Two pipes A and B can fill a cistern in 54 mintutes and 45 mintutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if pipe B is turned off after

Options

A. 5 minutes
B. 9 minutes
C. 10 minutes
D. 20 mintues

Answer

To solve the problem step by step, we will first determine the rates at which pipes A and B fill the cistern, then calculate how long pipe B can remain open while ensuring the cistern is filled in 30 minutes.

### Step 1: Determine the filling rates of pipes A and B
- Pipe A can fill the cistern in 54 minutes.
- Pipe B can fill the cistern in 45 minutes.

The rate of filling for each pipe can be calculated as follows:
- Rate of A = 1 unit / 54 minutes = \( \frac{1}{54} \) units per minute.
- Rate of B = 1 unit / 45 minutes = \( \frac{1}{45} \) units per minute.

### Step 2: Calculate the total amount of water filled in 30 minutes
Let’s denote the time for which pipe B is open as \( t \) minutes. Then, pipe A will be open for the entire 30 minutes, while pipe B will be open for \( t \) minutes.

The total amount of water filled by both pipes in 30 minutes can be expressed as:
- Water filled by A in 30 minutes = \( 30 \times \frac{1}{54} = \frac{30}{54} = \frac{5}{9} \) units.
- Water filled by B in \( t \) minutes = \( t \times \frac{1}{45} \) units.

### Step 3: Set up the equation for total filling
Since the total capacity of the cistern is 1 unit, we can set up the equation:
\[ \frac{5}{9} + \frac{t}{45} = 1 \]

### Step 4: Solve for \( t \)
To solve for \( t \), first isolate \( \frac{t}{45} \):
\[ \frac{t}{45} = 1 - \frac{5}{9} \]

Calculating the right side:
\[ 1 - \frac{5}{9} = \frac{9}{9} - \frac{5}{9} = \frac{4}{9} \]

Now substituting back:
\[ \frac{t}{45} = \frac{4}{9} \]

To find \( t \), multiply both sides by 45:
\[ t = 45 \times \frac{4}{9} = 5 \times 4 = 20 \text{ minutes} \]

### Conclusion
Pipe B should be turned off after 20 minutes to ensure the cistern is filled in 30 minutes.