Two vessels X and Y contain mixtures of milk and water. Vessel X has milk and water in the ratio $5 : 3$, while vessel Y has milk and water in the ratio $7 : 5$. The difference between the quantity of milk and water in vessel Y is equal to the quantity of water in vessel X. Also, the amount of water in vessel Y is 5 liters more than the amount of milk in vessel X. What is the total quantity of mixture in vessel X?

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**Solution:**

1. Let the quantities of milk and water in vessel X be $5x$ liters and $3x$ liters respectively.

2. Let the quantities of milk and water in vessel Y be $7y$ liters and $5y$ liters respectively.

3. According to the problem, the difference between milk and water in vessel Y equals the water in vessel X:
   $$7y - 5y = 3x \ 2y = 3x$$
   This implies the ratio:
   $$\frac{y}{x} = \frac{3}{2}$$

4. Let $y = 3a$ and $x = 2a$ for some positive number $a$.

5. It is also given that the water in vessel Y is 5 liters more than the milk in vessel X:
   $$5y = 5x + 5$$
   Substitute $y = 3a$ and $x = 2a$:
   $$5 	imes 3a = 5 	imes 2a + 5 \ 15a = 10a + 5 \ 15a - 10a = 5 \ 5a = 5 \ a = 1$$

6. Now, find the total mixture in vessel X:
   $$5x + 3x = 8x = 8 	imes 2a = 8 	imes 2 	imes 1 = 16 	ext{ liters}$$

**Answer:** $16$ liters

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