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

Answer

To solve the problem, we need to determine how long pipe B should be left open while both pipes A and B are filling the tank together, so that the tank is completely filled in 9 minutes.

### Step-by-Step Solution:

1. **Determine the filling rates of both pipes:**
- Pipe A can fill the tank in 12 minutes. Therefore, its rate of work (efficiency) is:
\[ \text{Efficiency of A} = \frac{1 \text{ tank}}{12 \text{ minutes}} = \frac{1}{12} \text{ tanks per minute} \]
- Pipe B can fill the tank in 16 minutes. Therefore, its rate of work (efficiency) is:
\[ \text{Efficiency of B} = \frac{1 \text{ tank}}{16 \text{ minutes}} = \frac{1}{16} \text{ tanks per minute} \]

2. **Calculate the combined efficiency of both pipes when opened together:**
- The combined efficiency when both pipes A and B are opened is:
\[ \text{Combined Efficiency} = \text{Efficiency of A} + \text{Efficiency of B} = \frac{1}{12} + \frac{1}{16} \]
- To add these fractions, we need a common denominator. The least common multiple (LCM) of 12 and 16 is 48.
\[ \frac{1}{12} = \frac{4}{48}, \quad \frac{1}{16} = \frac{3}{48} \]
- Thus,
\[ \text{Combined Efficiency} = \frac{4}{48} + \frac{3}{48} = \frac{7}{48} \text{ tanks per minute} \]

3. **Determine the total work done in 9 minutes:**
- If both pipes are opened for 9 minutes, the total work done by both pipes together is:
\[ \text{Total Work} = \text{Combined Efficiency} \times \text{Time} = \frac{7}{48} \times 9 = \frac{63}{48} = 1.3125 \text{ tanks} \]
- However, since we need only 1 tank to be filled, we need to determine how long pipe B should remain open.

4. **Let B be closed after 'x' minutes:**
- If pipe B is closed after 'x' minutes, then pipe A will work for the full 9 minutes, while pipe B will work for 'x' minutes.
- The work done by pipe A in 9 minutes is:
\[ \text{Work by A} = 9 \times \frac{1}{12} = \frac{9}{12} = \frac{3}{4} \text{ tanks} \]
- The work done by pipe B in 'x' minutes is:
\[ \text{Work by B} = x \times \frac{1}{16} = \frac{x}{16} \text{ tanks} \]

5. **Set up the equation for total work:**
- The total work done by both pipes must equal 1 tank:
\[ \frac{3}{4} + \frac{x}{16} = 1 \]

6. **Solve for 'x':**
- Rearranging the equation gives:
\[ \frac{x}{16} = 1 - \frac{3}{4} = \frac{1}{4} \]
- Multiplying both sides by 16:
\[ x = 16 \times \frac{1}{4} = 4 \text{ minutes} \]

### Conclusion:
Pipe B should be closed after 4 minutes for the tank to be completely filled in 9 minutes.