A vessel contains a total of 95 liters of a mixture of milk and water in the ratio of 15 : 4 respectively. If *X* liters of the mixture is removed from the vessel and then 18 liters of water is added to the remaining mixture, the new ratio of milk to water becomes 3 : 2. Find the value of *X*.

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**Solution:**

1. Calculate the initial quantities of milk and water in the mixture:
   - Total mixture = 95 liters
   - Ratio of milk to water = 15 : 4

2. Total parts = 15 + 4 = 19

3. Milk in mixture initially:
   $$ 95 	imes \frac{15}{19} = 75 	ext{ liters} $$

4. Water in mixture initially:
   $$ 95 	imes \frac{4}{19} = 20 	ext{ liters} $$

5. Let *X* liters of mixture be removed. Since the mixture is uniform, the amount of milk and water removed will be in the same ratio 15:4.

6. Milk left after removal:
   $$ 75 - X 	imes \frac{15}{19} $$

7. Water left after removal:
   $$ 20 - X 	imes \frac{4}{19} $$

8. After removing *X* liters, 18 liters of water is added. So, new water quantity:
   $$ 20 - X 	imes \frac{4}{19} + 18 $$

9. According to the problem, the new ratio of milk to water is 3 : 2:
   $$ \frac{75 - X 	imes \frac{15}{19}}{20 - X 	imes \frac{4}{19} + 18} = \frac{3}{2} $$

10. Cross-multiply and solve for *X*:

$$ 2 \left(75 - \frac{15X}{19}ight) = 3 \left(20 - \frac{4X}{19} + 18ight) $$

$$ 150 - \frac{30X}{19} = 3 \left(38 - \frac{4X}{19}ight) $$

$$ 150 - \frac{30X}{19} = 114 - \frac{12X}{19} $$

$$ 150 - 114 = \frac{30X}{19} - \frac{12X}{19} $$

$$ 36 = \frac{18X}{19} $$

$$ X = \frac{36 	imes 19}{18} = 38 $$

**Answer:** $38$ liters

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