Mixture A and Mixture B contain petrol and kerosene in the ratios of $5 : 4$ and $2 : 1$ respectively. From these, $20\%$ of mixture A and $50\%$ of mixture B are combined to form a new mixture C. If the quantity of petrol in mixture C is $90$ liters and the ratio of petrol to kerosene in mixture C is $30 : 19$, what was the initial quantity of mixture A?

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**Solution:**

1. Let the initial quantity of mixture A be $x$ liters and that of mixture B be $y$ liters.

2. Mixture A has petrol and kerosene in the ratio $5 : 4$, so petrol in A = $\frac{5}{9}x$ and kerosene in A = $\frac{4}{9}x$.

3. Mixture B has petrol and kerosene in the ratio $2 : 1$, so petrol in B = $\frac{2}{3}y$ and kerosene in B = $\frac{1}{3}y$.

4. From mixture A, $20\%$ is taken, so quantity taken from A = $0.2x$ liters.
   - Petrol from A in mixture C = $0.2x 	imes \frac{5}{9} = \frac{1}{9}x$
   - Kerosene from A in mixture C = $0.2x 	imes \frac{4}{9} = \frac{4}{45}x$

5. From mixture B, $50\%$ is taken, so quantity taken from B = $0.5y$ liters.
   - Petrol from B in mixture C = $0.5y 	imes \frac{2}{3} = \frac{1}{3}y$
   - Kerosene from B in mixture C = $0.5y 	imes \frac{1}{3} = \frac{1}{6}y$

6. Total petrol in mixture C = petrol from A + petrol from B = $\frac{1}{9}x + \frac{1}{3}y = 90$ liters (given).

7. The ratio of petrol to kerosene in mixture C is $30 : 19$, so:
   $$\frac{	ext{Petrol}}{	ext{Kerosene}} = \frac{30}{19}$$

Total kerosene in mixture C = kerosene from A + kerosene from B = $\frac{4}{45}x + \frac{1}{6}y$

Therefore,
   $$\frac{\frac{1}{9}x + \frac{1}{3}y}{\frac{4}{45}x + \frac{1}{6}y} = \frac{30}{19}$$

8. Substitute petrol total as 90 liters from step 6:
   $$\frac{90}{\frac{4}{45}x + \frac{1}{6}y} = \frac{30}{19}$$

9. Cross-multiplied:
   $$90 	imes 19 = 30 	imes \left(\frac{4}{45}x + \frac{1}{6}yight)$$
   $$1710 = 30 	imes \left(\frac{4}{45}x + \frac{1}{6}yight)$$

10. Simplify the right side:
    $$1710 = 30 	imes \left(\frac{4}{45}x + \frac{1}{6}yight) = 30 	imes \frac{4}{45}x + 30 	imes \frac{1}{6}y = \frac{120}{45}x + 5y = \frac{8}{3}x + 5y$$

11. So,
    $$\frac{8}{3}x + 5y = 1710$$

12. From step 6, petrol equation:
    $$\frac{1}{9}x + \frac{1}{3}y = 90$$
    Multiply both sides by 9:
    $$x + 3y = 810$$

13. Now solve the system:
    $$x + 3y = 810 \quad ...(i)$$
    $$\frac{8}{3}x + 5y = 1710 \quad ...(ii)$$

Multiply equation (i) by 5:
    $$5x + 15y = 4050$$

Multiply equation (ii) by 3:
    $$8x + 15y = 5130$$

14. Subtract the first from the second:
    $$8x + 15y - (5x + 15y) = 5130 - 4050$$
    $$3x = 1080$$
    $$x = 360$$

15. Substitute $x=360$ into equation (i):
    $$360 + 3y = 810$$
    $$3y = 450$$
    $$y = 150$$

**Answer:** The initial quantity of mixture A is $\boxed{360}$ liters.

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