A container initially holds 90 liters of milk and 18 liters of water. If *X* liters of the mixture are removed and then 6 liters of water are added, the new ratio of milk to water becomes $5:2$. Find the value of *X*.

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**Solution:**

1. Initially, the container has 90 liters of milk and 18 liters of water, so total mixture = $90 + 18 = 108$ liters.

2. When *X* liters of mixture is taken out, the quantity of milk and water removed will be in the same ratio as in the mixture, which is $90:18 = 5:1$.

3. Milk removed = $\frac{5}{6} 	imes X$ liters (since total parts = 5 + 1 = 6)

4. Water removed = $\frac{1}{6} 	imes X$ liters

5. After removal, milk left = $90 - \frac{5}{6}X$

6. Water left = $18 - \frac{1}{6}X$

7. Then 6 liters of water is added, so new water quantity = $18 - \frac{1}{6}X + 6 = 24 - \frac{1}{6}X$

8. According to the problem, the new ratio of milk to water is $5:2$, so:

$$\frac{90 - \frac{5}{6}X}{24 - \frac{1}{6}X} = \frac{5}{2}$$

9. Cross-multiply:

$$2 \left(90 - \frac{5}{6}Xight) = 5 \left(24 - \frac{1}{6}Xight)$$

10. Simplify:

$$180 - \frac{10}{6}X = 120 - \frac{5}{6}X$$

11. Rearranging terms:

$$180 - 120 = \frac{10}{6}X - \frac{5}{6}X$$

$$60 = \frac{5}{6}X$$

12. Multiply both sides by 6:

$$360 = 5X$$

13. Divide both sides by 5:

$$X = 72$$

**Answer:** $72$ liters

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