A mixture contains milk and water in a certain ratio. When 24 liters of milk is added to the mixture, the quantity of water becomes 30\% of the quantity of milk. However, if 24 liters of water is added instead of milk, the quantity of water becomes 14\frac{2}{7}\% less than the quantity of milk. Find the initial quantity of milk in the mixture.

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Accepted Answer

**Solution:**

1. Let the initial quantity of milk be *M* liters and the quantity of water be *W* liters.

2. According to the first condition, when 24 liters of milk is added, the water becomes 30\% of the milk:

$$W = 0.3 (M + 24)$$

3. According to the second condition, when 24 liters of water is added instead of milk, the water quantity becomes $14\frac{2}{7}\%$ less than the milk quantity.

Note that $14\frac{2}{7}\% = \frac{100}{7}\% = 0.142857 \approx 0.1429$.

So, water is $(100\% - 14\frac{2}{7}\%) = 85\frac{5}{7}\% = \frac{6}{7}$ of milk.

Therefore:

$$W + 24 = \frac{6}{7} M$$

4. From the first equation:

$$W = 0.3 M + 7.2$$

5. Substitute *W* from step 4 into the second equation:

$$0.3 M + 7.2 + 24 = \frac{6}{7} M$$

$$0.3 M + 31.2 = \frac{6}{7} M$$

6. Rearranging:

$$\frac{6}{7} M - 0.3 M = 31.2$$

Convert 0.3 to fraction: 0.3 = $\frac{3}{10}$

$$\frac{6}{7} M - \frac{3}{10} M = 31.2$$

Find common denominator 70:

$$\left(\frac{60}{70} - \frac{21}{70}ight) M = 31.2$$

$$\frac{39}{70} M = 31.2$$

7. Solve for *M*:

$$M = \frac{31.2 	imes 70}{39} = \frac{2184}{39} = 56$$

**Answer:** The initial quantity of milk in the mixture is **56 liters**.

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