Suppose an oblong figure,a, b, c, d, as here drawn (which is called arectanglein geometry), with the sidea b6 inches, and the sidea c4 inches. Dividea bandc d(which are equal) each into 6 inches by the pointsa, b, c, l, m, &c.; anda candb d(which are also equal) into 4 inches by the pointsf, g, h, x, y, andz. Joinaand l,bandm, &c., andfandx, &c. Then, the figurea b c dis divided into a number of squares; for a square is a rectangle whose sides are equal, and thereforeaa feis square, sinceaais of the same length asaf, both being 1 inch. There are also four rows of these squares, with six squares in each row; that is, there are 6 × 4, or 24 squares altogether. Each of these squares has its sides 1 inch in length, and is what was called in (215)a square inch. By the same reasoning, if one side had contained 6yards, and the other 4yards, the surface would have contained 6 × 4square yards; and so on.
235. Let us now suppose that the sides ofa b c d, instead of being a whole number of inches, contain some inches and a fraction. For example, leta bbe 3½ inches, or (114) ⁷/₂ of an inch, and leta ccontain 2½ inches, or ⁹/₄ of an inch. Drawa etwice as long asa b, anda ffour times as long asa c, and complete the rectanglea e f g. The rest of the figure needs no description. Then, sincea eis twicea b, or twice ⁷/₂ inches, it is 7 inches. And sincea fis four timesa c, or four times ⁹/₄ inches, it is 9 inches. Therefore, the whole rectanglea e f gcontains, by (234), 7 × 9 or 63 square inches. But the rectanglea e f gcontains 8 rectangles, all of the same figure asa b c d; and thereforea b c dis one-eighth part ofa e f g, and contains ⁶³/₈ square inches. But ⁶³/₈ is made by multiplying ⁹/₄ and ⁷/₂ together (118). From this and the last article it appears, that, whether the sides of a rectangle be a whole or a fractional number of inches, the number of square inches in its surface is the product of the numbers of inches in its sides. The square itself is a rectangle whose sides are all equal, and therefore the number of square inches which a square contains is found by multiplying the number of inches in its side by itself. For example, a square whose side is 13 inches in length contains 13 × 13 or 169 square inches.
236. EXERCISES.
What is the content, in square feet and inches, of a room whose sides are 42 ft. 5 inch. and 31 ft. 9 inch.? and supposing the piece fromwhich its carpet is taken to be three quarters of a yard in breadth, what length of it must be cut off?—Answer, The content is 1346 square feet 105 square inches, and the length of carpet required is 598 feet 6⁵/₉ inches.
The sides of a rectangular field are 253 yards and a quarter of a mile; how many acres does it contain?—Answer, 23.
What is the difference between 18square miles, and a square of 18 miles long, or 18miles square?—Answer, 306 square miles.
237. It is by this rule that the measure in (215) is deduced from that in (214); for it is evident that twelve inches being a foot, the square foot is 12 × 12 or 144 square inches, and so on. In a similar way it may be shewn that the content in cubic inches of a cube, or parallelepiped,[48]may be found by multiplying together the number of inches in those three sides which meet in a point. Thus, a cube of 6 inches contains 6 × 6 × 6, or 216 cubic inches; a chest whose sides are 6, 8, and 5 feet, contains 6 × 8 × 5, or 240 cubic feet. By this rule the measure in (216) was deduced from that in (214).
238. Suppose it required to find what 156 yards will cost, if 22 yards cost 17s.4d.This quantity, reduced to pence, is 208d.; and if 22 yards cost 208d., each yard costs ²⁰⁸/₂₂d. But 156 yards cost 156 times the price of one yard, and therefore cost
Again, if 25½ French francs be 20 shillings sterling, how many francs are in £20. 15? Since 25½ francs are 20 shillings, twice the number of francs must be twice the number of shillings; that is, 51 francs are 40shillings, and one shilling is the fortieth part of 51 francs, or ⁵¹/₄₀ francs. But £20 15s.contain 415 shillings (219); and since 1 shilling is ⁵¹/₄₀ francs, 415 shillings is
239. Such questions as the last two belong to the most extensive rule in Commercial Arithmetic, which is called theRule of Three, because in it three quantities are given, and a fourth is required to be found. From both the preceding examples the following rule may be deduced, which the same reasoning will shew to apply to all similar cases.
It must be observed, that in these questions there are two quantities which are of the same sort, and a third of another sort, of which last the answer must be. Thus, in the first question there are 22 and 156 yards and 208 pence, and the thing required to be found is a number of pence. In the second question there are 20 and 415 shillings and 25½ francs, and what is to be found is a number of francs. Write the three quantities in a line, putting that one last which is the only one of its kind, and that one first which is connected with the last in the question.[49]Put the third quantity in the middle. In the first question the quantities will be placed thus:
22 yds. 156 yds. 17s.4d.
In the second question they will be placed thus:
20s.£20 15s.25½ francs.
Reduce the first and second quantities, if necessary, to quantities of the same denomination. Thus, in the second question, £20 15s.must be reduced to shillings (219). The third quantity may also be reduced to any other denomination, if convenient; or the first and third may be multiplied by any quantity we please, as was done in thesecond question; and, on looking at the answer in (238), and at (108), it will be seen that no change is made by that multiplication. Multiply the second and third quantities together, and divide by the first. The result is a quantity of the same sort as the third in the line, and is the answer required. Thus, to the first question the answer is (238)
240. The whole process in the first question is as follows:[50]
The question might have been solved without reducing 17s.4d.to pence, thus:
The student must learn by practice which is the most convenient method for any particular case, as no rule can be given.
241. It may happen that the three given quantities are all of one denomination; nevertheless it will be found that two of them are of one, and the third of another sort. For example: What must an income of £400 pay towards an income-tax of 4s.6d.in the pound? Here the three given quantities are, £400, 4s.6d., and £1, which are all of the same species, viz. money. Nevertheless, the first and third are income; the second is a tax, and the answer is also a tax; and therefore, by (152), the quantities must be placed thus:
£1 : £400 ∷ 4s.6d.
242. The following exercises either depend directly upon this rule, or can be shewn to do so by a little consideration. There are many questions of the sort, which will require some exercise of ingenuity before the method of applying the rule can be found.
EXERCISES.
If 15 cwt. 2 qrs. cost £198. 15. 4, what does 1 qr. 22 lbs. cost?
Answer, £5 . 14 . 5 ¾ ¹⁸⁵/₂₁₇.
If a horse go 14 m. 3 fur. 27 yds. in 3ʰ 26ᵐ 12ˢ, how long will he be in going 23 miles?
Answer, 5ʰ 29ᵐ 34ˢ(²⁴⁶²/₂₅₃₂₇).
Two persons, A and B, are bankrupts, and owe exactly the same sum; A can pay 15s.4½d.in the pound, and B only 7s.(6¾)d.At the same time A has in his possession £1304. 17 more than B; what do the debts of each amount to?
Answer, £3340 . 8 . 3 ¾ ⁹/₂₅.
For every (12½) acres which one country contains, a second contains (56¼). The second country contains 17,300 square miles. How much does the first contain? Again, for every 3 people in the first, there are 5 in the second; and there are in the first 27 people on every 20 acres. How many are there in each country?—Answer, The number of square miles in the first is 3844⁴/₉, and its population 3,321,600; and the population of the second is 5,536,000.
If (42½) yds. of cloth, 18 in. wide, cost £59. 14. 2, how much will (118¼) yds. cost, if the width be 1 yd.?
Answer, £332. 5. (2⁴/₁₇).
If £9. 3. 6 last six weeks, how long will £100 last?
Answer, (65¹⁴⁵/₃₆₇) weeks.
How much sugar, worth (9¾d). a pound, must be given for 2 cwt. of tea, worth 10d.an ounce?
Answer, 32 cwt. 3 qrs. 7 lbs. ³⁵/₃₉.
243. Suppose the following question asked: How long will it take 15 men to do that which 45 men can finish in 10 days? It is evident that one man would take 45 × 10, or 450 days, to do the same thing, and that 15 men would do it in one-fifteenth part of the time which it employs one man, that is, in (450 ÷ 15) or 30 days. By this and similar reasoning the following questions can be solved.
EXERCISES.
If 15 oxen eat an acre of grass in 12 days, how long will it take 26 oxen to eat 14 acres?
Answer, (96¹²/₁₃) days.
If 22 masons build a wall 5 feet high in 6 days, how long will it take 43 masons to build 10 feet?
Answer, (6⁶/₄₃) days.
244. The questions in the preceding article form part of a more general class of questions, whose solution is called theDouble Rule of Three, but which might, with more correctness, be called the Rule ofFive, since five quantities are given, and a sixth is to be found. The following is an example: If 5 men can make 30 yards of cloth in 3 days, how long will it take 4 men to make 68 yards? The first thingto be done is to find out, from the first part of the question, the time it will take one man to make one yard. Now, since one man, in 3 days, will do the fifth part of what 5 men can do, he will in 3 days make ³⁰/₅, or 6 yards. He will, therefore, make one yard in ³/₆6 or (3 × 5)/30 of a day. From this we are to find how long it will take 4 men to make 68 yards. Since one man makes a yard in
Again, suppose the question to be: If 5 men can make 30 yards in 3 days, how much can 6 men do in 12 days? Here we must first find the quantity one man can do in one day, which appears, on reasoning similar to that in the last example, to be 30/(3 × 5) yards. Hence, 6 men, in one day, will make
From these examples the following rule may be drawn. Write the given quantities in two lines, keeping quantities of the same sort under one another, and those which are connected with each other, in the same line. In the two examples above given, the quantities must be written thus:
SECOND EXAMPLE.
Draw a curve through the middle of each line, and the extremities of the other. There will be three quantities on one curve and two on the other. Divide the product of the three by the product of the two, and the quotient is the answer to the question.
If necessary, the quantities in each line must be reduced to more simple denominations (219), as was done in the common Rule of Three (238).
EXERCISES.
If 6 horses can, in 2 days, plough 17 acres, how many acres will 93 horses plough in 4½ days?
Answer, 592⅞.
If 20 men, in 3¼ days, can dig 7 rectangular fields, the sides of each of which are 40 and 50 yards, how long will 37 men be in digging 53 fields, the sides of each of which are 90 and 125½ yards?
If the carriage of 60 cwt. through 20 miles cost £14 10s., what weight ought to be carried 30 miles for £5. 8. 9?
Answer, 15 cwt.
If £100 gain £5 in a year, how much will £850 gain in 3 years and 8 months?
Answer, £155. 16. 8.
245. In the questions contained in this Section, almost the only process which will be employed is the taking a fractional part of a sum of money, which has been done before in several cases. Suppose it required to take 7 parts out of 40 from £16, that is, to divide £16 into 40 equal parts, and take 7 of them. Each of these parts is
The process may be written as below:
Suppose it required to take 13 parts out of a hundred from £56. 13. 7½.
Let it be required to take 2½ parts out of a hundred from £3 12s.The result, by the same rule is
so that taking 2½ out of a hundred is the same as taking 5 parts out of 200.
EXERCISES.
Take 7⅓ parts out of 53 from £1 10s.
Take 5 parts out of 100 from £107 13s.4¾d.
Answer, £5. 7. 8 and ³/₂₀ of a farthing.
£56 3s.2d.is equally divided among 32 persons. How much does the share of 23 of them exceed that of the rest?
Answer, £24. 11. 4½ ½.
246. It is usual, in mercantile business, to mention the fraction which one sum is of another, by saying how many parts out of a hundred must be taken from the second in order to make the first. Thus, instead of saying that £16 12s.is the half of £33 4s., it is said that the first is 50 per cent of the second. Thus, £5 is 2½ per cent of £200; because, if £200 be divided into 100 parts, 2½ of those parts are £5. Also, £13 is 150 per cent of £8. 13. 4, since the first is the second and half the second. Suppose it asked, How much per cent is 23parts out of 56 of any sum? The question amounts to this: If he who has £56 gets £100 for them, how much will he who has 23 receive? This, by 238, is 23 × ¹⁰⁰/₅₆ or ²³⁰⁰/₅₆ or 41¹/₁₄. Hence, 23 out of 56 is 41¹/₁₄ per cent.
Similarly 16 parts out of 18 is 16 × ¹⁰⁰/₁₈, or 88⁸/₉ per cent, and 2 parts out of 5 is 2 × ¹⁰⁰/₅, or 40 per cent.
From which the method of reducing other fractions to the rate per cent is evident.
Suppose it asked, How much per cent is £6. 12. 2 of £12. 3? Since the first contains 1586d., and the second 2916d., the first is 1586 out of 2916 parts of the second; that is, by the last rule, it is ¹⁵⁸⁶⁰⁰/₂₉₁₆, or 54¹¹³⁶/₂₉₁₆, or £54. 7. 9½ per cent, very nearly. The more expeditious way of doing this is to reduce the shillings, &c. to decimals of a pound. Three decimal places will give the rate per cent to the nearest shilling, which is near enough for all practical purposes. For instance, in the last example, which is to find how much £6·608 is of £12·15, 6·608 × 100 is 660·8, which divided by 12·15 gives £54·38, or £54. 7. Greater correctness may be had, if necessary, as in theAppendix.
EXERCISES.
How much per cent is 198¼ out of 233 parts?—Ans.£85. 1. 8¾.
Goods which are bought for £193. 12, are sold for £216. 13. 4; how much per cent has been gained by them?
Answer, A little less than £11. 18. 6.
A sells goods for B to the amount of £230. 12, and is allowed a commission[51]of 3 per cent; what does that amount to?
Answer, £6 . 18. 4¼ ⁷/₂₅.
A stockbroker buys £1700 stock, brokerage being at £⅛ per cent; what does he receive?—Answer, £2. 2. 6.
A ship whose value is £15,423 is insured at 19⅔ per cent; what does the insurance amount to?—Answer, £3033. 3. 9½ ²/₅.
247. In reckoning how much a bankrupt is able to pay his creditors, as also to how much a tax or rate amounts, it is usual to find how many shillings in the pound is paid. Thus, if a person who owes £100 can only pay £50, he is said to pay 10s.in the pound. The rule is easily derived from the same reasoning as in 246. For example, £50 out of £82 is
or 12s.2½ ¹⁵/₄₁ in the pound.
248.Interestis money paid for the use of other money, and is always a per-centage upon the sum lent. It may be paid either yearly, half-yearly, or quarterly; but when it is said that £100 is lent at 4 per cent, it must be understood to mean 4 per cent per annum; that is, that 4 pounds are paid every year for the use of £100.
The sum lent is called theprincipal, and the interest upon it is of two kinds. If the borrower pay the interest as soon as, from the agreement, it becomes due, it is evident that he has to pay the same sum every year; and that the whole of the interest which he has to pay in any number of years is one year’s interest multiplied by the number of years. But if he do not pay the interest at once, but keeps it in his hands until he returns the principal, he will then have more of his creditor’s money in his hands every year, and if it were so agreed will have to pay interest upon each year’s interest for the time during which he keeps it after it becomes due. In the first case, the interest is calledsimple, and in the secondcompound. The interest and principal together are called theamount.
249. What is the simple interest of £1049. 16. 6 for 6 years and one-third, at 4½ per cent? This interest must be 6⅓ times the interestof the same sum for one year, which (245) is found by multiplying the sum by 4½, and dividing by 100. The process is as follows:
(82)100) 47,24 . 4 . 3(£47 . 4 . 10¹¹/₁₀₀
EXERCISES.
What is the interest of £105. 6. 2 for 19 years and 7 weeks at 3 per cent?
Answer, £60. 9, very nearly.
What is the difference between the interest of £50. 19 for 7 years at 3 per cent, and for 8 years at 2½ per cent?
Answer, 10s.(2½)d.
What is the interest of £157. 17. 6 for one year at 5 per cent?
Answer, £7. 17. 10½.
Shew that the interest of any sum for 9 years at 4 per cent is the same as that of the same sum for 4 years at 9 per cent.
250. In order to find the interest of any sum at compound interest, it is necessary to find the amount of the principal and interest at the end of every year; because in this case (248) it is the amount of bothprincipal and interest at the end of the first year, upon which interest accumulates during the second year. Suppose, for example, it is required to find the interest, for 3 years, on £100, at 5 per cent, compound interest. The following is the process:
When the number of years is great, and the sum considerable, this process is very troublesome; on which account tables[54]are constructed to shew the amount of one pound, for different numbers of years, at different rates of interest. To make use of these tables in the present example, look into the column headed “5 per cent;” and opposite to the number 3, in the column headed “Number of years,” is found 1·157625; meaning that £1 will become £1·157625 in 3 years. Now, £100 must become 100 times as great; and 1·157625 × 100 is 115·7625 (141); but (221) £·7625 is 15s.3d.; therefore the whole amount of £100 is £115. 15. 3, as before.
251. Suppose that a sum of money has lain at simple interest 4 years, at 5 per cent, and has, with its interest, amounted to £350; it is required to find what the sum was at first. Whatever the sum was, if we suppose it divided into 100 parts, 5 of those parts were added every year for 4 years, as interest; that is, 20 of those parts have been added to the first sum to make £350. If, therefore, £350 be divided into 120 parts, 100 of those parts are the principal which we want tofind, and 20 parts are interest upon it; that is, the principal is £(350 × 100)/150, or £291. 13. 4.
252. Suppose that A was engaged to pay B £350 at the end of four years from this time, and that it is agreed between them that the debt shall be paid immediately; suppose, also, that money can be employed at 5 per cent, simple interest; it is plain that A ought not to pay the whole sum, £350, because, if he did, he would lose 4 years’ interest of the money, and B would gain it. It is fair, therefore, that he should only pay to B as much as will,with interest, amount in four years to £350, that is (251), £291. 13. 4. Therefore, £58. 6. 8 must be struck off the debt in consideration of its being paid before the time. This is calledDiscount;[55]and £291. 13. 4 is called thepresent valueof £350 due four years hence, discount being at 5 per cent. The rule for finding the present value of a sum of money (251) is: Multiply the sum by 100, and divide the product by 100 increased by the product of the rate per cent and number of years. If the time that the debt has yet to run be expressed in years and months, or months only, the months must be reduced to the equivalent fraction of a year.
EXERCISES.
What is the discount on a bill of £138. 14. 4, due 2 years hence, discount being at 4½ per cent?
Answer, £11. 9. 1.
What is the present value of £1031. 17, due 6 months hence, interest being at 3 per cent?
Answer, £1016. 12.
253. If we multiply bya+b, or bya-b, when we should multiply bya, the result is wrong by the fraction
of itself: being too great in the first case, and too small in the second. Again, if we divide bya+b, where we should have divided bya, the result is too small by the fractionb/aof itself; while, if we divide bya-binstead ofa, the result is too great by the same fraction of itself. Thus, if we divide by 20 instead of 17, the result is ³/₁₇ ofitself too small; and if we divide by 360 instead of 365, the result is too great by ⁵/₃₆₅, or ¹/₇₃ of itself.
If, then, we wish to find the interest of a sum of money for a portion of a year, and have not the assistance of tables, it will be found convenient to suppose the year to contain only 360 days, in which case its 73d part (the 72d part will generally do) must be subtracted from the result, to make the alteration of 360 into 365. The number 360 has so large a number of divisors, that the rule of Practice (230) may always be readily applied. Thus, it is required to find the portion which belongs to 274 days, the yearly interest being £18. 9. 10, or 18·491.
13·878 = £13 . 17 . 7Answer.
But if the nearest farthing be wanted, the best way is to take 2-tenths of the number of days as a multiplier, and 73 as a divisor; sincem÷ 365 is 2m÷ 730, or (²/₁₀)m÷ 73. Thus, in the preceding instance, we multiply by 54·8 and divide by 73; and 54·8 × 18·491 = 1013·3068, which divided by 73 gives 13·881, very nearly agreeing with the former, and giving £13. 17. 7½, which is certainly within a farthing of the truth.
254. Suppose it required to divide £100 among three persons in such a way that their shares may be as 6, 5, and 9; that is, so that for every £6 which the first has, the second may have £5, and the third £9. It is plain that if we divide the £100 into 6 + 5 + 9, or 20 parts, the first must have 6 of those parts, the second 5, and the third 9. Therefore (245) their shares are respectively,
or £30, £25, and £45.
EXERCISES.
Divide £394. 12 among four persons, so that their shares may be as 1, 6, 7, and 18.—Answer, £12. 6. 7½; £73. 19. 9; £86. 6. 4½; £221. 19. 3.
Divide £20 among 6 persons, so that the share of each may be as much as those of all who come before put together.—Answer, The first two have 12s.6d.; the third £1. 5; the fourth £2. 10; the fifth £5; and the sixth £10.
255. When two or more persons employ their money together, and gain or lose a certain sum, it is evidently not fair that the gain or loss should be equally divided among them all, unless each contributed the same sum. Suppose, for example, A contributes twice as much as B, and they gain £15, A ought to gain twice as much as B; that is, if the whole gain be divided into 3 parts, A ought to have two of them and B one, or A should gain £10 and B £5. Suppose that A, B, and C engage in an adventure, in which A embarks £250, B £130, and C £45. They gain £1000. How much of it ought each to have? Each one ought to gain as much for £1 as the others. Now, since there are 250 + 130 + 45, or 425 pounds embarked, which gain £1000, for each pound there is a gain of £¹⁰⁰⁰/₄₂₄. Therefore A should gain 1000 ײ⁵⁰/₄₂₅pounds, B should gain 1000 ×¹³⁰/₄₂₅pounds, and C 1000 ×⁴⁵/₄₂₅pounds. On these principles, by the process in (245), the following questions may be answered.
A ship is to be insured, in which A has ventured £1928, and B £4963. The expense of insurance is £474. 10. 2. How much ought each to pay of it?
Answer, A must pay £132. 15. (2½).
A loss of £149 is to be made good by three persons, A, B, and C. Had there been a gain, A would have gained 4 times as much as B, and C as much as A and B together. How much of the loss must each bear?
Answer, A pays £59. 12, B £14. 18, and C £74. 10.
256. It may happen that several individuals employ several sums of money together for different times. In such a case, unless there be a special agreement to the contrary, it is right that the more time a sumis employed, the more profit should be made upon it. If, for example, A and B employ the same sum for the same purpose, but A’s money is employed twice as long as B’s, A ought to gain twice as much as B. The principle is, that one pound employed for one month, or one year, ought to give the same return to each. Suppose, for example, that A employs £3 for 6 months, B £4 for 7 months, and C £12 for 2 months, and the gain is £100; how much ought each to have of it? Now, since A employs £3 for six months, he must gain 6 times as much as if he employed it one month only; that is, as much as if he employed £6 × 3, or £18, for one month; also, B gains as much as if he had employed £4 × 7 for one month; and C as if he had employed £12 × 2 for one month. If, then, we divide £100 into 6 × 3 + 4 × 7 + 12 × 2, or 70 parts, A must have 6 × 3, or 18, B must have 4 × 7, or 28, and C 12 × 2, or 24 of those parts. The shares of the three are, therefore,
EXERCISES.
A, B, and C embark in an undertaking; A placing £3. 6 for 2 years, B £100 for 1 year, and C £12 for 1½ years. They gain £4276. 7 How much must each receive of the gain?
Answer, A £226. 10. 4; B £3432. 1. 3; C £617. 15. 5.
A, B, and C rent a house together for 2 years, at £150 per annum. A remains in it the whole time, B 16 months, and C 4½ months, during the occupancy of B. How much must each pay of the rent?[56]
Answer, A should pay £190. 12. 6; B £90. 12. 6; C £18. 15.
257. These are the principal rules employed in the application of arithmetic to commerce. There are others, which, as no one who understands the principles here laid down can fail to see, are virtually contained in those which have been given. Such is what is commonly called the Rule of Exchange, for such questions as thefollowing: If 20 shillings be worth 25½ francs, in France, what is £160 worth? This may evidently be done by the Rule of Three. The rules here given are those which are most useful in common life; and the student who understands them need not fear that any ordinary question will be above his reach. But no student must imagine that from this or any other book of arithmetic he will learn precisely the modes of operation which are best adapted to the wants of the particular kind of business in which his future life may be passed. There is no such thing as a set of rules which are at once most convenient for a butcher and a banker’s clerk, a grocer and an actuary, a farmer and a bill-broker; but a person with a good knowledge of theprincipleslaid down in this work, will be able to examine and meet his own future wants, or, at worst, to catch with readiness the manner in which those who have gone before him have done so for themselves.