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?

**Solution**

**Method 1: Equation Method**

Let the quantities of A and B in the original mixture be 4x and x litres.

According to the question, when 10 litres of the liquid B is poured into the jar

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**

First mixture contains A and B in the ratio of 4:1. Second mixture has only 10 litres of A. Resultant mixture is obtained by mixing these two.

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.

Now, 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 = quantity of second mixture = 10 litres.

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 in 1st jar = 10 litres

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