Question

A vessel has a milk solution in which milk and water are in the ratio 4:1. By addition of water to it, milk solution with milk and water in the ratio 4:3 was formed. On replacing 14 L of this solution with pure milk, the ratio of milk and water changed to 5:3. What is the volume of the water added?

• A. 12 L
• B. 32 L
• C. 24 L
• D. 14 L

Hint:

Use the method of alligation or equation substitution with ratios to systematically work through multi-step dilution and replacement problems.

Answer 1

The Correct Option is C

Let original quantity of solution be \( x \) litres. Initial ratio: Milk : Water = 4:1 → Milk = \( \frac{4x}{5} \), Water = \( \frac{x}{5} \) Let \( w \) litres of water be added. New total = \( x + w \)
Milk stays same \( = \frac{4x}{5} \), Water becomes \( \frac{x}{5} + w \)
New ratio = 4:3, so: \[ \frac{4x}{5} : \left(\frac{x}{5} + w\right) = 4:3 \Rightarrow \frac{4x}{5} = \frac{4}{7}(x + w) \] Cross-multiplied: \[ \frac{4x}{5} = \frac{4(x + w)}{7} \Rightarrow \frac{x}{5} = \frac{x + w}{7} \Rightarrow 7x = 5x + 5w \Rightarrow 2x = 5w \Rightarrow x = \frac{5w}{2} \] So total solution = \( x + w = \frac{5w}{2} + w = \frac{7w}{2} \) Milk = \( \frac{4x}{5} = \frac{4 \cdot \frac{5w}{2}}{5} = 2w \), Water = \( \text{total} - \text{milk} = \frac{7w}{2} - 2w = \frac{3w}{2} \) Now 14 L of solution is replaced with pure milk. So 14 L is removed in the current ratio 4:3 i.e.: \[ \text{Milk removed} = \frac{4}{7} \cdot 14 = 8 \text{ L}, \quad \text{Water removed} = \frac{3}{7} \cdot 14 = 6 \text{ L} \] Then, 14 L of pure milk is added: \[ \text{New milk} = 2w - 8 + 14 = 2w + 6
\text{New water} = \frac{3w}{2} - 6 \] Given new ratio = 5:3: \[ \frac{2w + 6}{\frac{3w}{2} - 6} = \frac{5}{3} \Rightarrow 3(2w + 6) = 5\left(\frac{3w}{2} - 6\right) \Rightarrow 6w + 18 = \frac{15w}{2} - 30 \] Multiply all terms by 2: \[ 12w + 36 = 15w - 60 \Rightarrow 96 = 3w \Rightarrow w = \boxed{32} \] \[ \boxed{\text{Water added} = 32 \text{ litres}} \]