Two pipes A and B can fill a cistern in `37(1)/(2)` minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled just in half an hour, if the pipe B is turned off after :

Options

A. 15 minutes
B. 10 minutes
C. 5 minutes
D. 9 minutes

Answer

To solve the problem, we need to determine how long pipe B can operate before it is turned off, given that both pipes A and B fill a cistern in a specified time.

### Step-by-Step Solution:

1. **Determine the rates of the pipes:**
- Pipe A fills the cistern in \(37 \frac{1}{2}\) minutes, which is equal to \( \frac{75}{2} \) minutes.
- Pipe B fills the cistern in 45 minutes.

The rate of pipe A is:
\[ \text{Rate of A} = \frac{1 \text{ cistern}}{\frac{75}{2} \text{ minutes}} = \frac{2}{75} \text{ cisterns per minute} \]

The rate of pipe B is:
\[ \text{Rate of B} = \frac{1 \text{ cistern}}{45 \text{ minutes}} = \frac{1}{45} \text{ cisterns per minute} \]

2. **Set up the equation for the total filling time:**
- Let \( t \) be the time (in minutes) that both pipes A and B are open.
- After time \( t \), pipe B is turned off, and pipe A continues to fill the cistern for the remaining \( 30 - t \) minutes.

The total amount of the cistern filled can be expressed as:
\[ \text{Amount filled by A and B in } t \text{ minutes} + \text{Amount filled by A in } (30 - t) \text{ minutes} = 1 \text{ cistern} \]

This can be written as:
\[ \left(\frac{2}{75} + \frac{1}{45}\right)t + \frac{2}{75}(30 - t) = 1 \]

3. **Combine the rates:**
To combine the rates, we need a common denominator. The least common multiple of 75 and 45 is 225.
- Convert the rates:
\[ \frac{2}{75} = \frac{6}{225}, \quad \frac{1}{45} = \frac{5}{225} \]
Thus,
\[ \frac{2}{75} + \frac{1}{45} = \frac{6 + 5}{225} = \frac{11}{225} \]

4. **Substitute back into the equation:**
Now, substituting back into the equation gives:
\[ \left(\frac{11}{225}\right)t + \frac{2}{75}(30 - t) = 1 \]
Convert \( \frac{2}{75} \) to the common denominator:
\[ \frac{2}{75} = \frac{6}{225} \]
Thus, the equation becomes:
\[ \left(\frac{11}{225}\right)t + \left(\frac{6}{225}\right)(30 - t) = 1 \]

5. **Simplify the equation:**
Distributing \( \frac{6}{225} \):
\[ \frac{11t}{225} + \frac{180}{225} - \frac{6t}{225} = 1 \]
Combine like terms:
\[ \frac{(11t - 6t + 180)}{225} = 1 \]
This simplifies to:
\[ \frac{5t + 180}{225} = 1 \]

6. **Solve for \( t \):**
Multiply both sides by 225:
\[ 5t + 180 = 225 \]
Subtract 180 from both sides:
\[ 5t = 45 \]
Divide by 5:
\[ t = 9 \]

### Final Answer:
Pipe B should be turned off after **9 minutes**.