CAT 2025

Slot 3

QA

Question & Solution

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.

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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: $$ \frac{3600}{60 + x} = \frac{15 \times 60}{19} $$

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

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

$$ 1140 = 900 + 15x $$

$$ 15x = 240 $$

$$ x = 16 $$

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3. Solution

Answer = 16 ✅

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