CAT 2025 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.
Question 9 : 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
Initial State
Total Volume: \(200L\)
Initial Acid Percentage: \(30\%\)
Initial Volume of Acid: \(200\times0.30=60L\)
Step 1: Replace \(20\%\) of the solution with water
Removed: \(20\%\) of the solution, which removes \(20\%\) of the existing acid.
Remaining Acid: \(60\times(1-0.20)=60\times0.80=48L\)
Added: \(20\%\) of \(200L=40L\) of pure water (0L of acid).
Acid Volume after Step 1: \(48L\)
Step 2: Replace \(10\%\) of the resulting solution with acid
Removed: \(10\%\) of the solution, which removes \(10\%\) of the existing acid.
Remaining Acid from mixture: \(48\times(1-0.10)=48\times0.90=43.2L\)
Added: \(10\%\) of \(200L=20L\) of pure acid.
Acid Volume after Step 2: \(43.2+20=63.2L\)
Step 3: Replace \(15\%\) of the solution with water
Removed: \(15\%\) of the solution, which removes \(15\%\) of the existing acid.
Remaining Acid: \(63.2\times(1-0.15)=63.2\times0.85=53.72L\)
Added: \(15\%\) of \(200L=30L\) of pure water (0L of acid).
Acid Volume after Step 3: \(53.72L\)
Final Concentration Calculation:
Final Percentage of Acid \(=\frac{\text{Final Volume of Acid}}{\text{Total Volume}}\times100\)
\(=\frac{53.72}{200}\times100=26.86\%\)
\(\approx27\%\)
Choice 4 is the correct answer.
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