A mixture has a total volume of 240 litres. In this mixture, water is 20% and the remaining 80% is milk. A quantity of the mixture is removed and replaced with pure water (assume perfect mixing). How many litres of the mixture should be removed and replaced so that the percentage of water becomes 40% in the final mixture?

Introduction / Context:
This problem tests replacement in a mixture where one component (water) is increased by removing some mixture and adding pure water. The key is to track how much water remains after removal and how much water is added back. Since the total volume stays constant, the final water amount must match 40% of 240 litres.

Given Data / Assumptions:

• Total mixture volume = 240 L
• Initial water = 20% of 240 = 48 L
• Initial milk = 240 - 48 = 192 L
• Remove x litres of mixture and replace with x litres pure water
• Final water percentage required = 40%

Concept / Approach:
After removing x litres, water removed = (initial water fraction)*x. Then we add x litres water. Final water = initial water - removed water + added water. Set this equal to 40% of total.

Step-by-Step Solution:

Verification / Alternative check:
If x=60: water removed = 0.20*60=12. Water left = 48-12=36. Add 60 water => water becomes 96. 96/240 = 40%. Verified.

Why Other Options Are Wrong:

Common Pitfalls:
Students sometimes treat the water removed as x instead of 0.20x, forgetting that the removed portion is a mixture. Another error is using 20% of remaining volume rather than 20% of x for the removed part. Keep the fractions consistent.

Final Answer:
60 litres