A certain liquid mixture in a container contains 25% milk and the remaining 75% is water. What percentage of this mixture must be removed and replaced with an equal quantity of pure milk so that, in the resulting mixture, the quantities of milk and water become equal?

Introduction:
This problem deals with replacing part of a mixture by pure milk in order to change the percentage composition. Initially, milk is in the minority. By removing some mixture and adding pure milk, we want the final mixture to have equal amounts of milk and water. This is a classic replacement percentage question in mixture and alligation.

Given Data / Assumptions:

• Initial mixture contains 25% milk and 75% water.
• A certain percentage of this mixture is removed.
• The same percentage volume is replaced with pure milk.
• After this operation, milk and water in the mixture are equal in quantity.
• We must determine the percentage of the mixture that is removed and replaced.

Concept / Approach:
For simplicity, assume the total initial mixture is 1 unit (or 100%). Then milk is 0.25 units and water is 0.75 units. Let the fraction of mixture removed be f. The removed portion contains milk and water in the same 25% to 75% ratio. After removing that portion, we add back the same fraction f, but now as pure milk. We then set the final milk amount equal to the final water amount and solve for f.

Step-by-Step Solution:
Step 1: Assume total mixture = 1 unit. Step 2: Initial milk = 0.25 units, initial water = 0.75 units. Step 3: Let fraction of mixture removed be f. Step 4: Milk removed = 0.25 * f, water removed = 0.75 * f. Step 5: After removal, milk = 0.25 - 0.25 * f, water = 0.75 - 0.75 * f. Step 6: The removed fraction f is replaced by pure milk. Step 7: New milk amount = (0.25 - 0.25 * f) + f = 0.25 + 0.75 * f. Step 8: New water amount = 0.75 - 0.75 * f. Step 9: In the final mixture, milk and water must be equal, so: 0.25 + 0.75 * f = 0.75 - 0.75 * f. Step 10: Rearrange: 0.75 * f + 0.75 * f = 0.75 - 0.25. Step 11: 1.5 * f = 0.50, so f = 0.50 / 1.5 = 1 / 3. Step 12: Convert f to percentage: f = 1 / 3 is approximately 33.33%.

Verification / Alternative check:
Suppose there are 100 litres initially. Then milk = 25 litres, water = 75 litres. Remove 33.33 litres of mixture (approximately 1 / 3 of the total). Milk removed = 25 * 1 / 3 ≈ 8.33 litres, water removed = 75 * 1 / 3 ≈ 25 litres. Remaining milk ≈ 16.67 litres, remaining water ≈ 50 litres. Add back 33.33 litres of pure milk, giving final milk ≈ 50 litres and water ≈ 50 litres, which are equal as required.

Why Other Options Are Wrong:
31.8%, 31%, and 29.85% do not give exactly equal quantities of milk and water when used in the above calculation.
40% overshoots, resulting in more milk than water in the final mixture rather than equal amounts.

Common Pitfalls:
A common mistake is to remove a fixed volume in litres without relating it proportionally to the total, which makes the percentage calculation inconsistent. Another frequent error is to equate percentages directly instead of actual quantities, or to forget that the total mixture remains the same after replacement. Careful algebra and assuming a simple total like 1 or 100 makes the algebra more manageable.

Final Answer:
The percentage of the mixture that must be removed and replaced with milk is 33.33%.