The vessel contains 60 liters of a mixture that contains 80% milk and the rest is water. Firstly, 20 liters of mixture are replaced by water, and then, 15 liters of mixture are removed from the resultant mixture. If x liters of water are added to the remaining mixture, then the ratio of milk to water becomes equal in the final mixture. Find the value of x.

Correct option is A

Given: 

Total mixture = 60 liters

percentage of milk = 80%

The ratio of milk to water in resultant mixture is same

Explanation:

$$\text{Quantity of milk in the initial mixture} = \frac{60}{100} \times 80 = 48 \, \text{litres} \\\text{Quantity of water in the initial mixture} = 60 - 48 = 12 \, \text{litres} \\\text{After first replacement,} \\\text{Quantity of milk} = 48 - 20 \times \frac{4}{5} = 32 \, \text{litres} \\\text{Quantity of water} = 12 - 20 \times \frac{1}{5} + 20 = 28 \, \text{litres} \\\text{Ratio of milk to water} = 32:28 = 8:7 \\\text{Now,} \\\frac{32 - 15 \times \frac{8}{15}}{28 - 15 \times \frac{7}{15} + x} = \frac{1}{1} \\24 = 21 + x \\x = 3$$​