CAT 2025 Slot1 Quant: A container holds 200 litres of a solution of acid and water,
February 20, 2026 2026-02-20 22:09CAT 2025 Slot1 Quant: A container holds 200 litres of a solution of acid and water,
Use Multipliers for concentration! Total volume remains 200L.
Initial Acid = $30\%$.
1. Replace 20% with Water: Acid remains $0.8$ of itself.
2. Replace 10% with Acid: Acid becomes $(0.9 \times \text{Current}) + 10\%$.
3. Replace 15% with Water: Final Acid = $0.85 \times \text{Result}$.
$[(30 \times 0.8 \times 0.9) + 10] \times 0.85 = [21.6 + 10] \times 0.85 = 31.6 \times 0.85 = \mathbf{26.86\%} \approx 27\%$.
Detailed Step-by-Step Solution
Initial State: Total = 200L. Acid = $30\%$ of 200 = 60L.
Step 1: Replace 20% (40L) with Water
When 40L of solution is removed, 20% of acid is also removed.
Acid remaining = $60 - (0.20 \times 60) = 48$ L.
Current Concentration = $48/200 = 24\%$.
Step 2: Replace 10% (20L) with Pure Acid
Remove 20L of solution: Acid removed = $10\%$ of 48 = 4.8L.
Acid remaining before addition = $48 - 4.8 = 43.2$ L.
Add 20L of pure acid: New Acid = $43.2 + 20 = 63.2$ L.
Current Concentration = $63.2/200 = 31.6\%$.
Step 3: Replace 15% (30L) with Water
Remove 30L of solution: Acid removed = $15\%$ of 63.2 = 9.48L.
Final Acid = $63.2 - 9.48 = 53.72$ L.
Final Percentage = $(53.72 / 200) \times 100 = 26.86\%$.
Conclusion: The nearest integer is $\mathbf{27}$.
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