CAT 2020

Slot 3

QA

Question & Solution

1. Concept Used

• Topic: Mixtures and Alligations (Weighted Average / Alligation Rule)
• Formula: $$\frac{\text{Quantity of Cheaper}}{\text{Quantity of Dearer}} = \frac{\text{Dearer Price} - \text{Mean Price}}{\text{Mean Price} - \text{Cheaper Price}}$$

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2. Calculation

Step 1: Treat the initial mixture of A and B as a single component.

Let the initial mixture of A and B (mixed in ratio 1:3) be called Component 1, with alcohol percentage x%.

This mixture is then doubled by adding solution A. This means equal volumes of Component 1 and solution A are combined (ratio 1:1), and the resulting mixture has 72% alcohol.

Applying the Alligation Rule between Component 1 (x%) and Solution A (60%), with mean = 72%:

$$\frac{\text{Quantity of A}}{\text{Quantity of Component 1}} = \frac{x - 72}{72 - 60} = \frac{1}{1}$$

$$x - 72 = 12 \implies x = 84%$$

So the initial mixture of A and B has 84% alcohol.

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Step 2: Find the percentage of alcohol in solution B.

Solution A has 60% alcohol. Solution B has y% alcohol. They are mixed in ratio A : B = 1 : 3, and the resulting mixture has 84% alcohol.

Applying the Alligation Rule:

$$\frac{\text{Quantity of A}}{\text{Quantity of B}} = \frac{y - 84}{84 - 60} = \frac{1}{3}$$

$$\frac{y - 84}{24} = \frac{1}{3}$$

$$y - 84 = 8 \implies y = 92%$$

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3. Solution

Answer = Option C ✅

The percentage of alcohol in solution B is 92%.