A vessel initially contains 2.5 liters of water and 10 liters of milk. Then, 20\% of the total contents of the vessel are removed. To the remaining mixture, $x$ liters of water is added to reverse the ratio of water to milk. After that, $y$ liters of milk is added to reverse the ratio of water to milk once again. Find the value of $y$.

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**Solution:**

1. Initially, the vessel contains $2.5$ liters of water and $10$ liters of milk, so total contents = $2.5 + 10 = 12.5$ liters.

2. After removing 20\% of the contents, the remaining quantity is:
   $$\frac{80}{100} \times 12.5 = 10 \text{ liters}$$

3. The ratio of water to milk remains the same after removal, so:
   - Water = $\frac{2.5}{12.5} \times 10 = 2 \text{ liters}$
   - Milk = $\frac{10}{12.5} \times 10 = 8 \text{ liters}$

4. The current ratio of water to milk is $2 : 8 = 1 : 4$.

5. To reverse the ratio, water should become four times the milk. Let the new quantity of water after adding $x$ liters be:
   $$2 + x = 4 \times 8 = 32$$
   So,
   $$x = 32 - 2 = 30 \text{ liters}$$

6. Now, the mixture has water = 32 liters and milk = 8 liters.

7. To reverse the ratio again, milk should become four times the water. Let $y$ liters of milk be added:
   $$8 + y = 4 \times 32 = 128$$
   Therefore,
   $$y = 128 - 8 = 120 \text{ liters}$$

**Answer:** $120$

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