6 litre is taken out from a vessel full of Kerosene and substituted by pure petrol. This process is repeated two more times. Finally the ratio of petrol and Kerosene in the mixture becomes `1701:27`. Find the volume of the original solution:

Options

A. 14 litre
B. 16 litre
C. 8 litre
D. 42 litre

Answer

To solve the problem step by step, we will analyze the process of replacing kerosene with petrol and derive the original volume of the solution.

### Step 1: Understand the Replacement Process
When 6 liters of kerosene is removed from a vessel and replaced with 6 liters of petrol, this process is repeated three times. We need to find the original volume of the solution based on the final ratio of petrol to kerosene.

### Step 2: Set Up the Final Ratio
The final ratio of petrol to kerosene after three replacements is given as:
\[ \text{Petrol} : \text{Kerosene} = 1701 : 27 \]

### Step 3: Calculate Total Volume
The total volume of the mixture can be calculated as:
\[ \text{Total Volume} = 1701 + 27 = 1728 \text{ units} \]

### Step 4: Determine the Ratio of Kerosene to Total Mixture
The ratio of kerosene in the total mixture after three replacements is:
\[ \frac{\text{Kerosene}}{\text{Total Mixture}} = \frac{27}{1728} \]

### Step 5: Find the Ratio After One Replacement
To find the ratio after one replacement, we take the cube root of the ratio of kerosene to the total mixture after three replacements:
\[ \text{Ratio after one replacement} = \sqrt[3]{\frac{27}{1728}} \]

### Step 6: Simplify the Ratio
Calculating the cube root:
\[ \sqrt[3]{27} = 3 \quad \text{and} \quad \sqrt[3]{1728} = 12 \]
Thus, the ratio after one replacement becomes:
\[ \frac{3}{12} = \frac{1}{4} \]

### Step 7: Relate the Ratio to the Original Volume
Let the original volume of the solution be \( V \). After one replacement, the volume of kerosene remaining is:
\[ \frac{1}{4} \text{ of } V \]
This means that the volume of kerosene remaining is \( \frac{1}{4}V \) and the volume of petrol is \( V - \frac{1}{4}V = \frac{3}{4}V \).

### Step 8: Set Up the Equation
Since we know that 6 liters were removed (which corresponds to the 3 units of petrol added), we can set up the equation:
\[ \frac{3}{4}V = 6 \text{ liters} \]

### Step 9: Solve for \( V \)
To find \( V \):
\[ V = \frac{6 \times 4}{3} = 8 \text{ liters} \]

### Conclusion
The original volume of the solution is:
\[ \boxed{8 \text{ liters}} \]