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

To solve this problem, we first need to understand the mixture composition and the effect of replacing part of it with water.

Initial Composition:

• The mixture is composed of 6 parts water and 10 parts milk.

This means initially, the total number of parts is \(6 + 10 = 16\) parts.

Goal:

• We want a new mixture that is half water and half milk.

This means after replacement, the mixture should have 8 parts water and 8 parts milk.

Steps to Solve:

1. Let's assume the total amount of the mixture is 1 unit (for simplicity).
2. Initially, water fraction = \(\frac{6}{16} = \frac{3}{8}\) and milk fraction = \(\frac{10}{16} = \frac{5}{8}\).
3. Let \(x\) be the fraction of the mixture drawn off and replaced with water.
4. After drawing off \(x\) of the mixture and replacing with water, the amount of water becomes \(\frac{3}{8}(1-x) + x\) and the amount of milk becomes \(\frac{5}{8}(1-x)\).
5. We want these to be equal: \(\frac{3}{8}(1-x) + x = \frac{5}{8}(1-x)\).

Equation:

\[ \frac{3}{8}(1-x) + x = \frac{5}{8}(1-x) \]

6. Simplifying the equation:
   \[ \frac{3}{8} - \frac{3}{8}x + x = \frac{5}{8} - \frac{5}{8}x \]
   \[ \left(\frac{3}{8} + x\right) - \frac{3}{8}x = \frac{5}{8} - \frac{5}{8}x \]
7. Collect terms concerning \(x\) and simplify:
   \[ x - \frac{3}{8}x + \frac{5}{8}x = \frac{5}{8} - \frac{3}{8} \]
8. Further simplify:
   \[ \frac{5}{8} = \frac{5}{8} + \frac{3}{8} \]
9. Simplifying, we get:
   \[ x = \frac{1}{5} \]

Conclusion:

This means that \(\frac{1}{5}\) of the mixture needs to be replaced with water to achieve a half water and half milk mixture. Thus, the correct answer is \(\frac{1}{5}\).

Answer 2

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 \( V \).