A 200 litre mixture contains 70% milk and the rest water. At first, 50 litres of the mixture are removed and replaced with 10 litres of water. Then 32 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 25:11, respectively. Find the value of 'x'.

Ratio of quantity of milk to that of water, in the initial mixture = 70:30 = 7:3 After removing 50 litres of the mixture and adding 10 litres of water, quantity of milk in the mixture = (200 − 50) × (7/10) = 105 litres quantity of water in the mixture = (200 − 50) × (3/10) + 10 = 45 + 10 = 55 litres So, ratio of quantity of milk to that of water, now = 105:55 = 21:11 After removing 32 litres of the mixture (in the same 21:11 proportion) and replacing with 'x' litres of milk, quantity of milk removed = 32 × (21/32) = 21 litres quantity of water removed = 32 × (11/32) = 11 litres So, milk left = 105 − 21 = 84 litres; water left = 55 − 11 = 44 litres New quantity of milk = 84 + x litres; water remains = 44 litres So, (84 + x):44 = 25:11 Or, 11 × (84 + x) = 25 × 44 Or, 924 + 11x = 1100 Or, 11x = 176 So, x = 16