CAT 2017

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Mixtures and Alligations
• Formula: $$\frac{\text{Proportion of milk in Bottle 1} \times V_1 + \text{Proportion of milk in Bottle 2} \times V_2}{V_1 + V_2} = \text{Required proportion of milk in final mixture}$$

---

2. Calculation

First, extract the proportion of milk from each bottle.

Bottle 1 has milk and water in ratio $7:2$, so the proportion of milk $= \frac{7}{9}$.

Bottle 2 has milk and water in ratio $9:4$, so the proportion of milk $= \frac{9}{13}$.

The desired final mixture has milk and water in ratio $3:1$, so the required proportion of milk $= \frac{3}{4}$.

Let the volumes of Bottle 1 and Bottle 2 mixed be $V_1$ and $V_2$ respectively. Let $V_1 : V_2 = X : 1$, i.e., we take $X$ parts of Bottle 1 and $1$ part of Bottle 2.

Setting up the milk equation:

$$\frac{7X}{9} + \frac{9 \times 1}{13} = \frac{3}{4} \left(X + 1\right)$$

Alternatively, using the milk-to-water ratio condition directly:

Total milk $= \frac{7X}{9} + \frac{9}{13}$

Total water $= \frac{2X}{9} + \frac{4}{13}$

Setting up the ratio condition $\frac{\text{Total milk}}{\text{Total water}} = \frac{3}{1}$:

$$\frac{7X}{9} + \frac{9}{13} = 3 \left(\frac{2X}{9} + \frac{4}{13}\right)$$

Multiplying both sides by $9 \times 13 = 117$:

$$13 \times 7X + 9 \times 9 = 3 \times 13 \times 2X + 3 \times 9 \times 4$$

$$91X + 81 = 78X + 108$$

$$91X - 78X = 108 - 81$$

$$13X = 27$$

$$X = \frac{27}{13}$$

Therefore, $V_1 : V_2 = \frac{27}{13} : 1 = 27 : 13$.

---

3. Solution

Answer = Option B ✅

The volumes of Bottle 1 and Bottle 2 should be combined in the ratio 27 : 13 to obtain a milk-to-water ratio of 3:1 in the final mixture.