A vessel contains 79.99 liters of a mixture of milk and water, where the quantity of milk is 50% more than that of water. From this vessel, 20.03 liters of the mixture is removed, and then 59.99 liters of another mixture of milk and water is added to the remaining mixture in the vessel. If in the final mixture, the quantity of milk is twice the quantity of water, find the approximate quantity (in liters) of water added in the new mixture.

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**Solution:**

1. Let us approximate the total mixture as 80 liters for ease of calculation.

2. Given that milk is 50% more than water, let the quantity of water be *w* liters.
   Then milk = *w + 0.5w = 1.5w* liters.

3. Total mixture = milk + water = *w + 1.5w = 2.5w = 80* liters.
   So, *w = 80 / 2.5 = 32* liters (water), and milk = *1.5 	imes 32 = 48* liters.

4. When 20 liters of mixture is removed, the ratio of milk to water remains the same.
   - Milk removed = *20 	imes rac{3}{5} = 12* liters (since milk:water = 3:2)
   - Water removed = *20 	imes rac{2}{5} = 8* liters

5. Remaining milk = *48 - 12 = 36* liters
   Remaining water = *32 - 8 = 24* liters

6. Now, 60 liters of a new mixture is added, containing *x* liters of milk and *(60 - x)* liters of water.

7. After addition, total milk = *36 + x* liters
   Total water = *24 + (60 - x) = 84 - x* liters

8. According to the problem, milk is twice the water in the final mixture:
   $$36 + x = 2(84 - x)$$

9. Simplify the equation:
   $$36 + x = 168 - 2x$$
   $$x + 2x = 168 - 36$$
   $$3x = 132$$
   $$x = 44$$ liters (milk in the added mixture)

10. Quantity of water added = *60 - 44 = 16* liters

**Answer:** 16 liters

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