Two pipes A and B can fill a tank in 12 and 16 min, respectively. If both the pipes are opened simultaneously , after how much time should B be closed so that the tank is full 9 min ?

Answer

To solve the problem step by step, we need to determine how long pipe B should remain open while both pipes A and B are filling the tank, so that the tank is full in a total of 9 minutes.

### Step-by-Step Solution:

1. **Determine the rates of filling for pipes A and B:**
- Pipe A can fill the tank in 12 minutes, so its rate is \( \frac{1}{12} \) of the tank per minute.
- Pipe B can fill the tank in 16 minutes, so its rate is \( \frac{1}{16} \) of the tank per minute.

2. **Let \( X \) be the time (in minutes) that pipe B is open.**
- This means that pipe A will be open for the entire 9 minutes, while pipe B will be open for \( X \) minutes.

3. **Calculate the work done by both pipes:**
- The work done by both pipes A and B together for \( X \) minutes is:
\[ \text{Work by A and B together} = (X \cdot \frac{1}{12}) + (X \cdot \frac{1}{16}) \]
- The work done by pipe A alone for the remaining \( 9 - X \) minutes is:
\[ \text{Work by A alone} = (9 - X) \cdot \frac{1}{12} \]

4. **Set up the equation for total work done:**
- The total work done by both pipes must equal 1 tank:
\[ (X \cdot \frac{1}{12}) + (X \cdot \frac{1}{16}) + (9 - X) \cdot \frac{1}{12} = 1 \]

5. **Combine the terms:**
- To combine the terms, we need a common denominator. The least common multiple of 12 and 16 is 48.
- Rewrite the equation:
\[ \frac{4X}{48} + \frac{3X}{48} + \frac{4(9 - X)}{48} = 1 \]
- This simplifies to:
\[ \frac{4X + 3X + 36 - 4X}{48} = 1 \]
- Which simplifies further to:
\[ \frac{3X + 36}{48} = 1 \]

6. **Clear the fraction by multiplying both sides by 48:**
\[ 3X + 36 = 48 \]

7. **Solve for \( X \):**
- Subtract 36 from both sides:
\[ 3X = 12 \]
- Divide by 3:
\[ X = 4 \]

### Conclusion:
Pipe B should be closed after 4 minutes to ensure that the tank is full in 9 minutes.