Vessel X initially contains a mixture of milk and water in the ratio $5:4$. When $9$ litres of this mixture is taken out, $20$ litres of another mixture from vessel Y, which has milk and water in the ratio $3:2$, is added to the remaining mixture in vessel X. After this operation, the ratio of milk to water in vessel X becomes $4:3$. What was the initial quantity of the mixture in vessel X?

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**Solution:**

1. Let the initial quantity of mixture in vessel X be $x$ litres.

2. Milk and water in vessel X are in the ratio $5:4$, so:
   - Milk in vessel X initially = $\frac{5}{9}x$
   - Water in vessel X initially = $\frac{4}{9}x$

3. When $9$ litres of mixture is removed from vessel X, the quantities of milk and water removed are:
   - Milk removed = $9 	imes \frac{5}{9} = 5$ litres
   - Water removed = $9 	imes \frac{4}{9} = 4$ litres

4. Remaining mixture in vessel X after removal:
   - Milk = $\frac{5}{9}x - 5$
   - Water = $\frac{4}{9}x - 4$

5. Mixture added from vessel Y is $20$ litres with milk and water in ratio $3:2$:
   - Milk added = $20 	imes \frac{3}{5} = 12$ litres
   - Water added = $20 	imes \frac{2}{5} = 8$ litres

6. After mixing, total milk and water in vessel X are:
   - Milk = $\left(\frac{5}{9}x - 5ight) + 12 = \frac{5}{9}x + 7$
   - Water = $\left(\frac{4}{9}x - 4ight) + 8 = \frac{4}{9}x + 4$

7. The new ratio of milk to water is $4:3$, so:
   $$\frac{\frac{5}{9}x + 7}{\frac{4}{9}x + 4} = \frac{4}{3}$$

8. Cross-multiplied:
   $$3\left(\frac{5}{9}x + 7ight) = 4\left(\frac{4}{9}x + 4ight)$$

9. Simplify:
   $$\frac{15}{9}x + 21 = \frac{16}{9}x + 16$$

10. Multiply both sides by 9 to clear denominators:
    $$15x + 189 = 16x + 144$$

11. Rearranged:
    $$189 - 144 = 16x - 15x$$
    $$45 = x$$

**Answer:** The initial quantity of mixture in vessel X is $45$ litres.

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