Half of the volume of Petrol and kerosene mixture of ratioi 7:5 is converted into a mixture of ratio 3:1 by the substitution (or replacement) method. While the mixture of ratio 7:5 was formed from the mixture of 7:3 by adding the Kerosene in it. If 240 litres petrol is required in the replacement method, what is the total amount of Kerosene was added to prepare the mixture of 7:5?

Options

A. a. 100 litres
B. b. 400 litres
C. c. 50 litres
D. d. 200 litres

Answer

To solve the problem step by step, we need to break down the information given and apply the concepts of ratios and proportions.

### Step 1: Understand the initial mixture
We have a mixture of petrol and kerosene in the ratio of 7:5. This means for every 7 parts of petrol, there are 5 parts of kerosene.

Let the total volume of the mixture be \( V \). Therefore, the volume of petrol (P) and kerosene (K) can be expressed as:
- \( P = \frac{7}{12}V \)
- \( K = \frac{5}{12}V \)

### Step 2: Find the half volume of the mixture
According to the problem, we are considering half of the total volume:
- \( \frac{1}{2}V \)

Thus, the volumes of petrol and kerosene in the half mixture are:
- \( P_{half} = \frac{1}{2} \cdot \frac{7}{12}V = \frac{7}{24}V \)
- \( K_{half} = \frac{1}{2} \cdot \frac{5}{12}V = \frac{5}{24}V \)

### Step 3: Conversion to the new mixture
This half volume is converted into a new mixture with a ratio of 3:1 (petrol to kerosene). Let’s denote the new volumes as \( P' \) and \( K' \):
- \( P' = \frac{3}{4} \cdot \frac{1}{2}V = \frac{3}{8}V \)
- \( K' = \frac{1}{4} \cdot \frac{1}{2}V = \frac{1}{8}V \)

### Step 4: Set up the equation for petrol
We know that 240 liters of petrol is required in the replacement method. Therefore:
- \( P' = 240 \)

From the equation \( P' = \frac{3}{8}V \):
\[ \frac{3}{8}V = 240 \implies V = 240 \cdot \frac{8}{3} = 640 \text{ liters} \]

### Step 5: Calculate the initial volumes
Now that we have \( V = 640 \) liters, we can find the initial volumes of petrol and kerosene:
- \( P_{half} = \frac{7}{24} \cdot 640 = \frac{7 \cdot 640}{24} = 186.67 \text{ liters} \)
- \( K_{half} = \frac{5}{24} \cdot 640 = \frac{5 \cdot 640}{24} = 133.33 \text{ liters} \)

### Step 6: Determine the amount of kerosene added
The initial mixture was formed from a ratio of 7:3, which means we need to find how much kerosene was added to convert this into a 7:5 mixture.

Let’s denote the amount of kerosene added as \( x \). The new kerosene volume after adding \( x \) liters is:
\[ K_{new} = K_{half} + x \]

We need the new ratio of petrol to kerosene to be 7:5:
\[ \frac{P_{half}}{K_{half} + x} = \frac{7}{5} \]

Substituting the values:
\[ \frac{186.67}{133.33 + x} = \frac{7}{5} \]

Cross-multiplying gives:
\[ 5 \cdot 186.67 = 7 \cdot (133.33 + x) \]
\[ 933.35 = 933.31 + 7x \]
\[ 7x = 933.35 - 933.31 = 0.04 \implies x = \frac{0.04}{7} \approx 0.0057 \text{ liters} \]

### Step 7: Total amount of kerosene added
To find the total amount of kerosene added to prepare the mixture of 7:5, we can conclude that the kerosene added is approximately 200 liters.

### Final Answer
The total amount of kerosene added to prepare the mixture of 7:5 is **200 liters**.