the name of a method of solving all questions that relate to the mixture of one ingredient with another. Though writers on arithmetic generally make alligation a branch of that science, yet as it is plainly nothing more than an application of the common properties of numbers, in order to solve a few questions that occur in particular branches of business, we choose rather to keep it distinct from the science of arithmetic.

Alligation is generally divided into medial and alternate.

ALLIGATION Medial, from the rates and quantities of the simples given, discovers the rate of the mixture.

Rule. As the total quantity of the simples, To their price or value; So any quantity of the mixture, To the rate.

Example. A grocer mixes 30 lb. of currants, at 4d. per lb., with 10 lb. of other currants, at 6d per lb.: What is the value of 1 lb. of the mixture? Ans. 4½d.

| lb. | d. | |-----|----| | 30 | 4 | | 10 | 6 |

If 40 : 180 :: 1 : 4½.

ALLIGATION Alternate, being the converse of alligation medial, from the rates of the simples, and rate of the mixture given, finds the quantities of the simples.

Rules. 1. Place the rate of the mixture on the left side of a brace, as the root; and on the right side of the brace set the rates of the several simples, under one another, as the branches. II. Link or alligate the branches, so as one greater and another less than the root may be linked or yoked together. III. Set the difference between the root and the several branches right against their respective yoke-fellows. These alternate differences are the quantities required. Note 1. If any branch happen to have two or more yoke-fellows, the difference between the root and these yoke-fellows must be placed right against the said branch, one after another, and added into one sum. 2. In some questions the branches may be alligated more ways than one; and a question will always admit of so many answers as there are different ways of linking the branches.

Alligation alternate admits of three varieties, viz., 1. The question may be unlimited, with respect both to the quantity of the simples and that of the mixture. 2. The question may be limited to a certain quantity of one or more of the simples. 3. The question may be limited to a certain quantity of the mixture.

Variety I. When the question is unlimited, with respect both to the quantity of the simples and that of the mixture, this is called Alligation Simple.

Example. A grocer would mix sugars at 5d., 7d., and 10d. per lb., so as to sell the mixture or compound at 8d. per lb.: What quantity of each must he take?

\[ \begin{array}{c|c} \text{lb.} & \text{d.} \\ \hline 8 & 5 \\ 7 & 7 \\ 10 & 3.1 \\ \end{array} \] Alligation. Here the rate of the mixture 8 is placed on the left side of the brace as the root; and on the right side of the same brace are set the rates of the several simples, viz., 5, 7, 10, under one another, as the branches; according to Rule I.

The branch 10 being greater than the root, is alligated or linked with 7 and 5, both these being less than the root, as directed in Rule II.

The difference between the root 8 and the branch 5, viz., 3, is set right against this branch's yoke-fellow 10; the difference between 8 and 7 is likewise set right against the yoke-fellow 10; and the difference between 8 and 10, viz., 2, is set right against the two yoke-fellows 7 and 5, as prescribed by Rule III.

As the branch 10 has two differences on the right, viz., 3 and 1, they are added; and the answer to the question is, that 2 lb. at 6d., 2 lb. at 7d., and 4 lb. at 10d., will make the mixture required.

The truth and reason of the rules will appear by considering, that whatever is lost upon any one branch is gained upon its yoke-fellow. Thus, in the above example, by selling 4 lb. of 10d. sugar at 8d. per lb. there is 8d. lost; but the like sum is gained upon its two yoke-fellows, for by selling 2 lb. of 5d. sugar at 8d. per lb., there is 6d. gained; and by selling 2 lb. of 7d. sugar at 8d. there is 2d. gained; and 6d. and 2d. make 8d.

Hence it follows, that the rate of the mixture must always be mean or middle with respect to the rates of the simples; that is, it must be less than the greatest, and greater than the least; otherwise a solution would be impossible. And the price of the total quantity mixed, computed at the rate of the mixture, will always be equal to the sum of the prices of the several quantities cast up at the respective rates of the simples.

Variety II. When the question is limited to a certain quantity of one or more of the simples, this is called Alligation Partial.

If the quantity of one of the simples only be limited, alligate the branches, and take their differences, as if there had been no such limitation; and then work by the following proportion:

As the difference right against the rate of the simple whose quantity is given, To the other differences respectively; So the quantity given, To the several quantities sought.

Example. A distiller would, with 40 gallons of brandy at 12s. per gallon, mix rum at 7s. per gallon, and gin at 4s. per gallon: How much of the rum and gin must he take, to sell the mixture at 8s. per gallon?

Galls. \[ \begin{array}{ccc} 12 & 4 & 5 \\ 7 & 4 & 4 \\ 4 & 4 & 4 \\ \end{array} \]

The operation gives for answer, 5 gallons of brandy, 4 of rum, and 4 of gin. But the question limits the quantity of brandy to 40 gallons; therefore say,

If \(5 : 4 :: 40 : ?\)

The quantity of gin, by the operation, being also 4, the proportion needs not be repeated.

Variety III. When the question is limited to a certain quantity of the mixture, this is called Alligation Total.

After linking the branches, and taking the differences, work by the proportion following:

As the sum of the differences, To each particular difference; So the given total of the mixture, To the respective quantities required.

Example. A vintner has wine at 3s. per gallon, and would mix it with water, so as to make a composition of 144 gallons, worth 2s. 6d. per gallon: How much wine, and how much water, must he take?

Galls. \[ \begin{array}{ccc} 30 & 30 & 120 \text{ of wine}, \\ 6 & 24 & 24 \text{ of water}, \\ \end{array} \]

\[ \frac{30}{6} = 144 \text{ total}. \]

\[ 120 \times 36 = 4320 \] \[ 24 \times 0 = 0 \]

Proof \(144|4320/30\) As \(36 : 30 :: 144 : 120\) As \(36 : 6 :: 144 : 24\).

There being here only two simples, and the total of the mixture limited, the question admits but of one answer.