A container contains 50 litres of milk. From this container, 10 litres of milk…

Concept: In a repeated-replacement (dilution) problem, when a fixed quantity is removed from a mixture and replaced by another substance, only the CURRENT proportion of the original substance is lost at each step, not the full quantity removed, because later removals already carry away some of the replacement substance too. For a container of volume V, if a quantity x is removed and replaced n times, the amount of the original substance remaining is V times (1 minus x/V) raised to the power n.

1. Initial volume V = 50 litres, all milk. Quantity removed each round x = 10 litres, so the removed fraction each round is 10/50 = 1/5, and the remaining fraction each round is 4/5.

2. After round 1: milk left = 50 times 4/5 = 40 litres (40 litres milk + 10 litres water, total still 50 litres).

3. In round 2, the 10 litres removed is now a mixture of milk and water in the ratio 40:10, so it still carries away the same 4/5 remaining-fraction relationship. Milk left = 40 times 4/5 = 32 litres.

4. Equivalently, milk left = 50 times (4/5)2 = 50 times 16/25 = 32 litres.

Cross-check: The water present in the container at the end should equal 50 − 32 = 18 litres. This must be less than the 20 litres poured in over the two rounds (10 litres each time), since part of the water added in round 1 gets carried out again when the diluted mixture is removed in round 2 — confirming the compounded (not simply additive) nature of the result.