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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. 100 x 6 Therefore (245) their shares are respectively, £- £ and

20

100 X 5

20

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Divide £394. 12 among four persons, so that their shares may be as 1, 6, 7, and 18.—Answer, £12.6.7-; £73. 19.9; £86. 6. 6.4/2/ 4-; £221. 19.3.

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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 1000 × 250 pound there is a gain of £- Therefore A should gain

1000 425 1000 X 130 425

425

pounds, and C

1000 × 45

pounds.

425

pounds, B should gain 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. 2A loss of £149 is to be made good by three persons, A, B, and C.

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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 is a special agreement to the contrary, it is right that the more time a sum is 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 6 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

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A, B, and C embark in an undertaking; A placing £3

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2 years, B £100 for 1 year, and C £12 for 1 years. They gain £4276. 7. How much must each receive of the gain?

Answer, A £226. 10. 4; B £3432. I 3; C £617. 15. 5.

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A, B, and C rent a house together for two years, at £150 per annum. A remains in it the whole time, B, 16 months, and C, 4 months, during the occupancy of B. How much must each pay of the Answer, A should pay £190. 12. 6; B £90. 12 6; C £18. 15.

rent?

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*

* This question does not at first appear to fall under the rule. A little thought will serve to shew, that what probably will be the first idea of the proper method of solution is erroneous.

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 the following: If 20 shillings be worth 25 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

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understands them need not fear that any ordinary question will be above his reach.

APPENDIX.

RULES FOR THE APPLICATION OF ARITHMETIC TO GEOMETRY.

258. The student should make himself familiar with the most common terms of Geometry, after which the following rules will present no difficulty. In them all, it must be understood, that when we talk of multiplying one line by another, we mean the repetition of one line as often as there are units of a given kind, as feet or inches, in another. In any other sense, it is absurd to talk of multiplying a quantity by another quantity. All quantities of the same kind should be represented in numbers of the same unit; thus, all the lines should be either feet and decimals of a foot, or inches and decimals of an inch, &c. And in whatever unit a length is represented, a surface is expressed in the corresponding square units, and a solid in the corresponding cubic units. This being understood, the rules apply to all

sorts of units.

259. To find the area of a rectangle. Multiply together the units in two sides which meet, or multiply together two sides which meet; the product is the number of square units in the area. Thus, if 6 feet and 5 feet be the sides, the area is 6 × 5, or 30 square feet. Similarly, the area of a square of six feet long is 6 × 6, or 36 square feet (234).

To find the area of a parallelogram. Multiply one side by the perpendicular distance between it and the opposite side; the product is the area required in square units.

To find the area of a trapezium.* Multiply either of the two sides

A four-sided figure, which has two sides parallel, and two sides not parallel.

which are not parallel, by the perpendicular let fall upon it from the middle point of the other.

To find the area of a triangle. Multiply any side by the perpendicular let fall upon it from the opposite vertex, and take half the product. Or, halve the sum of the three sides, subtract the three sides severally from this half sum, multiply the four results together, and find the square root of the product. The result is the number of square units in the area; and twice this, divided by either side, is the perpendicular distance of that side from its opposite vertex.

To find the radius of the internal circle which touches the three sides of a triangle. Divide the area, found in the last paragraph, by half the sum of the sides.

260. Given the two sides of a right-angled triangle, to find the hypothenuse. Add the squares of the sides, and extract the square root of the sum.

Given the hypothenuse and one of the sides, to find the other side. Multiply the sum of the given lines by their difference, and extract the square root of the product.

261. To find the circumference of a circle from its radius, very nearly. Multiply twice the radius, or the diameter, by 3.1415927, taking as many decimal places as may be thought necessary. For a rough computation, multiply by 22 and divide by 7. For a very exact computation, in which decimals shall be avoided, multiply by 355 and divide by 113. See (131, last example).

To find the arc of a circular sector, very nearly, knowing the radius and the angle. Turn the angle into seconds,* multiply by the radius, and divide the product by 206265. The result will be the number of units in the arc.

To find the area of a circle from its radius, very nearly. the square of the radius by 3.1415927.

Multiply

* The right angle is divided into 90 equal parts called degrees, each degree into 60 equal parts called minutes, and each minute into 60 equal parts called seconds. Thus, 2° 15′ 40′′ means 2 degrees, 15 minutes, and 40 seconds.

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