CAT 2025 Slot1 Quant: A container holds 200 litres of a solution of acid and water,

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 $\mathbf{27}$.