Vessel A contains 150 liters of a mixture of milk and water, where water makes up 40% of the mixture. From this vessel, $x$ liters of the mixture is removed such that the quantity of water in these $x$ liters is 12 liters. This $x$ liters of mixture is then poured into vessel B, which already contains $x$ liters of water. What is the difference between the quantity of milk and water (in liters) in the resulting mixture in vessel B?

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**Solution:**

1. Total mixture in vessel A = 150 liters.

2. Water content in vessel A = 40% of 150 = $0.40 	imes 150 = 60$ liters.

3. Milk content in vessel A = Total - Water = $150 - 60 = 90$ liters.

4. Let $x$ liters of mixture be taken out from vessel A. Given that water in $x$ liters = 12 liters.

5. Since water is 40% of the mixture in vessel A, water in $x$ liters should be $0.40 	imes x = 12$ liters.

6. From this, $x = \frac{12}{0.40} = 30$ liters.

7. Amount of milk in $x$ liters taken out = $x -$ water = $30 - 12 = 18$ liters.

8. Vessel B initially contains $x = 30$ liters of water.

9. After adding $x$ liters of mixture from vessel A, vessel B will have:
   - Water = initial water + water from mixture = $30 + 12 = 42$ liters
   - Milk = milk from mixture = 18 liters

10. Difference between quantity of milk and water in vessel B = $|42 - 18| = 24$ liters.

**Answer:** 24

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