A vessel of capacity 65 litres contains a mixture of milk and water in an unknown ratio. One-third of the mixture is removed and replaced by pure water. This operation is repeated once more (again removing one-third of the mixture and replacing with water). In the final mixture, the ratio of milk to water is 16 : 23. Find the initial quantities of milk and water in the vessel.

ATQ, Let initial milk = M litres, water = W litres. Total volume V = M + W = 65. Each time, 1/3 of the mixture is removed and replaced with water. Fraction of milk left after one replacement = (1 − 1/3) = 2/3. After two replacements, fraction left = (2/3)^2 = 4/9. So final quantity of milk = (4/9)M. Final total volume is still 65 L, so final water = 65 − (4/9)M. Given final ratio milk : water = 16 : 23: (4/9 M) / [65 − (4/9)M] = 16/23 Cross-multiply: 23 × (4/9)M = 16[65 − (4/9)M] (92/9)M = 1040 − (64/9)M Add (64/9)M both sides: (92/9 + 64/9)M = 1040 (156/9)M = 1040 (52/3)M = 1040 ⇒ M = 1040 × (3/52) = 1040/52 × 3 = 20 × 3 = 60 litres. Since total is 65 L: W = 65 − 60 = 5 litres. Answer: Initially milk = 60 L; water = 5 L (ratio 12 : 1).