Question

A vessel contains a mixture of milk and water in the ratio of 5:3 respectively. How much of the mixture must be siphoned off and replaced with water, so that the mixture may be half milk and half water?

• A. 1/4
• B. 1/7
• C. 1/5
• D. 1/3

Hint:

In replacement problems, always work with the proportion of each component and track how it changes step-by-step.

Answer 1

The Correct Option is C

Let the volume of the mixture be \( V \) litres. Initially, milk = \( \frac{5}{8}V \) litres and water = \( \frac{3}{8}V \) litres.
We want the final mixture to have milk = water = \( \frac{V}{2} \). This means milk should decrease from \( \frac{5}{8}V \) to \( \frac{1}{2}V \).
The quantity of milk removed when siphoning off \( x \) litres is \( \frac{5}{8}x \), because the ratio in the mixture remains the same while removing.
So the equation becomes: \[ \frac{5}{8}V - \frac{5}{8}x = \frac{1}{2}V \] Simplifying: \[ - \frac{5}{8}x = \frac{1}{2}V - \frac{5}{8}V \] \[ - \frac{5}{8}x = \frac{4}{8}V - \frac{5}{8}V = - \frac{1}{8}V \] Cancelling the negative sign: \[ \frac{5}{8}x = \frac{1}{8}V \] \[ x = \frac{V}{5} \] This means \(\frac{1}{5}\) of the mixture must be removed and replaced with water.