From a container, 6 liters milk was drawn out and was replaced by water. Again…

Concept: When a fixed volume d is repeatedly drawn from a mixture that starts as V liters of pure milk and replaced by water each time, the fraction of the original milk left after n such replacements follows the successive-replacement rule: remaining milk fraction = (1 − d/V) raised to the power n.

Application:

1. Let the total quantity of the mixture be k liters (initially pure milk).

2. Each operation withdraws 6 liters and replaces it with water, and this happens n = 2 times, so the remaining-milk fraction is (1 − 6/k)2.

3. The final ratio of milk to water is 9 : 16, so the total parts are 9 + 16 = 25, giving a remaining-milk fraction of 9/25.

4. Equating the two expressions: (1 − 6/k)2 = 9/25.

5. Taking the positive square root of both sides: 1 − 6/k = 3/5.

6. Solving for k: 6/k = 1 − 3/5 = 2/5, so k = 6 × 5/2 = 15.

So the total quantity of the mixture is 15 liters.

Cross-check:

With k = 15 L: after the first draw-and-replace, milk left = 15 − 6 = 9 L out of 15 L (fraction 3/5). The second draw removes 6 L of this diluted mixture, taking out 6 × 3/5 = 18/5 L of milk, leaving 9 − 18/5 = 27/5 L of milk. Milk : water = 27/5 : 48/5 = 27 : 48 = 9 : 16 — matching the given ratio and confirming k = 15.