User:Earthkingfather/sandbox

Alligation is a method of solving arithmetic problems related to mixtures of ingredients. There are two types of alligation, Alligation medial and Alligation alternate. Alligation medial involves finding the a weighted mean concentration, or mean price of a mixture of ingredients by mass. Alligation alternate is a procedure where the proportions of ingredients needed to achieve a target mean concentration or price are found.

There are two further variations on Alligation alternate, Alligation Partial and Alligation Total, These involve finding the precise quantities of ingredients needed to achieve a target mean concentration or price.

Alligation medial
Suppose you make a cocktail drink combination out of 1/2 Coke, 1/4 Sprite, and 1/4 orange soda. The Coke has 120 grams of sugar per liter, the Sprite has 100 grams of sugar per liter, and the orange soda has 150 grams of sugar per liter. How much sugar does the drink have? This is an example of alligation medial because you want to find the amount of sugar in the mixture given the amounts of sugar in its ingredients. The solution is just to find the weighted average by composition:


 * $${1\over2}\times 120 + {1\over4}\times 100 + {1\over4}\times 150 = 122.5$$ grams per liter

Alligation alternate
Alligation alternate may be thought of as the inverse of alligation medial. Given several numbers, to find multiples of those numbers, that would result in a given mean value. Such a problem does not have a unique answer, and there are several methods by which the problem may be solved.


 * 1)  Write the mean.
 * 2)  Write each of the [prices] in descending order beside the mean in a column.
 * 3)  Draw a line,connecting a number greater than the mean, with one or more numbers that is less than the mean.
 * 4)  Find the [absolute value of the difference] between the mean and each of the numbers. Write the differnce beside the numbers that are linked to the subtrahend, in a third column. Do not write this difference beside the subtrahend itself.
 * 5)  The numbers found in step (4) are the relative proportions of the quantities to be taken, to generate the mean. If more than one number results from step 4, add the numbers together.

Example 1
There are four solutions of alcohol. They have concentrations of 20%, 15%, 5% and 1%. In what proportions must the they be mixed, to give a mixture that has the average concentration of 13% ?

1. The concentrations are first written down, along with the mean:

| 20 13| 15  |  5   |  1

2. Then lines are drawn, connecting a term that is greater than the mean, to a price less than the mean. | 20───┐         13| 15──┐│   |  5──┘│      |  1───┘

3. The numbers written down in step (1) are one by one, subtracted from the mean (or, if greater than the mean, the mean subtracted from them). The result of the subtraction is written beside the number that was linked to the minuend in step 2, and not the minuend itself.

(Explanation) | 20───┐ 12(13-1)                  13| 15──┐│  8 (13-5)                |  5──┘│  2 (15-13)                    |  1───┘  7 (20-13)

In this example, 13 is subtracted from 20. The difference, seven, is written beside the number 1, which was linked to 20 in step 2. Likewise, the difference between 13 and 1, twelve is written beside 20, the number that it was linked to.

The numbers that are produced by step 3 are the relative proportions of the solutions. So, to achieve the desired concentration, one must mix 12 parts of the 20% with 8 parts of the 15% with 2 parts of 5% and 7 parts of 1%.

Linking the numbers in the following manner would also yield a correct answer.

| 20───┐ 8       13| 15──┐│ 12   |  5──│┘ 7   |  1──┘  2

Example 2
There five types of tea. Each costs $20, $12, $3, $2 and $1 per kilogram. In what proportions must the teas be mixed to give a mixture with an average cost of $8 per kilogram?

Owing to the odd number of terms, three terms are be linked to one another. Here 20 is linked to 1 and 2

| 20───┐─┐   | 12──┐│ │   8|  3──┘│ │     |  2───┘ │    |  1─────┘             The subtractions are performed as in the previous example. As two terms are linked to "20", so two differences are written beside it. | 20───┐─┐ 7, 6    (8-1),(8-2)    | 12──┐│ │ 5        (8-3)  8|  3──┘│ │ 14       (20-8)   |  2───┘ │ 12       (20-8)   |  1─────┘ 12       (20-8)

There are two differences beside "20".These are added up. The final sum is the answer.

(answer) | 20───┐─┐ 7,6 │13  | 12──┐│ │ 5   │5  8|  3──┘│ │ 14  │4   |  2───┘ │ 12  │12   |  1─────┘ 12  │2

Another answer may be obtained as follows:

| 20─────┐ 11    |   | 12──┐  │ 7      |  8|  3──│──┘ 12    8|   |  2──│──┘ 12     |   |  1──┘    4      |

Alligation partial
Alligation partial finds the exact quantities of ingredients needed to generate a mixture of a given concentration, when the quantity of one of the ingredients is already fixed. This is a matter of proportion.

Alligation Total
Alligation total is a variation on alligation alternate, when the total mass of the final mixture is fixed.

Suppose you like 1% milk, but you have only 3% whole milk and 0.5% low fat milk. How much of each should you mix to make an 8-ounce cup of 1% milk?

By performing alligation alternate, the respective proportions of milk are 0.5 parts of 3%, and 2 parts of 0.5%.

| 3.0──┐0.5 1|      │   | 0.5──┘2 As the total amount of milk in the final mixture is fixed at 8 ounces, finding the respective masses of milk, is simply a matter of finding the appopriate proportion of eight ounces.

The total amount, 8 ounces, is then divided by the sum {\displaystyle 2+{1 \over 2}={5 \over 2}} {\displaystyle 2+{1 \over 2}={5 \over 2}} to yield {\displaystyle 16 \over 5} {\displaystyle 16 \over 5}, and the amounts of the two ingredients are {\displaystyle {16 \over 5}\times {1 \over 2}={8 \over 5}} {\displaystyle {16 \over 5}\times {1 \over 2}={8 \over 5}} ounces whole milk and {\displaystyle {16 \over 5}\times 2={32 \over 5}} {\displaystyle {16 \over 5}\times 2={32 \over 5}} ounces low fat milk.

A general formula
A general formula that works for both alligation "alternate" and alligation "medial" is the following:
 * $$Aa + Bb = Cc.$$

In this formula, A is the volume of ingredient A and a is its mixture coefficient (i.e. a= 3%); B is volume of ingredient B and b is its mixture coefficient; and C is the desired volume C, and c is its mixture coefficient. Returning to example 4,
 * A(0.03) + B(0.005) = 8oz(0.01). We know B = (8oz-A), and so can easily solve for A and B to get 1.6 and 6.4oz, respectively. Using this formula you can solve for any of the 6 variables A,a,B,b,C,c, regardless of whether you're dealing with medial, alternate, etc.