Average, Mixture and Alligation Question 2 - CAT 2025
A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of this solution with water, then replaces 10% of the resulting solution with acid, and finally replaces 15% of the solution thus obtained, with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to:
Step-by-Step Explanation
Since every volume removed is replaced by an equal volume of either water or acid, the total volume of the container remains constant at 200 litres throughout the process. Let us track the absolute volume of acid at each step:
- Initial State:
Acid volume = 30% of 200 litres = 60 litres.
- Replacement 1 (20% solution replaced with water):
When 20% of the solution is removed, 20% of the existing acid is lost.
Acid remaining = 60 × (1 − 0.20) = 48 litres
- Replacement 2 (10% solution replaced with pure acid):
Removing 10% of the solution removes 10% of the acid:
Acid before refill = 48 × (1 − 0.10) = 43.2 litres
Now, 10% of the total volume (10% of 200 litres = 20 litres) of pure acid is added back:
Acid remaining = 43.2 + 20 = 63.2 litres
- Replacement 3 (15% solution replaced with water):
When 15% of the solution is removed, 15% of the acid is lost:
Final Acid volume = 63.2 × (1 − 0.15) = 63.2 × 0.85 = 53.72 litres
- Final Concentration:
Percentage of acid = (53.72 / 200) × 100 = 26.86% ≈ 27%
Correct Option: D
💡 FundaMakers Tip:
In multi-step replacement problems, only track the active component (acid). As long as the total volume stays constant, you can treat removals as simple fractional reductions (multiplying by 0.80, 0.90, etc.) and additions as direct numerical values based on the fixed total container capacity.