A milkman has two containers A and
B. Container A contains pure milk and container B contains the mixture of milk and water. The milkman replaces 6 litres of pure milk from container A with water. He attempts this process thrice such that the ratio of the milk to water in the final mixture becomes 343:169 in container
A. If the amount of mixture in container B is 50% more than the amount of pure milk present initially in container A and the ratio of the milk to water in container B is 5:4, then find the quantity of milk present in container
B.

Solution

Let the quantity of milk and water in the final mixture of container A be 343y litres and 169y litres, respectively. So, total quantity of pure milk initially = 343y + 169y = 512y litres By replacement formula, Final quantity of pure milk = Initial quantity of pure milk × [1 – amount of replacement done/amount of liquid in container A]n, where n is the number of attempts of replacement. 343y = 512y × [1 – 6/k]3, where k is the amount of pure milk in container A initially. 343y/512y = [1 – 6/k]3 (7/8)3 = [1 – 6/k]3 1 – 6/k = 7/8 6/k = 1/8 k = 48 litres So, Quantity of mixture in container B = 1.5 × 48 = 72 litres Quantity of milk present in container B = 5/9 × 72 = 40 litres