B 0: 01 1 ^ 21 9[ 4 I 3 1 TUE POCKET REFERENCE ED ^S A WOEK OF FEACTICAL UTILITY TEVCFTEK, THE MAN OF BUSINESS, AN1> HX3IAX10AL flTUBJSKT. By I. M. WILCOXSON; HUKK SfRlIfGS, HART COrNTT, KBNT* T L0UI8VII.liB, KY: MORTO.V & GRISWOt.D, PrYtE'S 1858. ARITHMETICAL VADE MECUM; POCKET REFERENCE- DieiGNED AS A WORK OF PRACTICAL UTILITY, FOl THE TEACHER, THE MAN OF BUSINESS, AND THE MATHEMATICAL STUDENT. By I. N. WILCOXSON, THEEE SPRINGS, HABT COUNTY, XT. LOUISVILLE : A. F. OOX, STBBBJEOTTPER AND rSINria. 1856, Entered according to act of Congress, in the yeai 1856, by I. N. WILCOXEN, In the Clerk's Office for the District of Kentucky sRie URU TO REV. J. P. MURREL, i rBINCIPAL OF CAUDEN SEMIKAET, IN TOKEN OF OCR HTOn APPRECIATION OF HIS LIFE AND CHARACTBB AS A TEACHER OF YOUTH; AND TO ED. PORTER THOMPSON, MT YOU.VQ FRIEND AND FELLO^-STCDENT, THIS LITTLE WORK IS BESrECTFULLY AND FRATERNALLY INSCRIBED, BY THE AUTHOR, PREFACE. It may be thought that the great number of Arithmetical works, now before the public, renders it unnecessary to multiply them ; but in adding another to the list, we deem it sufficient apology to remark, that the importance of mathematical science renders it justifiable in any one to present to the public every improvement that will enable calculators to arrive, with greater facility, from premises to conclusions. The increasing interest manifested in the cause of Education, and the labor of Rev. J. P. Murrel, Principal of Camden Seminary, together with that of the author of these pages, for many years, in abridging and simplifying the common methods of calculation, are sufficient inducements for the publication of the result of those efforts. The method of statement given, is better adapted to all questions that may arise, in business transactions, than any heretofore offered. We have arranged the common rules of Practical Arithmetic — Simple and Compound Proportion, Practice, Interest, Discount, Barter, Percentage, and those for all kinds of Mensura- tion — under the general head of Cause ana Effect ; and it will be perceived that the method ▼1 PREFACE of reasoning cmploj-ed, adapts itself to the most common capacity, and is applicable to the solution of all c^uestions that may arise under any of those individual heads. As the work is designed for the teacher, the business man, and the advanced mathematical student, and as our object has been to illustrate the nature and principles of calculation, we have given only a limited number of problems ; deeming it unnecessary to give more than a sufficiency to illustrate the rules, as competent teachers can readily give examples under every head, and to the business man questions will naturally arise. It will be observed that we have mostly used abstract numbers ; but as it is presumed that those who use the work will have previously become acquainted with both the Simple and Compound rules of Addition, Subtraction, Mul- tiplication, and Division, they can adapt the rules here given to the solution of all questions involving different denominations. Indulging the fond hope that this little work may prove itself worthy the consideration of those for whom it is designed, we now submit it to a generous public. I. N. WILCOXSON. Ifnrt County, Kentucky, August, 1855. INDEX. Dbdication— .... . Poge I Preface— . . . ... i Explanatory Articles — . . . 9 Explanation of Signs, . . . . 10 Cancellation — .... 10 Vulgar Fractions — 12 Examples of Different Kinds of Fractions, 12 Reduction of Vulgar Fractions, . 12 Addition of Vulgar Fractions, . 13 Subtraction of Vulgar Fractions, . . 14 Multiplication of Vulgar Fractions, 14 Division of Vulgar Fractions, . 14 Reduction of Complex Fractions, . U pEciMAL Fractions — . 16 Numeration Table, 18 Addition and Subtraction of Decimals, . 17 Multiplication of Decimals, 17 Division of Decimals, . 17 Reduction of Decimals, IS Factor Tables— .... . 18 Cause and Effect— 20 Simple Proportion, . 20 Definition of Terms, 21 Compound Proportion, . . . . 23 Practice, .... 26 Interest, .... 26 Interest Table, .... 27 Miscellaneous Examples, . . . 29 Percentage, 29 Barter, ..... 32 Discount, . . . . . 33 Measurement . . . , . Zi ym INDEX. Mensuration Table, . Pagi 3S Cord Wood, 35 Land Measure, 36 Carpeting, . 37 Crib and Box Measure, zr House Covering, 39 Brick Work, 41 Log and Plank Measure, . 42 Promiscuous Examples, 44 Mknscbation of Superfices— 45 Mensuration op Solids- 50 Square Root- > 52 Cube Boot— 63 Table of Squares and Cubes, . 55 Application of Square and Cube Root, with Mia- cellaneocs Matter— , , &7 FalliDg Bodies, • • • M EXPLANATORY ARTICLES. Article 1. In placing numbers on the line, place the individual numbers directly under each other. If fractions, place the numerator where you would place it were it a whole number, with its denominator on the opposite side. The solu- tion of fractional sums in Multiplication and Division will then be nothing more than those in whole numbers. Art. 2. Observe that we cannot add, sub- tract, multiply, or divide numbers of different denominations, without first reducing them to the same denomination ; also, that no ratio can exist between different denominations, but the abstract numbers rej)resenting them may have a ratio. Art. 3. To render verification easy, observe that subtraction is the converse of addition, and division of multiplication. Art. 4. A unit is the basis of all work ; the point from whence the analytical student reckons with facility ; for the value of a fraction is de- creased when multiplied by itself, while any number greater than one is increased. Unity is the result of any fraction divided by itself ; and, 1-^-1=1. 1x1=1. Art. 5. Cancel as much as possible in every Instance, as it will aflord every opportunity of 10 ARITHMETICAL shortening the work, and in no case of length- ening it. Explanation of Signs. The sign = equality, thus, 100 cents=$l ♦* + addition, " 3-f 3=6 " — ■ subtraction, *' 6—3=3 " '* X multiplication, " 3x3=9 " " -1. division, *' 9-7-3=3 or thus, I =3 319 =3 " " :::: proportion, " 2:4::3:6 Read 2 is to 4, as 3 is to 6. ** " J square root, or radical, showing that the square root is required, thus, ^9=3 " ** vinculum, showing that all under it is taken as one number, thus, ^100— ] 9=9 The exponent (^ ), shows that the num- ber over which it is placed must be squared, thus, 6* xk36 CANCELLATION. Art. 6. First draw a perpendicular line, which in all instances separates the dividend from the divisor, or the factors of the dividend from those VAUE MECLM. n of the divisor. The dividend, or its factors, is always placed on the right ; the divisor, or its factors, on the left. Art. 7. After having placed the sum on the line, divide the continued product on the right, by the continued product on the left. Art. 8. Much mechanical labor may be saved, 1st, By canceling all equal numbei-s of noughts on opposite sides of the line. 2d, By canceling all equal numbers of figures. 3d, If the product of any numbers on one side will equal any number, or the product of any numbers, on the opposite side, by canceling all such numbers. 4th, By canceling all numbers that are divisi- ble one by the other, * placing the quotient on the side of the greater. 5tli, By continuing to cancel any numbers, (on opposite sides of the line), that are divisible by any supposed number, using the quotients. 6t]i, The operation may be completed by (Ar- ticle 7). ^^ote. The answer always comes on the right. If the divisor is greater than the dividend, the answer will be a fraction. y. B. The preceding should be committed to memory, and applied throughout the work. *The divisions on tho lino must all be made without* remainder, until the last. 12 ARITHMETICAL VULGAR FRACTIONS. Art. 9. A fraction is a part of a unit. The, bottom is the divisor, or denominator, showing into how many parts a unit is divided. The top is the dividend, or numerator, showing how many of such parts are taken. Examples of the different hinds of Fractions, i-, |, I, etc., Proper fractions. r> if h 6tc., Improper fractions. I, |, !■ Complex fractions. 3 8) 12^, 6|, etc., Mixed numbers. Art. 10. 1. Reduce 6§ to an improper fraction. 6x3-|-2=20, numerator, Ans.%'. 3, denominator. 2. Reduce 12\ to an improper fraction. IS. '2 Art. 11. 1. Reduce V *^ ^ mixed number. 75h-4=18|. 2. Reduce V ^^ ^ mixed number. Ans. 7^ Art. 12. 1. Required to reduce y/o to its low est terms. (See Article 8.) -^\^-YAns. 2. Reduce ^V ^^ ^^s lowest terms. Ans. §. YADIS UECUM. 18 Art. 13. To find the least common denomi- nator of fractions. Place to the left all the prime numbers of the denominators and the least number of prime factors that with the prime numbers, any two or more of them multiplied together will equal any of the com- posite denominators, their continued product will be the least common denominator, 1. Required to find the least common denomi- nator of \, f, \, ^, Divisors. •< 211 = 1 812=1 4il = l {'Quotients. 61=1 8l3=l 2. Reduce | Com. denom. 24 8' I' h To. to a common denom- Ans. 120. Addition of Vulgar Fractions. inator. Art. 14 Rule. Find the least common denominator. For the numerators divide the common denomina- tor by each particidar denominator, and multiply the quotient by its numerator. 1. Add I, h t\, ' -V, and |. together. 6: 6 5= 1 "50^ 2: 2 1= 1 30 5 :10 9= 1 54 ^new numerators. :12 11= 1 55 24 : 5 2= 1 60i213=3H Ans, 14 ARITHMETICAL 2. Add I, L, I, I and | togetlier. Ans. 3||. Art. 15. Subtraction of Vulgar Fractions. Hule. Prepare the fraction as in Addition, and take the difference of the numerators. 1. From §, take y^- 4: 4 3= 1,9 :12 5= 3|5 12|4=i. Ans. 2. From 4, take fV. Ans, \. Art. 16. Multiplication of Vulgar Fractions. Rule. Place all the numerators on the right. QTid the denominators opposite. 1. Multiply ^ of 1 of 1 of f of 1 of I, by 14. A? m ?l^ m 1 1 2 Ans. 2. Multiply i of yV of VI of 1, by y\ of f of 12. Ans.\. Note. Of, between fractions always denotes multiplication. Art. 17. Division of Vulgar Fractions. Rule. Place the numerators of the dividend on the right, and the numerators of the divisor on the left. (See Art. 1.) VADE MECDM. 15 1. Divide ^ of f of Jy of H» ^7 /a of \ of j\. t\x% 1|4 I 4 -4n5. 2. Divide \ of ^i of | of ^f , by i- of y\. ^rw. 4. JVb/e. To understand multiplication and divi- iion of fractions, it is sufficient to observe Art, 9, 6 and 8. Art. 18. Reduction of Complex Fractions. Rule. Consider the numerator the dividend and the denominator the divisor, and proceed as in Division of Fractions. numerator. 1 - f denominator. 4 5 f of -?- 2. Reduce This is the same as |-r-|. 1 oi — y to a mixed number. ^2 12 1 XX 4 3 X^ xp 11 33 8^ Ant. 16 ARITHMETICAL 3. Kfiduce t\ of | of y|. Ans. 2\, DECIMAL FRACTIONS. Art. 19. Decimal Fractions are managed like whole numbers ; and as their denominators art always 10, 100, etc., as y%-, yV^, etc., we express them by placing a decimal point to the left of the numerator ; thus, .2, .25, etc. Noughts to the left decrease their value in a tenfold ratio. Numeration Table, ii el i .11 il 4444444444 • 444444444 Ascending, Descending. Kote. This table shows ''Jiat the value of figures is determined by their distance from th« decimal point. VADE MECDM. 17 Art. 20. Addition and subtractio i of Decimals. Rule. Place the decimal points directly under each other, and add, or subtract, as in whole numbers. Add .4, .06, 1.12, 10.002, together. .4 .06 1.12 10.002 11.582 Ans. From 3.856, take 2.412. 3.856 2.412 1.444 Ans. Art. 21. Multiplication of Decimals. Hule. Multiply as in whole numbers, and point off in the product as many figures for decimals as there are decimal places in both factors. Note. Point off from the right, and if thera are not enough figures in the product to supply the decimal place, prefix noughts. 1. Multiply 2.34 by .12. 2.34 .12 .2808 2. Multiply .275 by .25. Ans. .06875. Art. 22. Division of Decimals. Rule. Make an equal number of decimal placet in both factors, by annexing ciphers to eiiher, and divide as in whole numbers. 2 18 ARITHMETICAL Note. The answer will be in whole nnmbers. If decimals are required, annex cijihers to tl^ remainder, and continue the division. Divide 2.3421 by 21.1. 21.100012.3421 10.111 Ans. Art. 23. 1. What decimal is equivalent to f ? 4|3.00 1 .75 Ans. 2. What decimal is equivalent to |-? Ans. .125. Art. 24. 1. Reduce .75 to a vulgar fraction. 4 i'.pj31^ 3 I Ans. 2. Heduce .125 to a vulgar fraction. A3iu,\. FACTOR TABLES. Federal Money. 10 mills (•.) make 1 cent. marked et. 100 ceots 1 dollar. Avoirdupoin Weijjht. $ 16 drama {dr.) make 1 ounce, marked oz. J 6 ounces " 1 pound, " lb. 25 pounds " 1 quarter, ti qr. 4 quarters " 1 buudrud weight, " ewU 20 hundred « 1 ton, <« r. VADE MECCM. 19 Ling Meaaure, 12 inches (in.) maku 1 foot, marked ft. 3 feet " 1 yard, " yd. 5^ yards " 1 pole or perch, " p. 40 poles " 1 furlong, " fur. 8 furlongs « 1 mile, " J/. Land, or S({uare Measure. 144 square inche3( « J. ih) make 1 square foot, marked 'q.ft. 9 square foct " 1 square yard, " aq. n'l. 'H)]4 square yards " 1 square pole, " p. 40 square y in cause and efect, the actual numbers given need not be used, provided we use their proportionate numbers. If 48 lbs. of pork cost 144 cts., what will 115 lbs. cost ? Cause. Effect. Cause. Effect. 48 : 144 1 : 3 wv&m This is simply another fonn of Art. 8. It ia readily seen that 1 : 3, is as 48 : 144. Compound Proportion. Art. 31. In this system of statement, the philosophical idea is the only sure guide ; hence, in stating a question, the mind should rest on denominations only ; but after it is stated, we should look on the terms as abstract numbers. JV. B. \Yhen a correct statement is made, there will be the same number of elements, or factors, under the same letters, or in similar terms : as in the following example. If 2 men in 4 days can mow 5 acres of grass by working 10 hours per day, how many acres will 6 men mow in 2 days by working* 12 hours per day ? An^. 9 A. Cause. Effect. Cause. Effect. Men, 2 : Acres, b::Men, 6 : Acres [] Days, 4 : Days, 2 : Ifours, 10 : Hours, 12 : 24 ARITHMETICAL 4 \x^ ^ • $ M^ 3 I 9 Ans. Art. 32. There are many questions that appeaj to be in simple proportion, that are really in compound proportion ; the reason is, because of one terra in each couplet being the same. Example — If 4 men build a wall in 8 days, 6ow long will it require 6 men to build it ? Cause. Effect. Cause. Effect. ^ J^ 2 4 : 1 : : 6 : 1 ^ ^ 8 8 : [] ' 3 1 16=5^ Am, Note, Here one term in each couplet is one wall. Art. 33. Similar questions sometimes give rise to the apparent view of more requiring less. Money must be compounded with time before it can produce interest, as more money will require less time to produce the same interest ; but this is all apparent, for there is no such thing as more cause producing less effect. Art. 34. This difficulty arises from not being able to readily determine which is cause, and which is effect; but this (the only difficulty to be met with in this system of statement), is readily overcome, when we consider that all action of any nature must be cause, and that which is accomplished, or follows such action, must bo ^ect. VADE MECUM. 2* If 5 horses in 8 days consume 80 bushels of oats, how many bushels will 3 -gQ -j^q horses consume in 15 days ? ^i g Cause. Effect. Cause. Effect. Jf^ 3 5 : 80 : : 3 : [] , 8 : 15 : 1 90 Am, Note. Here it is evident that the action of the horses, multiplied by the days, in both couplets, must express the cause, and the consumption of the oats is the effect. Art. 35. There are many questions, however^ where it is indifferent which is taken for cause, and which for effect ; only, observe when one thing is taken for cause in the first couplet, a similar one must be taken for cause in the second couplet. If the transportation of 4 cwt. 12 miles cost 810, how far may 6 cwt. be carried for 815. Ans. 12 M. Cause. Effect. Cause. Effect. 10 : 4 : : 15 : 6. 12:: []. X0 12 12^. Or thus Or thus Cktuse. Effect. Cause. Effect. 4 : 10 : : 6 : 15. 12 : [] : Cause. Cause. Effect. Effect. 4 : 6 : : 10 : 15. 12: [] :: 9& AHITHMETIOAL Or thus : Effect. Effect. Cause. Cause, 15 : 10 : : 6 : 4. [] : 12. The same terms, multiplied together in each of the different statements, show that this method 16 strictly scientific. 1. If the wages of 6 men, 14 days, be $84, what will be the wages of 9 men for 16 days ? Ans. 8144. 2. If I lend 8400 to a friend for 16 months, how long ought he to lend me 81600 to return the favor ? (See Art. 32.) Ans. 4 mos. 3. If 2L yds. of cloth, 2| yds. wide, cost 83.35, how many yds., that is 1^ yds. wide, can I have for 8134.00 ? Ans. 160 yds. 4. If 11 men in 7 days, working 13 hours per day, dig a ditch that is 37| ft. long, 2| ft. wide, 3^ ft. deep, in how many days can 5l men, working 14 hours per day, dig another that is 18| ft. long, 14f ft. wide, and 10^ ft. deep? Ans. 120 J days. Kote. In stating all questions in Cause and Effect, see Art. 25. Practice. The preceding principles of Cause and Effect will apply to the solution of all questions arising in Practice, which is nothing more than Simple Proportion, having 1 for the first term. Interest. We omit the definitions of terms ■'^^^^ *- VADE MECUM. 2J treating of Interest, they being so common that the learner is supposed to be familiar with them. Art. 36. The system of Cause and Effect is very extensive and easy in its application. It covers every case that can arise under Interest. To find the interest, principal, time, or rate per cent., and thus dispense with five or six special rules, as found in almost every Arithmetic, w« will use the following Interest Table. CauHK. Effect. Cause. Effect. 100 : Ptate pr. ct. : : Principal : Interest, lyear:* Time : .^^" Make the blank lohere the table designates (he term, or factor, you wish tojind. "What is the interest of 8200, for 3 years, at 6 per cent.? ^2^ Cause. Effect. Cause. Eff'Ct. ^VPi 3 100 : 6 : : 200 : "[] ^ 6 1 : 3 : I IS36 Ans. Art. 37. What principal at interest for 24 months, at 6 per cent., will , -^ o gain84S? ^'^^ ^ Cause. Effect. Cause. Effect. r^\ 100 : 6 :: f] : 48 — 12 : 24 : |8400 An*. * Or the factors of a year. If the time be months, 12 ; if days, 12 and 30 ; if weeke, 52. etc llOO 12 ARITHMETICAL Art. 38. At what rate per cent, will $200, in 450 days, gain 815 interest ? Cause Eject. jj^0 Cause. Eject, 100 12 30 [] 200 450 15 X$ 6 6 pr. ct. An$. Art. 39. In wliat time will $160 gain $2 interest at 3 per cent ? Cause. Effect. Cause. J Wect. 100 : 3 :: 160 : 2 ]2 : [] : J0P 5 5 mos.Ans. Note. Many abridged rules miglit be given for tbe solution of interest questions ; we shall, however, give but few, as we are satisfied that those who make themselves acquainted with the preceding general principles, will be able to make their own abridgments. Art. 40. 1. When the time is months, and rate per cent. 6, to find the interest, multiply the principal by half the number of months. 2. When days, divide them by 60, and multi- ply the quotient by the principal. 3. When the time is months, and the rate per cent. 4, multiply the principal by ^ the number of months. 4. When the time is months, and the rate per cent. 3, multiply the principal by \ the number df months. VADE MECUM. 29 5. To find tlie interest of any jrincipal, for any time, at any rate per cent : Make a dividend of the principal, time, and rate per cent. If the time he months, the divisor is 12 ; if days, 12 and 30, etc. Note. When the time is given in different denominations — as months, days, etc. — it must first be reduced to the lowest denomination men- tioned, then placed on the line. Miscellaneous Examples. 1. What is the interest of $100, for 3 years, at 5 per cent. ? Ans. 815.00. 2. What is the interest of 821, for 1 year and 4 months, at 3 per cent. ? Ans. 84 cts. 3. What is the interest of 850, for 18 months, at 4 per cent. ? Ans. 83.00. 4. What is the interest of 884, for 6 months and 20 days, at 9 per cent. ? Ans. 84.20. 5.' What is the interest of 875, for 60 days, at 8 per cent. ? Ans. 81.00. 6. What is the interest of 846, for 90 days, at 6 per cent. ? Ans. 69 cts. Note. To save space in giving other problems, use the interest in the preceding, and find the priiLcipal, time, and rate per cent., alternately, by the Interest Table (Art. 36). Percentage. Art. 41. To find the amount for which an article must be sold, to gain or lose any given rate per cent. 30 ARITHMETICAL State thus: As 100 is to 100 with the gain per cent, added, or loss per cent, subtracted, so is the prime cost to the required price. 1. A merchant paid 44 cents per yard for cloth, for what must he sell it to gain 25 per cent. ? Cause. Effect. Cause. Effect. ^ ^W\ ^^ \\ 100 : 1004-25:: 44 : [] 1^. 55 cts. 2. Paid ^80 for a horse, for what must he be sold to gain 50 per cent. ? Ans. ^120. 3. Paid $90 for a horse, how must he be sold to lose 33 i- per cent. ? ? i^p 30 Caxish. Effect. Cause. Effect. I0pl200 100 : 100—33^ :: 90 : []. | I $60 Ans, 4. Paid 6100 for sheep, how must they be sold to lose 20 per cent. ? Ans. 880. Art. 42. Having the cost and selling price given to find the gain or loss per cent. State thus : 100 is to the required gain or loss per cent, as the prime cost to the difference between the prime cost and selling price. 1 . A merchant bought cloth at 48 cts. per yd., and sold it at 60 cts. what was the gain per cent. ^ Cause. Effect. Caute. Effect. 100 [] :: 48 : 60—48 JP0 25 .4.25 cts. VADE MEOUM. SI 2. Had he paid 60 cts., and gold it for 48 eta., what would have been the loss per cent. ? ^^1 X$ 2 Gaute. Ejfect. Caute. Effect. \ 100 : [] :; 60 : 60—48 U.20ct8. 3. A farmer paid 875 for a horse, and 8150 for a chaise; he sold the horse for 8100, and the chaise for 8125. What per cent, did he gain on the horse, and Jose on the chaise ? A \^^}^ per cent, gained. ( 16|i?cr cent, lost, Art. 43. Having the selling price of an article, and the rate per cent., gained or lost, given, to find the cost. State thus : 100 t* to 100 with the gain per cent, added, or loss per cent, subtracted, as the required cost to the the selling price. 1. Having sold a watch for 814, I thereby lost 30 per cent., what did it cost me ? Oaute, Effect. Cause. Effect. '^"jlO0 100 : 100—30 ;: [] : 14. 120 A ns. 2. If a farm be sold for 8220, and 10 per ct. is gained, what did it cost ? 100 m 100 : 100+10 :: [] 220. & Ans. Caute. Effect. Caute Effect. V^^ S2 ARITHMETICAL 3. I sold a horse for $40, and by so doing lost 20 per cent. ; whereas, I ought, in trading, to have cleared 30 per cent. How much was he sold ior unuer iiis real value ? V)0~20: 100 :: 40 : n=50 :[]: lO'J :100-f30:: 50 : []=65— 40=.^25^»# Or thus : J:pp,X?0 65 I ^j3 65—40=825 Am. Barter, Art. 44. (See Article 32). The term under- stood in Barter is one amount. *^ 1. How many bushels of apples must I have, at 25 cts. pr. bu., for 35 yds. of calico, at 20 cts. per yard ? ^ . ^ Cause. Effect, Cause, Effect. ^ ?^ ?^ J 1 ; : 35 : 1 W^ II ; 20 I 28 In.. 2. If I barter 12 bu. of flaxseed, at 65 cts. pr. bu., for cloth, at 60 cts. pr. yd. How many yards must I have ? Cause. Effect. Cause. Effect. P W 12 : 1 :: [] :: 1 65 • 60 :: 13 ^n# ^^ 13 VADE MECUM. 33 DiscouTit. Art. 45. State thus: As 100 is to the amount of 100 /or the given time, at the given rate per cent., 90 is the required present worth, to the given debt. Xote. Subtracting the present worth from the amount, will give the discount, 1, What is the discount of $436, for 18 months, at 6 per cent. ? Cause. Effect. Cause. Effect. ^^^ ^ 100 : 109 :: [] : 436. r; ^ ^^^^ '-■' \Pres. worth 8400 436—400=36 Ans. The following process is preferable for its brevity ; Make a dividend of the 2^rincipal and time; multiply one year (or its equivalent in months or days) by 100 ; divide by the rate per cent., and add the time (of the same denomination) to the quotient, for a divisor, N. B. The result will be the discount in the lame denomination of the sum given. To find the present worth, subtract the di§- count from the sum given. Take the preceding example : 12xlOO-^6=200. 200+18=218, divisor. PPM^ 2 18 I $36 discount, 3 84' ARITHMETICAL To prove discount. The interest of the present worth must equal the discount of the sum, for the same tm^. at- the same rate per cent. 836 interest=^'^Q discount. 2. What is the discount of ^80, for 200 days, at 12 per cent. ? 360x100-^12=3000. 3000+200=3200 c/^V'r. X^ ?^0P 80 5 $5 Ans. 3. What is the discount of $48, for 4 years, at 5 per cent.? Ans. $8. 4. What is the discount of 8218, for 9 mos., at 12 per cent. ? Ans. 818. 5. What is the discount of 81140, for 120 days, at 4 per cent. ? Ans. 815. 6. What is the difference between the interest of 8160, for 400 days, at 6 per cent, and the discount of the same sum, for the same time, at the same per cent.? Ans. 66| cts. Measurement. Art. 46. For all measurement, use the follow- ing table in stating — or rule, if Cause and EffetU Ise read alternately : VADE MECUM. 59r Mensuration Table. Cam St. KjftcL. Cause. Eject, Fioion! Of tue unit of measure. Unit of measure. Factors of the :: thing to be measured. Number of units.* S^ Make the blank in the term where the tablt designates the term or factor sought. Cord Wood. Art. 47. 1. Ilowmany cords are there in a pile of wood 80 feet long, 4 feet wide, ,8 feet deep ? 8 $0 20 luse. Eject. Cause. Eject 8 : 1 : : 80 []• 4 4 4 8 I 20 Ans. 2. How mncli will a pile of wood cost, that is 48 ft. long, 2\ ft. wide, and 16 ft. deep, at $1 per cord ? Ans. §15.00 3. How long must a pile of wood be to con- tain 13 cords, that is 6^ ft. wide, and 4 ft. deep ? X^\ 2 Cause. 8 4 4 Eject. 1 Cause. R 4 Eject. 13. 8 4 164 Ans. *The number of units is the thing, or answer, sought ; but If this be givcB some factor is sought. 86 ARITHMtllCATi 4. How wide must a pile of wood be, that ia 24 ft. long, and 16 ft. deep, to contain 12 cords ? Jins. 4 ft. Land Measure, Art. 48. 1. How many acres of land in a field 80 rods square? ^ ^^ 20 Cause. Eject. Cause. Effect. ^0 g0 2 4 : 1 :: 80 : []. . 40 : 80 : A^ Ans. 2. What is the area of a rectangular field 60 rods long, and 121 yards wide ? Cause. 4 40 5^ Effect. Cause. Effect. ^ ^p 1 : : 60 : []. XX 121 : 4 ^0 3 xp 11 ^ 33 Ans. S\, 3. How wide must a rectangular lot be that ie 24 rods long, to contain 3 acres ? 3 $fi Cause. Effect. Cause. Effect 4 : 1 : : 24 : 3. 40 : [] : 20 Ans. Nate. This principle is applicable to the meas- urement of lands of all shapes, as given under Mensuration. VADE MECCM. Carpeting. Art. 49. 1. How many yards of carpeting, t^at is ^ of a yard wide, will be required to cover a floor, 27 feet long, and 13 feet wide? ^P Cause. I^ect Cause. Effect. ^|13 1 : 1 : : 27 : [J- 4 5 13 : — 9 152 yds. ^. Xote. The 9 under theT^r*-^ cause is to reduce it to square feet, to be of the same denomination of the second cause. 2. How wide must a floor be, that is 18 feet long, to require 12 square yards to cover it ? Cause. Effect. Cause. Effect. 1 : 1 : : 18 : 12 1 [] : 9 X^ \Xf 6 6ft.^. 8. What will it cost to carpet a room that is 36 feet long, 10 feet wide — carpeting 1^ yards wide, and worth 37^ cents per yard ? Cause. E^ect. Cause. Effect. # j'p 1 : 1 : : 36 : tJ- ^ 75 1\ : 10 4 9 37i §12.00 A. Crib and Box Measure. Art. 50. 1. How many barrels of corn will a crib bold, that is 50 ft. long, 9 ft. wide, and 3 ft. 4 in. deep ? B8 i •; ^k : 1 : :50 : []. V5 ^ 3 5 9 3a4in : 10 ]20bbls. .1. 2. How many bushels of corn will a crib hold, that is 18 ft. 9 in. long, 8 ft. high, 7 ft. 6 in. wide ? Cause. 2k Effect. 1 Cause. 8 18ft9in 7ft6in Effect. [] ^12 150 bu. A. 3. How high must a crib be, to contain dGO bushels of corn, when it is 18 feet long, and 6 feet 3 inches wide ? Cause. Effect. Cause. Effect. ^^K^^ ^ . 2k : 1 :: 18 : 860 .mWn^ 6ft3in [] : ^ m 2 I8ft.^«« 4. How many bushels of wheat in a box, 10 feet long, 3 feet wile, and 2 feet 8 inches high ? Cau$e. Effect. 1 Cause. 10 : 3 : 2ft8in Effect. ^0 2 jr^i32 l64 ^«#. VAJJE ME CUM. d9 5. How lonsr must a box be, that is 4 feet 2 inches wide, and 50 bushels ? feet 1 inch dee}), to contain 4|5 Cause. Effect. i Canae. Effect. 4 ft. 2 in. : 50 2 ft. 1 in. : [] ^12 on 36 I'jft.A. 6. How many panes of glass in a box of 50 feet, 8 inches by 10 inches ? Cause. 10 ffect. Cause. Effect. 1 : : 50 1 2 I 2 '' n ip X^ 3 1^ 6 5p 90^. 7. How many feet of glass in a box, that con- tains 120 panes, 10 in. wide, and 12 in. long? ^^ 10 Cause. Effect. Cause. Effect. 10 : 1 ■■■■ [] : 120. 12 1 2 1 2 X^ jr^ 100 ^n5. y yote. The 12s under the second cause in the preceding examples are to reduce them to square inches, to be of the same denomination of the Jirsi cause. House Covering. Art. 51. 1. How many shingles will be re- quired tD cover a house, that is 24 ft. long, and 15 ft. wide? 40 ARITHMETCAL. ^ 12 Cau»e. 4 : 6 : Effect. 1 Oavse. Effect. J 12 ^ ? 12 n 5 4 2880 An^ Note* It is customary to allow the shingles to be 4 inches wide, and to show 6 inches, and the rafters to be | the width of the house, both making \. When this is the case, the process may be shortened by Multiplying the product of the length and width of the house, by 8. 2. How many shingles will be required to cover a building, that is 30 feet long, and 25 feet wide ? Ans. 6000. Art. 52. 1. How many boards will be re- quired to cover a house 36 ft. long, 24 ft. wide ; tiie boards 6 in. wide, and to show 18 inches? 'ause. 6 18 Effect. : 1 : : > Cause. 36 : 24 : Effect. ^$ [] Xf 2 X^ 4 24 4 1536 Ans 2. How much must I pay for boards 6 inches wide, and to show 15 inches, at 86 per 1000, to cover a house 24 feet long, and 20 feet wide ? YADA MEC Cause. Effect. Cause. Effect, 6 : 1 : : 24 : []. 15: 20 : lOOO : 6 : : 1 : 12 ja 'CM. 5 X^ 25JPPP X% 4 X% 4 24 2p 4 125 768 41 86 14c. 4m Brklc Work. Art. 53. 1. How many bricks, 8 inclies long, 4 inches wide, will be required to pave a walk 3 feet 4 inclies wide, and ^ of a mile long ? ^\X^ k \x^ X$\4p 10 1 3 11 40 Caiuc, Effect. Cause. Effect, 8 : 1 : ' \ : []• 4 3ft4in 12 'I 4 8 13200^. 2. How many bricks will be required to build tbe walls of a house 20 feet long, 15 feet wide, 16 feet high, and 8 inches thick, allowing \ for mortar, the brick to be 8 inches long, 4 inches wide, and 2^ inches thick ? 42 ARITHMETICAL Cause. Effecx. Cause. 8 . 1 :: 35 4 : 2 2| 16 8 5 6 I 2 12 Effect. []• $ 12 85 2 16 13440^. ^Vb^c. Adding tlie length and width together, and doubling the sum, gives one straight wall. Multiply that by the height and thickness and by f, (making a deduction of \ for mortar). N. B. Deductions must be made for windows and doors. Log and Plank Measure. Art. 54. 1. How many solid feet in a log, 21 inches in diameter, and 16 feet lorn Cans 12 12 12 Effect. Cause. Effect. 1 :: 21 : []. 21 16 1 2 ? x^ px^ 2 X^ X$ ^X7 11 x^ ^ 2 77 38L^ Note. To find the solid contents of a log, we first find the area of the end by Art. 68, and multiply by the length. VADE MECUM. 43 2. How many square feet of plank, for ceiling, flooring, etc. (1 inch thick), in a log, 24 inches in diameter, 20 feet long, allowing \ for saw- calf? 7ause. Eject. Cause. 12 : t— 1 24 12 : 24 H 20 2 1 a Xote. Dividin s:by2 Effect. ^124 ?0 4 384 ^n.9. ws away in getting the area of the end. (See Art. 61 ) 3. How many square feet of plank, 1 inch ihick, in a log 30 inches in diameter and 12 feet long, allowing ^ for saw-cut ? Ans. 360. 4. How many square feet of plank. 1 inch thick, in a log 48 inches in diameter, and 10 feet long, allowing ^ for the saw-cut ? Ans. 768. 5. How many square feet of sheeting plank, \ of an inch thick (including saw-cut), in a log, 12 feet long, and 7 inches in diameter ? Cause. 12 1 2 3 4 Effect. 1 Cause 12 7 7 vv Effect. [] 3U ? xm^ Xfl 11 2 X^ _ 154 51i- Ans. 6. To cut 7|- square feet off of a plank 6 inches wide, how many feet of it€ length must be taken ? u Cause, 12 I. ARITHMETICAL EJecL 1 Cause, I Eject, 1\, m ^15 !l5 Ans. Promiscuous Examples. 1. How many acres are tliere in a round field, 56 rods in diameter ? Cause, 4 40 Effect. 1 Cause. r;6 5(3 Effect. 4 m 7 xm 77 15| A, 2. If I send 12 bushels of wheat to mill, how many pounds of Hour will I get, allowing ^^ for toll, \ for bran, \ for shorts ; weight of wheat, 60 lbs. per bushel ? Cause, 12 8 6 Effect. Cause. : 11 : : 12 : : 7 : : 60 : : 5 : Effect. xi 11 4 ? 7 5 Jr? ^j3^p5 4 1925 48H^ 8. I wish to get 481i- lbs. of flour ; how many bushels of wheat must I send to mill, making tlie eame allowances as in the preceding example ? VADE 3IECCM. 45 Chuse. 12 8 6 Effect. 11 7 5 Cause. y 481V 12 ^>w. 4. How many barrels of corn in a field 240 hills long, by 160 wide, each hill to average 2 ears, 120 of which will make 1 bushel ? Cause. Effect. Cause. 120 : 1 : : IGO 5 ' ' : 240 2 Effect. []• m ^^p 2 jr0P 32 2 ^. 128 MENSUEATIOX OP SUPERFICES. Art. 55. To measure a square. ^ Hule. Multiply the side A B irdo A D. Let AB=12. AD=12. Then, 12X12=144. D 48 ARITHMETICAL Note. The area will be of the same denomi- nation that the sides are. Art. 56. To measure A B a rectangle. Rule. Multiply the length hy the width. D Let A B=40. A D=12. Then, 40x12=480. Art. 57. When the area and one side are given, to find the other. Rule. Divide the area [reduced to the same denomination as the side), hy the given side. In the last fi^ire, A B=40, and area=480, to find A D. 480-^0=12. Art. 58. To measure a ^ B rhombus. xT |\ \ ' \ Rzde. Multi'ply one side \ : \ hy the shortest distance be- \ l \ tween the sides. V U ^ LetAB=16. BE=12. Then, 16x12=192. Art. 59. To meas- ure a rhomboid. Rule. Multiply one D E C of the longer sides by the shortest distance between them. TADE JrECUM. 47 LetAB=40. A E=16. Then, 40x10=040. Art. 60. To measure a trapezoid. Eule. Multiijhj the half sum of the parallel sides by the shortest dis- tance between them. iT LetAB = 8 DC = 18. B E=10. Then, 18+8-MixlO=130. Art. 61. Tlie diagonal* of a square given to find the area. Ride. Multiply the diagonal hy half itself. Let B D=SO. Then, 80x40=3200. Art. 62. To measure a right- angled triangle. Ride. Multiply the base by ik€ perp)endicidar , and tale half. LetAC=16. BC=19. 19x16h-2=152. 48 ARITHMETICAL Art. 68. To measure an acute or obtuse-angled triangle. Rule. Multiply the base hy a perpendicular line from the vertex ^ to the base, and take half. LetAC=60. BD==24. 60x24-^2=720. I^ote. Take the longest side for the base. Art. 64. To measure any regular polygon. Rule. Multiply one side by the perpendicular distance from the center ; take half the product, and multiply the quotient by the number of sides. Let AB=15, ab=20, (No. of side 15x20-^2x8=1200. S). Art. 65. To measure a trapezium. Ride. Draio a diag- onal line, and calculate the two triangles by Art. C 63, t/ieir sum will be the area of the trapezium. VADE ME CUM. 49 Art. 66. To measure any irregU' lar figure. Rule. First cut it into triangles hy drawing diagonal lines. Calcu- late [by Article 63) the several triangles, and tJieir sum will he the area of tht figure. Art. 67. To measure a circle. Given, the diameter of a circle, to find the circumference. A Rule, Multiply the diameter byS'. Let AB=14. 14 XV =44. A^ote. When the circumference is given, to find the diameter, multiply the circumference by /o, (the converse of the preceding.) Let A a B b=44, to find A B. 44x/2=14. Art. 68. To find the area of a circle. Rule. Multiply the circumference by the dianu- ter, and take one-fourth. In the preceding 44x1 4 .' 1 =154, Or, multiply the square of the diameter by \\. Art. 69. To find the diameter of a circle equal to a square whose side is given. Rule. Multiply the side by 1.128. Let the side=10. Then, 10x1.128=11.280. 50 ARITHMETICAL Art. 70. To find tlie area of an ellipse. Bule. Multiply the pro- a duct of the transverse and conjugate diameters (A B and a b) by \\. Let A B=14. a b=10. Art. 71. To find the area of a globe. Rule. Multiply the circumfer- ence by the diameter ; or, multiply the square of the diameter by \- . Art. 72. To find the area of a cylinder, or round body of equal largeness from end to end. Rule. Multiply the circumference by the length. Art. 73. To find the area of a right cone. Rx.le. Multiply the circumference of the base by the slant height, and take half. MENSURATION OF SOLIDS. Art. 74. To find the solidity of a right-angular solid. Rule. Multijyly the length, breadth, and depth together. VADE MECUM. 51 Let tlie lengtli=20, widtli=12, ]ieiglit=10. 20x12x10=2400. Art. 75. To find the solidity of a cylinder or prism. Rule. Multiply the area of the base (or end), by the length. Xote. Find the area of the base by previous rules, according to its shape. Art. 76. To find the solidity of a solid wedge. Rule. Multiply the area of the base by half the perpendicular length. Art. 77. To find the solidity of a "pyramid or cone. d Rule. Miiltiply the area of the hase by one-third of the altitude. Let a b=14, c d=24. m 2 4! 22 14 ^4 2 11232^^5. Art. 78. To find the solidity of the frustrum of a pyramid, or cone. 52 ARITHMETICAL Rule. Add the areas of the upper, the lowers and the middle bases tor/ether, (the middle base it found by multiplying the upper andlower bases together, and extracting the square root of the pro- duct,) and multiply the sum by one-third of the altitude^ Art. 79. To find the solidity of a globe. Rule. Multiply the area by one-sixth of the diameter ; or, multiply the cube of the diameter by \\. Art. 80. To find the solidity of a spherical segment. d Rule. Add the square of the height to three times the square of the semi- diameter of the base, and midtiply the sum by the height, and by ll. Let a b=10, c d=4. 5X5X3+4X4=91. 91x4XH=190H^»*. SQUARE ROOT. Art. 81. — Rule. 1. Separate the given nmnber into periods of two f mires each, commencing with Knits. VADE MECDM. 53 2. Find the greatest root in the left hand period, and place it on the right. Subtract its square from the Jirst period, and to the remainder bring down the next period ; and make a dividend of the remainder , with the first figure of the period annexed. 3. Double the root for a divisor, and set the quotient in the root, and to the right of the divisor. 4. Multiply and subtract as in division, and proceed, as before, until all the periods are brought down, when periods of noughts may be annexed to obtain decimals. 1. What is the square root of 55225 ? 5'52'25 (235=roo/ 43)1 52 129 ^^^^ 23 25 2. What is the square root of 15625 ? ^.125. 3. What is the square root of 5'35.92'25 ? Ans. 23.15. CUBE ROOT. Art. 82. — Rule. 1. Separate the given nurriber into periods of three figures each, commencing with units. 54 ARITHMETICAL 2. Find the greatest root in the left hand period^ 2^lace it on the right, and subtract its cube from the left hand period. 3. To the remainder bring down the next period^ and make a dividend of the remainder and the first figure of the period annexed. 4. Multiply the square of the root by 3, for a iefedive divisor. Place the quotient in the root, and its square to the right of said divisor, supply- ing the pilace of tens with a cypher, if the square be less than ten. 5. Complete the divisor by adding to it the pro- duct of the last figure in the root by the rest, and iy 30. 6. Multiply and subtract as in division, and bring doivn the next period. 7. Find the next defective divisor, by adding to the last complete divisor the number which com- pleted it, ivith twice the square of the last figure in the root, and proceed until all the periods are brought down, when decimals may be found by annexing periods of noughts. 1. What is the cube root of 9663597 ? 9'663'597(213 2x2 X 3=1201 1X2X30 = 00 Complete div 1261 IXl X 2= 2 Defect. rfiyV....132309 21X3X30=1890 Complete div. ..l^^^l^ 1663 \2^\=subtrahend. U2597 A02hTi=subtrah€nd, VADE MECUM. 03 2. What is the cube root of 1953125 ? Ans. 125. 3. What is the cube root of 33131834.347032 ? Ans. 321.18. We shall facilitate the rules of square and cube root by the following properties : 1. The product of two square numbers is a square number. 2. The quotient of two square numbers is a square. 3. The product of two cube numbers is a cube. 4. The quotient of two cube numbers is a cube. Table of Sqn ares and Cubes. Nos. Squ'rs. Cubos. 1|2 3| 4| 5 1|4 9 161 25 l|8!27|64il25 6 7| 8 36 49| 64 216|843 512 9 10 81 100 729! 1000 (See Art. 30.) In order to work with speed and alacrity, attention and tact are necessaiy on the part of the learner. 1. What is the side of a square piece of land containing 360 acres ? Instead of multiplying by 160, and extracting the square root, according to the common method, Ave lemove a nought from one to the other, making them both squares, whose roots are 60 and 4, or 6 and 40. Maltiply the roots together. 60 X 4=240 Ana. §6 ARITHMETICAL 2. What is tlie square root of the product of 32 and 128 ? i'^ n 2 716=4x16=64^725. Here it will be observed that both numbers are divided by the factor 16, and the root of the product of the quotients multij)lied by the factor. N^ote. We should examine the table closely, so as to recognize a square or cube as soon as seen. The principle of removing noughts, or using factors, as explained in square root, is also applicable to cube root. VADE MECUM. W APPLICATION OF SQUARE AND CUBE ROOT, WITH MISCELLANEOUS MATTER. To find the area of a soalene triangle. Rule. From the half sum of the three sides, take the three sides severally, and extract the square root of the product of the three remainders and half sum. 1. Let AB=5 Let B C=6 Let AC =7 5+6+7-4-2=9.^^ &— 5=4, 9—6=3, 9—7=2. 4 X 3 X 2X9=^216=14.696+ Ans, 2. What is the area of a scalene triangle, whose sides are 13, 14 and 15, respectively ? Ans. 84. 3. Given, the area of a circle 1296, to find th« side of a square equal in area. V1296=36. ^^. 36. Given, the area of a circle, to find the diameter. Hide. Divide the area by \\, and extract the square root of the quotient. 58 AKITHMiniCAL In any right an- jjjled triangle, (see adjoining figure, right-angled at C), when one leg and the liypotheniise are given, to find tlic otluir leg. Kule. Multi/>Jy the suvi by the dif- fercnce, and extract the square root of tlie j^roduct. When the two legs are given, to find the hypothennse. Rule. To double their product, add the square of their difference, and extract the square root of the sum. 4. What is the hase of a riglit-angled triangle, whose hypothcnnsc is 10, and f)erpendicular 6 ? yr(J+6"xT0-^=8 Ans. 5. Given, the jterpendicnlar 3, and base 4, to find the bypothenuse. ^ax4xi-'+l-=5 Ans. Scholium 1. Tiie product of the snm and dif- ference of two numbers, is equal to the difference of their sfjuares. Scholium 2. Double the product of two num- bers, plus the sipiaic of their difference, is equal to the sum of their squares. YAM mcasc. 6. A. and B. start from the same point ; A travels due east 24 miles, and B. due north 1>> miles ; how far are they apart ? Ans. SO M. 7. Two men start from the same point ; one travels 15 miles due south, the other 25 milea south-east ; how far are they apart ? Ans. 20 M. 8. What is the mean proportional between 4 and 9 ? ^9x4=6^^5. 9. What is the area of a parallelogram, whose diagonal is 50, and the sides are as 3 to 4 ? 32+42=25 : 502 .. 3>^4 . []=1200 Ans. 10. What are the sides of a parallelogram, ♦vhose area is 1200, and proportional of sides as 3 to 4 ? Ans. 30 and 40. 3X4: 1200 ::oa. n.^gr,,}. 3X4: 1200 ::-i» : LJ=' ^9UU=30. 1^<^0. ^1600=40. 11. Given, the dimensions of a plank, 20 feet long, 18 inches wide at one end, and 6 at the other, to find how long the smaller end must be to contain half the number of square inches in the plank ? Solution. Draw A B D C, and pro- duce A B and C D to meet in O, mak- ing the triangle A C 30 feet in length, con- Uining 22^ square feet ; the triangle B D, 10 60 ARITHMETICAL feet in length, containing 2^- square feet. Thei in tlie triangle A C, we only have to obtain a distance from O, sufficient to contain half th« plank A B D C, plus 2| feet. 22L— 2V==20, No. of sq. ft. in the plank. 20-^-2=10, half the No. of sq. ft. in the plank. 104-2L=12L. 221 : 12L :: 30^ : []=500. ^500=22.36— 10=12.36, length of small end. Note. Areas are to each other as the squarei of their similar sides. 12. ^1 Rev. J. P. MuRREL. How long a line, do you suppose, Would just the amount of land inclose, That one can see on level ground. Just from the cent'r, by turning round j The eye about six feet in height, And nothing to obstruct the sight ? ^ution : Ans. 18| M. 6-r'2«=V3+6=3IlX2=6DxV=18| circumf. 13. If an eye be elevated 24 feet, how far, on level ground, can an object be seen ? 24-2=12. ^24+12=6 M. Ans, 14. If an object is seen 9 miles, how high is the eye elevated ? Ans. 54 ft. 92x1=54 ft. 15 By Rev. J. P. Murrel. Which will inclose the most ground, A fence made square, or one made round, Two pannels to each rod of land. Ten rails in eachj we understand ; VADE MECUM. 61 And ev'ry rail in each suppose To just an acre of land inclose. The next thing is to toll exact, How many acres in each tract? J j 1024000, No. acres in sq*re. -^^^- I 804571f, '* *' circle. j-i^^=area of 1 rod square, wliich takes 80 rails to fence it. i j ,-1^:80:: 80 : [1=1024000. Otaterrmus. -j^^ : 11 ::1024000: [J=804571f. Note. One side of the square equals the diameter of the circle. 16. In what time will triple itself, at 5 per cent ? 3— IXlOO-f any sum of money Ans. 40 yrs. •5=40. k 17. In what time will any sum of money double itself, at 6 per cent ? Ans. 16| yrs. 2— lXl00-r-6=16|. 18. If I pay 850 apiece for 7 mules, and the ■arae amount for horses at $70 apiece, and sell them at an average of $60 apiece ; do I gain or lose ? Ans. $20 gain. 19. If I purchase a number of pears at 3 cts. apiece, and pay the same amount for oranges at 6 cents apiece, and sr-U them all i\i an average of 4 cents, what do I gain or lose ? Ans. $0.00. 62 ARITHMETICAL 2yj. A horse in the midst of a meadow, suppose, Made fast to a stake by a cord from his nose, How long must this cord be, that feeding all 'round, Permits him to graze just two acres of ground? Ann. 10.0904+. 160x2-i-Vi=407.272727. 7407.272727=20.1809=diameter of circle. Take half for radius, or cord. 21. A snail, climbing a pole 20 feet high, ascends 8 feet per day, and falls back 4 feet at night; how many days will it take him to reach the top ? Ans. 4 days. Statement— 1 : 4 :: [] : 20—4=4. N^ote. When an example and the solution are given, the solution is applicable to all similar examples. 22. A. can do a piece of work in 4 days, B. in 6 days ; in what time can they both do it working together ? Ans. 2| days. 4x6-h4+6=2|. 23. As I was beating on the forest ground, Up starts a hare before my two grey hounds, The dogs being light of foot, did fairly rua Unto her fifteen rods, just twenty-one ; The distance that she started up before Was four-score, sixteen rods, just, and no more. Now, I would have you unto me declare, How far they ran before they caught the hare ? D 96 H Ans. 336 rmls. 21— 15 : 96 :: 21 : [J. =336. VADE MECUM. 68 24. The hour and minute hands are exactly together at 12 o'clock; at what time will they next be together ? 11 : 12 :: 1 : [].=lh. 5min. 27f\sec. 25. The hands of a clock are together between 5 and 6 ; what is the time ? 11 : 12 :: 5 : []=5h. 27min. 16yVsec. 26. The hour and minnte hands are directly opposite^he minute hand between 4 and 5, the hour between 10 and 11 ; what is the time ? Ans. lOh. 21m. 49J,-sec. ^oie. It is the same time past 10 as it would be past 4, were the hour and minute hands to- gether between 4 and 5. 27. The time past noon is equal to ^ of the time past midnight ; what is the time ? Ans. 6 o'clock. Denominator — numerator : numerator :: 12 : []. 2. The time past noon is equal to ^ of the time till midnight ; what is the time ? Ans. 3 o'clock. Denominator-|-numorator : numerator :: 12 : []. 2o. When first the marriage-knot was tied. Between my ■wife and me, Jly age did hcr-i as far exceed As three times three does three. But after ten. and half ten years, We man and wife had been, Her age came up as near to mine, As e'ght is to sixteen. 64 ARITHMETICAL Now, tyro, skilled in numbers, say, What were our ages on wedding-day ? _,. . j 3-r-3=l. 1x15=15. - Solution. I 9_^3^3^ 3x15=45. 29. A father gave to his son f of his whole estate ; to his daughter | of the remainder, and the remaining part to his widow. The son re- ceived ^75 more than the daughter. Required, the share of each. C 8450=son's share. Ans. i ^375=daughter's. ( ^225=widow's. 80. If a beam, 10 ft. long, 5 in. wide, and 2 in. deep, will bear up 100 lbs., how many lbs. will another support, that is 15 ft. long, 6 in. deep, and 3 in. wide, the support being at the end. Ans. 360 lbs. 2) 6 4 —10 : 100 : : 6 5 3 ■15 : [J. ? X^ 6 3 1'0J3 2O 360 To find the least common multiple of fractions. Hule. Find the least common multipW^ of tli% numerators, and divide it by the greatest common divisor of the denominators. * The process is the same as in Art. 13 VADE MEC'UM. 65 31. Find the least common multiple of |, f , and |. Ans. 12. The least common multiple is 12. The greatest common divisor is 1. Thus 12-^1=12. 32. By Ret. J. P. Murrel. Suppose two walls erect should stand, Across a street on either hand; Now, if a pole should stand upright, Close to the one of the same height, If foot be drawn twelve feet in street. And top slips down about two feet j Then if the top be turned to fall, So as to strike the other wall Below its summit just six feet — What are their heights, and width of street ? . J 37 ft. height of walls. ^"*- ( 82.2 ft. width of street. Sohction. 12^-|-2^-r-double the distance the pole slipjped down=37 ft., height. 37—6+37x37—31=408. V'408=20.2(nearly)+12=32.2, width. 33. A man has 60 lbs. of wool, worth 25 cts. per lb., which he wishes carded. The carder has 5 cents per lb. for carding, and takes the toll in wool before carding it ; how many lbs. must he take ? Ans. 10 lbs. 60x5-^-25+5=10. Falling Bodies. Rule. Multiply the square of the timt hy 16|*j feet, {the distance a body will fail in, ike first second) . 66 ARITHMETICAL 34. To what height must a stone be raised to require it 4 seconds to reach the ground ? Ans. 25Ti- ft, 4^X16 Jo=257i-. 35. What is the depth of a well, to the bot- tom of which a stone would be 3 seconds in falling ? Ans. 144| ft. 36. How high above the earth's surface must a body be raised to lose | of its weight ? Ans. 898.97. Rule. Denominator — numerator : denomina- tor :: square of the earth's semi-diameter r to the square of the distance from the center of tht earth. 3—1 : 3 :: 4000^ : 724000000=4898.97—4000 =898.97 miles. 37. By Rev. J. P. Murrel. If a man o'er head, in a balloon, Should fire a level gun at noon, How high is he, if ball and sound, At the same time, should strike the ground ? Ans. 15i^-|-mile8. Note. Sound flies 1142 feet per second. 11422-^16^2=^0. of feet. 38. Divide 100 apples between John and James, go that John will have ^ more than James. VADE MECUM. 67 S9. If the fourth of 20 bo 3, What will the fifth of 30 be ? 40. If the third of 7 be 3, What will the sixth of 20 be? Ans. 3|. Ans. 4f . 41 . Two pence is what part of f of 3 pence ? Ans. the whole. 42. What is nothing, twice yourself, and 50 ? A71S. 0. W. L. 43. Four-fifths of 815 are six-tenths of how many thirds of 821 ? Ans. 2|. 44. Four-sevenths of 42 are f of how many times that number of which i- of 9 is f ? Ans. 9. 45. If A., B., C. and D. start from the same point around a circular island, 80 miles in cir- cumference, in how many days will they next be together, if A. travels 4 miles per day, B. 8, C. 20, and D. 28? Ans. 20. Mule. Divide the distance romid the island by the greatest common divisor of distances traveled. 46. By Rev. J. P. Murrel. Suppose a horizontal plane, On which did stand a stalk of cane, The height of which I took quite neat. And found to be one hundred feet. Soon as I did this measure take, A blast of wind this cane did break. t$ ARITHMETICAL The top of it did strike the ground Some ten yards from the base, I found; One end of which did still remain, Just where the wind did break the cane. Now, can you all these measures take, And tell how high the cane did break ? A71S. 45L ft, 1002 _302 _^ioO X 2=45^. 47. Two men, A. and B., leave Iowa City. A. travels due east, 39 miles, to Berlin ; B. due soutli to St. Francisville, thence to Berlin, and says he has traveled 120 miles since he left Iowa City. They both start a due south course, and after having passed the latitude of St. Francis- ville, travel 83 miles to Jefferson Barracks. They then turn a due west course to a point ex- actly south of St. Francisville. They here part. B. passes directly to Iowa City. A. continues his course due west to Jefferson City, thence to St. Francisville, antl says, since he parted with B. the last time, he has traveled 100 miles. Returning to Jefferson City, he inquires how far he and B. are separated. Ans. 137.544-j-M. 48. What is the length of a trout, whose head is 3 inches long, his tail as long as his head and I of his body, and his body as long as his head and tail together ? Ans. 20 in. Head. I Body. I Tail. 3 i 3 I 3 I 4 i 4 I 3 49. Edwin bought 5 pears for 5 cents ; Charles bought 3 for 3 cents ; being afterwards joined by James, the three made a meal of the 8 pears. On leaving, James pays them 8 cents, of which VADE MECDM. 69 Charles claims 3 cents, as he furnished 3 pears. How, in equity, should the 8 cents be divided ? J j Charles, 1 ct. ^''^- ( Edwin, 7 cts. 50. By Rev. J. P. Mukrei,. In partnership, wo understand, Two brothers bought a piece of land, Two hundred acres when survey'd, And each four hundred dolhirs paid. One end being richer than the other, The elder said to the younger brother, I'll take the end that 's cot so poor. And pay a half a dollar more Per acre, if you will agree, To let that end belong to me. To which the younger thus replied, I will, if you '11 the land divide. T will, he said, and at it went ; Eut, after ho some days had spent, And found it did his soul perplex. He called on his surveyor next, Who labored hard, but could not quite Make land and money come out right ; Then threw it down with grief and pain. Declaring he'd ne'er try again. Since that, this sum has traveled round, To see if any could be found Who could this piece of land divide. As elder brother did decide ; Likewise how much each man must pay Per acre, in his own survey. Now, reader, as it's come to you, Take hold, and see what you can do ! {Elder brother's, 93J4-^cres. Younger brother's, 106^ acres, nearly. Price of elder brother's, 84.26.5+. Price of younger brother's, $3.76.5-j-. For the solution of the above : Hule. Find the cost of the whole number of acres, at the difference between the j^rices per acrt, 70 ARITHMETICAL which subtract from the amount paid for the whole land ; to the square of the remainder, add the product obtained by multiplying the cost of the whole number of acres {at the difference between the prices per acre), by four times the whole sum paid by him who paid least per acre ; extract tlie square root of the sum ; to the result add the remainder that was squared, and divide the su7/i by twice the whole number of acres, for the price per acre paid by him who paid least per acre. Having this, other requirements of the question are easily found. 51. V>y Rev. J. P. Mubrel. If round a point two wheels you start, By axlo kept five feet apart, The height of inner wheel complete, Supposed to be about three feet — To form a circle would require The outer wheel a little higher ; How much higher would you suppose, That inner might an acre inclose ? Ans. .1274+ft. 117.728 ft.=radius of one acre. 117.728 : 117.728+5 :: 3 : []. 3,1274+ft. height of larger wheel. 52. By Rev. J. P. Murkel. Suppose a cart drawn once around A level circular piece of ground, Diameter of wheels to be, In inches, each just sixty-three. Now, if the axle of this cart Should keep the wheels five feet apart, IIow many times will one turn round More than the other, once round this ground, If inner track should just inclose. Five thousand acres we'll suppose ? Ans. l\\. VADE MECUM. 71 5X 2xV=^^^^''^^"^^''^ ^^ peripherics. 63-r-12xV=^^''^""^'^*^''^''^^'^ of Avheels. 2 2 o_i_3^ 1 L2. 7 • 2 ■^2 1* Note. The No. of acres has nothing to do with ihe solution. 53. How wide must a walk he around a rec- tangular garden, 24 yds. long, and 16 yds. wide, to contain as much land as the garden ? c 1 ^. Ans. 4 yds. Solution. •' 24x16^-4=06. 24+16-^4=10. 10^ +96= ^196=14. 14—10=4, width of walk. 54. By Rev. J. P. Murbkl. Four men, A., B., C. and D., In partnership did buy A grindstone, which they did agree To grind away, all bat the eye. How deep in radius must each grind, To have an equal share ; Diameter fivo feet thoy find, And eye three inches square ? 4.00+in. 1st man. 4.7o+in. 2d man. 6.15+in. 3d man. 12.99+in.4th man. 72 ARITHMETICAL 602 xi-i-=2828 4 =areca of stone. 32 _j_32 y^ Li=i4 1^ =area of circle about the eye. 4)2814 Y =area to be ground away. 703V|-==area of one man's share. J 28284— 703LfIZI=52.+. 60—52^-2=4 ■=radius of first man's share. 2^ot€. For the other depths of radii proceed in the same manner. 55. Bought 2 watches for 820 each, and sold one at 25 per cent, gain, the other at 20 per cent, loss ; did I gain or lose ? Ans. $1 gain. 56. When 5 pears are worth 7 peaches, and 12 peaches 15 apples, and 21 apples 24 damsons, how many pears can I have for 36 damsons ? Ans. 18 pears. $ n fX ^ 18 18 pears. 57^ By Ret. J. P. Murrel. How large a field would be required. Inclosed by fence both staked and ridered ; Two pannels to each rod of land, Ten rails in each we understand, So that the fence may just inclose As many acres, we'll suppose. As rails, and stakes, and riders there, The field itself to be a square ? Ans. 1730560 acres. VADE MECUM. 78 53^ By Rev. N. C. DeWitt. A wealthy maa a daughter had, A son likewise — a sprightly lad — To whom he gave a piece of land, Which was a square, we understand ; On ev'ry side the distance found Was just four hundred rods of ground. The next thing is the daughter's share, Which must be round, and not a eqnare ; How long a line, do you suppose, Would just the daughter's land inclose, And she the same amount obtain That's in her brother's largo domain ? Ans. 1418 roda. Solution: 400x3.545=1418.000. 59, By KeV. J. P. MUKREL. A youth who lived a single life, Set out at last to hunt a wife, So to a house he did repair, To see if he could find one there. Directly after he stepped in. With Miss a courtship did begin, When she embraced the chance to find What were the powers of his mind. " My brother and myself," she said, "Were all the heirs my father had; And now, kind sir, please understand, We both were in a foreign land." Then with a plaintive voice she sighed, " Pa heard that one of us had died." Before she closed she added still, " While we were there Pa made his will, Which did for Ma and me provide, If I had lived and brother died : Two-thirds of his estate should be Secured to Ma, one-third to me. Had brother lived and I had died, For them the will did thus provide : Two-thirds were given to my brother, And only one-third left to mother. Now, as we both are living still. y. And must be governed by tlie will, 6 Ans. 74 ARITHMETICAL. Which does my mother's share express, About three hundred dollars less Than it would be if I had died, And they should by "the will abide. And now, kind sir, please calculate. What is the sum of Pa's estate ; Likewise how much each share will be, According to the will's decree ? Unless you do all these decide, Vl\ not consent to be your bride, {a^ire.) Mother's, 61800. Daughter's, $ 900. Brother's, ^SGOO. Solution: I Whole estate, 86300. 1=J>. 2=M. 4=B. 7-r-3=2^, had daughter died. 2 , all being alive. T:300::2:[]=:$1800M's. 00. By Ret. S. H. Hodges. A neighbor asked me for the time, And as I love to speak in rhyme, I told him it was after ten, And both the hands together then. Tray tell me now the time precise, By any means you can devise. Ans. lOh. 54min. 82/|Sev 61. By Rev. S. H. Hodges. If three-elevenths of the age Of one who is a noble sage. By twenty-eight should 1^ increased, And one year from the last released. VADE MECUM. 7* And one-eleventh of a year To the remainder added here, And of a year elevenths three Be now subtracted, yoa will sre,- You'll have one half the sage's yeara ; Now, what is all his life of cares ? A71S. 118 years. 62. By E. P. TnoMPSOx. Suppose a mirror sixteen inches wide, In inches long precisely twenty-four, The frame of which the owner does decide, In superfico must equal it — no mnre. How wide a frame, pray unto me declare, That each shall equal be, in inches square ? Arts. 4 inches. 63. % ^^^- J- P- ^Iphrel. A youth who lived a lonely life. Concluded he must have a wife ; He sought a fair one for his bride. Whose father just before had died. The fair one's heart and hand were gained, But something to be done remained — The maiden's mother must consent, Before the matter further went. On being asked, the mother said, "Why do you bother thus ray head ? — To you, sir, it is clearly known That I have just five daughters grown. Their fathers will says, ^ my first four Must have ticdve thousand, and no more; To my lant four I do declare Eleven tJious'and is their share ; To my last three and first I give Ten thoH'^nnd dollars, if they live ; As to my last, and my frst three, Nine thousand shall their portion be ; To all, except the third alono, I give ci'/ht thousand — now. I'm done.* Now," said the mother, "if you tell The third one's part, I'll give you Nell." Ans. S4500. 78 ARITHMETICAL Solution : 12000=A.B.C.D. 11000=B.C.D.E. 10000=A.C.D.E. 9000=A.B.C.E. 800Q==A.E.D.E . 4) 50000 =four times A.B.C.D.E. 12500=A.B.C.D.E. 8000=A.B.D.E. I 4500==C. (or Hell.) 64. Suppose a cone to stand upright, Which is one foot esa«t in height, How high, 'bove base must a line be, That will divide it equally ? Ans. 2.4 76 inches. ^12^-^2=9.5 24. 12—9.5 24=2.4 76. 65. By Rev. S. H. Hodges. A farmer has a square of fertile land. And in the center all his build- ings stand ; His land he has determined to divide Among twelve sons, as ho can best decide. Ho first proceeds to draw a circle round. With area as broad as all his ground ; And next he doth proceed to draw a square Whose angles in the circle's bound'ry are ; He then four radii doth draw complete. Which in four points with square and circle meet ; Eaoh radius in length is sixty poles, And this amount alone the sum controla. (/ \ \\ 3 /y VADE ATECCM. 77 The liaes now made divide the farmer's ground, So all the sons may have their portions round. Now, tell me how much land there is in all, And how much will within the circlo fall — How much within the inner square will be; And also each son's portion tell to me. But bear in mind, the first four each must take One-fourth the inner square, his part to make. Unto the second four the segments give, That they within the circle's rim may live. As to the last four, give them for their lots The outside sections — smallest, richest spots. In tlie wliole track, 90 acres In tlie circle, 70 y In tlie inner square, 45 First four sons, next the farmer's, each, 11 \ Second four, in the segments, have each 6 f Last four, in the outside sections, each, 4|j 66. Three men. A, B, and C, being employed to perform a certain piece of work for 8105, A and B are supposed to do j^ of it, A and 7—, and B and C f . They are paid proportion- ally ; please divide their pay for them as it should be. 67. A stick of timber 30 feet long, of uniform thickness and breadth, is to be lifted by 3 men, one at one end, and the other two holding a hand-spike near the other. How far from the end must they be placed that each of the three may raise an equal portion of the stick ? 68. A man 6 feet high traveled round the earth. How much further did his head go than his feet ? 78 AKITHMETICAL Am. By Ret. J. P. Muekel. There is a Quaker, Tre understand, "WTio for three sons laid off his land. And made three equal circles meet, So as to bound an acre neat. Just in the center of that acre, Is found the dwelling of the Quaker ; In centers of the circles round, A dwelling for each son is found : Now, can you tell, by skill and art. How many rods they live apart ? j Distance from son to son, 63 rods. \ father to son, 36.372 rods. Solution : ^160x6.20156=31.5(nearly)=S b. 31.5x^=63=8 S. 31.5-^2=151 =a b. 7|:L::15| :[]=9.093=a q. ^3]i2_i5.752_,27.279=S a. 27.279-1 9.093=36.372=q S. vaoe mecdm. 79 70. I^'Y E. p. Thompson. The author of this work I chanced to ask, " Tell me, sir, if you please, how old you are?" And he replied, " Let mc impose this task. Which, when performed, it will my years declare: If to my a;i;o one-fourth score added be, And of that sum the square root you extract, And add it to the sum, you then will see Th.'-t you will have just one score, ten, exact." You, who with numbers do your thoughts engage. Pray tell me, if you can, what is his age. Ana. — , 71. What number is that, to which, if you add the square root, the sum will be 42 ? Ans. 36. Ride. To the sum (42), add one-fourth, and extract the square root, from which take one-half for tli£ square root of the number, 72. By E. p. Thompson. "Pray," said a lover to his "gal," *' Will you not be my bride ? My heart, my hand, my all, are thine. Whatever may betide." To which, that she might test his skill. The pretty thing replied : **A gent wa^ wont to visit where Three sisters did reside; And go, ere starting there one day, Within his pockets wide He put of pears a number there, And did them thus divide : To Kate he gave one-hajf he had. And half a one beside ; The half then left, and half a pear, For Mol he did provide ; Half then remaining, plus one-half, For Puss he set aside ; He'd then one left. Now, tell me, sir,** The gentle maiden sighed. ''How many pears in all h^d he, To 'mong the three divide ? My answer must, shall be delayed, Till this you do decide. Ant. — 80 VADE MECUM. IlE(JOMMENDATIONS. I HAVE examined, with much interest, many portaona of Mr. I. N. Wilcoxson's "Arithmetical Vade Mecum," (in manuscript), and find myself decidedly favorable to the -vyork. It will doubtless prove, not only a convenient, but also a very useful companion, to all calculators into whose bands it may fall. I have been a teacher of youth, the most of my time, for about fourteen years ; during which time, T have exam- ined and used a considerable number of Arithmetics ; but, for conciseness, perspicuity, and practical utility, I feel constrained to praise the " Vade Mecum " more than all of them. The young authol of this new work deserves the thanks and patronage of many, for the great improvement he has made in the philosophy and practice of Arithmetic. I am strikingly impressed with the idea of oar author — by his untiring energy and iyitense study — becoming an instruc- tor of instructors t Let every instructor, merchant, mechanic, student, and farmer, procure a copy of this new work, study it, and apply its rules in practice, before he suffers himself to epeak or think against it. S. H. HODGES. Barren County, Kentucky. The following is from the pen of James G. Hardy, Lieu- tenant Governor of Kentucky : Glasgow, Ky., Sept. 29, 1855. I have briefly examined the "Arithmetical Vade Meeum," prepared by Isaac N. Wilcoxson, Esq. The principles of *^ Cause and Effect," or a separating the order of producing from the parts produced, are worthy the consideration and patronage of the public, and I have no hesitation in saying Ihat the work will be useful to business men. JAS. G. HARDY. A SUPPLEMENT ARITHMETICAL VADE MECUM: GIVING AN EXPLANATION AND SHOWING THE APPLICATION OP A NEW DIAGRAM; Bi' MEANS OF WHICH WE READILY APPLY CAUSE AND EFFECT TO NUMERICAL CALCULATION. There is but one rule by which every arith- metical question of a practical nature in arith, metic may be scientifically and philosophically analyzed, stated, and solved ; it being founded upon an axiom in natural philosophy, " That equal causes produce equal effects, and that effects are always proportionate to their causes.'^ There are but two primal principles — increase and decrease — in numerical calculation ; and.by^ the same statement, we increase or decrease ac- cording to the nature of the question. This may be learned by those of common capacity^ who are attentive, and understand the first prin- ciples of calculation, in a very few days. There is contained in the following pages of the Sup- plement the labor and experience of years, teaching and investigating, excluiively, the new system. I, owe many thanks to Ed. Porter Thompson,, now Principal of Rich Grove Seminary, a new and flourishing Institute on the turnpike seven /miles north of Glasgow, Kentucky, in a delight- ful neiorhb0r?hood, for much assistance rendered r ISAAC N. WILCOXSON. AuJtJUST 24, 3.858. SUPPLEMENT FIRST COUPLKT. SFX'OND COLPLET. 1st Proposition. 2nrl Proposition^ , — [means.] V Cause : Effect : : Cause : Effect. (Isi teiTQ.) (2ud terra.) {3rd term.) (4rh term.) [extremes.] — ' Question. What is a diagram ? — Ansicer. An arithmetical scheme. Q. What system is here used : — A. Dia- gramic system of applying cause and eff^ect to numerical calculation. Q. What is the ndel A. Rule : Cause : (is to) ej^ect : : (as) cause : (is to) effeft. Note — The prereding is the Ri i.e complete, but it may be read with black terms, as follows : Cause : (is to) effect : : (as) cause : [ — ]* (is t<» required) f-ffecf. * Is the blank, and shows the place of the required term, or factor. Bi ARITHMETICAL or, Cause : (ia to) effect : : [ — ] (as required) cause : (is to) ejlect. or, Cause : [ — ] (is to required) e^ect : : (as) cause : (i» to) ejhct, or, [ — ] (Required) cause i (is to) effect : : (as) cause : (is to) effect. (See Art. 29.) Note. — Sometimes there is only a factor of a term blank, or wanting ; but ihi« does not alter the rule, or statement, in the least ; it would be as the following example in all Uie terms. Cause : (is to) eject : : (as) cause : (is to) ( factor eject. c [ — ] ^'yquired fac. Q. What is a term ? — A, One member of a proportion, Q. How many terms in tho diagram ? — A, N'oTE. — There may be a multitude of factors ; there may he several in a term, as for example: If 3 men in 8 days Mrld a -wall 20 feet lomg, 12 feet high, and 2 feet thick ; how uiHiiy men would be required in 4 days to bnild another 24 itet long, 10 feet high, and 3 feet thick ? STATEMENT. c. E. C. E. Men 3 Days 8 length 20 : ; height 12 thickoess 2 (-) 4 : 24 10 5 Here 3 and 8 are factors of the first term ; 3x8=24 the term; 20, 12, and 2 are factors of the second term; 20xl2x 2=480 the term. i VADE MBOUM. BS Q. What aie they divided into ? — A. Means and extremes. (See Art. 26 A. V. M. for answers to follovy- ing questions.) Q. What is ratio 1 Q. What is a couplet 1 Q. What terms compared constitute the firsl couplet ? Q. What terms compared constitute the see" ond couplet ? Q. What are the means ? Q. What are the extremes ? -Q. What is a proportion ? Q. The product of the means must equal what ? (See Art. 27.) Note. — If 4 yds. of cloth cost Sl2, 6 yd«. will cost hovy much ? STATEMENT. C, E. C. E. Yds. 4 : $12 :: 6 : (—J After the statement is made, we consider the factors s« abstract numbers. In the first couplet the effect is 3 times as large as its cause ; so must the second efiect be 3 times S.S large as its cause — 6x3=18 second effect. Then 13x4= 12x6. The same reasoning applies to all the terms. Q. What is a proposition'? — A, A statement in terms, one clause of a question containing two t&ims, as 4 yds. of cloth cost $12. 86 ARITHMETICAL Q. How many propositions in every aritli- metical question ? — A. Two. Q. What are they! — A. First and second. Q. What kind are they ? — A. Complete and incomplete. Note. — The complete proposition is' a irjodel, anwxample, by which to arrange or complete the incomplete proposition. The complete governs the incomplete. Q. Where is the first proposition arranged ? — A. In the first couplet. Q. How ? — A. To suit convenience. Q. If time is given, where is it placed? — A. In the same term with that that can produce ac- tion or agency. Q. Where is the second proposition arranged? — A. In the second couplet. Q. How? — A. Similar to the first. Q. What is cause? — A. Action or agency. Q. What is effect? — A That which is ac- complished or follows action or agency. Q. What are the guides in stating the ques- tion? — A. The elements or denominations in the question. (See Art. 31.) Q. There must be the same number of ele- ments or factors where? — A. In similar terms. (See Art. 31. N. B.) I VADE MECUftl. 87 Q. What must be in similar terms? — A. The same number of elements ; also, the same num- ber of factors, though one may be a blank fac- tor. Q. ^Causes must always be how ? — A. Of the same or similar kind. Q. And must be reduced to what before placed on the line ? — A. To the same name or denomination. Q. Where is the ( — ) blank placed ? — A. In the place of the required term or factor ; or in that term similar to that in which is placed the same or similar element or name in the arrange- ment of the complete proposition. Note. — The observing student will notice that there is an action or agency that passes from the cause to the effect — a relationship shown — which will enable bim to readily separate the factors of the cause from those of the effect, The philosophical way of staling the question is to place the acting term as cause ;, but, to find the trEe result or answer, it is immaterial which is placed as cause, so that the second is like the first. Q. If the first or fourth term is blank, which are complete, means or extremes ? and where are they placed ? — A. The means are complete, and are placed upon the right as a dividend. Q. Which are incomplete ? and where are they placed ? — A. The extremes are incomplete, and are placed upon the left as a divisor. Q. If the blank is in either the 2nd or 3rd 88 ARITHMETICAL term, which are complete, means or extremes? and where are they placed ? — A, Extremes are complete and are placed upon the right as a dividend. Q. Which are incomplete? and where are they placed ] — A. The means are incomplete, and are placed upon the left as a divisor. Q. Then what is involved to find the true result or answer to the question ? — A. Nothing but simple multiplication and division. Q. What is cancellation used for? — A. To shorten the work of multiplication and division. INTEREST. (See Art. 36 A. V. M.) PERCENTAGE. Cause : effect : : cause : ej^ect. 100 : 100-\-gain "^ ct. : : cost price : se.i g price- or, 100— loss ^ ct. BARTER. Cause : effect lit commodity : 1 [ajnH) : : 2d comm^y : i (amH) Its jpTice cause : effect. its price. VADK MKCUM. DISCOUNT. 89 Cause : effect : : cause : effect. 100 : \00-{-interest of : : present : amount 1 OOybr given time worth and rate ^ ct. to he dis- counted. EXAMPLE. What is the present worth and the discount of $110 for 1 year and 8 months, at 6 per cent ? Ans. Pres. $100. Dis. $10. C. E. C. E. 100 : 110 : : (— ) : 110=100 pres. ly. 87n.=20w.X6-M2=10+100=110. MENSURATION TABLE. (See Art. 46 A. V. M.) SQUARE MEASURE. Cause : effect : : cause : Factors of : U/iit of : : Length : the unit of measure Width measure effect. No. of units. 90 ARITHMETFCAL SOLID OR CUBIC MEASURE Cause : Factors of : the unit of measure effect : : Unit of : : measure cause : Length : Width Depth effect. No. of units. Note. — Tn square measure put iecgth and width, atid in solid measure put length, width, and depth, under the sec- oud cause, and reduce the first cause to the same denomi- nation, then place it on the line for solution. Q. What is the first proposition in interest ? - — A. 100, 1 year, and rate per cent. Q. What is the second proposition 1 — A. Principal, time, and interest, Q. What is the first proposition in percent" age 1 — A. 100 and its amount. Q. How do you find that amount? — A. By adding the gain per cent to, or subtracting the loss per cent from, 100. Q. What is the second proposition in per- centage 1 — A. Cost and selling price. Q. What is the first proposition iti barter ? — A. First commodity, its price, and 1 (amount). Q. What is the second proposition? — A. Sec- ond commodity, its price, and 1 (amount). Q. What is the first proposition in discount? — ^. 100 and its amount. I VADE MECUM. 91 Q. How is that amount found? — A. By get- ting the interest on 100 for given time and rate per cent and adding it to the 100. Q. What is the second proposition in dis- count? — A. Present worth and amount tc be discounted. Q. What is the first proposition in measure- ment 1 — A, The unit of measure and its factors. Q,. What is the second proposition? — A. The factors of the thing to be rneasuied, and the number of units it contains. Note. — By the foregoing questions the investigator csin dearly see that there is only the one rale and principle necessary to analyze, state, and solve all questions coming up under the preceding specified rules. It may also be applied to any practical calculation in life. MISCELLANEOUS QUESTIONS. 1. Add together \l\l \\ and i. 3 2 2 3 6 5 30 3 2 24 4 1 9 9 i 28 12 11 33 2 1 18 36 142= Least com. denom. 2. Add together \ \ I \ li- =3U (See A. V. M.) Ans. 2%. 9J ARITHMETICAL 3. From j| take |. Ans. j|. 4. Multiply together ^i ^i | ^ -| || and |. Ans. 15, 5. Multiply together | | | | i; 3 1| and If. Ans. 2|. 6. Divide | by |, ^w* 2. 7. Divide i by |. ^7^5. |. Note. — Multiplying by a fraction decreases the multipli- cand ; and dividing by a fraction increases the divideEd. 8. Divide 4 by |. Ans. 12. 9. Divide | by 5. Ans. ji^ 10. Divide | of 4 by j| of 2, and multiply the quotient by 2| of |, and divide by 3i of |. Ans. 3. 11. If 8 hats cost $40, what v/ill 6 hats cost? Ans. $30. 12. If 200 ft> of pork cost $10, how many ib can I have for 840 ? Ans. 800 ft) 13. If a pole 7 feet high, at noon east a sha- dow 5 feet long, what is the height of a tree whose shadow measures 80 feet? Ans. 112 ft. 14. If 3 men in 10 days build 20 rods of fence, how long a time should be allowed 5 men to build 50 rods ? Ans. 15 days. 15. If $200 in 8 months gain $10 interest, how many months will it take the same prin- cipal to amount to $250 ? Ans. 40m, VADE MEUUM. 93 16. Suppose 1 woman in 2J days sews the seams of two pair of pants, each of which aver- age 4 1 seams 30 inches long and 12 stitches ta the inch, how many days should it take 5 women to make 12 pair each of which averages 4^ seams 48 incEes long and 15 stitches to the inch ? Ans. 5 days. 17. When 10 men in 6 days dig a trench 40 feet long, 5 feet wide, and S feet deep, that is of the hardness of 3, how many days ought it to take 80 men to dig another that is 400 feet long, 8 feet wide, 18 feet deep, and of the hard- ness of 4 ? Ans 36 days» "What is the interest of— 18. $235 for 30 days at 12 per ct. Ans. 2.35 19. $350 for 60 days at 6 per ct. Ans. 3.60 20. $700 for 90 days at 4 per ct. Ans. 7.0O 21. $360 for 2 days at 6 per ct. Ans. 12cts. 22. S120 for ly. 4m. 20d. at 6 per ct. Ans. 10.00 23. What principal at interest for 8 months and 10 days, at 6 per ct., will draw $15 interest? Ans. 360 00 24. At what rate per ct. will $900, in 1 year, 1 month and 10 days, draw $60 interest ? Ans. 6 per ct. 94 ARITHMETICAL 25. In what time will $1000, at 6 per ct., for domestic at 10 cts. per yard, how many yards mnst I have ? Ans. 10 yds. What is the present worth and discount of — 42. 8303 for Im. 15cl. at 8 per ct.] Ans. P. W. S300. D. $3. 43. $166 for ly. and 8m. at 6 per ct.? Ans. P. W. $15U. D. $15. 44. $i21.80 for 90 davs, at 6 per ct.? Ans. P. \V. S120. D. $1.80. 45. $100 f )r 1 year at 6 per ct.? Ans. P. W. $941?. D. 85||. 46. SI 00 for one year at 10 per ct.? Ans. P. W. $90|f. D. 9i. 47. A note of $500 was given, January I, 1857, and indorsed July 4, 1857, by 3115.331; another indorsement made December 25, 1857, of $211. 33|, what was due March 1, 1S58-. and if interest was paid oh it then, what is due to-day, interest at 6 per ct,? Due March 1, '58, $202. 16|. Due to-day, $ Note. — Bank Discount is simple interest calcalated npon amount, with " three days of grace " add«d to the time : this tak«n from the amount will leave the principal. J ^^ I. UC SOUTHERN REOO'.A, :PCiD/c< llililillllllllilll „„ B 000 012 943 7