Mixtures are formed when two or more quantities of different values are mixed together. Alligation is a practical method of solving arithmetic problems related to mixtures of ingredients.

There are two kinds of problems:

  1. To find the quantity of a mixture given the quantities of its ingredients. (Alligation Medial)

  2. To find the amount of each ingredient needed to make a mixture of a given quantity. (Alligation Alternate)

Chain Rule

  1. When a fraction has its numerator greater than the denominator, its value is greater than one. Let us call it greater fraction. Whenever a number (say x) is multiplied by a greater fraction, it gives a value greater than itself.

  2. When a fraction has its numerator less than the denominator, its value is less than one. Let us call it less fraction. Whenever a number (say x) is multiplied by a less fraction, it gives a value less than itself.

Important Facts and Formula

Allegation: It is the rule that enables us to find the ratio in which two of more ingredients at the given price must be mixed to produce a mixture of a desired price.

Mean Price: The cost price of a unit quantity of the mixture is called the mean price.

Rule of Allegation: If two ingredients are mixed then Quantity of Cheaper / Quantity of Dearer = (C.P. of Dearer - Mean Price) / (Mean Price - C.P. of Cheaper)

Cheaper quantity : Dearer quantity = (d - m) : (m - c)

Example 1: A sum of Rs 39 was divided among 45 boys and girls. Each girl gets 50 paise, whereas a boy gets one rupee. Find the number of boys and girls.

Average amount of money received by each = 39/45 = 13/15

Amount received by each girl = 50 paise = Rs 1/2

Amount received by each boy = Re 1

By alligation rule:

Number of boys / Number of girls

= (13/15 - 1/2) / (1 - 13/15)

= 11/4

∴ Number of boys = 11/(11+4) × 45 = 33

Number of girls = 45 – 33 = 12

Important Tip

Always identify the ingredients as cheaper and dearer to apply the alligation rule. In the alligation rule, the variables c, d & m may be expressed in terms of percentages (e.g. A 20% mixture of salt in water), fractions (e.g. two-fifth of the solution contains salt) or proportions (e.g. A solution of milk and water is such that milk : Water = 2 : 3). The important point is to remember is that c & d may represent pure ingredients or mixtures.

Mixing a Pure Component to a Solution

Example 2: A jar contains a mixture of two liquids A and B in the ratio 4 : 1. When 10 litres of the liquid B is poured into the jar, the ratio becomes 2: 3. How many litres of liquid A were contained in the jar?

Method 1: weighted average or equation method

Let the quantities of A & B in the original mixture be 4x and x litres. According to the question

4x / (x + 10) = 2 / 3

12x = 2x + 20

x = 2

The quantity of A in the original mixture = 4x = 4 × 2 = 8 litres.

Method 2: Alligation with composition of B

The average composition of B in the first mixture is 1/5.

The average composition of B in the second mixture = 1

The average composition of B in the resultant mixture = 3/5

Hence, applying the rule of Alligation

[1 – (3/5)]/[(3/5) – (1/5)] = (2/5)/(2/5) = 1

So, initial quantity of mixture in the jar = 10 litres.

And, quantity of A in the jar = (10 × 4)/5 = 8 litres.

Method 3: Alligation with percentage of B

The percentage of B in 1st mixture = 20%

The percentage of B in 2nd mixture = 100%

The percentage of B in Final Mixture = 60%

By rule of allegation

Volume 1st : Volume 2nd = (100% - 60%) : (60% - 20%)

V1 : V2 = 1 : 1

Volume of mixture 1st = 10 litres

Volume of A in mixture 1st = 80% of 10 litres = 8 litres

Removal and Replacement 

Suppose a container contains x units of liquid from which y units are taken out and replaced by water. After n operations the quantity of pure liquid = x(1 - y/x)n units.

Example 3: Nine litres of solution are drawn from a cask containing water. It is replaced with a similar quantity of pure milk. This operation is done twice. The ratio of water to milk in the cask now is 16 : 9. How much does the cask hold?

Let there be x litres in the cask. After n operations

Water left in vessel  / Original quantity of water = (1 – 9/x)n

(1 – 9/x)2 = 16/25

∴ x = 45 litres