A container is initially full of milk. One-third is removed and replaced with water. This removal-and-replacement by the same quantity of water is repeated 4 times. If 16 L of milk remain after the 4th operation, find the container’s capacity.

Introduction / Context:
This is a repeated dilution problem. Each cycle removes one-third of the current mixture and then replaces that amount with water. The milk fraction shrinks by the same retention factor every time. After several iterations, the amount of milk left is related to the initial capacity by a power of the retention fraction.

Given Data / Assumptions:

• Each operation removes 1/3 of the mixture and refills with water.
• Number of operations = 4.
• Milk remaining after 4th operation = 16 L.
• Let capacity be V litres.

Concept / Approach:
After removing 1/3, the milk retained is (1 − 1/3) = 2/3 of what it was before the removal. Repeating this 4 times means milk becomes V * (2/3)^4. Set this equal to the final milk and solve for V.

Step-by-Step Solution:
Milk after n operations = V * (2/3)^n. For n = 4: V * (2/3)^4 = V * (16/81) = 16. Thus V * 16/81 = 16 ⇒ V = 16 * 81 / 16 = 81 L.

Verification / Alternative check:
Retention factor check: After each step milk fraction multiplies by 2/3. Four steps give (2/3)^4 = 16/81 of original, consistent with the result.

Why Other Options Are Wrong:
76, 82, 85 L do not satisfy V*(16/81) = 16 when substituted.

Common Pitfalls:
Forgetting to raise the retention fraction to the power of the number of operations, or using 1/3 instead of 2/3 for retention.

Final Answer:
81 litres