Question

A container is filled with liquid,6 part of which are water and 10 part milk.How much of the mixture must be drawn off and replaced with water,so that the mixture may be half water and half milk?

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{7}\)
• C. \(\frac{1}{5}\)
• D. \(\frac{1}{8}\)

Answer 1

The Correct Option is C

Let the total volume of the mixture be \( V \). Initially, the ratio of water to milk is \( 6 : 10 \), meaning the total volume is divided into 6 parts water and 10 parts milk. Therefore, the amount of water is:

\[ \text{Water} = \frac{6}{16}V \quad \text{and} \quad \text{Milk} = \frac{10}{16}V \]

Let \( x \) be the amount of the mixture drawn off and replaced with water. After removing \( x \) parts of the mixture, the amount of water decreases by \( \frac{6}{16}x \) and the amount of milk decreases by \( \frac{10}{16}x \). When \( x \) parts of water are added back, the amount of water becomes:

\[ \frac{6}{16}V - \frac{6}{16}x + x \]

We want the final mixture to be half water and half milk, so we set up the equation where the amount of water equals the amount of milk:

\[ \frac{6}{16}V - \frac{6}{16}x + x = \frac{10}{16}V - \frac{10}{16}x \]

Simplifying the equation:

\[ \frac{6}{16}V - \frac{6}{16}x + x = \frac{10}{16}V - \frac{10}{16}x \]

\[ \frac{6}{16}V + \left(1 - \frac{6}{16}\right)x = \frac{10}{16}V - \frac{10}{16}x \]

\[ \frac{6}{16}V + \frac{10}{16}x = \frac{10}{16}V \]

Now, solving for \( x \):

\[ \frac{10}{16}x = \frac{10}{16}V - \frac{6}{16}V \]

\[ \frac{10}{16}x = \frac{4}{16}V \]

\[ x = \frac{4}{16}V \times \frac{16}{10} = \frac{V}{5} \]

Thus, the amount of mixture to be drawn off and replaced with water is \( \frac{1}{5} \) of the total volume \( V \).