The ratio of milk to water in mixture A is $9:5$. The quantity of milk in mixture B is $20\%$ more than that in mixture A, and the quantity of water in mixture A is $25\%$ more than that in mixture B. If the total quantity of water in both mixtures combined is $54$ liters, find the total quantity of mixture A.

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**Solution:**

1. Let the quantity of mixture A be $x$ liters.

2. Since the ratio of milk to water in mixture A is $9:5$, the quantity of milk in A is:
   $$\frac{9}{9+5} 	imes x = \frac{9}{14}x$$
   and the quantity of water in A is:
   $$\frac{5}{14}x$$

3. Let the quantity of milk in mixture A be $M_A = \frac{9}{14}x$ liters.

4. The quantity of milk in mixture B is $20\%$ more than in mixture A, so:
   $$M_B = M_A + 0.20 	imes M_A = 1.2 M_A = 1.2 	imes \frac{9}{14}x = \frac{10.8}{14}x$$

5. Let the quantity of water in mixture B be $W_B$ liters.

6. The quantity of water in mixture A is $25\%$ more than that in mixture B, so:
   $$W_A = W_B + 0.25 W_B = 1.25 W_B$$

7. From step 2, water in A is also:
   $$W_A = \frac{5}{14}x$$

8. Equate the two expressions for $W_A$:
   $$\frac{5}{14}x = 1.25 W_B \implies W_B = \frac{5}{14 	imes 1.25} x = \frac{5}{17.5} x = \frac{2}{7} x$$

9. The total water in both mixtures is given as $54$ liters:
   $$W_A + W_B = 54$$
   Substitute values:
   $$\frac{5}{14}x + \frac{2}{7}x = 54$$

10. Find common denominator and simplify:
    $$\frac{5}{14}x + \frac{4}{14}x = \frac{9}{14}x = 54$$

11. Solve for $x$:
    $$x = \frac{54 	imes 14}{9} = 54 	imes \frac{14}{9} = 54 	imes \frac{14}{9} = 6 	imes 14 = 84$$

12. Therefore, the total quantity of mixture A is $84$ liters.

**Answer:** $84 	ext{ liters}$

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