adding 201 to ' 50, and taking | of' it we luyte for the value of second horse $40. 118. I havd two horses, and,- a, .saddle worth $20. If I put the saddle on the first horse It makes him worth 14 times the value of the second horse; if I put it" on the seed ond it inak.es him Worth 1.^ times the first* . TV hat' is the value of each ? . ' a ( First horse $100. Ans' I Second' , $90. 119. There are two hoi'ses and a saddle worth $30. 'idle 274 SOLJJHOWS OF VARIOUS PROBLEMS. first hotse. witli* the saddle .is worth twice as much as the" seeond ; the second, with the saddle,«is worth only | of the first. What are the values ? ' . f 1st. IPorse $210. AUS'{-2d". . " ($120. 120. A gentleman lias two plantation^ and 8300 worth ofi'stock. 'If the stock are 'on the first plantation its value will-he 3 times (the second; if the stock are on the second plantation its Value will be f of the value of the first.' "What, is the value of each plantation ? ' ** f , (One §1200. \ (■ The other 8500. We here--close the solutions of Problems. "The student, ih reviewing the?above, will find it a very'pleasant as well as profitable exercise to make for himself similar problems "for solution. Indeed the requiring of students to oriyin- cde'problems has not, I think, received from teachers the attention that it deseVvcs. It is a field of exercise never failing, in its, harvest, o'f the very best results. «, MISCELLANEOUS. EXAMPLES. MISCELLANEOUS EXAMPLES. 1. Wliat part of $0 is £ of a unit? Ans. 6V 2. -Divide 50 cents betweeh A and B so that A shall have Gl cents more than B. A 284 cents. Ans. , ^ ,2].i centg_ 3.- A B and C are trading in partnership / A's capital was, in- 3 ihonths, and it was 540; B's was in 4 months, and it;Was $G0; C's was in 2 months and it was 8100 : tjiey gain 828 ;—what is, each partner's share of the gain ? f A's .86: Ans. < B's $1-2. (C's $10. 4. If 4 horses or 5 mules would consume a certain amount of provender in 14 days} in what time would 3 horses and 5 mules consuihe it?_ Ans. 8 days.. 5. Bought a box of cigars for $2.50, and by retailing them at 30 cents a dozen my gain is 20 per cent. p how many cigars in the box ? ■ Ans. 100. G. If I of a yard of silk be bought for G3'cents w^hat will be the price of 10 yards ? . . Ans. $8.10. i . 7. A buys of B two bags ohcotton weighing each 4501bs., at 10 V cents per lb. He pays him'$20,in Gash,1 and a note on-C given for $6Q which lias been - ojn. interest 8 months 276- MISCElLANEOUS EXAMPLES'. at 7 per cent.; and is to pay the balanee in coffee at. Gets, per pound; how many pounds of coffee wiH B receive ? Ans. 1051bs. . 8. A tree 45 feet high stands on the bank Of a stream 24 feet wide; what would be the length of a line to reach from the top of the tree to the opposite'bank." off the stream ? • ' ' '.Ans. 51 feet. 9." If 3 ifien iri 7 'days dig a.well 42 feet deep and 5 feet in diameter, what is the diameter .of a'well that 4 men in 8 days dig to the depth of 441 feet? » . 1 Ans. 6 feet. 10. Suppose a vessel holding ♦ 37 gallons of water to be full of water,-and to lxave two outlets by one of which it would, be emptied in 12 hours, by the other in six hours, and to have also an inlet by which it would be filled in 5 hours. Leaving all thr-e'e of these open, in what time ^ould the vessel be emptied of its contents? Ans. 20 hours. I 11. The weight of a cubic foot of lead being 709 Jibs what; would, be the diameter of. a'ball, of lead that would weigh lOlbs? ^ Ans. 3.59-fin. 12. A gambler in tlife first game loses ^ of his money; and in the second game wins three dollars; in -the third game losing -j of what he then had, he has 86 remaining. How much had he at first ? Ans. 827 '13. Let a ladder stand beside a wall of equal height with the ladder; if moving the lower end of the ladder 30 fecf ftom the base of th'e .wall brings the top of the ladder 10 feet below the top of the wall, wliaff is the height of the wall or ladder? ' % Ans. 50 feet. 14.- Out of a. cask of wine, which had leaked away, f part, 22 gallons were afterwards drawn, and th'e'cask was- then | full. IIow much does it'hold ? ■ Anfe. 40 gallons-, • MISCELLANEOUS EXAMPLES. 277 15. If a sphere -20 inches in diameter weighs 4001bS., what will be the weight of a segment of it 4 inches high ?' • . . Ans. 41.Gibs.. 16. A gentleman being asked the tipae replied that £ of ,the time, that it was past noon was equal to |'of the time it wanted tillmiidnight.- What was the time? Ans. 56min., i.7-§-|sec., after 5, P. M.v 17. What principal a.t interest at 7 per cent, will draw half a,cent a-day"? ' . Ans.. $25.71f. 18. I buy in market 16 sacks of coffee each weighing 160\b. at 10 cents a .pound ; I pay 6i cents "a sack, drayage ; and the freight is | of a cent a pound. Now how' many pounds niust I sell for a dollar, to gain 33j||A per pent ? . . Ans. 71bs. 19. A's flock is blessed with great increase, lie oWas ^ multitude of geese ; There's neighbor B'just by his side ' Whose flock has also multiplied. , Says A—"You give me forty-two, I'll have four tiihes as much as you."- ' Says B—"You give me forty-two And I'll have just as much as you." Let those who're skilled in numbers teach ' * 11 ow many geese belonged to each'.- - Ans ' I A hadi182. . ; S,|R " -98. 20. I have brandy worth 7-5 "cents, p'er gallon, arid I have a cylindrical vessel 2 feet and 6 indies deep, and-20 inches in diameter, full ofithe brandy. What is the worth of the brandy in .the vessel ? .... Ans. $S0.60 21. -The distance' from Atlanta to Newnan is 40 mihes, and A hires a carriage in Atlaiita for $8.00 to go to Newnan and badk, with the privilege of faking in three other-s. Fairburn he takes in B ; at Palmetto he.take§ it C ; and at Newnan hp. takes in D. Fr-ojur.Atlanta" to-Fairburn is 20 278 MISCELLANEOUS'EXAMPLES, miles; from Fairburn to Palmetto-6 miles.. Now how much must each pay for the use of the carriage ? - . • fA $8.76#. ' . . • .. J B 1.761- " • . Ans- ] c iM#; [D ■ "Xoo. • -V • 22. Required t,he length of one .side of the largest'cube that can be cut out of a sphere 120 inches in diameter ? . Ans. 69.28-j-in. 23. Suppose two small wheels one 12 inches in' diameter, the other .10, connected by an axle 15 inches long, be started on a level floor with sufficient impetus to bring ihem back to the place of starting, what will be the diamiker of tlfe .cir¬ cle made hy'the inner wheel on the floor ? Ans. 12ft. Gin. 24. A gentleman has a horse for which he asks' 8125 ; but he can fall 20 per cent and stilf make S3 J'per cent. What.did the horse cost him ?, ' ^ t Ans. 875. 25. John Smith had a note for 81080 discounted at a Bank on the 2flrd of December 1857 the note was due on the 10th. of January 1858. The percent being 7 what were the proceeds of tin; note ? Ans. 1075.80 26. A farmer carries some horses, oxen, and hogs to market; there are 4 tlfiies as many oxen as horses, aud twice as many hogs as horses and oxen together;—he sells ehe.h horse for as. many dollars as there are hogs, each ox for as many dollars as there are oxen, and each hog for us many dollars as there are horses ; aud received for the whole 82304. Required the number of each ? ^ ,8 horses. * ' ■ • Ans. -j 32 oxen. ' (_ 8.0 li'ogs. 27. Th<* periitreter ©fa oertain room is 140 feet, and the MiscELLisrEckrs EXAMPLES. -279 distance from one corner of the. room to an opposite corner diagonally is 50 feet. . Required the -dimensions of the room. A " Ans; 40 by 80 feet; 28. A, B, antrGhbuy^ grindstone 16 inches in diameter for 80 cents. B paid 25 per cent.more than C/and A paid | as much as both of -th,cm-. How much must each grind p'ff from the semidiameter to obtain his proportional shafe of the stone ? „ v- ",Ans. .A, 2in.; B, 2in.; 'C, 4in. 29. A gentleman dres leaving three sons, the eldest 20 years old, the next 10, an$ the youngest 18. lie leaves $3000, with directions that it be put out at interest imme¬ diately at 7 percent, and that as his sous "come of age " each shall receiver his portion, and that the shards of all shall be the same. Required the share that each will re¬ ceive. , Ansi $1143.15* 30. ,A, B, and' C are to share $60 in the proportion of -J, J, and respectively. ' Wliat will be the ^harc of each ? • ' (' A, $25.53^. Ans J B, 19.14 RR SCr 31. Suppose a comical vessel 12 inches high, and 10 inches Aide at the top ; what Will be the diameter of a sphere that will sink into it just half of its own diameter? . , Ans. OR* inches. 82. A find B h;iv£-agreed' to-feed their horses- together, each having but onqluhfce. - A furnishes 2,0 barrels jof com, B furnishes' 16 barrqls. C requests tliat they let him put his hpAjo in with theirs, which is granted. At the end of •the. year the'three ho'rsds have just consumed till thevc'o-rri, ■and'C pays $30 for thhftted of his horse. How is the ,$30 to be divided between- A ttnd B ? " * 1 ' Vns f A gets $20./ Ans- j- B « 2-8(J MISCELLANEOUS EXAMPLES, 33. The perimeter of a room is 114 fee't,* and 100 yards of carpet 32| incfes wide will just'covet the floor. Requir¬ ed the length and width of thd-room. • • . , Ans. Length, 30 feet; width, 27 feet. 34. Let $375. be divided between 7 men, 8 women, and 9 boys,'so-that each womairmay have three times as much 3s a boy, and half as much as a man. What will be the share of each ? f iSach man's share $30. " . Ans. < Each .woman's $15. (. Each Boy's * $5. 35. What is the* weight of a hqllow spherical shell of sil¬ ver-20 inches in diameter, the thickness of-the metal being '5 inihes, and a cubic foot of silver weighing 6551b. T Ans. 13891b., 4oz.-(-. 30. If through a sphere 50 inches in diatneter there be perforated a hole 14 inches in diameter, the axis of said hole exactly coinciding with the centre of the sphere,-, how many*solid inches Will bo left in .the sphere ? '' 1 Ans. 57905.9712cu. in. ' 3-7. Sow much longer w\ll it. take $40 to double itself at 10 per cpn-t. than at 12. per cent., simple interest? • . -Anfe: 1 year and 8 months. 38. a circle 21.'55 incheadn diameter be described ori a board, and three equal circles as large as Can be made with¬ in its circumference touching each other externally be drawn, how many square inches will be contained in' the space between the three smaller circles ? ' .' Ans'. 4.lD8sq. in. . 39. .If 49,6 Then', in 5} days of 11 hours each, dig,a ires eh 465 feet long, 3i feet wide, 2} feet deep,.and 7"degrees. of hardness, in how many days, of. 9 fours <*ach,. will 24 men dig a trench 337 J feet long, 5? feet wide,'3 2 feet deep, and ob 4 degrees of hardness ? . ~ *' Ans. 132 days. 'MISCELLANEOUS fcXAJJPLES. 281 . 40. A round log of wood 5 feet long contains 188494 cu- # f ° bic, inches. If a wheel were sawn off it, hotv many times would it turnover in going a mile —counting the circum¬ ference to th.e diameter as 22 to 7. ; Ans. 1Q08 times. . - 41. A and B can do a piece of work in 5T5T days, B and V in 61 days, A and C in 6 days; in what time would each of them perform the work alone ? f A, in 10 days. Ans. 1 B, in 12 days. (_ C, in 15 daysh ' 42. I have a "rectangular field containing 121 Acres, skirted all round by a slip of wood land 5 rods wide and containing 61 Acres. Required the length -and width of the'field. . • Ans.f 50 rods long, and 40 rods wide. • 43. My tailor informs me that it will take 101 square yards'of cloth to make me a full suit of clothes. ' The cloth - " • ' 1 'I am about,to purchase is'11 yards wide, and on sponging it will' shrink 5 per cent, in width and in length-. How nrany yards must I, purchase for my suit ? , •' • Anfs, Oyfjg-gyds. ' 44, A gentleman has a piece of land whose shape is a scalene triangle, the sides being 340, 50,0, and 560' rods re¬ spectively. He wishes to give, his son a square o,f land as largo as can be got in the triangle. Required the side of the soifs portion. ' Ans. 195 {5 rods. 45.' xV and R buy 200 acres of hand for $400, each paying equajly. A, in order, to get choice off the land,, agrees to take his at $2.25 per 'aerd Required how much lan'd each wilhget, and at what B's will be pe'r 'acre ? ,v .' ~ f A, , 881 Acres. Ans. B, J 11 1 Acres, b B's price per A, $1.30. 282 MISCELLANEOUS EXAMPLES. Note.—This problem is generally given wrong. We have nr> right nor reasbn to assume that B will take the rest at $1.75 per Acre. , ' . . , . 40. A butcher, wishing to buy some sheep, asked the owner how much he ,must give him for 20 ; the owner said -he might have 20 for 347.50 ; the butcher said it was too much; the owner then agreed to sell him 10, if ho would pay 1 cent for every different choice of 10 in 20 ; to which the butcher agreed. IIow much did the butcher make by trading that way ? Ans. He lost S1800.00. > 47. There is a conical wine glass 0 inches high, and 5 inches wide at the top, and which is 1 part filled with watey "What must be the diameter of a ball let.fall into the water that, shall be immersed by it ? . ' Ans. 2.445—j-in. ■ 48. Bought a fi.pantityof cloth for 8750, f of whh-h be¬ ing injured I hadf to sell at $1.25 per yard, and. by this I lost $1Q0 : wl;at must I sell tho rest at per yard to gain 5 per cent, on the whole ? \ n„ OOl O'J.ooji. 40. A father dying left his son a legacy, 1 of which he spent it* ,4 yemrs ; i of the remainder lasted him.5 years longer. Them dying.himself, hit money was put out at in¬ terest for'l year, at 7 per cent, till his oven son shcukl be¬ come of age,-4wlio received, when of age, $1210. ddequir- cd tho amount "of the legacy at fret. ' , . Am. $20000. 50,, The dimensions of a bushel measure gee IS} inched wide, and 8 inches deep; what sho;dd be the dimensions of a similar pleasure to poutajm S bushels? . j Ans. 37iu. wide; 16in. deep. FOREIGN COINS AND CURRENCIES. 28o APPENDIX A TABLE: . • Showing the Present Legal Value of Certain Foreign Coins and Currencies in. Dollars and Cents. •Pdund Sterling of Great Britain,. .• $4.84. Pound Sterling of British Prov., N. Scotia, N. Bruns¬ wick, 4.00 'Ounce of Sicily, ....... 2.40 Pagoda of India, and Star Pagoda of Madras, 1.94. Tael of China, 1.48 Millrea of Portugal, 1.X2 Specie Dollar of Norway and Sweden, 1.06 Specie Dollar of Denmark,'. 1.05 Dollar of Mexico, Peru, Chili, and Cen. America,... 1.00 Millrea of Azores, i.. .881 Ducat of Naples, " 80 Rix Dollar of Bremen, 1 7.8 f Ruble of Russia (Silver,) i ,75 Thaler of Bremen,.- i •. 71 Rix Dollar or Thaler of Prussia and N. Germany, • .69 Sicca Rupee of'Bengal or Bombay, : » <■ .50 Florin of Austrian Empire hnd of Augsburg, 481 Rupee of British India, '. .' *... .444 Florin of South Germany, and Florin or Guilder of „ • 'Netherlands, .. ^ .40 Mark -Banco of Hamburg, , 35 Franc of France and Belgium, and'Lira of Sardinia, .1S| Livre Tournois of France........ 1A} Leghorn Livre, and Lira of Tuscany and Lombardo- Yenetian Kingd., ( .16 Real Plate of Spain, 10' Real Vellort of Spain, & 05 19 2'84 tVEIGIITS. A TABLE: _ ' Showing The Weight of a (Tabic Foot of Certain Sub¬ stances in Pounds. ;Jb Air A- Cork 15.. Fur..,....'... .' 34f. Tallow 59. Water 62 Mahogany ' '. 661. Oak.... 734. Loose Earth or Sand &5. Common Soil 124. Bsick '.... 125. Strong Soil (. '. '. 127. Clay ., '. 135. Portland Stone .' \ 1571. Clay and Stones ' 160. Gray Granite. .' —. —'. '171 Marble 172. Porpliyny 179. Crown Glass .' -. I8O4. Iron 450J. Copper..: 7 486$. (Steel — ?■ 490. Pure Silver *. .'. 6541 Lead * 7091. Pure Gold 1203 f. Platina ?..., 12181- ERRATA. 285 ERRATA. On page 11. "Arithmas" should be arithmos. " ,u 18. In the second line "multiplier" should be multiplicand. " " 20. Ans. to Ex. 14 should be 18688000. " " 34. In Ex. 16, the mark D is left out in the Ans. u 36- Ex. 3, should read .divide by 8 instead of 9. " 55. Operation at middle of page should have the ,sign*-j- instead of X- " " 56. In operation near top of the page, and in the note at the bottdm wrong sign again. " u 63. Ex. 4, should read divide 7f by 4 « " 72. Ex.2, should read divide 11.97 by 5.7. " " 83. Ex. 5, should read, when corn is 50 cents per bushel the distiller sells 65 gallons of whisky for a certain sum ; how many gal¬ lons should he sell for the same sum-when corn is 80 cents per bushel? Ans. 40|gal. " " 118. In the operation 12.50 should be 1250. " " 139. In Ex. 8, 53/^ should be 53Ty2. ERRATA. 1 55. First Ans. on the page should have a deci¬ mal point at the left of it. 158. Fourth line from the bottom where it reads " and squared equals 36," it should read and'Q squared equals 3G. 191. At the top it should he, to find this average- 208. Prob. 3, should read, the four sides of a' piece of land in the form of a trapezium are as follows : the northern 25 rods, west¬ ern 65, southern 131, eastern G1J. 228. Last word on the page should be axes in¬ stead of "axis." THE END. o o ftl AND NEW ARRANGEMENTS. The undersigned is increasing his stock of Books and Stationery by every Steamer from the North, and is also in receipt of the New Publications of the principal publishing houses as soon as issued from the press. He offers for sale at low prices: One hundred reams Folio Post, Cap, Letter, Note and Commercial Note Paper. THE best qualities of Tissue and Drawing Paper, Music Pa¬ per, Bristol Board, Bill Paper, Blotting Paper, Envel¬ opes of all sizes and qualities; also, the CHEAPEST Blank Books of all qualities, together with all the sta¬ ple articles of Stationery usually found in a STORE such as Counting-house, School, Fancy and Pocket Inkstands, Portfolios, Back-gammon ana Chess Boards, Chess Men, Portmonaies, Pocket Books, Single and Double Slates, Porcelain Slates, Erasing Knives, Pa¬ per Folders, Pen racks, Steel and Quill Pens, Mucil¬ age, all the best brands of Writing, Copying and In¬ delible Inks, Visiting and Playing Cards—all warran- tedfgood, and at as low prices as can be Found IN THE CITY. Also, Bill Files, India Rubber, India Rubber Rings, Quills, Sealing Wax—Red, Black, Fancy and White, India Ink, Drawing Slates, Crayon, Crayon Holders, Black Sand, Sand Boxes, Bill Head Boxes, Calendars, &c. All of,which will be sold at the low¬ est prices by WILLIAM KAY, Agent, Whitehall St., Atlanta, Ga. Sgr* Please remember the location, nearest store to the General Passenger Depot, and all the Hotels. v&*