A mixture contains milk and water in the ratio of $4:1$. When $10$ liters of this mixture is removed, the quantity of water becomes $\frac{2}{3}$ of the quantity of milk. What was the initial quantity of water (in liters)?

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**Solution:**

1. Let the initial quantity of milk be $4x$ liters and water be $x$ liters, since the ratio is $4:1$.

2. Total initial mixture = $4x + x = 5x$ liters.

3. When $10$ liters of mixture is taken out, the quantity of milk and water removed will be in the same ratio $4:1$.

4. Milk removed = $\frac{4}{5} 	imes 10 = 8$ liters.

5. Water removed = $\frac{1}{5} 	imes 10 = 2$ liters.

6. Remaining milk = $4x - 8$ liters.

7. Remaining water = $x - 2$ liters.

8. According to the problem, after removal, water is $\frac{2}{3}$ of milk:
   $$x - 2 = \frac{2}{3} (4x - 8)$$

9. Multiply both sides by $3$:
   $$3(x - 2) = 2(4x - 8)$$

10. Simplify:
    $$3x - 6 = 8x - 16$$

11. Rearranging:
    $$8x - 3x = 16 - 6$$
    $$5x = 10$$

12. Solve for $x$:
    $$x = 2$$

13. Initial quantity of water = $x = 2$ liters.

**Answer:** $2$ liters

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