2, 2, and 5, will satisfy the question equally well; so that in Alligation Alternate the number of solutions are indefinite; all that we can do is to find the ratios of the quantities required.

NOTE.-In many cases the ingredients will admit of being connected in several ways, and then we shall obtain as many sets of ratios as there are methods of connecting them.

2. How many pounds of raisins at 4, 6, 8, and 10 cents per pound, must be mixed, so that a pound of the compound may be worth 7 cents?

In this question, the terms may be connected in seven distinct ways; therefore, we shall obtain seven sets of ratios, as follows:

3. How much wine, at 72 cents per gallon, and 48 cents per gallon, must be mixed together, that the composition may be worth 60 cents per gallon?

Ans. An equal quantity of each. 4. How many gallons of wine and water must be mixed together, so that the mixture may be worth 60 cents per

gallon, the water being considered of no value, and the wine with which it is mixed being worth 90 cents per gallon? Ans. 2 gallons of wine to 1 of water. 5. Having gold of 12, 16, 17, and 22 carats fine, what proportion of each kind must I take, to make a compound of 18 carats fine? Ans. 4, 4, 4, 9. 6. It is required to mix different sorts of grain, at 56, 62, and 75 cents per bushel, so that the mixture may be worth 60 cents per bushel. How much of each kind must Ans. 17, 4, 4. Besides the variety of answers which may be obtained by connecting the simples differently, an infinite number of solutions may be found, by combining the different ratios, as we will illustrate by the aid of the following question:

be taken ?

7. How much tea at 5 shillings, 6 shillings and 8 shillings per pound, must be mixed so that the mixture may be worth 7 shillings per pound?

If we compound only the 5 and 8 shilling teas, we must take them in the ratio of 1 to 2, since 7 shillings is 1 shilling less than 8 shillings, and 2 shillings greater than 5 shillings. Hence, any one of the compounds in the fol lowing group (A,) will be worth 7 shillings per pound. (1) (2) (3) (4) (5) (6)

If we now mix the 6 and 8 shilling teas, we see that it will be necessary to take equal quantities of each, since the average price is to be as much above 6 shillings as it is below 8 shillings. Hence, the following compound will also be worth 7 shillings per pound:

Now, it is obvious, we may combine any one of these last results with any one of the former results. Thus, if we combine (1) of group (A) with (1) of (B) we have

If we combine (1) of (A,) with (2) of (B,) we have,

Combining (2) of (A,) with (3) of (B,) we have,

Combining (5) of (A,) with (4) of (B,) we have,

The number of combinations which could be made in this way is unlimited; hence, the above class of questions in Alligation admit of an infinite number of answers.

CASE II.

When one of the ingredients is limited to a certain quantity.

1. A person wishes to mix 10 bushels of wheat, worth $1 per bushel, with rye, worth 70 cents per bushel, and oats worth 30 cents per bushel, so that the mixture may be worth 60 cents per bushel. How many bushels of rye

and oats must he use?

Proceeding, according to Case I., we find the propor tionate numbers to be 30, 30, and 50.

Hence,

So that he must make use of 10 bushels of rye, and 16 bushels of oats. Hence, this

RULE.

Find the proportionate quantities of each ingredient, by Case I., in the same manner as though there was no limitation; then, as the difference against the simple whose quantity is given, is to each of the other differences, so is the given quantity of that simple to the quantity required of each of the other simples.

Repeat this Rule.

2. A grocer has 90 pounds of tea, worth 90 cents per pound, which he wishes to mix with three other qualities, valued at 80 cents, 70 cents, and 60 cents per pound. How much must he take of these three kinds, so as to be able to sell the mixture at 85 cents per pound?

Ans. 10 pounds of each.

3. A merchant has 90 pounds of spice worth 86 cents per pound, which he wishes to mix with three other sorts which are worth 30, 40, and 50 cents per pound, respectively. How many pounds must be used so that the compound may be worth 55 cents per pound?

Ans. He must use 62 pounds of each.

MENSURATION.

143. For the reason of many of the rules which we shall give for measuring surfaces and solids, we shall refer to the principles of geometry. The reference being in all cases to the "Elements of Geometry."

PROBLEM I-To find the area of a rectangle.

Suppose ABCD to be a rectangle whose length is 5 feet, and width 3 feet.

If we divide this rectangle into portions of

D

A

B

one square foot each, by means of lines drawn parallel to the sides of the rectangle, we shall obtain 15 such squares; that is, the rectangle will contain 15 square feet. In this example there are 3 strips of 5 square feet in each, or 5 strips of 3 square feet each. So that the number of square feet is found by multiplying the number of feet in length by the number of feet in width..

Hence, to find the area of a rectangle we have this

RULE.

Multiply the length by the width, and the product will de note the number of squares of the same kind as the measure used in estimating the sides of the rectangle. If the sides of the rectangle are measured in feet, the product will be the