A 250 litre mixture contains 80% milk and the rest water. At first, 50 litres of the mixture are removed and replaced with 10 litres of water. Then 42 litres of the resultant mixture are removed and replaced with 'x' litres of milk such that the ratio of the quantity of milk to that of water in the final mixture becomes 4:1, respectively. Find the value of 'x'.

Ratio of quantity of milk to that of water, in the initial mixture = 80:20 = 4:1 After removing 50 litres of the mixture and adding 10 litres of water, quantity of milk in the mixture = (250 − 50) × (4/5) = 160 litres quantity of water in the mixture = (250 − 50) × (1/5) + 10 = 40 + 10 = 50 litres So, ratio of quantity of milk to that of water, now = 160:50 = 16:5 After removing 42 litres of the mixture (in the same 16:5 proportion) and replacing with 'x' litres of milk, quantity of milk removed = 42 × (16/21) = 32 litres quantity of water removed = 42 × (5/21) = 10 litres So, milk left = 160 − 32 = 128 litres; water left = 50 − 10 = 40 litres New quantity of milk = 128 + x litres; water remains = 40 litres ATQ, (128 + x):40 = 4:1 Or, 1 × (128 + x) = 4 × 40 Or, 128 + x = 160 So, x = 32. Hence, the correct answer is B.