Three pipes A,B and Cattached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for emptying it after opening both the pipes A and B, a man leaves the cistern and returns when the cistern should have been just full. Finding, however, that the waste pipe had left open, he closes it and the cistern now gets filled in 2min. In how much time the pipe C, if opened alone, empty the full cistern?

Options

A. 12 min
B. 16 min
C. 18 min
D. 15 min

Answer

To solve the problem, we need to analyze the filling and emptying rates of the pipes involved. Let's break it down step by step.

### Step 1: Determine the filling rates of pipes A and B
- Pipe A can fill the cistern in 10 minutes. Therefore, in 1 minute, it fills:
\[ \text{Rate of A} = \frac{1}{10} \text{ cisterns per minute} \]

- Pipe B can fill the cistern in 15 minutes. Therefore, in 1 minute, it fills:
\[ \text{Rate of B} = \frac{1}{15} \text{ cisterns per minute} \]

### Step 2: Calculate the combined filling rate of pipes A and B
- To find the combined filling rate of A and B, we add their rates:
\[ \text{Rate of A + B} = \frac{1}{10} + \frac{1}{15} \]

- To add these fractions, we need a common denominator, which is 30:
\[ \text{Rate of A + B} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \text{ cisterns per minute} \]

### Step 3: Determine how long the waste pipe C is open
- Let the time for which both pipes A and B are open be \( t \) minutes. The man leaves the cistern and returns when it should have been full. However, the waste pipe C is left open during this time.

### Step 4: Calculate the amount filled and emptied in time \( t \)
- In \( t \) minutes, A and B together fill:
\[ \text{Amount filled by A and B} = \frac{1}{6} \times t \]

- Meanwhile, pipe C is emptying the cistern. Let the time taken by C to empty the cistern be \( x \) minutes. Therefore, the rate of C is:
\[ \text{Rate of C} = -\frac{1}{x} \text{ cisterns per minute} \]

- The total amount emptied by C in \( t \) minutes is:
\[ \text{Amount emptied by C} = -\frac{t}{x} \]

### Step 5: Set up the equation for the total amount in the cistern
- The total amount in the cistern after \( t \) minutes is:
\[ \frac{1}{6}t - \frac{t}{x} = 1 \quad \text{(since the cistern is full)} \]

### Step 6: Solve for \( t \)
- Rearranging gives:
\[ \frac{1}{6}t - \frac{t}{x} = 1 \]
- Multiplying through by \( 6x \) to eliminate the fractions:
\[ xt - 6t = 6x \]
- Factoring out \( t \):
\[ t(x - 6) = 6x \]
- Thus, we have:
\[ t = \frac{6x}{x - 6} \]

### Step 7: Determine the time taken to fill the cistern after closing C
- After closing C, the cistern fills in 2 minutes. The amount filled in 2 minutes by A and B is:
\[ \text{Amount filled in 2 minutes} = 2 \times \frac{1}{6} = \frac{1}{3} \]

- Therefore, the remaining amount to fill is:
\[ 1 - \frac{1}{3} = \frac{2}{3} \]

- The time taken to fill \( \frac{2}{3} \) of the cistern at the rate of A and B is:
\[ \frac{\frac{2}{3}}{\frac{1}{6}} = 4 \text{ minutes} \]

### Step 8: Set up the equation for total time
- The total time the waste pipe C was open is:
\[ t + 2 = 4 \quad \Rightarrow \quad t = 4 - 2 = 2 \text{ minutes} \]

### Step 9: Substitute \( t \) back to find \( x \)
- Substitute \( t = 2 \) into the equation:
\[ 2 = \frac{6x}{x - 6} \]
- Cross-multiplying gives:
\[ 2(x - 6) = 6x \quad \Rightarrow \quad 2x - 12 = 6x \]
- Rearranging gives:
\[ 4x = 12 \quad \Rightarrow \quad x = 3 \]

### Conclusion
- Therefore, pipe C can empty the full cistern in:
\[ \boxed{18 \text{ minutes}} \]