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Successive Replacement in Mixture

CAT 2025 Slot 1 · Quantitative Ability · Medium · Mixtures and Alligations

This is a medium Quantitative Ability question from the CAT 2025 Slot 1 paper. It tests Mixtures and Alligations. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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A container holds 200200 litres of a solution of acid and water, having 30%30\% acid by volume. Atul replaces 20%20\% of this solution with water, then replaces 10%10\% of the resulting solution with acid, and finally replaces 15%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

  1. A.

    27

  2. B.

    25

  3. C.

    29

  4. D.

    23

Answer

A

Explanation

Initial acid volume =30% of 200=60 L= 30\% \text{ of } 200 = 60\text{ L}.

  1. Replace 20%20\% with water: Acid remaining =60×(1−0.20)=48 L= 60 \times (1 - 0.20) = 48\text{ L}.

  2. Replace 10%10\% with pure acid: Remove 10%10\% of solution (removes 4.8 L4.8\text{ L} acid), then add 10% of 200=20 L10\% \text{ of } 200 = 20\text{ L} pure acid. Acid =48−4.8+20=63.2 L= 48 - 4.8 + 20 = 63.2\text{ L}.

  3. Replace 15%15\% with water: Acid remaining =63.2×(1−0.15)=63.2×0.85=53.72 L= 63.2 \times (1 - 0.15) = 63.2 \times 0.85 = 53.72\text{ L}.

Final concentration =53.72200×100%=26.86%≈27%= \frac{53.72}{200} \times 100\% = 26.86\% \approx 27\%.

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