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: 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.