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21. A bankrupt owes $150000, to pay which he has the following property real estate worth $17500, merchandise worth $3750, personal property worth $20000, and sums due from various individuals amounting to $100000. If he collects all his dues, how much can he pay on every dollar of his debts?

22. A., B. and C. purchase a house for $10000, A. contributing $4000, B. $3600, and C. $2400. The house rents for $800, and the taxes and repairs amount to $50 a year. What income does each of the owners derive from the estate, and what per cent. of his investment?

23. Divide 7500 into 5 parts, in the proportions of, ,,, and .

24. In the distribution of a bankrupt's property, $763. 50 was divided among four of his creditors. A.'s bill was $250, B's. $300, C.'s $325, and D.'s $362.50. How much did each receive?

25. Four men were joint owners of a farm, their shares being in the proportion of 6, 2, 5, and 3, respectively. It is required to divide the annual rent, which is $1500, equitably among them.

CHAPTER XIII.

ALLIGATION.

When the values of a variety of ingredients are given, and it is desired to make a mixture of any fixed value, the quantity of each ingredient to be employed is determined by ALLIGATION.

When there is no limitation in the quantity of either ingredient, we may make the desired mixture by the fol lowing

RULE.

Having written the values of the ingredients in a perpendicular column, connect by a line each value that is less than the required average with one or more that is greater, and each value that is greater with one or more that is less.

Write the difference between each value and the average, op

posite the ingredient with which that value is connected, and the difference (or the sum of the differences, if there be more than one) opposite each ingredient will be the quantity of that ingredient required.

EXAMPLE FOR THE BOARD.

How much sugar, at 5cts., 7cts., 8cts., 10cts., and 12cts., must be mixed together, that the mixture may be worth 9cts. a pound?

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1st Ans. 4lb. at 5, 1 lb. at 7, 4lb. at 8, 7lb. at 10, 5lb. at 12.

2d Ans. 3lb. at 5, 3lb. at 7, 1lb. | 3d Ans. 1lb. at 5, 3lb. at 7, 3lb. at 8,
at 8, 1lb. at 10, 6lb. at 12.
4lb. at 10, 3lb. at 12.

We may obtain as many answers as there are different ways of connecting the numbers above, with those below the average. To prove the rule correct, let us examine the second of the above answers. If we were mixing sugars at 5 and 12cts. to sell the mixture at 9cts., we should gain 4cts. on every pound of the former, and lose 3cts. on every pound of the latter. Then, on 3lb. of the former we should gain 12cts. and on 4lb. of the latter we should lose 12cts.; therefore, if we mix these quantities, we shall neither gain nor lose by selling the mixture at 9cts. In the same way it may be shown that 3lb. at 7cts. and 2lb. at 12cts., 1lb. at 8cts. and 1lb. at 10cts. may be sold at the average of 9cts., and the same reasoning will prove the truth of each of the other

answers.

1. How much tea at 50cts. 75cts. 90cts. and $1.00, must be taken to form a mixture worth 80cts.

2. A jeweller has gold of 16, 17, 18, 20, 22, and 24 carats fine. What proportion of each must he take to make a mixture 21 carats fine? Pure gold is 24 carats fine, or 34.

3. How much grain, at 50cts. 75cts. $1.00, and $1.10 per bushel, will make a mixture worth 90cts. a bushel?

4. How much water must be mixed with wine at $1.50 and $2.00 a gallon, to make the whole worth $1.00 per gallon?

5. What quantity of raisins, at 10cts. 18cts. and 20cts. per lb. must be mixed together, to fill a cask containing 150lb. and to be worth 19cts. a lb. [After obtaining the

proportions by Alligation, find the exact quantities by Fellowship.]

6. It is required to mix sugar at 7cts. 8cts. 10cts. and 12cts. per lb. in such manner as to form a mixture of 2cwt. 3qr., worth 11cts. per lb.

7. Mix tobacco at 8cts. 10cts. 12cts. and 16cts. so as to make 100 pounds worth 11cts. a pound.

EXAMPLE FOR THE BOARD.

A farmer wishes to mix 10 bushels of barley at 50cts., 4 bushels of oats at 45cts., and 16 bushels of rye at 75cts. with wheat at $1.25 and corn at 90cts. a bushel, so that the mixture may be worth $1.00 per bushel.

621.00 90

25

25

1.25

We We may regard the limited quantities as a single ingredient of 30 bushels, worth 62cts. a bushel. Proceeding in the usual way, we find that 25 bushels at 62cts., 25 at 90cts., and 48 at $1.25, would give us 38+10 a mixture of the desired average value. But as we have 30 bushels at 62cts., we must take or of these proportionate quantities, and we have 30 bushels at 90cts. and 573bu. at $1.25, for the Answer.

8. How much water must be mixed with 40 gallons of syrup, at 50cts. a gallon, to make the whole worth 37 cts. a gallon?

9. A grocer has 10 gallons of wine at 75cts., 12gal. at 90cts., and 8gal. at $1.00, with which he would mix brandy at $1.25, and water, so as to make a mixture worth 95cts, a gallon. How much of each must he take?

10. If a cubic inch of gold weighs 11.11oz. and a cubic inch of silver 6.04oz., what quantity is contained in a mass of 63oz., of which 1 cubic inch weighs 8.35oz.?

11. How much molasses, at 50 cents, and water, must be mixed with 15 gallons at 37 cents, 28 gallons at 25 cents, and 19 gallons at 33 cents, to make a mixture worth 31 cents a gallon?

12. A grocer has an order for 150lb. of tea, at 90 cents per lb., but having none at that price, he would mix some at 75 cents, some at 87 cents, and some at $1.00 per pound. How much of each sort must he take?

CHAPTER XIV.

PERMUTATION AND COMBINATION.

PERMUTATION shows the number of changes that can be made in the order of a given number of things. PROBLEM I.

To find the number of changes that can be made of any given number of things, all different from each other. How many changes may be made in the position of 4 persons at table?

If there were but two persons, a and b, they could sit in but two positions, ab and ba. If there were three, the third could sit at the head, in the middle, or at the foot, in each of the two changes, and there could then be 1×2×3=6 changes. If there were 4, the fourth could sit as the 1st, 2d, 3d, or 4th, in each of these 6 changes, and there would then be 1×2×3×4=24 changes.

RULE.

Multiply together the series of numbers 1, 2, 3, &c., up to the given number, and the product will be the number sought.

1. How many variations may be made in the order of the 9 digits?

2. How many changes may be made in the position of the letters of the alphabet?

3. How long a time will be required for 8 persons to seat themselves at table in every possible order, if they eat 3 meals a day?

PROBLEM II.

Any number of different things being given, to find how many changes can be made out of them by taking a given number of the things at a time.

If we have five things, each one of the 5 may be placed before each of the others, and we thus have 5×4 permutations of 2 out of 5. If we take 3 at a time, the third thing may be placed as 1st, 2d, and 3d, in each of these permutations, and we have 5X4×3 permutations of 3 out of 5. For a similar reason we have 5×4×3×2 permutations of 4 out of 5, &c.

RULE.

Take a series of numbers, commencing with the number of things given, and decreasing by 1, until the number of terms is

equal to the number of things to be taken at a time: the product of all the terms will be the answer required.

4. How many changes can be rung with 8 bells, taking 5 at a time?

5. How many numbers of 4 different figures each, can be expressed by the 9 digits?

6. In how many different ways may 10 letters of the alphabet be arranged?

PROBLEM III.

To find the number of permutations in any given number of things, among which there are several of a kind. How many permutations can be made of the letters in the word terrier?

If the letters were all different, the permutations, according to Problem I. would be 1×2×3×4×5X6X7 5040. But the permutations of the three r's would, if they were all different, be 1×2×3, which could be combined with each of the other changes; the number must therefore be divided by 1×2×3. For the same reason it must also be divided by 1×2, on account of the 2 e's. 1X2×3×4×5×6×7 Then the true number sought is 1×2×3×1X2

RULE.

=420.

Take the natural series, from 1 up to the number of things of the first kind, and the same series up to the number of things of each succeeding kind, and form the continued product of all the series.

By the continued product divide the number of permutations of which the given things would be capable if they were all different, and the quotient will be the number sought.

7. How many changes can be made in the order of the letter's in the word Philadelphia ?

8. How many different numbers can be made, that will employ all the figures in the number 119089907343?

9. How many permutations can be made with the letters in the word Cincinnati ?

COMBINATION shows in how many ways a less number of things may be chosen from a greater.

If we have ten articles, each may be combined with every one of the nine remaining ones, and therefore we may have 10×9 permutations of 2 out of 10. But each combination will evi

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