Vessel A contains a mixture of milk and water in the ratio of $3 : 1$ respectively. From vessel A, $20$ liters of this mixture is taken out and added to vessel B, which initially contains $x$ liters of milk and $30$ liters of water. After this addition, the ratio of milk to water in vessel B becomes $11 : 7$. Find the value of $x$.

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**Solution:**

1. The ratio of milk to water in vessel A is $3 : 1$, so in 20 liters of mixture taken from vessel A:
   - Milk = $\frac{3}{4} 	imes 20 = 15$ liters
   - Water = $\frac{1}{4} 	imes 20 = 5$ liters

2. Vessel B initially contains $x$ liters of milk and $30$ liters of water.

3. After adding the 20 liters from vessel A, the total milk in vessel B becomes:
   $$x + 15$$ liters

4. The total water in vessel B becomes:
   $$30 + 5 = 35$$ liters

5. The new ratio of milk to water in vessel B is given as $11 : 7$, so:
   $$\frac{x + 15}{35} = \frac{11}{7}$$

6. Cross-multiplying:
   $$7(x + 15) = 11 	imes 35$$
   $$7x + 105 = 385$$

7. Subtract 105 from both sides:
   $$7x = 280$$

8. Divide both sides by 7:
   $$x = 40$$

**Answer:** $40$

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