CAT 2025

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Mixtures and Alligations — Successive Replacement
• Formula: $$\text{Acid after replacement} = \text{Current Acid} \times (1 - \text{fraction removed}) + \text{Pure substance added}$$

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2. Calculation

The total volume of the solution always remains 200 litres throughout all replacements (since equal volume is removed and added back each time).

Initial Setup: Initial acid $$= 30% \times 200 = 0.30 \times 200 = 60 \text{ L}$$

Step 1 — Replace 20% of solution with water: When 20% of the mixture is removed, 20% of the acid is also removed. No acid is added back (only water is added). $$\text{Acid remaining} = 60 \times (1 - 0.20) = 60 \times 0.80 = 48 \text{ L}$$

Step 2 — Replace 10% of the resulting solution with pure acid: First, 10% of the current mixture is removed, taking away 10% of the acid present. Then, pure acid is added back to restore volume to 200 L. Volume of pure acid added $$= 10% \times 200 = 20 \text{ L}$$. $$\text{Acid after removing 10%} = 48 \times (1 - 0.10) = 48 \times 0.90 = 43.2 \text{ L}$$ $$\text{Acid after adding pure acid} = 43.2 + 20 = 63.2 \text{ L}$$

Step 3 — Replace 15% of the solution with water: When 15% of the mixture is removed, 15% of the acid is removed. Only water is added back. $$\text{Acid remaining} = 63.2 \times (1 - 0.15) = 63.2 \times 0.85 = 53.72 \text{ L}$$

Final Concentration: $$\text{Acid %} = \frac{53.72}{200} \times 100 = 26.86% \approx 27%$$

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3. Solution

Answer = Option D (27%) ✅

The final percentage of acid by volume in the solution after all three replacements is approximately 26.86%, which is nearest to 27%.