CAT 2025 SLOT-3 QA Modern Math Question

1. Concept Used

Topic: Mixtures and Alligations — Replacement Method
Formula: After two-step replacement, alcohol remaining in A = $ \frac{3600}{60 + x} $ litres, where $ x $ is the volume exchanged.

2. Calculation

Let the volume taken out from Vessel A in the first step be $ x $ litres.

Step 1 — Transfer from A to B:

Vessel A initially has 60 litres of pure alcohol. After removing $ x $ litres, Vessel A has $ (60 - x) $ litres of alcohol.

Vessel B initially has 60 litres of pure water. After adding $ x $ litres of alcohol, Vessel B has a total of $ (60 + x) $ litres, with $ x $ litres of alcohol and $ 60 $ litres of water.

Step 2 — Transfer from B to A:

After thorough stirring, the concentration of alcohol in Vessel B is: $ \frac{x}{60 + x} $

When $ x $ litres are taken out from Vessel B, the alcohol removed = $ x \times \frac{x}{60 + x} = \frac{x^2}{60 + x} $ litres.

This mixture is poured back into Vessel A.

Alcohol in Vessel A after the entire process: $ = (60 - x) + \frac{x^2}{60 + x} $ $ = \frac{(60 - x)(60 + x) + x^2}{60 + x} $ $ = \frac{3600 - x^2 + x^2}{60 + x} $ $ = \frac{3600}{60 + x} $

Setting up the equation using the given ratio:

The total volume in Vessel A is restored to 60 litres. The ratio of alcohol to water in A is $ 15 : 4 $, so the fraction of alcohol in A is: $ \frac{15}{15 + 4} = \frac{15}{19} $

Therefore, the alcohol in Vessel A equals: $ \frac{15}{19} \times 60 $

Setting the two expressions equal: -> $ \frac{3600}{60 + x} = \frac{15 \times 60}{19} $

-> $ \frac{60}{60 + x} = \frac{15}{19} $

-> $ 60 \times 19 = 15 \times (60 + x) $

-> $ 1140 = 900 + 15x $

-> $ 15x = 240 $

-> $ x = 16 $

3. Solution

Answer = 16 ✅

The volume initially taken out from Vessel A is 16 litres.

CAT 2025Slot 3QAQuestion & SolutionQues & Sol

Question

Solution

1. Concept Used

2. Calculation

3. Solution