4. 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. $7,163cts.

5. A grocer would mix 12cwt. of sugar at 10 dollars per cwt. with 3cwt. at 83 dollars per cwt. and Scwt. at 74 dollars per cwt. ; what will a cwt. of this mixture be worth? Ans. $8,95cts. 6mills. +

6. If 16 gallons of wine at 1 dollar 25 cents, and 4 gallons of water, be mixed with 40 gallons of wine at 3 dollars per gallon; what will the mixture be worth per gallon? Ans. $2,33 cts.

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

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.

Write the difference between the mixture rate, and that of each of the simple, opposite to the rates, with which they are respectively linked.

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. 15s. 19s. and 22s. per gall. so that the mixture may be worth 18s. per gallon; what quantity of each must be taken?

2. 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. & 18 at 4s. 8d. 3. A goldsmith has gold of 18 carats fine, 16, 19, 22 and 24; how much must he take of each to make it 21 carats fine? 3oz. of 16, 1oz. of 18, 1oz. of 19,

Ans. {30z. of 20, 107501. 1824 Carats fine:

4. It is required to mix wine at 80 cents, wine at 70 cents, cider at 10cts. and water together, so that the mixture may be worth 50cts. per gallon.

Ans. 9gals. of each sort of wine, 5 of cider & 5 of water.

CASE 2.-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, to the required quantity of each.

EXAMPLES.

1. How many gallons of water at Octs. per gallon, must be mixed with wine worth 60cts. per gallon so as to fill a cask of 100 gallons, and that a gallon may be afforded at 50 cts.?

Ans. 16 gallons of water, and 833 of wine. 2. How much gold of 15, of 17, of 18 and 22 carats fine, must be mixed together to form a composition of 40 ounces of 20 carats fine?

Ans. 5oz. of 15, 17 and 18, and 25oz. of 22. 3. Wine at 3s. 6d. and at 5s. 9d. per gallon, is to be mixed, so that a hogshead of 63 gallons may be sold for £12 12s.; how many gallons must be taken of each? Ans. 14gals. at 5s. 9d. and 49gals. at 3s. 6d.

CASE 3.-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.

1. A grocer would mix teas at 1 dollar 20cts. 66cts. and 1 dollar per pound, with 20 pounds at 40 cents per pound; how much of each sort must he take to make the composition worth 80 cents per pound?

120 40 40 40 20: 20 at 1,20

:

2. How much wine at 80cts. at 88 and 92 per gallon, must be mixed with 4 gallons at 75cts. per gallon, so that the mixture may be worth 86 cents per gallon?

Ans. 4gals. at 80cts. 84 at 88 and 84 at 92.

3. With 95 gallons of wine at 8s. per gallon, I mixed other wine at 6s. 8d. per gallon, and some water; then I found it stood me in 6s. 4d. per gallon ;-I demand how much wine at 6s. 8d. I took, and how much water.

Ans. 95 gallons wine at 6s. 8d. and 30 gallons water.

POSITION.

POSITION is a method of performing such questions as cannot be resolved by the common direct rules, and is of two kinds, Single and Double.

SINGLE POSITION.

Single Position teaches to resolve those questions whose results are proportioned to their suppositions. RULE.-1. Take any number and perform the same operations with it, as are described to be performed in the question..

2. Then say, as the result of the operation is to the position,so is the result in the question to the number required.

EXAMPLES.

1. A's age is double that of B, and B's is triple that of C, and the sum of all their ages is 140; what is the age of each? Suppose A's age to be 48

Then will B's=48=24

And C's 28

2. A certain sum of money is to be divided between 4 persons in such a manner that the first shall have of it, the second, the third, and the fourth the remainder, which is 28 dollars; what is the sum ? 3. A person, after spending 60 dollars left; what had he at first? 4. What number is that which being increased by 1, 3, and of itself, the sum will be 125?

Ans. $112. and of his money, had

Ans. $144.

Ans. 60.

5. A person lent his friend a sum of money, to receive interest for the same at 6 per cent. per annum, simple interest; at the end of three years he received for principal and interest 383 dollars 50 cents; what was the sum lent ? Ans. 325 dollars.

6. A cistern is supplied with three cocks, A, B, and C : A can fill it in 1 hour, B in 2, and C in 3; in what time will it be filled by all of them together? Ans. r hour.

DOUBLE POSITION.

DOUBLE POSITION teaches to resolve questions by making two suppositions of false numbers.*

* Questions in which the results are not proportional to their positions, belong to this rule; such are those, in which the number sought is increased or diminished by some given number, which is no known part of the number required.

RULE.-1. Take any two convenient numbers, and proceed with each according to the conditions of the question.

2. Find how much the results are different from the result in the question.

3. Multiply each of the errours by the contrary suppo

sition.

4. If the errours be alike, divide the difference of the products by the difference of the errours, and the quotient will be the answer.

5. If the errours be unlike, divide the sum of the products by the sum of the errours, and the quotient will be the answer.

NOTE. The errours are said to be alike, when they are both too great, or both too little; and unlike, when one is too great, and the other too little.

EXAMPLES.

1. A lady bought cambric for 40 cents a yard, and India cotton at 20 cents a yard; the whole number of yards she bought was 8, and the whole cost 2 dollars; how many yards had she of each sort?

Suppose 4 yards of cambric, value $1,60cts. Then she must have 4 yards of cotton, value 80

Sum of their values, 2,40

So that the first errour is +40

Again, suppose she had 3 yards of cambric, $1,20cts. Then she must have 5 yards of India cotton 1,00

Sum of their values, 2,20

So that the second errour is +20

Then 40-20-20=difference of the errours.
Also 4×20=80=product of the first supposition and
second errour.

And 3×40=120=product of the second supposition and first errour.

And 120-80=40=their difference.
Whence 40+20-2 yards of cambric.
And 8-2-6 yards of India cotton.

Ans.

2. A and B have both the same income; A saves of his yearly; but B, spending 50 dollars a year more than