A 20 L vessel has milk and water in the ratio 3 : 2. Ten litres are removed and replaced with pure milk, and this process is repeated once more. Find the final milk : water ratio.

Introduction / Context:
This is a successive replacement (removal and refill) problem. Removing a fixed volume of well-mixed solution takes out components in proportion to their current fractions. Replacing with pure milk changes the composition before the next removal.

Given Data / Assumptions:

• Total volume stays 20 L.
• Initial ratio milk : water = 3 : 2 ⇒ milk = 12 L, water = 8 L.
• Remove 10 L of mixture each time; refill 10 L pure milk; repeat twice in total.

Concept / Approach:
At each removal, amounts removed are proportional to current composition. After removal, add 10 L milk to restore volume to 20 L, then repeat once more. Keep track of milk and water after each step.

Step-by-Step Solution:
Initial: milk 12, water 8. First removal (10 of 20): removes 6 milk and 4 water ⇒ remains milk 6, water 4. Add 10 L milk ⇒ milk 16, water 4. Second removal (10 of 20): composition milk 16/20, water 4/20 ⇒ removes 8 milk and 2 water ⇒ remains milk 8, water 2. Add 10 L milk ⇒ final milk 18, water 2 ⇒ ratio 18 : 2 = 9 : 1.

Verification / Alternative check:
Using fraction retention: milk fraction retained each removal = (1 − 10/20) = 1/2, but because we add back pure milk between removals, stepwise tracking as shown is more transparent here.

Why Other Options Are Wrong:
5:3 and 4:1 do not match the increased milk dominance after two milk refills; 1:4 is impossible since milk is being added, not removed.

Common Pitfalls:
Forgetting that removal is proportional to current composition or failing to update before the second cycle. Always recalculate before each step.

Final Answer:
9:1