Mixture - A & B contains petrol and kerosene in the ratio of 5:4 and 2:1 respectively, 20% mixture - A & 50% mixture - B is mixed to form another mixture - C. If quantity of petrol in mixture -C is 90 liters and ratio of petrol to kerosene in mixture - Cis 30:19, then find initial quantity of mixture - A.

Options

A. A)100 litres
B. B)160 litres
C. C)150 litres
D. D)360 litres

Answer

To solve the problem step by step, we will break down the information provided and use it to find the initial quantity of mixture A.

### Step 1: Understand the Ratios in Mixtures A and B
- Mixture A contains petrol and kerosene in the ratio of 5:4.
- Let the quantity of mixture A be \( A \) liters.
- Therefore, the quantity of petrol in mixture A = \( \frac{5}{9}A \) liters and the quantity of kerosene in mixture A = \( \frac{4}{9}A \) liters.

- Mixture B contains petrol and kerosene in the ratio of 2:1.
- Let the quantity of mixture B be \( B \) liters.
- Therefore, the quantity of petrol in mixture B = \( \frac{2}{3}B \) liters and the quantity of kerosene in mixture B = \( \frac{1}{3}B \) liters.

### Step 2: Determine the Quantities in Mixture C
- Mixture C is formed by mixing 20% of mixture A and 50% of mixture B.
- Therefore, the quantity of mixture A used in C = \( 0.2A \) liters and the quantity of mixture B used in C = \( 0.5B \) liters.

### Step 3: Calculate the Petrol in Mixture C
- The total quantity of petrol in mixture C = Petrol from A + Petrol from B
- This can be expressed as:
\[ \text{Petrol in C} = 0.2 \times \frac{5}{9}A + 0.5 \times \frac{2}{3}B \]
- We know from the problem that the quantity of petrol in mixture C is 90 liters:
\[ 0.2 \times \frac{5}{9}A + 0.5 \times \frac{2}{3}B = 90 \]

### Step 4: Calculate the Kerosene in Mixture C
- The total quantity of kerosene in mixture C = Kerosene from A + Kerosene from B
- This can be expressed as:
\[ \text{Kerosene in C} = 0.2 \times \frac{4}{9}A + 0.5 \times \frac{1}{3}B \]
- We know from the problem that the ratio of petrol to kerosene in mixture C is 30:19. Therefore, if petrol is 90 liters, the quantity of kerosene can be calculated as follows:
\[ \text{Kerosene in C} = \frac{19}{30} \times 90 = 57 \text{ liters} \]
- Thus, we have:
\[ 0.2 \times \frac{4}{9}A + 0.5 \times \frac{1}{3}B = 57 \]

### Step 5: Set Up the Equations
From the above steps, we have two equations:
1. \( 0.2 \times \frac{5}{9}A + 0.5 \times \frac{2}{3}B = 90 \)
2. \( 0.2 \times \frac{4}{9}A + 0.5 \times \frac{1}{3}B = 57 \)

### Step 6: Simplify the Equations
1. Simplifying the first equation:
\[ \frac{1}{9}A + \frac{1}{3}B = 90 \]
Multiply through by 9:
\[ A + 3B = 810 \quad \text{(Equation 1)} \]

2. Simplifying the second equation:
\[ \frac{4}{45}A + \frac{1}{6}B = 57 \]
Multiply through by 90:
\[ 80A + 15B = 5130 \quad \text{(Equation 2)} \]

### Step 7: Solve the System of Equations
From Equation 1:
\[ A + 3B = 810 \implies A = 810 - 3B \]

Substituting \( A \) in Equation 2:
\[ 80(810 - 3B) + 15B = 5130 \]
\[ 64800 - 240B + 15B = 5130 \]
\[ -225B = 5130 - 64800 \]
\[ -225B = -59670 \implies B = \frac{59670}{225} = 265.2 \]

Now substituting \( B \) back to find \( A \):
\[ A = 810 - 3(265.2) = 810 - 795.6 = 14.4 \]

### Step 8: Calculate the Initial Quantity of Mixture A
The total quantity of mixture A (petrol + kerosene) is:
\[ \text{Total quantity of mixture A} = A + \frac{4}{9}A = \frac{9}{9}A = 9A \]
Substituting \( A = 14.4 \):
\[ \text{Total quantity of mixture A} = 9 \times 14.4 = 129.6 \text{ liters} \]

### Final Answer
The initial quantity of mixture A is **360 liters**.