A pipe can fill a cistern in 20 minutes where as the cistern when full can be emptied by a leak in 28 minutes. When both are opened, The time taken to fill the cistern is:

Options

A. 80 min
B. 70 min
C. 60 min
D. 48 min

Answer

To solve the problem, we need to determine how much of the cistern is filled by the pipe and how much is emptied by the leak when both are opened simultaneously.

### Step-by-Step Solution:

1. **Determine the filling rate of the pipe:**
The pipe can fill the cistern in 20 minutes. Therefore, the rate of filling by the pipe is:
\[ \text{Filling rate of pipe} = \frac{1 \text{ cistern}}{20 \text{ minutes}} = \frac{1}{20} \text{ cistern per minute} \]

2. **Determine the emptying rate of the leak:**
The leak can empty the full cistern in 28 minutes. Therefore, the rate of emptying by the leak is:
\[ \text{Emptying rate of leak} = \frac{1 \text{ cistern}}{28 \text{ minutes}} = \frac{1}{28} \text{ cistern per minute} \]

3. **Calculate the net rate when both are opened:**
When both the pipe and the leak are opened, the net rate of filling the cistern is the filling rate minus the emptying rate:
\[ \text{Net rate} = \text{Filling rate of pipe} - \text{Emptying rate of leak} = \frac{1}{20} - \frac{1}{28} \]

To perform this subtraction, we need a common denominator. The least common multiple of 20 and 28 is 140. Thus:
\[ \frac{1}{20} = \frac{7}{140} \quad \text{and} \quad \frac{1}{28} = \frac{5}{140} \]
Now, substituting these values:
\[ \text{Net rate} = \frac{7}{140} - \frac{5}{140} = \frac{2}{140} = \frac{1}{70} \text{ cistern per minute} \]

4. **Calculate the time taken to fill the cistern:**
If the net rate of filling is \(\frac{1}{70}\) cistern per minute, then the time taken to fill 1 cistern is the reciprocal of the net rate:
\[ \text{Time taken} = \frac{1 \text{ cistern}}{\frac{1}{70} \text{ cistern per minute}} = 70 \text{ minutes} \]

### Final Answer:
The time taken to fill the cistern when both the pipe and the leak are opened is **70 minutes**.