6. If I mix 27 bushels of wheat, at 5s. 6d. the bushel, with the same quantity of rye, at 4s. per bushel, and 14 bushels of barley at 2s. 8d. per bushel : what is the worth of a bushel of the mixture? Ans. 4s. 3 d.§.

7. A grocer mingled 3 cwt. of sugar, at 56s. per cwt., 6 cwt. at £1 17 4 per cwt., and 3 cwt. at £3 14 8 per cwt. : what is 1 cwt. of this mixture worth? Ans. £2 11 4.

8. A mealman has flour of several sorts, and would mix 3 bushels at 3s. 5d. per bushel, 4 bushels at 5s. 6d. per bushel, and 5 bushels at 4s. 8d. per bushel: what is the worth of a bushel of this mixture? Ans. 48. 7d..

9. A vintner mixes 20 gallons of Port, at 5s. 4d. per gallon, with 12 gallons of white wine, at 5s. per gallon, 30 gallons of Lisbon, at 6s. per gallon, and 20 gallons of Mountain, at 4s. 6d per gallon: what is a gallon of this mixture worth? Ans. 5s. 3 d..

10. A farmer mingled 20 bushels of wheat, at 5s. per bushel, and 36 bushels of rye, at 3s. per bushel, with 40 bushels of barley, at 2s. per bushel : I desire to know the worth of a bushel of this mixture ? Ans. 3s.

11. A person mixing a quantity of oats, at 2s. 6d. per bushel, with the like quantity of beans, at 4s. 6d. per bushel, would be glad to know the value of 1 bushel of that mixture Ans. 3s. 6d.

12. A refiner, having 12 lb. of silver bullion of 6 oz. fine, would melt it with 8 lb. of 7 oz. fine, and 10 lb. of 8 oz. fine required the fineness of 1 lb. of that mixture?

Ans. 6 oz. 18 dwt. 16 grs.

13. If with 40 bushels of corn, at 4s. per bushel, there are mixed 10 bushels, at 6s. per bushel, 30 bushels at 5s. per bushel, and 20 bushels, at 3s. per bushel : what will 10 bushels of that mixture be worth? Ans. £2 3.

ALLIGATION ALTERNATE.

ALLIGATION ALTERNATE is the method of finding what quantity of any number of simples, whose rates are given, will compose a mixture of a given rate: so that it is the reverse of Alligation Medial, and may be proved by it.

RULE. 1. Write the rates of the simples in a column under each other.

2. Connect or link with a continued line the rate of each simple, which is less than that of the compound, with one, or any number, of those that are greater than the compound, and each greater rate with one or any number of the less.

3. Write the difference between the mixture rate and that of each of the simples, opposite the rates with which they are linked.

4. Then if only one difference stand against any rate, it will be the quantity belonging to that rate; but if there be several, their sum will be the quantity.

EXAMPLES.

1. A merchant would mix wines at 14s. 19s. 15s. and 22s. per gallon, so that the mixture may be worth 188. the gallon what quantity of each must be taken ?

NOTE. Questions in this rule admit of a great variety of answers, according to the manner of linking them.

2. How much wine at 6s. per gallon, and at 4s. per gallon, must be mixed together, that the composition may be worth 5s. per gallon? Ans. 1 qt. or 1 gal. of each, &c.

3. How much corn, at 2s. 6d., 3s. 8d., 4s., and 4s. 8d. per bushel, must be mixed together, that the compound may be worth 3s. 10d. per bushel?

Ans. 12 at 2s. 6d., 12 at 3s. 8d., 18 at 4s., and 18 at 4s. 8d. 4. A goldsmith has gold of 17, 18, 22, and 24 carats fine; how much must he take of each to make it 21 carats fine? Ans. 3 of 17, 1 of 18, 3 of 22, and 4 of 24.

5. It is required to mix brandy at 8s., wine at 7s., cider at 1s., and water, together, so that the mixture may be worth 58. per. gallon?

Ans. 9 gals. brandy, 9 of wine, 5 of cider, and 5 of water.

When the whole composition is limited to a certain quantity.

RULE. Find an answer, as before, by linking; then say, As the sum of the quantities, or differences, thus determined, is to the given quantity, so is each ingredient found by linking, to the required quantity of each.

EXAMPLES.

6. How many gallons of water must be mixed with wine worth 3s. per gallon, so as to fill a vessel of 100 gallons, and that a gallon may be afforded at 2s. 6d.?

Ans. 83 gals. of wine, and 16 of water.

7. A grocer has currants at 4d., 6d., 9d. and 11d. per lb. and he would make a mixture of 240 lbs. so that it might be afforded at 8d. per lb.: how much of each sort must he take?

Ans. 72 lbs. at 4d., 24 at 6d., 48 at 9d. and 96 at 11d. 8. How much gold of 15, of 17, of 1c and of 22 carats fine, must be mixed together to form a composition of 40 oz. of 20 carats fine?

Ans. 5 oz. of 15, of 17 and of 18, and 25 oz. of 22. When one of the ingredients is limited to a certain quantity. RULE. Take the difference between each price and the mean rate, as before; then,

As the difference of that simple, whose quantity is given, is to the rest of the differences severally, so is the quantity given to the several quantities required.

EXAMPLES.

9. How much wine, at 5s., at 5s. 6d., and at 6s. the gal. must be mixed with 3 gallons, at 4s. per gallon, so that the mixture may be worth 58. 4d. the gallon?

8+2=10 8-2-10 16-4-20 16-4-20

3 : 3

3 : 6
6

10 : 20 :: 3

Ans. 3 gals. at 5s., 6 at 5s. 6d., and 6 at 6s.

10. A grocer would mix teas at 12s., 10s., and 6s., with 20 lbs. at 4s. per lb.: how much of each sort must he take to make the composition worth 8s. per lb.?

Ans. 20 lbs. at 4s., 10 at 6s., 10 at 10s., and 20 at 12s.

11. How much gold of 15, of 17, and of 22 carats fine, must be mixed with 5 oz. of 18 carats fine, so that the composition may be 20 carats fine?

Ans. 5 oz. of 15 carats fine, 5 oz. of 17, and 25 of 22.

POSITION

Is a rule which, by false or supposed numbers, taken at pleasure, discovers the true one required. It is divided into two parts; SINGLE and DOUBle.

SINGLE POSITION

Is by using one supposed number, and by working with it as the true one, you find the real number required, by the following

RULE. As the total of the errors is to the given sum, so is the supposed number to the true one required.

PROOF. Add the several parts of the result together, and if it agrees with the given sum it is right.

EXAMPLES.

1. A schoolmaster, being asked how many scholars he had, said, If I had as many, half as many, and one quarter as many more, I should have 264: how many had he?

2 A person, after spending

and of his money, had 60 Ans. 144 dols.

per

dollars left; what had he at first? 3. A certain sum of money is to be divided between 4 sons, in such a manner, that the first have of it, the second 4, the third, and the fourth the remainder, which is $28; what was the sum? Ans. 112 dols.

4. A person lent his friend a sum of money unknown, to receive interest for the same, at 6 per cent. per annum, simple interest, and at the end of 5 years he received for principal and interest 644 dollars 80 cents; what was the sum lent? Ans. 496 dols.

DOUBLE POSITION

Is by making use of two supposed numbers, which, if both prove false, are, with their errors, to be thus disposed: RULE. 1. Place each error against its respective position. 2. Multiply them crosswise.

3. If the errors are alike, that is, both greater or both less than the given number, divide the difference of the products by the difference of the errors, and the quotient is the answer; but if the errors be unlike, divide the sum of the products by the sum of the errors, and the quotient will be the answer.

EXAMPLES.

1. B. asked C. how much his horse cost; C. answered, that if he cost him three times as much as he did, and $15