In a vessel, water constitutes 25\% of the milk. When 10 liters of the mixture is removed from the vessel and then 25 liters of milk is added, the quantity of milk becomes 400\% more than the quantity of water in the resulting mixture. What was the initial quantity of milk (in liters) in the vessel?

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**Solution:**

1. Let the initial quantity of milk be $4x$ liters.

2. Since water is 25\% of milk, the quantity of water is:
   $$4x 	imes \frac{25}{100} = x 	ext{ liters}$$

3. Therefore, the total initial mixture quantity is:
   $$4x + x = 5x 	ext{ liters}$$

4. When 10 liters of mixture is removed, the amount of milk and water removed will be in the same ratio as in the mixture:
   - Milk removed: $$10 	imes \frac{4x}{5x} = 8 	ext{ liters}$$
   - Water removed: $$10 	imes \frac{x}{5x} = 2 	ext{ liters}$$

5. After removal, the quantities left are:
   - Milk: $$4x - 8$$
   - Water: $$x - 2$$

6. Then, 25 liters of milk is added, so new milk quantity becomes:
   $$4x - 8 + 25 = 4x + 17$$

7. According to the problem, the milk quantity is 400\% more than water, which means milk is 5 times water:
   $$4x + 17 = 5(x - 2)$$

8. Simplify the equation:
   $$4x + 17 = 5x - 10$$
   $$17 + 10 = 5x - 4x$$
   $$27 = x$$

9. Initial quantity of milk:
   $$4x = 4 	imes 27 = 108 	ext{ liters}$$

**Answer:** 108

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